## Inverse Trigonometric Functions Class 12 Important Extra Questions Maths Chapter 2

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 2 Inverse Trigonometric Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Maths Chapter 2 Important Extra Questions Inverse Trigonometric Functions

### Inverse Trigonometric Functions Important Extra Questions Very Short Answer Type

Question 1.
Find the principal value of sin-1 ( $$\frac { 1 }{ 2 }$$ )
Solution:
$$\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$
Hence, the principal value of sin-1 ( $$\frac { 1 }{ 2 }$$ ) is $$\frac{\pi}{6}$$

Question 2.
What is the principal value of:
cos-1 (cos $$\frac{2 \pi}{3}$$ + sin-1 (sin $$\frac{2 \pi}{3}$$ ) ?
Solution:

Question 3.
Find the principal value of:
tan-1 (√3)- sec-1 (-2). (A.I.C.B.S.E. 2012)
Solution:

Question 4.
Evaluate : tan -1 ( 2 cos (2 sin-1 ( $$\frac{1}{2}$$ )))
Solution:

Question 5.
Find the value of tan-1(√3) – cot-1(-√3). (C.B.S.E. 2018)
Solution:
tan-1(√3) – cot-1(—√3)
$$=\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right)=-\frac{\pi}{2}$$

Question 6.
If sin-1 ( $$\frac { 1 }{ 3 }$$ ) + cos-1 x = $$\frac{\pi}{2}$$, then find x.(C.B.S.E. 2010C)
Solution:
sin-1 ( $$\frac { 1 }{ 3 }$$ ) + cos-1 x = $$\frac{\pi}{2}$$
⇒ x = 1/3
[sin-1 x + cos-1 x = $$\frac{\pi}{2}$$

Question 7.
If sec-1 (2) + cosec-1 (y) = $$\frac{\pi}{2}$$ , then find y.
Solution:
sec-1 (2) + cosec-1 (y) = $$\frac{\pi}{2}$$
⇒ y = 2 [∵ sec-1 x + cosec-1 x= $$\frac{\pi}{2}$$ ]

Question 8.
Write the value of sin [ $$\frac{\pi}{3}$$ – sin -1 ( $$\frac { -1 }{ 2 }$$ ) ]
Solution:
sin [ $$\frac{\pi}{3}$$ – sin-1 ( $$\frac { -1 }{ 2 }$$ ) ]
= sin [ $$\frac{\pi}{3}$$ + sin-1 ( $$\frac { 1 }{ 2 }$$ ) ]
[∵ sin-1 (-x) = -sin-1x]
= sin ( $$\frac{\pi}{3}+\frac{\pi}{6}$$ ) = sin $$\frac{\pi}{2}$$ = 1

Question 9.
Prove the following:

Solution:

Question 10.
If tan-1 x + tan-1y = $$\frac{\pi}{4}$$ , xy < 1, then write the value of the x + y + xy (A.I.C.B.S.E. 2014)
Solution:
We have tan-1 x + tan-1y = $$\frac{\pi}{4}$$

Hence x + y + xy = 1.

Question 11.
Prove that: 3 sin-1 x = sin-1(3x – 4x2);
x ∈ [$$\frac { -1 }{ 2 }$$ , $$\frac { 1 }{ 2 }$$] (C.B.S.E 2018)
Solution:
To prove: 3 sin-1 x = sin-1(3x – 4x2)
Put sin-1 x = θ
so that x = sin θ.
RHS = sin-1 (3 sin θ – 4 sin3 θ)
= sin-1 (sin 3θ) = 3θ = 3 sin-1x = LHS.

### Inverse Trigonometric Functions Important Extra Questions Short Answer Type

Question 1.
Express sin-1 ( $$\frac{\sin x+\cos x}{\sqrt{2}}$$ )
Where $$-\frac{\pi}{4}$$ < x < $$\frac{\pi}{4}$$, in the simples form.
Solution:

Question 2.
Prove that :
$$\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$$   (A.I.C.B.S.E. 2019; C.B.S.E. 2010)
Solution:

Question 3.
Prove that :
sin -1 $$\frac{8}{17}$$ + cos -1 $$\frac{4}{5}$$ = cos-1 $$\frac{36}{77}$$ (A.I.C.B.S.E. 2019)
Solution:
L.H.S = sin-1 $$\frac{8}{17}$$ + cos-1 $$\frac{4}{5}$$
= tan-1 $$\frac{8}{15}$$ + tan-1$$\frac{3}{4}$$

= R.H.S

Question 4.
Solve the following equation:
$$\tan ^{-1}\left(\frac{x+1}{x-1}\right)+\tan ^{-1}\left(\frac{x-1}{x}\right)=\tan ^{-1}(-7)$$ (A.I.C.B.S.E. 2019; C)
Solution:

⇒ 2x2 – 8x + 8 = 0
⇒ x2 – 4x + 4 = 0
⇒ (x – 2)2 = 0.
Hence, x = 2.

Question 5.
Solve the following equation:
2 tan-1(sin x) = tan-1 (2 sec x), x ≠ $$\frac{\pi}{2}$$   (C.B.S.E. (F) 2012)
Solution:
2 tan-1(sinx) = tan-1 (2 secx)

⇒ tan -1 (2sec x tan x) = tan-1 (2sec x)
⇒ 2 sec x tan x = 2 sec x
tan x = 1 [∵ sec x ≠ 0 ]
Hence, x= $$\frac{\pi}{2}$$

Question 6.
Solve the following equation
cos (tan-1 x) = sin ( cot -1$$\frac { 3 }{ 4 }$$ )
(A.I.C.B.S.E. 2013)
Solution:
We have:
cos (tan-1 x) = sin ( cot -1$$\frac { 3 }{ 4 }$$ )
cos (tan-1 x) = sin ( sin -1$$\frac { 4 }{ 5 }$$ )
cos (tan-1 x) = $$\frac { 4 }{ 5 }$$
tan-1 x = cos-1$$\frac { 4 }{ 5 }$$
⇒ tan-1 x = tan -1$$\frac { 3 }{ 4 }$$
Hence x = $$\frac { 3 }{ 4 }$$

Question 7.
Prove that
3cos-1 x = cos-1 (4x3 – 3x), x ∈ [ $$\frac { 1 }{ 2 }$$ , 1 ]
Solution:
Put x = cos θ in RHS
As 1/2 ≤ x ≤ 1
RHS = cos-1 (4cos3 θ – 3cos θ),
= cos-1 (cos 3θ) = 3θ = 3cos-1 x = L.H.S

### Inverse Trigonometric Functions Important Extra Questions Long Answer Type 1

Question 1.
prove that $$\frac { 1 }{ 2 }$$ ≤ x ≤ 1, then
$$\cos ^{-1} x+\cos ^{-1}\left[\frac{x}{2}+\frac{\sqrt{3}-3 x^{2}}{2}\right]=\frac{\pi}{3}$$ (CBSE. Sample paper 2017 – 18)
Solution:

Question 2.
Find the value of :
$$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$$
Solution:

Question 3.
Prove that :
tan -1( $$\frac{1}{2}$$ ) + tan -1 ( $$\frac{1}{5}$$ ) + tan -1( $$\frac{1}{8}$$ ) = $$\frac{\pi}{4}$$ (C.B.S.E. 2013: A.I.C.B.S.E. 2011)
Solution:

Question 4.

Solution:

Question 5.

Solution:
Put cos -1 (a/b) = θ so that cos θ = a/b.

Question 6.

Solution:

Question 7.

( A.I.C.B.S.E. 2016)
Solution:

Question 8.
$$\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$$ (C.B.S.E. 2016)
Solution:

Question 9.
Solve for x : 2 tan-1 (cos x ) = tan-1(2 cosec x). (C.B.S.E. 2016)
Solution:
2 tan-1 (cos x ) = tan-1 (cos x) + tan-1 (cos x)

= tan-1 (2cot x cosec x) …(1)
Now 2 tan-1(cos x) = tan-1(2 cosec x)
⇒ tan-1(2 cot x cosec x) = tan-1 (2cosec x)
[Using (1)]
⇒ 2 cot x cosec x = 2 cosec x
⇒ cotx cosecx = cosec x
⇒  sin x = tan x sin x
⇒either sin x = 0 or tan x = 1.
Hence, x = nπ ∀ n ∈ Z or x
= πm + $$\frac{\pi}{4}$$ ∀ m ∈ Z

Question 10.
If tan-1 $$\left(\frac{x-2}{x-4}\right)$$ + tan-1 $$\left(\frac{x+2}{x+4}\right)$$ = $$\frac{\pi}{4}$$ find the value of ‘x’   (A.I.C.B.S.E. 2014)
Solution:
We have

⇒ 2x2 – 16 = -12
⇒ 2x2 = 16 – 12
⇒ 2x2 = 4
⇒ x2 = 2
Hence x = ±√2

Question 11.
If tan-1 $$\left(\frac{x-3}{x-4}\right)$$ + tan-1 $$\left(\frac{x+3}{x+4}\right)$$ = $$\frac{\pi}{4}$$ find the value of ‘x’ (A.I.C.B.S.E. 2017)
Solution:

⇒ 2x2 – 24 = -7
⇒ 2x2 = 17
⇒ 2x2 = 4
⇒ x2 = 17/2
Hence x = $$\pm \sqrt{\frac{17}{2}}$$

Question 12.
Prove that : $$\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$$
x∈ [0,1] (C.B.S.E. 2019C)
Solution:
Let tan -1 √x = θ
So that √x = tan θ i.e. x = tan2θ

Question 13.
Solve for x : tan -1 (2x) + tan -1 (3x) = $$\frac{\pi}{4}$$ (C.B.S.E. 2019)
Solution:
The given equation is
tan -1 (2x) + tan -1 (3x) = $$\frac{\pi}{4}$$ …. (1)

⇒ 5x = 1 – 6x2
⇒ 6x2 + 5x – 1 = 0
⇒ x = -1 or x = $$\frac{1}{6}$$
Hence, x = $$\frac{1}{6}$$
[∵ x = -1 does not satisfy (1)]

Question 14.
Solve : tan-1 4x + tan-1 6x = $$\frac{\pi}{4}$$ (C.B.S.E. 2019)
Solution:
We have tan-1 4x + tan-1 6x = $$\frac{\pi}{4}$$

⇒ 10x = 1 – 24x2
⇒ 24x2 + 10x – 1 = 0
⇒ 24x2+ 12x – 2x – 1 = 0
⇒ 12x(2x + 1) – 1 (2x + 1) = 0
⇒ (2x+1)(12x – 1) = 0
⇒ 2x + 1 = 0 or 12x – 1 = 0
x = $$-\frac{1}{2}$$ or x = $$\frac{1}{2}$$.
Hence, x = $$-\frac{1}{2}$$ or $$\frac{1}{12}$$
As x = $$-\frac{1}{2}$$ does not satisfy the given equation.
Hence x = $$-\frac{1}{12}$$

Question 15.
If (tan-1 x )2 + (cot-1 x )2 = $$\frac{5 \pi^{2}}{8}$$ , then find ‘x’ (C.B.S.E. 2015)
Solution:
We have
(tan-1 x )2 + (cot-1 x )2 = $$\frac{5 \pi^{2}}{8}$$
Put tan-1 x = t so that cot-1x = $$\frac{\pi}{2}$$ – t
∴ (1) become : t2 + ($$\frac{\pi}{2}$$ – t)2 = $$\frac{5 \pi^{2}}{8}$$

When tan-1x = $$\frac{3 \pi}{4}$$
then x = tan $$\frac{3 \pi}{4}$$ = -1.
When tan-1 x = $$-\frac{\pi}{4}$$,
then x = tan($$-\frac{\pi}{4}$$) = -1.
Hence, x = -1.

Question 16.
Write : tan -1$$\frac{1}{\sqrt{x^{2}-1}}$$ , |x| > 1 in the simplest form.
Solution:
Put x = sec θ

Question 17.
Prove that

(C.B.S.E. 2012)
Solution:

Question 18.
Prove that :
tan-1x + tan-1 $$\frac{2 x}{1-x^{2}}$$ = tan -1$$\frac{3 x-x^{3}}{1-3 x^{2}}$$ (N.C.E.R.T
Solution:
Put x = tan θ so that θ = tan -1 x

= 3θ = 3 tan-1x = tan-1x + 2tan-1x
= tan-1x + tan-1$$\frac{2 x}{1-x^{2}}$$
= L.H.S.

Question 19.
Simplify : tan-1 [ $$\left[\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right]$$]
If $$\frac{a}{b}$$ tan x > -1    (N.C.E.R.T)
Solution:

Question 20.
Prove that tan -1 [ $$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$ ]
= $$\frac{\pi}{4}+\frac{1}{2}$$ cos-1 x ; $$-\frac{1}{\sqrt{2}}$$ ≤ x ≤ 1. (C.B.S.E. 2019C)
Solution:

Question 21.
Show that

(N.C.E.R.T.A.I. CBSE 2014)
Solution:

Question 22.
cos[tan-1{sin (cot-1x)}] = $$\sqrt{\frac{1+x^{2}}{2+x^{2}}}$$
(A.I. C.B.S.E. 2010)
Solution:
Put cot-1 = θ so that x = cot θ
∴ sin θ = $$\frac{1}{\sqrt{1+x^{2}}}$$

Question 23.
Find the value of sin (2 tan-1 1/4) + cos (tan-1 2√2)   (C.B.S.E. Sample Paper 2018 – 2019)
Solution:
Put tan-1 1/4 = θ so that θ = 1/4
Now, sin 2θ = $$\frac{2 \tan \theta}{1+\tan ^{2} \theta}$$

To evaluate cos(tan-1 2√2) :
Put tan-12√2 = Φ
So that tan Φ = 2√2
cos Φ = 1/3
Hence, sin (2tan-1(1/4)) + cos (tan-12√2)
= sin 2θ + cos Φ = $$\frac{8}{17}+\frac{1}{3}=\frac{41}{51}$$
[Using (1) and (2)]

Question 24.
Solve for x:
tan-1(x – 1) + tan-1x + tan-1(x + 1) = tan-1(3x).    (A.I.C.B.S.E. 2016)
The given equation is:
tan-1(x —1) + tan-1(x) + tan-1(x + 1) = tan-13x
tan-1(x— 1)+tan-1(x+ 1) = tan-13x – tan-1

⇒ 1 + 3x2 = 2 – x2
⇒ 4x2 = 1
⇒ x2 = 1/4
⇒ x = $$\pm \frac{1}{2}$$
Hence, x = 0, $$\pm \frac{1}{2}$$

## Matrices Class 12 Important Extra Questions Maths Chapter 3

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 3 Matrices. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Maths Chapter 3 Important Extra Questions Matrices

### Matrices Important Extra Questions Short Answer Type

Question 1.
Write the element a23 of a 3 x 3 matrix A = [aij] whose elements atj are given by : $$\frac{|i-j|}{2}$$
Solution:
We have [aij] = $$\frac{|i-j|}{2}$$
∴ a23 = $$\frac{|2-3|}{2}=\frac{|-1|}{2}=\frac{1}{2}$$

Question 2.
For what value of x is
$$\left[\begin{array}{lll} 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=0$$ ? (C.B.S.E. 2019(C))
We have
$$\left[\begin{array}{lll} 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=0$$

[1 + 4 + 1 2 + 0 + 0 0 + 2 + 2] $$\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]$$ = 0
[6 2 4 ]$$\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]$$ = 0
⇒ [0 + 4 + 4x] =0
⇒ [4 + 4x] = [0]
⇒ 4 + 4x = 0.
Hence, x = -1.

Question 3.
Find a matrix A such that 2A – 3B + 5C = 0,
Where B = $$\left[\begin{array}{ccc} -2 & 2 & 0 \\ 3 & 1 & 4 \end{array}\right]$$ and C = $$\left[\begin{array}{ccc} 2 & 0 & -2 \\ 7 & 1 & 6 \end{array}\right]$$
Solution:
Here, 2A – 3B + 5C = 0
⇒ 2A = 3B – 5C

Question 4.
If A = $$\left(\begin{array}{l} \cos \alpha-\sin \alpha \\ \sin \alpha-\cos \alpha \end{array}\right)$$ , then for what value of ‘α’ is A an identity matrix? (C.B.S.E. 2010)
Solution:
Here A = $$\left(\begin{array}{l} \cos \alpha-\sin \alpha \\ \sin \alpha-\cos \alpha \end{array}\right)$$
Now A = I = $$\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right)$$ when
cos α = 1 and sin α = 0.
Hence, α = 0.

Question 5.
Find the values of x, y, z and t, if:
$$2\left[\begin{array}{ll} x & z \\ y & t \end{array}\right]+3\left[\begin{array}{rr} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]$$
Solution:
We have :

⇒ 2x + 3 = 9 …………. (1)
2z – 3 = 15 …………. (2)
2y = 12 …………. (3)
2t + 6 = 18 …………. (4)

From (1), ⇒ 2x = 9 – 3
⇒ 2x = 6
⇒ x = 3.

From (3) 2y = 12
⇒ y = 6.

From (2), ⇒ 2z – 3 = 15
⇒ 2z = 18
⇒ z = 9.

From (4), 2t + 6 = 18
⇒ 2t = 12
⇒ t = 6.
Hence, x = 3,y = 6, z = 9 and t = 6.

Question 6.
If A = $$\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]$$, then find (A2 – 5A). (CBSE 2019)
Solution:
We have A = $$\left[\begin{array}{rrr} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]$$
Then A2 = AA

Question 7.
If A = $$\left[\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right]$$ and I = $$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$ , find k so that A2 = 5A + b kI (C.B.S.E. Sample Paper 2018-2019)
Solution:

⇒ 8 = 15 + k and 3 = 10 + k
⇒ k = -1 and k = -7.
Hence, k – -7.

Question 8.
If A and B are symmetric matrices, such that AB and BA are both defined, then prove that AB – BA is a skew symmetric matrix. (A.I.C.B.S.E. 2019)
Solution:
Since A and B are symmetric matrices,
∴ A’ = A and B’ = B …(1)
Now,(AB – BA)’= (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB [Using (1)]
= – (AB – BA).
Hence, AB – BA is a skew-symmetric matrix.

Question 9.
For the matrix A = $$\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]$$ find (A + A’) and show that it is a symmetric matrix. (A.I.C.B.S.E. 2019)
Solution:

Now (A + A’)’ = $$\left[\begin{array}{cc} 4 & 8 \\ 8 & 14 \end{array}\right]$$ = (A + A’)
Hence (A + A’) is symmetric

Question 10.
If the matrix A = $$\left[\begin{array}{ccc} 0 & a & -3 \\ 2 & 0 & -1 \\ b & 1 & 0 \end{array}\right]$$ is skew symmetric, find the values of ‘a’ and ‘b’: (C.B.S.E. 2018)
Solution:

Comparing, 2 = -a ⇒ a = -2
and – 3 = -b ⇒ b = 3.
Hence, a = -2 and b = 3.

### Matrices Important Extra Questions Long Answer Type 1

Question 1.
Find the values of a, b, c and d from the following equation :
$$\left[\begin{array}{cc} 2 a+b & a-2 b \\ 5 c-d & 4 c+3 d \end{array}\right]=\left[\begin{array}{cc} 4 & -3 \\ 11 & 24 \end{array}\right]$$ (N.C.E.R.T)
Solution:
We have
$$\left[\begin{array}{cc} 2 a+b & a-2 b \\ 5 c-d & 4 c+3 d \end{array}\right]=\left[\begin{array}{cc} 4 & -3 \\ 11 & 24 \end{array}\right]$$
Comparing the corresponding elements of two given matrices, we get:
2a + b = 4 …(1)
a-2b = – 3 …(2)
5c-d = 11 …(3)
4c + 3d = 24 …(4)
Solving (1) and (2):
From (1),
b = 4 – 2a …(5)
Putting in (2), a – 2 (4 – 2a) = – 3
⇒ a – 8 + 4a = -3
⇒ 5a = 5
⇒ a = 1.
Putting in (5),
b = 4 – 2(1) = 4 – 2 = 2.
Solving (3) and (4):
From (3),
d = 5c- 11 …(6)
Putting in (4),
4c+ 3 (5c- 11) = 24
⇒ 4c + 15c – 33 = 24
⇒ 19c = 57
⇒ c = 3.
Putting in (6),
d = 5 (3) – 11 = 15 – 11 = 4.
Hence, a = 1, b = 2, c = 3 and d = 4.

Question 2.
If $$\left[\begin{array}{rrr} 9 & -1 & 4 \\ -2 & 1 & 3 \end{array}\right]$$ = A + $$\left[\begin{array}{rrr} 1 & 2 & -1 \\ 0 & 4 & 9 \end{array}\right]$$ then find the matrix A. (C.B.S.E. 2013)
Solution:

Comparing:
9 = a11 + 1 – 1 = a12 + 2,
4 = 113 – 1, -2 = a21
1 = a22 + 4, and 3 = a23 + 9
a11 = 8, a12 = – 3,
a13 = 5, a21 = -2
a22 = – 3, and a23 = – 6.
Hence, A = $$\left[\begin{array}{rrr} 8 & -3 & 5 \\ -2 & -3 & -6 \end{array}\right]$$

Question 3.
If A = $$=\left[\begin{array}{rr} 2 & 2 \\ -3 & 1 \\ 4 & 0 \end{array}\right]$$ B = $$\left[\begin{array}{ll} 6 & 2 \\ 1 & 3 \\ 0 & 4 \end{array}\right]$$find the matrix C such that A + B + C is a zero matrix.
Solution:

Comparing :
8 + c11 = 0 ⇒ c11 = -8,
4 + C12 = 0 ⇒ C12 = -4,
– 2 + C21 = 0 ⇒ C21 = 2
4 + C22 = 0 ⇒ C22 =- 4,
4 + c31 = 0 ⇒ C31 = -4
and 4 + c32 = 0 ⇒ C32 = -4.
Hence, C = $$\left[\begin{array}{rr} -8 & -4 \\ 2 & -4 \\ -4 & -4 \end{array}\right]$$

Question 4.
If A = $$\left[\begin{array}{rr} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{array}\right]$$ B = $$\left[\begin{array}{rr} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{array}\right]$$ then find the matrix ‘X’, of order 3 x 2, such that 2A + 3X = 5B. (N.C.E.R.T.)
Solution:
We have : 2A + 3X = 5B
⇒ 2A + 3X-2A = 5B-2A
⇒ 2A-2A + 3X = 5B-2A
⇒ (2A – 2A) + 3X = 5B – 2A
⇒ O + 3X = 5B – 2A
[ ∵ – 2A is the inverse of2A]
⇒ 3X = 5B – 2A.
[ ∵ O is the additive identity]
Hence, X = $$\frac{1}{3}$$(5B – 2A)

Question 5.
If A is a square matrix such that A2 = A, then write the value of 7A – (I + A)3, where I is an identity matrix. (A.I.C.B.S.E. 2014)
Solution:
(I + A)2 = (I + A) (I + A)
= II + IA + AI + AA
= I + A + A + A2
= I + 2A + A [∵ A2 = A]
= I + 3A …(1)
∴ (I + A)3 = (I + A)2 (I + A)
= (I + 3A) (I + A) [Using (1)]
= II + IA + 3AI + 3AA
= I + A + 3A + 3A2
= I + A + 3A + 3A [∵ A2 = A]
= I + 7 A …(2)
Hence, 7A – (I + A)3 = 7A – (I + 7A)
[Using (2)]
= -I.

Question 6.
Compute the indicated product:
$$\left[\begin{array}{lll} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right]$$ (NCERT)
Solution:

Number of columns of A = Number of rows of B = 3.
Thus AB is defined and is a 3 x 3 matrix.

Question 7.
IF A = , show that A2 – 6A2 + 7A + 2I = 0   (N.C.E.R.T)
Solution:

Question 8.
Prove the following by the principle of Mathematical Induction:
If A = $$\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]$$
than An = $$\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]$$ (N.C.E.R.T)
Solution:
Set I. When n = 1

Thus the result is true when n = 1.
Step II. Let us assume that the result is true for any natural number m, where 1 ≤ m ≤ n

Thus the result is true when n = m+ 1.
Hence, by Mathematical Induction, the required result is true for all n ∈ N.

Question 9.
If A = $$\left[\begin{array}{r} -2 \\ 4 \\ 5 \end{array}\right]$$, B = [1 3 -6], then verify that (AB)’ = B’A’ (N.C.E.R.T.)
Solution:
We have :
A = $$\left[\begin{array}{r} -2 \\ 4 \\ 5 \end{array}\right]$$, B = [1 3 -6]

From (1) and (2), (AB)’ = B’A’
which verifies the result.

Question 10.
By using elementary transformations, find the inverse of the matrix A = $$\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]$$
(N.C.E.R.T.)
Solution:
(By Elementary Row Transformations)
We know A = I2A

### Matrices Important Extra Questions Long Answer Type 2

Question 1.
Two farmers Ram Kishan and Gurucharan Singh cultivate only three varieties of rice namely Basmati, Permal and Naura. The sale (in ?) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B :

Find :
(i) What were combined sales in September and October for each farmer in each variety?
(ii) What was the change in sales from September to October?
(iii) If both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety in October. (N.C.E.R.T.)
Solution:
(i) Combined sales in September and October :

(ii) Change in sales from September to October:

(iii) 2% of B = $$\frac{2}{100}$$ x B = 0.02 B

Hence, Ram Kishan receives ₹ 100, ₹ 200 and ₹ 120 as profit and Gurucharan Singh receives ₹ 400, ₹200 and ₹ 200 in each variety of rice in the month of October.

Question 2.
Three schools A, B and C organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold handmade fans, mats and plates from recycled meterial at a cost of ₹25, ₹100 and ₹50 each. The number of articles sold are given below :

Find the funds collected by each school separately by selling the above articles. Also, find the total fund collected for the purpose.
(C.B.S.E. 2015)
Solution:
Quantity matrix, A = $$\left[\begin{array}{lll} 40 & 25 & 35 \\ 50 & 40 & 50 \\ 20 & 30 & 40 \end{array}\right]$$
Cost Matrix B = [laatex]\left[\begin{array}{c}
25 \\
100 \\
50
\end{array}\right][/latex]
Total fund = Quantity matrix x Cost matrix

Fund collected by school A = ₹5,250
Fund collected by school B = ₹7,750
Fund collected by school C = ₹5,500 and total fund collected = 5250 + 7750 + 5500
= ₹18,500.

Question 3.
If A = $$\left(\begin{array}{rr} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)$$, find α satisfing 0 < α < when A + AT = √2I2
Where AT is transpose of A. (A.I.C.B.S.E. 2016)
Solution:

Question 4.
If $$\left(\begin{array}{cc} a+b & 2 \\ 5 & b \end{array}\right)$$ = $$\left(\begin{array}{ll} 6 & 5 \\ 2 & 2 \end{array}\right)^{\prime}$$ (C.B.S.E. 2010C)
Solution:

Comparing:
a + b – 6 ………….. (1)
and b = 2 ………….(2)
Putting the value of b from (2) in (1),
a + 2 = 6.
Hence, a = 4.

Question 5.
Express the matrix A as the sum of a symmetric and a skew-symmetric matrix, where :
A = $$\left[\begin{array}{rrr} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{array}\right]$$ (A.I.C.B.S.E. 2010)
Solution:
We have

Question 6.
Find the inverse of the following matrix, using elementary operations :
A = $$\left[\begin{array}{ccc} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{array}\right]$$  (C.B.S.E. 2019)
Solution:
We know that A + I3A\

Question 7.
Find the inverse of the following matrix, using elementary transformation:
$$\left[\begin{array}{rrr} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{array}\right]$$  (C.B.S.E. Sample Paper 2017-18)
Solution:
We know that A = IA

Question 8.
If A = $$\left[\begin{array}{rrr} 2 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & -1 \end{array}\right]$$ , find the inverse of A, using elementary row transformations and hence solve the following equation:
XA = [1 0 1 ] (C.B.S.E. Sample Paper 2017-18)
Solution:
(i) We know that A = I3 A

Here, x = 0, y = 1, z = 0

## Semiconductor Electronics: Materials, Devices and Simple Circuits Class 12 Important Extra Questions Physics Chapter 14

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

## Class 12 Physics Chapter 14 Important Extra Questions Semiconductor Electronics: Materials, Devices and Simple Circuits

### Semiconductor Electronics Important Extra Questions Very Short Answer Type

Question 1.
Draw the energy band diagram for a p-type semiconductor.
The energy level diagram is shown below.

Question 2.
Draw the voltage-current characteristic of a p-n junction diode in forwarding bias and reverse bias.
The characteristics are as shown.

Question 3.
Draw the voltage-current characteristic for a Zener diode.
The V-l characteristic of the Zener diode is as shown.

Question 4.
Draw the energy band diagram for n-type semiconductor.
The diagram is as shown.

Question 5.
An ac input signal of frequency 60 Hz is rectified by an
(i) Half wave and an
The output frequency remains the same in a half-wave rectifier, i.e. 60 Hz.

(ii) Full-wave rectifier. Write the output frequency in each case.
The output frequency becomes twice the input frequency in the case of the full-wave rectifier, i.e. 120 Hz.

Question 6.
Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor.
The ratio is one.

Question 7.
What is the depletion region in a p-n junction?
It is a thin layer between p and n sections of the p-n junction which is devoid of free electrons and holes.

Question 8.
Name an impurity which when added to pure silicon makes it a
(i) p-type semiconductor
Boron, aluminum, etc.

(ii) n-type semiconductor.
Phosphorous, antimony, etc.

Question 9.
Which type of biasing gives a semiconductor diode very high resistance?
Reverse biasing.

Question 10.
Identify the biasing in the figure given below.

Forward biasing.

Question 11.
Draw the circuit symbol of (a) photodiode, and (b) light-emitting diode.
The circuit symbols are as shown below.

Question 12.
What is the function of a photodiode? (CBSE AI 2013C)
It functions as a detector of optical signals.

Question 13.
When a p-n junction diode is forward biased, how will its barrier potential be affected? (CBSEAI 2019)
Potential barrier decreases in forwarding bias.

Question 14.
Name the junction diode whose l-V characteristics are drawn below: (CBSE Delhi 2017)

Solar cell.

Question 15.
How does the width of the depletion region of a p-n junction vary if the reverse bias applied to it decreases?
With the increase in the reverse bias, the depletion layer increases.

Question 16.
How does the width of the depletion region of a p-n junction vary if the reverse bias applied to it decreases?
If the reverse bias decreases, the width of the depletion layer also decreases.

Question 17.
Why is the conductivity of n-type semiconductors greater than that of p-type semiconductors even when both of these have the same level of doping?
It is because in n-type the majority carriers are electrons, whereas in p-type they are holes. Electrons have greater mobility than holes.

Question 18.
How does the conductance of a semiconducting material change with rising in temperature?
Increases with an increase in temperature.

Question 19.
How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers?
It is because it contains an equal number of electrons and protons and is made by doping with a neutral impurity.

Question 20.
How is the bandgap, Eg, of a photodiode related to the maximum wavelength, λm, that can be detected by it?
Eg = $$\frac{h c}{\lambda_{m}}$$

Question 21.
Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. How does it affect the
(i) Width of the depletion layer
Junction width will be small and

(ii) Junction field?
The junction field will be high.

Question 22.
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction? (NCERT Exemplar)
No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite.

### Semiconductor Electronics Important Extra Questions Short Answer Type

Question 1.
Draw a labeled circuit diagram of a full-wave rectifier using a p-n junction.
The diagram is as shown.

Question 2.
What is a solar cell? How does it work? Give one of its uses.
It is a p-n junction used to convert light into electrical energy. In such a diode, one region either the p-type or the n-type is made so thin that light falling on the diode is not absorbed appreciably before reaching the junction. The thin region in the solar cell is called the emitter and the other is called the base. The magnitude of current depends upon the intensity of light reaching the junction. A solar cell can be used to charge storage batteries during the daytime, which can be used during the night.

These are used as power supplies for satellites and space vehicles.

Question 3.
Draw the output signal in a p-n junction diode when a square input signal of 10 V as shown in the figure is applied across it. (CBSE AI 2019)

The diode will conduct only when it is forward biased. Therefore, till the input voltage is + 5 V, we will get an output across R, accordingly the output waveform shown in the figure.

Question 4.
The following diagrams, indicate which of the diodes are forward biased and which are reverse biased.

(a) Forward biased.
(b) Reverse biased.
(c) Forward biased,
(d) Reverse biased.

Question 5.
Mention the important considerations required while fabricating a p-n junction diode to be used as Light-Emitting Diode (LED). What should be the order of bandgap of an LED if it is required to emit light in the visible range? (CBSE Delhi 2013)
The important considerations are

• It should be heavily doped.
• The diode should be encapsulated with a transparent cover so that emitted light can come out.

The semiconductor used for the fabrication of visible LEDs must at least have a bandgap of 1.8 eV.

Question 6.
In the given circuit diagram shown below, two p-n junction diodes D1 and D2 are connected with a resistance R and a dc battery E as shown. Redraw the diagram and indicate the direction of flow of appreciable current in the circuit. Justify your answer.

The redrawn diagram showing the flow of appreciable current is shown below.

Here diode D2 is forward biased, hence it conducts. Therefore appreciable current will pass through it. However, diode 0, is reverse biased, hence negligible current will flow through it.

Question 7.
The diagram below shows a piece of pure semiconductor S in series with a variable resistor R and a source of constant voltage V. Would you increase or decrease the value of R to keep the reading of ammeter (A) constant when semiconductor S is heated? Give reason.

When a semiconductor is heated, its resistance decreases. As a result, the total resistance of the circuit will decrease. In order to maintain constant current flow, the total resistance of the circuit must remain constant. Hence, the external resistance has to be increased to compensate for the decrease of resistance of the semiconductor.

Question 8.
Two semiconductor materials X and Y showed in the figure are made by doping germanium crystal with indium and arsenic respectively. The two are joined end to end and connected to a battery as shown,
(i) Will the junction be forward or reverse biased?
(ii) Sketch a V-l graph for this arrangement.

Material X is p-type and material Y is n-type.
(i) The junction is reverse biased.
(ii) For the V-l graph
The characteristics are as shown.

Question 9.
Draw the output waveform across the resistor (figure). (NCERT)

It is a half-wave rectifier, therefore only the positive cycle will be rectified. Thus the output waveform is as shown.

Question 10.
(i) Name the type of a diode whose characteristics are shown in figure
(a) and figure (b).
(ii) What does point P in figure (a) represent?
(iii) What do the points P and Q in figure (b) represent? (NCERT Exemplar)

(i) Zener junction diode and solar cell.
(ii) Zener breakdown voltage
(iii) P-open circuit voltage.
Q-short circuit current

Question 11.
A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuated between 3 V and 7 V, what should be the value of R, for safe operation (figure)? (NCERT Exemplar)

Given P = 1 W, Vz = 5V, Vs = 7 V, Rs = ?
We know that

Question 12.
Two material bars A and B of the equal area of the cross-section are connected in series to a dc supply. A is made of usual resistance wire and B of an n-type semiconductor.
(i) In which bar is the drift speed of free electrons greater?
(ii) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected? Justify each answer. (CBSE Sample Paper 2018-19)
(i) Drift speed in B (n-type semiconductor) is higher.
Reason: Since the two bars A and B are connected in series, the current through each is the same.
Now l = neAvd
Or
vd = $$\frac{1}{n e A}$$ ⇒ vd ∝ $$\frac{1}{n}$$ (As l and A are same).

As n is much lower in semiconductors, drift velocity will be more.

Question 13.
Explain how the width of the depletion layer in a p-n junction diode changes when the junction is (i) forward biased and (ii) reverse biased. (CBSE Delhi 2018C)
The width of the depletion region in a p-n junction diode decreases when it is forward biased because the majority of charge carriers flow towards the junction. While it increases when the junction diode is reverse biased because the majority of charge carriers move away from the junction.

### Semiconductor Electronics Important Extra Questions Long Answer Type

Question 1.
Define the terms ‘potential barrier’ and ‘depletion region’ for a p – n junction diode. State how the thickness of the depletion region will change when the p-n junction diode is (i) forward biased and (ii) reverse biased.
Potential barrier: The potential barrier is the fictitious battery, which seems to be connected across the p-n junction with its positive terminal in the n-region and the negative terminal in the p-region.

Depletion region: The region around the junction, which is devoid of any mobile charge carriers, is called the depletion layer or region.

1. When the p-n junction is forward biased, there is a decrease in the depletion region.
2. When the p-n junction is reverse biased, there is an increase in the depletion region.

Question 2.
Explain (i) forward biasing and (ii) reverse biasing of a p-n junction diode.
(i) A p-n junction is said to be forward-biased if its p-type is connected to the positive terminal and its n-type is connected to the negative terminal of a battery.
(ii) A p-n junction is said to be reverse-biased if its n-type is connected to the positive terminal and its p-type is connected to the negative terminal of a battery. The diagrams are as shown.

Question 3.
Draw V-l characteristics of a p-n junction diode. Answer the following questions, giving reasons:
(i) Why is the current under reverse bias almost independent of the applied potential up to a critical voltage?
(ii) Why does the reverse current show a sudden increase at the critical voltage? Name any semiconductor device which operates under the reverse bias in the breakdown region. (CBSEAI 2013)
The characteristics are as shown.

(i) This is because even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction.

(ii) As the reverse bias voltage is increased, the electric field at the junction becomes significant. When the reverse bias voltage V = Vz critical voltage, then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to the n-side. These electrons account for the high current observed at the breakdown.

Zener diode operates under the reverse bias in the breakdown region.

Question 4.
Draw the energy band diagrams of (i) n-type and (ii) p-type semiconductor at temperature T > 0 K.
In the case of n-type Si semiconductors, the donor energy level is slightly below the bottom of the conduction band, whereas in p-type semiconductors, the acceptor energy level is slightly above the top of the valence band. Explain what role do these energy levels play in conduction and valence bands. (CBSE AI 2015 C)
For energy bands
(i) The energy level diagram is shown below.

(ii) The diagram is shown as

In the energy band diagram of n-type Si semiconductor, the donor energy level EA is slightly below the bottom Ec of the conduction band and electrons from this level move into the conduction band with a very small supply of energy. At room temperature, most of the donor atoms get ionized but very few (-10-12) atoms of Si get ionized. So the conduction band will have most electrons coming from the donor impurities.

Similarly, for p-type semiconductors, the acceptor energy level EA is slightly above the top Ev of the valence band. With the very small supply of energy, an electron from the valence band can jump to the level EA and ionize the acceptor negatively. Alternately, we can also say that with a very small supply of energy, the hole from level EA sinks down into the valence band. Electrons rise up and holes fall down when they gain external energy.

Question 5.
Give reasons for the following:
(i) High reverse voltage does not appear across an LED.
It is because the reverse breakdown voltage of LED is very low, i.e. nearly 5 V.

(ii) Sunlight is not always required for the working of a solar cell.
Because solar cells can work with any light whose photon energy is more than the bandgap energy.

(iii) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of about 5 V. (CBSE Delhi 2016C)
The heavy doping of p and n sides of the p-n junction makes the depletion region very thin, hence for a small reverse bias voltage, the electric field is very high.

Question 6.
State the reason why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of a wavelength of 400 nm incident on it. (CBSE Al 2017C)
It is easier to observe the change in the current with the change in the light intensity if a reverse bias is applied. Thus photodiode is used in the reverse bias mode even when the current in the forward bias is more the energy gap (Eg) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons.

Due to the electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows.

Given λ = 400 nm,
Energy of photon
E = $$\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9} \times 1.6 \times 10^{-19}}$$ = 3.105 eV

Since the bandgap is lesser than this energy, therefore it will be able to detect the wavelength.

Question 7.
Explain the two processes involved in the formation of a p-n junction diode. Hence define the term ‘barrier potential’. (CBSE Delhi 2017C)
The two processes are
(i) Diffusion and
(ii) Drift

• Diffusion: The holes diffuse from the p-side to the n-side and electrons diffuse from the n-side to the p-side.
• Drift: The motion of charge carriers due to the applied electric field which results in the drifting of holes along the electric field and of electrons opposite to the electric field.

The potential barrier is the fictitious battery that seems to be connected across the junction with its positive end on the n-type and the negative end on the p-type.

Question 8.
Explain briefly how a photodiode operates.
A Photodiode is again a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on the diode. It is operated under reverse bias. When the photodiode is illuminated with light (photons) with energy (hv) greater than the energy gap (Eg) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode.

Question 9.
Name the p-n junction diode which emits spontaneous radiation when forward biased. How do we choose the semiconductor, to be used in these diodes, if the emitted radiation is to be in the visible region?
The p-n junction diode, which emits spontaneous radiation when forward biased, is the “light-emitting diode” or LED.

