Category: CBSE Class 12

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, \(\pm \frac{1}{2}\)

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 = [a_ij] whose elements atj are given by : \(\frac{|i-j|}{2}\)
Solution:
We have [a_ij] = \(\frac{|i-j|}{2}\)
∴ 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))
Answer:
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 + A^T = √2I₂
Where A^T 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

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.
Answer:
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.
Answer:
The characteristics are as shown.

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

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

Question 5.
An ac input signal of frequency 60 Hz is rectified by an
(i) Half wave and an
Answer:
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.
Answer:
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.
Answer:
The ratio is one.

Question 7.
What is the depletion region in a p-n junction?
Answer:
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
Answer:
Boron, aluminum, etc.

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

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

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

Answer:
Forward biasing.

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

Question 12.
What is the function of a photodiode? (CBSE AI 2013C)
Answer:
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)
Answer:
Potential barrier decreases in forwarding bias.

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

Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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, λ_m, that can be detected by it?
Answer:
E_g = \(\frac{h c}{\lambda_{m}}\)

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
Answer:
Junction width will be small and

(ii) Junction field?
Answer:
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)
Answer:
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.
Answer:
The diagram is as shown.

Question 2.
What is a solar cell? How does it work? Give one of its uses.
Answer:
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)

Answer:
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.

Answer:
(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)
Answer:
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.

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.

Answer:
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.

Answer:
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)

Answer:
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)

Answer:
(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)

Answer:
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)
Answer:
(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 = neAv_d
Or
v_d = \(\frac{1}{n e A}\) ⇒ v_d ∝ \(\frac{1}{n}\) (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)
Answer:
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.
Answer:
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.
Answer:
(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)
Answer:
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 = V_z 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)
Answer:
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 E_A 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 E_A is slightly above the top E_v of the valence band. With the very small supply of energy, an electron from the valence band can jump to the level E_A 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.
Answer:
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.
Answer:
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)
Answer:
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)
Answer:
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 (E_g) 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)
Answer:
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.
Answer:
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 (E_g) 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?
Answer:
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.
Answer:
The diode used is Zener diode

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

(c) Briefly explain one use of this device.
Answer:
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.
Answer:
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.
Answer:
(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.
Answer:

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

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 l_z so that the total current (l_L + l_z) remains constant.

(c) Draw its l-V characteristics. (CBSE 2019C)
Answer:
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)
Answer:
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 (h_w) greater than the energy gap (E_g) 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)
Answer:
The circuit diagram is as shown.

The two ends S₁ and S₂  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 R_L. The input is applied across the primary and the output is ob¬tained across the load resistance R_L. 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 R_L.

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 R_L.

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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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 (V_i) is applied to the Zener diode (whose breakdown voltage is V_z as shown in the figure. If the applied voltage V_i > V_z, then the Zener diode is in the breakdown condition. As a result of a wide range of values of load (R_L), 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.
Answer:
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 h_v > E_g). 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)
Answer:
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)
Answer:
When the photodiode is illuminated with light (photons) with energy (h_v) greater than the energy gap (E_g) 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)
Answer:
(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)
Answer:

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.
Answer:
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)
Answer:
(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)
Answer:
(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)
Answer:
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.
Answer:
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)
Answer:
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)
Answer:
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, D_2, 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)
Answer:
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)

Answer:
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.

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.
Answer:
The figure is as shown.

Question 2.
How would you explain the propagation of light on the basis of Huygen’s wave theory?
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
The wavefront is as shown.

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

Question 8.
State the conditions which must be satisfied for two light sources to be coherent.
Answer:
(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.
Answer:
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.
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
Heatwaves are transverse In nature.

Question 14.
Define the term ‘wavefront’. (CBSE AI 2014C)
Answer:
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)
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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)
Answer:
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?
Answer:
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?
Answer:
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)
Answer:
Using the expression
μ = tan i_p = 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?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
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)
Answer:
Given μ = 1.5, i_P = ?
Using the relation μ = tan i_P we have
l_p = tan^-1 (1.5) = 56.3°

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

Question 30.
What is the shape of the wavefront on earth for sunlight? (NCERTExemplar)
Answer:
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)
Answer:

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)
Answer:
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?
Answer:

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)
Answer:
(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)
Answer:
When the reflected ray and the refracted ray are perpendicular then μ = tani_p where i_p 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.
Answer:
Let l_o 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 = l_ocos 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 = l_o cos² θ cos² (90° – θ) = l_o cos² θ sin² θ = (l_o /4)sin² 2θ

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

Question 6.
Is energy conserved in interference? Explain.
Answer:
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 l_max = 4a² and intensity at the minima l_min = 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.
Answer:
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?
Answer:
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.
Answer:
(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 i_p = 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)
Answer:
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)
Answer:
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 = l_o 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)
Answer:
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)
Answer:
Polaroid sunglasses are preferred over coloured sunglasses because they reduce the intensity of light.

