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NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:

1. work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket,
2. work done by the gravitational force in the above case,
3. work done by friction on a body sliding down an inclined plane,
4. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
5. work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer:
1. We know that: W = \(\overrightarrow{\mathrm{F}}, \overrightarrow{\mathrm{S}}\) = FS cos θ
‘Positive’
Reason: Force is acting in the direction of displacement (θ = 0°)
2. ’Negative’
Reason: Force is acting in the opposite direction to displacement (θ =180°)
3. ’Negative’
Reason: Force of friction is opposite to the displacement (θ = 180°)
4. ‘Positive’
Reason: The body mover in the direction of force applied (θ = 0°)
5. ‘Negative’
Reason: The resistive force opposes the motion (θ = 0°)

Question 2.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
(a) work done by the applied force in 10 s
(b) work done by friction in 10 s
(c) work done by the net force on the body in 10 s
(d) change in kinetic energy of the body in 10 s and interpret your results.
Answer:
Here M = 2 kg; u = 0; μ= 0.1; applied force, F = 7N, t = 10s
Force of friction, f= μMg = 0.1 x 2 x 9.8 = 1.96 N
.’. net force under which body moves, F’ = F- f = 7 – 1.96 = 5.04 N

Question 3.
Given in Fig. are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.


Answer:
(a) We know that Total energy E = KE + PE, kinetic energy can never be negative. In the region between x = 0 & x = a.
Potential energy is ‘0’. So, kinetic energy y is positive. In region x > a the potential energy has a value greater than ‘E’. So kinetic energy will be negative in this region. Hence the particle cannot be present in the region x > a.

(b) Here PE > E, the total energy of the object and as such the kinetic energy of the object would be negative. Thus object cannot be present in any region on the graph.

(c) Here x = 0 to x = a & x > b, the P E is more then E so, K E is negative. The particle cannot be present in these portions.

(d) The object cannot exist in the region between
x = \(\frac{-b}{2}\) to x = \(\frac{-a}{2}\) & x = \(\frac{-a}{2}\) to x = \(\frac{-b}{2}\)
Because in this region P E > E.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by V (x) = Kx² /2, where A is the force constant of the oscillator. For A = 0.5 N nr¹, the graph of V (x) versus x is shown in Fig. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Answer:

Question 5.
Answer the following :
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere ?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?
(d) In Fig.

1. The man walks 2 m carrying a mass of 15 kg on his hands. In Fig.
2. He walks the same distance pulling the rope behind him. The rope goes over a, pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater ?
   

Answer:
(a) Heat energy required for the burning of the casing of a rocket is obtained by a rocket. Since the work is done against the friction, the kinetic energy of the rocket decreases continuously and this work against friction reappears as heat energy.

(b) This is because of the conservative nature of the gravitational force. Work done by the gravitational force in a closed path is zero.

(c) As an artificial satellite gradually loses its energy due to dissipation against atmospheric resistance, its potential decreases rapidly. As a result, the kinetic energy of the satellite slightly increases i.e. its speed increases progressively.

(d) In fig (i) the force applied by the man is perpendicular to the direction of movement of mass, i.e. θ = 90°
W = Fs cos θ = Fs cos 90° = 0
In figure (ii) the force applied is in direction of the movement of mass i.e. θ = o°
∴ W = Fs cos θ
= Mgs cos θ
= 15 × 9.8 × 2 × 1
= 294 J.

Question 6.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/ remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
The rate of change of total momentum of many-particle system is proportional to the external force/sum of the internal forces on the system.
(c) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy total linear momentum!total energy of the system of two bodies.
Answer:
(a) Work done by conservative force is equal to the negative of potential energy. When work done is positive, potential energy decreases.
(b)  Kinetic energy, because friction does work against motion of the body.
(c) External force, because in many-particle systems, the internal forces in the system cancel each other and hence cannot change the net momentum of the system.
(d) In inelastic collision, total energy and linear momentum are conserved. However, kinetic energy decreases.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.

1. In an elastic collision of two bodies, the momentum and energy of each body are conserved.
2. The total energy of a system is always conserved, no matter what internal and external forces on the body are present.
3. Work done in the motion of a body over a closed loop is zero for every force in nature.
4. In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Answer:

1. False, the momentum and energy of each body are conserved.
2. False, the external force on the system may increase or decrease the total energy of the system.
3. False, for the nonconservative forces (friction) the work done in closed-loop is not zero.
4. True, usually in an inelastic collision the final kinetic energy is always less than the initial kinetic energy of the system.

Question 8.
Answer carefully, with reasons :
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e., when they are in contact) ?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls ?
(c) What are the answers to (a) and (b) for an inelastic collision ?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic ? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Answer:
(a) In this case total kinetic energy is not conserved because when the bodies are in contact during elastic collision even, the kinetic energy is converted into internal energy.
(b) Yes, because total momentum conserves as per law of conservation of momentum.
(c) In inelastic collision, K.E. is not conserved but linear momentum is conserved.
(d) It is a case of elastic collision because in this case the forces are of conservative nature

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(1) t^m
(2) t
(3) t^3/2
(4) t²
Answer:

Question 10.
A body is moving uni-directionally under the influence of a source of constant power. Its displacement in time t is proportional to
(1) t^m
(2) t
(3) t^3/2
(4) f²
Answer:

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by \( \vec { F }= \) \( \hat {i}+{ 2j }+{ 3K }\)
where \( \hat {i}+{ j }+{ K }\)are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?
Answer:

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 x 10^-31 kg, proton mass = l.67 x 10^-27 kg, 1 eV = 1.60 x 10^-19 J).
Answer:
Let _υe = Speed of electron
_υp = speed of the proton
mg = mass of electron
and mp = mass of proton

Question 13.
A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^-1?
Answer:

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 ms^-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision ? Is the collision elastic or inelastic ?
Answer:
In all types of collisions, momentum is conserved. Let us check the conservation of kinetic energy.As the wall is too heavy, the recoiling molecule produces no velocity in the wall. l%m is mass of the gas molecule and M is mass of wall, then total K.E. after collision,

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m³ in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump ?
Answer:

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless. table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following (Fig.) is a possible result after collision ?

Answer:
Collision is elastic, so K.E. of the system is conserved.

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.
Answer:

Since the collision is elastic, therefore A would come to rest and B would begin to move with the velocity of A to conserve the linear momentum.

Question 18.
The bob of a pendulum is released from a horizontal position A as shown in Fig. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point B, given that it dissipated 5% of its initial energy against air resistance ?
Answer:

P.E. of the bob at position A = mgh = m x 9.8 x 1.5
Since 5% of energy is lost when reach at B, so K.E. at the lowermost point

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the trolley’s floor at the rate of 0.05 kgs^_1. What is the speed of the trolley after the entire sandbag is empty ?
Answer:
The system of trolley and sandbag is moving at a uniform speed. Clearly, the system is not being acted upon by external force. If the sand leaks out, even when no external force acts. So there shall no change in the speed of the trolley.

Question 20.
A particle of mass 0.5 kg travels in a straight line with velocity υ = ax^3/2, where a = 5 m^-1/2 s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2m?
Answer:
Here m = 0.5 kg

Question 21.
The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?
(b) What is the kinetic energy of the air ?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m², υ= 36 km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced ?
Answer:
(a) Volume of wind flowing per sec = Aυ
Mass of wind flowing per sec = Aυp
Mass of air passing in time t = Aυpt

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass 0.5 m, 1000 times. Assume that the potential energy lost each time she lowers the mass is dissipated,
(a) How much work does she do against the gravitational force ?
(b) Fat supplies 3.8 x 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up ?
Answer:
Here m = 10 kg, h = 0.5 m, n = 1000
(a) Work done against gravitational force = W = n mgh
= 1000 x (10 x 9.8 x 0.5)
= 49000 J

Question 23.
A large family uses 8 kW of power,
(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?
(b) Compare this area to that of the roof of a house constructed on a plot of size 20 m x 15 m with a permission to cover upto 70%.
Answer:
(a) Energy incident per square meter = 200 W
Let A be the area needed to supply 8 kW
.’. Energy incident on area A = (200 A) W
Energy converted into useful electrical energy

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceding by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Answer:
By using law of conservation of momentum,
m₁u₁+ m₂u₂ ⁼ (m₁ +m₂)υ

Question 25.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig.). Will the stones reach the bottom at the same time ? Will they reach there with the same speed ? Explain. Given θ₁ = 30°,θ₂ = 60°, and h = 10 m, what are the speeds and times taken by the two stones?

Answer:
Here, K.E. at bottom = P.E. at top

Since vertical height of both planes is same, so they will reach the bottom with same speed. Acceleration of a body sliding down an inclined plane, a = g sin θ
Let t be the time taken by stone 1 to travel AB distance.

Question 26.
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^-1 as shown in Fig. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has negligible mass and the pulley is frictionless.

Answer:
From fig, R = mg cos θ
Force of friction, F = μR = μ mg cosθ
Net force on the block down the incline
= mg sinθ – F = mg sinθ – (a mg cosθ = mg (sinθ – μcos θ)
Distance moved, x = 10 cm = 0.1 m
In equilibrium, Work done = P.E. of stretched spring

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms^-1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?
Answer:
Potential energy of bolt = mgh = 0.3 × 9.8 × 3 = 8.82 J.
Since the bolt does not rebound, the while energy is converted into heat. Since the value of acceleration due to gravity is the same in all inertial systems, the answer will not change even if the elevator is stationary.

Question 28.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg bins on the trolley from one end to the other (10 m away) with a speed of 4 ms^-1 relative to the trolley in a direction opposite to the trolley’s motion and jumps out of the trolley. What is the final speed of the trolley ? How much has the trolley moved from the time the child begins to run ?
Answer:
Initial total momentum = p_i = (m₁ + m₂) u₁

Question 29.
Which of the following potential energy curves in Fig. cannot possibly describe the elastic collision of two billiard balls ? Here r is the distance between centers of the balls.

Answer:
During short time of collision, the kinetic energy converts into potential energy. Since potential energy of a system of two masses varies inversely as the distance between them i.e., as 1/r, all the potential energy curves except the one shown in fig (v) cannot describe an elastic collision.

Question 30.
Consider the decay of a free neutron at rest: n → p + e^–. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the (β-decay of a neutron or a nucleus (Fig.) [Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of (3-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin 1/2 (like e^–, p or n), but is neutral, and either massless or having an extremely small mass (compared to electron’s mass) and which interacts very weakly with matter. The correct decay process of neutron is : n —> p + e^– + v]

Answer:
If the decay of a neutron (inside the nucleus) into proton and electron is according to the given scheme, then the available energy in the decay must be carried by the electron coming out of the nucleus and therefore the emitted electrons should always possess a fixed value of kinetic energy. However the graph shows that the emitted electron can have any value of energy between zero and the maximum value. Therefore, the given decay mode cannot account for the observed continuous energy spectrum in the β decay.

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NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion

Question 1.
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed.
(b) a cork of mass 10 g floating on water.
(c) a kite skilfully held stationary in the sky.
(d) a car moving with a constant velocity of 30 km/h on a rough road.
(e) a high-speed electron in space far from all gravitating objects, and free of electric and magnetic fields.
Answer:
(a) (As the drop of rain is falling with constant speed, therefore, according to Newton’s first law of motion, the net force on the drop of rain is zero, or since v = u = constant, therefore, F =0.
(b) As the cork is floating on water, its weight is balanced by the upthrust due to water. Hence net force on cork is zero.
(c) Since kite is held stationary, in accordance with first law of motion, the net force on the kite is zero.
(d) Since car is moving with a constant velocity, the net force on the car is zero.
(e) Since the high speed electron in space is far from all gravitating objects and free of electric and magnetic fields, the net force on electron is zero.

Question 2.
A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble

1. during its upward motion,
2. during its downward motion,
3. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of say 45° with the horizontal direction? Ignore air resistance.

Answer:

1. When the pebble is moving upward the force acting on it is a gravitational force in a downward direction.
   F = mg = 0.05 × 10 = 0.5 N
2. Even in this case F = mg = 0.5 N in downward direction.
3. Since there is no force other than gravitational force acting on a pebble, during the whole process F = mg = 0.5 N. Note that the pebble moves in opposite direction because of its initial velocity. The situation remains the same for pebble thrown at an angle.

Question 3.
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg.
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h,
(c) just after it is dropped from the window of a train accelerating with I ms^-2,
(d) lying on the floor of a train which is accelerating with 1 ms^-2, the stone being at rest relative to the train. Neglect air resistance throughout.
Answer:
(a) In this case F = weight of stone = Mg = 0.1 x 9.8 = 0.98 N, (vertically downwards)

(b) Since no force acts on the stone due to the motion of the train,
∴ F = 0.98 N (vertically downwards)

(c) The stone will experience an additional force F’ (along horizontal) ie.,
F’ = Ma = 0.1 x 1= 0.1 N
As the stone is dropped, the force F’ no longer acts and the net force acting on the stone is
F = Mg = 0.98 N (vertically downwards)

(d) In this case, the weight of the stone is balanced by the normal reaction. The net force on the stone is given by F’ = Ma = 0.1 x 1 = 0.1 N (horizontally)

Question 4.
One end of a string of length is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed υ the net force on the particle (directed towards the center) is :

T is the tension in the string. [Choose the correct alternative].
Answer:
(i) The centripetal force necessary for the particle to move in a circular path is provided by the tension in the string. Hence net force on the particle is nothing but tension T in the string.

Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^-1. How long does the body take to stop?
Answer:
Here M = 20 kg, F = -50 N [retarding force]


Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms^-1 to 3.5 ms^-1 in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:

Mass of body, M = 3 kg
Force acting on the body, F = Ma = 3 x 0.06 = 0.18 N
Since the applied force increases the speed of the body; it acts in the direction of motion.

Question 7.
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
Answer:
Here, F₁ = 8 N and F₂ = 6 N
The magnitude of the resultant force acting on the body is given by

Question 8.
The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
Answer:

0 = 10 + 4a                                                              .                ‘
=>a = – 2.5 m s^-2
∴ Magnitude of retarding force = 465 x 2.5 = 1162.5 N

Question 9.
A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer:
Given m = 20000 kg
a = 5ms^-2 (against gravity)
since the rocket has to move upwards against gravity the total initial thrust of the blast is given by
F = ma + mg
= m (a + g) = 20000 (5 + 9.8)
= 296 × 10⁵ N.

Question 10.
A particle of mass 0-40 kg moving initially with a constant speed of 10 ms^–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the particle at that time to be x = 0, and predict its position at f = -5s, 25 s, 100 s.
Answer:
Here m = 0.40 kg, h = 10 ms^–1, F = – 8N (retarding force)

Question 11.
A truck starts from rest and accelerates uniformly with 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the
(a) velocity, and
(b) acceleration of the stone at t = 11s? (Neglect air resistance).
Answer:
(a) Here, u = 0, a = 2.0 ms^-2, t = 10 s, g = 9.8 m s^-2
Velocity along horizontal direction, υ_x = u + at = 0 + 2 x 10 = 20 ms^–1
After Is (11 s – 10 s), velocity along vertical direction, υ_y =u + gt = 0 + 9.8 X 1 = 9.8 m s^–1
Therefore, resultant velocity,

Question 12.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms^-1. What is the trajectory of the bob if the string is cut when the bob is
(1) at one of its extreme positions,
(2) at its mean position?
Answer:
1. When the bob is at one of its extreme positions its velocity is zero. Hence if the string is cut, it will fall straight down due to gravitational force.

2. At the mean position the bob has a horizontal velocity of 1 m/s. If the string is cut, bob is acted by vertical gravitational force = a = 9.8 ms^-2. Hence bob will behave like a projectile and follows a parabolic path.

Question 13.
A man of mass 70 kg stands on a weighing scale in a lift which is moving

1. upwards with a uniform speed of 10 ms^-1,
2. downwards with a uniform acceleration of 5 ms^-2,
3. upwards with a uniform acceleration of 5 ms^-2, What would be the readings on the scale in each case?
4. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Answer:
The weighing machine measures the reaction R which is nothing but the apparent weight.
1. when the lift is moving upwards with uniform speed.
R = mg = 70 × 9.8 = 686 N.

2. When lift moves downwards with an acceleration of 5m/s²
R = m (g – a) = 70 (9.8 – 5) = 336 N.

3. When lift moves upwards with with an acceleration of 5m/s²
R = m (g + a) = 70 (9.8 + 5) = 1036 N.

4. If the lift falls down freely under gravity
R = m (g – g) = 0.

Question 14.
Figure shows the position-time graph of a particle of mass 4 kg. What is the
(1) force on the particle for t < 0, t > 4 s, 0 < f < 4 s ?
(2) (impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).
Answer:

1. From the position time graph we can see that the particle is in rest during t < 0 and t > 4. Hence net force on it is zero t < 0, t > 4 s. During 0 < t < 4; the graph has a constant slope i.e, particle
has uniform velocity = 3/4 = 0.75 m/s.
Hence net force is zero.