The visible tight is from 400 nm to 700 nm and the corresponding energy is between 2.8 eV to 1.8 eV. Therefore, the energy gap of the semiconductor to be used in LED, in order to have the emitted radiation be in the visible region, should be 1.8 eV. Phosphorous doped gallium arsenide and gallium phosphide are two such suitable semiconductors.

Question 10.
The figure shows the V-l characteristic of a semiconductor diode designed to operate under reverse bias. (CBSE Al 2019)

(a) Identify the semiconductor diode used.
The diode used is Zener diode

(b) Draw the circuit diagram to obtain the given characteristics of this device.
The circuit diagram is as shown.

(c) Briefly explain one use of this device.
The Zener diode can be used as a voltage regulator in its breakdown region. The Zener voltage remains constant even when the current through the Zener diode changes.

Question 11.
With the help of a diagram, show the biasing of a light-emitting diode (LED). Give its two advantages over conventional incandescent lamps.
The biasing of a light-emitting diode (LED), has been shown below.

Two main advantages of LED over conventional incandescent lamps are as follows:

1. Low operational voltage and less power consumption.
2. Fast action and no warm-up time required.
3. The bandwidth of emitted light is 100 A to 500 A or in other words, it is nearly (but not exactly) monochromatic.
4. Long life and ruggedness.
5. Fast on-off switching capability.

Question 12.
(a) Writetheprincipleofasemiconductor device which is used as a voltage regulator.
(a) Zener diode is used as a voltage regulator
Principle: It is based on the Principle that when breakdown voltage V2 takes place, there is a large change in the reverse current even with the insignificant change in the reverse bias voltage.

(b) With the help of a circuit diagram explain its working.

Working: If the reverse voltage across a Zener diode is increased beyond the breakdown voltage Vz, the current increases sharply and large current lz flows through the Zener diode and the voltage drop across Rs increases maintaining the voltage drop across RL at constant value Vo = Vz.

On the other hand, if we keep the input voltage constant and decrease the load resistance RL, the current across the load will increase. The extra current cannot come from the source because drop-in Rs will not change as the Zener is within its regulating range. The additional load current will pass through the Zener diode and is known as Zener current lz so that the total current (lL + lz) remains constant.

(c) Draw its l-V characteristics. (CBSE 2019C)
l-V Characteristics of Zener diode

Question 13.
With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram.
Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason? (CBSE Delhi 2015)
It is fabricated with a transparent window to allow light to fall on the diode. It is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode.

When the photodiode is illuminated with light (photons) with energy (hw) greater than the energy gap (Eg) of the semiconductor, then electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to the electric field of the junction, electrons and holes are separated before they recombine.

The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows. The magnitude of the photocurrent depends on the intensity of incident light (photocurrent is proportional to incident light intensity). The diagram is as shown.

It is easier to observe the change in the current with a change in the light intensity if a reverse bias is applied. Thus photodiode is used in the reverse bias mode even when the current in the forward bias is more.

Question 14.
Draw the circuit diagram of a full-wave rectifier and explain its working. Also, give the input and output waveforms. (CBSE Delhi 2019)
The circuit diagram is as shown.

The two ends S1 and S of a center-tapped secondary of a transformer are connected to the P sections of the two diodes D1 and D2 respectively. The n-sections of the two diodes are joined together and their com¬mon junction is connected to the central tap C of the secondary winding through a load resistance RL. The input is applied across the primary and the output is ob¬tained across the load resistance RL. The arrows show the direction of the current.

Assume that the end A of the secondary is positive during the first half cycle of the supply voltage. This makes diode D1 forward biased and diode D2 reverse biased. Thus diode D1 conducts and an output is obtained across the load RL.

During the second half cycle of the supply voltage, the polarities of the secondary windings reverse. A becomes negative and B becomes positive with respect to the central terminal C. This makes diode D2 forward biased. Hence it conducts and an output is obtained across RL.

The input-output waveforms are as shown.

Question 14.
Draw the circuit diagram to show the use of a p-n junction diode as a half-wave rectifier. Also show the input and the output voltages, graphically. Explain its working.
The diagram is as shown.

The input and output waveforms are as shown.

A p-n junction diode is used as a half-wave rectifier. Its work is based on the fact that the resistance of the p-n junction becomes low when forward biased and becomes high when reversing biased. These characteristics of a diode are used in rectification.

Question 15.
Distinguish between conductors, insulators, and semiconductors on the basis of the band theory of solids.
The diagrams are as shown.

Metals: A distinguishing character of all conductors, including metals, is that the valence band is partially filled or the conduction and the valence band overlap. Electrons in states near the top of the filled portion of the band have many adjacent unoccupied states available, and they can easily gain or lose small amounts of energy in response to an applied electric field. Therefore these electrons are mobile and can contribute to electrical and thermal conductivity. Metallic crystals always have partially filled bands figure (ii).

Insulators: In the case of insulators, there is a large energy gap of approximately 6 eV depending upon the nature of the crystal. Electrons, however, heated, find it difficult or practically impossible to jump this gap and thus never reach the conduction band. Thus electrical conduction is not possible through an insulator figure (iii).

Semiconductors: There is a separation between the valence band and the conduction band. The energy gap is of the order of 1 eV (0.67 eV for germanium and 1.12 eV for silicon). At absolute zero the electrons cannot gain this energy. But at room temperature, these electrons gain energy and move into the conduction band where they are free to move even under the effect of a weak electric field figure (i).

Question 16.
What is a Zener diode? How is it symbolically represented? With the help of a circuit diagram, explain the use of the Zener diode as a voltage stabilizer.
It is a special diode made to work only in the reverse breakdown region.
Symbol:

The figure below shows the use of the Zener diode in providing a constant voltage supply.

This use of the Zener diode is based on the fact that in the reverse breakdown (or Zener) region, a very small change in voltage across the Zener diode produces a very large change in the current through the circuit. If voltage is increased beyond Zener voltage, the resistance of the Zener diode drops considerably. Consider that the Zener diode and a resistor R, called dropping resistor, are connected to a fluctuating voltage supply, such that the Zener diode is reverse biased.

Whenever voltage across the diode tends to increase, the current through the diode rises out of proportion and causes a sufficient increase in the voltage drop across the dropping resistor. As a result, the output voltage lowers back to the normal value. Similarly, when the voltage across the diode tends to decrease, the current through the diode goes down out of proportion, so that the voltage drop across the dropping resistor is much less and now the output voltage is raised to normal.

Question 17.
Explain briefly with the help of a circuit diagram how V-l characteristics of a p-n junction diode are obtained in (i) forward bias and (ii) reverse bias.
Forward biased characteristics: A p-n junction is said to be forward-biased if its p-type is connected to the positive terminal and its n-type is connected to the negative terminal of a battery shows a circuit diagram that is used to study the forward characteristics of a p-n junction. The p-n junction is forward biased. Different readings are taken by changing the voltage and noting the corresponding milliammeter current.

Practically no current is obtained till the applied voltage becomes greater than the barrier potential. Above the potential barrier voltage, even a small change in potential causes a large change in current.

Reverse biased characteristics: In reverse biased characteristics, instead of a milliammeter, a microammeter is used. The voltage across the p-n junction is increased and the corresponding current is noted.

In the reverse bias, the diode current is very small. As the voltage has increased the current also increases. At a certain voltage, the current at once becomes very large. This voltage is called Breakdown voltage or Zener voltage. At this voltage, a large number of covalent bonds break releasing a large number of electrons and holes. Hence a large current is obtained. The characteristics are as shown.

Question 18.
Explain how the heavy doping of the p and n sides of a p-n junction diode helps in internal field emission (or Zener breakdown), even with a reverse bias voltage of a few volts only. Draw the general shape of the V-I characteristics of a Zener diode. Discuss how the nature of these characteristics led to the use of a Zener diode as a voltage regulator.
Consider a p-n junction where both p- and n-sides are heavily doped. Due to the high dopant densities, the depletion layer junction width is small and the junction field will be high. Under large reverse bias, the energy bands near the junction and the junction width decrease. Since the junction width is < 10-7 m, even a small voltage (say 4 V) may give a field as large as 4 × 10-7 Vm-1. The high junction field may strip an electron from the valence band which can tunnel to the n-side through the thin depletion layer. Such a mechanism of emission of electrons after a certain critical field or applied voltage V is termed as internal field emission which gives rise to a high reverse, current, or breakdown current.

The general shape of the V-l characteristics of a Zener diode is as shown in the figure.

Suppose an unregulated dc input voltage (Vi) is applied to the Zener diode (whose breakdown voltage is Vz as shown in the figure. If the applied voltage Vi > Vz, then the Zener diode is in the breakdown condition. As a result of a wide range of values of load (RL), the current in the circuit or through the Zener diode may change but the voltage across it remains unaffected by the load. Thus, the output voltage across the Zener is a regulated voltage.

Question 19.
(i) Describe briefly with the help of a necessary circuit diagram, the working principle of a solar cell.
Solar Cell: A solar cell is a junction diode that converts solar energy into electrical energy. In a solar cell, the n-region is very thin and transparent so that most of the incident light reaches the junction. The thin region is called the emitter and the other base. When light is incident on it, it passes through the crystal onto the junction. The electrons and holes are generated due to light (with hv > Eg). The electrons are kicked to the n-side and holes to the p-side due to the electric field of the depletion region. Thus p-side becomes positive and the n-side becomes negative giving rise to a photo-voltage. Thus it behaves as a cell.

(ii) Why are Si and GaAs preferred materials for solar cells? Explain. (CBSE AI 2011C)
Si and GaAs are preferred for solar cell fabrication due to the fact that their bandgap is ideal. Further, they have high electrical conductivity and high optical absorption.

Question 20.
Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it is used to detect optical signals. (CBSE Delhi 2013)
When the photodiode is illuminated with light (photons) with energy (hv) greater than the energy gap (Eg) of the semiconductor, then – electron-hole pairs are generated due to the absorption of photons. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. Due to the electric field of the junction, electrons and holes are separated before they recombine. The direction of the electric field is such that electrons reach the n-side and holes reach the p-side. Electrons are collected on the n-side and holes are collected on the p-side giving rise to an emf. When an external load is connected, current flows.

The magnitude of the photocurrent depends on the intensity of incident light incident on it. This helps in detecting optical signals.

Question 21.
(i) Explain with the help of a diagram, how depletion region and potential barrier are formed in a junction diode.
(ii) If a small voltage is applied to a p-n junction diode, how will the barrier potential be affected when it is (i) forward biased and (ii) reverse biased? (CBSE AI2015)
(i) Since there is an excess of electrons in the n-type and excess of holes in the p-type, on the formation of a p-n junction the electrons from the n-type diffuse into the p-region, and the holes in the p-type diffuse into the n-region.

The accumulation of electric charge of opposite polarities in the two regions across the junction establishes a potential difference between the two regions. This is called the potential barrier or junction barrier. The potential barrier developed across the junction opposes the further diffusion of the charge carriers from p to n and vice versa. There is a region on either side of the junction where there is a depletion of mobile charges and has only immobile charges. The region around the junction, which is devoid of any mobile charge carriers, is called the depletion layer or region.

(ii) (a) In forwarding bias the potential barrier decreases.
(b) In reverse bias the potential barrier increases.

Question 22.
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction. (CBSE Delhi 2017)

1. Drift and
2. diffusion.

The n-type has an excess of electrons and the p-type has an excess of holes. When a p-n junction has formed the electrons from the n-type diffuse into the p-region and the holes in the p-type diffuse into the n-region. These diffusing electrons and holes combine near the junction. Each combination eliminates an electron and a hole. This results in the n-region near the junction becoming positively charged by losing its electrons and the p-region near the junction becoming negatively charged by losing its holes.

This accumulation of electric charge of opposite polarities in the two regions across the junction establishes a potential difference between the two regions. This is called the potential barrier or junction barrier.
The potential barrier developed across the junction opposes the further diffusion of the charge carriers from p to n and vice versa. As a result, a region develops on either side of the junction where there is a depletion of mobile charges and has only immobile charges. The region around the junction which is devoid of any mobile charge carriers is called the depletion layer or region.

Question 23.
(i) State briefly the processes involved in the formation of the p-n junction explaining clearly how the depletion region is formed.
As we know that n-type semi-conductor has more concentration of electrons than that of a hole and a p-type semi-conductor has more concentration of holes than an electron. Due to the difference in concentration of charge carriers in the two regions of the p-n junction, the holes diffuse from p-side to n-side, and electrons diffuse from n-side to p-side. When an electron diffuses from n to p, it leaves behind it an ionized donor on the n-side. The ionized donor (+ve charge) is immobile as it is bound by the surrounding atoms. Therefore, a layer of positive charge is developed on the n-side of the junction. Similarly, a layer of negative charge is developed on the p-side.

Hence, a space-charge region is formed on either side of the junction, which has immobile ions and is devoid of any charge carrier, called depletion layer or depletion region.

The potential barrier is the fictitious battery which seems to be connected across the junction with its positive end on the n-type and the negative end on the p-type.

(ii) Using the necessary circuit diagrams, show how the V – l characteristics of a p-n junction are obtained in
(a) Forward biasing
(b) Reverse biasing
How these characteristics are made use of in rectification? (CBSE Delhi 2014)
(a) p-n junction diode under forwarding bias: The p-side is connected to the positive terminal and the n-side to the negative terminal. Applied voltage drops across the depletion region. Electron in n-region moves towards the p-n junction and holes in the p-region move towards the junction. The width of the depletion layer decreases and hence, it offers less resistance. Diffusion of majority carriers takes place across the junction. This leads to the forward current.

(b) p-n junction diode under reverse bias: Positive terminal of the battery is connected to the n-side and negative terminal to p-side. Reverse bias supports the potential barrier. Therefore, the barrier height increases and the width of the depletion region also increases. Due to the majority of carriers, there is no conduction across the junction. A few minority carriers across the junction after being accelerated by the high reverse bias voltage. This constitutes a current that flows in opposite direction, which is called reverse current.

For V-l curves

A p-n junction diode is used as a half-wave rectifier. Its work is based on the fact that the resistance of the p-n junction becomes low when forward biased and becomes high when reversing biased. These characteristics of the diode are used in rectification.

Question 24.
(i) Explain with the help of a suitable diagram, the two processes which occur during the formations of a p-n junction diode. Hence define the terms (i) depletion region and (ii) potential barrier.
(ii) Draw a circuit diagram of a p-n junction diode under forwarding bias ‘ and explain its working. (CBSE 2018C)
(i) The two important processes are diffusion and drift.
Due to the concentration gradient, the electrons diffuse from the n-side to the p side, and holes diffuse from the n-side to the n side.

Due to the diffusion, an electric field develops across the junction. Due to the field, an electron moves from the p-side to the n-side; a hole moves from the n-side to the p-side. The flow of the charge carriers due to the electric field is called drift.

(a) Depletion region: It is the space charge region on either side of the junction that gets depleted of free charges is known as the depletion region.
(b) Potential Barrier: The potential difference that gets developed across the junction and opposes the diffusion of charge carriers and brings about a condition of equilibrium, which is known as the barrier potential.

(ii) The circuit diagram is as shown

Working:
In the forward bias condition, the direction of the applied voltage is opposite to the barrier potential. This reduces the width of the depletion layer as well as the height of the barrier. A current can, therefore, flow through the circuit. This current increases (non¬linearly) with the increase in the applied voltage.

Numerical Problems

Question 1.
(i) Three photodiodes D1, D2, and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm? (CBSE Delhi 2019)
λ = 600 nm
The energy of a photon of wavelength λ
E = $$\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9} \times 1.6 \times 10^{-19}}$$ = 2.08 eV

The bandgap energy of diode D2 (= 2eV) is less than the energy of the photon. Hence diode D2 will not be able to detect light of wavelength 600 nm.

(ii) Why photodiodes are required to operate in reverse bias? Explain.
A photodiode when operated in reverse bias can measure the fractional change in minority carrier dominated reverse bias current with greater ease than when forward biased.

Question 2.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency? (NCERT)
The output frequency of the half-wave rectifier is the same as the input frequency, while that of the full-wave rectifier is double that of the input. Therefore the frequency is 50 Hz for half-wave and 100 Hz for full-wave.

Question 3.
A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm? (NCERT)
Given λ = 6000 nm,
Energy of photon

Since the bandwidth is greater than this energy, it will not be able to detect the wavelength.

Question 4.
Three photodiodes D1, D2, and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV, and 3 eV, respectively. Which ones will be able to detect light of wavelength 600 nm? (NCERT Exemplar)
The energy of an incident light photon is given by

For the incident radiation to be detected by the photodiode, the energy of the incident radiation photon should be greater than the bandgap. This is true only for D2. Therefore, only D2 will detect this radiation.

Question 5.
If each diode in the figure has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the current l1 l2, l3, and l4? (NCERT Exemplar)

nswer:
Current l3 is zero as the diode in that branch is reverse biased.

Resistance in the branches AB and EF is each (125 +25) Ω = 150 Ω
As AB and EF are identical paraLleL branches, their effective resistance is 150/2 = 75 Ω

Therefore net resistance in the circuit Is
Rnet =75 + 25 = 100 Ω

Therefore current l1 is
l1 = 5/100 = o.05 A

As resistance of branches AB and EF is equal, the current l1 will be equally shared by the two, hence
l2 = l4 = 0.05/2 = 0.025A
Hence l1 = 0.05 A, l2 = l4 = 0.025 A, l3 = 0

Question 6.
Assuming the ideal diode, draw the output waveform for the circuit given in the figure. ExplaIn the waveform. (NCERT Exemplar)

When input voltage Will is greater than 5 V, the diode wilt becomes forward biased and will conduct. When the input is Less than 5 V, the diode will be reverse biased and will not conduct, i.e. open circuit, hence the output is as shown.

## Wave Optics Class 12 Important Extra Questions Physics Chapter 10

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 10 Wave Optics. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

## Class 12 Physics Chapter 10 Important Extra Questions Wave Optics

### Wave Optics Important Extra Questions Very Short Answer Type

Question 1.
Sketch the refracted wavefront emerging from convex tens, If a plane wavefront is an incident normally on it.
The figure is as shown.

Question 2.
How would you explain the propagation of light on the basis of Huygen’s wave theory?
To explain the propagation of light we have to draw a wavefront at a later instant when a wavefront at an earlier instant is known. This can be drawn by the use of Huygen’s principle.

Question 3.
Draw the shape of the reflected wavefront when a plane wavefront is an incident on a concave mirror.
The reflected wavefront is as shown.

Question 4.
Draw the shape of the refracted wavefront when a plane wavefront is an incident on a prism.
The shape of the wavefront is as shown.

Question 5.
Draw the type of wavefront that corresponds to a beam of light diverging from a point source.
The wavefront formed by the light coming from a very far off source is a plane and for a beam of light diverging from a point, a wavefront is a number of concentric circles.

Question 6.
Draw the type of wavefront that corresponds to a beam of light coming from a very far off source.
The wavefront is as shown.

Question 7.
Name two phenomena that establish the wave nature of light.
Interference and diffraction of light.

Question 8.
State the conditions which must be satisfied for two light sources to be coherent.
(a) Two sources must emit light of the same wavelength (or frequency).
(b) The two light sources must be either in-phase or have a constant phase difference.

Question 9.
Draw an intensity distribution graph for diffraction due to a single-slit.
The intensity distribution for a single-slit diffraction pattern is as shown.

Question 10.
Name one device for producing plane polarised light. Draw the graph showing the variation of intensity of polarised light transmitted by an analyser.
Nicol prism can be used to produce plane polarised light. The graph is as shown.

Question 11.
State Huygens’ principle of diffraction of light. (CBSE AI 2011C)
Huygens principle states that
(a) Each point on a wavefront is a source of secondary waves which travel out with the same velocity as the original waves.
(b) The new wavefront is given by the forward locus of the secondary wavelets.

Question 12.
In what way is a plane polarised tight different from an unpolarised light? (CBSE AI 2012C)
Plane polarized light vibrates 1n only one plane.

Question 13.
Which of the following waves can be polarised: (i) Heatwaves (ii) Sound waves? Give a reason to support your answer. (CBSE Delhi 2013)
Heatwaves are transverse In nature.

Question 14.
Define the term ‘wavefront’. (CBSE AI 2014C)
It Is defined as the locus of all points In a medium vibrating in the same phase.

Question 15.
Define the term ‘coherent sources’ which are required to produce interference pattern in Young’s double-slit experiment. (CBSE Delhi 2014C)
Two sources that are In phase or have a constant phase difference are called coherent sources.

Question 16.
What change would you expect if the whole of Young’s double-slit apparatus were dipped into the water?
The wavelength λ, of light In water, is less than that in air. Since the fringe width β is directly proportional to the wavelength of light, therefore, the fringe width will decrease.

Question 17.
When light travels from a rarer to a denser medium, it loses some speed. Does the reduction in speed Imply a reduction in the energy carried by the light wave?
No, the energy carried by a wave depends upon the amplitude of the wave and not on Its speed of propagation.

Question 18.
If one of the slits say S1, is covered then what changes occur in the Intensity of light at the centre of the screen?
The intensity 1s decreased four times because l ∝ 4a² where a is the amplitude of each wave.

Question 19.
How does the angular separation between fringes in a single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? (CBSE AI 2012)
No change.

Question 20.
What is the effect on the interference fringes in Young’s double-slit experiment If the separation between the screen and slits Is Increased?
The fringe width Increases.

Question 21.
How does the Intensity of the central maximum change If the width of the slit Is halved in a single-slit diffraction experiment?
The width of the central maxima is doubled and the intensity is reduced to one-fourth of Its original value.

Question 22.
The polarising angle of a medium Is 60°, what is the refractive index of the medium? (CBSE Delhi 2019)
Using the expression
μ = tan ip = tan 60° = 1.732.

Question 23.
In the wave picture of light intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light?
For a given frequency intensity of light in the photon, the picture is determined by the number of photons crossing a unit area per unit time.

Question 24.
How does the Intensity of the central maximum change if the width of the slit is halved in the single-slit diffraction experiment?
The width of the central maximum is doubled and the intensity is reduced to one-fourth of its original value.

Question 25.
What would happen If the path difference between the interfering beams that is S2P – S1P became very large?
If the path difference becomes very large it may exceed the coherent length. Thus the coherence of the waves reaching P is lost and no interference takes place.

Question 26.
In Young’s double-slit experiment, what would happen to the intensity of the maxima and the minima if the size of the hole illuminating the two coherent holes were gradually Increased?
The fringe width will decrease and finally, there will be general illumination on the screen.

Question 27.
What is the Brewster angle for air to glass transition? (Refractive index of glass =1.5.) (NCERT)
Given μ = 1.5, iP = ?
Using the relation μ = tan iP we have
lp = tan-1 (1.5) = 56.3°

Question 28.
Is Huygen’s principle valid for longitudinal sound waves? (NCERT Exemplar)
Yes

Question 29.
Consider a point at the focal point of a convergent lens. Another convergent lens of the short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image? (NCERT Exemplar)
Spherical.

Question 30.
What is the shape of the wavefront on earth for sunlight? (NCERTExemplar)
Spherical with a huge radius as compared to the earth’s radius so that it is almost a plane.

Question 31.
Draw a graph showing the intensity distribution of fringes due to diffraction at a single-slit. (CBSE 2018C)

### Wave Optics Important Extra Questions Short Answer Type

Question 1.
How can one distinguish between an unpolarised and linearly polarised light beam using polaroid? (CBSE Delhi 2019)
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

Question 2.
What is meant by plane polarised light? What type of waves shows the property of polarisation? Describe a method of producing a beam of plane polarised light?

1. The light that has its vibrations restricted in only one plane is called plane polarised light.
2. Transverse waves show the phenomenon of polarization.
3. Light is allowed to pass through a polaroid. The polaroid absorbs those vibrations which are not parallel to its axis and allows only those vibrations to pass which are parallel to its axis.

Question 3.
Write the Important characteristic features by which the Interference can be distinguished from the observed diffraction pattern. (CBSE AI 2015)
(a) In the interference pattern the bright fringes are of the same width, whereas in the diffraction pattern they are not of the same width.
(b) In interference all bright fringes are equally bright while in diffraction they are not equally bright.

Question 4.
State Brewster’s law. The value of Brewster’s angle for the transparent medium is different for the light of different colours. Give reason. (CBSE Delhi 2016)
When the reflected ray and the refracted ray are perpendicular then μ = tanip where ip is the polarising angle or Brewster angle.

Brewster’s angle depends upon the refractive index of the two media in contact. The refractive index in turn depends upon the wavelength of light used (different colours) hence Brewster’s angle is different for different colours.

Question 5.
Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids.
Let lo be the intensity of polarised light after passing through the first polarizer P1. Then the intensity of light after passing through the second polarizer P2 will be l = locos 2θ, where θ is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – θ). Hence the intensity of light emerging from P3 will be
l = lo cos² θ cos² (90° – θ) = lo cos² θ sin² θ = (lo /4)sin² 2θ

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 6.
Is energy conserved in interference? Explain.
Yes, energy is conserved in interference. Energy from the dark fringes is accumulated in the bright fringes. If we take
l = 4a²cos²$$\frac{\phi}{2}$$, then intensity at bright points is lmax = 4a² and intensity at the minima lmin = 0. Hence average intensity in the pattern of the fringes produced due to interference is given by
Ī = $$\frac{I_{\max }+I_{\min }}{2}=\frac{4 a^{2}+0}{2}$$ = 2a²

But if there is no interference then total intensity at every point on the screen will be l = a² + a² = 2a², which is the same as the average intensity in the interference pattern.

Question 7.
An incident beam of light of intensity lo is made to fall on a polaroid A. Another polaroid B is so oriented with respect to A that there is no light emerging out of B. A third polaroid C is now introduced midway between A and B and is so oriented that its axis bisects the angle between the axes of A and B. What is the intensity of light now between (i) A and C (ii) C and B? Give reasons for your answers.
Polaroids A and B are oriented at an angle of 90°, so no light is emerging out of B. On placing polaroid C between A and B such that its axis bisects the angle between axes of A and B, then the angle between axes of polaroids A and B is 45° and that of C and B also 45°.
(a) Intensity of light on passing through Polaroid A or between A and C is l1 = $$\frac{l_{0}}{2}$$
(b) On passing through polaroid C, intensity of light between C and B becomes
l2 = l1 cos² θ = $$\frac{l_{0}}{2}$$ × cos² 45° = $$\frac{l_{0}}{4}$$

Question 8.
One of the slits of Young’s double-slit experiment is covered with a semi¬transparent paper so that it transmits lesser light. What will be the effect on the interference pattern?
There will be an interference pattern whose fringe width is the same as that of the original. But there will be a decrease in the contrast between the maxima and the minima, i.e. the maxima will become less bright and the minima will become brighter.

Question 9.
Light from a sodium lamp is passed through two polaroid sheets P1 and P2 kept one after the other. Keeping P1, fixed, P2 is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P1.
An experimentalist records the following data for the intensity of light coming out of P2 as a function of the angle θ.

I = Intensity of beam falling on P1
(a) One of these observations is not in agreement with the expected theoretical variation of I, identify this observation and write the correct expression.
(b) Define the Brewster angle and write the expression for It in terms of the refractive index of the medium.
(a) The observation $$\frac{1}{\sqrt{2}}$$ is not correct. It should be 1/2.
(b) It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan ip = μ

Question 10.
How will the interference pattern in Young’s double-slit experiment get affected, when
(a) distance between the slits S1, and S2 reduced and
(b) the entire set-up is immersed in water? Justify your answer in each case. (CBSE Delhi 2011C)
We know that fringe width β of the dark or bright fringes is given by β = $$\frac{D \lambda}{d}$$ where d is the distance between the slits.
(a) When the distance between the slits, i. e. d is reduced then p will increase. The interference pattern will thus become broader.
(b) When the entire set up is immersed in water, the pattern will become narrow due to the decrease in the wavelength of light. The new wavelength λ’ = λ/n, hence β’= β/n

Question 11.
Discuss the intensity of transmitted light when a Polaroid sheet is rotated between two crossed polaroids? (NCERT)
Let l0 be the intensity of polarised light after passing through the first polarizer P1. Then the intensity of light after passing through the second polarizer P2 will be l = lo cos² θ

where θ is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – θ). Hence the intensity of light emerging from P3 will be

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 12.
A Polaroid (I) is placed In front of a monochromatic source. Another Polaroid (II) is placed in front of this Polaroid (I) and rotated till no light passes. A third Polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain. (NCERT Exemplar)
Only in the special case when the pass axis of (III) is parallel to fill or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Question 13.
(a) Good quality sunglasses made of polaroids are preferred over ordinary coloured glasses. Explain why.
(CBSE AI2019)
Polaroid sunglasses are preferred over coloured sunglasses because they reduce the intensity of light.

(b) How is plane polarised light defined?
Plane polarised light: Light in which vibrations of electric field vector are restricted to one plane containing the direction of propagation.

(c) A beam of plane polarised light is passed through a polaroid. Show graphically, a variation of the intensity of the transmitted light with the angle of rotation of the Polaroid.
The graphical variation is as shown.

### Wave Optics Important Extra Questions Long Answer Type

Question 1.
Define the term wavefront. Using Huygen’s wave theory, verify the law of reflection.
Or
Define the term, “refractive index” of a medium. Verify Snell’s law of refraction when a plane wavefront is propagating from a denser to a rarer medium. (CBSE Delhi 2019)
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM. Let us consider the situation when one end A of was front strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

Question 2.
(a) Sketch the refracted wavefront for the incident plane wavefront of the light from a distant object passing through a convex lens.

(b) Using Huygens’s principle, verify the laws of refraction when light from a denser medium is incident on a rarer medium.
Refraction from denser to the rarer medium: Let XY be plane refracting surface separating two media of refractive index μ1 and μ21 > μ2)

Let a plane wavefront AB incident at an angle i. According to Huygen’s principle, each point on the wavefront becomes a source of secondary wavelets and

Time is taken by wavelets from B to C = Time taken by wavelets from A to D

(c) For yellow light of wavelength 590 nm incident on a glass slab, the refractive index of glass Is 1.5. Estimate the speed and wavelength of yellow light Inside the glass slab. (CBSE 2019C)
Given λ = 590 nm, μ = 1.5
Velocity of light inside glass slab.
∴ v = $$\frac{C}{\mu}=\frac{3 \times 10^{8}}{1.5}$$ = 2 × 108 ms-1

Wavelength of yellow light inside the glass slab.
λ1 = $$\frac{\lambda}{\mu}=\frac{290}{1.5}$$ = 393.33 nm

Question 3.
(a) State the postulates of Huygen’s wave theory.
The postulates are
All points on a given wavefront are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate outward with speed characteristic of waves in that medium.

(b) Draw the type of wavefront that corresponds to a beam of light (i) coming from a very far off the source and (ii) diverging from a point source.
After some time has elapsed, the new position of the wavefront is the surface tangent to the wavelets or the envelope of the wavelets in the forward direction.

Question 4.
What is meant by the diffraction of light? Obtain an expression for the first minimum of diffraction.
The divergence of light from its initial line of travel when it passes through an opening or an obstacle is called diffraction or the phenomenon of bending of light around the sharp corners and spreading into the regions of the geometrical shadow is called diffraction.

Consider that a monochromatic source of light S, emitting light waves of wavelength λ, is placed at the principal focus of the convex lens L1. A parallel beam of light, i.e. a plane wavefront, gets incident on a narrow slit AB of width ‘a’ as shown in the figure.

The diffraction pattern is obtained on a screen Lying at a distance D from the slit and at the focal plane of the convex lens L2.

Consider a point P on the screen at which wavelets travelLing in a direction making angle O with CO are brought to focus by the lens. The wavelets from different parts of the slit do not reach point P in phase, although they are initially in phase. It is because they cover unequal distances in reaching point R The waveLets from points A and B will have a path difference equal to BN.

From the right angLed ANB, we have
BN = AB sin θ or BN = a sin θ …(1)

Suppose that the point P on the screen is at such a distance from the centre of the screen that BN = λ. and the angle θ = θ1.

Then, equation 1 gives
λ = a sin θ1 or sin θ1 = $$\frac{λ}{a}$$

Such a point one screen will be the position of the first secondary minimum.

Question 5.
Describe an experiment to show that light waves are transverse in nature.
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is Eo. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse In nature.

Question 6.
Derive an expression for the width of the central maxima for diffraction of light at a single-slit. How does this width change with an increase in the width of the slit?
The width of the central maxima for diffraction of light at a single-slit is the distance from the 1st diffraction minima on one side of the central maxima to the 1st diffraction minima on another side of the central maxima.

If a be the width of slit then for 1st diffraction minima, we have
sin θ = ± λ
or
sin θ = θ = ± $$\frac{λ}{a}$$

Angular width of central maxima
= 2θ = ± $$\frac{2λ}{a}$$

If D be the distance between the slit and the screen, then linear width ‘x’ of the central maxima is given by
x = d × 2θ = D × $$\frac{2λ}{a}$$ = $$\frac{2Dλ}{a}$$

As the width (a) of the slit is increased, the linear width of central maxima goes on decreasing because x ∝ $$\frac{1}{a}$$

Question 7.
(a) Sketchthegraphshowingthevariation of the intensity of transmitted light on the angle of rotation between a polarizer and an analyser.
The graph is as shown below

(b) A ray of light is incident at an angle of incidence ip on the surface of separation between air and a medium of refractive index µ, such that the angle between the reflected and refracted ray is 90°. Obtain the relation between ip and µ.
Suppose an unpolarised light beam is an incident on a surface as shown in the figure below. The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially polarized.

Now suppose the angle of incidence, i is varied until the angle between the reflected and the refracted beam is 90°. At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle ip.

An expression can be obtained relating the polarising angle to the index of refraction n, of the reflecting surface. From the figure, we see that at the polarising angle
ip + 90° + r = 180°
or
r = 90° – ip.

Using Snell’s law we have n = $$\frac{\sin i}{\sin r}=\frac{\sin i_{p}}{\sin r}$$

Now sin r = sin (90° – ip) = cos ip, therefore the above expression becomes
n = $$\frac{\sin i_{p}}{\cos i_{p}}$$ = tan ip

Question 8.
Describe Young’s double-slit experiment to produce an interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width. (CBSE Delhi 2011)
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S1 and S2, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r2 – r1 = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = $$\frac{Y}{D}$$ …(3)

Substituting this in equation (2) we have
d $$\frac{Y}{D}$$ = m λ …(2)

rewriting the above equation we have
y = $$\frac{mDλ}{d}$$ …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
yo = 0 …….. position of the central bright fringe

y1 = $$\frac{Dλ}{d}$$ ……. position of the first bright fringe

y2 = $$\frac{2Dλ}{d}$$ ……. position of the second bright fringe

ym = $$\frac{mDλ}{d}$$ …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’o = $$\frac{Dλ}{2d}$$, ……. position of the first dark fringe,
y’1 = $$\frac{3Dλ}{2d}$$ ……. position of the second dark fringe
y’2 = $$\frac{5Dλ}{2d}$$ ……. position of the third dark fringe and,
y’m = $$\frac{(2 m+1) D \lambda}{2 d}$$ ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = $$\frac{Dλ}{d}$$ …(10)

Question 9.
(a) Explain, with the help of a diagram, how plane polarised light Is obtained by scattering.
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Between two polaroids placed In crossed position a third Polaroid is introduced. The axis of the third Polaroid makes an angle of 30° with the axis of the first polaroid. Find the intensity of transmitted light from the system assuming / to be the intensity of polarised light obtained from the first polaroid. (CBSE A! 2011C)
Let lo be the intensity of light passing through the first polaroid.

The intensity of light passing through the middle polaroid whose axes are inclined at 30° to the first polaroid by Malus law is
l’ = lo cos² 30° = lo × 3/4 = 3lo/4

The intensity of light passing through the system is, therefore, (for the second crystal θ = 60°)
l” = l’cos² 60° = 3lo/4 × 1 /4
or
l” = 3lo/16

Question 10.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
Interference will be sustained if there is a constant phase difference between the two interfering waves. This is possible if the two waves are coherent.

(b) In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference Is λ, is K units. Find out the intensity of light at a point where the path difference is λ/3. (CBSE Delhi 2012)
Intensity at any point on the screen
l = l1 + l2 + 2$$\sqrt{1_{1}1_{2}}$$ cos Φ

Let l0 be the intensity of either source, then l1= l2 = lo
When p = λ, Φ = 2π

Question 11.
Use Huygen’s principle to explain the formation of the diffraction pattern due to a single-slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this affect the size and intensity of the central diffraction band? (CBSE Delhi 2012)
(a) The arrangement is shown below in the figure.

When a plane wavefront WW’ falls on a single-slit AB, each point on the unblocked portion ADB of wavefront sends out secondary, wavelets in all the directions. For secondary waves meeting at point O, the path difference between waves is zero and hence the secondary waves reinforce each other giving rise to the central maximum at the symmetrical point O.

Consider secondary waves travelling in a direction making an angle θ with DO and reaching the screen at point P. Obviously, path difference between the extreme secondary waves reaching point P from A and B.
= BC = AB sin θ = a sin θ.

If this path difference a sin θ = λ, then point P will be minimum Intensity. In this situation, wavefront may be supposed to consist of two equal halves AD and BD and for every point on AD, there will be a corresponding point on DB having a path difference λ/2. Consequently, they nullify the effect of each other and point P behaves as the first secondary minimum. In general, if path difference a sin θ = nλ where n = 1, 2, 3, …. then we have secondary minima corresponding to that angle of diffraction θn.

However, if for some point P1 on the screen secondary waves BP1 and AP1 differ In path by 3λ/2 then point P1 will be the position of the first secondary maxima. Because in this situation, wavefront AB may be divided into three equal parts such that path difference between corresponding points on the first and second will be λ/2 and they will nullify. But secondary waves from the third part remain as such and give rise to the first secondary maxima, whose Intensity will be much less than that of central maxima.

In general, if path difference a sin θ = (2n + 1)θ/2, where n = 1, 2, 3, …… then we have nth secondary maxima corresponding to these angles.

The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle θ subtended by the first minima on either side of the central maxima. Now sin θ = $$\frac{\lambda}{a}$$. Since θ is small a therefore it can be replaced by tan θ, hence sin θ = $$\frac{\lambda}{a}=\frac{y}{L} or y = [latex]\frac{L \lambda}{\mathrm{a}}$$.

This gives the distance of the first secondary minima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = $$\frac{2L \lambda}{\mathrm{a}}$$

(b) The size reduces by half according to the relation: size = X/d. Intensity becomes twice the original intensity.

Question 12.
(a) Using the phenomenon of polarization, show how the transverse nature of light can be demonstrated.
Light is a transverse wave. This can be shown with the help of this simple experiment. The figure shows an unpolarized light beam incident on the first polarising sheet, called the polariser where the transmission axis is indicated by the straight line on the polariser.

The polariser can be a thin sheet of tourmaline (a complex boro-silicate). The Light, which is passing through this sheet, is polarised vertically as shown, where the transmitted electric vector is Eo. A second polarising sheet called the analyser intercepts this beam with its transmission axis at an angle θ to the axis of the polariser. As the axis of the analyser is rotated slowly, the intensity of Light received beyond It goes on decreasing.

When the transmitting axis of the analyser becomes perpendicular to the transmission axis of the polariser, no beam is obtained beyond the analyser. This means that the anaLyser has further polarised the beam coming from the polariser. Since only transverse waves can be polarised, therefore, this shows that light waves are transverse in nature.

(b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity l0 is Incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 30° with that of P1. Determine the intensity of light transmitted through P1 P2 and P3. (CBSE AI 2014)
The intensity of light passing through P1 is half of the light falling on it. Therefore, the light coming out of P1 is l0o/2 Now light coming out of P3 is

and light coming out of polariser P2 is

Question 13.
What does a Polaroid consist of? Show using a simple Polaroid that light waves are transverse in nature. The intensity of light coming out of a Polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why. (CBSE AI 2015)
A polaroid consists of long-chain molecules aligned in a particular direction.

Let the light from an ordinary source (like a sodium lamps pass through a polaroid sheet P1 it is observed that its intensity is reduced by half. Now, let an identical piece of polaroid P2 be placed before P1 As expected, the light from the lamp is reduced in intensity on passing through P22 alone. But now rotating P1 has a dramatic effect on the light coming from P2. In one position, the intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90° from this position, P1 transmits nearly the full intensity emerging from P2 shown in the figure.

Since only transverse waves can be polarised, therefore, this experiment shows that light waves are transverse in nature. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarized light wave is an incident on such a polaroid then the light wave will get linearly polarized with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the Polaroid.

Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P it is observed that its intensity is reduced by half. Rotating P has no effect on the transmitted beam and transmitted intensity remains constant as there is always an electric vector that oscillates in a direction perpendicular to the direction of the aligned molecules.