(b) How is plane polarised light defined?
Answer:
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.
Answer:
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)
Answer:
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 1^st 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.
Answer:

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

(b) A ray of light is incident at an angle of incidence i_p 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 i_p and µ.
Answer:
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 i_p.

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
i_p + 90° + r = 180°
or
r = 90° – i_p.

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

Now sin r = sin (90° – i_p) = cos i_p, 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)
Answer:
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
y_o = 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

y_m = \(\frac{mDλ}{d}\) …….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.
Answer:
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)
Answer:
Let l_o 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’ = l_o cos² 30° = l_o × 3/4 = 3l_o/4

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

Question 10.
(a) Why are coherent sources necessary to produce a sustained interference pattern?
Answer:
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)
Answer:
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 = l_o
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)
Answer:
(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 θ_n.

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.
Answer:
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 E_o. 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)
Answer:
The intensity of light passing through P₁ is half of the light falling on it. Therefore, the light coming out of P₁ is l0_o/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)
Answer:
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)
Answer:
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 = l_o 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)
Answer:
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.
Answer:
(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)
Answer:
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.
Answer:
The ray diagram is as shown.


(b) Draw the sketches to differentiate between plane wavefront and spherical wavefront. (CBSE AI 2016)
Answer:
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)
Answer:
(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 P_1. 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)
Answer:
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.
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.
Answer:
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
l_A = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_A = l + l + 2\(\sqrt{l l}\)cos Φ = 4l

Now intensity at the screen when the phase difference is Φ = 90° Is
l_B = l₁ +l₂ + 2\(\sqrt{l_{1} l_{2}}\)cos Φ
or
l_B = 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?
Justify your answer In each case.
Answer:
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)
Answer:
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)
Answer:
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.
Answer:
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 y_n
so y_n = 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 = y_n – y_n-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)
Answer:
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 = 4l_o cos² Φ/2, where l_o = a².
Answer:
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)
Answer:
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.
Answer:
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)
Answer:
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.
Answer the following questions:
(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.
Answer:

(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)
Answer:
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) Good quality sun-glasses made of polaroids are preferred over ordinary coloured glasses. Justify your answer,
Answer:
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)
Answer:
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 = l_o 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.
Answer:
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) l_o = 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.
Answer:
The observation \(\frac{l_{0}}{2 \sqrt{2}}\) is not correct. It should be l_o/4.

(b) Define the Brewster angle and write the expression for it in terms of the refractive index of the medium.
Answer:
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 i_p = 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.
Answer:
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 i_c = sin^-1 (ωt i_B)
Answer:
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)
Answer:
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)
Answer:
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 (i_p).

Question 33.
Answer the following questions:
(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?
Answer:
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?
Answer:
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?
Answer:
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)
Answer:
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.
Answer:
(a) The intensity of the linearly polarised light would be l_o/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)
Answer:
We have l = l_o 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.
Answer:
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?
Answer:
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)
Answer:
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 = l_o 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.
Answer:
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
y_o0 = 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

y_m = \(\frac{mDλ}{d}\) …….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.
Answer:

(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.
Answer:
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;
Answer:
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;
Answer:
The separation of the fringes (and also angular separation) decreases.