2. at t = 0 u = 0 v = 0.75
impulse = change in momentum
= M (v – u) = 4 (0.75 – 0)
= 3 kg m/s
at t = 4, u = 0.75 v = 0
impulse = 4 (0 – 0.75) = – 3 kg m/s.

Question 15.
A horizontal force of 600 N pulls two masses 10 kg and 20 kg (lying on a frictionless table) connected by a light string. What is the tension in the string ? Does the answer depend on which mass end the pull is applied ?
Answer:


Question 16.
Two masses 8 kg and 12 kg are connected at the two ends, of a light in extensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
Answer:

Let M₁ = 8 kg and M₂ = 12 kg
Let T be the tension in the string and the acceleration with which the system moves is a

Question 17.
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must be emitted in opposite directions.
Answer:
Let M = Initial mass of the nucleus
M₁ and M₂ are masses after disintegration and υ₁ and υ₂ are their respective velocities.
Now, initial momentum of the nucleus = M x 0 = 0

Here, a negative sign shows that the products will be emitted in opposite directions.

Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer:

Impulse = change in momentum
Initial momentum of each ball = 0.05 × 6 = 0.3 kg m/s
Final momentum of each ball = 0.05 × (-6) = -0.3 kg m/s
Impulse = 0.6 kgm/s (in magnitude).

Question 19.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Answer:
Initially, gun and shell are at rest, thus Initial momentum = 0
Final momentum = momentum of bullet + momentum of gun = m_bv_b + m_gv_g By applying law of conservation of momentum,
Initial momentum of system = Final momentum of system
0 = m_bv_b + m_gv_g
m_bv_b =- m_gv_g

Negative sign shows that gun moves backward as the shell moves forward.

Question 20.
A batsman deflects ability an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Answer:
Suppose the point O is the position of bat. AO line shows the path along which the ball strikes the bat with velocity u and OB is the path showing deflection such that


Question 21.
A.stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?
Answer:
M = 0.25 kg, r = lm

Question 22.
If in Q. 21 the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks :
(a) the stone jerks radially outwards,
(b) the stoneflies oft tangentially from the instant the string breaks.
(c) the stoneflies oft at an angle with the tangent whose magnitude depends on the speed of the particle?
Answer:
(b) Stoneflies off tangentially from the instant the string breaks due to inertia of direction.

Question 23.
For ordinary terrestrial experiments, which of the observers below are inertial and which non-inertial:

1. a child revolving in a “giant wheel”,
2. a driver in a sports car moving with a constant high speed of 200 km/h on a straight road,
3. the pilot of an aeroplane which is taking oft,
4. a cyclist negotiating a sharp turn,
5. The guard of a train which is slowing down to stop at a station?

Answer:

1. In accordance with Newton’s 1^st law of motion since there is no external agent the horse cannot pull cart.
2. The passenger continues to move forward when a speeding bus breaks because of their inertia of motion. Hence they are thrown forward from their seats.
3.  A lawn mover is pulled or pushed by applying a force at an angle. When it is pushed, the normal force (N) must be more than its weight, for equilibrium in the vertical direction. This results in greater friction and hence greater applied force to move. It is just opposite while pulling.
4. The ball will have a large momentum. If the player tries to stop it instantaneously, the time of contact is low which results in a large impulse which may hurt his hand. Hence he tries to move his hands backward which increases the time of contact hence reducing the impulse.

Question 24.
Explain why
(a) a horse cannot pull a cart and run in empty space,
(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,
(c) it is easier to pull a lawnmower than to push it,
(d) a cricketer moves his hands backward while holding a catch.
Answer:
(a) A horse on earth pushes the earth with its feet. According to Newton’s third law of motion, the earth exerts a reaction equal to push on the horse. Hence, horse moves forward and pulls the cart. No reaction is available in empty space and hence horse cannot pull the cart and run in empty space.
(b) Due to inertia of motion.
(c) During pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight.
(d) By increasing time to stop the ball or decreasing its momentum to zero, force is reduced.

Question 25.
The figure shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle? What is the magnitude of each impulse?

Answer:
This graph can be of a ball rebounding between two walls situated at positions 0 cm and 2 cm. The ball is rebounding from one wall to another, time and again every 2s with uniform speed.

Question 26.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms^-2. What is the net force on the man ? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt ? (Mass of the man = 65 kg.)

Answer:
Acceleration of belt, a = 1 ms^-2, μ_s = 0.2
Net force on the man = mass of man x a = 65 x 1 = 65 N (y man is stationary w.r.t. belt)
The direction of this force is opposite to the direction of motion of the belt.
If a’ is the acceleration of the belt up to which the man can continue to be stationary relative to the belt, then
ma’ = maximum value of static friction
ma’ = μ_sR
ma’ =μ_s mg or a’= μ_s g
∴ d = 0.2 x 9.8 = 1.96 ms^-2

Question 27.
A stone of mass m tied to the end of a string is revolved in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are :
[Choose the correct alternative]

Here T₁, T₂ ( υ₁ and υ₂)) denote the tension in the string (and the speed of the stone) at the lowest and the highest point respectively.
Answer:
(a)

Question 28.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force op the helicopter due to the surrounding air.
Answer:
(a) Force on the floor by the crew and passengers = apparent weight = m (a + g)
= 300 x (15 + 10) = 7500 N (vertically downwards)
(b) Action of the rotor of the helicopter on the surrounding air= apparent weight of helicopter, crew and passengers = (M + m) (a + g) = (1000 + 300) x (15 + 10) = 32500 N (vertically downwards)
(c) Applying Newton’s III rd law of motion, we find that the force on the helicopter due to the surrounding air is equal and opposite to the action of rotor on the surrounding air = 3500 N (vertically upward).

Question 29.
A steam of water flowing horizontally with a speed of 15 ms^-1 gushes out of a tube of cross-sectional area 10^-2 m², and hits at a vertical wall near by. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
Answer:
Volume of water striking the wall per second = υ x A = 15 x 10^–2 = 0-15 m³s^-1
Mass of water hitting wall per second, M = p x υ x A = 1000 x 0.15 = 150 kg s^_1
Initial momentum of water per second = Mυ = 150 x 15 = 2250 kg m s^–1
Final momentum of water per second = 0    ( ∴ there is no rebound of water)
Magnitude of force = Rate of change of momentum = 2250 N.

Question 30.
Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m kg. Give the magnitude and direction of
(a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top.
(b) the force on the 7^th coin by the eighth coin,
(c) the reaction of the 6^th coin on the 7^th
Answer:
(a) The seventh coin will experience force equal to the sum of weights of the three coins above it. As each count is of m kg, therefore
force on seventh coin = 3m kgf = 3 mg N
where   g = acceleration due to gravity.
(b) The eighth coin supports the weights of two coins above it. Therefore, the force on seventh coin due to the eighth coin will be equal to the sum of the weights of the eighth coin and the two coins above it.
Therefore, force on seventh coin due to eight coin = m + 2m = 3m kgf = 3mg N
(c) The sixth coin experiences force equal to weight of th.e four coins above it. Hence, reaction due to sixth coin on the seventh coin = 4 m kgf = 4 mgf.

Question 31.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop ?
Answer:

Question 32.
A train rounds an unbanked circular bend of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose ? The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?
Answer:
(1) The centripetal force is provided by the lateral force acting due to outer rails on the wheels of the train.
(2) The outer rails will wear out faster as train exerts force (reaction) on them.
(3)

Question 33.
A block of mass 25 kg is raised by a 50 kg man in two different ways as slown in figure. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?
Answer:

In I^st case, man applies an upward force of 25 kg wt., (same as the weight of the block). According to Newton’s third law of motion, there will be a downward reaction on the floor.The action on the floor by the man
= 50 kg wt. + 25 kg wt.
= 75 kg wt = 75 x 9.8 = 735 N.
In case II, the man applies a downward force of 25 kg wt. According to Newton’s III^rd law, the reaction is in the upward direction.
In this case, action on the floor by the man
= 50 kg wt. – 25 kg wt. = 25 kg wt = 25 X 9.8 = 245 N
∴ Man should adopt the second method.

Question 34.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break : the monkey
(a) climbs up with an acceleration of 6 ms^–2
(b) climbs down with an acceleration of 4 ms^–2
(c) climbs up with a uniform speed of 5 ms^–1
(d) falls down the rope nearly freely under gravity ?
(Ignore the mass of the rope.)
Answer:

Maximum tension the rope can stand = 600 N, g = 10 ms^-2
Mass of monkey = M = 40 kg
(a) Here a = 6 ms^–2 (upwards)
∴ apparent weight of the monkey = M (g + a) = 40 (10 + 6) = 640 N
Since the maximum permissible tension is 600 N, thus the rope will break.

(b) Here a 4ms^–2(downwards)
∴ apparent weight of the monkey = tension in the rope
= M(g – a) = 40 (10 – 4) = 240 N
Thus rope will not break.

(c) Here uniform velocity = 5ms, so a = 0
∴ tension in the rope = M(g + a) = 40(10 + 0) = 400 N
Thus the rope will not break.

(d) When the monkey falls down freely under gravity, it is in the state of weightlessness. As there will be no tension in the string, the rope will not break.

Question 35.
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a
table against a rigid partition (Fig.). The co-efficient of friction between the bodies
and the table is 015. A force of 200 N is applied horizontally at A. What are
(a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the partition is removed ? Does the answer to (b) change, when the
bodies are in motion ? Ignore the difference between μ_s and μ_k
Answer:
(a) Limiting force of friction, F₁ = μR= μ (m_A+ m_B) g = O.15(5 + 10) 9.8 = 22.05 N (towards left)
Net force on the partition = F – F_f = 200 — 22.05 = 177.95 N
:. Reaction of partition = -177.95 N
(b) Force of friction on A, F_A = μm_Ag = 0.15 x 5x 9.8 = 7.35 N
;. Net force exerted by A on B
=(F-F_A) =200-7.35=19265 N

Question 36.
A block of mass 15 kg is placed on a long trolley. The co-efficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms^-2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by
(a) a stationary observer on the ground,
(b) an observer moving with the trolley.
Answer:
(a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N
Force of friction, F_f – μmg = 0.18 x 15 x 9.8 = 26.46 N
i.e..force experienced by block is less than the friction, so the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.

(b) The observer moving with trolley has an accelerated motion i.e. he forms non inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.

Question 37.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig.The co-efficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall off the truck ? (Ignore the size of the box).

Answer:


Question 38.
A disc revolves with a speed of 33\( \frac { 1}{ 3 } \)  rev/min, and has a radius of 15 cm. Two coins are placed 4 cm and 14 cm away from the center of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record ?
Answer:
The coin can only revolve with the record when the force of friction is enough to provide the centripetal force. If this force is riot enough then the coin slips on the record.
Here  R = mg and the centripetal force is given by mrω²
Therefore, to prevent slipping, the condition should be μmg ≥ mm
or μg ≥ rω²

Thus, coin placed at 4 cm continues to rotate with the record.

Question 39.
You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m ?
Answer:
At the point top of the loop, the equation of motion is given by
(1)
Where R is Normal reaction and r is radius of loop. Minimum possible speed at uppermost point is when R = 0

Question 40.
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is
0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck of the wall (without falling) when the floor is suddenly removed?
Answer:
The horizontal reaction R of the wall on the man provides the necessary centripetal force

The frictional force/acting upwards balances the weight mg of the man.
The man will remain stuck to the wall after the floor is removed

Question 41.
A thin circular wire of radius R rotates about its vertical diameter with an angular frequency co. Show that a small bead on the wire remains at its lowermost point for ω≤ \(\sqrt { g/r } \). What is the angle made by the radius vector joining the center to the bead with the vertical downward direction for ω= \(\sqrt { 2g/r } \)? Neglect friction.
Answer:
Let the radius vector joining the bead to the center of the wire make an angle 6 with the vertical downward direction.



We hope the NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion, help you. If you have any query regarding . NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line.

Question 1.
In which of the following examples of motion, can the body be considered approximately a point object:

1. A railway carriage moving without jerks between two stations.
2. A monkey sitting on top of a man cycling smoothly on a circular track.
3. A spinning cricket ball that turns sharply on hitting the ground.
4. A tumbling beaker that has slipped off the edge of the table.

Answer:

1. This is an example of uniform linear motion. Hence the carriage can be considered as a point object.
2. The monkey also undergoes motion smoothly on the circular track. Hence it can be heated as a point object.
3. The ball is spinning and undergoes changes in the plane of its motion when it hits the ground. Various parts of the ball experience different forces when the ball hits the ground. It is a rigid body and cannot be treated as a point object.
4. Different parts of the beaker experience the different magnitudes of force during its motion. Hence it cannot be treated as a point object.

Question 2.
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig.
Choose the correct entries in the brackets below:
(a) \(\frac { A }{ B } \) lives closer to the school than \( \frac { B }{ A } \)
(b) \(\frac { A }{ B } \) starts from the school earlier than \( \frac { B }{ A } \)
(c) \( \frac { A }{ B } \) walks faster than \( \frac { B }{ A } \)
(d) A and B reach home at the (same/different) time.
(e) \( \frac { A }{ B } \) overtakes \( \frac { B }{ A } \) on the road (once/twice).
Answer:
(a) A lives closer to school than B, because B has to cover higher distances [OP < OQ],
(b) A starts earlier form school than B, because t = 0 for A but for B, t has some finite time.
(c) As slope of B is greater than that of A, thus B walks faster than A.
(d) A and B reach home at the same time.
(e) At the point of intersection (i.e., X), B overtakes A on the roads once.

Question 3.
A woman starts from her home at 9.00 am, walks at a speed of 5 km h^_1 on a straight road up to her office 2-5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h^_1. Choose suitable scales and plot the x-t graph of her motion.
Answer:
Distance covered while walking = 2.5 km
Speed while walking = 5 km h^_1
Time taken to reach office while walking =\( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{2} \)
If O is taken as the origin for both time and distance then at t = 9-00 AM, x = 0 and at t 9-30 AM, x = 2.5 km OA is the x-t graph of the motion when the woman walks from her home to office. She stays in the office from 9-30 AM to 5-00 PM and is represented by the straight line AB.
Now time taken to return home = \( \frac { 2.5 }{ 5 } \)= \( \frac { 1 }{ 10 } \)
h = 6 minutes
So at 5. 06 PM, x = 0
This is represented by the line BC in the graph.

Question 4.
A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Answer:
Distance travelled in 5s = 5 m
Distance travelled in 8s=5-3=2m
Distance travelled in 13 s = 2 + 5 = 7m
Distance travelled in l6s = 7- 3 =4m
Distance travelled in 21s = 4 + 5 = 9m
Distance travelled in 24s = 9- 3 =6m
Distance travelled in 29s = 6 + 5 = 1m
Distance travelled in 32s=11-3 = 8m
Distance travelled in 37s=8 + 5 =13m
Since each step requires one second of time therefore total time is 37 seconds. The graph is as shown.

Question 5.
A jet airplane travelling at the speed of 500 km h^_1 ejects its products of combustion at the speed of 1500 km h^_1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?
Answer:
Speed of the exhaust with respect to an observer on the ground = Speed of exhaust with respect to the plane – Speed of plane with respect to the ground, (minus sign because the plane and exhaust move in opposite directions)
= (1500 – 500) km h^-1
= 1000 km h^-1

Question 6.
A car moving along a straight highway with a speed of 126 km h^_1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?
Answer:
u = 126 km h ¹ = 126 x\( \frac { 5 }{ 18 } \)=35 ms^_1
S = 200 m and υ = 0
υ²– u² = 2 a S
∴ 0 – (35)² = 2a x 200

Question 7.
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h^-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m s^-2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Answer:

∴ Original distance between trains = (Sb – S_A)-((L_A + L_B) = (2250 – 1000) – (800) = 450 m.