Question 14.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted Intensity be maximum? (CBSE Delhi 2015)
Let l0 be the intensity of polarised light after passing through the first polarizer P1. Then the intensity of light after passing through the second polarizer P2 will be
l = lo cos² θ

where θ is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – θ). Hence the intensity of light emerging from P3 will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 15.
Distinguish between unpolarised and linearly polarised light. Describe with the help of a diagram how unpolarised light gets linearly polarised by scattering. (CBSE Delhi 2015)
The two lights will be allowed to pass through a polariser. When the polarizer is rotated in the path of these two light beams, the intensity of light remains the same in all the orientations of the polariser, then the light is unpolarised. But if the intensity of light varies from maximum to minimum then the light beam is a polarised light beam.

It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

Question 16.
(a) Explain how a diffraction pattern is formed due to interference of secondary wavelets of light waves from a slit.
(a) Diffraction at single slit: Consider monochromatic plane wavefront WW’ incident on the slit AB as shown in Fig. Imagine the slit to be divided into a large number of very narrow strips of equal width parallel to the slit. When the wavefront reaches the slit, each narrow strip parallel to the slit can be considered to be a source of Huygen’s secondary wavelets.

Central Maxima. Let us first consider the effect of all the wavelets at point O. All wavelets cover the same distance and reach 0 in the same phase. The wavelets superimpose constructively and give maximum intensity at O.

Position of secondary minima. Consider the intensity at a point P at an angle θ with the original direction.
Draw AL ⊥ BP.
In ΔABL

or
sin θ = $$\frac{λ}{a}$$

For second minima, BL = 2A
or a sin θ = 2λ

Similarly, for nth minima a sin θ = nλ
or sin θ = $$\frac{nλ}{a}$$ , where n = 1, 2, ……

Position of secondary maxima: If path difference BL = $$\frac{3λ}{2}$$, then slit AB can be divided into three equal parts and the path difference between wavelets from corresponding points in the first two parts will be $$\frac{λ}{2}$$ and hence cancels each other’s effect and produces destructive Interference and the third path will produce its maxima at P called first secondary maxima.

Similarly, if BL = $$\frac{5λ}{2}$$, we get second secondary maxima and so on.

For nth maxima
a sin θ = (2n + 1)$$\frac{λ}{a}$$
or
sin θ = $$\frac{(2 n+1) \lambda}{2 a}$$

Thus we find that the intensity of the central fringe is maximum whereas that of other fringes fall off rapidly in either direction from the centre of the fringe pattern.

(b) Sodium light consists of two wavelengths, 5900 Å and 5960 Å. If a slit of width 2 × 10-4 m is Illuminated by sodium light, find the separation between the first secondary maxima of the diffraction pattern of the two wavelengths on a screen placed 1.5m away. (CBSE 2019C)
The separation between the first secondary maxima of the diffraction pattern of two wavelengths is:

Question 17.
(a) Derive Snell’s law on the basis of Huygen’s wave theory when light Is travelling from a denser to a rarer medium.
The ray diagram is as shown.

(b) Draw the sketches to differentiate between plane wavefront and spherical wavefront. (CBSE AI 2016)
Spherical wavefront and plane wavefront

Question 18.
The figure drawn here shows the geometry of path differences for diffraction by a single-slit of width a.

Give appropriate ‘reasoning’ to explain why the intensity of light is
(a) Maximum at the central point C on the screen.
(b) (Nearly) zero for point P on the screen when θ = λ / a.
Hence write an expression for the total linear width of the central maxima on a screen kept at a distance D from the plane of the slit. (CBSE Delhi 2016C)
(a) At central point C, the angle is zero, all path differences are zero. Hence all the parts of the slit contribute to the same phase. This gives the maximum intensity at point C.
(b)

From figure
NP – LP = NQ= a sin θ = aθ When θ = λ/a

Then path difference NP – LP = aθ = λ.
Hence MP – LP = NP – MP = λ/2

It implies that the contribution from corresponding points in two halves of the slit has a phase difference of π. Therefore, contributions from two halves cancel each other in pairs, resulting in a zero net intensity at point P on the screen. Half angular width of central maxima = λ/a Half linear width = λD/a

Linear width of central maxima = 2λD/a

Question 19.
Two polaroids, P1 and P2, are ‘set-up’ so that their ‘pass axis is ‘crossed’ with respect to each other. AQ third Polaroid, P3 is now introduced between these two so that its ‘pass axis makes an angle with the ‘pass axis of P1. A beam of unpolarized light, of Intensity I, Is Incident on P1 If the Intensity of light that gets transmitted through this combination of three polaroids, Is I’, find the ratio $$\left(\frac{I^{\prime}}{I}\right)$$ when θ equals: (i) 30°, (ii) 45° (CBSE Delhi 2016C)
The diagram is as shown.

The intensity of unpolarized light is given as l.
It becomes half after passing through Polaroid P1 (l/2)
Using Malus law for the intensity of light passing through P3 we have
l1 = $$\left(\frac{l}{2}\right)$$ cos²θ

This intensity l1 is Incident on P2, hence the intensity of light coming out of P2 will be

Question 20.
What is the effect on the Interference pattern observed In Young’s double-slit experiment In the following cases:
(a) Screen is moved away from the plane of the slits,
(b) Separation between the slits is Increased and
(c) Widths of the slits are doubled. Give the reason for your answer.
The fringe width is given by the expression
β = $$\frac{Dλ}{d}$$

(a) When D Is Increased, the fringe width Increases.
(b) When d 1s Increased the fringe width decreases.
(c) If the width w of the slits Is changed then Interference occurs only If $$\frac{1}{w}$$ > $$\frac{1}{d}$$ remains satisfied, where d is the distance between the slits.

Question 21.
Two slits In Young’s double-slit experiment are illuminated by two different lamps emitting light? Will you observe the Interference pattern? Justify your answer. Find the ratio of Intensities at two points on a screen In Young’s double-slit experiment, when waves from two slits have a path difference of (i) 0 and (ii) λ/4.
The two sources act as Independent sources of light and hence can never be coherent. In a light, source light is produced by billions of atoms under proper excitation condition and each atom acts independently of the other atoms. Thus there Is no coherence ‘ between these two Independent sources hence no interference.

The phase difference corresponding to the two paths are Φ = 0 and Φ = π/2

Now intensity at the screen when the phase difference is Φ = 0 is
lA = l1 +l2 + 2$$\sqrt{l_{1} l_{2}}$$cos Φ
or
lA = l + l + 2$$\sqrt{l l}$$cos Φ = 4l

Now intensity at the screen when the phase difference is Φ = 90° Is
lB = l1 +l2 + 2$$\sqrt{l_{1} l_{2}}$$cos Φ
or
lB = l + l + 2$$\sqrt{l l}$$cos 90° = 2l
Therefore, ratio of intensities is
$$\frac{l_{A}}{l_{B}}=\frac{4l}{2l}$$ = 2

Question 22.
(a) In a single-slit diffraction pattern, how does the angular width of the central maximum vary, when
(i) the aperture of the slit Is Increased?
(ii) distance between the slit and the screen is decreased?
The angular width of the central maxima in a single-slit diffraction pattern is given by 2θ = $$\frac{2λ}{a}$$ where λ is the wavelength of light and ‘a’ the slit width.
(i) When the aperture of the slit is increased the angular width decreases.
(ii) When the distance between the slit and the screen is decreased, the angular width will remain the same but the linear width will increase.

(b) How Is the diffraction pattern different from the interference pattern obtained In Young’s double¬slit experiment? (CBSE Delhi 2011C)
The difference is shown in the table:

 Diffraction Interference 1. It is due to interference between the wavelets starting from two parts of the same wavefront. 1. It is a superposition of the two waves starting from two coherent sources. 2. The intensity of the consecutive bands goes on decreasing. 2. All bright fringes have the same intensity. 3. Fringes have poor contrast. 3. Fringes have good contrast. 4. Diffraction fringes are not of the same width. 4. Fringes may or may not be of the same width.

Question 23.
(a) Can two independent monochromatic light sources be used to obtain a steady interference pattern? Justify your answer. (CBSE 2019C)
No, because the phase difference between the light waves from two independent sources keeps on changing continuously and for a steady interference pattern, the phase difference between the waves should remain constant with time. Hence two independent monochromatic light sources cannot produce a steady interference pattern.

(b) In Young’s double-slit experiment, explain the formation of interference fringes and obtain an expression for the fringe width.
Interference of light

Let S1, S2 be the two fine slits illuminated by a monochromatic source S of wavelength λ.

The intensity of light at any point P on the screen at a distance D from the slit depends upon the path difference between S2P and S1P.
∴ Path difference = S2P – S1P
= (S2A + AP) – S1P

Path difference = S2A = d sin θ
Since θ is small, sin θ can be replaced by tan θ.
∴ Path difference = d tan θ = d$$\frac{y}{D}$$

Constructive interference [Bright Fringes]. For bright fringes, the path difference should be equal to integral multiple of A.
∴ $$\frac{dy}{D}$$ = nλ
or
y = n$$\frac{λD}{d}$$

For nth fringe, let us write y as yn
so yn = n$$\frac{λD}{d}$$

The spacing between two consecutive bright fringes is equal to the width of a dark fringe.
∴ Width of the dark fringe.
β = yn – yn-1 = $$\frac{nλD}{d}$$ – (n – 1)$$\frac{λD}{d}$$ = $$\frac{λD}{d}$$ …(i)

Destructive interference. [Dark Fringes].
For dark fringes, the path difference should be an odd multiple of λ/2.
∴ $$\frac{d}{D}$$y = (2n + 1)λ/2
or
y’ = $$\frac{D}{d}$$(2n + 1)$$\frac{λ}{2}$$ = $$\frac{(2 n+1) \lambda D}{2 d}$$

For nth fringe, writing y as yn, we get
y’n = $$\frac{(2 n+1) \lambda D}{2 d}$$.

The spacing between two consecutive dark fringes is equal to the width of a bright fringe.
∴ Width of the bright fringe.

From Eqs. (i) and (ii), we find that dark and bright fringes are of same width given by
β = $$\frac{λD}{d}$$.

(c) In an interference experiment using monochromatic light of wavelength A, the intensity of light of point, where the path difference is X, on the screen is K units. Find out the Intensity of light at a point when path difference is λ/4. (CBSE 2019C)
The intensity of light on the screen where the waves meet having phase difference Φ is:

Question 24.
(a) Two monochromatic waves emanating from two coherent sources have the displacements represented by y1 = a cos ωt and y2 = a cos (ωt + Φ), where Φ is the phase difference between the two displacements. Show that the resultant intensity at a point due to their superposition is given by l = 4lo cos² Φ/2, where lo = a².
Let the displacements of the waves from the sources S1 and S2 at a point on the screen at any time t be given by y1 = a cos ωt and y2 = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore the intensity at that point is
l = A² = 4a² cos² $$\frac{Φ}{2}$$

(b) Hence obtain the conditions for constructive and destructive interference. (CBSE AI 2014C)
Bright fringes: For bright fringes I = max, therefore Φ = 0° or cos Φ = +1
or
Φ = 2 n π
Dark fringes: For dark fringes l = 0,
therefore Φ = π or cos Φ = -1
or
Φ = (2n + 1) π

Question 25.
A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction pattern on the screen placed parallel to the plane of the slit. Use Huygens’ principle to explain that
(a) the central bright maxima is twice as wide as the other maxima.
The width of the central maxima is the distance between the first secondary minima on either side of the centre of the screen. The width of the central maxima is twice the angle 6 subtended by the first minima on either side of the central maxima.

Now sin θ = $$\frac{λ}{a}$$. Since θ is small there, a fore it can be replaced by tan 0, hence
tan θ = $$\frac{λ}{a}$$ = $$\frac{y}{L}$$
or
y = $$\frac{Lλ}{a}$$. This gives the distance of the first secondary min¬ima on both sides of the centre of the screen. Therefore, the width of the central maxima is 2y, hence
2y = $$\frac{2Lλ}{a}$$

(b) the Intensity falls as we move to successive maxima away from the centre on either side. (CBSE Delhi 2014C)
The maxima in diffraction pattern are formed at (n +1 /2) λ/a, with n = 2, 3, etc. These become weaker with increasing n, since only one third, one- fifth, one-seventh, etc., of the slit contributes in these cases.

Question 26.
(a) In a double-slit experiment using the light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.

(b) Light of wavelength 5000 A propagating 1n air gets partly reflected from the surface of the water. How will the wavelengths and frequencies of the reflected and refracted light be affected? (CBSE Delhi 2015)
No change in the wavelength and frequency of reflected light. In the case of refracted light, there is no change in frequency but the wavelength becomes 1/1.33 times the original wavelength.

Question 27.
A Polaroid sun-glass limits the light entering the eye, thus providing a soothing effect.

(b) Two polaroids and P2 are placed In crossed positions. A third Polaroid P3 is kept between P1 and P2 such that the pass axis of P3 is parallel to that of P1. How would the Intensity of light l2 transmitted through P2 vary as P3 Is rotated? Draw a plot of Intensity l2 Vs the angle ‘θ’, between pass axes of P1 and P3. (CBSE AI 2015C)
Let l0 be the intensity of polarised light after passing through the first polarizer P1 Then the intensity of light after passing through the second polarizer P3 will be
l = lo cos² θ
where θ is the angle between pass axes of P1 and P3. Since P1 and P2 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – θ).

Hence the intensity of light emerging from P2 will be

Therefore, the transmitted intensity will be maximum when θ = π/4

For graph

Question 28.
In a single-slit diffraction pattern, how does the angular width of central maximum change, when
(a) slit width is decreased,
(b) distance between the slit and screen is increased, and
(c) light of smaller visible wavelength is used? Justify your answer in each case.
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by
w = $$\frac{2Dλ}{a}$$.

(a) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ $$\frac{1}{a}$$

(b) Increase in distance between the slit and the screen does not affect the angular width of diffraction maxima. However, linear width of the maxima
w = $$\frac{2Dλ}{a}$$ will increase.

(c) If the light of a smaller visible wavelength is used, the angular width is decreased because x ∝ λ.

Question 29.
Light, from a sodium lamp, is passed through two polaroid sheets P1 and P2 kept one after the other. Keeping P1 fixed, P2 is rotated so that its ‘pass axis can be at different angles, θ, with respect to the pass-axis of P1
An experimentalist records the following data for the Intensity of light coming out of P2 as a function of the angle θ.

(a) lo = Intensity of beam falling on P1 One of these observations is not in agreement with the expected theoretical variation of l, Identify this observation and write the correct expression.
The observation $$\frac{l_{0}}{2 \sqrt{2}}$$ is not correct. It should be lo/4.

(b) Define the Brewster angle and write the expression for it in terms of the refractive index of the medium.
It is the angle of incidence at which the refracted and the reflected rays are perpendicular to each other. It is related to the refractive index as tan ip = n

Question 30.
(a) Light, from a monochromatic source, is made to fall on a single-slit of variable width. An experimentalist records the following data for the linear width of the principal maxima ‘ on a screen kept at a distance of 1 m from the plane of the slit.

Use any two observations from this data to estimate the value of the wavelength of light used.
The width of the central maxima is given by the expression β = $$\frac{2Lλ}{a}$$
or
λ1 = $$\frac{βa}{2L}$$

Using the values of the first observation we have
λ1 = $$\frac{6 \times 10^{-3} \times 0.1 \times 10^{-3}}{2 \times 1}$$ = 0.3 × 10-6

Using the values of the second observation we have
λ1 = \frac{3 \times 10^{-3} \times 0.2 \times 10^{-3}}{2 \times 1} = 0.3 × 10-6

Thus the wavelength of light used is λ = 0.3 × 10-6 m

(b) Show that the Brewster angle iB for a given pair of transparent media is related to their critical angle ic through the relation ic = sin-1 (ωt iB)
We know that

Question 31.
For a single-slit of width, “a” the first minimum of the interference pattern of monochromatic light of wavelength λ occurs at an angle of λ/a. At the same angle λ/a, we get a maximum for two narrow slits separated by a distance ‘a’. Explain. (CBSE Delhi 2014)
The path difference between two secondary wavelets is given by nλ = a sin θ. Since θ is a very small sin θ = 0. So, for the first-order diffraction n = 1, the angle is λ/a. Now we know that θ must be very small θ = 0 (nearly) because of which the diffraction pattern is minimum.

Now for interference case, for two interfering waves of intensity l1 and l2 we must have two slits separated by a distance. We have the resultant intensity,
l = l1 + l2 + 2$$\sqrt{l_{1} l_{2}}$$ cos θ

Since θ = 0° (nearly) corresponding to angle λ/a so cos θ = 1 (nearly)
So,
l = l1 + l2 + 2$$\sqrt{l_{1} l_{2}}$$ cos θ
l = l1 + l2 + 2$$\sqrt{l_{1} l_{2}}$$

We see the resultant intensity is the sum of the two intensities, so there is a maxima corresponding to the angle λ/a.

This is why, at the same angle of λ/a, we get a maximum for two narrow slits separated by a distance “a”.

Question 32.
Show using a proper diagram of how unpolarised light can be linearly polarized by reflection from a transparent glass surface. (CBSE AI 2018, Delhi 2018)
An ordinary beam of light, on reflection from a transparent medium, becomes partially polarised. The degree of polarisation increases as the angle of incidence is increased. At a particular value of the angle of incidence, the reflected beam becomes completely polarised. This angle of incidence is called the polarising angle (ip).

Question 33.
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
The size reduces by half according to the relation: size = λ/d. Intensity increases fourfold.

(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot.

(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily? (NCERT)
For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to the wavelength. If the size of the obstacle/aperture is much too large compared to the wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about 5 × 10-7 m, while sound waves of, say, 1 kHz frequency have a wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot.

Question 34.
(a) When an unpolarized light of intensity l0 is passed through a polaroid, what is the intensity of the linearly polarised light? Does it depend on the orientation of the polaroid? Explain your answer.
(a) The intensity of the linearly polarised light would be lo/2.
No, it does not depend on the orientation.
Explanation: The polaroid will let the component of the unpolarized light, parallel to its pass axis, to pass through it irrespective of its orientation.

(b) A plane polarised beam of light is passed through a polaroid. Show graphically the variation of the intensity of the transmitted light with an angle of rotation of the polaroid incomplete one rotation. (CBSE Delhi 2018C)
We have l = lo cos² θ
∴ The graph is as shown below

Question 35.
(a) If one of two identical slits producing interference in Young’s experiment is covered with glass so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the Interference pattern.
As intensity is directly proportional to the square of the amplitude.

(b) What kind of fringes do you expect to observe if white light is used Instead of monochromatic light?
The central fringe will be white and the remaining will be coloured fringes of different width in the VIBGYOR sequence.

Question 36.
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum? (CBSE Delhi 2015)
Let l0 be the intensity of polarised light after passing through the first polarizer P1 Then the intensity of light after passing through the second polarizer P2 will be
l = lo cos² θ
where 0 is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – θ). Hence the intensity of light emerging from P3 will be

The transmitted intensity will be maximum when 2θ = π/2 or θ = π/4

Question 37.
Derive the expression for the fringe width In Young’s double-slit experiment.
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S1 and S2, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r2 – r1 = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = $$\frac{Y}{D}$$ …(3)

Substituting this in equation (2) we have
d $$\frac{Y}{D}$$ = m λ …(2)

rewriting the above equation we have
y = $$\frac{mDλ}{d}$$ …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
yo0 = 0 …….. position of the central bright fringe

y1 = $$\frac{Dλ}{d}$$ ……. position of the first bright fringe

y2 = $$\frac{2Dλ}{d}$$ ……. position of the second bright fringe

ym = $$\frac{mDλ}{d}$$ …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’o = $$\frac{Dλ}{2d}$$, ……. position of the first dark fringe,

y’1 = $$\frac{3Dλ}{2d}$$ ……. position of the second dark fringe

y’2 = $$\frac{5Dλ}{2d}$$ ……. position of the third dark fringe and,

y’m = $$\frac{(2 m+1) D \lambda}{2 d}$$ ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, a width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = $$\frac{Dλ}{d}$$ …(10)

Question 38.
(a) Using Huygens’ principle, draw a diagram to show the propagation of a wave-front originating from a monochromatic point source.

(b) Describe diffraction of light due to a single-slit. Explain the formation of a pattern of fringes obtained on the screen and plot showing a variation of intensity with path difference in single-slit diffraction.
Let us discuss the nature of the Fraunhofer diffraction pattern produced by a single-slit. Let us examine the waves coming from the various portions of the slit, as shown in the figure. According to Huygens principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the slit can interfere with the light of another portion, and the resultant intensity on the screen will depend on the direction of θ. The secondary waves coming from the different parts of the slit interfere to produce either maxima or minima, thereby giving rise to a diffraction pattern.

For the diagram

2y = $$\frac{2Lλ}{a}$$

Question 39.
What is the effect on the interference fringes in Young’s double-slit experiment due to each of the following operations:
(a) the screen is moved away from the plane of the slits;
The angular separation of the fringes remains constant (= λ/d). The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the slits

(b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength;
The separation of the fringes (and also angular separation) decreases.

(c) the separation between the two slits is Increased
The separation of the fringes (and also angular separation) decreases.

(d) the source slit is moved closer to the double-slit plane;
Let s be the size of the source and S its distance from the plane of the two slits. For interference, fringes to be seen the
condition $$\frac{s}{S}<\frac{\lambda}{d}$$ should be satisfied otherwise interference patterns produced by different parts of the source overlap and no fringes are seen. Thus as S decreases (i.e. the source slit is brought closer), the interference pattern gets less and less sharp and when the source is brought too close for this condition to be valid the fringes – disappear. Till this happens, the fringe separation remains fixed.

(e) the width of the source slit is Increased;
Same as in (d). As the source slit width increases fringe pattern gets less and less sharp. When the source slit is so wide that the condition $$\frac{s}{S}<\frac{\lambda}{d}$$ is not satisfied the interference pattern disappears.

(f) the monochromatic source is replaced by a source of white light? (In each operation take all parameters other than the one specified to remain unchanged.)
The interference patterns due to different component colours of white light overlap (incoherently). The central bright fringes for different colours are in the same position. Therefore, the central fringe is white. For a point P for which S2P – S1P= λb /2, where (λb = 400 nm) represents the wavelength for the blue colour, the blue component will be absent and the fringe will appear red in colour. Slightly farther away from where S2Q – S1Q = λb = λr/2 where λr (800 nm) is the wavelength for the red colour, the fringe will be predominantly blue. Thus, the fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen.

Question 40.
What is the diffraction of light? Draw a graph showing the variation of intensity with the angle in a single-slit diffraction experiment. Write one feature which distinguishes the observed pattern from the double-slit interference pattern. How would the diffraction pattern of a single-slit be affected when:
(i) the width of the slit is decreased?
(ii) the monochromatic source of light is replaced by a source of white light?
Diffraction is the bending of light around obstacles or openings. It is a consequence of the wave nature of light. For diffraction to take place the obstacle should be of the order of the wavelength of light.

The intensity distribution for the single-slit diffraction pattern is as shown.

The intensity of all bright fringes is the same in Young’s interference pattern, but in diffraction, the intensity of bright fringes falls off on both sides of the central fringe.
We know that the angular width of the central maximum of the diffraction pattern of a single-slit is given by = $$\frac{2Dλ}{a}$$
(i) If slit width ‘a’ is decreased, the angular width will increase because
x ∝ $$\frac{1}{a}$$

(ii) When monochromatic light is replaced by white light, all the seven wavelengths form their own diffraction pattern, so coloured fringes are formed. The first few fringes are visible but due to overlapping the clarity of fringes decreases as the order increases.

Question 41.
(a) Define a wavefront. Using Huygens’ geometrical construction, explain with the help of a diagram how the plane wavefront travels from the instant t1 to t2 in the air.
(a) Wavefront: It is the locus of the medium or points of a medium that is in the same phase of disturbance
First, consider a plane wave moving through free space as shown in the figure. At t = 0, the wavefront is indicated by the plane labelled AA’. In Huygens’s construction, each point on this wavefront is considered a point source. For clarity, only a few points on AA’ are shown. With these points as sources for the wavelets, we draw circles each of radius cΔt, where c is the speed of light in free space and Δt is the time of propagation from one wavefront to the next. The surface drawn tangent to these wavelets is the plane BB’, (Wavefront at a later time t). Here A1 A2 = B1 B2 = C1C2 = Cτ

(b) A plane wavefront is incident on a convex lens. Explain, with the help of the diagram, the shape of the refracted wavefront formed. (CBSE AI 2019)
Spherical wavefront

Question 42.
(a) In Young’s double-slit experiment, derive the condition for
(i) constructive Interference and
(ii) destructive Interference at a point on the screen.
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S1 and S2, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a wavelength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r2 – r1 = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = $$\frac{Y}{D}$$ …(3)

Substituting this in equation (2) we have
d $$\frac{Y}{D}$$ = m λ …(2)

rewriting the above equation we have
y = $$\frac{mDλ}{d}$$ …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
yo = 0 …….. position of the central bright fringe

y1 = $$\frac{Dλ}{d}$$ ……. position of the first bright fringe

y2 = $$\frac{2Dλ}{d}$$ ……. position of the second bright fringe

ym = $$\frac{mDλ}{d}$$ …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’o = $$\frac{Dλ}{2d}$$, ……. position of the first dark fringe,

y’1 = $$\frac{3Dλ}{2d}$$ ……. position of the second dark fringe

y’2 = $$\frac{5Dλ}{2d}$$ ……. position of the third dark fringe and,

y’m = $$\frac{(2 m+1) D \lambda}{2 d}$$ ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = $$\frac{Dλ}{d}$$ …(10)

(b) A beam of light consisting of two wavelengths, 800 nm and 600 nm, is used to obtain the interference fringes in Young’s double-slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (CBSE Al 2012)
Let at a distance y from central maxima the bright fringes due to wavelengths l1 and l2 coincide first time. For this to happen, nl1 = (n + 1)l2, where n is an integer.
or
$$\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{800 \times 10^{-9}}{600 \times 10^{-9}}=\frac{4}{3}$$ solving for n we have n = 3

It means that at distance nth (3th) maxima for wavelength λ1 is just coinciding with (n + 1)th (4th) maxima for wavelength l2

Question 43.
(a) How does an unpolarized light incident on a Polaroid get polarised? Describe briefly, with the help of a necessary diagram, the polarisation of light by reflection from a transparent medium.
(b) Two polaroids ‘A’ and ‘B’ are kept in a crossed position. How should a third Polaroid ‘C’ be placed between them so that the intensity of polarised light transmitted by Polaroid B reduces to 1/8th of the intensity of unpolarized light incident on A? (CBSE AI 2012)
(a) The polariser allows only those vibrations of light to pass which are parallel to the pass axis of the polariser and blocks the remaining vibrations. Thus the light vibrations are restricted to only one plane.

Suppose an unpolarised light beam is an incident on a surface as shown in (figure a). The beam can be described by two electric field components, one parallel to the surface (the dots) and the other perpendicular to the first and to the direction of propagation (the arrows). It is found that the parallel components reflect more strongly than the other component, this results in a partially polarised beam. Furthermore, the refracted ray is also partially. polarised. Now suppose the angle of incidence 6, is varied until the angle between the reflected and the refracted beam is 90° (fig b). At this particular angle of incidence, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised.

(b) Let l0 be the intensity of unpolarized light incident on Polaroid A. Then lo/2 will be the intensity of polarised light after passing through the first polariser A. Then the intensity of light after passing through second polariser B will be l = (lo/2)cos²θ

where θ is the angle between pass axes of A and B. Since A and B are crossed the angle between the pass axes of B and C will be (π/2 – θ). Hence the intensity of light emerging from C will be

Therefore, we have
$$\frac{l_{0}}{8}=\frac{l_{0}}{8}$$ sin² 2θ or sin 2θ = 1 or 2θ = 90°
i.e. θ = π/4

Therefore, the transmitted intensity will be maximum when θ = π/4

Question 44.
(a) Describe any two characteristic features which distinguish between interference and diffraction phenomena. Derive the expression for the intensity at a point of the interference pattern in Young’s double-slit experiment.
(b) In the diffraction due to a single slit experiment, the aperture of the slit is 3 mm. If monochromatic light of wavelength 620 nm is incident normally on the slit, calculate the separation between the first order minima and the 3rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. (CBSE Delhi 2019)
(a) (i) Interference pattern has a number of equally spaced bright and dark bands, while the diffraction pattern has a central bright maximum which is twice as wide as the other maxima.

(ii) In interference pattern, the intensity of all bright fringes is the same, while in diffraction pattern intensity of bright fringes goes on decreasing with the increasing order of the maxima.

Let the displacements of the waves from the sources S1 and S2 at a point P on the screen at any time t be given by y1 = a cos cot and y2 = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves.

By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is
A = 2a cos (Φ/2) …(4)

Therefore the intensity at that point is
l = A² = 4a² cos²$$\frac{Φ}{2}$$ …(5)

(b) Given a = 3mm = 3 × 10-3m, λ = 620nm = 620 × 10-9 m, D = 1.5 m Distance of first-order minima from centre of the central maxima = XD1 = λD/a

Distance of third order maxima from centre of the central maxima XD3 = 7λD/2a

Distance between first order minima and third order maxima = XD3 – XD1

Question 45.
(a) In Young’s double-slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence deduce the expression for the fringe width.
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits SD1 and S2, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r2 – r1 = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = $$\frac{Y}{D}$$ …(3)

Substituting this in equation (2) we have
d $$\frac{Y}{D}$$ = m λ …(2)

rewriting the above equation we have
y = $$\frac{mDλ}{d}$$ …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
yo = 0 …….. position of the central bright fringe

y1 = $$\frac{Dλ}{d}$$ ……. position of the first bright fringe

y2 = $$\frac{2Dλ}{d}$$ ……. position of the second bright fringe

ym = $$\frac{mDλ}{d}$$ …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’o = $$\frac{Dλ}{2d}$$, ……. position of the first dark fringe,

y’1 = $$\frac{3Dλ}{2d}$$ ……. position of the second dark fringe

y’2 = $$\frac{5Dλ}{2d}$$ ……. position of the third dark fringe and,

y’m = $$\frac{(2 m+1) D \lambda}{2 d}$$ ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = $$\frac{Dλ}{d}$$ …(10)

(b) Show that the fringe pattern on the screen is actually a superposition of single-slit diffraction from each slit.
It is a broader diffraction peak in which there appear several fringes of smaller width due to the double-slit interference pattern. This is shown below:

(c) What should be the width of each slit to obtain 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, for the green light of wavelength 500 nm, if the separation between two slits is 1 mm? (CBSE AI 2015)
Given 10 b = width of central maxima in diffraction pattern λ = 500 nm, d= 1 mm,
Now 10$$\frac{Dλ}{d}$$ = $$\frac{Dλ}{a}$$
or
a = $$\frac{d}{5}=\frac{1}{5}$$ = 0.2 mm

Question 46.
(a) Define a wavefront. How is it different from a ray?
A wavefront is defined as the locus of all adjacent points at which the phase of vibration of a physical quantity associated with the wave is the same.

It is in two dimensions while a ray is in one dimension.

(b) Depict the shape of a wavefront in each of the following cases.
(i) Light diverging from a point source,

(ii) Light emerging out of a convex lens when a point source is placed at its focus.

(c) Using Huygen’s construction of secondary wavelets, draw a diagram showing the passage of a plane wavefront from a denser into a rarer medium. (CBSE AI 2015C)
The diagram is as shown

Question 47.
(a) Why does unpolarised light from a source show a variation in intensity when viewed through a Polaroid which is rotated? Show with the help of a diagram how unpolarised light from sun got linearly polarised by scattering.
(b) Three identical Polaroid sheets P1, P2 and P3 are oriented so that the pass axis of P2 and P3 are inclined at angles of 60° and 90° respectively with the pass axis of P1. A monochromatic source S of unpolarised light of intensity l0 is kept in front of the polaroid sheet P! as shown in the figure. Determine the intensities of light as observed by the observer at O when Polaroid P3 is rotated with respect to P2 at angles θ = 30° and 60°. (CBSE AI 2016)

(a) When light passes through a Polaroid it absorbs all vibrations which are not parallel to its pass axis.
It is found that when unpolarised sunlight is incident on a small dust particle or air molecules it is scattered in all directions. The light scattered in a direction perpendicular to the incident light is found to be completely polarised.

The electric vector of incident light has components both in the plane of the paper as well as perpendicular to the plane of the paper. The electrons under the influence of the electric vector acquire motion in both directions. However, radiation scattered by the molecules in perpendicular direction contains only components represented by dots (.), i.e. polarised perpendicular to the plane of the paper.

(b) Let l0 be the intensity of light before passing P1 Intensity of light coming out of P1 will be lo/2.

Then the intensity of light after passing through second poloroid (analyser) is
l2 = $$\frac{l_{0}}{2}$$cos²60° = $$\frac{l_{0}}{8}$$

When P3 is rotated with respect to P2 at an angle of 30°, then angle be¬tween the pass axis of P2 and P3 will be θ = 30°+ 30° = 60°.
Hence l3 = l2 cos²60° = $$\frac{l_{0}}{8} \times \frac{1}{4}=\frac{l_{0}}{32}$$
or
θ = 30° – 30° = 0°

Therefore, l3 = l2 cos² 0° = l2 = lo/8
When P3 is rotated with respect to P2 at an angle of 60°, then the angle between the pass axis of P2 and P3 will be
θ = 30° + 60° = 90°

Question 48.
(a) Derive an expression for path difference in Young’s double-slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S1 and S2, which are separated by a distance d as shown in the figure. Let us assume that the source of tight is monochromatic having a waveLength X. Under these conditions, the waves emerging from S and S have the same frequency and amplitude and are In phase.

If δ is the path difference between the waves from slits S2 and S1 at point P then
δ = r2 – r1 = d sin θ …(1)

where it is assumed that r1 and r2 are parallel, which is approximately true because D is much greater than d.

If this path difference is either zero or some Integral multiple of the wavelength, the waves are in phase at P and constructive Interference results. Therefore, the condition for bright fringes, or constructive interference, at P is given by
δ = d sin θ = m λ …(2)

Here m is called the order number. In order to find the position of maxima measured vertically from O to P, i.e. y we assume that D >> d, that is, the distance from the slits to the screen is much larger than the distance between the two slits. (In practice D = 1 m and d = a few millimetre) Under these conditions, θ is small, and so the approximation sin θ ≈ tan θ. Therefore, from triangle POQ in the figure, we find that
sin θ = tan θ = $$\frac{Y}{D}$$ …(3)

Substituting this in equation (2) we have
d $$\frac{Y}{D}$$ = m λ …(2)

rewriting the above equation we have
y = $$\frac{mDλ}{d}$$ …(4)
This gives the distance of the mth maxima from the centre 0 of the screen. For m = 0, 1, 2, 3, 4, ……. the values of y are
yo = 0 …….. position of the central bright fringe

y1 = $$\frac{Dλ}{d}$$ ……. position of the first bright fringe

y2 = $$\frac{2Dλ}{d}$$ ……. position of the second bright fringe

ym = $$\frac{mDλ}{d}$$ …….position of the mth bright fringe

The spacing between any two bright fringes gives the width of a dark fringe. Therefore,

This gives the width of the dark fringe. Similarly, when the path difference is an odd multiple of λ/2, the waves arriving at P will be 180° out of phase and will give rise to destructive interference. There¬fore, the condition for dark fringes, or de¬structive interference, at P is given by

This gives the distance of the mth minima from the centre O of the screen. For m is equal to 0, 1, 2, 3, ……. we have
y’o = $$\frac{Dλ}{2d}$$, ……. position of the first dark fringe,
y’1 = $$\frac{3Dλ}{2d}$$ ……. position of the second dark fringe
y’2 = $$\frac{5Dλ}{2d}$$ ……. position of the third dark fringe and,
y’m = $$\frac{(2 m+1) D \lambda}{2 d}$$ ……. position of the mth dark fringe

Now the spacing between two consecutive dark fringes gives the width of the bright fringe, therefore, the width of the bright fringe is

From equations 5 and 9 we find that the fringe width of the bright and the dark fringe is the same, therefore, the fringe width is given by
β = $$\frac{Dλ}{d}$$ …(10)

(b) The Intensity at the central maxima in Young’s double-slit experiment is l0. Find out the intensity at a point where the path difference is λ/6, λ/4 and λ/3. (CBSE AI 2016)
Let l be the intensity of light coming out of each slit. Since waves at the central maxima are in phase, therefore, we have

Question 49.
(a) Define a wavefront. Using Huygens’ Principle, verify the laws of reflection at a plane surface.
The wavefront is a locus of points that oscillate in the same phase.

Consider a plane wavefront AB incident obliquely on a plane reflecting surface MM. Let us consider the situation when one end A of wavefront strikes the mirror at an angle i but the other end B has still to cover distance BC. The time required for this will be t = BC/c.

According to Huygen’s principle, point A starts emitting secondary wavelets and in time t, these will cover a distance c t = BC and spread. Hence, with point A as centre and BC as radius, draw a circular arc. Draw tangent CD on this arc from point C. Obviously, the CD is the reflected wavefront inclined at an angle ‘r’. As incident wavefront and reflected wavefront, both are in the plane of the paper, the 1st law of reflection is proved.

To prove the second law of reflection, consider ΔABC and ΔADC. BC = AD (by construction),
∠ABC = ∠ADC = 90° and AC is common.

Therefore, the two triangles are congruent and, hence, ∠BAC = ∠DCA or ∠i = ∠r, i.e.the angle of reflection is equal to the angle of incidence, which is the second law of reflection.
Or
The refractive index of medium 2, w.r.t. medium 1 equals the ratio of the sine of the angle of incidence (in medium 1) to the sine of the angle of refraction (in medium 2), The diagram is as shown.

From the diagram

(b) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? Explain.
Size of central maxima reduces to half, (Size of central maxima = 2λD/d) Intensity increases.
This is because the amount of light, entering the slit, has increased and the area, over which it falls, decreases.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the obstacle. Explain why. (CBSE AI 2018, Delhi 2018)
This is because of the diffraction of light. Light gets diffracted by the tiny circular obstacle and reaches the centre of the shadow of the obstacle.
β = $$\frac{Dλ}{d}$$.

(i) When D is decreased, the fringe width decreases.
(ii) When d is Increased, the fringe width decreases.

Question 50.
The following figure shows an experiment set up similar to Young’s double-slit experiment to observe interference of light.

Here SS2 – SS1 = λ/4
Write the condition of (i) constructive,
(ii) destructive Interference at any point P in terms of path difference Δ = S2P – S1P.

Does the central fringe observed in the above set up tie above or below O? Give reason in support of your answer.
Yellow light of wavelength 6000°A produces fringes of width 08 mm in Young’s double-slit experiment. What will be the fringe width if the light source is replaced by another monochromatic source of wavelength 7500° A and separation between the slits is doubled?
Given SS2 – SS1 = λ/4
Now path difference between the two waves from slits S1 and S2 on reaching point P on screen is

(a) For constructive interference at point P both difference Δx = nλ.
Therefore

where n = 0, 1, 2, 3, …

(b) For destructive interference at point P path difference

For central bright fringe, putting n = 0 in equation (1), we get
$$\frac{y d}{D}=-\frac{\lambda}{4}$$
or
y = – $$\frac{\lambda D}{4 d}$$

The -ve sign indicates that the central bright fringe will be observed below centre O of the screen, at distance below it.
Given λ1 = 6000°A, β1 = 0.8 mm,
λ2 = 7500°A, β2 = ?, d2 = 2d1

Using the expression
β = $$\frac{Dλ}{d}$$ we have
$$\frac{\beta_{2}}{\beta_{1}}=\frac{\lambda_{2} d_{1}}{\lambda_{1} d_{2}}=\frac{7500}{6000} \times \frac{d_{1}}{2 d_{1}}$$

Solving we have β2 = 0.5 mm

Question 51.
(a) There are two sets of apparatus of Young’s double-slit experiment. Inset A, the phase difference between the two waves emanating from the slits does not change with time, whereas in set B, the phase difference between the two waves from the slits changes rapidly with time. What difference will be observed in the pattern obtained on the screen in the two setups?
(b) Deduce the expression for the resultant intensity in both the above- mentioned setups (A and B), assuming that the waves emanating from the two slits have the same amplitude A and same wavelength λ.
Or
(a) The two polaroids, in a given setup, are kept ‘crossed’ with respect to each other. A third polaroid, now put in between these two polaroids, can be rotated. Find an expression for the dependence of the intensity of light I, transmitted by the system, on the angle between the pass axis of the first and the third polaroid. Draw a graph showing the dependence of l on θ.
(b) When an unpolarized light is an incident on a plane glass surface, find the expression for the angle of incidence so that the reflected and refracted light rays are perpendicular to each other. What is the state of polarization, of reflected and refracted light, under this condition?
(a) Set A: Stable interference pattern, the positions of maxima and minima, does not change with time.
Set B: Positions of maxima and minima will change rapidly with time and an average uniform intensity distribution will be observed on the screen.