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

(d) the source slit is moved closer to the double-slit plane;
Answer:
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;
Answer:
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.)
Answer:
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= λ_b /2, where (λ_b = 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 = λ_b = λ_r/2 where λ_r (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?
Answer:
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.
Answer:
(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 = C_τ

(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)
Answer:
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.
Answer:
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
y_o = 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

y_m = \(\frac{mDλ}{d}\) …….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)
Answer:
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 (3^th) maxima for wavelength λ₁ is just coinciding with (n + 1)^th (4^th) 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)
Answer:
(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 l_o/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 = (l_o/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 3^rd order maxima on one side of the screen. The distance between the slit and the screen is 1.5 m. (CBSE Delhi 2019)
Answer:
(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 = X_D1 = λ_D/a

Distance of third order maxima from centre of the central maxima X_D3 = 7λ_D/2a

Distance between first order minima and third order maxima = X_D3 – X_D1

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.
Answer:
Consider a point P on the viewing screen; the screen is Located at a perpendicular distance D from the screen containing the slits S_D1 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
y_o = 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

y_m = \(\frac{mDλ}{d}\) …….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.
Answer:
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)
Answer:
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?
Answer:
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,
Answer:

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

(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)
Answer:
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)

Answer:
(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 l_o/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 = l_o/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.
Answer:
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
y_o = 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

y_m = \(\frac{mDλ}{d}\) …….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)
Answer:
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.
Answer:
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 1^st 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.
Answer:
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)
Answer:
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?
Answer:
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?
Answer:
(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\(\frac{Φ}{2}\) – 4l_o cos²\(\frac{Φ}{2}\)
Since Φ = 0, l.e. there is no phase difference, hence l = 4 l_o

Inset B, the intensity will be given by the average intensity
l = 4l_o\(\left(\cos ^{2} \frac{\phi}{2}\right)\) = 2l_o
Or
(a) Let P₁ and P₂ be the two crossed Polaroids and P₃ be the polaroid kept between the two. Let l_o, 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 = l_o 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 i_B.

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

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

Now sin r = (90° – i_B), 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 θ_p
• Intensity of light coming out of a polariser l = l_o 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 l_max 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)
Answer:
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)
Answer:

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)
Answer:
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)
Answer:

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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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°.
Answer:
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)
Answer:
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
θ_max = 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
θ_min = 6.397°

Therefore, angular separation is
θ_min – θ_max = 6.397 – 1.047 = 5.35°

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)
Answer:
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)
Answer:
Nylon-6, 6

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

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

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

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

Question 7.
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymerisation reaction? (CBSE AI 2010)
Answer:
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)
Answer:
Buna – S < Polythene < Nylon 6.

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

Question 10.

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

Question 11.
Identify the Monomers of the following polymer:

Answer:
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
Answer:
PVC : CH₂ = CH – Cl (Vinyl chloride)

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

Question 2.
Mention two important uses of each of the following : (CBSE Delhi 2011)
(i) Bakelite
Answer:
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
Answer:
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
Answer:
The monomer of Nylon-6:

(ii) Polypropene (CBSE Delhi 2012)
Answer:
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
Answer:
Phenol and formaldehyde

(ii) Neoprene
Answer:
2-Chloro-1,3-butadiene(chloroprene)

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

(ii) Nylon-6, 6
Answer:
Hexamethylenediamine and adipic acid.

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

(ii) Buna-N
Answer:
1, 3-butadiene and acrylonitrile

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)
Answer:
(i) Nylon-6,6
Hexamethylenediamine: NH₂ – (CH₂)₆ – NH₂ and Adipic acid

(ii) Glyptal

(iii) Buna – S
CH₂ = CH — CH = CH₂
1, 3 Butadiene

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:

Answer:
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?
Answer:
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)
Answer:
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
Answer:
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
Answer:
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.

(iii) Addition polymers (CBSE 2012)
Answer:
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
Answer:
Buna-S:
1, 3-Butadiene: CH₂ = CH – CH == CH₂
Styrene: C₆H₅CH = CH₂

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

(iii) Nyton-6, 6 (CBSE Delhi 2013)
Answer:
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
Answer:
Bakelite:

• Phenol:
  
• Formaldehyde: HCHO

(ii) Nylon-6
Answer:
Nylon-6:
Caprolactam

(iii) Polythene (CBSE Delhi 2013)
Answer:
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)
Answer:
(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
Answer:
Chloroprene CH₂ = C — CH = CH₂

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

(c) PVC (CBSE 2019C)
Answer:
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)
Answer:
(a) Natural rubber: 2-methyl-1,3-butadiene

(b) PVC: vinyl chloride
CH₂ = CH – Cl

(c) Nylon-6,6:
HOOC-(CH₂)₄ – COOH (Adipic acid)

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)
Answer:
(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)
Answer:
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
Answer:
Polystyrene: Styrene:

(ii) Dacron
Answer:
Dacron:

• Ethylene glycol: HOCH₂CH₂OH
• Terephthalic acid:
  

(iii) Teflon
Answer:
Teflon:
Tetrafluoroethene: F₂C = CF₂

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

(b) Terylene
Answer:

(c) Buna-N
Answer:

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)
Answer:
(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?
Answer:
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.
Discuss the mechanism of free radical addition polymerisation.
Answer:
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.