Question 8.
On a two-lane road, car A is travelling at a speed of 36 km h^_1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^_1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Answer:

Question 9.
Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^_1 in the direction A to B notices that a bus goes past him every 18 min, in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Answer:
Let υ_b = speed of each bus and υ_c = speed of cyclist
Relative speed of buses plying in the same direction of motion of cyclist = υ_b – υ_c
The buses plying in the direction of motion of the cyclist go past him after every 18 minutes i.e.,  \( \frac { 18 }{ 60 } \) h
.’. Distance covered by each bus is (υ_b – υ_c) x \( \frac { 18 }{ 60 } \)
Since a bus leaves after every T minute therefore distance is also equal to υ_b x \( \frac { T }{ 60 } \)
.’. (υ_b – υ_c) x \( \frac { 18 }{ 60 } \) = υ_b x \( \frac { T }{ 60 } \)
Relative velocity of the buses plying opposite to the direction of motion of the cyclist is υ_b + υ_c after every 6 minutes.
.’. Distance covered by each bus is (υ_b + υ_c) x  \( \frac { 6 }{ 60 } \)

Question 10.
A player throws a ball upwards with an initial speed of 294 ms^_1.
(a) What is the direction of acceleration during the upward motion of the ball?
(b) What are the velocity and acceleration of the ball at the highest point of its motion?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically
downward direction to be the positive direction of the x-axis, and give the signs of position, velocity, and acceleration of the ball during its upward and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g=98 m s^_2 and neglect air resistance).
Answer:
(a) The ball is under the influence of acceleration due to gravity which always acts vertically downwards.
(b) Velocity at the highest point = zero
Acceleration at highest point = g = 98 ms^_2 (vertically downwards)

Question 11.
Read each statement below carefully and state with reasons and examples, if it is true or false:
A particle in one-dimensional motion

1. with zero speed at an instant may have non-zero acceleration at that instant
2. with zero speed may have non-zero velocity. ‘
3. with constant speed must have zero acceleration.
4. with a positive value of acceleration must be speeding up.

Answer:

1. True. Consider a ball thrown vertically, upward. At the highest point, the speed is zero but the acceleration of the ball is non-zero ( 9.8ms^-2 vertically downwards). Acceleration does not depend on instantaneous speed.
2. False. Since the magnitude of velocity is speed, a body with zero speed must have zero velocity.
3. True. In the case of a body rebounding with the same speed, the acceleration at the time of impact is infinite, which is not practical physically.
4. False. This depends on the chosen positive direction. The statement is true when the direction of motion and acceleration is along the chosen positive direction.

Question 12.
A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one-tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Answer:

Question 13.
Explain clearly, with examples, the distinction between:
(a) the magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
(b) the magnitude of average velocity over an interval of time, and the average speed over the same interval [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Answer:
(a) Magnitude of displacement over an interval of time may be zero, whereas the total length of the path covered by the particle over the same interval is not zero. For example, consider a particle moving along a straight line from point A to point B distant S from each other and then back to point A in time interval t as shown in fig.

In this case, magnitude of displacement of the particle over an interval of time t = 0.
Total length of the path covered by the particle over the interval of time t = AB + BA = 2 S
(b)

(c) In both the cases (a) and (b), the second quantity is greater than the first quantity e. The total length of path > magnitude of displacement and average speed > magnitude of average velocity. If the direction of motion of a particle along a straight line does not change, the the magnitude of displacement of the particle over a time interval = Total length of the path covered by the particle over the same time interval and magnitude of velocity = speed of the particle.

Question 14.
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h^-1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h^-1. What is the
(a) the magnitude of average velocity, and
(b) the average speed of the man over the interval of time
(1) 0 to 30 min,
(2) 0 to 50 min,
(3) 0 to 40 min?
[Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and hot as the magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]
Answer:

 (3) In 0-40 min
In 0 – 30 min. man goes from home to market with a sped of 5 km h’. In the next 10 mm, the man goes from the market towards home with speed of 75 km h^-1 Distance travelled by man in these 10 min = speed x time = 75 km h^-1x \( \frac { 10 }{ 60 } \) h

Question 15.
In 13 and 14, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider the instantaneous speed and the magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Answer:
The instantaneous speed is always equal to the magnitude of the instantaneous velocity because for very small instants of time the length of the path is equal to the magnitude of displacement.

Question 16.
Look at the graphs (a) to (d) (Fig.) carefully and state, with reasons, which of these cannot possibly represent one- dimensional motion of a particle.

Answer:
(a) From the graph, we see that for certain instants of time, the particle has 2 positions, which is not possible. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(b) From the graph, we see that for certain instants of time, the particle has 2 velocities, which is not possible. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(c) From the graph, we see that for some instants of time, the particle has negative speed. But speed is always a non-negative quantity. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

(d) From the graph, we see that for a time interval the total path length is decreasing. But total path length is always a nondecreasing quantity. Hence this graph cannot possibly represent the one-dimensional motion of a particle.

Question 17.
Figures show the x-t plot of the one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.

Answer:
For t < 0, x = 0, so particle is at rest and not moving in a straight line,
For t > 0, a particle can move on a parabolic path if its acceleration is constant.

Therefore, it is not correct to say from graph that the particle moves in a straight line for r < 0 and on a parabolic path for t > 0.
For the graph, a suitable physical context can be the particle thrown from the top of a tower at the instant t = 0.

Question 18.
A police van moving on a highway with a speed of 30 km he fires a bullet at their car speeding away in the same direction with a speed of 192 km h^-1. If the muzzle speed of the bullet is 150 ms^-1, with what speed does the bullet hit their car? (Note: Obtain that speed which is relevant for damaging their car).
Answer:

Question 19.
Suggest a suitable physical situation for each of the following graphs.

Answer:
(a) consider a ball which is pushed at some time t < 0 towards a wall. Upon rebounding from this wall it hits the opposite wall and comes to a stop. If x = 0 for the initial position, then this context may have the given x -t graph.

(b) Consider a ball thrown vertically upwards, with the vertically upward direction chosen as the positive direction. Each time it hits the ground, it loses a fraction of its velocity and finally comes to rest. The ball may be represented in the given v -t graph in this context.

(c) Consider a cricket ball moving with a uniform velocity which is hit by the bat and then turns back. In this case, the a-t graph for the ball may be similar to the one given above.

Question 20.
The figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 14). Give the signs of position, velocity, and acceleration variables of the particle at t = 0.3 s, 1.2 s, -1.2 s.

Answer:

(1) At t = 0-3 s, x is -ve. Velocity = slope of x -t graph.
Since slope of x -t is negative, so velocity is negative. In simple harmonic motion, the direction of acceleration is opposite to the direction of displacement of the particle, so acceleration is positive.
(2) At t = 1.2 s, x = + ve. v is also +ve as slope of x -t graph is + ve. Acceleration a is -ve
(3) At t = -1.2 s, x =-ve. so a is +ve, v =Δx/Δt = +ve as both Δx and Δt are negative.

Question 21.
The figure gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.
Answer:
The magnitude of the slope of the x – t graph is highest in 3 and least in 2. Hence, the average speed is greatest in 3 and least in 2. Also, the sign of the slope of the x -t graph is positive for 1 and 2 and negative for 3. Therefore sign of the average velocity ‘V’ is:-
v > 0 for intervals 1 and 2
v < 0 for intervals 3.

Question 22.
The figure gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of o and a in the three intervals. What are the accelerations at points A, B, C and D?

Answer:
Acceleration magnitude is greatest in 2 because slope of υ -t graph at this interval is maximum.
Average speed is greatest in 3.
υ>0  in 1, 2 and 3\a > 0 in 1, a < 0 in 2, a = 0 in 3.
Acceleration is zero at A, B, C and D because slope of  υ-t graph at these points is zero.

Question 23.
A three-wheeler starts from rest, accelerates uniformly with 1 m s^-2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1, 2, 3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Answer:

Question 24.
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 m s^_1. How much time does the ball take to return to his hands ? If the lift starts moving up with a uniform speed of 5 m s^_1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ?
Answer:

When lift is moving upward with uniform velocity, the initial velocity of the ball will remain 49 ms^-1 only w.r.t lift. Thus the time take up by ball will be 10 s.

Question 25.
On a long horizontally moving belt (Fig.), a child runs to and fro with a speed of 9 km h^_1 (with respect to the belt) between ms father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h^_1. For an observer on a stationary platform outside, what is the
(a) speed of the child running in the direction of motion of the belt?
(b) speed of the child running opposite to the direction of motion of the belt?
(c) time is taken by the child in (a) and (b)?
Which of the answers alter if motion is viewed by one of the parents?

Answer:

Question 26.
Two stones are thrown up simultaneously from the edge of a cliff 200 m high with initial speeds of 15 ms^-1 and 30 ms^-1. Verify that the graph shown in Fig. correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 10 ms^-2. Give the equations for the linear and curved parts of the plot.

Answer:

For maximum separation, t = 8 s
So maximum separation is 120 m
After 8 seconds, only the second stone would be in motion. Its motion is described by eqn.(ii) So, the graph is in accordance with the quadratic equation.

Question 27.
The speed-time graph of a particle moving along a fixed direction is shown in
Fig. Obtain the distance traversed by the particle between
(a) t = 0 s to 10 s.
(b) t = 2 s to 6 s.
What is the average speed of the particle over the intervals in (a) and (b)?

Answer:
(a) Distance travelled from t = 0 to t = 10 s
= area under speed time graph,
= \( \frac { 1 }{ 2 } \) x 10 x 12 = 60m
(b) Average speed over the interval from t = 0 to t = 10 s is \( \frac { 60 }{ 10 } \) =6 ms^-1
(c) In order to calculate distance from t = 2 s to t = 6 s, let us first determine separately the distance covered from t
= 2 s to r = 5 s and the distance covered form t = 5 s to t = 6 s, then add.
12
(i) Acceleration =\( \frac { 12 }{ 5 } \) = 24 ms2
Velocity at the end of 2 s = 24 X 2 = 4.8 ms^-1.

Question 28.
The velocity-time graph of a particle in one-dimensional motion is shown in Fig.
Which of the following formulae are correct for describing the motion of the particle over the time-interval t₁ to t₂ :


Answer:
(a) This formula is not correct as it is applicable only if a is constant. In time interval t₁ to t₂, a is not constant.
(b) This formula is not correct as it is applicable only if a is In time-interval t₁ to t₂, a is not constant.
(c) and (d) are correct. They represent the definitions of υ_av and υ_av.
(d) This formula is not correct as such formula does not contain u_av and a_av.
(e) This formula is correct because of the area under υ-t graph = displacement of a particle.

We hope the NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 3 Motion in a Straight Line, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement

Question 1.
Fill in the blanks.
(a) The volume of a cube of side 1 cm is equal to……… m³.
(b) The surface area of solid cylinder of radius 2.0 cm and height 10.0 cm is equal to……. (mm)².
(c) A vehicle moving with a speed of 18 km h^-1 covers………… m in 1 s.
(d) The relative density of lead is 11.3. Its density is…….. g cm^-3 or…………. kg m^-3.
Answer:

Question 2.
Fill in the blanks by suitable conversion of units
(a) 1 kg m² s^-2 = …………. g cm² s^–2
(b) 1 m =……………. ly
(c) 3 m s^-2 = ………. km h^–2
(d) G = 6.67 x 10-¹¹ N m² (kg)^–2 =……………. (cm)³ s^–2 g¹.
Answer:

Question 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1 J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude of 4.2 α^-1 β^-2 γ² in terms of the new units.
Answer:

 

Question 4.
Explain this statement clearly:
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.
Answer:

1. Atoms are very small objects when compared to a cricket ball.
2. A jet plane moves with great speed when compared to a car.
3. The mass of Jupiter is much larger than that of Earth.
4. The air inside this room contains a large number of molecules when compared to the number of objects in the room.
5. No change necessary.
6. No change necessary.

Question 5.
A new unit of length is chosen such that the speed of light in a vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Answer:
New unit of length = 3 × 10⁸ ms^-1
Distance between the Earth and the sun
= \((8 \min 20 \mathrm{s}) \times 3 \times 10^{8} \mathrm{ms}^{-1}\)
= 500 × 3 × 10⁸ ms^-1
∴Distance between the Earth and the Sun in terms of the new units
=\(\frac{500 \times 3 \times 10^{8}}{3 \times 10^{8}}\)
= 500 new units.

Question 6.
Which of the following is the most precise device for measuring length :
(a) vernier calipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure the length to within a wavelength
Answer:

(c) Least count of optical instrument w 6000 A
(average wavelength of visible light as 6000 A) = 6 x 10^-7 m
∴ (c) is the most precise instrument.

Question 7.
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and, finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate of the thickness of the hair?
Answer:

Question 8.
Answer the following :
(a) You are given a thread and a meter scale. How will you estimate the diameter of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier calipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer:
(a) It if; done by winding a known number of turns over a pencil, turns touching each other closely. Then the length occupied by every single turn will be equal to the diameter of the thread.
(b) Yes, because the least count of the screw gauge is given by

i.e. least count of screw gauge is inversely proportional to the number of divisions on a circular scale. So with the increase in a number of the divisions on the circular scale, the least count will improve. Thus the accuracy of the screw gauge will increase.
(c) Increasing the number of observations, increases the reliability as the mean error is also reduced. The best possible value is

Question 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected onto a screen and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?
Answer:
Given Area of the house in photograph = 1.75 cm²
Area of house on screen = 1.55 m² = 1.55 x 10⁴ cm²

Question 10.
State the number of significant figures in the following:

1. 0.007 m²
2. 2.64 x 10⁴ kg
3. 0.2370 gem^-3
4. 6.320 J
5. 6.032 N nr^-2
6. 0.0006032 m²
   

Answer:

1. 1
2. 3
3. 4
4. 4
5. 4
6. 4

Question 11.
The length, breadth, and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Answer:
Length l = 4.234 m,
Breadth b = 1.005 m,
Thickness t = 2.01 cm = 0.0201 m
Area = 2 × (lb + bt + It)
= 2 × (4.234 × 1.005 + 1.005 × 0.0201 + 4.234 × 0.0201)
= 8.72 m²
(Rounding off to 3 significant figures)
Volume = lbt
= 4.234 × 1.005 × 0.0201
= 8.55 × 10^-2 m³

Question 12.
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is
(a) the total mass of the box
(b) the difference in the masses of the pieces to correct significant figures ?
Answer:
(a) Total mass of the box = (2.3 + 0.0217 + 0.0215) kg = 2.3442 kg
Since the least number of significant figure is 2, therefore, the total mass of the box = 2.3 kg.
(b) Difference of mass = 2.17 – 2.15 = 0.020 g
Since there are two significant figures so the difference in masses to the correct significant figures is
0.020 g.

Question 13.
A physical quantity P is related to four observables a, b, c and d as follows :

The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 2%, respectively. What is the percentage error in the quantity P ? If the value of P calculated using the above relation turns out to be 3.763, to what value should you round off the result ?
Answer:

Question 14.
A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion :
(a) y = a sin 2π \( \frac { t}{ T } \)
(b)   y  = a sin υt
(c) y = \( \frac { a}{ T } \) sin \( \frac { t}{ a } \)
(d)   y  = (a√2) (sin 2π \( \frac { t}{ T } \)  + cos 2π \( \frac { t}{ T } \)  )
(a = maximum displacement of the particle, υ= speed of the particle, T = time period of motion). Rule out the wrong formulas on dimensional grounds.
Answer:
The argument of a trigonometric function i. e., angle is dimensionless

Question 15.
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein).A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

Guess where to put the missing c.
Answer:
On rearranging, we have

Since the left-hand side is dimensionless, so the right-hand side should be dimensionless. This will be so ……………

Question 16.
The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A : 1 A = 10^–10 m. The size of a hydrogen atom is about 0.5 A. What is the total atomic volume in m³ of a mole of hydrogen atoms?
Answer:
Volume of one hydrogen atom = \( \cfrac {4}{ 3 } \)
_{\( \cfrac {4}{ 3 } \) X} 3.14 x (0.5 x l(T^-10)m³ = 5-23 x 10^-31 m³.
According to Avogadro’s hypothesis, one mole of hydrogen contains 6 023 x 10²³ atoms.
∴ The atomic volume of 1 mole of hydrogen atoms = 6.023 x 10²³ x 5.23 x 10^–31 = 3.15 x 10^–7 m³.

Question 17.
One gram mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of the hydrogen molecule to be about 1 A). Why is this ratio so large?
Answer:

 

This high ratio is because of intermolecular spaces in gas being much larger than the size of the molecules.

Question 18.
Explain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses, etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hilltops, the Moon, the stars, etc.) seem to be stationary. (In fact, since you are aware that you are moving, these distant objects seem to move with you).
Answer:
Objects nearer to the eye subtend a greater angle in the eye than distant objects. When we move, the change in this angle is less for distant objects than for near objects. So the distant objects seem stationary but nearer objects seem to move in the opposite direction.