(b) Let the displacements of the waves from the sources S1 and S2 at a point on the screen at any time t be given by y1 = a cos ωt and y2 = a cos (ωt + Φ), where Φ is the constant phase difference between the two waves. By the superposition principle, the resultant displacement at point P is given by

Thus the amplitude of the resultant displacement is A = 2a cos (Φ/2)

Therefore, the intensity at that point is l = A² = 4a² cos²$$\frac{Φ}{2}$$ – 4lo cos²$$\frac{Φ}{2}$$
Since Φ = 0, l.e. there is no phase difference, hence l = 4 lo

Inset B, the intensity will be given by the average intensity
l = 4lo$$\left(\cos ^{2} \frac{\phi}{2}\right)$$ = 2lo
Or
(a) Let P1 and P2 be the two crossed Polaroids and P3 be the polaroid kept between the two. Let lo, be the intensity of polarised light after passing through the first Polaroid P1. Then the Intensity of light after passing through the second Polaroid P2 will be,
l = lo cos² θ
where θ is the angle between pass axes of P1 and P2. Since P1 and P2 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – 0). Hence the intensity of light emerging from P22 will be

The transmitted intensity will be maximum when π = π/4. The graph is as shown

(b) The ray diagram is as shown.

When light Is incident at a certain angle called Brewster angle, the reflected beam is completely polarised with its electric vector parallel to the surface, while the refracted beam is partially polarised. The angle of incidence at which this occurs is called the polarising angle iB.

From figure we see that at the polaris-ing angle
iB + 90° + r = 180°
or
r = 90° – iB.

Using Snell’s law we have
μ = $$\frac{\sin i_{\mathrm{B}}}{\sin r}$$

Now sin r = (90° – iB), therefore the above expression becomes

Nature of polarisation: Reflected light and linearly polarised

Numerical Problems:
Formulae for solving numerical problems

• Fringe width is given by β = $$\frac{Dλ}{d}$$
• Brewster’s Law: μ = $$\frac{\sin \theta_{p}}{\cos \theta_{p}}$$ = tan θp
• Intensity of light coming out of a polariser l = lo cos² θ
• If l1 and l2 are the intensities of light emitted by the two sources, w1 and w2 be the widths of the two slits, a1 and a2 be the amplitudes of the waves from the two slits then,
$$\frac{l_{1}}{l_{2}}=\frac{w_{1}}{w_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}$$
• lf lmax and lmin be the intensities of light at the maxima and the minima the
$$\frac{l_{\max }}{l_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}$$

Question 1.
Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of laser light that produces interference fringes separated by 8.1 mm using the same pair of slits. (CBSE Al 2011C)
Given λ1 = 630 nm, β1 = 7.2 mm, β2 = 8.1 mm, λ2 =?

We know that β = $$\frac{Dλ}{d}$$, for the same value of D and d we have β2 ∝ λ,
Therefore, we have

Question 2.
What is the speed of light in a denser medium of polarising angle 30°? (CBSE Delhi 2019)

Question 3.
Laser light of wavelength 640 nm incident on a pair of slits produces an interference pattern in which the bright fringes are separated by 7.2 mm. Calculate the wavelength of another source of light that produces interference fringes separated by 8.1 mm using the same arrangement. Also, find the minimum value of the order (n) of the bright fringe of the shorter wavelength which coincides with that of the longer wavelength. (CBSE AI 2012C)
Given λ1 = 640 nm, λ2 = ?, b1 = 7.2 mm and b2 = 8.1 mm

Using the expression β = $$\frac{Dλ}{a}$$ we have

Now n fringes of shorter wavelength will coincide with (n – 1) fringes of Longer wavelength
n1λ1 = n2λ2
or
n × 640 = (n – 1) × 720
Solving for n we have n =9

Question 4.
Yellow light (λ = 6000 Å) illuminates a single-slit of width 1 × 10 m. Calculate
(a) the distance between the two dark tines on either side of the central maximum, when the diffraction pattern is viewed on a screen kept 1.5 m away from the slit;
(b) the angular spread of the first diffraction minimum. (CBSE AI 2012C)

Question 5.
Find the ratio of Intensities of two points P and Q on the screen in Young’s double-slit experiment when the waves from sources S1 and S2 have a phase difference of (i) π/3 and (11) π/2.
Intensity at the screen when the phase difference is Φ = π/3 is

Now intensity at the screen when the phase difference is Φ = 90° is

Therefore, ratio of intensities is
$$\frac{l_{p}}{l_{Q}}=\frac{3l}{2l}=\frac{3}{2}$$

Question 6.
In young’s double-slit experiment, two sifts are separated by 3 mm distance and illuminated by the light of wavelength 480 nm. The screen is at 2 m from the plane of the slits. Calculate the separation between the 8th bright fringe and the 3N dark fringes observed with respect to the central bright fringe.
Using the formuLa

Question 7.
Two coherent sources have Intensities In the ratio 25: 16. Find the ratio of the Intensities of maxima to minima, after the interference of light occurs.
In the given problem

Question 8.
The ratio of intensities of maxima and minima in an interference pattern is found to be 25: 9. Calculate the ratio of light intensities of the sources producing this pattern.
Given

Question 9.
In Young’s double-slit experiment using the light of wavelength 400 nm, Interference fringes of width X are obtained. Th. the wavelength of light is Increased to 600 nm and the separation between the slits Is halved. If on. wants the observed fringe width on the screen to b. the same in the two cases, find the ratio of the distance between the screen and the plan. of the Interfering sources In the two arrangements.
Let D1 be the distance between the screen and the sources when a tight of wavelength 400 nm Is used.

β = $$\frac{Dλ}{a}$$
or
X = $$\frac{D_{1} \times 400 \times 10^{-9}}{d}$$ …(i)

Let D2 be the distance between the screen and the sources to obtain the same fringe width, when light of wavelength 600 nm is used. Then,
X = $$\frac{D_{2} \times 600 \times 10^{-9}}{d}$$ ….(ii)

From the equations (i) and (ii), we have
$$\frac{D_{1}}{D_{2}}=\frac{600 \times 10^{-9}}{400 \times 10^{-9}}$$ = 1.5

Question 10.
In Young’s slit experiment, Interference fringes are observed on a screen, kept at D from the slits. If the screen Is moved towards the slits by 5 × 10-2 m, the change in fringe width Is found to be 3 × 10-5 m. If the separation between the slits is 10-3 m, calculate the wavelength of the light used.
Given ΔD = 5 × 10-2-2 m, Δβ = 3 × 10-5-5 m, d = 10-3 m,

Using the relation β = $$\frac{Dλ}{a}$$ we have for the two cases
β1 = $$\frac{D_{1} \lambda}{d}$$ and
β2 = $$\frac{D_{2} \lambda}{d}$$ subtracting we have
β1 – β2 = $$\frac{λ}{d}$$(D1 – D2)
or

Question 11.
A slit of width ‘a’ is illuminated by monochromatic light of wavelength 700 nm at Normal Incidence. Calculate the value of ‘a’ for the position of
(a) the first minimum at an angle of diffraction of 30°.
(b) first maximum at an angle of diffraction of 30°.
Given X = 700 nm = 7 × 10-7 m

Question 12.
Estimate the angular separation between . first order maximum and third order minimum of the diffraction pattern due to a single-slit of width 1 mm, when light of wavelength 600 nm is incident normal on it. (CBSE AI 2015C)
Given d = 1 mm = 10-3 m,
λ = 600 nm = 6 × 10-7 m,
For first order maxima
d sin θ = n λ
or
sin θ = $$\frac{n \lambda}{d}=\frac{6 \times 10^{-7}}{10^{-3}}$$ = 6 × 10-4
or
θmax = 1.047°
Now for minima we have
d sin θ = (2n+1 )$$\frac{λ}{2}$$

For third-order minima we have n = 3
Therefore we have

sin θ = (2 × 3 + 1)$$\frac{λ}{2d}$$ = $$\frac{7λ}{2d}$$
= $$\frac{7 \times 6 \times 10^{-7}}{2 \times 10^{-3}}$$
= 21 × 10-4
or
θmin = 6.397°

Therefore, angular separation is
θmin – θmax = 6.397 – 1.047 = 5.35°

## Polymers Class 12 Important Extra Questions Chemistry Chapter 15

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 15 Polymers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Chemistry Chapter 15 Important Extra Questions Polymers

### Polymers Important Extra Questions Very Short Answer Type

Question 1.
Define the term ‘homopolymerisation’ giving an example. (CBSE Delhi 2012)
The polymers formed from one type of monomers and having same repeating units are called homopolymers. For example polyvinyl chloride (PVC), polythene.

Question 2.
Give one example of condensation polymer. (CBSE 2013)
Nylon-6, 6

Question 3.
Which of the following is a natural polymer? (CBSE 2014)
Buna-S, Proteins, PVC
Proteins.

Question 4.
Based on molecular forces what type of polymer is neoprene? (CBSE 2014)
Elastomer.

Question 5.
Which of the following is a fibre?
Nylon, Neoprene, PVC (CBSE 2014)
Nylon

Question 6.
Draw structure for the polymer used for the manufacture of non-stick utensils. (CBSE2019C)

Question 7.
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymerisation reaction? (CBSE AI 2010)
The monomers must be bifunctional i.e. contain two functional groups.

Question 8.
Arrange the following polymers in the increasing order of tensile strength. Nylon 6, Buna-S, Polythene. (CBSE Sample Paper 2010)
Buna – S < Polythene < Nylon 6.

Question 9.
What type of reaction occurs in the formation of Nylon 6,6 polymer? (CBSE Sample Paper 2019)
Condensation

Question 10.

Is a homopolymer or R copolymer? (CBSE 2019C)
Homopolymer

Question 11.
Identify the Monomers of the following polymer:

Buta-1,3-diene, styrene.
It is obtained by the polymerisation of buta-1, 3-diene and styrene in the ratio of 3: 1 in the presence of sodium.

### Polymers Important Extra Questions Short Answer Type

Question 1.
Draw the molecular structures of the monomers of
(i) PVC
PVC : CH2 = CH – Cl (Vinyl chloride)

(ii) Teflon (CBSE 2010)
Teflon : F2C = CF2 (Tetrafluoroethene)

Question 2.
Mention two important uses of each of the following : (CBSE Delhi 2011)
(i) Bakelite
Uses of bakelite:

• Used for making combs, fountain pens, barrels, electrical switches and handles of utensils.
• Soft bakelites are used for making glue for binding wooden laminated planks in varnishes.

(ii) Nylon 6
Uses of Nylon 6:

• It is used for making tyre cords.
• It is used for making fabrics and ropes.

Question 3.
Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6
The monomer of Nylon-6:

(ii) Polypropene (CBSE Delhi 2012)
The monomer of polypropene:

Question 4.
Differentiate between thermoplastic and thermosetting polymers. Give one example of each. (CBSE Delhi 2010, 2012) Answer:
Thermoplastics when heated become soft and more or less fluid. These can be moulded into any desired shape. The thermoplastics ‘ can be processed again and again.

The intermolecular forces in thermoplastics are intermediate between those of elastomers and fibres.

On the other hand, thermosetting plastics on heating become hard and insoluble masses. These cannot be moulded into the desired shape and cannot be reprocessed. The intermolecular forces in thermosetting are strong and there are cross-links that hold the molecules in place so that heating does not allow them to move freely.

The common examples are:

1. Thermoplastics: Polythene, polystyrene and polyvinyl chloride.
2. Thermosetting: Bakelite and melamine formaldehyde.

Question 5.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Bakelite
Phenol and formaldehyde

(ii) Neoprene

Question 6.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Terylene
Ethylene glycol, terephthalic acid

(ii) Nylon-6, 6

Question 7.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Teflon
Tetrafluoroethylene

(ii) Buna-N

### Polymers Important Extra Questions Long Answer Type

Question 1.
Write the structures of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
OR
(i) Is it a homopolymer or copolymer? Give reason.
(ii) Write the monomers of the following polymer:

(iii) What is the role of the Sulphur in the vulcanization of rubber? (CBSE Delhi 2019)
(i) Nylon-6,6
Hexamethylenediamine: NH2 – (CH2)6 – NH2 and Adipic acid

(ii) Glyptal

(iii) Buna – S
CH2 = CH — CH = CH2

OR
(i) It is a homopolymer because it has only one monomer.

(iii) Sulphur makes the rubber hard, tough with greater tensile strength and non-sticky by forming sulphur crosslinks.

Question 2.
(a) Write the names of monomers of the following polymer:

Ethylene glycol and terephthalic acid or Ethane-1,2- diol and Benzene 1, 4- dicarboxylic acid.

Terylene: It is a polymer of ethylene glycol (ethane-1,2-diol) and terephthalic acid (benzene-1, 4-dicarboxylic acid). It is obtained by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide, [Zn(OOCCH3)2 + Sb2O3] catalyst. It is known as terylene or dacron.

(b) What does part 6,6 mean in the polymer Nylon-6,6?
It represents 6 carbon atoms present in both the monomer units.

Nylon-6,6: The monomer units of nylon-6,6 are hexamethylenediamine and adipic acid. It is prepared by condensation of hexamethylene diamine with adipic acid under high pressure and high temperature.

(c) Give an example of a Biodegradable polymer. (CBSE 2019C)
Poly-p-hydroxybutyrate-co-p-hydroxy valerate (PHBV).

It is a copolymer of 3-hydroxybutyric acid and 3-hydroxypentanoic add, in which the monomer units are joined by ester linkages.

Question 3.
Explain the following terms giving a suitable example for each :
(i) Elastomers
Elastomers: These are polymers in which the polymer chains are held together by weak intermolecular forces. Because of the presence of weak forces, the polymer can be easily stretched. However, a few cross-linked are also introduced in the chains which impart the property of regaining the original positions after the stretching force is released. A common example is vulcanised rubber.

(ii) Condensation polymers
Condensation polymers: A polymer formed by the condensation of two or more than two monomers is called condensation polymer. In this type, each monomer generally contains two functional groups, and condensation takes place by the loss of molecules such as H20, HCl, etc. Common examples are nylon-6, nylon-6,6, bakelite, terylene and alkyl resins.

Addition polymers: A polymer formed by the direct addition of repeated monomers is called an addition polymer. In this type, the monomers are unsaturated compounds and are generally derivatives of ethene. In addition, polymers have the same empirical formula as their monomers. For example, the addition of polymers polyethene or polypropylene are obtained as:

Question 4.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Buna-S:
1, 3-Butadiene: CH2 = CH – CH == CH2
Styrene: C6H5CH = CH2

(ii) Neoprene
Neoprene:
Chloroprene: CH2 = C – CH = CH2

(iii) Nyton-6, 6 (CBSE Delhi 2013)
Nylon 6, 6:

• Adipic acid: HOOC (CH2)4 COOH
• Hexamethylenediamine: NH2(CH2)6NH2

Question 5.
Write the names and structures of the monomers of the following polymers:
(i) Bakelite
Bakelite:

• Phenol:
• Formaldehyde: HCHO

(ii) Nylon-6
Nylon-6:
Caprolactam

(iii) Polythene (CBSE Delhi 2013)
Polythene: Ethene: CH2 = CH2.

Question 6.
(i) What is the role of benzoyl peroxide in the polymerisation of ethene?
(ii) Identify the monomers in the following polymer:

(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Nylon-6, 6, Polythene, Buna-S
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE 2016)
(i) Benzoyl peroxide acts as a free radical generator. In the presence of benzoyl peroxide, phenyl free radical is formed which initiates the chain initiation step.

(iii) Buna-S < Polythene < Nylon 6, 6
OR
Mechanism
(i) Chain initiation step:

(ii) Chain propagating step:

(iii) Chain termination step:

Question 7.
Write the names and structures of the monomers of the following polymers:
(a) Neoprene
Chloroprene CH2 = C — CH = CH2

(b) Bakelite
C6H5OH(Phenol) and HCHO (formaldehyde)

(c) PVC (CBSE 2019C)
Polyvinyl chloride CH2 = CH – Cl

Question 8.
Write the structures of monomers used to obtain the following polymers:
(a) Natural rubber
(b) PVC
(c) Nylon-6,6
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE AI 2019)

(b) PVC: vinyl chloride
CH2 = CH – Cl

(c) Nylon-6,6:

NH2 — (CH2)6 — NH2
(Hexamethylenedlamine)
OR
Chain Initiation Step:

Chain propagation:

Chain termination step

Question 9.
Write the structures of monomers used to obtain the following polymers:
(a) Buna-S
(b) Glyptai
(c) Nylon-6
OR
(a) Arrange the following polymers in increasing order of their intermolecular forces: Polyvinylchloride, Neoprene, Terylene
(b) Write one example of each of
(i) Natural polymer
(ii) Thermosetting polymer
(c) What is the significance of numbers 6,6 in the polymer Nylon-6,6? (CBSE Al 2019)
(a) Buna-S
CH2 = CH – CH = CH2 (1,3-Butadiene) and C6H5CH = CH2 (Styrene)

(b) Glyptal
HO – CH2 — CH2 — OH (Ethylene glycol) and

(c) Nylon-6

OR
(a) Neoprene < Polyvinyl chloride (PVC) < Terylene
(b) (i) Isoprene
(ii) Bakelite
(c) The numbers ‘6, 6’ signifies that both the monomers of nylon-6, 6 have six carbon atoms.

Question 10.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester. (CBSE AI 2013)
Polymers that are degraded by micro-organisms within a suitable period so that the polymers and their degraded products do not cause any serious effects on the environment are called biodegradable polymers. For example,

Poly β – hydroxybutyrate – co – β – hydroxy valerate (PHBV)

Question 11.
Write the names and structures of the monomers of the following polymers:
(i) Polystyrene
Polystyrene: Styrene:

(ii) Dacron
Dacron:

• Ethylene glycol: HOCH2CH2OH
• Terephthalic acid:

(iii) Teflon
Teflon:
Tetrafluoroethene: F2C = CF2

Question 12.
Write the names and structures of the monomers of the following polymers:
(a) Teflon
Tetrafluoroethylene, CF2 = CF2

(b) Terylene

(c) Buna-N

Question 13.
Write the structures of monomers used to obtain the following polymers:
(a) Neoprene
(b) PHBV
(c) Bakelite
OR
(a) Arrange the following polymers in decreasing order of their intermolecular forces:
Bakelite, Polythene, Buna-S, Nylon-6,6
(b) Write the monomers of the following polymer:

(c) What is the structural difference between high-density polythene (HDP) and low-density polythene (LDP)? (CBSE AI 2019)
(a) Neoprene:
Chloroprene

(b) PHBV

(c) Bakelite

OR
(a) Buna-S < potythene < Bakelite < Nylon-6,6

(b) HO — CH2 — CH2 — OH and

(c) Low-density polythene (LDP) consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore, it has low density.

High-density polythene (HDP) consists of linear chains and therefore, the molecules can be closely packed. Hence, it has a high density.

Question 14.
What are high density and low-density polythene?
High-density polythene is obtained by heating ethene at about 333-343 K under a pressure of 6-7 atm in the presence of a catalyst such as triethylaluminium and titanium tetrachloride (known as Zeigler-Natta catalyst)

This polymer consists of linear chains and therefore the molecules can be closely packed in space. It, therefore, has a high density (0.97g/cm3) and a higher melting point (403K). It is harder, tougher and has greater tensile strength than low-density polythene.

Low-density polythene (LDP) is prepared by free radical addition and hydrogen atom abstraction. It consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore it has low density (0.92 g/cm3) and low melting point (384 K). Low-density polythene is transparent. It has moderate tensile strength and high toughness. It is chemically inert.

Question 15.
A large number of unsaturated compounds such as alkenes or dienes and their derivatives are polymerised by this process. The polymerisation takes place through the generation of an initiator, which is a molecule that decomposes to form free radicals. The commonly used initiator is t-butyl peroxide.

For example, most of the commercial addition polymers are obtained from alkenes and their derivatives,

The general model of polymerisation is:

Chain initiation step:

Chain Propagating Step:

Chain terminating step

Question 16.
List some important differences between natural rubber and vulcanised rubber.
Differences between natural rubber and vulcanised rubber:

 Natural rubber Vulcanized rubber 1. Natural rubber is soft and sticky. 1. Vulcanized rubber is hard and non-sticky. 2. It has low tensile strength. 2. It has a high tensile strength. 3. It has low elasticity. 3. It has high elasticity. 4. It can be used over a narrow range of temperature (from 10°C to 60°C). 4. It can be used over a wide range of temperature (-40°C to 100°C). 5. It has low wear and tears resistance. 5. It has high wear and tears resistance. 6. It is soluble in solvents like ether, carbon- tetrachloride, petrol etc. 6. It is insoluble in all the common solvents.

Question 17.
(a) What is the difference between two notations: nylon-6 and nylon-6,6? Give their synthesis.
Nylon-6 has only one compound having 6-carbon atoms while nylon-6,6 refers to a polymer obtained from 6-carbon atoms of dicarboxylic acid (adipic acid) and 6-carbon atoms of diamine (hexamethylene diamine).

Nylon-6 is synthesised from caprolactam as

Nylon-6,6 is synthesised from hexamethylenediamine and adipic acid.

(b) How is Buna-S prepared?
Buna-S is obtained by copolymerisation of styrene and 1, 3-butadiene.

Question 18.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
1, 3-Butadiene: CH2 = CH — CH = CH2
Styrene: C6H5CH = CH2

(ii) Buna-N
1, 3-Butadiene: CH2 = CH – CH = CH2
Acrytonitnie: CH2 = CH — CN

(iii) Dacron

• Ethylene glycol:
• Terephthalic acid:

(iv) Neoprene.
Chloroprene:

## Aldehydes, Ketones and Carboxylic Acids Class 12 Important Extra Questions Chemistry Chapter 12

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Chemistry Chapter 12 Important Extra Questions Aldehydes, Ketones and Carboxylic Acids

### Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone. (CBSE Delhi 2012)
butanone < propanone < propanal < ethanal

Question 2.
Which of the following compounds would undergo the Cannizzaro reaction? Benzaldehyde, Cyclohexanone, 2- Methylpentanal. (CBSE Sample Paper 2019)
Benzaldehyde

Question 3.
Draw the structure of semicarbazone of cyclopentanone. (CBSE 2019C)
OR
Draw the structure of the product formed when propanal is treated with zinc amalgam and concentrated hydrochloric acid.

OR

Question 4.
Write the IUPAC name of the following: (CBSE AI 2012)

Pent-2-enal.

Question 5.
Write the structure of 3-methylbutanal. (CBSE Delhi 2013)

Question 6.
Write the structure of the p-Methylbenzal- dehyde molecule. (CBSE Delhi 2013, 2014)

Question 7.
Draw the structure of the compound whose IUPAC name is 4-chloropentan 2-one. (CBSE Delhi 2008, CBSE AI 2011, 2013, 2014)

Question 8.
Draw the structural formula of a 1-phenyl propane-2-one molecule. (CBSE 2010)

Question 9.
Draw the structure of 3-methylbutanal. (CBSE 2011, CBSE Delhi 2013)

Question 10.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:
ethanal, propanal, propanone, butanone. (CBSE 2012)
butanone < propanone < propanal < ethanal

Question 11.
Write the IUPAC name of (CBSE 2012)
Ph — CH = CH — CHO.
3-Phenyl prop-2-en-al.

Question 12.
Rearrange the following compounds in the increasing order of their boiling points:
CH3 – CHO, CH3 – CH2 – OH, CH3 – CH2 – CH3 (CBSE 2013)
CH3CH2CH3 < CH3CHO < CH3CH2OH

Question 13.
Ethanol is soluble in water. Why? (CBSE 2013)
Ethanol is soluble in water because of hydrogen bonding between the polar carbonyl group of ethanal and water molecules:

Question 14.
Write the IUPAC name of the compound (CBSE Delhi 2014)

3-Hydroxybutanoic acid

Question 15.
Write the IUPAC name of the compound (CBSE Delhi 2014)

Question 16.
Write the IUPAC name of the compound (CBSE Delhi 2014)

3-Aminobutanal

Question 17.
Write the IUPAC name of the following compound: (CBSE AI2019)

But-3-en-2-one

Question 18.
Write the IUPAC name of the following compound: (CBSE AI 2019)

1-Phenylbutan-2-one.

### Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Short Answer Type

Question 1.
Write structures of compounds A and B in each of the following reactions: (CBSE Delhi 2019)

Question 2.
What is Tollen’s reagent? Write one use of this reagent. (CBSE 2010)
The ammoniacal solution of silver nitrate is called Tollen’s reagent. It is used as an oxidizing reagent and helps to distinguish between aldehydes and ketones. Aldehydes are oxidized by Tollen’s reagent whereas ketones are not.

Question 3.
Name the reagents used in the following reactions: (CBSE Delhi 2015)

Lithium aluminium hydride, LiAlH4

Alkaline potassium permanganate (KMnO4), KOH

Question 4.
Distinguish between
C6H5CH = CH – COCH3 and C6H5CH = CH CO CH2 CH3 (CBSE AI 2016)
Heat both the compounds with NaOH and l2. C6H5CH = CHCOCH3 gives yellow ppt of iodoform. C6H5CH = CHCOCH2CH3 does not give yellow ppt. of iodoform.

Question 5.
Arrange the following in the increasing order of their reactivity in nucleophilic addition reactions.
CH3CHO, C6H5CHO, HCHO (CBSE AI 2016)
C6H5CHO < CH3CHO < HCHO

Question 6.
Write the product in the following reaction: (CBSE AI 2017)

Question 7.
Write the structure of the products formed: (CBSE Sample Paper 2019)
(a) CH3CH2COOH
CH3 – CH(Cl) – C00H

(b) C6H5COCl
C6H5CHO

Question 8.
Name the reagents in the following reactions: (CBSE AI 2015)

Lithium aluminium hydride, LiAlH4

Alkaline potassium permanganate (KMnO4), KOH

CH3 Mg Br, H3O+

Cl2, P (Hell-Volhard-Zelinsky reaction)

Question 9.
Predict the organic products of the following reactions:

Question 10.
Describe how the following conversions are carried out:
(i) Toluene to benzoic acid

(ii) Bromobenzene to benzoic acid (CBSE AI 2013)

(iii) Ethylcyanide to ethanoic acid

(iv) Butan-1-ol to butanoic acid (CBSE Delhi 2010)

Question 11.
Write the structures of A, B, C, and D in the following reaction: (CBSE AI 2016)

Question 12.
Aromatic carboxylic acids do not undergo Friedel Crafts reaction. Explain. (CBSE Al 2018)
Aromatic carboxylic acids do not undergo Friedel Crafts reaction because -COOH group is deactivating and the catalyst aluminum chloride (Lewis acid) gets bonded to the carboxyl group.

Question 13.
pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain. (CBSE AI 2018)
Due to the presence of a strong electron-withdrawing (-NO2) group in 4-nitrobenzoic acid, it stabilizes the carboxylate anion and hence strengthens the acid. Therefore, 4-nitrobenzoic acid is more acidic than benzoic acid and its pKa value is lower.

### Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Long Answer Type

Question 1.
Complete the following reactions:

Or
Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH)2.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCI
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO4. (CBSE Delhi 2019)

Or

Question 2.
Predict the products of the following B reactions: (CBSE Delhi 2015)

Question 3.
Predict the products of the following reactions: (CBSE 2015)

Question 4.
What happens when
(a) Salicylic acid Is treated with (CH3CO)2O/H+?
Write a chemical equation in support of your answer. (CBSE AI 2019)
Salicylic acid is treated with acetic anhydride In the presence of H+ to give aspirin.

(b) Phenol is oxidized with Na2Cr2O7/H+?
Phenol in the presence of acidified sodium dichromate gives benzoquinone.

(C) Anisole Is treated with CH3Cl/anhydrous AlCl3?

Question 5.
Draw the structures of the following compounds:
(i) 3-Methylbutanal

(ii) 4-Chlocopentan-2-one

(iii) 4-Methylpent-3-en-2-one

(iv) p-Methyl benzaldehyde (CBSE AI 2014)

Question 6.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene (CBSE Delhi 2017)
Propanone to propene

(ii) Propanal to butanone
Propanal to butanone

(iii) Benzaldehyde to benzophenone
Benzaldehyde to benzophenone

(iv) Benzaldehyde to 3-phenyl propan – 1-ol
Benzaldehyde to 3- phenyl propan -1 -ol

(v) Benzaldehyde to α-hydroxyphenyl acetic acid
Benzaldehyde to α-hydroxyphenyl acetic acid

(vi) Ethanol to 3-hydroxybutanal
Ethanol to 3-hydroxybutanal

Question 7.
Convert the following:
(i) Ethanal to propanone. (CBSE AI 2018)
Ethanal to propanone

(ii) Ethanal to lactic acid.
Ethanal to lactic acid

(iii) Ethanal to 2-hydroxy-3-butenoic acid.
Ethanal to 2-hydroxy-3-butenoic acid

(iv) Acetaldehyde to formaldehyde.
Acetaldehyde to formaldehyde

(v) Formaldehyde to acetaldehyde.
Formaldehyde to acetaldehyde

(vi) Acetaldehyde to crotonic acid.
Acetaldehyde to crotonic acid

Question 8.
A compound ‘X’ (C2H4O) on oxidation gives ‘Y’ (C2H4O2). ‘X’ undergoes haloform reaction. On treatment with HCN ‘X’ forms a product ‘V which on hydrolysis gives 2-hydroxy propanoic acid.
(i) Write down structures of ‘X’ and ‘Y’.
(ii) Name the product when ‘X’ reacts with dil NaOH.
(iii) Write down the equations for the reactions involved. (CBSE Sample Paper 2007)
Compound ‘X’ (C2H4O) is oxidized to ‘Y’ (C2H4O2). Since it undergoes a haloform reaction, it must be acetaldehyde.
(i) X = CHCHO Y = CH3COOH
On treatment with HCN, X gives cyanohydrin which on hydrolysis gives 2-hydroxypropanoic acid.

(ii) When ‘X’ reacts with dil. NaOH, undergoes an aldol condensation reaction forming aldol which on heating gives but-2-enal.

(iii) Equations for reactions

Other equations are given above.

Question 9.
Complete the following reactions:

(CBSE Sample Paper 2017 – 2018)

(CBSE Delhi 2013)

(CBSE AI 2013)

Question 10.
(i) An organic compound (A) has a characteristic odor. On treatment with NaOH, it forms two compounds (B) and (C). Compound (B) has molecular formula C7H80 which on oxidation gives back (A). The compound (C) is a sodium salt of an acid. When (C) is treated with soda lime it yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write the sequence of reactions involved.
The compound A is C6H5CHO, benzaldehyde having a characteristic odor. The reactions are:

(ii) Complete each synthesis by filling the missing starting materials, reagents, or products. (X and Z).
(a) C6H5CHO + CH3CH2CHO NaOH X

(b) CH3(CH2)9 COOC2H5 CH3(CH2),CHO

(iii) How will you bring about the following conversions in not more than two steps?
(a) Toluene to benzaldehyde

(b) Ethylcyanide to 1-phenylpropanoid. (CBSE Sample Paper 2011)

Question 11.
(i) How do you convert the following?
(a) Ethanal to propanone
(b) Toluene to benzoic acid
OR
(ii) (A), (B), and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCI gives the same product (D).
(a) Write the structures of (A), (B), (C), and (D).
(b) Out of (A), (B), and (C) isomers, which one is least reactive towards the addition of HCN? (CBSEAI 2018)

OR
(ii) (a) Out of A, B, and C, (A) and (C) give positive Tollens’ test, and therefore, these are aldehydes. (B) does not give Tollens’ test and therefore, it is a ketone, with a group because it gives positive iodoform test. Thus, the three isomers are:

Question 12.
Predict the products of the following reactions:

(CBSE Delhi 2015)

(CBSE Delhi 2015)

(CBSE AI 2015)

(CBSE AI 2015)

(CBSE AI 2017)

(CBSE AI 2017)

Question 13.
Write the structures of A, B, C, D, and E in the following reactions: (C8SE Delhi 2016)

Question 14.
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanone.
(ii) Pentan-2-one and Pentan-3-one.
(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4 -Dinitrobenzoic acid, 4- Methoxybenzoic acid.
OR
Compare the reactivity of benzaldehyde and ethanal towards nucleophilic addition reactions. Write the cross aldol condensation product between benzaldehyde and ethanal. (CBSE Sample Paper 2019)
(a) (i)

 Experiments Ethanal Propanone 1. Tollen’s Test: Warm the organic compound with freshly prepared ammoniacal silver nitrate solution (Tollen’s reagent). A bright silver mirror is produced. No silver mirror is formed. 2. Fehling’s Test: Heat the organic compound with Fehling’s reagent. A reddish-brown precipitate is obtained. No precipitate is obtained. Anyone test

Acetaldehyde (Ethanal) and Acetone (Propanone)
(i) Acetaldehyde gives silver mirror with Tollen’s reagent.

Acetone does not give this test.

(ii) Acetaldehyde gives red ppt with Fehling solution.

(ii) Pentan-2-one and Pentan-3-one

 Experiment Pentan-2-one Pentan-3-one Iodoform Test: The organic compound is heated with iodine in presence of sodium hydroxide solution. A yellow precipitate is obtained. No yellow precipitate is obtained.

Pentan-3-one and Pentan-2-one
(i) Pentan-2-one forms yellow ppt with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
CH3COCH2CH2CH3 + 3I2+ 4NaOH → CH3CH2CH2COONa + CHI3 + 3H20 + 3NaI

(ii) Pentan-2-one gives white ppt with sodium bisulphite while pentan-3-one does not.

Or any other suitable test.

(b) 4- Methoxybenzoic add < Benzoic acid < 4- Nitrobenzoic acid < 3,4-Dinitrobenzoic acid Effect of substituents on acidic strength of acids. The substituents have a marked effect on the acidic strength of carboxylic acids. The nature of substituents affects the stability of the conjugate base (carboxylate ion) and hence affects the acidity of the carboxylic acids.

In general, electron-withdrawing groups (EWG) increase the stability of the carboxylate ion by delocalizing the negative charge and hence increase the acidity of the carboxylic acid. Conversely, electron-donating groups (EDG) decrease the stability of the carboxylate ion by intensifying the negative charge and hence decrease the acidity of the carboxylic acid.
OR
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in ethanol. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence less reactive than ethanal.

Aromatic Aldehydes and Ketones
In general, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogs. For example, benzaldehyde is less reactive than aliphatic aldehydes. This can be easily understood from the resonating structures of benzaldehyde as shown below:

It is clear from the resonating structures that due to the electron releasing (+I effect) of the benzene ring, the magnitude of the positive charge on the carbonyl group decreases, and consequently it becomes less susceptible to the nucleophilic attack. Thus, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic aldehydes and ketones.

However, amongst aromatic aldehydes and ketones, aromatic aldehydes are more reactive than alkyl aryl ketones which in turn are more reactive than diaryl ketones. Thus, the order of reactivity of aromatic aldehydes and ketones is:

Question 15.
(i) Illustrate the following name reaction giving a suitable example:
(a) Clemmensen reduction
(b) Hell-Volhard-Zelinsky reaction

(ii) How are the following conversions carried out?
(a) Ethylcyanide to ethanoic acid
(b) Butan-1-ol to butanoic acid
(c) Benzoic acid to m-bromobenzoic acid .
OR
(i) Illustrate the following reactions giving a suitable example for each.
(a) Cross aldol condensation
(b) Decarboxylation

(ii) Give simple tests to distinguish between the following pairs of compounds:
(a) Pentan-2-one and Pentan-3-one
(b) Benzaldehyde and Acetophenone
(c) Phenol and Benzoic acid (CBSE Delhi 2012)
(i) (a) Clemmensen reduction

(b) Hell-Volhard-Zelinsky reaction

(ii)

OR
(i) (a) Cross Aldol condensation

(b) Decarboxylation

(ii) (a) Pentan-2-one and pentan-3-one can be distinguished by iodoform test

Pentan-3-one does not give this test.

(b) Benzaldehyde and acetophenone: Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent).

Acetophenone does not react with Tollen’s reagent.

(c) Phenol and Benzoic acid: Benzoic acid reacts with NaHC03 to give effervescence due to the evolution of CO2.

Phenol does not give effervescence.

Question 16.
(i) Write a suitable chemical equation to complete each of the following transformations:
(a) Butan-1-ol to butanoic acid
(b) 4-Methylacetophenone to benzene-1, 4-dicarboxylic acid.

(ii) An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Toilen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
OR
(i) Give chemical tests to distinguish between (a) Propanol and propanone
(b) Benzaldehyde and acetophenone

(ii) Arrange the following compounds in increasing order of their property as indicated:
(a) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(b) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)

(ii) The given compound forms a 2,4-DNP derivative. Therefore, it is an aldehyde or ketone. Since it reduces Tollen’s reagent, it must be an aldehyde. The compound undergoes Cannizzaro’s reaction, so it does not contain hydrogen. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid, which means that it must be containing an alkyl group at 2-position with respect to the CHO group on the benzene ring.
The molecular formula suggests it should be
(2- Ethytbenza(dehyde)

(i) (a) Propanone gives iodoform test but propanol does not. When propanone is heated with aqueous sodium carbonate and iodine solution, a yellow precipitate of iodoform is obtained.

Propanol does not give this test.

(b) Acetophenone forms yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test). Benzaldehyde does not react.

(ii) (a) Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(b) 4-Methoxybenzoic acid < Benzoic add < 3, 4-Dinitrobenzoic acid.
(c) (CH3)2CHCOOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

Question 17.
(i) Although phenoxide ion has more resonating structures than carboxylate ion, the carboxylic acid is a stronger acid than phenol. Give two reasons.
(ii) How will you bring about the following conversions?
(a) Propanone to propane
(b) Benzoyl chloride to benzaldehyde
(c) Ethanal to but-2-enal
OR
(i) Complete the following reactions:

(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Benzoic acid and Phenol (CBSE AI 2013)
(i) The carboxylic acids are stronger acids than phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrid structures of carboxylate ion and phenoxide ion.

The resonance hybrids may be represented as:

(a) The electron charge in the carboxylate ion is more dispersed in comparison to the phenoxide ion since there are two electronegative oxygen atoms in the carboxylate ion as compared to only one oxygen atom in the phenoxide ion.

(b) Carboxylate ion is stabilized by two equivalent resonance structures in which negative charge is on more electronegative 0 atoms. But phenoxide ion has non-equivalent resonance structures in which the negative charge is also on a less electronegative carbon atom.

Therefore, the carboxylate ion is relatively more stable as compared to the phenoxide ion. Thus, the release of H+ ions from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

OR

(ii) (a) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

(b) Benzoic acid reacts with NaHC03 to give effervescence due to the liberation of CO2.
C6H5COOH + NaHCO3 → C6H5COONa + H20 + CO2
Phenol does not give effervescence.
Phenol gives violet color with FeCl3 solution but benzoic acid does not give such color.

Question 18.
(i) How will you convert the following:
(a) Propanone to Propan-2-ol
(b) Ethanal to 2-hydroxypropanoic acid
(c) Toluene to benzoic acid

(ii) Give a simple chemical test to distinguish between:
(a) Pentan-2-one and Pentan-3-one
(b) Ethanal and Propanal
OR
(i) Write the products of the following reactions:

(ii) Which acid of each pair shown here would you expect to be stronger? (CBSE AI 2013)
(a) F – CH2 – COOH or Cl – CH2 – COOH OH

(ii) (a) Pentan-2-one forms yellow ppt. with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.

(b) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.

OR

(ii) (a) FCH2COOH
(b) CH3COOH

Question 19.
(a) Predict the main product of the following reactions:

(b) Give a simple chemical test to distinguish between

(c) Why is alpha (a) hydrogen of carbonyl compounds acidic in nature?
OR
(a) Write the main product formed when propanal reacts with the following reagents:
(i) 2 moles of CH3OH in presence of dry HCI
(ii) Dilute NaOH
(iii) H2N – NH2 followed by heating with KOH in ethylene glycol

(b) Arrange the following compounds in increasing order of their property as indicated:
(i) F – CH2COOH, O2N – CH2COOH, CH3COOH, HCOOH – acid character
(ii) Acetone, Acetaldehyde, Benzaldehyde, Acetophenone — reactivity towards the addition of HCN (CBSE AI 2019)

(b) On adding NaOH/l2 and heat, acetophenone gives yellow ppt of iodoform but benzophenone does not.