The addition occurs as

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.
Answer:
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.
Answer:
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?
Answer:
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
Answer:
1, 3-Butadiene: CH₂ = CH — CH = CH₂
Styrene: C₆H₅CH = CH₂

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

(iii) Dacron
Answer:

• Ethylene glycol:
  
• Terephthalic acid:
  

(iv) Neoprene.
Answer:
Chloroprene:

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)
Answer:
butanone < propanone < propanal < ethanal

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

OR

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

Answer:
Pent-2-enal.

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

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

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

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

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

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

Question 11.
Write the IUPAC name of (CBSE 2012)
Ph — CH = CH — CHO.
Answer:
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)
Answer:
CH₃CH₂CH₃ < CH₃CHO < CH₃CH₂OH

Question 13.
Ethanol is soluble in water. Why? (CBSE 2013)
Answer:
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)

Answer:
3-Hydroxybutanoic acid

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

Answer:

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

Answer:
3-Aminobutanal

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

Answer:
But-3-en-2-one

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

Answer:
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)

Answer:

Answer:

Question 2.
What is Tollen’s reagent? Write one use of this reagent. (CBSE 2010)
Answer:
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)

Answer:
Lithium aluminium hydride, LiAlH₄

Answer
Alkaline potassium permanganate (KMnO₄), KOH

Question 4.
Distinguish between
C₆H₅CH = CH – COCH₃ and C₆H₅CH = CH CO CH₂ CH₃ (CBSE AI 2016)
Answer:
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)
Answer:
C₆H₅CHO < CH₃CHO < HCHO

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

Answer:

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

(b) C₆H₅COCl
Answer:
C₆H₅CHO

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

Answer:
Lithium aluminium hydride, LiAlH₄

Answer:
Alkaline potassium permanganate (KMnO₄), KOH

Answer:
CH₃ Mg Br, H₃O⁺

Answer:
Cl₂, P (Hell-Volhard-Zelinsky reaction)

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

Answer:

Answer:

Answer:

Answer:

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

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

(iii) Ethylcyanide to ethanoic acid
Answer:

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

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

Answer:

Question 12.
Aromatic carboxylic acids do not undergo Friedel Crafts reaction. Explain. (CBSE Al 2018)
Answer:
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.
pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain. (CBSE AI 2018)
Answer:
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 pK_a 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)
Answer:



Or

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

Answer:

Answer:

Answer:

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

Answer:

Answer:

Answer:

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)
Answer:
Salicylic acid is treated with acetic anhydride In the presence of H⁺ to give aspirin.

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

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

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

(ii) 4-Chlocopentan-2-one
Answer:

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

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

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

(ii) Propanal to butanone
Answer:
Propanal to butanone

(iii) Benzaldehyde to benzophenone
Answer:
Benzaldehyde to benzophenone

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

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

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

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

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

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

(iv) Acetaldehyde to formaldehyde.
Answer:
Acetaldehyde to formaldehyde

(v) Formaldehyde to acetaldehyde.
Answer:
Formaldehyde to acetaldehyde

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

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Sample Paper 2017 – 2018)
Answer:

(CBSE Delhi 2013)
Answer:

(CBSE AI 2013)
Answer:

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.
Answer:
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
Answer:

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

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

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

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)
Answer:

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)
Answer:

(CBSE Delhi 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2015)
Answer:

(CBSE AI 2017)
Answer:

(CBSE AI 2017)
Answer:

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

Answer:

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)
Answer:
(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)
Answer:
(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)
Answer:

(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)

Answer:
(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)
Answer:
(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

Answer:

(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)
Answer:

(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
Answer:
(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)
Answer:
(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
Answer:
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
Answer:
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)

Answer:

Answer:

Answer:

Question 23.
(i) Give chemical tests to distinguish between the following:
Benzoic acid and ethyl benzoate.
Answer:
(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:

Answer:

Answer:

(CBSE 2011)
Answer:

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)
Answer:

(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 pK_a 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)
Answer:

(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)
Answer:

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 pK_a 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) pK_a 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
Answer:
(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

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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
Four.