Question 19.
The principle of ‘parallax’ is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≈3 x 10¹¹ m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1 (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1 (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters ?
Answer:

Question 20.
The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs ? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?
Answer:
As we know, 1 light year = 9.46 x 10¹⁵ m
∴ 4.29 light years = 4.29 x 9.46 x 10¹⁵ = 4.058 x 10¹⁶ m

Question 21.
Precise measurements of physical quantities are a need of modern times. For example, to ascertain the speed of an enemy fighter plane, one must have an accurate method to find its positions at closely separated instants of time. Only then we can hope to shell it with an antiaircraft gun. This was the actual motivation behind the discovery of radar in World War II. Think of different examples in modern science where precision measurements of length, time, mass etc. are needed. Also, wherever you can, give a quantitative idea of the precision needed.
Answer:

• Length: In, sports, sending satellites
• Mass: To add proper proportions of different salts for preparing medicines, the mass of the satellite should be accurately measured.
• Time: For the study of various chemical reactions, the study of various activities in the universe.

Question 22.
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity) :
(a) The total mass of rain-bearing clouds over India during the Monsoon
(b) The mass of an elephant
(c) The wind speed during a storm
(d) The number of strands of hair on your head.
(e) The number of air molecules in your classroom.
Answer:
(a) Firstly to calculate the total rain in India, we can get an estimate of it and then knowing the weight of water we can estimate the weight of clouds.
(b) To estimate the mass of an elephant, we take a boat of known base area A. Measure the depth of boat in water. Let it be x₁ Therefore, volume of water displaced by the boat, V₁ – AX₁
Move the elephant into this boat. The boat gets deeper into water. Measure the depth of boat now into the water. Let it be x₂.
∴ Volume of water displaced by boat and elephant V₂ = Ax₂
∴ Volume of water displaced by the elephant V = V₂ – V₁ = A(x₂ – X₁)
If p is the density of water, then the mass of elephant = mass of water displaced by it.
= V ρ = A(x₂ – x₁)ρ
(c) The pressure generated by the wind can give us an estimation of its speed.
(d) Estimating no. of hairs per cm² area of the head, we can estimate the total no. of hairs on our head,
( ∴ we can estimate the area of our head)
(e) We can get the density of air and hence an estimation of no. of molecules in 1 cm³ can be made by which we can estimate no. of molecules in our room.

Question 23.
The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be? In the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: a mass of the Sun = 2.0 x 10³⁰ kg. radius of the Sun = 7.0 x 10⁸ m.
Answer:

Mass density of Sun is in the range of mass densities of solid/liquids and not gases.

Question 24.
When the planet Jupiter is at a distance of 824.7 million kilometers from the Earth, its angular diameter is measured to be 35-72 of arc. Calculate the diameter of Jupiter.
Answer:
Given angular diameter θ = 35.72 = 35.72 x 4.85 x 10^-6 rad
= 173.242 x 10^-6 = 1.73 x 10^-6 rad
∴Diameter of Jupiter, D = θ x d = 1.73 x 10^-4 x 824.7 x 10⁹ m
= 1426.731 x 10⁵ = 1.43 x 10⁸ m

Question 25.
A man walking briskly in rain with speed v> must slant his umbrella forward making an angle with the vertical. A student derives the following relations between θ and v: tan θ = υ and checks that the relation has a correct limit: as υ→0, θ→0, as expected. (We are assuming there is no strong wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct? If not, guess the correct relation.
Answer:
According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.
Here υ = tan θ i.e. [L¹ T^-1] = dimensionless, which is incorrect.
Correcting die L.H.S, we get
υ/u = tan θ, where u is the velocity of rain.

Question 26.
It is claimed that two cesium clocks if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time- interval of 1 s?
Answer:
Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Question 27.
Estimate the average mass density of a sodium atom assuming its size to about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m^-3. Are the two densities of the same order of magnitude? If so, why?
Answer:
The volume of 1 sodium atom,

Both the densities are of the order of 10³ i.e. the atoms are tightly packed. They belong to the solid phase.

Question 28.
The unit of length convenient on the nuclear scale is a fermi: 1 f = 10^-15 m. Nuclear sizes obey roughly the following empirical relation: r = r₀A^1/3 where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in 2.20.
Answer:

Question 29.
A LASER is a source of very intense, monochromatic, and a unidirectional beam of light. These properties of laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the lunar orbit around the Earth?
Answer:

Question 30.
A SONAR (sound navigation and ranging) used ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between the generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450 m s^-1).
Answer:

Question 31.
The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3’0 billion years to reach us?
Answer:
Given t = 3 x 10⁹ years = 3 x 10⁹ x 365.25 x 24 x 60 x 60 s
c = 3 x 10⁵ km s^-1 (velocity of e.m. waves)
∴ d = c x t = 3 x 10⁵ x 3 x 10⁹ x 365.25 x 24 x 60 x 60
= 2840184 x 10¹⁶ km = 2.84 x 10²² km.

Question 32.
It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and the information, you can gather from examples 1 and 2, determine the approximate diameter of the moon.
Answer:
Here CD = diameter of Sun
AB = diameter of moon
and E = Position of earth
As Δ CDE and ABE are similar,

Question 33.
A great physicist of this century (PA.M. Dirac) loved playing with numerical values of fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, the mass of the electron, the mass of a proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (≈15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?
Answer:
Trying out with basic constants of atomic physics (speed of light c, the charge on electron e, the mass of electron m_e, the mass of proton m_p) and universal gravitational constant G, we can arrive at a quantity which has the dimensions of time. One such quantity is

We hope the NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 2 Units and Measurement, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Physics Chapter 1 Physical World

These Solutions are part of NCERT Solutions for Class 11 Physics. Here we have given NCERT Solutions for Class 11 Physics Chapter 1 Physical World

Question 1.
Some of the most profound statements on the nature of science have come from Albert Einstein, one of the greatest scientists of all time. What do you think did Einstein mean when he said: “The most incomprehensible thing about the world is that it is comprehensible”?
Answer:
Probably Einstein meant that science can marvelously explain through simple theories of the various natural phenomena which become comprehensible to us. In fact, it is unthinkable that complex natural phenomenon can be so comprehensible with scientific analysis which for an ordinary person is incomprehensible.

Question 2.
“Every great physical theory starts as a heresy and ends as a dogma”. Give some examples from the history of science d(f the validity of this incisive remark.
Answer:
Heresy is .something which is not established, whereas dogma means established view e.g. there was a heresy that inertia of a body depended upon its energy. But Einstein gave a simple equation E = mc², relation between mass and energy. This is a dogma in physics. Another heresy is that in ancient times Ptolemy postulated that the earth is stationary and entire heavenly bodies move around. But the dogma is that earth itself moves around the Sun.

Question 3.
“Politics is the art of the possible”. Similarly, “Science is the art of the soluble”. Explain this beautiful aphorism on the nature and practice of science.
Answer:
Politicians would make anything possible by their sheer words. But the majority of things may not be possible in practice. Whereas science can make us understand the phenomena around us. e.g. total solar eclipse shows an interesting aspect of solar temperature. Its chromosphere temperature is about 6000 K. But as we go out towards the rim, it first falls and then suddenly rises to a million kelvin or higher. Science is concerned to provide an explanation or find a solution to this riddle. The repeated practice of science allows us to hypothesis, make calculations, experiment with these and then predict the possible solution.

Question 4.
Though India now has large base in science and technology, which is fast expanding, it is still a long way from i. realizing its potential of becoming a world leader in science. Name some important factors, which in your view hindered the advancement of science in India.
Answer:

• Lack of education
• Lack of scientific attitude in the students
• Money plays the key role
• Lack of practice etc.

Question 5.
No physicist has ever “seen” an atom. Yet, all physicists believe in the existence of atoms. An intelligent but superstitious man advances this analogy to argue that ‘ghosts’ exist even though no one has ‘seen’ one. How will you refute his argument?
Answer:
It is simply a superstition that ghosts exist. There is not even single authentic evidence that ghosts exist. There are many examples to prove this fact. Atomic power plants, atomic bombs, atomic clocks, etc. exist because atoms exist in nature. Thus there is no correlation between the two parts of the statement.

Question 6.
The shells of crabs found around a particular location in Japan seem mostly to resemble the legendary face of a Samurai. Given below are two explanations of this observed fact. Which of these strikes you as a scientific explanation?
(a) A tragic sea accident several centuries ago drowned a young Samurhi. As a tribute to this bravery, nature through its inscrutable ways immortalized his face by imprinting it on the crab shells in that area.
(b) After the sea tragedy, fishermen in the area, in a gesture of honor to their dead hero, let free any crab shell caught by them which accidentally had a shape resembling the face of a Samurai. Consequently, the particular shape of the crab shell survived longer, and therefore in course of time the shape was genetically propagated. This is an example of evolution by artificial selection.
Answer:
Statement (b) is scientific.

Question 7.
The industrial revolution in England and Western Europe more than two centuries ago was triggered by some key scientific and technological advances. What were their advances?
Answer:
Prior to 1750 AD when the Industrial revolution happened, simple tools and machines were used. But industrial revolution brought new machinery. Some of the outstanding contributions of the industrial revolution were

1. Steam engine
2. Blast furnace which converts low-grade iron into steel
3. Cotton gin separates the seed from cotton three hundred times faster than by hand etc.

Question 8.
It is often said that the world is witnessing now a second industrial revolution, which will transform society as radically as did the first. List some key contemporary areas of science and technology, which are responsible for this revolution.
Answer:
The key areas which are responsible for revolution are

1. Superfast computers
2. Biotechnology
3. Development of superconducting materials at room temperature etc.

Question 9.
Write in about 1000 words a fiction piece based on your speculation on the science and technology of the twenty-second century.
Answer:
Let us imagine a spaceship moving towards a distant star, 500 light-years away. Suppose this is propelled by current fed into the electric motor consisting of superconducting wires. In space, suppose there is a particular region which has such a (the high temperature that destroys the superconducting property of the electric wires of the motor. At this stage, another spaceship filled with matter and anti-matter comes to the rescue of the first ship, and the first ship continues its onward journey.

Question 10.
Attempt to formulate your ‘moral’ views on the practice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous consequences for human society. How, if at all, will you resolve your dilemma?
Answer:
Scientific discovery reveals the truth of nature. Therefore any discovery, good or bad for mankind must be made public. The discovery which appears to be dangerous today may become useful to mankind later on. In order to avoid the misuse of scientific technology, we must build up a strong public opinion. Thus scientists should do two things

1. To discover truth and
2. To prevent its misuse.

Question 11.
Science, like any knowledge, can be put to good or bad use, depending on the user. Given below are some of the applications of science. Formulate your views on whether the particular application is good, bad or something that cannot be so clearly categorized.

1. Mass vaccination against smallpox to curb and finally eradicate this disease for the population. (This has already been successfully done in India).
2. Television for the eradication of illiteracy and for mass communication of news and ideas.
3. Prenatal sex determination
4. Computers for an increase in work efficiency
5. Putting artificial satellites into orbits around the Earth
6. Development of nuclear weapons
7. Development of new and powerful techniques of chemical and biological warfare. Purification of water for drinking.
8. Plastic surgery
9. Cloning

Answer:

1. Good
2. Good
3. Bad
4. Good
5. Good
6. Bad
7. Good
8. Cannot clearly categorize
9. Cannot clearly categorize

Question 12.
India has had a long and unbroken tradition of great scholarship — in mathematics, astronomy, linguistics, logic, and ethics. Yet, in parallel with this, several superstitious and obscurantistic attitudes and practices flourished in our society and unfortunately continue even today — among many educated people too. How will you use your knowledge of science to develop strategies to counter these attitudes?
Answer:
In order to popularise scientific explanations of the everyday phenomenon, mass media like radio, television, and newspapers should be used.

Question 13.
Though the law gives women equal status in India, many people hold unscientific views on a woman’s innate nature, capacity and intelligence, and in practice give them a secondary status and role. Demolish this view using scientific arguments, and by quoting examples of great women in science and other spheres; and persuade yourself and others that, given equal opportunity, women are on par with men.
Answer:
The nutrition contents of pre-natal and post-natal diet contribute a lot towards the development of the human mind. If equal opportunities are afforded to both men and women then the female mind will be as efficient as the male mind.

Question 14.
“It is more important to have beauty in the equations of physics than to have them agree with experiments”. The great British physicist P.A.M. Dirac held this view. Criticize the statement. Look out for some equations and results in this book which strike you as beautiful.
Answer:
Generally, it is considered that physics is a dry subject and its main aim is to give qualitative and quantitative treatment i.e. any derived relation or equation must be verified through experimentation. It is felt that the truth of an equation is more important than the simplicity, wonderfulness, symmetry, or beauty of the equation. But frankly, if a relation is true to experimentation and simultaneously it is simple, interesting, symmetrical, wonderful, or beautiful, it will certainly add to the charm of the relation.

Question 15.
Though the statement quoted above may be disputed, most physicists do have a feeling that the great laws of physics are at once simple and beautiful. Some of the notable physicists, besides Dirac, who have articulated this feeling are : Einstein, Bohr, Heisenberg, Chandrasekhar, and Feynman. You are urged to make special efforts to get access to the general books and writings by these and other great masters of physics. (See the Bibliography at the end of this book). Their writings are truly inspiring.
Answer:
General books on Physics make interesting reading. Students are advised to consult a good Library. ‘Surely you are joking, Mr. Feynman’ by Feynman is one of the books that would amuse the students. Some other interesting books are: Physics for the inquiring mind by EM Rogers; Physics, Foundations, and Frontiers by G. Gamow; Thirty years that shook Physics by G. Gamow; Physics can be Fun by Perelman.

Question 16.
Textbooks on science may give you the wrong impression that studying science is dry and all too serious and that scientists are absent-minded introverts who never laugh or grin. This image of science and scientists is patently false. Scientists, like any other group of humans, have their share of humorists and may have led their lives with a great sense of fun and adventure, even as they seriously pursued their scientific work. Two great physicists of this genre are Gamow and Feynman. You will enjoy reading their books listed in the Bibliography.
Answer:
True, scientists like any other group of humans have their share of humorists; lively, jovial, fun-loving, adventurists people. Some of them are absent-minded introverts too. Students are advised to go through books by two great Physicists, Feynman and Gamow to realize this view.

We hope the NCERT Solutions for Class 11 Physics Chapter 1 Physical World, help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 1 Physical World, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry.

Question 1.
Define environmental chemistry.
Answer:
Environmental chemistry is the branch of science which deals with the chemical changes in the environment. It includes our surroundings such as air, water, soil, forests, sunlight etc.

Question 2.
Explain tropospheric pollution in 100 words.
Answer:
Tropospheric Pollution
Tropospheric pollution is caused by both inorganic and organic gases. Gases like oxides of nitrogen, oxides of sulphur, oxides of carbon, H₂S, HCN, HC1 etc. constitute inorganic pollutants.
Organic pollutants include mercaptants, hydrocarbons, formaldehyde, alcohol, certain organic acids chlorinated hydrocarbon etc.
Some of these are released as such in the atmosphere and are known as primary pollutants. Some others are formed in the atmosphere as a result of chemical reactions.
These are called secondary pollutants. A few examples are : Ozone, chlorofluorocarbons (CFCs), formaldehyde, acrolein, methyl isocyanate etc. Let us briefly study some of the tropospheric pollutants.

Question 3.
Carbon monoxide gas is more dangerous than carbon dioxide, why ?
Answer:
Carbon monoxide (CO) when inhaled reacts with haemoglobin (Hb) to form complex carboxy haemoglobin (CoHb). The compound formed is not in a position to transport the inhaled oxygen to the various parts of the body. On the other hand, the presence of carbon dioxide can lead only to green house effect causing global warming.

Carbon dioxide
Carbon dioxide, is present in air in about 0.03% by volume. It is being constantly released into the atmosphere by the combustion of fossil fuels such as coal and oil for energy.
In addition to this, volcanic erruptions and decomposition of lime stone release carbon dioxide in the atmosphere. However, it is also taken away from the atmosphere regularly by green plants and forests because the plants need carbon dioxide for photosynthesis,
Thus carbon dioxide cycle is working round the clock which maintains its percentage in the atmosphere. However, with the deforestation that has taken place, there is an increased build up of the gas in the atmosphere.

Question 4.
Which gases are responsible for green house effect ? Name them.
Answer:
The green house effect is caused by the following gases which are capable of trapping heat energy.
(i) carbon dioxide
(ii) methane
(iii) ozone
(iv) chlorofluorocarbon compounds (CFC’s)
(v) water vapours.