(c) α – hydrogen of carbonyl compounds is acidic because of the electron-withdrawing nature of the carbonyl group.
OR

(b) (i) CH3COOH < HCOOH < FCH2COOH < NO2-CH3 COOH
(ii) Acetophenone < Benzaldehyde < Acetone < Acetaldehyde

Question 20.
(a) An organic compound with the molecular formula C7H60 forms a 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes the Cannizzaro reaction. On oxidation, it gives benzoic acid. Identify the compound and state the reactions involved.
(b) Give chemical tests to distinguish between the following pair of compounds:
(i) Phenol and propanol
(ii) Benzoic acid and benzene
OR
(a) Predict the products of the following:

(b) Arrange the following in increasing order of acidic character: (CBSE 2019C)
HCOOH, CF3COOH, ClCH2COOH, CCl3COOH
(a) Compound = Benzaldehyde or C6H5CHO
Reaction
Reaction with 2,4-DNP

With Tollens reagent
RCHO + 2[Ag(NH3)2 ]+ + 30H → RCOO + 2 Ag + 2H2O + 4NH3 (where R = – C6H5)
Cannizzaro

(b) (i) FeCl3 test: Add 2 drops of neutral FeCl3 to both the compounds separately. Phenol gives violet color.

(ii) Sodium bicarbonate test: Add NaHCO3 to both the compounds: Benzoic acid reacts with NaHCO3 to give effervescence due to liberation of CO2.
C6H5COOH + NaHCO3 → C6H5COONa + H2O + CO2
OR
(a) A = CH3COOH
B = CH3COCl
C = CH3CONH2
D = CH3NH2

(b) HCOOH < ClCH2COOH < CCl3COOH < CF3COOH

• the electron-withdrawing substituents disperse the negative charge on carboxylate ion and stabilize it and thus, increase acidity.
• the electron-releasing substituents intensify the negative charge of the carboxylate ion, destabilize it and thus, decrease the acidity.

Question 21.
(i) Give chemical tests to distinguish between:
(a) Propanal and propanone
(b) Benzaldehyde and acetophenone

(ii) How would you obtain:
(a) But-2-enal from the ethanal
(b) Butanoic acid from butanol
(c) Benzoic acid from ethylbenzene? (CBSE 2011)
(i) (a) Propanal gives a silver mirror with Tollen’s reagent.

Propanone does not give this test.

(b) Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent). Acetophenone does not react.

Question 22.
(i) Describe the following giving linked chemical equations:
(a) Cannizzaro reaction
Aldehydes that do not contain any a-hydrogen atom (e.g. benzaldehyde, formaldehyde) undergo self- oxidation and reduction reaction on treatment with cone, solution of caustic alkali. In this reaction, one molecule is oxidized to acid while another molecule is reduced to an alcohol.

(b) Decarboxylation
The process of removal of a molecule of C02 from a carboxylic acid is called decarboxylation. It is usually carried out by heating a mixture of carboxylic acid or its sodium salt with soda lime (NaOH + CaO).

(ii) Complete the following chemical equations: (CBSE Delhi 2011)

Question 23.
(i) Give chemical tests to distinguish between the following:
Benzoic acid and ethyl benzoate.
(a) When treated with NaHCO3 solution, benzoic acid gives brisk effervescence while ethyl benzoate does not

(b) Ethyl benzoate on boiling with an excess of NaOH gives ethyl alcohol which on heating with iodine gives yellow ppt. of iodoform.

(ii) Complete each synthesis by giving missing reagents or products in the following:

(CBSE 2011)

Question 24.
(i) Write the products formed when CH3CHO reacts with the following reagents:
(a) HCN
(b) H2N – OH
(C) CH3CHO in the presence of dilute NaOH
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzoic acid and Phenol
(b) Propanal and Propanone
OR
(i) Account for the following:
(a) Cl – CH2COOH is a stronger acid than CH3COOH.
(b) Carboxylic acids do not give reactions of the carbonyl group.
(ii) Write the chemical equations to illustrate the following name reactions:
Rosenmund reduction
(iii) Out of CH3CH2—CO—CH2 – CH3 and CH3CH2 — CH2 — CO — CH3, which gives iodoform test? (CBSE 2014)

(ii) (a) Add neutral FeCl3 in both the solutions, phenol gives violet color with FeCl3 solution, but benzoic acid does not give such color.
(b) Add an ammoniacal solution of silver nitrate (Tollen’s reagent) in both the solutions, propanal gives silver mirror whereas propanone does not.

OR
(0) Chlorine is an electron-withdrawing atom (-I effect). It withdraws the electrons from carbon to which it is attached and therefore, this effect is transmitted throughout the chain. As a result, electrons are withdrawn more strongly towards oxygen of 0—H bond and promotes the release of the proton.

Consequently, ClCH2COOH is a stronger acid than CH3COOH.

(b) Carboxylic acids do not give the characteristic reactions of the carbonyl group (>C = 0) as given by aldehydes and ketones. In carboxylic acids, the carbonyl group is involved in resonance, as follows:

Therefore, it is not a free group. But no resonance is possible in aldehydes and ketones. They give the characteristic reactions of the group.

(ii) Rosenmund reduction. Acid chlorides are converted to corresponding aldehydes by catalytic reduction. The reaction is carried out by passing through a hot solution of the acid chloride in the presence of palladium deposited over barium sulfate (partially poisoned with sulfur or quinoline).

The poisoning of palladium catalyst decreases its activity and it does not allow the further reduction of an aldehyde into alcohol.

(iii) CH3CH2CH2COCH3 gives iodoform test

Question 25.
(i) Write the structures of A, B, C, and D in the following reactions:

(ii) Distinguish between:
(a) C6H5 – CH = CH – COCH3 and C6H5 – CH = CH – CO CH2CH3
(b) CH3CH2COOH and HCOOH

(iii) Arrange the following in the increasing order of their boiling points:
CH3CH2OH, CH3COCH3, CH3COOH
OR
(i) Write the chemical reaction involved in the Etard reaction.
(ii) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
CH3 – CHO, C6H5COCH3, HCHO
(iii) Why pKa of Cl-CH2-COOH is lower than the pKa of CH3COOH?
(iv) Write the product in the following reaction.

(v) A and B are two functional isomers of compound C3H6O. On heating with NaOH and l2, isomer A forms a yellow precipitate of iodoform whereas isomer B does not form any precipitate. Write the formulae of A and B. (CBSE 2016)

(ii) (a) Heat both the compounds with NaOH and l2, C6H5 CH = CHCOCH3 gives yellow ppt. of iodoform. C6H5CH = CHCOCH2CH3 does not give yellow ppt. of iodoform.
(b) Add ammoniacal AgNO3 solution (Tollen’s reagent), HCOOH gives silver mirror while CH3CH2COOH does not.

(iii) CH3COCH3 < CH3CH2OH < CH3COOH
OR
(i) Etard’s reaction

(ii) C6H5COCH3 < CH3CHO < HCHO

(iii) Chlorine is an electron-withdrawing group (-I inductive effect).
It withdraws the electrons from the C to which it is attached and this effect is transmitted throughout the chain. As a result, the electrons are withdrawn more strongly towards 0 of the O-H bond and promote the release of the proton. Consequently, Cl — CH2COOH is more acidic than acetic acid.

Therefore, pKa of ClCH2COOH is less than that of CH3COOH.

(iv) CH3 CH2 CH = CH CH2 CN CH3 CH2 CH = CH CH2 CHO

(v) Two isomers are CH3COCH3 and CH3CH2CHO
Since A forms yellow ppt. of iodoform on heating with NaOH and l2, it is CH3COCH3.
CH3COCH3 + 3I2 + 4NaOH → CHI3 + CH3COONa + 3Nal + 3H2O
B does not give iodoform, it is CH3CH2CHO.
A: CH3COCH3 B: CH3CH2CHO

Question 26.
Write the structures of A, B, C, D, and E in the following reactions:

OR
(i) Write the chemical equation for the reaction involved in the Cannizzaro reaction.
(ii) Draw the structure of the semicarbazone of ethanal.
(iii) Why pKa of F-CH2-COOH is lower than that of Cl – CH2 – COOH?
(iv) Write the product in the following reaction:
CH3 – CH = CH – CH2CN
(v) How can you distinguish between propanal and propanone? (CBSE Delhi 2016)

OR
(i) Cannizzaro’s reaction: Aldehydes that do not contain a-hydrogen undergo self-oxidation and reduction on treatment with cone. KOH.

(iii) Fluorine is more electronegative than Cl and therefore, will have a greater electron-withdrawing effect (-I effect). As a result, the carboxylate ion will be stabilized to a larger extent and therefore fluoro acetic acid will be a stronger acid than chloroacetic acid. Hence, the pKa of FCH2COOH will be lower than that of ClCH2COOH.

(iv) CH3 — CH = CH — CH2 CN CH3 CH = CH CH2 CHO

(v) Propanal gives red ppt with Fehling solution but propanone does not.

Question 27.
(i) Give reasons :
(a) HCHO is more reactive than CH3-CHO towards the addition of HCN. (CBSE 2018C)
(b) pKa of O2N—CH2—COOH is lower than that of CH3—COOH.
(c) Alpha hydrogen of aldehydes and ketones is acidic in nature.

(ii) Give simple chemical tests to distinguish between the following pairs of compounds :
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
OR
(i) Write the structure of the product(s) formed :
(a) CH3 — CH2 — COOH
(b) C6H5COCl
(c) 2HCHO

(ii) How will you bring the following conversions in not more than two steps :
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid
(i) (a) Due to + I effect of the methyl group is CH3CHO.
(b) Due to – I effect of the nitro group in nitroacetic acid.
(c) Due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) (a) Add NaOH and l2 to both the compounds and heat, ethanal gives yellow ppt of iodoform.
(b) Add NaOH and l2 to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.
OR

## The p-Block Elements Class 12 Important Extra Questions Chemistry Chapter 7

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 7 The p-Block Elements.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Chemistry Chapter 7 Important Extra Questions The p-Block Elements

### The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
Which one of PCl4+ and PCl4 is not likely to exist and why? (CBSE Delhi 2012)
PCl4 is not likely to exist because lone pair of PCl3 can be donated to Cl+ and not to Cl.

Question 2.
Of PH3 and H2S which is more acidic and why? (CBSE Delhi 2012)
H2S is more acidic because of weak S-H bond. PH3 behaves as a Lewis base because of the presence of lone pair of electrons on P.

Question 3.
Which is a stronger reducing agent, SbH3 or BiH3 and why? (CBSE Al 2012)
BiH3 is a stronger reducing agent than SbH3. This is because BiH3 is less stable than SbH3 because of larger size of Bi than Sb.

Question 4.
What is the covalency of nitrogen in N2O5? (CBSE Delhi 2013)
Four.

Question 5.
Name two poisonous gases which can be prepared from chlorine gas. (CBSE AI 2013)

1. Phosgene (COCl2)
2. Mustard gas (ClCH2CH2SCH2CH2Cl)

Question 6.
What is the basicity of H3PO3 and why? (CBSE AI 2013)
H3PO3 contains two P-OH bonds and therefore can give 2H+ ions. Its basicity is two.

Question 7.
Why does ammonia act as a Lewis base? (CBSE 2014)
Nitrogen atom in NH3 has one lone pair of electrons which is available for donation. Therefore,it acts as a Lewis base.

Question 8.
What is the oxidation number of phosphorus in H3PO2 molecule? (CBSE Delhi 2010)
+1

Question 9.
Out of white phosphorus and red phosphorus, which one is more reactive and why? (CBSE 2015)
White phosphorus is more reactive because of angular strain in the P4 molecules where the angles are only 60°.

Question 10.
What is the basicity of H3PO4? (CBSE Delhi 2015)
Basicity of H3PO4 is three because it has three ionisable hydrogen atoms.

Question 11.
On heating Pb(NO3)2 a brown gas is evolved which undergoes dimerisation on cooling. Identify the gas. (CBSE 2016)
NO2.

2NO2 → N2O4

Question 12.
Solid PCl5 is ionic in nature. Give reason. (CBSE AI 2016)
PCl5 is ionic in the solid state because it exists as [PCl4]+ [PCl6] in which the cation is tetrahedral and anion is octahedral.

Question 13.
Write the structural difference between white P and red P. (CBSE Delhi 2014)
White P consists of P4 units in which four P atoms lie at the corners of a regular tetrahedron with ∠PPP = 60°. Red P also consists of P4 tetrahedra units but it has polymeric structure consisting of P4 tetrahedra linked together by covalent bonds.

Question 14.
What is the basicity of H3PO2 acid and why? (CBSE AI 2012)
H3PO2 has the structure:

It has only one ionisable hydrogen and therefore, its basicity is one.

Question 15.
Name the promoter used In Haber’s process. (CBSE Sample Paper 2017-18)
Molybdenum.

Question 16.
NF3 is an exothermic compound whereas NCl3 is not. Explain. (CBSE AI 2011, 2012)
NF3 is an exothermic compound while NCl3 is not because:

• The bond dissociation enthalpy of F2 is lower than that of Cl2.
• Size of F is small as compared to Cl and therefore F forms stronger bonds with nitrogen releasing large amount of energy.

Therefore, overall energy is released during the formation of NF3 and energy is absorbed during the formation of NCl3.

Question 17.
N-N single bond is weaker than P-P single bond. (CBSE Delhi 2014)
N-N single bond is weaker than P-P single bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

Question 18.
What happens when orthophosphorous acid is heated? (CBSE Sample Paper 2017-18)
Phosphoric acid and phosphine are formed.

Question 19.
Write the order of thermal stability of the hydrides of group 16 elements. (CBSE Delhi 2017)
The thermal stability of the hydrides of group 16 elements decreases from H2O to H2Te as:
H2O > H2S > H2Se > H2Te

Question 20.
The two O – O bond lengths in ozone molecule are equal. Why? (CBSE AI 2013, CBSE Delhi 2014)
Ozone is a resonance hybrid of two structures and therefore, the two O-O bond lengths are equal.

Question 21.
SF6 is not easily hydrolysed. (CBSE Delhi 2005)
OR
SF6 is kinetically inert substance. Explain. (CBSE Al 2011, CBSE Delhi 2011)
SF6 is chemically inert and therefore, does not get hydrolysed. Its inert nature is due to the presence of stearically protected sulphur atom which does not allow thermodynamically favourable hydrolysis reaction.

Question 22.
Which of the following compounds has a lone pair of electrons at the central atom? (CBSE Sample Paper 2011)
H2S2O8, H2S2O7, H2SO3, H2SO4.
H2SO3

Question 23.
Ozone is thermodynamically unstable. Explain. (CBSE Sample Paper 2011)
Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (∆H is -ve) and increase in entropy (∆S is +ve). These two factors reinforce each other resulting negative ∆G (∆G = ∆H – T∆S) for its conversion to oxygen.

Question 24.
Which neutral molecule would be isoelectronic with ClO? Is that molecule a Lewis base? (CBSE Delhi 2008)
ClO is isoelectronic with ClF. Yes, it is a Lewis base.

Question 25.
Which compound of xenon has distorted octahedral shape? (CBSE Sample Paper 2012)
XeF6 has distorted octahedral shape.

Question 26.
Iodide ions can be oxidised by oxygen in acidic medium. Give chemical equation to support this. (CBSE Sample Paper 2011)
Iodide can be oxidised by oxygen in acidic medium:
4I(aq) + O2(g) + 4H+(oq) > 2l2(s) + 2H2O(l).

Question 27.
What happens when XeF6 undergoes complete hydrolysis? (CBSE Sample Paper 2017-18)
XeO3 is formed.
XeF6 + 3H2O → XeO3 + 6HF

Question 28.
Which xenon compound is isostructural with lCl4? (CBSE Sample Paper 2011)
XeF4

Question 29.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

### The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
What inspired N. Bartlett for carrying out reaction between Xe and PtF6? (CBSE Delhi 2013)
In 1962, N. Bartlett noticed that platinum hexafluoride PtF6, is a powerful oxidising agent which combines with molecular oxygen to form ionic compound, dioxygenyl hexafluoroplatinate (V), O2+ [PtF6].
O2(S) + PtF6(g) → O2+ [PtF6]
This indicates that PtF6 has oxidised O2 to O2+. Now, oxygen and xenon have some similarities:

• The first ionisation energy of xenon gas (1170 kJ mol-1) is fairly close to that of oxygen (1177 kJ mol-1).
• The molecular diameter of oxygen and atomic radius of xenon are similar (4Å). These prompted Bartlett to carry out the reaction between Xe and PtF6.

Question 2.
Calculate the number of lone pairs on central atom in the following molecule and predict the geometry. (CBSE Sample Paper 2019)
XeF4 has square planar structure. It involves sp3d2 hybridisation of Xenon in which two positions are occupied by lone pairs.

Structure of XeF4 (Square planar)

Question 3.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE AI 2013)
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X, e.g. ClF, ICl3, BrF5, IF7, etc.

Question 4.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalides. Give reason. (CBSE 2013, CBSE Delhi 2015)
Nitrogen belongs to second period (n = 2) and has only s and p-orbitals. It does not have d-orbitals in its valence shell and therefore, it cannot extend its octet. That is why nitrogen does not form pentahalides.

Question 5.
Why does NO2 dimerise? Explain. (CBSE 2014, CBSE Delhi 2014)
NO2 contains odd number of valence electrons. It behaves as a typical molecule. In the liquid and solid state, it dimerises to form stable N2O4 molecule, with even number of electrons. Therefore, NO2 is paramagnetic, while N2O4 is diamagnetic in which two unpaired electrons get paired.

Question 6.
Draw the structure of O3 molecule. (CBSE Delhi 2010)
Ozone has angular structure as:

It is resonance hybrid of the following structures:

Question 7.
Complete the following equations: (CBSE 2014)
(i) Ag + PCl5
(ii) CaF2 + H2SO4
(i) 2Ag + PCl5 → 2AgCl + PCl3
(ii) CaF2 + H2SO4 → CaSO4 + 2HF

Question 8.
Write balanced chemical equations for the following processes:
(i) XeF2 undergoes hydrolysis.
(ii) MnO2 is heated with cone. HCl.
OR
Arrange the following in order of property indicated for each set:
(i) H2O, H2S, H2Se, H2Te – increasing acidic character
(ii) HF, HCl, HBr, HI – decreasing bond enthalpy (CBSE Delhi 2019)
(i) 2XeF2 + 2H2O → 2Xe + 4HF + O2
(ii) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
OR
(i) H2O < H2S < H2Se < H2Te
(ii) HF > HCI > HBr > HI

Question 9.
Explain the following giving an appropriate reason in each case.
(i) O2 and F2 both stabilise higher oxidation states of metals but O2 exceeds F2 in doing so.
(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach. (CBSE 2012)
(i) This is because fluorine can show only -1 oxidation state, whereas oxygen can show -2 oxdiation state as well as positive oxidation states such as +2, +4 and +6.

(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach because xenon is a noble gas and its octet is complete. Therefore, according to Valence Bond theory, xenon should not combine with fluorine to form fluorides due to its stable outermost shell configuration.

Question 10.
Draw the structures of the following molecules:
(i) XeOF4
(ii) H3PO3 (CBSE 2013)
(i)

Structure of XeOF4

(ii)

Structure of H3PO3

Question 11.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE 2013)
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X,
e.g. ClF, ICl3, BrF5, IF7, etc.

Question 12.
Draw the structures of the following molecules: (CBSE 2013)
(i) N2O5
(ii) XeF2
(i)

(ii)

Structure of XeF2 (Linear)

Question 13.
What happens when
(i) PCl5 is heated?
(ii) H3PO3 is heated?
Write the reactions involved. (CBSE Delhi 2013)
(i) On heating, PCl5 sublimes and decomposes on strong heating:

(ii) On heating, H3PO3 decomposes into phosphoric acid and phosphine.
4H3PO3 → 3H3PO4 + PH3

Question 14.
Complete the following chemical equations:
(i) Ca3P2 + H2O →
(ii) Cu + H2SO4(conc.) →
OR
Arrange the following in the order of property indicated against each set:
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H2O, H2S, H2Se, H2Te – increasing acidic character. (CBSE Delhi 2014)
(i) Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3
(ii) Cu + 2H2SO24;(conc.) → CuSO4
+ SO2 + 2H2O
OR
(i) HI < HBr < HCl < HF
(ii) H2O < H2S < H2Se < H2Te

Question 15.
Complete the following equations:
(i) P4 + H2O →
(ii) XeF4 + O2F2 → (CBSE 2014)
(i) P4 + H2O → no reaction
(ii) XeF4 + O2F2 → XeF6 + O6

Question 16.
Draw the structures of the following:
(i) XeF2
(ii) BrF3
(i)

Structure of XeF2 (Linear)

(ii)

Structure of BrF3 (T Shape)

Question 17.
(a) Draw the structure of XeF4.
(b) What happens when CaF2 reacts with cone. H2SO4? Write balanced chemical equation. (CBSE 2019C)
(a)

(b) HF is formed.
CaF2 + H2SO4 → 2HF + CaSO4

Question 18.
(a) Draw the structure of H2S2O7.
(b) What happens when carbon reacts with cone. H2SO4? Write balanced chemical equation. (CBSE 2019C)
(a)

(b) Carbon dioxide is formed.
C + 2H2SO4(conc.) → CO2 + 2 SO2 + 2H2O

Question 19.
Draw the structures of the following molecules: (CBSE 2014)
(i) XeO2
(ii) H2SO4
(i)

(ii)

Question 20.
Write the structure of the following: (HPO3)3 (CBSE 2016)
(HPO3)3

Question 21.
Why is N2 less reactive at room temperature? (CBSE Al 2015, H.P.S.B. 2015, 2016, Meghalaya S.B. 2017)
In molecular nitrogen, there is a triple bond between two nitrogen atoms (NsN) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 22.
Phosphine has lower boiling point than ammonia. Give reason. (CBSE Al 2008,
CBSE Delhi 2013)
Ammonia exists as associated molecule due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH3, phosphine (PH3) molecules are not associated through hydrogen bonding in liquid state.
This is because of low electronegativity of P than N. As a result, the boiling point of PH3 is lower than that of NH3.

Question 23.
Draw structures of the following:
(a) XeF4
(b) S2O82- (CBSE AI 2019)
(a)

(b)

Question 24.
(i) Draw the structure of phosphinic acid (H3PO2).
(ii) Write a chemical reaction for its use as reducing agent. (CBSE Sample Paper 2011)
(i) Phosphinic acid, H3PO2.

(ii) H3PO2 reduces Ag+ ion to Ag which shows its reducing nature.
H3PO2 + 4AgNO3 + 2H2O → 4Ag + 4HNO3 + H3PO4

Question 25.
Among the hydrides of Group 15 elements, which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character? (CBSE AI 2018)
(a) PH3
(b) NH3
(c) NH3
(d) BiH3

Question 26.
The bond angles (O-N-O) are not of the same value in NO2 and NO2+. Give reason. (CBSE Delhi 2012)
In NO2 there is one electron on N. Therefore, in NO2+, there is no electron on N atom and there is a lone pair of electrons on N atom in NO2.

Question 27.
H2PO2 is a stronger reducing agent than H3PO3. Give reason. (CBSEAI 2014, 2016)
The reducing character of the acid is due to H atoms bonded directly to P atom. In H3PO2, there are two P-H bonds whereas in H3PO3, there is one P-H bond. Therefore, H3PO2 is a stronger reducing agent than H2PO3.

Question 28.
Bi(V) is a stronger oxidising agent than Sb(V). Give reason. (CBSE Delhi 2014)
On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

Question 29.
Dioxygen is a gas while sulphur is a solid at room temperature. Why?
(CBSE AI 2013, 2018)
Oxygen exists as a stable diatomic molecule and is, therefore, a gas. On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has strong tendency to form multiple bonds with itself and forms strong O=O bonds rather than O-O bonds.

On the other hand, sulphur-sulphur double bonds (S=S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O=O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule.
Therefore, oxygen exists as a diatomic gas and sulphur exists as S8 solid.

Question 30.
Draw structures of the following:
(a) H2S2O7
(b) HClO3 (CBSE Al 2019)
(a) H2S2O7

(b) HClO3

Question 31.
Complete and balance the following equations:
(a) C + H2SO4 (cone.) →
(b) XeF2 + PF5
OR
Write balanced chemical equations involved in the following reactions:
(a) Fluorine gas reacts with water.
(b) Phosphine gas is absorbed in copper sulphate solution. (CBSE AI 2019)
(a)

(b)

OR

(a) 2F2 + 2H2O → 4HF + O2

(b)

Question 32.
Sulphuric acid has low volatility. Give chemical reactions in support of this. (CBSE Sample Paper 2011)
Sulphuric acid has low volatility and decomposes the salts of volatile acids forming its own salts:
2MX + H2SO4 → 2HX + M2SO4
(M = metal, X = F, Cl, NO3)
e.g.,NaCl + H2SO4 → NaHSO4 + HCl
KNO3 + H2SO4 → KHSO4 + HNO3.

Question 33.
Electron gain enthalpies of halogens are largely negative. Why? (CBSE AI 2017)
The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns2np6).

Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have largely negative electron gain enthalpies.

Question 34.
Why does R3P = 0 exist but R3N = 0 does not (R = alkyl group). (CBSE AI 2014)
R3N = 0 does not exist because nitrogen cannot have covalency more than four. Moreover, R3P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bond whereas nitrogen cannot form dπ-pπ bond.

Question 35.
Write any two oxoacids of sulphur and draw their structures. (CBSE Al 2019)
(i) Sulphuric acid

(ii) Peroxodisulphuric acid

Question 36.
Complete and balance the following equations:
(a) NH3 (excess) + Cl2
(b) XeF6 + 2H2O →
OR
Write balanced chemical equations involved in the following reactions:
(a) XeF4 reacts with SbF5.
(b) Ag is heated with PCl5. (CBSE AI 2019)
(a)
8NH3 + 3Cl2 → 6NH4Cl + N2
excess
(b) XeF6 + 2H2O → XeO2F2 + 4HF

OR

(a) XeF4 + SbF5 → [XeF3]+[SbF6]
(b) 2Ag + PCl5 → 2AgCl + PCl3

Question 37.
Complete and balance the following equations:
(a) S + H2SO4 (cone.) →
(b) PCl3 + H2O →
Write balanced chemical equations involved in the following reactions:
(a) Chlorine gas reacts with cold and dilute NaOH
(b) Calcium phosphide Is dissolved in water. (CBSE AI 2019)
(a) S + 2H2SO4(conc.) → 3SO2 + 2H2O
(b) PCl3 +3H2O → H3PO3 + 3HCl
OR
(a) Cl2 + ZNaOH → NaCl + NaOCl + H2O
(b) Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

Question 38.
Bond enthalpy of fluorine Is lower than that of chlorine. Why? (CBSE AI 2018)
Fluorine atoms are smalL and internuclear distance between two F atoms In F2 molecule is also small (1 .43Å). As a result, electron-electron repulsions among the lone pairs on two fluorine atoms in F2 molecule are large. These Llarge electron-electron repulsions weaken the F-F bond. However, the electron-electron repulsions among lone pairs in Cl2 molecule are less because of Larger Cl atoms and larger Cl-Cl distance (1.99Å). Therefore, the bond enthalpy of fluorine is lower than that of chlorine.

### The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
Give reasons for the following:
(a) Dioxygen is a gas but sulphur a solid.
(b) NO (g) released by jet aeroplanes is slowly depleting the ozone layer.
(c) Interhalogens are more reactive than pure halogens. (CBSE Al 2019)
(a) Due to small size and high electronegativity, oxygen atom forms pπ-pπ double bond, O = O. The intermolecular forces in oxygen are weak van der Waal’s forces and therefore, oxygen exists as a gas. On the other hand, sulphur does not form stable pπ-pπ bonds and does not exist as S2. It is linked by single bonds and forms polyatomic complex molecules having eight atoms per moledule (S8) and has puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

(b) Nitrogen oxide (NO) gas emitted from the exhaust of jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen.
NO + O3 → NO2 + O2
Since jet aeroplanes fly in the stratosphere, near ozone layer, these are responsible for the depletion of ozone layer.

(c) This is because covalent bond between dissimilar atoms in interhalogens (X- Y) is polar and weaker than between similar atoms (X-X and Y-Y). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping between similar atoms.

Question 2.
Give reasons for the following:
(i) Where R is an alkyl group, R3P = 0 exists but R3N = 0 does not.
(ii) PbCl4 is more covalent than PbCl2.
(iii) At room temperature, N2 is much less reactive. (CBSE AI 2013)
(i) R3N = 0 does not exist because nitrogen cannot have covalency more than four. But R3P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bonds whereas nitrogen cannot form dπ-pπ bond.

(ii) Pb4+ has smaller size than Pb2+. As a result, it can polarise the Cl ion to greater extent than Pb2+. Consequently, PbCl4 is more covalent than PbCl2.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 3.
Give reasons for the following:
(i) Oxygen is gas but sulphur is a solid,
(ii) O3 acts as a powerful oxidising agent.
(iii) BiH3 is the strongest reducing agent amongst all the hydrides of Group 15 elements. (CBSE Al 2013)
(i) Oxygen exists as a stable diatomic molecule and is, therefore, a gas.
On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has good tendency to form multiple bonds with itself and forms strong o = o bonds than o – o bonds. On the other hand, sulphur- sulphur double bonds (S = S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O = O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule. Therefore, oxygen exists as a diatomic gas and sulphur exists as S8 solid.

(ii) Ozone acts as a powerful oxidising agent because it has higher energy content and decomposes readily to give atomic oxygen as:
O3 → O2 + O
Therefore, ozone can oxidise a number of non-metals and other compounds. For example,
PbS (s) + 4O3 (8) → PbSO4 (s) + 4O2 (g)
2I (aq) + H2O(l) + O3(g) → 2OH (aq) + I32(s) + O2(g)

(iii) The stability of group 15 hydrides decreases from NH3 to BiH3. Therefore, BiH3 is most unstable hydride among group 15 hydrides because of lowest M-H bond strength. As a result, BiH3 is the strongest reducing agent.

Question 4.
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalphy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bonds lengths in ozone molecule are identical. (CBSE Al 2013)
(i) Nitrogen does not have vacant d-orbitals in its valence shell. Therefore, it cannot extend its valency beyond 3. Thus, it cannot form pentahalides.

(ii) The less negative electron gain enthalpy of fluorine as compared to chlorine is due to very small size (72 pm) of fluorine atom. As result, there are strong inter- electronic repulsions in the relatively small 2p sub-shell of fluorine and therefore, the incoming electron does not feel much attraction. Thus, its electron gain enthalpy is small.

(iii) Ozone is resonance hybrid of two structures and therefore, the two oxygen-oxygen bonds are identical.

Question 5.
How would you account for the following:
(i) H2S is more acidic than H2O.
(ii) The N-O bond in NO2 is shorter than the N-O in NO3.
(iii) Both O2 and F2 stabilise higher oxidation states but the ability of oxygen to stabilise the higher oxidation states exceeds that of fluorine. (CBSE 2011)
(i) Since size of S is more than that of O, the distance between central atom and hydrogen atoms is more in H2S than in H2O. Therefore, bond dissociation enthalpy of H2S is less and bond cleavage is more easy than in H2O. Therefore, H2S is more acidic than H2O.

(ii) In the resonance structures of NO2, two bonds are sharing a double bond, while in NO3, three bonds are sharing a double bond.

As a result, the bond in NO2 will be shorter than in NO3.

(iii) The larger tendency of oxygen to stabilise the higher oxidation state is because of the tendency of oxygen to form multiple bonds with metal.

Question 6.
How would you account for the following:
(i) NF3 is an exothermic compound but NCl3 is not.
(ii) The acidic strength of compounds increases in the order:
PH3 < H2S < HCl
(iii) SF6 is kinetically inert. (CBSE 2011)
(i) NF3 is an exothermic compound while NCl3 is an endothermic compound because
(a) The bond dissociation enthalpy of F2 is lower as compared to Cl2.
(b) Size of F is small as compared to Cl and therefore, F forms stronger bond with N releasing large amount of energy.
Therefore, overall energy is released during the formation of NF3 and energy is absorbed during the formation of NCl3.

(ii) The bond dissociation enthalpy decreases in the order:
P-H > S-H > Cl-H
Therefore, acidic character follows the order:
PH3 < H2S < HCl

(iii) SF6 is kinetically inert because sulphur is sterically protected by six F atoms and it is a coordinatively saturated compound. Therefore, it does not undergo thermodynamically favourable hydrolysis reaction.

Question 7.
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H2SO4
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
(a) (i) XeOF4

(ii) H2SO4

(b) White phosphorus consists of P4 units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Red phosphorus also consists of P4 tetrahedra units, but it has polymeric structure consisting of P4 tetrahedra linked together by covalent bond.

Question 8.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N-N single bond is weaker than P-P single bond.
(iii) Noble gases have very low boiling points. (CBSE Delhi 2014)
(i) On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

(ii) N-N single bond is weaker than single P-P bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

(iii) Noble gases are monoatomic gases and are held together by weak vander Waals forces. Therefore, these are liquified at very low temperatures. Hence they have low boiling points.

Question 9.
Give reasons for the following: (CBSE 2014)
(i) (CH3)3 P = 0 exists but (CH3)3 N = 0 does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H3PO2 is a stronger reducing agent than H3PO3.
(i) (CH3)3 N = 0 does not exist because
nitrogen cannot have covalency more than four. Moreover, (CH3)3 P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ – pπ bonds whereas nitrogen cannot form dπ – pπ bonds.

(ii) This is due to small size of oxygen atom so that its electron cloud is distributed over a small region of space and therefore, it repels the incoming electron. Hence the electron gain enthalpy of oxygen is less negative than sulphur.

(iii) In H3PO3, there are two P-H bonds whereas in H3PO3, there is one P-H bond. Therefore, H3PO3 is stronger reducing agent than H3PO3.

Question 10.
Assign reason for the following:
(i) H3PO2 is a stronger reducing agent than H3PO4.
(ii) Sulphur shows more tendency for catenation than oxygen.
(iii) Reducing character increases from HF to HI. (CBSE 2016)
(i) The reducing character of the acid is due to H atoms bonded directly to P atom. In H3PO2, there are two P-H bonds whereas in H3PO3, there is one P-H bond. Therefore, H3PO2 is a stronger reducing agent than H3PO3.

(ii) Sulphur shows more tendency for catenation than oxygen because of stronger S-S bonds than O-O bonds.

(iii) Due to decrease in bond enthalpy down the group, the thermal stability decreases from HF to HI.
As a result of decrease in stability from HF to HI, the reducing character increases down the group as:
HF < HCl < HBr < HI

Question 11.
What happens when
(i) (NH4)2Cr2O7 is heated?
(ii) PCl5 is heated?
(iii) H3PO3 is heated? Write the equations involved. (CBSE AI 2015, CBSE Delhi 2008, 2013, 2017)
(i) On heating (NH4)2Cr2O7, nitrogen gas is evolved.

(ii) On heating, PCl5 first sublimes and then decomposes on strong heating:

(iii) On heating, H3PO3 disproportionates to give orthophosphoric acid and phosphine.

Question 12.
(a) Suggest a quantitative method for estimation of the gas which protects us from U.V. rays of the sun.
(b) Nitrogen oxides emitted from the exhaust system of supersonic jet aeroplanes slowly deplete the concentration of ozone layer in upper atmosphere. Comment. (CBSE Sample Paper 2011)
(a) The gas which protects us from U.V. rays of the sun is ozone. It reacts with I ions to give iodine as:
O3 + 2I + H2O → O2 + l2 + 2OH
l2 liberated is titrated against sodium thiosulphate solution and amount of O3 can be estimated.

(b) The release of nitrogen oxides (NOx) into stratosphere by the exhaust system of supersonic jet aeroplanes deplete the concentration of O2 because NO reacts with O3 to give O2.
NO (g) + O2 (g) → NO2 (S) + O2 (s)
Therefore, NO is slowly depleting the concentration of ozone.

Question 13.
Give reasons for the following:
(a) Acidic character decreases from N2O3 to Bi2O3.
(b) All the P-Cl bonds in PCl5 are not equivalent.
(c) HF is a weaker acid than HCl in an aqueous solution. (CBSE Al 2019)
(a) On moving down the group, metallic character increases. Therefore acidic character of oxides decreases from nitrogen to bismuth.

(b) PCl5 has trigonalbipyramidal geometry.

In this geometry, axial bonds are larger than equatorial bonds because of large repulsions at axial positions. Therefore, all P-Cl bonds in PCl5 are not equivalent.

(c) The H-F bond has high bond dissociation enthalpy than H-Cl bond. Therefore, H-F bond is stronger and can be dissociated into H+ and F ions with difficulty. Hence, it is weaker acid than HCl.

Question 14.
Give reasons for the following:
(a) O-O single bond is weaker than S-S single bond.
(b) Tendency to show -3 oxidation state decreases from Nitrogen (N) to Bismuth (Bi).
(c) Cl2 acts as a bleaching agent. (CBSE AI 2019)
(a) O-O single bond is weaker than S-S single bond because oxygen has smaller size and has high inter-electronic repulsion.

(b) Tendency to show -3 oxidation state decreases because metallic character and size increase down the group.

(c) Cl2 acts as a bleaching agent in the presence of moisture
Cl2 + H2O → 2HCl + [O]
Coloured matter + [O] → Colourless matter
The nascent oxygen is responsible for the bleaching action of Cl2.

Question 15.
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H2SO4
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
(a) (i) XeOF4

(ii) H2SO4

(b) White phosphorus consists of P4 units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Red phosphorus also consists of P4 tetrahedra units, but it has polymeric structure consisting of P4 tetrahedra linked together by covalent bond.

Question 16.
Account for the following:
(i) PCl5 is more covalent than PCl3.
(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(iii) The two O-O bond lengths in the ozone molecule are equal. (CBSE Delhi 2014)
(i) In PCl5, the oxidation state of P is +5 while in PCl3, it is +3. As a result of higher oxidation state of the central atom, PCl5 has larger polarising power and can polarise the chloride ion (Cl) to a greater extent than in the corresponding PCl3. Since larger the polarisation, larger is the covalent character; PCl5 is more covalent than PCl3.

(ii) Iron reacts with HCl to form ferrous chloride with the evolution of hydrogen gas.
Fe + 2HCl → FeCl2 + H2

(iii) The two O-O bond lengths in ozone molecule are equal and are intermediate between single and double bonds. This is because of resonance between two structures (a) and (b) and actual molecule is resonance hybrid of these two structures:

Question 17.
Account for the following:
(a) Moist SO2 decolourises KMnO4 solution.
(b) In general, interhalogen compounds are more reactive than halogens (except fluorine).
(c) Ozone acts as a powerful oxidising agent. (CBSE Sample Paper 2019)
Moist sulphur dioxide behaves as a reducing agent. It reduces MnO4 to Mn2+.
5 SO2 + 2MnO4 + 2H2O → 5 SO42- + 4H+ + 2Mn2+

(b) X-X’ bond in inter halogens is weaker than X-X bond in halogens except F-F bond. This is due to the fact that the overlapping of orbitals of two dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(c) Due to the ease with which it liberates atoms of nascent oxygen, Ozone acts as a powerful oxidising agent.
O3 → O2 + O

Question 18.
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of Fe(lll) salt?
(a) (i) Sodium chlorate and sodium chloride are formed.

(ii) Fe (III) salt is reduced to Fe (II) salt.
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+

(i) What is the basicity of H3PO3 and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points? (CBSE Delhi 2011)
(i) H3PO3 is dibasic and has basicity of
two because it has two P-OH bonds which are ionisable. The third H atom is linked to P and is non-ionisable.

H3PO3 ⇌ HPO32- + 2H+

(ii) Fluorine does not play the role of a central atom in interhalogen compounds because it is highly electronegative. Moreover, it has only one electron less than the octet and does not have vacant d-orbitals in its valence shell. Therefore, it can form only one bond with other halogen atoms and cannot act as central atom in interhalogen compounds.

(iii) Noble gases have very low boiling points because only weak van der Waals’ forces are present between the atoms of the noble gases in the liquid state.

Question 19.
(a) Give reasons for the following:
(i) Bond enthalpy of F2 is lower than that of Cl2.
(ii) PH3 has lower boiling point than NH3.
(a) (i) Due to small size of F atom, there are strong repulsions between the non-bonding electrons of F atoms in the small sized F2 molecule. Therefore, bond enthalpy of F2 is lower than relatively larger Cl2 molecule in which repulsions between non-bonding electrons are less.

(ii) Ammonia exists as associated molecules due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH3, phosphine (PH3) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of PH3 is lower than that of NH3.

(b) Draw the structures of the following molecules:
(i) BrF3
(ii) (HPO3)3
(iii) XeF4
(i)

BrF3 (T shape)

(ii)

(HPO3)3

(iii)

XeF4

OR

(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
(a) (i) Helium alongwith oxygen (helium- oxygen mixture) is used by deep sea divers in preference to nitrogen-oxygen mixture because of its low solubility in blood.