Question 5.
Name two poisonous gases which can be prepared from chlorine gas. (CBSE AI 2013)
Answer:

1. Phosgene (COCl₂)
2. Mustard gas (ClCH₂CH₂SCH₂CH₂Cl)

Question 6.
What is the basicity of H₃PO₃ and why? (CBSE AI 2013)
Answer:
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)
Answer:
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)
Answer:
+1

Question 9.
Out of white phosphorus and red phosphorus, which one is more reactive and why? (CBSE 2015)
Answer:
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)
Answer:
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)
Answer:
NO₂.

2NO₂ → N₂O₄

Question 12.
Solid PCl₅ is ionic in nature. Give reason. (CBSE AI 2016)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
Molybdenum.

Question 16.
NF₃ is an exothermic compound whereas NCl₃ is not. Explain. (CBSE AI 2011, 2012)
Answer:
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)
Answer:
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)
Answer:
Phosphoric acid and phosphine are formed.

Question 19.
Write the order of thermal stability of the hydrides of group 16 elements. (CBSE Delhi 2017)
Answer:
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)
Answer:
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)
Answer:
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.
Answer:
H₂SO₃

Question 23.
Ozone is thermodynamically unstable. Explain. (CBSE Sample Paper 2011)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
XeO₃ is formed.
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 28.
Which xenon compound is isostructural with lCl₄^–? (CBSE Sample Paper 2011)
Answer:
XeF₄

Question 29.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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 →
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
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
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(i) P₄ + H₂O → no reaction
(ii) XeF₄ + O₂F₂ → XeF₆ + O₆

Question 16.
Draw the structures of the following:
(i) XeF₂
(ii) BrF₃
Answer:
(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)
Answer:
(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)
Answer:
(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
Answer:
(i)

(ii)

Question 20.
Write the structure of the following: (HPO₃)₃ (CBSE 2016)
Answer:
(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)
Answer:
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)
Answer:
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) S₂O₈^2- (CBSE AI 2019)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
(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)
Answer:
(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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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°.
Answer:
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)
Answer:
(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.
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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 (NO_x) 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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
Moist sulphur dioxide behaves as a reducing agent. It reduces MnO₄^– to Mn²⁺.
5 SO₂ + 2MnO₄^– + 2H₂O → 5 SO₄^2- + 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?
Answer:
(a) (i) Sodium chlorate and sodium chloride are formed.

(ii) Fe (III) salt is reduced to Fe (II) salt.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^{2- + 4H+}

(b) Answer the following:
(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)
Answer:
(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 ⇌ HPO₃^2- + 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.
Answer:
(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
Answer:
(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.
Answer:
(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)
Answer:
(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.
Answer:
(a) (i) In the vapour state, sulphur exists as S2 molecules. S2 molecule has two unpaired electrons in anti-bonding molecular orbitals (π^x and π^y) 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
Answer:
(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.
Answer:
(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.
Answer:
(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)
Answer:
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 →
Answer:
(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) SO₃^2- and the angle O – S – O
(ii) ClF₃ and the angle F – Cl – F
(iii) XeF₂ and the angle F – Xe – F
Answer:
(i) SO₃^2-

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 →
Answer:
(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)
Answer:
(i) H₃PO₂

(ii) H₂S₂O₇

Question 22.
(a) Draw the molecular structures of following compounds:
(i) XeF₆
(ii) H₂S₂O₈
Answer:
(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.
Answer:
(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 ) →
Answer:
(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)
Answer:
(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
Answer:
(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.
Answer:
(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
Answer:

(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)
Answer:
(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)
Answer:
(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)
Answer:
(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)
Answer:
(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.
Answer:
(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
Answer:
(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.
Answer:
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.
Answer:
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)
Answer:
(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.
Answer:
(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
Answer:
(i)

T – Shaped molecule

(ii)

Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
Answer:
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?
Answer:
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?
Answer:
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?
Answer:
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF₂ + PF₅ →
Answer:
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.
Answer:
(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
Answer:
(i)

(ii)