Question 5.
Statues and monuments in India are affected by acid rain. How ?
Answer:
The statues and monuments are mainly made from marble which is chemically calcium carbonate (CaCO₃). Acid rain has vapours of sulphuric acid dissolved in it. When it comes in contact with the various statues or monuments, the acid reacts chemically with calcium carbonate. As a result of the chemical reaction, the material of the statues are slowly eaten up.
It is also called corrosion. Thus, acid rain is a threat to our precious historical monuments and statues.
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 6.
What are smogs ? How are classical and photochemical smogs different ?
Answer:
Air pollution is commonly caused in big and industrial cities in the form of smog which is quite often termed as smoke- smog. This may be either classical or photochemical in nature.

Question 7.
Write chemical reactions involved during the formation of photochemical smog.
Answer:

Question 8.
What are the harmful effects of the photochemical smog ? How can they be controlled ?
Answer:
Harmful Effects of Photochemical Smog.
The main constituents of photochemical smog are :
ozone, oxides of nitrogen, acrolein, formaldehyde and peroxyacetylnitrate (PAN). These are responsible for the harmful effects. A few out of these are listed.
(i) Ozone and nitric oxide cause irritation in nose as well as in throat. Their high concentration usually causes headache and chest pain.
(ii) The gases which constitute photochemical smog, usually cause dryness of throat, cough and are responsible for breathing problems.
(iii) Photochemical smog causes substantial damage to plant life.
(iv) It also results in corrosion of metals, building materials, rubber and painted surface etc.
How to control Photochemical Smog
Following measures can check the pollution caused by photochemical smog to some extent.
(i) Use of catalytic converters in the engines of automobiles will check the release of both N02 and certain hydrocarbons known as primary precursors. This will automatically check the formation of secondary precursors such as ozone and PAN which are harmful.
(ii) Certain plants like pinus, pyrus, vitis quercus etc. are capable of causing the metabolism of the oxides of nitrogen which are quite harmful. Their plantation will definitely help in checking the spread of these gases in the atmosphere.

Question 9.
What are the reactions involved for ozone layer depletioh in stratosphere ?
Answer:
Chlorofluorocarbons such as freon etc. present in the stratosphere are involved in the chemical reaction with ozone. These are of free radical nature and carried in the presence of U. V. radiations.

Since ozone takes part in the chemical reaction, there is a gradual depletion of ozone layer which is taking place.

Question 10.
What do you understand by ozone hole ? What are its consequences ?
Answer:
Ozone hole implies destruction of the ozone layer by the harmful ultraviolet (UV) radiations. The depletion will virtually result in creating some sort of holes in the blanket of ozone which surrounds us. As a result, the harmful radiations will cause skin cancer, loss of sight and will also affect our immune stystem.

The depletion of ozone layer also called ozone hole was noticed in the year 1980 by the scientists working in Antarctica region in the South Pole. Special conditions prevailing in the region were responsible for the depletion of ozone. During summer, nitrogen dioxide and methane react with chlorine monoxide and chlorine atoms (as free radicals)
CI\(\dot { O } \) (g) + NO₂(g) → ClONO₂(?)
Chlorine nitrate
\(\dot { C } \)l (g)+ CH₄(g) → \(\dot { C } \)H₃ (g) + UCl(g)

Under the conditions, \(\dot { C } \)l are not in a position to react with ozone and thus, the depletion of ozone layer is checked. During winter, special types of clouds known as polar stratospheric clouds get formed over Antarctica region. They provide a surface over which chlorine nitrate react with water vapours and also with hydrogen chloride gas.
ClONO₂(g) + H₂O (g) → HOCl(g) + HNO₃(g)
ClONO₂(g) + HCl(g) → Cl₂(g) + HNO₂(g)
During spring season, sunlight returns and its warmth cleaves both HOCl and Cl₂ to form free radicals. In other words, they undergo photolysis.

Cl₂(g)→ 2\(\dot { C } \)l(g)
The \(\dot { C } \)l flee radicals initiate chain reaction with ozone resulting in the depletion of ozone layer or forming ozone hole.

Question 11.
What are the major causes of water pollution ? Explain.
Answer:
1. Organic Pollutants. The list of organic pollutants is very large. It includes manures wastes from food processing, rags, paper discards, decaying plants etc. and other organic wastes. All of them cause pollution of water.

• These are decomposed by aerobic bacteria into carbon dioxide, nitrates, sulphates, phosphates etc. but they take up dissolved oxygen from water.
• As a result, the oxygen contents in water decrease considerably.
  This causes the death of aquatic animals particularly the fish.
• It may be noted that anaerobic bacteria do not need oxygen from the decomposition of the organic matter. However, some toxic gases like hydrogen sulphide, ammonia, phosphine, methane etc. are produced.
• These are quite often noticed in the sewage wastes.
• We know that the presence of oxygen in water is quite essential for the aquatic life.
• The source of oxygen in water is natural aeration or photosynthesis carried by water plants during day time in the presence of sun light.

The quantity of the oxygen consuming wastes in water can be determined in terms pf biological oxygen demand (B.O.D.).
It may be defined as: the amount of oxygen in milligrams dissolved in water needed to break down the organic matter present in one litre of water for five days at 20°C.

2. Industrial wastes. The compounds of lead, mercury, cadmium, nickel, cobalt, zinc etc. which are the products of chemical reactions carried in the industrial units also pollute water to a large extent and are responsible for many diseases.
Mercury leads to minamata disease,’and lead poisoning leads to various types of deformaties. In addition to this, these chemical substances become apart of soil. They harmfully affect the plant growth and soil biota.
Both ground water and water bodies are polluted due to the chemical reactions known as leaching.

3. Fertilizers. These are the chemical substances which are added to the soil to provide the essential minerals containing N, P, S etc.
The common fertilizers are calcium ammonium nitrate, urea, triple super phosphate, potassium sulphate, potassium nitrate etc. However, a certain part of these fertilizers react with water chemically (known as leaching) and pollute the underground water.
When this water is used for drinking purposes containing potassium nitrate in particular, it harms the respiratory system. Moreover the presence of extra minerals in water is also harmful to many crops.

Question 12.
Have you ever observed any water pollution in your area ? What measures would you suggest to control the same.
Answer:
Control of Water Pollution.
We have seen that the two major sources of water pollution are : sewage and industrial wastes. They should be removed from water before it is put to use.
Treatment of Sewage. Following measures must be taken to check pollution by sewage.
(i) Sewage must be churned by machines so that the large pieces may break into smaller ones and may get mixed thoroughly. The churned sewage is passed into a tank with a gentle slope. Heavier particles settle and the water flowing down is relatively pure.
(ii) Water must be sterilised with the help of chlorination. It kills microbes of sewage fungus as well as some pathogens, spores or cytes. Chlorination is very essential particularly in rainy season.
(iii) Treatment of water with alum, lime etc. also helps in its purification.
Treatment of industrial waste. The treatment of industrial waste depends upon the nature of the pollutants present. In order to ascertain it, the pH of the medium is first determined and the waste is then neutralised with the help of suitable acids or alkalies.
The chemical substances present in the industrial waste products dissolved in water can be precipitated by suitable chemical reactions and removed later on from water. Quite recently, photocatalysis and ion-exchangers have been developed for the treatment of industrial wastes.

Question 13.
What do you understand by biochemical oxygen demand (B.O.D.) ?
Answer:
It may be defined as :
the amouitt of oxygen in milligrams dissolved in water needed to break down the organic matter present in one litre of water fob five days at 20° C.
Pure water contains B.O.D. upto 3 ppm. In case, this level is more, it will suggest the presence of organic waste in water which consume oxygen.

Question 14.
Do you observe any soil pollution in your neighbourhood ? What efforts will you make in controlling the soil pollution ?
Answer:
Faulty Agricultural Practices. In the present era, the major thrust is to get more yield of the crop and on intensive farming. This employs the use of a lot of fertilizers, pesticides, weedicides etc. All of them are chemical substances and from the soil they pass to the ground water and are harmful to the aquatic animals. Moreover, water develops foul smell, bad taste and also acquires some brown colour.
Control of Soil Pollution
In order to control soil pollution, the following measures are necessary :

(i) Use of manures. Manure is a semi-decayed organic matter which is added to the soil to maintain its fertility. These
are mostly prepared from animal dung and other farm refuse. These are much better than the commonly used fertilizers. •

(ii) Use of bio-fertilizers. These are organisms which are inoculated in order to bring about nutrient enrichment of the soil e.g., nitrogen fixing bacteria and blue-green algae.

(iii) Proper sewerage system. A proper sewerage system must be employed and sewerage recycling plants must be installed in all towns and cities.

(iv) Salvage and recycling. Rag pickers remove a large number of waste articles such as paper, polythene, card board, rags, empty bottles and metallic articles. These are subjected to recycling and this helps in checking soil pollution.

Question 15.
What are pesticides and herbicides ? Explain giving examples.
Answer:
Pollution by Pesticides
We have so far discussed the pollution resulting from air, water and soil. In addition to these, pesticides are the major pollutants. These are the chemical substances which contaminate our food as well as drinking water. In fact, their major role is to kill or block the reproductive processes in organism which are not needed and, thus, save the soil from pests. These have been classified in three types.
Insecticides.
Insecticides are the chemical substances which destroy the bacterias causing malaria and yellow fever. Moreover, they also protect the crops from various insects.
The best known insecticide is D.D.T. (dichlorodiphenyltrichloroethane). Since it is an organo chloro coippound, the chlorine acts as a toxic to insects. It also does not dissolve in water and there is no danger of causing water pollution.
However, major disadvantage is because of its non-biodegradable nature. It gets accumulated in the environment and has many harmful effects. It is no longer being used and is replaced by a better insecticides like BHC.

Question 16.
What do you understand by green chemistry ? How will it help in decreasing environmental pollution ?
Answer:
We have studied that the major cause of environment pollution is the release of toxic chemicals which are formed as a result of the processes and reactions carried at various levels. In other words, chemists are mainly responsible for polluting the atmosphere although they manufacture products that are source of our comfort. This has forced them to change their outbook. Since 1990, a new concept called Green Chemistry has been introduced.

• By Green Chemistry we mean the production of substances of daily use by chemical reactions which neither employ toxic chemicals nor release the same to the atmosphere.
• No doubt this is altogether a new field but some success has been achieved. Efforts have been made to carry the reactions in the presence of ultraviolet sunlight (known as photochemistry) and with sound waves (called sonochemistry).
• Microwaves have been used to carry reactions which neither need toxic solvents nor release such vapours into the atmosphere.
• Automobile engines have been fitted with catalytic converters which prevent the release of the vapours of hydrocarbons and oxides of nitrogen into the atmosphere.
• These are the real culprits since they form poisonous substances such as formaldehyde, acrolein and peroxyacetyl nitrate.
• Carbon dioxide has replaced of chlorofluorocarbons as blowing agents in the manufacture of polystyrene foam sheets.
• This has checked the release of chemicals into the environment which cause depletion of ozone layer.
• Similarly, some enzymes have been employed as biocatalysts in the manufacture of certain antibiotics such as ampicillin and amoxycillin.

A few noteworthy measures under the fold of Green Chemistry to check pollution are mentioned :

(a) Dry cleaning of clothes. A commonly used dry cleaning solvent is tetrachloroethene (Cl₂C = CCl₂). It pollutes water and is also carcinogenic. This has been replaced by some other detergents which contain liquid carbon dioxide. These do not pollute water and give better results. These days, hydrogen peroxide is being used in the laundries for removing stains from clothes.

(b) Bleaching of paper. Chlorine gas was used earlier for bleaching of paper. It releases poisonous fumes in the atmosphere. At present, hydrogen peroxide is also used for this purpose.

(c) Synthesis of chemicals. Ionic catalysts in the form of Pd²⁺ and Cu²⁺ salts have been employed for the preparation of acetaldehyde (ethanal) from etIrene by carrying oxidation with oxygen. The yield of ethanal is excellent (about 90%)

This is just the beginning and the results are encouraging. We are sure, our scientists particularly the chemists will be successful in developing new techniques as well as chemicals to minimise pollution. All countries are giving importance to the study of biotechnology which has a major role in checking pollution. It is the duty of every individual to keep the environment free from pollution.

Question 17.
What would happen if green house gases were totally missing in earth’s atmosphere ? Discuss.
Answer:
The solar energy which is radiated back from the earth surface is absorbed by different green house gases that we have listed. As a consequence of it, the atmosphere around the surface of the earth becomes warm. This helps in the growth of vegetation and also supports life. In the absence of this effect, there will be no life of both plant and animal on the surface of the earth.

Question 18.
A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you can find an abundance of phytoplankton. Suggest a reason for the fish kill.
Answer:
Phytoplankton growth occurs in water because of the presence of organic matter like leaves, grass, trash etc. in water. It is likely to consume a lot of oxygen dissolved in water which is of course very much essential for the life of sea animals particularly fish.
If the level of dissolved oxygen in water is below 6 ppm, this means that the oxygen is not sufficiently available to the variety of fish living in water. They are likely to perish or die. This might have happened in this particular case.

Question 19.
How can domestic waste be used as manure ?
Answer:
Domestic waste consists of both biodegradable and non-biodegradable components. The latter consisting of plastic, glass, metal scrap etc. is separated from it. The biodegradable portion which consists of organic matter can be converted into manures by suitable methods.

Question 20.
For your agriculture field or garden, you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Answer:
For the healthy growth of plants and grass in the garden, compost is periodically required. Compost producing pit is usually created nearly provided space is available. Difficulties do arise in urban areas due to paucity of space.
The pits generally give foul smell and flies roam about. This is very bad for health. In order to check it, the pits must be properly covered.
The waste materials such as glass articles, plastic bags, old newspapers etc. must be handed over to the vendors regularly. These are ultimately sent to recycling plants without creating pollution problems.

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 14 Environmental Chemistry, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 14 Environmental Chemistry, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 13 Hydrocarbons

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 13 Hydrocarbons.

Question 1.
How will you account for the formation of ethane during chlorination of methane ?
Answer:
The chlorination of methane proceeds by free radical mechanism. The methyl free radicals (CH₃) are converted to ethane during chain termination step.
H₃\(\dot { C } \) + \(\dot { C } \)H₃ → CH₃—CH₃
Chlorination. In the chlorination of methane, all the four hydrogen atoms present in the molecule get replaced by one by to form a mixture of different substituted products.

Question 2.

Answer:

Question 3.

Answer:

Question 4.
Write the IUPAC names of the products obtained by the ozonolysis of the following compounds :
(i) Pent-2-ene
(ii)3, 4-dimethylhept-3-ene
(iii) 2-Ethylbut-l-ene
(iv) 1-Phenylbut-l-ene.
Answer:

Question 5.
An alkene ‘A’ upon ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of ‘A’.
Answer:
The double bond is present between the carbon atoms of the two carbonyl compounds that are formed by ozonolysis.

Question 6.
An alkene ‘A’ contains three C—C, eight C—H and one C—C (JI) bonds. Upon ozonolysis ‘A’ gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of ‘A’.
Answer:
The aldehyde (product of ozonolysis) with molar mass 44 u is CH₃CH=0. Since two moles of the same aldehyde (propanal) are formed from the alkene ‘A’, the formula of alkene is :

Question 7.
Propanal and pentan-3-one are the ozonolysis products of an alkene. What is the structural formula of alkene ?
Answer:

Question 8.
Write the chemical equations for the combustion of the following hydrocarbons :
(a) Butane
(b) Pentene
(c) Hexyne
(d) Toluene.
Answer:
By definition, combustion is for one molecule (one mole) of the substance. The combustion equations may be written as :

Question 9.
Draw cis and trans structures of hex-2-ene. Which isomer will have higher boiling point and why ?
Answer:

Cis isopier will have higher boiling point because of greater magnitude of dipole-dipole intractions as compared to the trans isomer.

Question 10.
Why is benzene extra-ordinary stable though it contains three double bonds ?
Answer:
It is on account of resonance shown by benzene. Moreover, there is delocalisation of π-electron charge in benzene.

Question 11.
What are tjhe necessary conditions for a system to be aromatic ?
Answer:
We have stated in our earlier discussion that benzene and arenes are aromatic in nature. In general, if a compound is to be aromatic, it must fulfil the following conditions : ,
(i) The compound must be cyclic in nature with atleast one or more double bonds in the ring.
(ii) Contrary to unsaturation as suggested by the molecular formula, it must behave like saturated compounds i.e., must resist addition and take part in the electrophilic substitution reactions.
(iii) The compound must be capable of exhibiting resonance.
(iv) The most essential criteria for the aromatic character is that the compound must obey Huckel’s rule. According the rule, a cyclic compound will behave as aromatic compound if it contains (4n + 2) π electrons, where n may be 1, 2, 3, …. etc.