(ii) Fluorine is the most electronegative element and therefore, it shows an oxidation state of -1 only. It does not show any positive oxidation state.

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

(b) Draw the structures of the following molecules.
(i) XeF2
(ii) H2S2O8 (CBSE Delhi 2013)
(i)

XeF2 (Linear)

(ii)

Question 20.
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
(a) (i) In the vapour state, sulphur exists as S2 molecules. S2 molecule has two unpaired electrons in anti-bonding molecular orbitals (πx and πy) and hence shows paramagnetism.

(ii) The N-N bond is weaker than P-P bond because of high interelectronic repulsions of non-bonding electrons due to small N-N bond length (109 pm).

(iii) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is negative) and increase in entropy (ΔS positive). These two factors make large negative ΔG for its conversion to oxygen.

(b) Write the name of gas released when Cu is added to
(i) dilute HNO3 and
(ii) conc. HNO3
(i) With dilute HNO3; NO (nitric oxide)
3Cu + 8HNO3(dil) → 3Cu(NO3)2 + 2NO + 4H2O

(ii) With cone. HNO3: NO2 (nitrogen dioxide)
Cu + 4HNO3(conc) → Cu(NO3)2 + 2NO2 + 2H2O

OR

(a) (i) Write the disproportionation reaction of H3PO3.
(ii) Draw the structure of XeF4.
(i) 4H3PO3 → 3H3PO4 + PH3

(ii)

(b) Account for the following:
(i) Although fluorine has less negative electron gain enthalpy, yet F2 is strong oxidising agent.
(ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.
(i) Fluorine is the strongest oxidising agent, although it has less negative electron gain enthalpy. This is because of its smaller size, large bond dissociation enthalpy of F2 and high exothermic hydration enthalpy of small F ion.

(ii) On moving down the group, the metallic character increases. Therefore, the acidic character of oxides decreases down the group from N2O3 to Bi2O3

(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. (CBSE Delhi 2019)
SO2 decolourises pink violet colour of acidified potassium permanganate solution.
2KMnO4 + 5SO2 + 2H2O → K2SO4 + 2MnSO4 + 2H2SO4

Question 21.
(a) Complete the following chemical reaction equations:
(i) P4 + SO2Cl4
(ii) XeF6 + H4O →
(i) P4 + 10SO2Cl2 → 4PCl5 + 10SO2

(ii) XeF6 + H2O → XeOF4 + 2HF

(b) Predict the shape and the asked angle (90° or more or less) in each of the following cases:
(i) SO32- and the angle O – S – O
(ii) ClF3 and the angle F – Cl – F
(iii) XeF2 and the angle F – Xe – F
(i) SO32-

The O-S-O angle is more than 90°.

(ii) ClF3

The F-Cl-F bond angle is equal to 90°.

(iii)
XeF2

F-Xe-F bond angle is equal to 180° (greater than 90°).

OR

(a) Complete the following chemical equations:
(i) NaOH + Cl2
(hot and cone.)
(ii) XeF4 + O2F2
(i) 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
(Hot and conc.)

(ii) XeF4 + O2F2 → XeF6 + O2

(b) Draw the structures of the following molecules:
(i) H3PO2
(ii) H2S2O7 (CBSE 2012)
(i) H3PO2

(ii) H2S2O7

Question 22.
(a) Draw the molecular structures of following compounds:
(i) XeF6
(ii) H2S2O8
(i)

(ii)

(b) Explain the following observations:
(i) The molecules NH3 and NF3 have dipole moments which are of opposite directions.
(ii) All the bonds in PCl5 molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
(i) Both NH3 and NF3 have pyramidal shape with one lone pair on N atom.

The lone pair on N is in opposite direction to the N-F bond moments and therefore, it has very low dipole moment (about 0.234 D). But ammonia has high dipole moment because its lone pair is in the same direction as the N-H bond moments.

(ii) PCl5 has trigonal bipyramidal structure in which there are three P-Cl equatorial bonds and two P-Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90° while the three equatorial bonds are being repelled by two bond pairs at 90°. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm).

(iii) In vapour state, sulphur partly exists as S2 molecule and S2 molecule like O2 has two unpaired electrons in anti-bonding π* molecular orbitals. Therefore, it is paramagnetic.

OR

(a) Complete the following chemical equations:
(i) XeF4 + SbF5
(ii) Cl2 + F2 (excess ) →
(i) XeF4 + SbF5 → XeF4.SbF5 → [XeF3]+ [SbF6]

(ii)

(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in NO2 and NO2+. (CBSE Delhi 2012)
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) The stablity of +5 oxidation state decreases down the group because of inert pair effect. Therefore, the +5 oxidation state of Bi is less stable than that of Sb.

(iii) In NO2, there is one electron on N while in NO2+ there is no electron and in NO2+, there is a lone pair of electrons on N as shown below:

NO2+ molecule is linear and has bond angle of 180°. The repulsion by single electron is less as compared to repulsion by a lone pair of electrons. Therefore, bond pairs in NO2 are forces more closer than in NO2.

Question 23.
(a) Draw the structure of the following molecule: H3PO2
(a)

(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is nonpolar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) Though electronegativity of F is more than O, yet water forms more extensive hydrogen bonding than HF. In H2O, each oxygen is tetrahedrally surrounded by two covalent bonds and two hydrogen bonds, (-O….H). In HF there is only one hydrogen bond (-F….H).

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

OR

(a) Draw the structures of the following molecules:
(i) N2O5
(ii) HClO4

(a)
(i)

(ii)

(b) Explain the following observations:
(i) H2S is more acidic than H2O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (CBSE 2012)
(i) The size of S is more than that of O. Therefore, the distance between S and H, i.e. S-H bond length, is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in H2S than O-H bond in water. Therefore, H2S will be more acidic than H2O.

(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of – 1 only. It does not show any positive oxidation state.

(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

Question 24.
Write balanced equations for the following reactions:
(a) P4 + NaOH + H2O → CBSE Delhi 2009)
(b) As4 + Cl2(excess) →
(c) P4O10 + H2O →
(d) Ca3P2 + H2O → (CBSE Delhi 2008, 2014)
(e) POCl3 + H2O →
(f) HgCl2 + PH3 → (CBSE AI 2010)
(g) Ag + PCl5 → (CBSE AI 2014)
(a) P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
(b) As4 + 10Cl4(excess) → 4AsCl5
(c) P4O10 + 6H4O → 4H3PO4
(d) Ca3P2 + 6H2O → 2PH3 + 3 Ca(OH)2
(e) POCl3 + 3H2O → H3PO4 + 3HCl
(f) 3HgCl2 + 2PH3 → Hg3P2 + 6HCl
(g) 2Ag + PCl5 → 2 AgCl + PCl3

Question 25.
Complete the following reactions
(i) NaOH + Cl2
hot and conc.
(ii) NH3 + Cl2 (excess) → (CBSE Delhi 2017)
(iii) NaNO2 + HCl →
(iv) F2(g) + H2O(l) → (CBSE Delhi 2008)
(v) K2CO3 + HCl →
(vi) Cl2 + H2O → (CBSE Delhi 2017)
(vii) F2 + 2Cl → (CBSE Delhi 2017)
(i) 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
hot and conc.
(ii) NH3 + 3Cl2 (excess) → NCl3 + 3HCl
(iii) 2NaNO2 + 2HCl → 2NaCl + H2O + NO2 + NO
(iv) 2F2(g) + 2H2O(l) → 4H+(aq) + 4F (aq) + O2(g)
(v) K2CO3 + 2HCl → 2KCl + CO2 + H2O
(vi) 2Cl2 + 2H2O → 4HCl + O2
(vii) F2 + 2Cl → 2F + Cl2

Question 26.
Complete the following reactions:
(i) XeF4 + SbF5 → (CBSE Delhi 2012)
(ii)

(CBSE Delhi 2012, CBSE Al 2012, 2014)
(iii) XeF2 + H2O →
(iv) XeF4 + H2O →
(v) XeF6 + H2O → (CBSE Delhi 2012)
(vi) XeF6 + 2H2O → (CfiSE Delhi 2017)
(vii) XeF6 + 3H2O → (CBSE Delhi 2017)
(i) XeF4 + SbF5 → [XeF3]+ [SbF6]
(ii)

(iii) XeF2 + H2O → 2Xe + 4HF + O2
(iv) XeF4 + H2O → 4Xe + 2XeO3 + 24HF + 3O2
(v) XeF6 + H2O → XeOF4 + 2HF
(vi) XeF6 + 2H2O → XeO2F2 + 4HF
(vii) XeF6 + 3H2O → XeO3 + 6HF

Question 27.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl5 is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
(a) (i) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is -ve) and increase in entropy ΔS is +ve). These two factors reinforce each other resulting in negative ΔG (ΔG = ΔH – TΔS) for its conversion to oxygen.

(ii) PCl5 has trigonal bipyramidal structure and is not very stable. It splits up into more stable tetrahedral and octahedral structures which are stable as
PCl5 ⇌ [PCl4]+ [PCl6]
Therefore, it exists as ionic.

(iii) Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

(b) Draw the structure of
(i) BrF5
(ii) XeF4
(i)

Square Pyramidal

(ii)

OR

(i) Compare the oxidising action of F2 and Cl2 by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
F2 is stronger oxidising agent than Cl2. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F2 than for Cl2. Hence, F2 is stronger oxidising agent than Cl2.

(ii) Write the conditions to maximise the yield of H2SO4 by contact process.
Conditions for maximum yield of H2SO4 by Contact Process:
(a) Low temperature (optimum temperature 720 K)
(b) High pressure (optimum pressure 2 bar)
(c) Presence of catalyst (V2O5 catalyst).

(iii) Arrange the following in the increasing order of property mentioned:
(a) H3PO3, H3PO4, H3PO2 (Reducing character)
(b) NH3, PH3, ASH3, SbH3, BiH3 (Base strength) (CBSE 2016)
(a) H3PO2 > H3PO3 > H3PO4
(b) NH3 > PH3 > ASH3 > SbH3 > BiH3

Question 28.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(a) (i) In gaseous state, hydrogen halides are covalent. But in aqueous solution, they ionise and behave as acids. The acidic strength of these acids decreases in the order:
HI > HBr > HCl > HF
Thus, HF is the weakest acid and HI is the strongest acid among these hydrogen halides.

The above order of acidic strength is reverse of that expected on the basis of electronegativity. Fluorine is the most electronegative halogen, therefore, the electronegativity difference will be maximum in HF and should decrease gradually as we move towards iodine through chlorine and bromine.

Thus, HF should be most ionic in nature and consequently it should be strongest acid. Although many factors contribute towards the relative acidic strengths, the major factor is the bond dissociation energy. The bond dissociation energy decreases from HF to HI so that HF has maximum bond dissociation energy and HI has the lowest value.

Since H-I bond is weakest, it can be dissociated into H+ and I ions readily while HF can be dissociated with maximum difficulty. Thus, HI is the strongest acid while HF is the weakest acid among the hydrogen halides.

(ii) Oxygen molecule is held by weak van der Waals forces because of the small size and high electronegativity of oxygen. On the other hand, sulphur molecules do not exist as S2 but form polyatomic molecules having eight atoms per molecule (S8) linked by single bonds. Therefore, S atoms are strongly held together by intermolecular forces and its melting point is higher than that of oxygen. Hence, there is large difference in melting and boiling points of oxygen and sulphur.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation enthalpy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(b) Draw the structures of the following:
(i) CIF3
(ii) XeF4
(i)

T – Shaped molecule

(ii)

Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
White phosphorus is most reactive of all the allotropes because it is unstable due to angular strain on P4 molecule with bond angle of 60°.

(ii) How are the supersonic jet aeroplanes responsible for the depletion of ozone layers?
Nitrogen oxide emitted from exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near ozone layer, they are responsible for the depletion of ozone layer.

(iii) F2 has lower bond dissociation enthalpy than Cl2. Why?
Because of small size of fluorineatoms, there are strong electron-electron repulsions between the lone pairs of electrons on F atoms. Hence, bond dissociation enthalpy of F2 is lower than that of Cl2.

(iv) Which noble gas is used in filling balloons for meteorological observations?
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF2 + PF5
XeF2 + PF5 → [XeF]+ [PF6]

Question 29.
(a) Account for the following:
(i) Interhalogens are more reactive than pure halogens.
(ii) N2 is less reactive at room temperature.
(iii) Reducing character increases from NH3 to BiH3.
(i) Interhalogen compounds are more reactive than component halogens.
This is because covalent bond between dissimilar atoms in interhalogen compounds is polar and weaker than between similar atoms in halogens (except F-F). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(ii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(iii) The reducing character of hydrides of group 15 depends upon the stability of the hydride. On going down the group, the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases.

The greater unstability of a hydride, the greater is its reducing character. Since the stability of the group 15 hydrides decreases from NH3 to BiH3, hence reducing character increases.

(b) Draw the structures of the following:
(i) H4P2O7 (Pyrophosphoric acid)
(ii) XeF4
(i)

(ii)

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.
Phosphine, PH3
P4 + 3NaOH + 3H2O → 3NaH2PO3 + PH3

(b) Which noble gas has the lowest boiling point?
Helium

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Fluorine has lower bond dissociation enthalpy of F-F bond than Cl-Cl bond of Cl2 and high enthalpy of hydration because of smaller size of F ion. As a result it has greater tendency to accept electron in solution and is stronger oxidising agent than chlorine.

(d) What happens when H3PO3 is heated?

(e) Complete the equation:
PbS + O3 → (CBSE 2015)
PbS + 4O3 → PbSO4 + 4O2

Question 30.
(a) Account for the following observations:
(i) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
(ii) Chlorine water is a powerful bleaching agent.
(iii) Bi(V) is a stronger oxidising agent than Sb(V).
(a) (f) S atom in SF4 is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF6 is protected by six F atoms. Thus attack by water molecules cannot take place easily.

(ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.
Cl2 + H2O → 2HCl + O

(iii) Due to inert pair effect Bi(V) can accept a pair of electrons to form more stable Bi (III) (+3 oxidation state of Bi is more stable than its +5 oxidation state).

(b) What happens when
(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2.
(ii) XeF6 undergoes partial hydrolysis.
(Give the chemical equations involved).
(i) Phosphorus undergoes disproportionation reaction to form phosphine gas.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2

(ii) On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
XeF6 + H2O → XeOF4 + 2HF

OR

(a) What inspired N.Bartlett for carrying out reaction between Xe and PtF6?
N. Bartlett first prepared a red compound O2+PtF6. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF6.

(b) Arrange the following in the order of property indicated against each set:
(i) F2, I2, Br2, Cl2 (increasing bond dissociation enthalpy)
(ii) NH3, ASH3, SbH3, BiH3, PH3 (decreasing base strength)
(i) I2 < F2 < Br2 < Cl2
(ii) NH3 > PH3 > ASH3 > SbH3 > BiH3

(c) Complete the following equations:
(i) Cl2 + NaOH (cold and dilute) →
(ii) Fe3+ + SO2 + H2O →
(CBSE Sample Paper 2018)
(i) 2NaOH + Cl2 → NaCl + NaOCl + H2O
(ii) 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+

Question 31.
(a) Give reasons:
(i) H3PO3 undergoes disproportionation reaction but H3PO4 does not.
(ii) When Cl2 reacts with excess of F2, ClF3 is formed and not FCl3.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
(i) In +3 oxidation state phosphorus tends to disproportionate to higher and lower oxidation states / Oxidation state of P in H3PO3 is +3 so it undergoes disproportionation but in H3PO4 it is +5 which is the highest oxidation state.

(ii) F cannot show positive oxidation state as it has highest electronegativity/ Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding due to pπ-pπ bonding whereas sulphur shows catenation. Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic and held by strong intermolecular forces.

(b) Draw the structures of the following:
(i) XeF4
(ii) HClO3
(i)

(ii)

OR

(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas Intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
(i) A = NO2, B = N2O4
(ii)

(iii) NO2 contains an odd electron. So it dimerises to give N2O4.

(b) Arrange the following In the decreasing order of their reducing character:
HF, HCl, HBr, HI
HI > HBr > HCl > HF

Complete the following reaction:
XeF4 + SbF5 →, (CBSE 2018)
(C) XeF4 + SbF5 → [XeF3] [SbF6]

Question 32.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF2 undergoes hydrolysis?
(a) 2NaOH + Cl2 → NaCl + NaOCl + H2O
(cold and dilute)
(b) 2XeF2(s) + 2H2O (l) → 2Xe (g) + 4HF(aq) + O2(S)

(ii) Assign suitable reasons for the following:
(a) SF6 Is inert towards hydrolysis.
(b) H3PO3 is diprotic.
(C) Out of noble gases only Xenon is known to form established chemical compounds.
(a) Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are present as -OH groups, thus behaves as dibasic acid.

(c) Xe has least ionisation energy among the noble gases and hence it forms chemical compounds particularly with O2 and F2.

OR

(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2.
F2 is stronger oxidising agent than Cl2. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F2 than for Cl2.
Hence, F2 is stronger oxidising agent than Cl2.

(ii) Complete the following reactions :
(a) Cu + HNO3(dilute) →
(b) Fe3+ + SO2 + H2O →
(c) XeF4 + O2F2 → (CBSE 2018)
(a) 3Cu + 8 HNO3 (dilute) → 3Cu(NO3)2 + 2NO + 4H3O
(b) 2Fe3+ + SO3 + 2H3O → 2 Fe2+ + SO42- + 4H+
(c) XeF4 + O2F2 → XeF6 + O2

Question 33.
A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified KMnO4 (aq.) solution and reduces Fe3+ to Fe2+. Identify ‘A’ and ‘B’ and write the reactions involved. (CBSE 2019C)
A = S8 / Sulphur
S8 + 8 O2 → 8SO2 / S + O2 → SO2
B = SO2

Decolourises KMnO4
2KMnO4 + 5 SO2 + 2H4O → 2H2SO4 + 2MnSO4 + K2SO4 / 2MnO4 + 5SO2 + 2H2O → 4H+ + 2Mn2+ + 5SO42-
Reduces Fe3+ to Fe2+
2Fe3+ + SO2 + 2 H2O → 2 Fe2+ + SO42- + 4H+

OR

(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength:
H2O, H2S, H2Se, H2Te
H2Te > H2Se > H2S > H2O

(b) Which one of PCl4+ and PCl4 is not likely to exist and why?
PCl4- is not likely to exist because lone pair on P in PCl3 can be donated to Cl+ and not to Cl. Phosphorus has 10e which cannot be accommodated in sp3 orbitals.

(c) Which aliotrope of sulphur is thermally stable at room temperature?
Rhombic sulphur is thermally stable at room temperature. Its melting point is 385.8 K. All other varieties of sulphur change into this form on standing. It has low thermal and electrical conductivity.

(d) Write the formula of a compound of phosphorus which is obtained when cone. HNO3 oxidises P4.
HNO3 oxidises phosphorus to phosphoric acid. H3PO4 is formed
P4 + 2OHNO3 → 4H3PO4 + 2ONO2 + 4H2O

(e) Why does PCl3 fume in moisture?
PCl3 fumes in moist air and reacts with water violently to form phosphorus acid. It hydrolyses in the presence of moisture to give fumes of HCl.
PCl3 + 3H2O → H3PO3 + 3HCl

## The d-and f-Block Elements Class 12 Important Extra Questions Chemistry Chapter 8

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 8 The d-and f-Block Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Chemistry Chapter 8 Important Extra Questions The d-and f-Block Elements

### The d-and f-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
How would you account for the increasing oxidising power in the series:
V02+ < Cr207 2-< Mn04? (CBSE Sample Paper 2010)
This is due to the increasing stability of the lower species to which they are reduced.

Question 2.
Which metal in the first transition series exhibits a +1 oxidation state most frequently and why? (CBSE Delhi 2013)
Copper has electronic configuration 3cf104s1. It can easily lose one (4s1) electron to give stable 3d10 configuration.

Question 3.
The magnetic moments of a few transition metal ions are given below:

 Metal ion Magnetic moment (BM) Sc3+ 0.00 Cr2+ 4.90 Ni2+ 2.84 Ti3+ 1.73

(atomic no. Sc = 21, Ti = 22, Cr = 24, Ni = 28)
Which of the given metal ions:
(i) has the maximum number of unpaired electrons?
Cr2+

(ii) forms colourless aqueous solution?
Sc3+

(iii) exhibits the most stable +3 oxidation state? (CBSE Sample Paper 2017-18)
Sc3+

Question 4.
Based on the data, arrange Fe2+, Mn2+ and Cr2+ in the increasing order of stability of +2 oxidation state:
Cr3+/Cr2+ = – 0.4 V, E°Mn3+/Mn2+ = 1.5 V, E°Fe3+/Fe2+ = 0.8 V (CBSE Sample Paper 2011)
As the value of reduction potential increases, the stability of +2 oxidation state increases. Therefore, correct order of stability is Cr3+ | Cr2+ < Fe3+ | Fe2+ < Mn3+ | Mn2+.

Question 5.
(i) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
Manganese

(ii) Name the element which shows only +3 oxidation state. (CBSEAI2013)
Scandium

Question 6.
Identify the oxoanion of chromium which is stable in an acidic medium. (CBSE Sample Paper 2017-18)
Cr2072-

Question 7.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (CBSE Delhi 2015)
This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

Question 8.
Name an important alloy that contains some of the Ianthanoid metals. (CBSE AI 2013)
Misch metal

Question 9.
Identify the Ianthanoid element that exhibits a +4 oxidation state. (CBSE Sample Paper 2017-18)
Cerium

Question 10.
Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number. (CBSE Delhi 2017)
Cr2072-

### The d-and f-Block Elements Important Extra Questions Short Answer Type

Question 1.
Assign reasons for the following:
(i) Cu(I) Is not known In an aqueous solution.
Because of the Lesser hydration enthalpy of Cu(I), It Is unstable In an aqueous solution and therefore, It undergoes disproportionation.

(ii) Actinolds exhibit a greater range of oxidation states than lanthanoids. (CBSE AI 2011)
Lanthanoids show Limited number of oxidation states, such as + 2, + 3 and + 4 + 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show a principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5, + 6 and + 7 and neptunium (Z = 94) shows oxidation states of + 3, +4, + 5, + 6 and + 7. This is because of the small energy difference between 5f and 6d orbitals.

Question 2.
Give reasons:
(a) MnO is basic whereas Mn2O7    is acidic in nature.
When a metal is in a high oxidation state, its oxide Is acidic and when a metaL is in a low oxidation state its oxide is basic.
The oxides of manganese have the following behavior:

(b) Transition metals form alloys. (CBSE 2019C)
The transition metals are quite similar in size and, therefore, the atoms of one metal can substitute the atoms of other metal in its crystal lattice. Thus, on cooling a mixture solution of two or more transition metals, solid alloys are formed which is shown in Fig.

Question 3.
Explain why the Eθ value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+. (CBSE AI 2008, 2010, 2018)
Mn2+ has 3d5 electronic configuration. It is stable because of the half-filled configuration of the d-subshell. Therefore, Mn has a very high third ionization enthalpy for the change from d5 to d4 and it is responsible for a much more positive Eθ value for Mn3+/Mn2+ couple in comparison to Cr3+/Cr2+ and Fe3+/ Fe2+ couples.

Question 4.
Complete the following chemical equations:
(i) Cr2O72++ H+ + I
Cr2O72+ + 6I + 14H+ → 2Cr3+ + 3I2 + 7H2O

(ii) MnO4+ NO2+ H+ → (CBSE Delhi 2012)
2MnO4+ 5N02+ 6H+ → 2Mn2+ + 5NO3+ 3H2O

Question 5.
(i) Which metal in the first transition series (3d series) exhibits +1 oxidatIon state most frequently and why?
Copper, because it has [Ar] 3d104s1 electronic configuration. After losing one electron it gets completely filled electronic configuration (3d10).

(ii) Which of the following cations are colored in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25) (CBSE Delhi 2013)
Sc3+: [Ar]  V3+: [Ar] 3d2
Ti4+: [Ar]  Mn2+: [Ar] 3d5
V3+ and Mn2+ cations are colored because they contain partially filled d-orbitals.

Question 6.
Why do transition elements exhibit higher enthalpies of atomization? (CBSE Delhi 2008, 2012)
The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

Question 7.
(a) Actinoid contraction is greater than lanthanoid contraction. Give reason,
This is due to poorer shielding by 5f electrons in actinoids as compared to shielding by 4f electrons in lanthanoids. In the Ianthanoid series, as we move from one element to another, the nuclear charge increases by one unit, and one electron are added. The new electrons are added to the same inner 4f-subshells. However, the 4f-electrons shield each other from the nuclear charge quite poorly because of the very diffused shapes of the f-orbitals.

The nuclear charge, however, increases by one at each step. Hence, with increasing atomic number and nuclear charge, the effective nuclear charge experienced by each 4f-electron increases. As a result, the whole of the 4f-electron shell contracts at each successive element, though the decrease is very small.

The actinoid contraction is due to the imperfect shielding of one 5felectron by another in the same subshell. Therefore, as we move along the series, the nuclear charge and the number of 5f-electrons increase by one unit at each step. However, due to imperfect shielding of 5f orbitals, the effective nuclear charge increases which results in contraction of the size.

(b) Out of Fe and Cu, which has a higher melting point and why? (CBSE 2019C)
Fe has a higher melting point. The metallic bond is formed due to the interaction of electrons in the outermost orbitals. The strength of bonding is related to the number of unpaired electrons. Fe has more unpaired electrons leading to stronger metallic bonding. So, it has a higher melting point.

Question 8.
How would you account for the following:
(i) Cr2+ is reducing in nature while with the same d-orbital configuration (d4), Mn3+ is an oxidizing agent.
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.

(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. (CBSE 2011)
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to the lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

Question 9.
When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in an acidic solution to give a purple compound (B). An alkaline solution of compound (B) oxidizes Kl to compound (C), whereas an acidified solution of compound (B) oxidizes Kl to (D). Identify (A), (B), (C), and (D). (CBSE Delhi 2019)
When MnO2 is fused with KOH in the presence of KNO3, as an oxidizing agent, it gives a dark green compound (A).

Question 10.
State reasons for the following:
(i) Cu(I) is not stable in an aqueous solution.
Cu+ ion is not stable in an aqueous solution because of its less negative enthalpy of hydration than that of Cu2+ ion.

(ii) Unlike Cr3+, Mn2+, Fe3+ and the subsequent other M2+ ions of the 3d series of elements, the Ad and the 5d series metals generally do not form stable atomic species. (CBSE 2011)
Because of lanthanoid contraction, the expected increase in size does not occur.

Question 11.
Assign reasons for each of the following:
(i) Transition metals generally form colored compounds.
Most of the transition metal ions are colored both in the solid-state and in aqueous solutions. The color of these ions is attributed to the presence of an incomplete (n – 1) d-subshell. The electrons in these metal ions can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite the electrons to higher energy states within the same d-subshell corresponds to the energy of certain colors of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain color is absorbed causing the promotion of d-electrons. This is known as d-d transitions. The remaining colors of white light are transmitted and the compound appears colored.

(ii) Manganese exhibits the highest oxidation state of +7 among the 3d-series of transition elements. (CBSE Delhi 2011)
Manganese has the electronic configuration [Ar] 3d5 4s2. It can lose seven electrons due to the participation of 3d and 4s electrons and therefore, exhibits the highest oxidation state of +7 in its compounds.

Question 12.
What are the transition elements? Write two characteristics of the transition elements. (CBSE Delhi 2015)
Transition elements are those elements that have incompletely filled (partly filled) d-subshell in their ground state or in any one of their oxidation states.

Characteristics:

1. Transition elements show variable oxidation states.
2. They exhibit catalytic properties.

Question 13.
Though both Cr2+ and Mn3+ have d4 configurations, yet Cr2+ is reducing while Mn3+ is oxidizing. Explain why? (CBSE AI 2008, 2012, CBSE Delhi 2012)
E° value for Cr3+/Cr2+ is negative (-0.41V), this means that Cr3+ ions are more stable than Cr2+. Therefore, Cr2+ can readily lose electrons to undergo oxidation to form Cr3+ ion, and hence Cr (II) is strongly reducing. On the other hand, the E° value for Mn3+ Mn2+ is positive (+1.57V), this means that Mn3+ ions can be readily reduced to Mn2+ and hence Mn (III) is strongly oxidizing.

Question 14.
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state? (CBSE Delhi 2012)
The electronic configuration of Mn2+ is [Ar] 3d5 which is half-filled and hence it is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron1 cannot be easily removed. In the case of Fe2+, the electronic configuration is 3d6. Therefore, Fe2+ can easily lose one electron to acquire a 3d5 stable electronic configuration.

Question 15.
The E°M2+|M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behavior. Why? (CBSE Sample Paper 2012, CBSE AI2012)
The E°M2+|M value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy.

Question 16.
Chromium is a typical hard metal while mercury is a liquid. (CBSE Sample Paper 2011)
In chromium, the M-M interactions are strong due to the presence of six unpaired electrons in the 3d and 4s subshell. On the other hand, in mercury, all the electrons in the 5d and 6s subshell are paired and therefore, the M-M interactions are weak. Therefore, chromium is a typical hard metal while mercury is a liquid.

Question 17.
Silver is a transition metal but zinc is not. (CBSE Sample Paper 2011)
According to the definition, transition elements are those which have partially filled d-subshell in their elementary state or in one of the oxidation states. Silver (Z = 47) can exhibit a +2 oxidation state in which it has an incompletely filled d-subshell (4d9 configuration). Hence, silver is regarded as a transition element.

On the other hand, zinc (Z = 30) has the configuration 3d10 4s2. It does not have partially filled d-subshells in its elementary form or in a commonly occurring oxidation state (Zn2+: 3d10). Therefore, it is not regarded as a transition element.

Question 18.
Name the ox metal anions of the first series of the transition metals in which the metal exhibits an oxidation state equal to its group number. (CBSE Delhi 2017)
MnO4: Oxidation state of Mn = +7 (equal to its group number)
CrO42-: Oxidation state of Cr = +6 (equal to its group number)

Question 19.
Give reasons:
(a) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising.
Because Cr is more stable in the +3 oxidation state due to the t2g3 configuration whereas Mn is more stable in the +2 oxidation state due to the half-filled 3d5 configuration.

E° value for Cr3+/Cr2+ is negative (-0.41 V); this means that Cr3+ ions are more stable than Cr2+. Therefore, Cr2+ can readily lose electrons to undergo oxidation to form a Cr3+ ion, and hence Cr (II) is strongly reducing. On the other hand, the E° value for Mn3+/ Mn2+ is positive (+1.57V), which means that Mn3+ ions can be readily reduced to Mn2+ and hence Mn (III) is strongly oxidizing.

(b) The d1 configuration is very unstable in ions. (CBSE 2019C)
After the loss of ns electrons, d1 electron can easily be lost to give a stable configuration. Therefore, the elements having d1 configuration are either reducing or undergo disproportionation.

Question 20.
When chromite ore FeCr2O4 is fused, with NaOH in presence of air, a yellow-colored compound (A) is obtained, which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KCI forms an orange colored crystalline compound (C).
(i) Write the formulae of the compounds (A), (B), and (C).
(ii) Write one use of the compound (C). (CBSE Delhi 2016)
OR
Complete the following chemical equations:
(i) 8MnO4+ 3S2O32-+ H2O →
(ii) Cr2O72-+ 3Sn2+ + 14H+

(ii) Potassium dichromate (C) is used as a powerful oxidizing agent in redox titrations in the laboratories.
OR
(i) 8MnO4+ 3S2O32-+ H2O → 8MnO2 + 6SO42- + 2OH
(ii) Cr2O7 2+ + 3Sn2+ +14H+ → 2Cr3+ + 3Sn4+ + 7H2O

Question 21.
Why do the transition elements have higher enthalpies of atomization? In 3d series (Sc to Zn), which element has the lowest enthalpy of atomization and why? (CBSE 2015)
The transition elements have high enthalpies of atomization because they have a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

Zinc has the lowest enthalpy of atomization because it has no unpaired electrons and hence weak metallic bonding.

### The d-and f-Block Elements Important Extra Questions Long Answer Type

Question 1.
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series. (CBSE AI 2011)
(i) Some transition metals and their compounds act as good catalysts for various reactions. This is due to their ability to show multiple oxidation states. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc.

The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. For example, during the conversion of SO2 to SO3, V2O5 is used as a catalyst. Solid V2O; absorbs a molecule of SO2 on the surface forming V2O4 and the oxygen is given to SO2 to form S03. The divanadium tetraoxide is then converted to V2O5 by reaction with oxygen:

(ii) The second and third transition series metals show a great tendency to form strong M-M bonds than the first transition series elements. For example, rhenium easily forms ReCl84- which contains a strong Re-Re bond. It does not have a manganese analog.

Similarly, both Niobium and tantalum form M6X12n+ species and has no vanadium analogue.

Question 2.
Explain the following observations giving an appropriate reason for each.
(i) The enthalpies of atomization of transition elements are quite high.
The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interaction and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series)
The heavier transition metals of second and third transition series metals show a great tendency to form strong M-M bonds than first transition series elements. For example, rhenium easily forms ReCl84- which contains a strong Re-Re bond. It does not have a manganese analog. Similarly, both niobium and tantalum form M6X12n+ species and has no vanadium analog.

(iii) Mn2+ is much more resistant than Fe2+ towards oxidation. (CBSE Delhi 2012)
Mn2+ is quite stable because it has a stable half-filled d5 electronic configuration and therefore, cannot be easily oxidized. On the other hand, Fe2+ has a less stable d6 electronic configuration and can be easily oxidized to a stable d5 configuration.

Question 3.
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as a catalyst.
OR
Complete the following chemical equations:
(i) Cr2072- + 6Fe2+ + 14H+
(ii) 2Cr042- + 2H+
(iii) 2Mn04 + 5C2042- + 16H+ → (CBSE Delhi 2013)
(i) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n – 1) d-electrons. This is because of the very small difference between the energies of (n -1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons. The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons.

In general, the oxidation state increases up to the middle and then decreases.

(ii) Due to lanthanoid contraction, the increase in radii from the second to third transition series vanishes. Therefore, Zr and Hf have almost the same radii.

(iii) Catalytic properties of transition metal ions. Some transition metals and their compounds act as good catalysts for various reactions. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc. For example, iron-molybdenum is used as a catalyst in Haber’s process for the manufacture of NH3. V205 is used for the oxidation of S02 to S03 in the Contact process for the manufacture of H2S04.

The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. For example, during the conversion of SO2 to SO3, V2O5 is used as a catalyst. Solid V2O5 absorbs a molecule of SO2 on the surface forming V2O4 and the oxygen is given to SO2 to form SO3. The divanadium tetraoxide is then converted to V2O5 by reaction with oxygen:

OR
(i) 6Fe2+ + Cr2072- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
(ii) 2CrO42- + 2H+ → Cr2O72-+ H2O
(iii) 2MnO4 + 5C2O42 + 16H+ → 2Mn2+ + 8H2O + 10CO2

Question 4.
Give reasons for the following:
(i) Transition elements and their compounds act as catalysts.
Transition metals and their compounds act as good catalysts because of their ability to show multiple oxidation states and form complexes, for example, Fe, Co, Ni, V, Cr, V2O5, etc.

(ii) E° value for (Mn2+|Mn) is negative whereas for (Cu2+| Cu) is positive.
The negative value of E° (Mn2+| Mn) is due to the stability of the half-filled d-subshell in Mn2+(3d5). The positive E°(Cu2+|Cu) is due to the high ionization enthalpy and high enthalpy of atomization of copper. Therefore, the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

(iii) Actinoids show irregularities in their electronic configuration. (CBSE Delhi 2019)
Actinoids have irregularities in the electronic configuration because of almost equal energy of 5f, 6d and 7s orbitals. Therefore, there are some irregularities in the filling of 5f, 6d, and 7s orbitals. The electron may enter either of these orbitals.

Question 5.
(a) Complete the following chemical reactions:
(i) Na2Cr2O7 + KCl →
(ii) 2MnO4+ 5 S032- + 6 H+
(i) Na2Cr2O7 + KCl → K2Cr2O7 + 2 NaCl
(ii) 5 S032- + 2MnO4 + 6 H+ → 2Mn2+ + 3H2O + 5SO42-

(b) How does the colour of Cr2O72- change when treated with an alkali? (CBSE2019C)
The orange color of Cr2O72- change to yellow due to the formation of chromate ion.
Cr2O72- + 2OH → 2Cr2O72-  + H2O (Yellow)

Question 6.
(a) Write chemical equations involved in the preparation of KMn04 from Mn02.
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3MnO42-+ 4H+ → 2MnO+ MnO2 + 2H2O (or any other correct equation of preparation)

(b) Actinoids show wide range of oxidation states. Why? (CBSE 2019C)
Lanthanoids show limited number of oxidation states, such as + 2, + 3 and + 4 (+ 3 is the principal oxidation state). This is because of large energy gap between 5d and 4f subshells. On the other hand, actinoids also show principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5 and + 6 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7.

This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 7.
How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds.
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
The metallic radii of the third (5d) series of transition metals are virtually the same as those of corresponding group members of the (4d) series due to the phenomenon of lanthanoid contraction. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic numbers is known as lanthanoid contraction.

(iii) Lanthanoids form primarily +3 ions, white the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical. (CBSE Delhi 2012)
Lanthanoids form primarily +3 ions, while actinoids usually have higher oxidation states, +4 or even +6 in their compounds because in actinoids the 5f, 6d, and 7s energy levels are of comparable energies.

Therefore, all these three subshells can participate.

Question 8.
How would you account for the following:
(i) Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained.
All lanthanoids exhibit a common stable oxidation state of +3. In addition, some lanthanoids show +2 and +4 oxidation states also in solution or in solid compounds. These are shown by these elements which by doing so attain the stable f (empty f-subshell), f (half-filled f-subshell), and f4 (filled f-subshell) configurations. For example,
(a) Ce and Tb exhibit +4 oxidation states. Cerium (Ce) and terbium (Tb) attain f and f configurations respectively when they get +4 oxidation state, as shown below:
Ce4+: [Xe] 4f°
Tb4+: [Xe] 4 f°

(b) Eu and Yb exhibit + 2 oxidation states.
Europium and ytterbium get f and f4 configurations in +2 oxidation state as shown below:
EU2+ : [Xe] 4 f7
Yb2+: [Xe] 4f14

(ii) The E°M2+/M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behavior.
The E°(M2+/M) value for copper is positive and this shows that it is the least reactive metal out of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

(iii) The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series. (CBSE 2012)
The metallic radii of the third transition series elements are virtually the same as those of corresponding members of the second transition series because of the lanthanoid contraction. Due to the presence of lanthanoids between the second and third transition series, the expected increase in radii vanishes. Consequently, the pairs of elements Zr-Hf, Nb-Ta, Mo-W, etc have almost similar sizes.

Question 9.
Explain the following observations:
(i) Many of the transition elements are known to form interstitial compounds.
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29).
There is a gradual increase in density from Ti to Cu because as we move along a transition series from left to right, the atomic radii decrease due to an increase in effective nuclear charge. Therefore, the atomic volume decreases but at the same time atomic mass increases. Therefore, density (mass/volume) increases.

(iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. (CBSE 2012)
Lanthanoids show limited a number of oxidation states, such as + 2, + 3, and + 4 (+ 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5 and + 6 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7. This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 10.
Following are the transition metal ions of 3d series.
Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidizing agent and why?
(iii) Which ion is colorless and why? (CBSE AI 2017)
(i) Cr3+ because of the half-filled t2g3 configuration.
(ii) Mn3+ due to stable d5 configuration of Mn2+
(iii) Ti4+ because it has no unpaired electrons.