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.
Answer:
Phosphine, PH₃
P₄ + 3NaOH + 3H₂O → 3NaH₂PO₃ + PH₃

(b) Which noble gas has the lowest boiling point?
Answer:
Helium

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Answer:
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?
Answer:

(e) Complete the equation:
PbS + O₃ → (CBSE 2015)
Answer:
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).
Answer:
(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).
Answer:
(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?
Answer:
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)
Answer:
(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)
Answer:
(i) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(ii) 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 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.
Answer:
(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
Answer:
(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?
Answer:
(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
Answer:
HI > HBr > HCl > HF

Complete the following reaction:
XeF₄ + SbF₅ →, (CBSE 2018)
Answer:
(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?
Answer:
(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.
Answer:
(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.
Answer:
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)
Answer:
(a) 3Cu + 8 HNO₃ (dilute) → 3Cu(NO₃)₂ + 2NO + 4H₃O
(b) 2Fe³⁺ + SO₃ + 2H₃O → 2 Fe²⁺ + SO₄^2- + 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)
Answer:
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⁺ + 5SO₄^2-
Reduces Fe³⁺ to Fe²⁺
2Fe³⁺ + SO₂ + 2 H₂O → 2 Fe²⁺ + SO₄^2- + 4H⁺

OR

Answer the following:
(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength:
H₂O, H₂S, H₂Se, H₂Te
Answer:
H₂Te > H₂Se > H₂S > H₂O

(b) Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why?
Answer:
PCl^4- 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?
Answer:
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.
Answer:
HNO₃ oxidises phosphorus to phosphoric acid. H₃PO₄ is formed
P₄ + 2OHNO₃ → 4H₃PO₄ + 2ONO₂ + 4H₂O

(e) Why does PCl₃ fume in moisture?
Answer:
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

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)
Answer:
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)
Answer:
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?
Answer:
Cr²⁺

(ii) forms colourless aqueous solution?
Answer:
Sc³⁺

(iii) exhibits the most stable +3 oxidation state? (CBSE Sample Paper 2017-18)
Answer:
Sc³⁺

Question 4.
Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state:
E°_{Cr³⁺/Cr²⁺} = – 0.4 V, E°_{Mn³⁺/Mn²⁺} = 1.5 V, E°_{Fe³⁺/Fe²⁺} = 0.8 V (CBSE Sample Paper 2011)
Answer:
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).
Answer:
Manganese

(ii) Name the element which shows only +3 oxidation state. (CBSEAI2013)
Answer:
Scandium

Question 6.
Identify the oxoanion of chromium which is stable in an acidic medium. (CBSE Sample Paper 2017-18)
Answer:
Cr₂0₇^2-

Question 7.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (CBSE Delhi 2015)
Answer:
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)
Answer:
Misch metal

Question 9.
Identify the Ianthanoid element that exhibits a +4 oxidation state. (CBSE Sample Paper 2017-18)
Answer:
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)
Answer:
Cr₂0₇^2-

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.
Answer:
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)
Answer:
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.
Answer:
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)
Answer:
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 E^θ value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺. (CBSE AI 2008, 2010, 2018)
Answer:
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 E^θ 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^– →
Answer:
Cr₂O₇²⁺ + 6I^– + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O

(ii) MnO₄^– + NO₂^– + H⁺ → (CBSE Delhi 2012)
Answer:
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?
Answer:
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)
Answer:
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)
Answer:
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,
Answer:
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)
Answer:
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.
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g 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)
Answer:
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)
Answer:
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.
Answer:
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)
Answer:
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.
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
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)
Answer:
MnO₄^–: Oxidation state of Mn = +7 (equal to its group number)
CrO₄^2-: 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.
Answer:
Because Cr is more stable in the +3 oxidation state due to the t_2g³ 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)
Answer:
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^–+ 3S₂O₃^2-+ H₂O →
(ii) Cr₂O₇^2-+ 3Sn²⁺ + 14H⁺ →
Answer:


(ii) Potassium dichromate (C) is used as a powerful oxidizing agent in redox titrations in the laboratories.
OR
(i) 8MnO₄^–+ 3S₂O₃^2-+ H₂O → 8MnO₂ + 6SO₄^2- + 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)
Answer:
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)
Answer:
(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 ReCl₈^4- 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.
Answer:
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)
Answer:
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 ReCl₈^4- 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)
Answer:
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) Cr₂0₇^2- + 6Fe²⁺ + 14H⁺ →
(ii) 2Cr0₄^2- + 2H⁺ →
(iii) 2Mn0₄^– + 5C₂0₄^2- + 16H⁺ → (CBSE Delhi 2013)
Answer:
(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⁺ + Cr₂0₇^2- + 14H+ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O
(ii) 2CrO₄^2- + 2H+ → Cr₂O₇^2-+ H₂O
(iii) 2MnO₄ ^–+ 5C₂O₄^2– + 16H+ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 4.
Give reasons for the following:
(i) Transition elements and their compounds act as catalysts.
Answer:
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.
Answer:
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)
Answer:
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 S0₃^2- + 6 H⁺ →
Answer:
(i) Na₂Cr₂O₇ + KCl → K2Cr2O7 + 2 NaCl
(ii) 5 S0₃^2- + 2MnO₄^– + 6 H⁺ → 2Mn²⁺ + 3H₂O + 5SO₄^2-

(b) How does the colour of Cr₂O₇^2- change when treated with an alkali? (CBSE2019C)
Answer:
The orange color of Cr₂O₇^2- change to yellow due to the formation of chromate ion.
Cr₂O₇^2- + 2OH^– → 2Cr₂O₇^2-  + H₂O (Yellow)

Question 6.
(a) Write chemical equations involved in the preparation of KMn04 from Mn02.
Answer:
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄^2-+ 4H⁺ → 2MnO^–+ MnO₂ + 2H₂O (or any other correct equation of preparation)

(b) Actinoids show wide range of oxidation states. Why? (CBSE 2019C)
Answer:
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.
Answer:
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.
Answer:
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)
Answer:
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.
Answer:
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.
Answer:
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)
Answer:
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.
Answer:
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).
Answer:
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)
Answer:
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)
Answer the following:
(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)
Answer:
(i) Cr³⁺ because of the half-filled t_2g³ 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)
Answer:
(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)
Answer:
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)
Answer:
(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) 3Mn0₄^2- + 4 H⁺ →
Answer:
(i) 5NO₂– + 2MnO4^– + 6H+ → 2Mn²⁺ + 5NO₃^– + 3H₂O
(ii) 3MnO₄^2-+ 4H+ → 2MnO₄^– + MnO₂ + 2H₂O

(b) Name a member of the lanthanoid series which shows a +4 oxidation state. (CBSE 2019C)
Answer:
Cerium / Ce

Question 15.
Give reasons:
(i) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
Answer:
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.
Answer:
Due to the higher number of unpaired electrons.

(iii) Sc³⁺ is colorless in an aqueous solution whereas Ti³⁺ is colored. (CBSE 2018)
Answer:
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) →
Answer:
(a) 8MnO₄^– (aq) + 3S₂O₃2-  (aq) + H₂O (l) → 8MnO₂ (s) + 6SO₄2- (aq) + 2OH- (aq)
(b) Cr₂O₇^2- (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).
Answer:
(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.
Answer:
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)
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g 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) + MnO^–4 (aq) + H⁺ (aq) →
(b) Cr₂O₇^2- (aq) + I^– (aq) + H⁺ (aq) →
Answer:
(a) 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) → Mn²⁺ (aq) + 5Fe³⁺ (aq) + 4H₂O(l)
(b) Cr₂O₇^2- (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.
Answer:
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.
Answer:
Cr²⁺ is reducing as its configuration changes from d⁴ to d³, the latter having a half-filled t_2g 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)
Answer:
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) + S₂O₃^2- (aq) + H₂O(l) →
(b) Cr₂O₇^2- (aq) + Fe²⁺ (aq) + H⁺ (aq) →
Answer:
(a) 8MnO₄^– + 3S₂O₃^2- + H₂O → 8MnO₂ + 6SO₄^2-+ 2OH-
(b) Cr₂O₇^2- + 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.
Answer:
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.
Answer:
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)
Answer:
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) 2CrO₄^2- + 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)
Answer:
(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) 2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + 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)
Answer:
(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)
Answer:
(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⁺ + Cr₂O₇^2- + 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) Cr₂O₇^2- + 14H⁺ + 6I^–
Or
The elements of 3d transition series are given as:
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(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)
Answer:
(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) Cr₂O₇^2- + 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.