Question 12.
Explain why the following systems are not aromatic ?

Answer:

Question 13.
How will you convert benzene into
(i) p-chloronitrobenzene
(ii) m-chloronitrobenzene
(iii) p-nitrotoluene
(iv) Acetophenone ?
Answer:

Question 14.
Inthealkane, CH₃—CH₂—C(CH₃)₂—CH₂—CH₂(CH₃)₂,identify 1°,2°,3° and4° carbon atoms.
Answer:

Question 15.
What is the effect of branching of alkane chain on the its boiling point ?
Answer:
Branching of carbon atom chain decreases the boiling point of alkane.
Boiling points. Alkanes are non-polar molecules and the only attractive forces in their molecules are weak van der Waals’ forces. Therefore, the members of the family are low boiling in nature. The addition of each carbon atom (or CH₂ group) in the chain increases the boiling point nearly by 30 K. The boiling points of some normal alkanes are given below :

Question 16.
Addition of HBr to propene yields 2-bromopropane whde in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Answer:
Mechanism of reaction. The addition of HBr in the presence of organic peroxide follows free radical mechanism.

Question 17.
Write the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekule structure of benzene ?
Answer:

All these products can be possible only in case, there are three double bonds in the ring in the alternate positions. The products of ozonolysis support Kekule structure.

Question 18.
Arrange benzene, n-hexane and ethyne in decreasing order of acidic strength. Also give reason for this behaviour.
Answer:

The acidic character is linked with percentage s-character. Greater the ^-character, more is the electronegativity of the carbon atom and more will be the acidic character.

Question 19.
Why does benzene undergo electrophilic substitution easily and nucleophilic substitution with difficulty ?
Answer:
Benzene has high electron density due to the presence of three ^-electron pairs representing double bonds. Although the electron charge is very much delocalised due to resonance, still electrophile attack leading to electrophilic substitution is possible. However, benzene does not respond to nucleophilic substitution because nucleophile prefers to attack a centre of low electron density.
Role of catalyst in Electrophilic Substitution Reactions.
In the monosubstitution reactions of benzene discussed earlier in the properties of arenes, we have seen that a catalyst is always present which may be either a Lewis acid (Ferric salt or Anhydrous aluminium chloride) or a proton acid (sulphuric acid).

The catalyst is needed to help in generating the electrophile (E⁺) from the attacking molecule. In fact, the ^-electrons in the benzene are delocalised and are not in a position to cause the polarisation of the attacking molecule.

The catalyst helps in its polarisation which may be illustrated by the chlorination of benzene. The catalyst FeCl₃ is a Lewis acid and causes the heterolysis of the chlorine molecule.

Question 20.
How will you convert following into benzene ?
(i) Ethyne
(ii) Ethene
(iii) Hexane
Answer:

Question 21.
Write the structures of all the alkenes which upon hydrogenation give 2-methylbutane.
Answer:

Question 22.
Arrange the following sets of compounds in order of their decreasing relative reactivity with an electrophile and assign reason.
(a) Chlorobenzene, 2, 4-dinitrochlorobenzene, p-nitrochlorobenzene.
(b) Toluene, pH₂CC₆H4NO₂, pO₂NC₆H₄NO₂
Answer:
(a) The correct order of decreasing reactivity towards electrophilic substitution is :

Nitro (NO₂) group is a deactivating group. Its presence on the benzene ring will deactivate it towards electrophile attack since electrophile seeks a centre of high electron density. Thus, more the number of nitro groups present, lesser will be the reactivity of the compound towards electrophilic substitution.

(b) The correct order of decreasing reactivity is :

Methyl group is an activating group while the nitro group is deactivating in nature. In the light of this, the decreasing order of reactivity towards electrophilic attack is justified.

Question 23.
Out of benzene, /n-dinitrobenzene and toluene, which will undergo nitration most easily and why ?

Answer:
Nitration of benzene involves the electrophile attack of NO₂ (nitronium ion) on the ring. Since CH₃ group has +I effect, it activates the ring and electrophilic substitution readily takes place. On the other hand, it is most difficult in m-dinitrobenzene because of deactivating nature of nitro groups. Thus, toluene will undergo nitration most readily.

Question 24.
Suggest the name of another Lewis acid instead of anhydrous aluminium chloride which can be used during ethylation of benzene.
Answer:
Anhydrous ferric chloride (FeCl₃) is another Lewis acid which can be used. It helps in generating electrophile (C₂H₅⁺). Even stannic chloride (SnCl₄) and boron trifluoride (BF₃) can be used.

Question 25.
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms ? Illustrate your answer by taking an example.
Answer:
In order to prepare alkane with odd number of carbon atoms, two different haloalkanes are needed ; one with odd number land the other with even number of carbon atoms. For example, bromoethane and 1-bromopropane will give pentane as a result of the reaction.

But side products will also be formed when the members participating in the reaction react separately. For example, bromoethane. will give butane and 1-bromopropane will give rise to hexane.

Thus, mixture of butane, pentane and hexane will be formed. It will be quite difficult to separate the individual components from the mixture.

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 13 Hydrocarbons, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 13 Hydrocarbons, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry : Some Basic Principles and Techniques

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 12 Organic Chemistry : Some Basic Principles and Techniques

Question 1.
What are the hybridised states of carbon atoms in the following compounds ?
(1) CH₂=C=O
(2) CH₃CH=CH₂
(3) (CH₃)₂ C=O
(4) CH=CH₂ CN
(5) C₆H₆
Answer:

Question 2.
Indicate sigma (σ) and pi (Π) bonds in the following molecules :
(1) C₆H₆ (2) C₆H₁₂  (3)   CH₂C₁₂  (4) CH₃NO₂  (5)  HCONHCH3  (6)  CH₂=C=CH₂
Answer:

 

Question 3.
Write bond line formulas for :
(1) Isopropyl alcohol (2) 2, 3-Dimethylbutanal (3) Heptan-4-one
Answer:

Question 4.
Give the IUPAC name of the following compounds :

Answer:

Question 5.
Which of the following represents the correct IUPAC name of the compounds concerned ?
(a) 2, 2-Dimethylpentane or 2-Dimethylpentane
(b) 2, 3-Dimethylpentane or 3, 4-Dimethylpentane
(c) 2, 4, 7-Trimethyloctane or 2, 5, 7-Trimethyloctane.
(d) But-3-yn-l-ol or But-4-ol-l-yne
Answer:

Question 6.
Draw the formulas of the first five members of each homologous series beginning with the following compounds :
(a) HCOOH (b)CH₃COCH₃ (c) CH₂ = CH₂
Answer:
(a) HCOOH : CH₃COOH, CH₃CH₂COOH, CH₃CH₂CH₂COOH, CH₃CH₂CH₂CH₂COOH
(b) CH₃COCH₃ : CH₃COCH₂CH₃, CH₃COCH₂CH₂CH₃, CH₃COCH₂CH₂CH₂CH₃ CH₃CO(CH₂)₄CH₃.
(c) CH₂ = CH₂ : CH₃CH=CH₂, CH₃CH₂CH=CH₂, CH₃CH₂CH₂CH = CH₂, CH₃CH₂CH₂CH₂CH = ch₂

Question 7.
Give the condensed and bond line formulas for the following compounds :
(1) 2,2, 4-Trimethylpentane (2) 2-Hydroxy-l, 2, 3-propanetricarboxylic acid (3) Hexanedial
Answer:

Question 8.
Identify the functional groups in the following compounds :

Answer:

1. Aldehydic, methoxy, phenolic
2. Amino, N-diethylaminopropanoate (Ester)
3. Ethylenic double bond, nitro

Question 9.
Which is expected to be more stable : O₂NCH₂CH₂O^– or CH₃CH₂O^– and why ?
Answer:
O₂NCH₂CH₂O^– is more stable because —NO₂ group with -I effect disperses the negative charge on the anion. On the other hand, CH₂CH₂– group with +1 effect increases the magnitude of the negative charge on the anion and thus destabilises it.

Question 10.
Give the various resonating structures associated with the following molecules :
(1) C₆H₅OH (2) C₆H₅NO₂ (3) CH₃CH=CHCHO (4) C₆H₅CHO (5) C₆H₅C H₂ (6) CH₃CH = CH C H₂
Answer:

 

 

Question 11.
Explain why does an alkyl group act as an electron donor when attached to a Π-electron system.
Answer:
Alkyl group has no lone pair of electrons but it acts as an electron donor when attached to a n-electron system because of hyper conjugation. Let us illustrate by toluene in which methyl (CH₃)group is attached to a benzene ring containing three pi-elecrons in the alternate positions. The various resonating structures are as follows :

Question 12.
What are nucleophiles and electrophiles. Explain with examples.
Answer:
Electrophiles: The name electrophiles means electron loving. Electrophiles are electron deficient. They may be positive ions or neutral molecules.
Ex: H⁺, Cl⁺, Br⁺, NO₂⁺, R₃C⁺, RN₂⁺, AlCl₃, BF₃

Nucleophiles: The name nucleophiles means ‘nucleus loving’ and indicates that it attacks the region of low electron density (positive centres) in a substrate molecule. They are electron rich they may be negative ions or neutral molecules.
Ex: Cl^– Br^–, CN^–, OH^–, RCR₂^–, NH₃, RNH₂, H₂O, ROH etc.

Question 13.
Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles.
(a) CH₃COOH + OH^– → CH₃COO^– + H₂O
(b) CH₃COCH₃+ NC→  CH₃C(CN)OHCH₃
(c) C₆H₆ + CH₃CO→ C₆H₅COCH₃
Answer:
(a) OH^– (nucleophile) (b) NC^– (nucleophile) (c) CH₃C⁺O (electrophile)

Question 14.
Classify the following reactions in one of the reaction type studied in this unit
(1) H₃CH₂Br + SH^– → CH₃CH₂SH + Br^–
(2) (CH₃)₂C = CH₂ + HCl→ (CH₃)₂C(Cl)CH₃
(3) (CH₃)₃CCH₂OH + HBr→ (CH₃)₂CBr CH₂CH₃
(4) CH₃CH₂Br + HO^–→ CH₂ = CH₂ + H₂O + Br^–
Answer:
(1) Nucleophilic’ substitution
(2) Electrophilic addition
(3) Rearrangement of carbocation intermediate formed followed by nucleophilic substitution.
(4) Elimination.

Question 15.
What is the relation between the following pairs ?

Answer:
These are position isomers.
(1) These are geometrical isomers.
(2) These are resonating structures since they differ in position of the electron pairs and not of atoms.

Question 16.
Classify each of the following as homolysis or heterolysis. Identify the reaction intermediates produced ; as free radical, carbocation and carbanion.

Answer:

Question 17.
Explain the terms inductive and electromeric effects. Which electron displacement effect explains the following correct order of the acidity of the carboxylic acids ?
(a) CH₃COOH > Cl₂CHCOOH > ClCH₂COOH
(b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃COOH
Answer:
Electronic displacements in covalent bonds present in the organic molecules may occur either due to the presence of an atom or group of different electronegativity or under the influence of some outside attacking species also called attacking reagent.

The electromeric effect may be defined as :

The temporary effect which operates in the organic compounds having multiple bonds i.e. double or triple bonds under the influence of an outside attacking species. As a result, one pi electron pair of the multiple bond gets completely transferred to one of the bonded atoms which is usually more electronegative.

The electromeric effect is shown by a curved arrow representing the electron transfer originating from the center of the multiple bond and pointing towards one of the atoms which is more electronegative.The effect can be illustrated by the attack of H⁺ ion on the molecule of alkene.

(a) The order of acidity is explained by -I effect of the chlorine atoms. Greater the magnitude of -I effect, easier will be the release of H⁺ from O-H bond and stronger will be acid.

(b) The order of acidity is explained by +I effect of alkyl groups. As the number of alkyl groups increases, the magnitude of +1 effect also increases. As a result, the release of H⁺ from O-H bond becomes more and more difficult and acidic strength decreases.

Question 18.
Give a brief description of the principle of the following processes taking one example in each case.
(1) Filtration (2) Recrvstallisation (3) Distillation (4) Chromatography
Answer:
(1) Filtration
The hot saturated solution is filtered preferably through a fluted filter paper placed in a glass funnel (Fig. 12.12). The use of the fluted filter paper makes the process of filtration rapid. If the organic compound to be purified has a tendency to crystallise out during filtration, then a hot water funnel is used for filtration (Fig 12.13). The jacket of the hot water funnel is heated from outside and this keeps the solution hot in the glass funnel. This will prevent the formation of crystals during filtration. The insoluble impurities will remain on the filter paper and a clear solution gets collected in the beaker or a dish placed below the funnel.

(2) Removal of colour (Recrystallisation). Some times the crystals are coloured due to the presence of some coloured impurities. In such a case, the coloured crystals are redissolved in the same solvent and the saturated solution is prepared. Before filtering the hot solution, a pinch of animal charcoal is added and the solution is boiled for one to two minutes. It is then filtered as before. Thecharcoal adsorbs all the coloured impurities. From the solution, pure crystals can be recovered as described earlier. The process is known as recrystallisation.

(3) Distillation
Distillation is the process of converting a liquid into vapours upon heating and then cooling the vapours back to the liquid state.
The process of simple distillation is used to purify those organic liquids which are quite stable at their boiling points and the impurities present are non-volatile. Liquids such as benzene, toluene, ethanol, acetone, chloroform, carbon tetrachloride can be purified by simple distillation.

Procedure. The impure liquid is taken in the distillation flask which is fitted with a .condenser called Liebig’s condenser and a thermometer. The flask is heated on a water bath, sand bath or even’directly depending upon the boiling point of the liquid to be distilled. As a result, the liquid changes to its vapours when its boiling point temperature is attained. The vapours then pass through the condenser around which water is circulated as shown in the figure. The vapours condense to form the liquid which is collected in the receiver. The non-volatile impurities are left behind in the distillation flask and can be recovered.

Most of the liquids start bumping when heated. In order to check this, a few pieces of unglazed porcelain or glass beeds are added in the flask.

The process of distillation can also be used to separate a liquid mixture in which the two components present differ in the boiling points ranging from 30 to 50 K e.g., a mixture of diethyl ether (b.p. 308 K) and benzene (b.p. 353 K). When the liquid mixture is heated in the distillation flask, only the vapours of the low boiling liquid will be formed at its boiling point while the high boiling liquid will not change to the vapour state. The vapours of the low boiling liquid will escape and after getting condensed will be collected in the receiver. The high boiling component left in the distillation flask can be recovered from it.

(4) Fractional Distillation
Sometimes, we come across a mixture of two liquids which differ in their boiling point temperatures by 10-20 K. The process of simple distillation will fail here because the vapours of both the liquids will be formed simultaneously and the distillate collected in the receiver will contain both of them. In such cases, the process of fractional distillation is used which employs specially designed columns called fractionating columns. The purpose of the column is to obstruct the movement of the vapours as they rise up. The column actually used depends upon the nature of the liquid mixture.
Procedure. The apparatus used is the same as in the distillation except a fractionating column. Upon heating the vapours of both the liquids will be formed.

As they rise into the column, the vapours of the high boiling liquid will have a tendency to condense first. In doing so, they release heat which is taken up by the vapours of the low boiling liquid. As a result, the latter remains in the vapour state. The vapours which escape the distillation flask are of low boiling liquid. They get condensed by the water condenser and the liquid formed is collected in the receiver. The vapours of the high boiling liquid fall back in the distillation flask. Thus, the separation of the two liquids can be done.

Sometimes, some vapours of the high boiling liquid also escape the distillation flask in case the difference in the boiling points of the two liquids is very small. In such cases the process of fractional distillation is repeated again.

In the laboratory, a mixture of methyl alcohol (b.p. = 338 K) and acetone (b.p = 329 K) can be separated by fractional distillation. The principle of the fractional distillation employed to separate petroleum into different fractions is also the same but since it has to be done on a commercial scale, a number of fractionating columns are used to separate different fractions.

Question 19.
Describe the method which can be used to separate two compounds with different solubilities in the solvent S.
Answer:
The separation can be done with the help of fractional crystallisation. Choice of the solvent. Inorganic compounds are mostly water soluble. The organic compounds, on the other hand, are generally soluble in organic solvents which may be different for different compounds. In order to make the proper choice of the solvent, the following points must be kept in mind.

• The organic solid must dissolve in the solvent upon heating and must get separated when the hot solution is cooled.
• The impurities should not normally dissolve in the solvent. If at all they dissolve, they should be soluble to such a small extent that they may remain in the mother liquor which gets separated from the crystals.
• The solvent must not react chemically with the organic compound.