Question 11.
Write chemical equations for the following reactions:
(i) Oxidation of nitrite ion by MnO4 in acidic medium.
(ii) Acidification of potassium chromate solution.
(iii) Disproportionation of manganese
(vi) in acidic solution. (CBSE Sample Paper 2011)
(i) 5NO2– + 2MnO4+ 6H+ → 2Mn2+ + 3H2O + 5NO3
(ii) 2K2CrO44 + 2H+ → K2Cr2O7 + 2K+ + H2O
(iii) 3MnO42- + 4H+ → 2MnO4 + MnO2 + 2H2O

Question 12.
The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series. (CBSE AI 2012, CBSE Delhi 2012)
Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 93) and plutonium (Z = 94) show oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 13.
Explain the following:
(i) Out of Sc3+, Co2+, and Cr3+ ions, only Sc3+ is colorless in aqueous solutions.
(Atomic no.: Co = 27; Sc = 21 and Cr = 24)
(iI) The E°Cu2+/Cu for copper metal is positive (+0.34), unlike the remaining members of the first transition series
(iiI) La(OH)3 is more basic than Lu(OH)3. (CBSE Sample paper 2018)
(i) Co2+: [Ar]3d7, Sc3+: [Ar]3d°
Cr3+: [Ar]3d3
Co2+ and Cr3+ have unpaired electrons. Thus, they are colored in an aqueous solution. Sc3+ has no unpaired electron. Thus it is colorless.
(ii) Metal copper has a high enthalpy of atomization and enthalpy of ionization. Therefore the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.
(iii) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with the increase in atomic size. Thus covalent character between lanthanoid ion and OH’ increases from La3+ to Lu3+. Thus the basic character of hydroxides decreases from La(OH)3 to Lu(OH)3

Question 14.
(a) Complete the following chemical reactions:
(i) 2MnO4+ 5 NO2+ 6 H+
(ii) 3Mn042- + 4 H+
(i) 5NO2– + 2MnO4 + 6H+ → 2Mn2+ + 5NO3 + 3H2O
(ii) 3MnO42-+ 4H+ → 2MnO4+ MnO2 + 2H2O

(b) Name a member of the lanthanoid series which shows a +4 oxidation state. (CBSE 2019C)
Cerium / Ce

Question 15.
Give reasons:
(i) E° value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.
The comparatively high value for Mn shows that Mn2+(d5) is particularly stable/Much larger third ionization energy of Mn (where the required change is from d5 to d4).

(ii) Iron has a higher enthalpy of atomization than copper.
Due to the higher number of unpaired electrons.

(iii) Sc3+ is colorless in an aqueous solution whereas Ti3+ is colored. (CBSE 2018)
The absence of unpaired d-electron in Sc3+ whereas in Ti3+ there is one unpaired electron or Ti3+ shows the d-d transition.

Question 16.
(i) Complete the following chemical equations for reactions:
(a) MnO4(aq) + S2O32- (aq) + H2O (I) →
(b) Cr2O7(aq) + H2S(g) + H+(aq) →
(a) 8MnO4 (aq) + 3S2O32-  (aq) + H2O (l) → 8MnO2 (s) + 6SO42- (aq) + 2OH- (aq)
(b) Cr2O72- (aq) + 3H2S(g) + 8H+ (aq) → 2Cr3+(aq) + 3S(s) + 7H2O

(ii) Give an explanation for each of the following observations:
(a) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction).
(a) This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

(b) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to a lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

(c) With the same d-orbital configuration (d4) Cr2+ ion is a reducing agent but the Mn3+ ion is an oxidizing agent. (CBSE Delhi 2009)
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.

Question 17.
(i) Complete the following chemical reaction equations:
(a) Fe2+ (aq) + MnO4 (aq) + H+ (aq) →
(b) Cr2O72- (aq) + I (aq) + H+ (aq) →
(a) 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O(l)
(b) Cr2O72- (aq) + 6I (aq) + 14H+ (aq) → 2Cr3+(aq) + 3I2 (s) + 7H2O(l)

(ii) Explain the following observations:
(a) Transition elements are known to form many interstitial compounds.
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(b) With the same d4 d-orbital configuration Cr2+ ion is reducing while the Mn3+ ion is oxidizing.
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.

(c) The enthalpies of atomization of the transition elements are quite high. (CBSE Delhi 2009)
The transition metals have strong interactions of electrons in the outermost orbitals and therefore, have high melting and boiling points. These suggest that the atoms of these elements are held together by strong forces and have high enthalpy of atomization.

Question 18.
(i) Complete the following chemical equations:
(a) MnO4(aq) + S2O32- (aq) + H2O(l) →
(b) Cr2O72- (aq) + Fe2+ (aq) + H+ (aq) →
(a) 8MnO4 + 3S2O32- + H2O → 8MnO2 + 6SO42-+ 2OH-
(b) Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ +6Fe3+ + 7H2O

(ii) Explain the following observations:
(a) La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions.
In La3+ there are no f-electrons and in Lu3+ the 4f sub-shell is complete (4/14). Therefore, there are no unpaired electrons and consequently, d-d transitions are not possible. Hence, La3+ and LU3+ do not show any color in solutions.

(b) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
Among the divalent cations in the first
transition series, Mn (Z = 25) exhibits the maximum paramagnetism because it has the maximum number of unpaired electrons (3d5): five unpaired electrons.

(c) Cu+ ion is not known in aqueous solutions. (CBSE 2010)
In an aqueous solution, Cu+ undergoes disproportionation changing to Cu2+ ion.
2Cu+ → Cu2+ + Cu

Question 19.
(i) Give reasons for the following:
(a) Mn3+ is a good oxidizing agent.
(b) E°M2+/M values are not regular for first-row transition metals (3d series).
(c) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn207.
(ii) Complete the following equations:
(a) 2CrO42- + 2H+
(b) KMnO4
OR
(i) (a) Why do transition elements show variable oxidation states?
(b) Name the element showing a maximum number of oxidation states among the first series of transition metals from Sc(Z = 21) to Zn (Z = 30).
(c) Name the element which shows only the +3 oxidation state.
(ii) What is lanthanoid contraction? Name an important alloy that contains some of the lanthanoid metals. (CBSE2013)
(i) (a) Mn3+ (3d4) on changing to Mn2+ (3d5) becomes stable, half-filled configuration has extra stability. Therefore, Mn3+ can be easily reduced and acts as a good oxidizing agent.

(b) E°(M2+/M) values are not regular in the first transition series metals because of irregular variation of ionization enthalpies (IE1 + IE2) and the sublimation energies.

(c) Among transition elements, the bonds formed in +2 and +3 oxidation states are mostly ionic. The compounds formed in higher oxidation states are generally formed by sharing of d-electrons. Therefore, Mn can form MnO4- which has multiple bonds also, while fluorine cannot form multiple bonds.

(ii) (a) 2CrO42- + 2H+ → Cr2O72- + H2O
(b) 2KMnO4 K2MnO4 + MnO2 + O2
OR
(i) (a) The transition elements show variable oxidation states because their atoms can lose a different number of electrons. This is due to the participation of inner (n -1). d-electrons in addition to outer ns electrons because the energies of the ns and (n – 1) d-subshells are almost equal.

(b) Manganese

(c) Scandium

(ii) The steady decrease in atomic and ionic sizes of lanthanide elements with increasing atomic numbers is called lanthanide contraction. In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. For example, the ionic radii decrease from Ce3+ (111 pm) to Lu3+ (93 pm).

Cause of lanthanoid contraction. As we move through the lanthanoid series, 4f-electrons are being added, one at each step. The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to the shape of f-orbitals. The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f shell. The sum of the successive reductions gives the total lanthanoid contraction. The important alloy is mischmetal.

Question 20.
(i) How do you prepare:
(a) K2MnO4 from MnO2?
(b) Na2Cr2O7 from Na2CrO4?
(ii) Account for the following:
(a) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(b) The enthalpy of atomization is lowest for Zn in the 3d series of the transition elements.
(c) Actinoid elements show a wide range of oxidation states.
OR
(a) Name the element of 3d transition series which shows a maximum number of oxidation states. Why does it show so?
(b) Which transition metal of 3d series has a positive E°(M2+/M) value and why?
(c) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(d) Name a member of the lanthanoid series which is well known to exhibit a +2 oxidation state.
(e) Complete the following equation:
Mn04+ 8H+ +5e → (CBSE Delhi 2014)
(i) (a) When Mn02 is fused with caustic potash (KOH) or potassium carbonate in the presence of air, a green mass of potassium manganate is formed.
2MnO2 + 4KOH + O2 4 2K2MnO4 + 2H2O
Or
2MnO2 + 2K2CO2 + O2 → 2K2MnO4 + 2CO2.

(b) Na2Cr04 is acidified with dilute sulphuric acid to get Na2Cr2O7.
2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

(ii) (a) Electronic configuration of Mn2+ is [Ar]Bd5 which is half-filled and hence is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron cannot be easily removed. In the case of Fe2+, the electronic configuration is 3d6. Therefore, Fe2+ can easily lose one electron to acquire a 3d5 stable configuration. Thus, Mn2+ is more stable than Fe2+ towards oxidation to + 3 states.

(b) The high enthalpies of atomization of transition elements are due to the participation of electrons (n – 1) d-orbitals in addition to ns electrons in the interatomic metallic bonding. In the case of zinc, no electrons from 3d-orbitals are involved in the formation of metallic bonds. On the other hand, in all other metals of 3d series electrons from d-orbitals are always involved in the formation of metallic bonds.

(c) Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f and 6d orbitals.
OR
(a) Manganese (Z = 25) shows the largest number of oxidation states because it has the maximum number of unpaired electrons. It shows oxidation states from +2 to +7, i.e. +2, +3, +4, +5, +6 and +7.

(b) The E°(M2+|M) value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

(c) Mn3+ is a stronger oxidizing agent than Cr3+ because Mn3+(3d4) changes to stable half-filled (3d5) configuration while in the case of Cr3+ (3d5) is stable (half-filled) and it cannot be readily changed to Cr2+ (3d4).

(d) Europium.

(e) MnO4 + 8H+ + 5e → Mn2+ + 4H2O.

Question 21.
(a) Account for the following:
(i) Manganese shows the maximum number of oxidation states in 3d series.
(ii) E° value for Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+.
(iii) Ti4+ is colorless whereas V4+ is colored in an aqueous solution.
(b) Write the chemical equations for the preparation of KMn04 from Mn02. Why does the purple color of acidified permanganate solution decolorize when it oxidizes Fe2+ to Fe3+?
OR
(a) Write one difference between transition elements and p-block elements with reference to variability of oxidation states.
(b) Why do transition metals exhibit higher enthalpies of atomization?
(c) Name an element of the lanthanoid series which is well known to show a +4 oxidation state. Is it a strong oxidizing agent or a reducing agent?
(d) What is lanthanoid contraction? Write its one consequence.
(e) Write the ionic equation showing the oxidation of Fe(ll) salt by acidified dichromate solution. (CBSE Al 2019)
(a) (i) Manganese has the electronic configuration: 3d5 4s2. It has a maximum number of unpaired electrons and hence shows maximum oxidation states.
(ii) Mn2+ has 3d5 electronic configuration. It is stable because of the half-filled configuration. Therefore, Mn3 easily gets reduced to Mn2+. Thus, E°(M3+/Mn2+) is positive. On the other hand, Cr3+ is more stable in the +3 oxidation state due to stable t2g3 configuration.
(iii) Ti4+ (3d°) does not have any d-electrons. Therefore, there are no d-d transitions whereas V4+, (3d1) has one electron in d-subshell and d-d transitions are possible and hence it is colored.

(b) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3K2MnO4 + 4HCL → 2KMnO + MnO2 + 2H2O

When acidified KMnO4 oxidizes Fe2+ to Fe3+ its purple color gets decolorized due to the formation of Mn2+ from
MnO4– ion.
MnO4– + 5Fe2+ + 8W → Mn2+ + 5Fe3+ + 4H2O
Or
(a) Transition elements, in general, show variable oxidation states which differ by 1 unit, whereas p-block elements show variable oxidation states which differ by 2 units.

(b) Transition eLements have unpaired d-electron and therefore have strong metallic bonding between atoms. Hence, they have high enthalpies of atomization.

(c) Cerium. It is a strong oxidizing agent.

(d) The regular decrease In atomic and ionic radii of the Lanthanides with increasing atomic number is catted Lanthanoid contraction.

Consequence: 5d series elements have almost the same size as the 4d series.

(e) 6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

Question 22.
(i) Account for the following:
(a) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(b) Zirconium and Hafnium exhibit similar properties.
(c) Transition metals act as catalysts.
(ii) Complete the following equations:
(a) 2MnO2 + 4KOH + O2
(b) Cr2O72- + 14H+ + 6I
Or
The elements of 3d transition series are given as:
Sc Ti V Cr Mn Fe Co Ni Cu Zn
(a) Write the element which is not regarded as a transition element. Give reason.
(b) Which element has the highest m.p.?
(c) Write the element which can show an oxidation state of +1.
(d) Which element is a strong oxidizing agent in the +3 oxidation state and why? (CBSE 2016)
(i) (a) Manganese shows the highest oxidation state of +7 with oxygen but +4 with fluorine. This is because oxygen has a tendency to form multiple bonds and hence stabilize the high oxidation state.
(b) Due to lanthanoid contraction, Zr and Hf show similar properties.
(c) The transition metals act as catalysts. This is due to their ability to show multiple oxidation states.

(ii) (a) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(b) Cr2O72- + 14H+ + 6I → 2Cr3+ + 7H2O + 3I2
Or
(a) Zn because of not having partially filled d-orbitals in its ground state or ionic state.
(b) Chromium, Cr
(c) Copper, Cu
(d) Mn, because Mn2+ has extra stability due to half-filled d-subshell.
Mn: [Ar] 3d5 4s2; Mn2+: [Ar]3d5
Mn3+ has four electrons (3d4) in 3d subshell and requires only one electron to get a half-filled 3d subshell and therefore, acts as a strong oxidizing agent.

## Haloalkanes and Haloarenes Class 12 Important Extra Questions Chemistry Chapter 10

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 10 Haloalkanes and Haloarenes.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Chemistry Chapter 10 Important Extra Questions Haloalkanes and Haloarenes

### Haloalkanes and Haloarenes Important Extra Questions Very Short Answer Type

Question 1.
Define ambident nucleophile with an example. (CBSE Delhi 2019)
The nucleophile which has two sites through which it can attack is called ambident nucleophile, for example, nitrite ion (NO2). It can attack either through 0 (-O -N = O) or through N

Question 2.
What happens when bromine attacks CH2 = CH – CH2 -C ≡ CH? (CBSE AI2012)
The Colour of bromine gets discharged.

Question 3.
What happens when CH3 – Br is treated with KCN? (CBSE Delhi 2013)
Ethane nitrile is formed.
CH3Br + KCN → CH3C ≡ N + KBr (Ethane nitrile)

Question 4.
Write the IUPAC name of

4-Bromo-4-methylpent-2-ene

Question 5.
What happens when ethyl chloride is treated with aqueous KOH? (CBSE Delhi 2013)
Ethyl alcohol is formed.
CH3CH2Cl + KOH(aq) → CH3CH2OH + KCl Ethyl alcohol

Question 6.
Write the IUPAC name of (CH3)2CH.CH(Cl)CH3. (CBSE Delhi 2013)
2-Chloro-3-methyl butane

Question 7.
Which compound in the following pair undergoes a faster SN1 reaction? (CBSE Delhi 2013)

Question 8.
Write the IUPAC name of the following compound: (CBSE AI 2013)

3-Chloro-2, 2-dimethylbutane

Question 9.
Which aerosol depletes the ozone layer? (CBSE AI 2013)
Chlorofluoro compounds of methane (CCl2F2) called freons.

Question 10.
Write the IUPAC name of the following compound: (CBSE AI 2013)

2-Bromo-4-chloropentane

Question 11.
Why is CH2 = CH – CH2 – Cl more easily hydrolysed than CH3 – CH2 – CH2 – Cl? (CBSE AI 2019)
CH2 = CH – CH2 – Cl gives allylic carbocation as intermediate, which is resonance stabilised.

Therefore, CH2 = CH – CH2 – Cl is more readily hydrolysed.

Question 12.
Write the IUPAC name of the following compound:
CH2 = CHCH2Br (CBSE 2011)
3-Bromoprop-1-ene

Question 13.
Write the IUPAC name of the following compound: (CH2 )3CCH2Br (CBSE 2011)
1 -Bromo 2,2-dimethylpropane

Question 14.
Write the IUPAC name of the following: (CBSE 2012)

3-Bromo- 2-methylpropene

Question 15.
Identify the chiral molecule in the following pair: (CBSE 2014)

2-Chlorobutane is a chiral molecule.

Question 16.
Which would undergo SN2 reaction faster in the following pair and why? (CBSE Delhi 2015)

CH3CH2Br would undergo SN2 reaction faster because of less steric hindrance.

Question 17.
Out and which is more reactive towards SN1 reaction and why? (CBSE Delhi 2016)
is more reactive.
The SN1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.

Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the SN1 reaction.

Question 18.
Write the structure of the product formed when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether. (CBSE Al 2018)

Question 19.
Which of the following two reactions is SN2 and why? (CBSE AI2016)

Reaction (i) is SN2 because it proceeds by inversion of configuration.

Question 20.
Why is chloroform kept in dark coloured bottles? (CBSE AI 2019)
Chloroform reacts with atmospheric oxygen in presence of light to produce an extremely poisonous gas called phosgene.

Chloroform is stored in dark coloured bottles so that it is not exposed to light and the formation of phosgene is prevented.

Question 21.
Write one stereochemical difference between SN1 and SN2 reactions. (CBSE Delhi 2019)
SN1 reaction forms a racemic mixture while SN2 reaction results in inversion of configuration.

Question 22.
A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl. Which one of these is more easily hydrolysed? (CBSE Delhi 2010)
CH3CHClCH2CH3

Question 23.
Write the structure of 2, 4-dinitrochlorobenzene. (CBSE Delhi 2017)

Question 24.
Write the structure of 1-Bromo-4- chlorobut-2-ene. (CBSE Delhi 2017)

Question 25.
Write the structure of 3-Bromo-2- methylprop-1 -ene. (CBSE Delhi 2017)

Question 26.
Write IUPAC name of the given compound: (CBSE Delhi 2019)

4-Chlorobenzene sulphonic acid

Question 27.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? (CBSE Sample paper 2018)
Benzyl chloride;
Due to resonance, stable benzyl carbocation is formed.

### Haloalkanes and Haloarenes Important Extra Questions Short Answer Type

Question 1.
Which one of the following compounds will undergo hydrolysis at a faster rate by SN1 mechanism? Justify. (CBSE Sample Paper 2019)

The following compound will undergo SN1 faster:

Greater the stability of the carbocation, greater will be its ease of formation from the corresponding halide and faster will be the rate of reaction. The benzylic carbocation formed gets stabilised through resonance.

CH3CH2CH2Cl forms 1° carbocation, which is less stable than benzylic carbocation.

Question 2.
How would you differentiate between SN1 and SN2 mechanisms of substitution reactions? Give one example of each. (CBSE 2010)
The SN1 reaction occurs in two steps and the reaction is of the first order.
The SN2 reaction occurs in one step and the reaction is of second order. These can also be distinguished as:

In SN1 reaction retention of the configuration takes place while in SN2 reaction inversion of the configuration takes place.

SN2 reaction:

SN1 reaction:

Question 3.
(i) State one use each of DDT and iodoform.
DDT is used as an insecticide for killing mosquitoes and other insects. chloroform is used as an antiseptic.

(ii) Which compound in the following couples will react faster in SN2 displacement and why?
(a) I -Bromopentane or 2-bromopentane
1-Bromopentane (1°) will react faster than 2-bromopentane (2°) by SN2 displacement because It is Less stericaLly hindered in the transition state.

(b) I-Bromo-2-methyl butane or 2-Bromo-2-methyl butane (CBSE Delhi 2010)
1 -Bromo-2-methyl butane (1°) reacts faster than 2-bron,o-2-methyl butane (2°) because being primary, it will have Less steric hindrance.

Question 4.
Although chlorine Is an electron-withdrawing group, yet It Is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why Is It so? (CBSE 2012)
Chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in eLectrophiLic aromatic substitution reactions due to eLectron reLeasing resonance effect (+ R-effect).

As can be seen from the above resonating structures that -Ct increases electron density at ortho- and para-positions due to the + R-effect. Therefore, the attack of electrophile takes pLace at ortho- and paro- positions and not at meta- positions.

Question 5.
(i) Which alkyl halide from the following pair Is chiral and undergoes a faster SN2 reaction? (CBSE Delhi 2014)

is chiral

(ii) Out of SN1 and SN2 which reaction occurs with
(a) Inversion of configuration
SN2

(b) Racemisation
SN1

Question 6.
Draw the structure of major moon halo product In each of the following reactions: (CBSE Delhi 2014)

Question 7.
Propose the mechanism of the reaction taking place when
(i) (-) – 2 – Bromooctane reacts with sodium hydroxide to form (+) – octane-2-ol.

(ii) 2-Bromopentane is heated with (ale.) KOH to form alkenes. (CBSE Sample Paper 2011)

Question 8.
Out of and which is more reactive towards SN1 reaction
is more reactive.
The SN1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.

Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the SN1 reaction.

### Haloalkanes and Haloarenes Important Extra Questions Long Answer Type

Question 1.
(i) Out of (CH3)3C-Br, and (CH3)3C-I, which one is more reactive towards SN1 and why?
(CH3)2 C-I, because it can readily form stable carbocation because the C – l bond is weaker than the C – Br bond.

(ii) Write the product formed when p-nitro chlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.

(iii) Why are dextro- and laevorotatory isomers of Butan-2-ol difficult to separate by fractional distillation? (CBSE Delhi 2019)
Dextro- and laevorotatory isomers of butan-2-ol are enantiomers of each other and both have the same boiling point. Hence, these cannot be separated by fractional distillation.

Question 2.
Give reasons for the following:
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
Ethyl iodide undergoes SN2 reaction faster than ethyl bromide because I- ion is a better leaving group than Br- ion.

(ii) (±) 2 – Butanol is optically inactive.
(±) 2 – Butanol represents a racemic mixture of (+) 2-butanol and (-) 2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, the (±) compound is optically inactive.

(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH3 — X. (CBSE 2013)
In halobenzene, there is delocalisation of electrons due to resonance. For example, chlorobenzene is considered to be a resonance hybrid of the following structures:

It is evident that the contribution of structures III, IV and V imparts a partial double bond character to the carbon-chlorine bond. Therefore, the C-X bond length in halobenzene is less than the C-X bond length in CH3-X, which has only a single C-X bond.

Question 3.
(i) Draw the structures of major moon halo products in each of the following reactions:

(ii) Which halogen compound in each of the following pairs will react faster in SN2 reaction:
(a) CH3Br or CH3I
CH3I

(b) (CH3)3C – Cl or CH3 – Cl (CBSE 2014)
CH3Cl

Question 4.
How would you convert the following:
(i) Prop-1-ene to 1-fluoropropane
(ii) Chlorobenzene to 2-chlorotoluene
(iii) Ethanol to propanenitrile
OR
Write the main products when
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
(iii) methyl chloride is treated with AgCN.

OR

Question 5.
Give reasons:
(i) C – Cl bond length in chlorobenzene is shorter than C — Cl bond length in CH3 — Cl.
In chlorobenzene, there is delocalisation of electrons due to resonance.
It is considered to be a resonance hybrid of the following structures.

The contribution of structure III, IV and V imparts a partial double bond character to the C – Cl bond.

This is confirmed by X-ray analysis which shows that the C – Cl bond length in chlorobenzene is 1.69 whereas the C – Cl bond length in ethyl chloride is 1.82 A. There is no resonance in CH3Cl and hence no partial double bond character.

(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
In chlorobenzene, the C of C-Cl bond is sp2-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp3-hybridised.

Therefore, the sp2-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp3-hybrid C of cyclohexyl chloride. As a result, the sp2 hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp3 hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride. Thus, chlorobenzene is less polar than cyclohexyl chloride.

(iii) SN1 reactions are accompanied by racemisation in optically active alkyl halides. (CBSE 2016)
The SN1 reactions proceed through the formation of carbocations. In the case of optically active alkyl halides, the product formed is a racemic mixture. This is because the intermediate carbocation is planar species, therefore, the attack of the nucleophile, OFT ion can take place from both the faces (front and rear) with equal ease. As a result, a 50: 50 mixture of the two enantiomers (laevo and dextro) is formed. Thus, the product formed is a racemic mixture (±) which is optically inactive.

For example, hydrolysis of optically active 2-bromobutane results in the formation of a racemic mixture (±)-butan-2-ol.

Question 6.
The following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(i) Write the compound which is most reactive towards the SN2 reaction.
1-Bromopentane

(ii) Write the compound which is optically active.
2-Bromopentane

(iii) Write the compound which is most reactive towards the SN2 elimination reaction.
2-Bromo-2-methyl butane

Question 7.
Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (CBSE Delhi 2016)
In chlorobenzene, the C of C-Cl bond is sp2-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp3-hybridised.

Therefore, the sp2-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp3-hybrid C of cyclohexyl chloride. As a result, the sp2 hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp3 hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride.

Thus, chlorobenzene is less polar than cyclohexyl chloride. In other words, the magnitude of negative charge (δ-) is less on the Cl atom of chlorobenzene than in cyclohexyl chloride. Further, due to the delocalisation of the lone pair of electrons of the Cl atom over the benzene ring due to resonance, the C-Cl bond in chlorobenzene acquires some double-bond character. On the other hand, the C-Cl bond in cyclohexyl chloride is a pure single bond.

Since dipole moment is a product of charge and distance, therefore, chlorobenzene has a lower dipole moment than cyclohexyl chloride due to the lower magnitude of charge (δ-) on Cl atom and small C-Cl distance.

(ii) alkyl halides, though polar, are immiscible with water,
Alkyl halides are polar molecules and therefore, their molecules are held together by dipole-dipole forces. On the other hand, the molecules of H20 are held together by hydrogen bonds. When alkyl halides are added to water, the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide-alkyl halide molecules and water-water molecules. Hence, alkyl halides are immiscible (not soluble) in water.

(iii) Grignard reagents should be prepared under anhydrous conditions? (CBSE Sample Paper 2011)
Grignard reagents are very reactive. They react with the moisture present in the apparatus or the starting materials (RX or Mg).
RMgX + HOH → R-H + Mg(OH)X
Therefore, Grignard reagents must be prepared in anhydrous conditions.

Question 8.
Which one of the following compounds will undergo a faster hydrolysis reaction by the SN1 mechanism? Justify your answer.

OR
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism. (CBSE Sample paper 2018-19)
C6 H5CH2Cl will undergo SN1 reaction faster.

The carbocation formed by C6 H5CH2Cl gets stabilised through resonance.
Greater the stability of carbocation, greater will be its ease of formation from the respective halide.

OR

The following isomer will exhibit enantiomerism:

IUPAC name: 1,2-Dichloropropane.

Question 9.
(i) Identify the chiral molecule in the following pair:
and

(ii) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.

(iii) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1- methylcyclohexane with alcoholic KOH. (CBSE 2018)

Question 10.
(a) Out of and which one is more reactive towards SN2 reaction and why?
because it is a primary halide and has less steric hindrance and therefore, undergoes SN2 action faster.

(b) Out of and which one is more reactive towards nucleophilic substitution reaction and why?
is more reactive because of the presence of electron-withdrawing – NO2 group.

(c) Out of and which one is optically active and why? (CBSE AI 2019)
is optically active because of the presence of chiral carbon. OH

Question 11.
Write the products of the following reactions:
(i) CH3 CH2 -CH = CH2 + HCl →

(ii) Me2CHCH2OH
Me2CHCH2OH MeCHCH2Cl

(iv) Me2C = CMe2

(v) (CBSE AI 2016)

Question 12.
(i) In the following pairs of the halogen compounds, which would undergo SN2 faster?
SN2 reaction proceeds through the formation of a transition state involving the bonding of carbon to five atoms or groups. The reactivity is decided by the stability of the transition state on the basis of steric hindrance. The reactivity follows the order: CH3 > 1° > 2° > 3° halide.
Therefore,

is primary alkyl halide and hence undergoes SN2 reaction faster.

I will undergo SN2 reaction faster because iodine is a better leaving group due to its large size and hence it will be released at a faster rate in the presence of incoming nucleophile.

(1 -Bromo-2, 2-dimethylpentane) reacts faster because it is 1 ° alkyl halide.

(CBSE AI 2009)
(2-Methyl-1-bromopropane) reacts faster because it has lesser steric hindrance in the transition state.

CH3 CH2 Br would undergo SN2 reaction faster because of less steric hindrance.

(ii) Which one of the following pairs undergoes SN1 substitution reaction faster and why?
SN1 reaction occurs through the formation of a carbocation. Therefore, the greater the stability of the carbocation, the faster is the rate of an SN1 reaction. Since the stability of carbocation follows the order 3° > 2° > 1° > CH3, therefore,

Out of or reacts faster because it invoLves the formation of 3° -carbocation while involves the formation of V -carbocation

(CBSE AI 2009)
Out of or reacts faster because it forms 2° carbocation while involves 1° carbocation.

(CBSE AI 2015)
undergoes SN1 reaction faster. This is because the carbocation derived from CH3(CH3)3 C Br is 3° carbocation and is more stable than carbocation (1°) obtained from CH3 CH2 Br.

Question 13.
(i) How would you convert the following:
(a) Prop-1-ene to 1-fluoropropane

(b) Chlorobenzene to 2-chlorotoluene

(ii) Write the main products when
(a) n-butyl chloride is treated with alcoholic KOH.

(b) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.

(c) methyl chloride is treated with AgCN. (CBSE AI2015)

Question 14.
Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
The molecular formulae of isomers of C4H9Br are CH3

Since the rate of reaction of the compound ‘A’ (C4H9Br) with aqueous KOH depends upon the concentration of compound ‘A’ only, therefore, the reaction occurs by SN1 mechanism and compound ‘A’ is tertiary bromide, i.e. 2-Bromo-2-methylpropane.
(CH3)3CBr + KOH(aq) → (CH3 )3COH + KBr rate = k[(CH3)3CBr]

(ii) Out of these two compounds, which one will be converted to the product with an inverted configuration. (NCERT Exemplar)
Since compound ‘B’ is optically active and is an isomer of compound ‘A’ (C4H9Br), therefore, compound ‘B’ must be 2-bromobutane. Since the rate of reaction of compound ‘B’ with aqueous KOH depends upon the concentration of compound ‘B’ and KOH, therefore, the reaction occurs by SN2 mechanism and the product of hydrolysis will have inverted configuration.

Compound ‘B’ will be converted with an inverted configuration.

Question 15.
(i) Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.

It is clear from resonating structures that there is an increase in electron density at ortho and para positions. Therefore, the functional group ‘X’ is ortho and para directing.

(ii) Classify the following compounds as primary, secondary and tertiary halides.
(a) 1-Bromobut-2-ene
Primary

(b) 4-Bromopent-2-ene
secondary

(c) 2-Bromo-2-methylpropane (NCERT Exemplar)
tertiary

Question 16.
tert-Butylbromide reacts with aq. NaOH by SN1 mechanism while n-butyl bromide reacts by an SN2 mechanism. Why?
In general, the SN1 reaction proceeds through the formation of a carbocation. The tert-butyl bromide readily loses Br ion to form stable 3° carbocation.
Therefore, it reacts with aqueous KOH by SN1 mechanism as:

On the other hand, n-butyl bromide does not undergo ionisation to form n-butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by an SN2 mechanism, which occurs in one step through a transition state involving the nucleophilic attack of OH ion from the backside with simultaneous expulsion of Br ion from the front side.

SN1 mechanism follows the reactivity order as 3° > 2° > 1° while the SN2 mechanism follows the reactivity order as 1° >2° >3°
Therefore, tert-butyl bromide (3°) reacts by SN1 mechanism while n-butyl bromide (1°) reacts by the SN2 mechanism.

## Biomolecules Class 12 Important Extra Questions Chemistry Chapter 14

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 14 Biomolecules. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

## Class 12 Chemistry Chapter 14 Important Extra Questions Biomolecules

### Biomolecules Important Extra Questions Very Short Answer Type

Question 1.
Write the structure of the product obtained when glucose is oxidised with nitric acid. (CBSE 2012)
Saccharic acid (or glucaric acid) is formed.

Question 2.
Write a reaction that shows that all the carbon atoms in glucose are linked in a straight chain. (CBSE 2012)
When glucose is heated with HI and red P at 100°C for a long period, it gives n-hexane.

The formation of n-hexane suggests that all six carbon atoms in glucose are arranged in a straight chain.

Question 3.
What are the hydrolysis products of sucrose? (CBSE 2019C)
Glucose and fructose

Question 4.
Write the name of the linkage joining two amino acids. (CBSE 2013)

Question 5.
What are the hydrolysis products of lactose? (CBSE2019C)
Lactose on hydrolysis gives β – D – galactose and β – D – glucose.

Question 6.
What are three types of RNA molecules which perform different functions? (CBSE Delhi 2013)
(i) messenger RNA: m – RNA
(ii) ribosomal RNA: r – RNA
(iii) transfer RNA: t – RNA

Question 7.
What type of bonding helps in stabilising the a-helix structure of proteins? (CBSE Delhi 2013)
Hydrogen bonding between -NH and -C = 0 groups of peptide bonds.

Question 8.
What is a glycosidic linkage? (CBSE Delhi 2013)
Two monosaccharide units are linked to each other by a bond called glycosidic linkage.

Question 9.
Which of the two components of starch is water-soluble? (CBSE Delhi 2014)
Amylose

Question 10.
Which component of starch is a branched polymer of a-glucose and insoluble in water? (CBSE Delhi 2014)
Amylopectin

Question 11.
What are the products of hydrolysis of maltose? (CBSE 2014)
Glucose

Question 12.
What is the basic structural difference between glucose and fructose?
OR
Write the products obtained after hydrolysis of lactose. (CBSE Delhi 2019)
Glucose contains an aldehyde group (-CHO) and is aldose, which is present at the end of the carbon chain, i.e. C1.

Fructose contains a keto group (>C = 0) and is called ketose, which is present at a carbon atom next to the terminal, i.e. C2.
OR
Hydrolysis of lactose gives β – D – glucose and β – D – galactose.

Question 13.
What is the difference between a glycosidic linkage and a peptide linkage?
OR
What is the difference between Nucleotide and Nucleoside? (CBSE AI 2019)
Glycosidic linkage is the linkage that joins two monosaccharides through an oxygen atom (-0-). Peptide linkage is the linkage that joins two amino acids through -CONH- bond.
OR
The molecules in which one of the nitrogen bases (purines or pyrimidine) is bonded with a sugar molecule (ribose or deoxyribose) is a nucleoside. The nucleoside linked with the phosphate group is called a nucleotide.

Question 14.
What is meant by the inversion of sugar? (CBSE Sample Paper 2011)
The change of specific rotation of sugar from dextrorotatory to laevorotatory is called inversion of sugar.

Question 15.
Glucose does not give 2, 4-DNP test and Schiff’s test. Why? (CBSE Sample Paper 2011)
Glucose has a cyclic structure in which the -CHO group is not free because it forms a hemiacetal linkage with the -OH group at C-5. Therefore, it does not give 2, 4-DNP test although it has -CHO group.

Question 16.
Write any two reactions of glucose that could not be explained by an open-chain structure of the glucose molecule.
The open-chain structure of a glucose molecule cannot explain the following:

1. Glucose does not react with sodium bisulphite (NaHS03) to form an addition product though it has an aldehyde group,
2. Glucose does not give the Schiffs test and 2,4-DNP test like other aldehydes.

Question 17.
Write the product when D-glucose reacts with the cone. HNO3.

Question 18.
(i) Which vitamin deficiency causes rickets?
Vitamin D

(ii) Name the base that is found in the nucleotide of RNA only. (CBSE Sample Paper 2017-18)
Uracil

Question 19.
Write the Zwitter ion structure of glycine. (CBSE Sample Paper 2011)

Question 20.
Name a carbohydrate present in the liver, muscles and brain. (CBSE 2019C)
Glycogen

Question 21.
Name the species formed when an aqueous solution of amino acid is dissolved in water. (CBSE Sample Paper 2019)
Zwitter ion/dipolar ion

Question 22.
Name the unit formed by the attachment of a base to the 1′ position of sugar in a nucleoside. (CBSE Sample Paper 2019)
Nucleotide

### Biomolecules Important Extra Questions Short Answer Type

Question 1.
Name the bases present in RNA. Which one of these is not present in DNA? (CBSE Delhi 2011)

• The bases present in RNA are guanine, adenine, cytosine and uracil.
• Uracil is not present in DNA. Instead, it contains thymine.

Question 2.
What is meant by biocatalysts? (CBSE Delhi 2012)
Biocatalysts. Biocatalysts or enzymes are produced by living cells which catalyse the biochemical reactions occurring in living cells.

Question 3.
Name one fibrous protein and one globular protein.
Fibrous: Keratin Globular: Albumin

Question 4.
What is the chemical name of vitamin A and which disease is caused by its deficiency?
Retinol, deficiency disease: Xerophthalmia.

Question 5.
What is the chemical name of vitamin C and which disease is caused by its deficiency?
Ascorbic acid, deficiency disease: Scurvy.

Question 6.
Which enzyme is present in saliva? What is its function?
The enzyme present in the saliva is amylase. It hydrolyses starch into maltose.

Question 7.
What is the difference between DNA and RNA on the basis of the bases they contain?
Both DNA and RNA contain two bases derived from purine: guanine and adenine and one base derived from pyrimidine: cytosine. However, they have a fourth different base; DNA contains thymine whereas RNA contains uracil.

Question 8.
Shanti, a domestic helper of Mrs Anuradha, fainted while mopping the floor. Mrs Anuradha immediately took her to the nearby hospital where she was diagnosed to be severely ‘anaemic’. The doctor prescribed an iron-rich diet and multivitamins supplement to her. Mrs Anuradha supported her financially to get the medicines. After a month, Shanti was diagnosed to be normal.

(i) What values are displayed by Mrs Anuradha?
Mrs Anuradha has shown generosity and a caring attitude for the health and life of her domestic helper. Timely help, kindness and financial support saved her domestic helper from a serious health problem.

(ii) Name the vitamin whose deficiency causes ‘pernicious anaemia’.
B12

(iii) Give an example of a water-soluble vitamin. (CBSE 2013)
Vitamin C

### Biomolecules Important Extra Questions Long Answer Type

Question 1.
Differentiate between the following:
(i) Amylose and Amylopectin
(iii) Fibrous proteins and Globular proteins
OR
Write chemical reactions to show that the open structure of D-glucose contains the following:
(i) Straight chain
(ii) Five alcohol groups
(iii) Aldehyde as carbonyl group (CBSE Delhi 2019)
(i) Amylose and Amylopectin:

 Amylose Amylopectin 1. It consists of a long chain of a-D- glucose. 1. It consists of branched polymeric chains of a-D- glucose. 2. It is water-soluble. 2. It is water-insoluble.

(iii) Fibrous and Globular proteins:

 Fibrous Globular 1. The polypeptide chains run parallel and are held by hydrogen and sulphide bonds. 1. The polypeptide chains coil around to give a spherical shape. 2. These are insoluble in water. 2. These are soluble in water. 3. For example, Keratin in hair. 3. For example, egg albumin.

OR
(i) Glucose on prolonged heating with HI and red P at 100 °C gives n-hexane, which shows the straight-chain structure of glucose.

(ii) Glucose reacts with acetic anhydride in the presence of anhydrous zinc chloride to form glucose pentaacetate. The formation of pentaacetate confirms the presence of five -OH groups in the glucose molecule.

(iii) Glucose reacts with hydrogen cyanide forming glucose cyanohydrin. This shows that glucose contains a carbonyl group (>C=0).

Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on treatment with a mild oxidising agent like Br2 water. This indicates that the carbonyl group present in glucose is an aldehyde (-CHO) group.

Question 2.
Define the following with a suitable example in each:
(i) Oligosaccharides
(ii) Denaturation of protein
(iii) Vitamins
OR
Write the reactions involved when D-glucose is treated with the following reagents:
(i) Br2 water
(ii) H2N-OH
(iii) (CH3CO)20 (CBSE Delhi 2019)
(i) Oligosaccharides. These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis, for example, disaccharides such as sucrose, lactose, maltose (C12H22O16), trisaccharides such as raffinose (C18H32O16).

(ii) Deriaturatlon of proteins: A process that changes the physical and biological properties of proteins without affecting the chemical composition of the protein is called denaturation. It is caused by certain physical change like change In temperature or chemical change like change In pH presence of electrolytes, etc., for example, coagulation of albumin present in the white of an egg.

(iii) Vitamins: The organic compounds which cannot be produced by the body and must be supplied in small amounts in the diet to perform specific biological functions for the normal health, growth and maintenance of the body are called vitamins, for example, vitamin A, B, C, D, etc.
OR

Question 3.
Define the following terms with a suitable example of each:
(a) Anomers
Anomers: Carbohydrates that differ in configuration at the glycosidic carbon (i.e. C1 in aldoses and C2 in ketoses) are called anomers. For example, α-D-glucose and β-D- glucose.

(b) Essential amino acids
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet are called essential amino acids. For example, valine, leucine, etc.

(c) Denaturation of protein (CBSE AI 2019)
A process that changes the native conformation of a protein is called denaturation. The denaturation can be caused by changes in pH, temperature, presence of salts of certain chemical agents. The denatured protein will lose its biological activity. During denaturation, the protein molecule uncoils from an ordered and specific conformation into a more random conformation and protein precipitates from the solution. For example, when an egg is boiled in water, the globular proteins present in it change to a rubber-like insoluble mass.