The solvents commonly used an of organic nature such as ethyl alcohol, benzene, chloroform, ether, carbon tetrachloride etc. Even water can be used in some cases.

Question 20.
What is the difference between distillation, distillation under reduced pressure and steam distillation ?
Answer:
Distillation is employed in case of volatile liquids associated with non-volatile impurities.
Distillation under reduced pressure is carried to purify liquids which decompose at their boiling point temperatures. Steam distillation is done for the steam volatile liquids associated with water immiscible impurities.
Steam Distillation

This process is used to purify the impure liquid by passing steam and is applicable under the following conditions.

• The liquid must be steam volatile but the impurities present must be non-volatile.
• The liquid must not be miscible with water.
• The liquid must possess sufficiently high vapour pressure at the boiling point temperature of water (373 K).

With the help of steam distillation, the liquids which have boiling points higher than the boiling point of water (373 K) can be distilled at lower temperature under reduced pressure inside the flask. Thus, steam distillation is comparable to distillation under reduced pressure. This can be explained with the help of Dalton’s law of partial pressures as given ahead :

According to the law, the pressure exerted by a gaseous mixture in a container is equal to sum of the partial pressures of the constituent gases provided they do not react chemically, i.e., P = P₁ + P₂.
Where P = Total pressure of the mixture (atmospheric pressure)
Pi = Vapour pressure of steam
P₂ = Vapour pressure of the liquid vapours

It is quite obvious that the vapour pressure of the liquid inside the flask is less than P which is atmospheric pressure. Thus, the liquid boils under reduced pressure and gets distilled along with vapours of steam at a temperature lower than its boiling point temperature.

We can also calculate the ratio of the masses of the organic liquid and water (steam) which escape from the distillation flask as distillate. According to ideal gas equation, PV = nRT. At constant temperature and volume,

Procedure. The liquid to be purified is mixed with small quantity of water and put in a flask which has two delivery tubes fitted to it, one going into a liquid and the other remaining near the cork. The delivery tube which dips into the liquid is connected to a steam generator and the other one is joined to a Liebig’s condenser which opens into a receiver. The liquid in the flask is heated and steam is bubbled through it. As a result, the liquid changes into vapours while the impurities remain in the flask. The vapours of the liquid along with the steam escape from the flask. These vapours get condensed and are collected in the receiver. The separation of the liquid from water can be done with the help of a separating funnel. The use of a separating funnel has been discussed under differential extraction.

• An impure sample of aniline can be purified by carrying out the steam distillation. The process of steam distillation can also be used to separate a mixture of two organic substances one of which is steam volatile while the other is not. For example, if steam is passed through a mixture of o-nitrophenol and p-nitrophenol taken in the distillation flask, then steam will carry along with it vapours of o-nitrophenol which is low boiling liquid leaving behind p-nitrophenol in the flask which has comparatively high boiling point.
• Fragrance of flowers is due to the presence of some steam. Volatile organic compounds called essential oils present in perfumes, cosmetics etc. These are insoluble in water at room temperature but are miscible in the vapour phase. In other words, these are steam volatile. These can be isolated with the help of steam distillation.

Question 21.
Describe the chemistry of Lassaigne’s test.
Answer:
Lassaigne’s Test,
Nitrogen in an organic compound is detected mainly by Lassaigne ’ s test which is described as follows:

(a) Preparation of Lassaigne’s extract. A small piece of dry sodium metal is heated gently in a fusion tube till it melts to a shining globule. At this stage, a small amount of organic substance is added and the tube is heated strongly for two to three minutes. The red hot tube is plunged into distilled water contained in a china dish. The contents of dish are boiled for couple of minutes, cooled and filtered. The filtrate is known as sodium extract or Lassaigne’s extract.

(b) Test for nitrogen. The Lassaigne’s extract is usually alkaline because excess of sodium reacts with water to form sodium hydroxide. If not, it may be made alkaline by the addition of a few drops of a dilute solution of sodium hydroxide. To a part of the extract, a small amount of a freshly prepared ferrous sulphate solution is added and the contents are warmed. A few drops of ferric chloride solution are then added to the contents and the resulting solution is acidified with dilute hydrochloric acid. The appearance of a bluish green colour due to the formation of ferric ferrocyanide (prussian blue) confirms the presence of nitrogen in the organic compound.

(c) Chemistry of the test. During fusion, carbon and nitrogen present in the organic compound combine with sodium to form sbdium cyanide.

If organic compound contains both nitrogen and sulphur, then they combine with sodium metal to form sodium thiocyanate also called sodium sulphocyanide. This compound will react with ferric chloride to form ferric thiocyanate (or sulphocyanide) which is blood red in colour.

Question 22
Differentiate between the principle of estimation of nitrogen in an organic compound by
(1) Duma’s method
(2) Kjeldahl’s method.
Answer:
In Duma’s method, nitrogen evolved from the organic compound is measured and then estimated. In Kjeldahl’s method, nitrogen present in the organic compound is converted into ammonia which then is estimated volumetrically.
Duma’s Method and Kjeldahl’s Method
Duma’s method can be employed to estimate nitrogen in all organic compounds.
Principle : A known mass of the given organic compound is heated strongly with excess of cupric oxide in an atmosphere of CO₂. Carbon and hydrogen are oxidised to CO₂ and H₂O respectively while nitrogen present in the compound is set free. A small amount of the oxides of nitrogen if formed are reduced back to nitrogen by passing over hot reduced copper gauze.


The gaseous mixture is passed through concentrated KOH solution which absorbs both CO₂ and H₂O vapours. Nitrogen (N₂) is not absorbed by KOH and gets collected over it. The volume of nitrogen evolved is noted and from this the amount of nitrogen or its percentage can be calculated by applying suitable calculations.
Apparatus. The apparatus used to estimate nitrogen by Duma’s method is shown in the figure 12.28.

It consists of a combustion tube which is a long hard glass tube open at both ends. It contains in it a roll of oxidised copper gauze to prevent the backward diffusion of the products of combustion. This is followed by small amount of weighed organic compound (about 2 g) mixed with excess of cupric oxide. A layer of coarse cupric oxide covers about half of the combustion tube as shown in the figure. At the end of the tube there is a roll of reduced copper gauze which has been kept to reduce any oxides of nitrogen formed in the oxidation reactions back to nitrogen. In order to collect the evolved nitrogen, the combustion tube is connected to a Schiff’s nitrometer which is a graduated tube containing mercury at its bottom. Mercury acts as a seal and does not allow any liquid to flow back irrthe combustion tube. The nitrometer contains in it about 40 % aqueous KOH solution which absorbs both CO₂ and H₂O vapours evolved in the combustion reaction. However, nitrogen cannot be absorbed and it get collected over it in the nitrometer. A reservoir attached to the nitrometer helps to record the volume of nitrogen at the atmospheric pressure. The combustion tube is kept in a furnace where it can be heated.

Procedure : The apparatus is fitted as shown in the Figure 12.28. To start with, the tap of the nitrometer is opened and a carbon dioxide formed by heating sodium bicarbonate is passed through the combustion tube in order to expel any air or oxygen present in the tube. After sometime, the tap of nitrometer is closed and the reservoir is raised in order to completely fill the nitrometer tube with KOH solution. The combustion tube is now heated in the furnace. Both CO₂ and H₂O vapours evolved are absorbed by KOH while nitrogen which is set free gets collected over KOH and its level is, therefore, pushed downwards. Towards the end of the experiment, a strong current of carbon dioxide is passed through the combustion tube to remove last traces of nitrogen if present.
The apparatus is cooled and nitrometer tube is disconnected. The volume of nitrogen is noted after levelling i.e., by keeping the level of KOH in the nitrometer and in the reservoir, same. This will give the volume of nitrogen at the atmospheric pressure which can be recorded from a barometer. The room temperature and the corresponding aqueous tension are also recorded.

Question 23.
Describe the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.
Answer:
All these elements if present in the organic compound, are estimated by Carius Method.
Estimation of Halogens
Principle : In carius method, the halogen present in an organic compound is converted into the corresponding silver halide (AgX). From the mass of the organic compound taken and that of silver halide formed, the percentage of halogen in the compound can be calculated.                                                                                                                              ‘

Procedure: About 5 mL of fuming nitric acid and 0-5 g of silver nitrate are taken in the carius tube made up of hard glass. It is about 50 cm long and is closed at one end. A small amount of the organic compound to be estimated is taken in a small tube which is also placed carefully in the carius tube. The tube is now sealed and is placed in an outer jacket made of iron.

It is heated in a furnace to 550 to 560 K for nearly six hours. Under the reaction conditions, carbon and hydrogen present in the compound are oxidised to CO₂ and H₂O vapours respectively. Halogen gets converted into silver halide which is precipitated. The high pressure developed inside the tube is released by softening the sealed end with a small flame. A hole gets created through which the gases escape. The end of the tube is then cut off and the contents are transferred into a beaker. The precipitate of silver halide is filtered, washed, dried and is then weighed.

Calulations

Question 24.
Explain the principle of paper chromatography.
Answer:
Principle of Chromatography     

The technique of chromatography is based on the difference in the rates at which the components of a mixture are adsorbed on a suitable adsorbent. The material on which the various components are adsorbed is called stationary phase. It is of porous nature and the common substances which can be used in preparing the stationary phase are alumina, silica gel, calcium carbonate or activated charcoal. The mixture to be separated is dissolved in a suitable medium; may be a liquid or a gas. It constitutes the moving phase. The moving phase is made to run on the stationary phase and the separation is based on the principle that the components of the mixture present in moving phase move at different rates through the stationary phase.

Question 25.
Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens ?
Answer:
Problem can arise in case the organic compound which also’contains nitrogen and sulphur in addition to halogens. Both will form NaCN and Na₂S on fusion with sodium metal and will react with silver nitrate solution to give precipitates.

These precipitates will interfere’with the precipitate of silver halide formed due to halogens. Nitric acid is added to destroy both these compounds to form volatile gases.

 

Question 26.
Why is an organic compound fused with sodium for testing nitrogen, halogens and sulphur ?
Answer:
On fusing with sodium, these elements present in the compound are converted into their sodium salts (NaCN, NaX and Na₂S) which are water soluble. From the solution, these elements can be detected by suitable tests.
Test for nitrogen. The Lassaigne’s extract is usually alkaline because excess of sodium reacts with water to form sodium hydroxide. If not, it may be made alkaline by the addition of a few drops of a dilute solution of sodium hydroxide. To a part of the extract, a small amount of a freshly prepared ferrous sulphate solution is added and the contents are warmed. A few drops of ferric chloride solution are then added to the contents and the resulting solution is acidified with dilute hydrochloric acid. The appearance of a bluish green colour due to the formation of ferric ferrocyanide (prussian blue) confirms the presence of nitrogen in the organic compound.

Question 27.
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
Answer:
The separation can be done by the process of sublimation. Camphor being volatile in nature will undergo sublimation. Calcium sulphate will remain as the residue as it is non-volatile in nature.
Sublimation
The process of sublimation is applicable to purify those solids which sublime i.e., they directly pass to the vapour state upon heating without passing through the liquid state and the vapours upon cooling give back the solid again. But the impurities associated with them are non-volatile.

The process can be used to purify substances like naphthalene, camphor, benzoic acid, etc. The impure sample is taken in a China dish and is covered by a perforated porcelain plate. An inverted glass funnel is placed over the dish and its stem‘is plugged with cotton. The dish is heated gently when the volatile substance changes to the vapours. These vapours pass through the perforations of the plate and get collected on the inner cold surface of the funnel. This is known as sublimate. The non-volatile impurities remain on the dish. The sublimate can be removed from the funnel.

For example, an impure sample of naphthalene can be purified by sublimation.
Naphthalene is collected as sublimate on the inner surface of the inverted funnel and the impurities are left as residue in the China dish.

Question 28.
An organic liquid vaporises at a temperature below its boiling point in steam distillation. Assign reason.
Answer:
Steam distillation is actually distillation under reduced pressure. The vapour pressures of both water vapours and organic liquid placed in the distillation flask become equal to the atmospheric pressure. This means that both of them will vaporise at a temperature which is less than their normal boiling point temperatures.

Question 29.
Carbon tetrachloride does not give a white precipitate upon heating with silver nitrate solution. Is it correct ?
Answer:
Yes it is correct. Carbon tetrachloride (CCI₄) is a completely non-polar covalent compound whereas silver nitrate is ionic in nature. Therefore, they are not expected to react and a white precipitate of silver chloride will not be formed.

Question 30.
A solution of potassium hydroxide is used to absorb carbon dioxide evolved during the estimation of carbon in an organic compound. Explain.
Answer:
Carbon dioxide reacts with KOH present in the solution to form soluble potassium carbonate and can be estimated.
2KOH + CO₂→K₂CO₃ + H₂O

Question 31.
It is not advisable to use sulphuric acid in place of acetic acid for acidification while testing sulphur by lead acetate test. Assign reason.
Answer:
Lead acetate will react with sulphuric acid to give white precipitate of lead sulphate. This will interfere with the detection of the test for sulphur.

Acetic acid (CH₃COOH) will not interfere in the detection of sulphur.

Question 32.
An organic compound contains 69% carbon and 4-8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0-20 g of this compound is subjected to complete combustion.
Answer:



Question 33.
0.50 g of an organic compound was kjeldahlished. The ammonia evolved was passed in 50 cm³ of in H₂SO₄. The residual acid required 60 cm³ of N/2 NaOH solution. Calculate the percentage of nitrogen in the compound.
Answer:

Question 34.
O.3780 g of an organic compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine in the compound.
Answer:

Question 35.
In an estimation of sulphur by Carius method, 0.468 of an organic sulphur compound gave 0.668 g of barium sulphate. Find the percentage of sulphur in the compound.
Answer:

 

Question 36.
In the organic compound CH2 = CH—CH₂—CH₂—OCH, the CH—CH₂ bond is formed by the interaction of a pair of hybridised orbitals :
(a) sp – sp²
(b) sp – sp³
(c) sp² – sp³
(d) sp³ – sp³.
Answer:

Question 37.
In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
(a) Na₄[Fe(CN)₆]
(b) Fe₄[Fe(CN)₆]₃
(c) Fe₂[Fe(CN)₆]
(d) Fe₃[Fe(CN)₆]₄.
Answer:
(b)   is the correct answer.

Question 38.
Which of the following carbocation is most stable ?

Answer:
(b) is the most stable since it is a tertiary carbocation.

Question 39.
The best and latest technique for isolation, purification and separation of organic compounds is :
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography 
Answer:
(d)   is the correct answer.

Question 40.
The following reaction is classified as :
CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI
(a) electrophilic substitution
(b)  nucleophilic substitution
(c)  elimination
(d) addition
Answer:
(b) It is a nucleophile substitution reaction. KOH (aq) provides OH^– ion for the nucleophile attack.

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NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements

Question 1.
Discuss the pattern of variation in oxidation states of :
(1) B to Tl
(2) C to Pb.
Answer:
(1) B to Tl. The common oxidation states are +1 and +3. The stability of
+ 3 oxidation state decreases from B to Tl.
+ 1 oxidation state increases from B to Tl.
Oxidation States. The elements of the boron family (Group 13) have ns²p^l confi’guration. This means that they have three valence electrons available for bond formation. By losing these electrons, they are expected to show + 3 oxidation states in their compounds. However, the following trends are observed in the oxidation states of these elements.

• The first two elements boron and aluminium show only +3 oxidation state in the compounds but the remaining elements gallium, indium and thallium also exhibit + 1 oxidation state in addition to + 3 oxidation state i. e., they show variable oxidation states.
• The stability of +3 oxidation state decreases from aluminium onwards and in case of last element thallium, + 1 oxidation is more stable than +3 oxidation state which means that TICl is more stable than TlCl₃.

Explanation. The above trend is explained with the help of the phenomenon of inert pair effect.
It represents the reluctance or inertness of the valence s-electrons of heavier elements of p-block to take part in the bond formation because of ineffective shielding of these electrons from the attraction of nucleus by the intervening d and f electrons.