Question 4.
Define the following terms with a suitable example of each:
(a) Tertiary structure of the protein
The tertiary structure of proteins: Tertiary structure of proteins arises because of the folding, coiling or bending of polypeptide chains producing a three-dimensional structure. This structure gives the overall shape of proteins. For example, fibrous proteins such as silk, collagen and α-keratin have large helical structures

(b) Essential amino acids
Essential amino acids: The amino acids which cannot be synthesised in our body and must be supplied through our diet are called essential amino acids, for example, valine, leucine, etc.

(c) Disaccharides (CBSE AI 2019)
Disaccharides: The carbohydrates which on hydrolysis give two same or different monosaccharides are called disaccharides. Sucrose is an example of a disaccharide, which on hydrolysis gives glucose and fructose.

Question 5.
Describe what do you understand by primary structure and secondary structure of proteins. (CBSE Delhi 2011)
Primary structure:
The sequence in which the amino acids are linked in one or more polypeptide chains of a protein is called the primary structure.

The primary structure is usually determined by its successive hydrolysis with enzymes or mineral acids.

Secondary structure: The secondary structure gives the manner in which the polypeptide chains are folded or arranged. Therefore, it gives the shape or conformation of the protein molecule. This arises from the plane geometry of the peptide bond and hydrogen bond between the —C=O and N—H groups of different peptide bonds. Pauling and Corey studied that there are two common types of structures:

(i) α – Helix structure: In this structure, the formation of hydrogen bonding between amide groups within the same chain causes the peptide chains to coil up into a spiral structure (Fig. 1). This is called the a-helix.

(ii) β-pleated sheet structure: In this structure, all polypeptide chains are stretched out to nearly maximum extension and then laid side by side in a zigzag manner to form a flat sheet. Each chain is held to the two neighbouring chains by a hydrogen bond. These sheets are stacked one upon another to form a three-dimensional structure called
β-pleated sheet structure (Fig. 2).

Question 6.
ExplaIn what is meant by the pyranose structure of glucose. (CBSE 2011)
The six-membered cyclic structure of glucose is catted pyranose structure similar to the pyran heterocyclic compound. For structure,

Question 7.
Write the main structural difference between RNA and DNA. Of the four bases, name those which are common to both DNA and RNA. (CBSE 2011)
Structural differences:

 DNA RNA (i) It has a double-stranded a-helix structure in which two strands are coiled spirally in opposite directions. (i) It has a single-stranded a-helix structure. (ii) The sugar molecule present in DNA is 2-deoxyribose. (ii) The sugar molecule present in RNA is ribose. (iii) Nitrogenous base uracil is not present. (iii) Nitrogenous base thymine is not present.

Common bases present in both: adenine, cytosine and guanine.

Question 8.
(i) Deficiency of which vitamin causes night blindness? (CBSE DelhI 2014)
(ii) Name the base that is found in the nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does It suggest about the structure of glucose?
(i) Deficiency of Vitamin A causes night blindness.
(ii) Uracil
(iii) Glucose reacts with HI to give n-hexane.

This suggests that alt the six carbon atoms in glucose are arranged in a straight chain structure of glucose.

Question 9.
(i) DefIciency of which vitamin causes rickets? (CBSE Delhi 2014)
Vitamin D.

(ii) Give an example for each of fibrous protein and globular protein.
Fibrous protein: Keratin Globular protein: Albumin

(iii) Write the product formed on reaction of D-glucose with Br2 water.

Question 10.
(i) Deficiency of which vitamin causes scurvy? (CBSE DelhI 2014)
Vitamin C.

(ii) What type of linkage is responsible for the formation of proteins?

(iii) Write the product formed when glucose is treated with HI.

Question 11.
Define the following terms: (CBSE 2014)
Glycosidic linkage: The condensation of hydroxyl groups of two monosaccharides to form a link between them is called glycosidic linkage.

(ii) Invert sugar
Invert sugar: The sugar which on hydrolysis with dilute adds or enzymes gives mixture having specific rotation opposite to the original is called invert sugar. For example, sucrose is inverted sugar.

(iii) Oligosaccharides
Oligosaccharides: These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis. These are further classified as disaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units present in their molecules. For example, Disaccharides: Sucrose, lactose, maltose. All these have the molecular

Formula C12H22O11.
Trisaccharide: Raffinose (C18H32O16).
Tetrasaccharides: Stachyose (C24H42O21 ).

Question 12.
Define the following terms: (CBSE 2014)
(i) Nucleotide
Nucleotide: A unit formed by the combination of a nitrogen-containing heterocyclic base, a pentose sugar and a phosphoric acid group.

(ii) Anomers
Anomers: The anomers are the isomers formed due to the change in the configuration of the -OH group at C-1 of glucose. For example, α-and β-forms of glucose are anomers.

(iii) Essential amino acids.
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet for the growth of the body are called essential amino acids.

Question 13.
Differentiate between the following:
(a) Fibrous protein and Globular protein
Difference between fibrous protein and globútar protein:

 Fibrous Globular (i) The polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. 1. The polypeptide chains colt around to give a spherical shape. (ii) These are insoluble in water. 2. These are soluble in water. (iii) For example, keratin in hair 3. For example, albumin in egg.

(b) Essential amino acids and Non-essential amino acids
Difference between essential amino acids and non-essential amino acids:

 Essential amino acids Non-essential amino acids These are not synthesìsed in our body and must be supplied in the diet. These are synthesised in our body and not required in our diet. For example, valine. For example, alanine

(C) Amylose and Amylopectin (CBSE AI 2019)
Difference between amylose and amylopectin:

 Amylose Amylopectin It consists of branched polymeric chains of α – D – glucose. It consists of a long straight chain of α – D – glucose. It is water-insoluble. It is water-soluble.

Question 14.
(i) Which one of the following is a polysaccharide: (CBSE 2015, 2014)
Starch, Maltose, Fructose, Glucose
Starch

(ii) What is the difference between native protein and denatured protein?
Native protein is the protein found in a biological system with a unique three-dimensional structure and biological activity.
Denatured protein is the protein which loses its biological activity by certain physical or chemical treatment such as changes in pH, temperature, presence of certain salts or chemical agent, known as denaturation.

(iii) Write the name of the vitamin responsible for the coagulation of blood.
Vitamin K.

Question 15.
(i) Write one reaction of D-glucose which cannot be explained by its open-chain structure. (CBSE 2016)
Despite having an aldehydic (-CHO) group, glucose does not react with NaHS03 to form an addition product.

(ii) What type of linkage is present in nucleic acids?

(iii) Give one example each for water-soluble vitamins and fat-soluble vitamins.
Water-soluble vitamin: vitamin B Fat-soluble vitamin: vitamin A

Question 16.
What is essentially the difference between the α-form of D-glucose and β-form of D-glucose? Explain.
α-form and β-form of glucose differ in the orientation of —H and —OH groups around the C1 atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C1 are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|D = +111° and the β-form of glucose has m.p. 423 K and |α|D = +19.2°.

Question 17.
(a) What happens when D-glucose is treated with the following reagents:
(i) HI
When glucose is treated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight line.

(ii) cone. HN03
Glucose on treatment with nitric acid gives a dicarboxylic acid, saccharic acid.

(b) What is the basic structural difference between starch and cellulose? (CBSE 2019C)
Starch consists of two components: amylose and amylopectin. Amylose is a long linear polymer of 200-1000 a-D-(+)-glucose units held by C1 – C4 glycosidic linkages. It is soluble in water. Amylopectin is a branched-chain polymer of a-D-(+)-glucose linkages whereas branching occurs by C1 – C6 glycosidic linkage. It is insoluble in water.

On the other hand, cellulose is a straight-chain polysaccharide composed only of (β – D – (+)-glucose units which are formed by the glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

Question 18.
(1) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar. (CBSE Delhi 2016)
Glucose, Galactose

(ii) Why Vitamin C cannot be stored in our body?
Vitamin C is water-soluble and hence cannot be stored in our body.

(iii) What is the difference between a nucleoside and a nucleotide?
A nucleoside contains only two basic components of nucleic acids, namely a pentose sugar and a nitrogenous base.

A nucleotide contains all the three basic components of nucleic acids, namely a phosphoric acid group, a pentose sugar and a nitrogenous base.

Question 19.
Define the following with an example of each: (CBSE 2018)
(i) Polysaccharides
(ii) Denatured protein
(iii) Essential amino acids
OR
(i) Write the product when D-glucose reacts with the cone. HNO3.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between a-helix and p-pleated structures of proteins.
(i) Carbohydrates that give a large number of monosaccharide units on hydrolysis or a large number of monosaccharide units joined together by glycosidic linkage, e.g. starch, glycogen, cellulose.
(ii) Proteins that lose their biological activity or proteins in which secondary and tertiary structures are destroyed, e.g. curdling of milk.
(iii) Amino acids cannot be synthesised in the body. e.g. Valine / Leucine
OR
(i) Saccharic acid / COOH-(CHOH)4-COOH
(ii) Due to the presence of carboxyl and amino group in the same molecule or due to formation of zwitterion or dipolar ion.
(iii) a-helix has intramolecular hydrogen bonding while p-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in p-pleated all polypeptide chains are stretched and arranged side by side.

Question 20.
Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids. (CBSE 2018C)
Due to the formation of zwitterion.

(ii) The two strands of DNA are complementary to each other.
The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

(iii) Reaction of glucose indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose.

Glucose gets oxidised to gluconic acid on reaction with a mild oxidising agent like Bromine water.

Question 21.
What is essentially the difference between α-glucose and β-glucose? What is meant by the pyranose structure of glucose?
α-form and β-form of glucose differ in the orientation of -H and -OH groups around C, atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C1 are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|D = + 111° and the β-form of glucose has m.p. 423 K and |α|D = + 19.2°.

The α-D-glucose and β-D-glucose can be drawn in a simple six-membered ring form called pyranose structures. These resemble pyran which is a six-membered heterocyclic ring containing five carbon atoms and one oxygen atom.

These are known as pyranose structures and are shown below:

Question 22.
Which sugar is called invert sugar? Why is it called so?
Sucrose is called invert sugar. The sugar obtained from sugar beet is a colourless, crystalline and sweet substance. It is very soluble in water and its aqueous solution is dextrorotatory having [α]D = + 66.5°.

On hydrolysis with dilute acids or enzyme invertase, cane sugar gives an equimolar mixture of D-(+)-glucose and D-(-)-fructose.

So, sucrose is dextrorotatory but after hydrolysis, gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)- glucose. Therefore, the resultant solution upon hydrolysis is laevorotatory in nature with a specific rotation of (-39.9°). Since there is a change in the sign of rotation from Dextro before hydrolysis to Laevo after hydrolysis, the reaction is called Inversion reaction and the mixture (glucose and fructose) is called invert sugar.

Question 23.
How do you explain the presence of an aldehydic group in a glucose molecule?
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin.

Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid-containing six carbon atoms.

This indicates that the carbonyl group present in glucose is an aldehydic group.

Question 24.
Describe the term D- and L-configuration used for sugars with examples.
The sugars are divided into two families: the D-family and L-family which have definite configurations. These configurations are represented with respect to glyceraldehyde as the standard. The glyceraldehyde may be presented in two forms:

The D-configuration has —OH attached to the carbon adjacent to —CH2OH on the right while L-configuration has —OH attached to the carbon adjacent to —CH2OH on left.

The sugars are calLed D- or L- depending upon whether the configuration of the molecule is related to D-glyceraldehyde or L-glyceraldehyde. It has been found that all naturally occurring sugars beLong to D-series, e.g. D-glucose, D-ribose and D-fructose.

## Nuclei Class 12 Important Extra Questions Physics Chapter 13

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 13 Nuclei. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

## Class 12 Physics Chapter 13 Important Extra Questions Nuclei

### Nuclei Important Extra Questions Very Short Answer Type

Question 1.
What will be the ratio of the radii of two nuclei of mass numbers A1 and A2?
The ratio is $$\frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}$$

Question 2.
Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities?
The densities of both nuclei are equal as they do not depend upon mass number.

Question 3.
A nucleus of mass number A has a mass defect Δm. Give the formula, for the binding energy per nucleon of this nucleus.
The formula is E = $$\frac{\Delta m \times c^{2}}{A}$$

Question 4.
Write the relation between half-life and decay constant of a radioactive sample.
The relation is T1/2 = $$\frac{0.693}{\lambda}$$

Question 5.
Write the nuclear decay process for β-decay of 1532P.
The process is

Question 6.
State the relation between the mean life (τ) of a radioactive element and its decay constant λ.
The two are related as τ = 1 / λ.

Question 7.
Write any two characteristic properties of nuclear force. (CBSE AI 2011)

1. Do not obey inverse square law and
2. Spin-dependent.

Question 8.
How is the radius of a nucleus related to its mass number? (CBSE AI 2011C, AI 2013C)
The radius R of the nucleus and mass number A is related as R = RoA1/3, where Ro is a constant.

Question 9.
A nucleus undergoes β – decay. How does
(i) the mass number,
(ii) atomic number change? (CBSE Delhi 2011C)
During β – decay
(i) the mass number remains the same,
(ii) atomic number increases by one.

Question 10.
Define the activity of a given radioactive substance. Write its SI unit. (CBSE AI 2013)
The rate of disintegration in a radioactive substance is known as its activity. SI unit is becquerel (Bq).

Question 11.
Why is it found experimentally difficult to detect neutrinos in nuclear β-decay? (CBSE AI 2014)
They are very difficult to detect since they can penetrate a large quantity of matter (even earth) without any interaction.

Question 12.
Four nuclei of an element undergo fusion to form a heavier nucleus, with the release of energy. Which of the two — the parent or the daughter nucleus – would have higher binding energy per nucleon? (CBSE AI 2018, Delhi 2018)
Daughter nucleus.

Question 13.
Why is nuclear fusion not possible In the laboratory?
Because temperature as high as 107 K cannot be sustained in the laboratory.

Question 14.
Why is the penetrating power of gamma rays very large?
Because they have high energy and are neutral.

Question 15.
Can it be concluded from beta decay that electrons exist inside the nucleus?
No, the beta particle although an electron is actually created at the instant of beta decay and ejected at once. It cannot exist inside the nucleus as its de-Broglie wavelength is much larger than the dimensions of the nucleus.

Question 16.
Why is the ionizing power of α – parties greater than that of γ-rays?
Because α – particles are heavy particles and their speed is comparatively small, so they collide more frequently with atoms of the medium and ionize them.

Question 17.
When a nucleus undergoes alpha decay, is the product atom electrically neutral in beta decay?
No, in alpha decay, the atomic number decreases by 2 hence the atom is left with 2 extra orbital electrons. It, therefore, has a double negative charge. In beta decay, the atom is left with a net positive charge.

Question 18.
You are given two nuclides 37X and 34Y. Are they the isotopes of the same element? Why?
Yes, because an atomic number of both nuclides is 3.

Question 19.
The variation of the decay rate of two radioactive samples A and B with time is shown in the figure. Which of the two has a greater decay constant? (NCERT Exemplar)

The decay constant of A is greater than that of B but it does not always decay faster than B.

Question 20.
Does the ratio of neutrons to protons in a nucleus increase, decrease, or remain the same after the emission of an alpha particle? (NCERT Exemplar)
The ratio of neutrons to protons in a nucleus increases after the emission of an alpha particle.

Question 21.
Which property of nuclear force explains the approximate constancy of binding energy per nucleon with mass number A for nuclei in the range 30 < A < 170? (CBSE2019C)
The short-range nature of the nuclear force explains the approximate constancy of binding energy per nucleon with mass number A in the range 30 < A < 170.

Question 22.
Draw a graph showing the variation of decay rate with a number of active nuclei. (NCERT Exemplar)
The graph is as shown.

Question 23.
Which sample, A or B, shown in the figure has a shorter mean-life? (NCERT Exemplar)

B has shorter mean life as λ is greater forB.

Question 24.
Why do stable nuclei never have more protons than neutrons? (NCERT Exemplar)
Protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so that an excess of neutrons, which produce only attractive forces, is required for stability.

Question 25.
Why does the process of spontaneous nuclear fission occur in heavy nuclei? (CBSE 2019C)
Because heavy nuclei contain a large number of protons that exert strong repulsive forces on one another.

### Nuclei Important Extra Questions Short Answer Type

Question 1.
Draw the curve showing the binding energy/nucleon with a mass number of different nuclei. Briefly state, how nuclear fusion and nuclear fission can be explained on the basis of this graph.
The diagram is as shown.

Light nuclei have a small value of binding energy per nucleon, therefore to become more stable they fuse to increase their binding energy per nucleon.

A very heavy nucleus, say A 240, has Lower binding energy per nucLeon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightLy bound. This implies energy would be released in the process.

Question 2.
Define decay constant for a radioactive sample. Which of the following radiations α, β, and γ rays
(i) are similar to X-rays,
(ii) are easily absorbed by matter, and
(iii) are similar in nature to cathode rays?
The decay constant is defined as the reciprocal of that time duration for which the number of nuclei of the radioactive sample decays to 1 / e or 37 % of its original value.
(i) Gamma
(ii) Alpha
(iii) Beta

Question 3.
State the law of radioactive decay.
Plot a graph showing the number of undecayed nuclei as a function of time (t) for a given radioactive sample having a half-life T1/2.
Depict In the plot the number of undecayed nuclei at (i) t = 3T1/2 and (ii) t = 5 T1/2 (CBSE Delhi 2011)
The number of nuclei disintegrating per second is proportional to the number of nuclei present at the time of disintegration and is independent of alt physical conditions like temperature, pressure, humidity, chemical composition, etc.

The plot is as shown.

Question 4.
Draw a plot of the potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces. (CBSE AI 2012)

For r > r0 (attraction), For r < ro (repulsion)

1. Strong attractive force (stronger than the repulsive electric force between the protons)
2. Are short-range forces.

Question 5.
(a) Write the relation for binding energy (BE) (in MeV) of a nucleus of mass ZAM atomic number (Z) and mass number (A) in terms of the masses of its constituents – neutrons and protons.
The required expression is
ΔE = (Zmp + (A – Z)mn – M) × 931 MeV

(b) Draw a plot of BE/A versus mass number A for 2 ≤ A ≤ 170. Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei. (CBSE Delhi 2014C)

Since the binding energy of the smaller nuclei like hydrogen is less, therefore they fuse together to form helium in order to increase their binding energy per nucleon and become stable. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

Question 6.
If both the number of neutrons and the number of protons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain. (CBSE AI2016C)
We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). Now, since proton number and neutron number are conserved in a nuclear reaction the total rest mass of neutrons and protons is the same on either side of a reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right-hand side.

The difference in these binding energies appears as the energy released or absorbed in a nuclear reaction. Since binding energy contributes to mass, we say that the difference in the total mass of nuclei on the two sides gets converted into energy or vice-versa.

Question 7.
State two properties of nuclear forces. Write the relation between half-life and decay constant of a radioactive nucleus. (CBSE AI 2017C)

1. They are saturated forces.
2. They are charge-independent.

The required relation is
T = $$\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}$$

Question 8.
(a) Draw a graph showing the variation of binding energy per nucleon (BE/A) vs mass number A for the nuclei in 20 ≤ A ≤ 170.

Since the binding energy of the smaller nuclei like hydrogen is less, therefore they fuse together to form helium in order to increase their binding energy per nucleon and become stable. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

(b) A nucleus of mass number 240 and having binding energy/nucleon 7.6 MeV splits into two fragments Y, 1 of mass numbers 110 and 130 respectively. If the binding energy/ nucleon of Y, 1 is equal to 8.5 MeV each, calculate the energy released in the nuclear reaction. (CBSE Al 2017C)
Energy released per fission
= (110 + 130) × 8.5 – 240 × 7.6
= 240 × (8.5 – 7.6) MeV
= 240 × 0.9
= 216.0 MeV

Question 9.
Explain with the help of an example, whether the neutron-proton ratio in a nucleus increases or decreases due to beta decay.
Consider the following decay

Number of neutrons before beta decay
= 234-90 = 144
Number of neutrons after beta decay
= 234-91 =143
Number of protons before beta decay
= 90
Number of protons after beta decay
= 91
Neutron-proton ratio before beta decay
= $$\frac{144}{90}$$ = 1.6

Neutron-proton ratio after beta decay
= $$\frac{143}{91}$$ = 1.57

Thus neutron-proton ratio decreases during beta decay.

Question 10.
How is the size of a nucleus experimentally determined? Write the relation between the radius and mass number of the nucleus. Show that the density of the nucleus is independent of its mass number. (CBSE Delhi 2011C)
The size of the nucleus can be determined by the Rutherford experiments on alpha particles scattering. The distance of the nearest approach is approximately the size of the nucleus. Here it is assumed that only coulomb repulsive force caused scattering. With alpha rays of 5.5 MeV, the size of the nucleus was found to be less than 4 × 10-14 m. By doing scattering experiments with fast electrons bombarding targets of different elements, the size of the nuclei of various elements determined accurately.

The required relation is
R = RoA1/3, where Ro = 1.2 × 10-15 m

The density of a nucleus of mass number A and radius R is given by

which is independent of the mass number A.

Question 11.
(a) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This leads to the saturation of forces in a medium or a large-sized nucleus, i.e. nuclei for which A is 30 < A < 170, which is the reason for the constancy of the binding energy per nucleon.

(b) Show that the density of nucleus over a wide range of nuclei is constant- independent of mass number A. (CBSE AI 2012)
The size of the nucleus can be determined by the Rutherford experiments on alpha particles scattering. The distance of the nearest approach is approximately the size of the nucleus. Here it is assumed that only coulomb repulsive force caused scattering. With alpha rays of 5.5 MeV, the size of the nucleus was found to be less than 4 x 10-14 m. By doing scattering experiments with fast electrons bombarding targets of different elements, the size of the nuclei of various elements determined accurately.

The required relation is
R = RoA1/3, where Ro = 1.2 × 10-15 m

The density of a nucleus of mass number A and radius R is given by

which is independent of the mass number A.

Question 12.
A radioactive nucleus ‘A’ decays as given below:
.
If the mass number and atomic number of A1 are 180 and 73 respectively, find the mass number and an atomic number of A and A2.
For A : Z = 72 and A = 180
For A2: Z = 71 and A = 176

Question 13.
The sequence of stepwise decay of a radioactive nucleus is. If the nucleon number and atomic number for D2 are 176 and 71 respectively, what are the corresponding values of D and D3? Justify your answer in each case.
For D: A = 180, Z = 72
For D3: A = 172, Z = 69

During alpha decay mass number decreases by 4 and the atomic number decreases by 2, while during beta decay the mass number remains the same, and the atomic number increases by 1.

Question 14.
Write symbolically the nuclear β+ decay process of 611C. Is the decayed product X an isotope or isobar of611C?
Given the mass values m (611C) = 11.011434 u and m (X) = 11.009305 u. (CBSE AI 2015)
Estimate the Q – value in this process.
The required equation is

X is an isobar
Mass defect = m(C) – m(X)
= (11.011434- 11.009305) u = 0.002129 u

Therefore Q = Δm × 931.5 MeV
= 0.002129 × 931.5 = 1.98 MeV

Question 15.
Two radioactive samples, X, Y have the same number of atoms at t = 0. Their half¬lives are 3 h and 4 h respectively. Compare the rates of disintegration of the two nuclei after 12 hours. (CBSE AI 2017C)
Let N0 be the nuclei present in X and Y at t = 0. Given Tx = 3 h and Ty = 4 h, t = 12 h.
The number of nuclei present in X and Y after 12 hours is

Question 16.
A radioactive sample has the activity of 10,000 disintegrations per second after 20 hours. After the next 10 hours, its activity reduces to 5,000 dis. sec-1. Find out its half-life and initial activity. (CBSE Delhi 2017C)
Since activity reduces to half in 10 hours from 10000 dis. sec-1 to 5000 dis. sec-1, therefore half-life of the sample will be 10 years.

Question 17.
Why is the energy of the beta particles emitted during beta decay continuous?
The phenomenon of beta decay arises due to the conversion of a neutron in the nucleus into a proton, electron, and an anti-neutrino. Because the energy available in beta decay is shared by the electron and the anti-neutrino in all possible ratios as they come out of the nucleus, therefore the beta ray energy spectrum is continuous in nature.

Question 18.
Explain, how radioactive nuclei can emit β-particles even though atomic nuclei do not contain these particles. Hence explain why the mass number of a radioactive nuclide does not change during β-decay.
Beta-particles (or electrons) as such are not present inside a nucleus. However, in the case of a radioactive nuclide, sometimes a neutron decays into a proton, an electron, and an antineutrino as given by the following equation:

where mass and charge of antineutrino particle is zero. Out of the particles formed, the proton remains within the nucleus itself but electron along with antineutrino comes out of the nucleus. It is this electron that is being emitted as a beta-particle.

As in the process of β-emission, one proton is produced in the nucleus at the expense of a neutron and the mass number of both is the same, hence the mass number of the nuclide remains unchanged during p-decay.

Question 19.
Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence: A → B → C. Here B is an intermediate nucleus that is also radioactive. Considering that there are No atoms of A initially, plot the graph showing the variation of the number of atoms of A and B versus time. (NCERT Exemplar)
At t = 0, NA = No while NB = 0. As time increases, NA falls off exponentially, while the number of atoms of B increases, becomes maximum, and finally decays to zero  ∞ (following exponential decay law).

Hence the graph is as shown.

### Nuclei Important Extra Questions Long Answer Type

Question 1.
Define the terms: half-life period and decay constant of a radioactive sample. Derive the relation between these terms.
The half-life is the time required for the number of radioactive nuclei to decrease to one-half the original number.

The decay constant is defined as the reciprocal of that time duration for which the number of nuclei of the radioactive sample decays to 1 / e or 37% of its original value.

To get the relation for half life T and decay constant λ we set N = $$\frac{N_{0}}{2}$$ and t = T in the equation N = No e-λt, obtaining $$\frac{1}{2}$$ = e-λt

Taking the logarithm of both sides and solving for T we have
T = $$\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}$$

Question 2.
(a) Draw a graph showing the variation of the potential energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is (i) attractive, and (ii) repulsive.
Graph showing the variation of potential energy U (in MeV) of a pair of nucleons as a function of their separation r (in fm) is shown here.

1. In the graph region AD (r > ro) shows the region where nuclear force is strongly attractive.
2. The region DE (r < ro) shows the region where nuclear force is strongly repulsive.

(b) Write two characteristic features of nuclear force which distinguish it from the Coulomb force.
Two characteristics of nuclear forces which distinguish it from Coulomb’s force are

• It is charge Independent.
• It is an extremely short-range force and does not obey the inverse square law.

Question 3.
Prove that the Instantaneous rate of change of the activity of a radioactive substance is Inversely proportional to the square of Its half-life.
The activity of a radioactive substance is
A = $$\frac{dN}{dt}$$.

We know that the number of nuclei of a radioactive substance Left behind after time t is given by N = Noe-λt

Differentiating the above relation with respect to time we have
A = $$\frac{d N}{d t}=\frac{d}{d t}$$Noe-λt = – Noλe-λt

Differentiating the above equation with respect to time we have

Therefore the Instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of Its half Life.

Question 4.
(a) Deduce the expression N = Noe-λt the law of radioactive decay.
(b) (i) Write symbolically the process expressing the β+ decay of, 1122Na Also write the basic nuclear process underlying this decay.
(ii) Is the nucleus formed in the decay of the nucleus 1122Na an Isotope or isobar? (CBSE Delhi 2014)
(a) Let N0 be the number of nuclei present in a freshly separated sample of a radioactive substance. Let after time t the number of nuclei left behind be N. Let dN number of nuclei disintegrate in a small time interval dt. Then by the – decay law,

where λ  is a constant of proportionality.

Question 5.
(a) Complete the following nuclear reactions:

(b) Write the basic process Involved in nuclei responsible for (i) β and (ii) β+ decay.
The basic nuclear process underlying β decay is the conversion of the neutron to proton
n → p + e + v
while for v+ decay, it is the conversion of a proton into a neutron
p → n + e+ + v

(c) Why is it found experimentally difficult to detect neutrinos? (CBSE AI 2015 C)
Neutrinos are neutral particles with very small (possibly, even zero) mass compared to electrons. They have only weak interaction with other particles. They are, therefore, very difficult to detect, since they can penetrate a large quantity of matter (even earth) without any interaction.

Question 6.
(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (B.E./A) versus the mass number A.
For the graph

From the plot, we note that

• During nuclear fission: A heavy nucleus in the larger mass region (A > 200) breaks into two middle-level nuclei, resulting in an increase in B.E./ nucleon. This results in the release of energy,
• During nuclear fusion: Light nuclei in the lower mass region (A < 20) fuse to form a nucleus having higher B.E. / nucleon. Hence Energy gets released.

(b) A radioactive Isotope has a half-life of 10 years. How long will It take for the activity to reduce to 3.125%? (CBSE AI2018)
3.125% means that the number of nuclei decays to 1/32 of its original value.
Therefore,
$$\frac{N}{N_{0}}=\frac{1}{32}=\left(\frac{1}{2}\right)^{5}$$

Now we know that
N = N0$$\left(\frac{1}{2}\right)^{t / T}$$

Therefore we have
$$\left(\frac{1}{2}\right)^{5}=\left(\frac{1}{2}\right)^{t / T}$$

Therefore
t = 5T = 5 × 10 = 50 years

Question 7.
Group the following six nuclides into three pairs of (?) isotones, (ii) isotopes, and (iii) isobars: 612C, 23He, 80198Hg, 13H, 79197Au 614C. How does the size of the nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.
(a) Isotones: 80198Hg, 79197Au
(6) Isotopes: 612C , 614C
(c) Isobars: 23He, 13H

The size of a nucleus depends upon its mass number as R = R0 A1/3

The nuclear density is given by the expression

The calculations show that the nuclear density is independent of the mass number.

Question 8.
Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10,000 disintegrations/s and 5,000 disintegration/s after 20 hr and 30 hr respectively from start. Calculate the half-life and an initial number of nuclei at t = 0. (CBSE Delhi 2019)
The decay constant of a radioactive element is the reciprocal of the time in which the number of its nuclei reduces to 1 /e of its original number.

We have R = λN
R(20hrs) = 100o0 = λN20
R(30hrs) = 5000 = λN30
$$\frac{N_{20}}{N_{30}}$$ = 2

This means that the number of nuclei, of the given radioactive nucleus, gets halved in a time of (30 – 20) hours = 10 hours
Half-life = 10 hours

This means that in 20 hours (= 2 half-Lives), the original number of nuclei must have gone down by a factor of 4.
Hence rate of decay at t = 0
λ N0 = 4 λ N20
R0 = 4 × 10,000 = 40,000 disintegration per second

Question 9.
(a) Write the relation between half-life and an average life of a radioactive nucleus.
The relation is τ = 1 .44T1/2

(b) In a given sample two isotopes A and B are initially present in the ratio of 1:2. Their half-lives are 60 years and 30 years respectively. How long will it take so that the sample has these isotopes in the ratio of 2:1? (CBSE Delhi 2019)

Question 10.
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction:

Using the data m(21H) = 2.014102 u,
m(13H) = 3.016949u, m(24He) = 4.002603 u,
mn = 1.008665 u,1 u = 931.5 MeV/c2 (CBSE Delhi 2015)
The distinction is shown in the table below.

 Nuclear Fission Nuclear Fusion 1. It is the splitting of a heavy nucleus into two or tighter unstable nuclei. 1. It is the combining of two light nuclei into a heavier nucleus. 2. It may or may not be a chain reaction. 2. It is always a chain reaction. 3. It is independent of temperature. 3. It is temperature-dependent. 4. It can be controlled. 4. It can’t be controlled. 5. Tremendous amount of energy is released. 5. Energy released per unit mass is seven times the energy released during fission. 6. By-products are harmful. 6. By-products are not harmful. 7. Example of reaction – The atom bomb. 7. Example of reaction – Reaction in stars, hydrogen born

In both reactions, there is a mass defect that is converted into energy.
Now energy released in the reaction

Question 11.
(a) Draw a plot showing the variation of the potential energy of a pair of nucleons as a function of their separation. Mark the regions where the nuclear force is (a) attractive and (b) repulsive.

For r > ro, the force is attractive For r < ro, the force is repulsive

(b) In the nuclear reaction

determine the values of a and b. (CBSE Delhi 2018 C)
We have,
1 + 235 = a + 94 + 2 × 1
∴ a = 236 – 96 = 140

Also
0 + 92 = 54+ 6 + 2 × 0
∴ b = 92 – 54 = 38

Question 12.
Binding energy per nucleon versus mass number curve is as shown. ZAS, Z1A1w, Z2A2X, and Z3A3Y, are four nuclei indicated on the curve.

Based on the graph:
(а) Arrange X, W, and S in the increasing order of stability.
(a) S, W, and X

(b) Write the relation between the relevant A and Z values for the following nuclear reaction. S → X + W
The equation is
Z = Z1 + Z2 and A = A1 + A2

(c) Explain why binding energy for heavy nuclei is low. (CBSE Sample Paper 2018-19)
Reason for low binding energy: For heavier nuclei, the Coulomb repulsive force between protons increases considerably and offsets the attractive effects of the nuclear forces. This can result in such nuclei being unstable.

Question 13.
(a) Derive the law of radioactive decay,
viz. N = Noe-λt
Let N0 be the number of nuclei present in a freshly separated sample of a radioactive substance. Let after time t the number of nuclei left behind be N. Let dN number of nuclei disintegrate in a small time interval dt. Then by the – decay law,

where λ is a constant of proportionality.

(b) Explain, giving necessary reactions, how energy is released during
(i) fission
Nuclear Fission: It is a process in which a heavy nucleus splits up into two Lighter nucLei of nearly equal masses. It is found that the sum of the masses of the product nuclei and particles is less than the sum of the masses of the reactants, i.e. there is some mass defect. This mass defect appears as energy. One such fission reaction is given below:

The Q value of the above reaction is about 200 MeV. The sum of the masses of Ba, Kr, and 3 neutrons is less than the sum of the masses of U and one neutron.

(ii) fusion
Nuclear Fusion: It is the process in which two light nuclei combine together to form a heavy nucleus. For fusion very high temperature of l is required. One such fusion reaction is given below:

The Q value of this nuclear reaction is 24 MeV. It is the energy equivalent of the mass defect in the above reaction. The energy released in fusion is much less than in fission but the energy released per unit mass infusion is much greater than that released in fission.

Question 14.
(a) Distinguish between isotopes and isobars, giving one example for each.
(b) Why is the mass of a nucleus always less than the sum of the masses of its constituents? Write one example to justify your answer.
Or
(a) Classify the following six nuclides into (i) isotones, (ii) Isotopes, and (iii) isobars: (CBSEAI2019)

(b) How does the size of a nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.
(a) Isotopes have the same atomic number while isobars have the same mass number
Examples of isotopes 612C, 614C
Examples of isobars 23He, 13H

(b) Mass of a nucleus is less than its constituents because in the bound state some mass is converted into binding energy which is energy equivalent of mass defect e.g., the mass of 1860 nucleus is less than the sum of masses of 8 protons and 8 neutrons
Or
(a) Isotones: 80198Hg, 79197Au
(6) Isotopes: 612C , 614C
(c) Isobars: 23He, 13H

(b) The radius of the nucleus is given by
R = RoA1/3

Volume of the nucleus $$\frac{4}{3}$$πR3 = $$\frac{4}{3}$$πRo3 A

If m is the average mass of the nucleon then the mass of the nucleus M = mA
Hence nuclear density

Which is independent of the A i.e., the size of the nucleus.

Numerical Problems:

• Radius of the nucleus R = RoA1/3
• Mass defect, Δm = Z mp + (A – Z) mn– M
• Energy released ΔE = Δm × 931 MeV
or
ΔE = [Z mp + (A – Z) mn – M] × 931 MeV
• Binding energy per nucleon BE/N = ΔE/A
• Relation between original nuclei (N) and nuclei left (No) after time t N = Noe-λt
• Relation between decay constant (λ) and half-life (T) = $$\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}$$
• Half-life is also given by the expression N = No$$\left(\frac{1}{2}\right)^{n}$$ where n = t/T
• The average life is given by
τ = $$\frac{1}{\lambda}=\frac{T}{\ln 2}=\frac{T}{0.693}$$ = 1.44 T
• Activity is given by A = -λN = $$\frac{0.693N}{T}$$

Question 1.
Calculate the binding energy per nucleon of Fe5626 Given mFe = 55.934939 u, mn = 1.008665 u and mp = 1.007825 u
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z mp + (A – Z)mn – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.

Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 1023 nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10-3 g are

Question 4.
The half-life of 238 92U is 4.5 × 109 years. Calculate the activity of 1 g sample of 92238U.
Given T = 4.5 × 109 years.
Number of nuclei of U in 1 g
= N = $$\frac{6.023 \times 10^{23}}{238}$$ = 2.5 × 1021

Therefore activity

Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Given λ = 0.3456 day-1
or
T1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = Noe-λt
or
N = No e-0 3456 × 4
or
N = N0 e-1 3824 = No × 0.25

Therefore the percentage of undecayed is

Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Given N/No = 6.25 %, t = 16 days, λ = ?

Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day-1

Question 7.
A radioactive substance decays to 1/32th of its initial value in 25 days. Calculate its half-life.
Given t = 25 days, N = No / 32, using

Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Given T1/2 = 30 s, N = 3No / 4, λ = ?, t = ?
(i) Decay constant
λ = $$\frac{0.693}{T_{1 / 2}}=\frac{0.693}{30}$$ = 0.0231 s-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Using N = Noe-λt we have

Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given Nox = NoY, TX = 1 h, TY = 2 h, therefore
$$\frac{\lambda_{X}}{\lambda_{Y}}=\frac{2}{1}$$ = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are Rx and Ry respectively, then

Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.

The mass of the star is 5 × 1032 kg and it generates energy at the rate of 5 × 1030 watt. How long will it take to convert all the helium to carbon at this rate?

As 4 × 10-3 kg of He consists of 6.023 × 1023 He nuclei so 5 × 1032 kg He will contain
$$\frac{6.023 \times 10^{23} \times 5 \times 10^{32}}{4 \times 10^{-3}}$$ = 7.5 × 1058 nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10-13 J of energy
So all nuclei in the star will produce
E = $$\frac{7.27 \times 1.6 \times 10^{-13}}{3}$$ × 7.5 × 1058
= 2.9 × 1046 J

As power generated is P = 5 × 1030 W, therefore time taken to convert all He nuclei into carbon is
t = $$\frac{E}{P}=\frac{2.9 \times 10^{46}}{5 \times 10^{30}}$$ = 5.84 × 1015 s
or
1.85 × 108 years

Question 11.
Radioactive material is reduced to (1/16)th of its original amount in 4 days. How much material should one begin with so that 4 × 10-3 kg of the material is left after 6 days?
N = No / 16, t = 4 days,
N = 4 × 10-3 kg,
t = 6 days

To calculate half-life of the material we have

Now using the expression 4 × 10-3 = No$$\left(\frac{1}{2}\right)^{6 / 1}$$
Solving we have No = 0.256 kg

Question 12.
Two different radioactive elements with half-lives T1 and T2 have N1 and N2 (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant.
The activity of a radioactive sample is given by the relation
A = – λN

Therefore the ratio of activity of these two radioactive elements is
$$\frac{A_{1}}{A_{2}}=\frac{-\lambda_{1} N_{1}}{-\lambda_{2} N_{2}}=\frac{T_{2} N_{1}}{T_{1} N_{2}}$$

Question 13.
Given the mass of the iron nucleus as 55.85 u and A = 56. Find the nuclear density? (NCERT)
Given mFe = 55.85 u = 9.27 × 10-26kg

The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these Neutron stars has been compressed to such an extent that they resemble a big nucleus.

Question 14.
We are given the following atomic masses: 92238U = 238.05079 u, 24He = 4.00260 u, 90234Th = 234.04363 u 11H = 1.00783 u, 91237Pa =237.05121 u Here the symbol Pa is for the element protactinium (Z = 91). (a) Calculate the energy released during the alpha decay of 92238U. (b) Show that cannot spontaneously emit a proton. (NCERT)
(i) The alpha decay of 92238Uis given by

The energy released in this process is given by
Q= (Mu – MTh – MHe) × 931.5 MeV

Substituting the atomic masses as given in the data we find that
Q = (238.05079 – 234.04363 – 4.00260) × 931.5 MeV ⇒ Q = 4.25 MeV.

(ii) If 29®U spontaneously emits a proton, the decay process would be

The Q for this process to happen is Q = (Mu – Mpa – MH) × 931.5 MeV
Q = (238.05079 – 237.05121 – 1.00783) × 931.5 MeV ⇒ Q = – 7.68 MeV

Thus the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to the 92238U nucleus to make it emit a proton.

Question 15.
The half-life of 90Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? (NCERT)