As a result of the inert pair effect, the electron pair representing the valence 5-electrons is more exposed to the nucleus than the p- electrons. In other words, these are held tightly by the nucleus and are not readily available for the bond formation. However, valence p-electrons are available for the same. The inert pair effect becomes more or more predominant as we go down the group because of effect of increased nuclear charge outweighs the effect of increased atomic size. In other words, the valence 5-electrons become more and more reluctant to be available for bond formation. As a result, the valence ^-electrons will be more available accounting for + 1 oxidation state and + 3 oxidation state will not be shown by these elements so easily. The inert pair effect is maximum in the last element thallium (Tl) in boron family since it is the heaviest with maximum atomic number (Z = 81). Therefore, it shows mainly + 1 oxidation state or we can say that TICl exists while TlCl₃ is rather unstable. Keeping this in mind, the relative stabilities of M ⁺ (monovalent) and M?⁺ (trivalent) cations follow the order:
_B3+ > _Al3+ > _Ga3+ > _In3+ > _Tl3+
B⁺ < Al⁺ < Ga+ < In+ < TI+
The inert pair is not restricted only to the members of boron family. It is also present in the heavier members of group 14 (Sn and Pb) and group 15 (Sb and Bi). Please note that in the 5-block, the group 1 and 2 elements show only one oxidation state which is the same as the group valency. In the p-block, the elements have tendency to show variable oxidation states which are different from the group valencies. The elements presents in group 16 and group 17 also exhibit variable oxidation states ion not because of inert pair effect but due to the presence of vacant ri-orbitals in the valence shells of their atoms to which the electrons can be promoted from s- and p-orbitals also present in the valence shell. We shall study in detail, the variable oxidation states of these elements in the next class.

(2) C to Pb. The common oxidation states are +4 and +2. The stability of
+4 oxidation state decreases from C to Pb
+2 oxidation state increases from C to Pb.
Down the family, tendency to show +4 oxidation state decreases while +2 oxidation state increases. This is explained on the basis of inert pair As a result, the valence -electrons because of their greater penetration into the nucleus are not so easily available for bond formation or these are reluctant to participate in the bond formation while the valence -electrons are easily available. The inert pair effect is more prominent in tin and lead because of very high atomic numbers. As a result, both of them normally show + 2 oxidation states i.e., as Sn²⁺ and Pb²⁺ ions. However, if sufficient energy is available, the valence electrons also become available to account for +4 oxidation states of these elements (Sn⁴⁺ and Pb⁴⁺ ions).

Question 2.
How can you explain higher stability of BCl₃ as compared to TlCl₃ ?
Answer:
In case of boron (B) atom, the inert pair effect is negligible. This means that all the three valence electrons (2s²   \( { p }_{ x }^{ 1 } \) )are available for bonding with chlorine atoms. Therefore, BCl₃ is quite stable. However, in case of thallium (Tl), the valence 5-electrons (6s²) are experiencing maximum inert pair effect. This means that only valence p-electron (6p¹) is available for bonding. Under these circumstances, TlCl is very much stable while TlCl₃ is comparatively little stable.

Question 3.
Why does BF₃ behave as Lewis acid ?
Answer:
BF₃ behaves as a Lewis acid because central boron atom has only six electrons (three pairs) after sharing with the electrons of the F atoms. It is an electron deficient compound and, therefore, behaves as a Lewis acid.
Boron Halides. The halides of boron are covalent in nature because the central boron atom, as stated earlier, is very small in size and cannot part with the valence electrons to give a trivalent B³⁺ Thus, BF₃, BCl₃, BBr₃ and BI₃ are all covalent in nature.
In these halides, the central B atom is sp² hybridised with a vacant 2p orbital (1 .v² 2s¹ 2px^l2pyⁱ2pz°). The hybridised orbitals are directed towards the three corners of an equilateral triangle and are involved in covalent bond formation with the half filled p-orbitals of the halogen atoms (ns²p⁵).

The empty 2p orbital lies perpendicular to the hybridised orbitals. Since it is empty, it can accept an electron pair from electron donor species (Lewis bases) resulting in co-ordinate or dative bond formation. With three shared pairs of electrons on the central boron atom, boron halides are Lewis acids i.e., electron deficient molecules and take part in Lewis acid – base reactions. In the compounds thus formed, boron atom undergoes a change in state of hybridisation from sp² to sp³. These are tetrahedral in nature.

Question 4.
Consider the compounds BCl₃ and CCl₄, How will they behave towards water ?
Answer:
In BCl₃ (B atom is sp² hybridised), the B atom has incomplete octet and unhybridised 2p-orbital which can take up electron pair from the H₂O molecule to form addition product.

In this way, a Cl atom has been replaced by OH group on reacting with water. Similarly, the other two Cl atoms will also be replaced by the OH groups as follows :

This shows that boron trichloride has undergone hydrolysis. But this is not possible with carbon tetrachloride (CCl₄). The carbon atom has a complete octet and there is no scope of forming addition product with H₂O molecules. As a result, carbon tetrachloride does not get hydrolysed. When added to water, it even does not mix and forms a separate oily layer.

Question 5.
Is boric acid a proton acid ? Explain.
Answer:
Boric acid is not a proton acid. It is a Lewis acid and accepts electron pair from hydroxyl ion of H₂O molecule.
B(OH)₃ + 2HOH →[B(OH)₄]^– + H₃O⁺

Question 6.
Explain what happens when boric acid is heated ?
Answer:
Preparation of Boric Acid. Boric acid can be prepared by the following methods.
(a) From borax. A hot and concentrated solution containing borax is boiled with hydrochloric or sulphuric acid. The solution upon concentration and cooling gives crystals of boric acid.


(b) From Colemanite. Sulphur dioxide gas is passed through the hot concentrated solution of mineral colemanite made in water. The solution upon concentration followed by cooling gives crystals of boric acid. Calcium bisulphite remains in solution as it is highly soluble in water.

(c) From boron compounds by hydrolysis. Certain boron compounds upon boiling with water (upon hydrolysis) give boric acid.
BCl₃ + 3H₂O→ H₃BO₃ + 3HCl
BN + 3H₂O → H₃BO₃ + NH₃

Question 7.
Describe the shapes of BF₃ and BH₄  Assign the hybridisation of boron atom in these species.
Answer:
BF₃ [H] : \(\frac { 1 }{ 2 } \) [3 + 3, – 0 + 01 = \(\frac { 6 }{ 2 } \) = sp² (trigonal planar)
BH4 [H] : \(\frac { 1 }{ 2 } \) [3 + 4 – 0 + 1] = \(\frac { 8 }{ 2 } \) = sp³ (tetrahedral)

Question 8.
Write reactions which justify amphoteric nature of aluminium.
Answer:
Aluminium can react both with acids and bases. It is therefore, amphoteric in nature. For example,
2Al (s) + 6HCl (dil) →2AlCl₃ (aq) 4- 3H₂ (g)
2Al (s) + 2NaOH (aq) + 6H₂O (l) → 2Na⁺[Al(OH)₄]^– (aq) + 3H₂ (g)

Question 9.
What are electron deficient compounds ? Are BCl₃ and SiCl₄ electron deficient ? Explain.
Answer:
Electron deficient compounds are the compounds in which the central atom in their molecules has urge or tendency to take up one or more electron pairs. The electron deficient compounds are also
called Lewis acids.
Yes, both BCl₃ and SiCl₄ are electron deficient. Whereas B atom has a vacant 2p orbital, Si atom at the same time has vacant 3d-orbitals. Both these atoms can take up electron pairs from electron donor species.

Question 10.
Write resonating structures for CO₃^2- and HCO^–₃.
Answer:

 Question 11.
What is the state of hybridisation of carbon in
(a) CO^2-
(b) diamond
(c) graphite ?
Answer:
(a) CO^2-₃ (sp²), (b) Diamond (sp³), (c) Graphite (sp²).

Question 12.
Explain the difference in properties of diamond and graphite on basis of their structures.
Answer:
Diamond

1. It generally occurs in the free state.
2. It is the hardest substance known.
3. Carbon atoms are sp³ hybridised.
4. It is a bad conductor of heat and electricity.
5. It is transparent with high refractive index (2-42).
6. It has very high melting point (4000 K or more).

Graphite

1. It occurs in the free state but can also be prepared artificially.
2. It is soft and greasy.
3. Carbon atoms are sp² hybridised.
4. It is a good conductor of heat and electricity.
5. It is opaque.
6. It has comparatively low melting point (1800 K

Question 13.
Rationalise the given statements and give chemical reaction :
lead(II) chloride reacts with Cl₂ to give PbCl₄.
lead(IV) chloride is highly unstable towards heatlead is known not to form an iodide, Pbl₄.
Answer:
(a) Lead (Pb) in 4-2 oxidation state i.e., Pb (II) is more stable than in 4- 4 oxidation state i.e., Pb (IV). This means that Pb (II) chloride will not react with chlorine to form Pb (IV) chloride
PbCl₂(s) + Cl₂(g)  → PbCl₄(g)
(b) Lead in (II) oxidation state is more stable than in lead in (IV) oxidation state.Therefore, lead (IV) chloride is highly unstable to heat. It decomposes upon heating to form lead (II) chloride.

(c) Lead is not known to form Pbl₄ because L ion being a powerful reducing agent reduces Pb⁴⁺ ion to Pb²⁺ ion in solution. Thus, Pbl₂ is generally formed.

Question 14.
Suggest reasons why the B-F bond lengths in BF₃ (130 pm) and BF^–₄ (143 pm) differ ?
Answer:

Difference in bond lengths is due to difference in the state of hybridisation of boron in the two fluorides.

Question 15.
If B-Cl bond has a dipole moment, why does BCI₃ have zero dipole moment ?
Answer:
B-Cl bond-has a certain dipole moment because it is of polar nature. But BCl₃ has zero dipole moment since the molecule is symmetrical (planar) just like BF₃ in which bond polarities cancel out.

Question 16..
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF₃ is bubbled through it. Give reason.
Answer:
Aluminium trifluoride (AlF₃) is insoluble in anhydrous HF because of its covalent nature. However, it forms a complex compound on reacting with NaF which is water soluble.

The complex gets cleaved when vapours of BF₃ are bubbled into the aqueous solution. As a result, aluminium trifluoride is again precipitated.

Question 17.
Suggest a reason as to why CO is poisonous in nature.
Answer:
In the lungs, oxygen combines with haemoglobins present in red blood cells to form oxyhaemoglobin

Oxyhaemoglobin travels to different parts of the body through blood stream and releases oxygen which passes on to the various tissues in the body. Carbon monoxide is highly poisonous or toxic in nature. Its poisonous character is due to its tendency to combine with haemoglobin present in blood to form carboxy haemoglobin which is not in a position to carry the inhaled oxygen to different parts in the body. This will lead to suffocation and ultimately to death.
Haemoglobin + CO —> Carboxyhaemoglobin

We may conclude that carbon monoxide reduces the oxygen carrying capacity of haemoglobin.

Question 18.
How is excess content of CO₂ responsible for global warirfing ?
Answer:
We know that CO₂ is very much essential for plants to carry photosynthesis. The gas is produced during various types of combustion reactions and is released into the atmosphere. It is taken up by plants as pointed above. Thus, a carbon dioxide cycle works in the atmosphere and its percentage remains nearly constant. However, over the years, combustion reactions have enormously increased. As a result, CO₂ gas is now present in excess in the atmosphere. Like methane, it also behaves like a green house gas and absorbs heat radiated by the earth. Some of the heat is released into the atmosphere while the rest is radiated back to earth. This has resulted in global warming over the years and has brought about major climatic changes.

Question 19.
Explain the structure of diborane and boric acid.
Answer:

Question 20.
What happens when     
(a) Borax is heated strongly
(b) Boric acid is added to water
(c) Aluminium is treated with dilute NaOH
(d) BF₃ is reacted with ammonia ?
Answer:
(a) When powdered borax is heated strongly in the flame of bunsen burner, it forms a colourless transparent glassy (glass-like) bead made of sodium meta borate and boric anhydride.

(b) It dissolves in water as it is electron deficient in nature.

(c) Aluminium dissolves in NaOH solution to form a soluble complex, liberating hydrogen gas.

(d) BF₃ (Lewis acid in nature) forms an addition compound with NH₃ (Lewis base in nature).

Question 21.
Explain the following reactions :
(a) Silicon is heated with methyl chloride at high temperature in the presence of copper
(b) Silicon dioxide is treated with hydrogen fluoride
(c) CO is heated with ZnO
(d) Hydrated alumina is treated with aqueous NaOH solution.
Answer:
(a) Silicon upon heating with methyl chloride to about 300°C in the presence of copper catalyst reacts as follows :

The compound upon hydrolysis leads to the formation of silicon polymers.
(b) On reacting silicon dioxide with hydrogen fluoride, silicon tetrafluoride (SiF₄) is formed.
SiO₂ + 4HF → SiF₄ + 2H₂O
SiF₄ reacts further with hydrogen fluoride to form hydrofluorosilicic acid.
SiF₄ + 2HF → H₂SiF₆
(c) ZnO is reduced to Zn by CO which is a strong reducing agent.

(d) The two compounds react upon heating under pressure to form a soluble complex.
Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) —> 2Na[Al(OH)₄](aq)

Question 22.
Give reasons :
(1) Con. HNO₃ can be transported in aluminium coptainer.
(2) Graphite is used as lubricant.
(3) Diamond is used as an abrasive.
(4) Aluminium alloys are used to make aircraft body.
(5) Aluminium utensils should not be kept in water overnight.
(6) Aluminium wire is used to make transmission cables.
Answer:

1. Cone. HNO₃ initially reacts with aluminium to form aluminium oxide (Al₂O₃) which forms a protective coating inside the container. The metal becomes passive and does not react with the acid any more. Therefore, the acid can be safely stored in aluminium container.
2. Graphite is used as lubricant because of its soft and greesy nature. This is probably due to the presence of layers in the arrangement of carbon atoms in graphite.
3. Diamond is used as an abrasive because of its extremely hard nature (hardest substance known). Actually, the carbon atoms in diamond are very closely packed. This is responsible for the hardness of diamond.
4. Alloys of aluminium Magnalium and Duralumin both of which contain about 95 % of the metal are used for making aircraft body. Actually, both of them are light, tough and quite resistant to corrosion. Therefore, they are used in making bodies of air crafts.
5. Although the metal as such is not affected by water, but when kept overnight, it may be slowly affected by moisture in the presence of oxygen (air).
   2Al (s) + O₂(g) + H₂O(l) → Al₂O₃(s) + U₂(g)
6. The metal is not affected by air and moisture (resistant to corrosion) and also because of its good conductivity, it is used to make transmission cables.

Question 23.
Explain why is there a phenomenal decrease in ionization enthalpy from carbon to silicon ?
Answer:
This is because of large atomic size of silicon (atomic radius 118 pm) than that of carbon (atomic radius = 77 pm). As a result, the electrons are less closely attracted to the nucleus in silicon atom as compared to carbon atom. The ionization enthalpy of silicon is, therefore, less. For example, A,H] of silicon is
786 kJ mol^-1 .

Question 24.
How would you explain the lower atomic radius of Ga as compared to Al ?
Answer:
The atomic radius of gallium is less than that of aluminium. Down the group, the atomic radius is expected to increase. The decrease in atomic radius of gallium (135 pm) as compared to that of aluminium (143 pm) may be attributed to the presence of ten elements of the first transition series (Z = 21 to 30) which have electrons in the 3d Since d-orbitals have large size than the p-orbitals, the intervening electrons do not have sufficient shielding effect to counter the increase in the nuclear charge. Therefore, the effective nuclear charge in case of Ga (Z=31) is more than the expected value. This decreases its atomic radius which otherwise is expected to increase. The ionic radii of these elements increase regularly as is evident from their values.

Question 25.
What are allotropes ? Sketch the structure of two allotropes of carbon namely diamond and graphite. What is the impact of structure on physical properties of the two allotropes ?
Answer:
Elemental carbon exists in a number of allotropic forms which are both crystalline and amorphous in nature. A few important out of these are briefly discussed. Carbon exists in two types of allotropic forms. These are crystalline and amorphous.

Crystalline Allotropic Forms
Diamond and graphite are the two crystalline forms but they differ in the arrangement of the carbon atoms and hence have different physical properties.

Diamond    
Diamond is the hardest crystalline form of carbon and is present in different parts of world as a constituent of hard rocks. Diamonds are mainly found in South Africa, Brazil, Australia, British Guiana etc. In India, diamonds are mainly found in Golconda and Panna. In nature, diamonds occur as transparent octahedral crystals with curved surfaces. Their beauty is revealed only when these are properly cut and polished.

Structure of Diamond. In diamond, all the carbon atoms are sp³ hybridised and are arranged tetrahedrally in space. Each carbon atom is linked to three other carbon atoms by covalent bonds. This results in a three dimensional network as shown in the Fig. 11.11.1 Each C—C bond length is 154 pm. As the carbon atoms in diamond are very closely packed in space, it is, therefore, very hard with density equal to 3-5 g cm³ and has a very high melting point (4000 K or even more). Because of high refractive index (2-42), it produces maximum total internal reflection responsible for its bluish green colour. Since all the four valence electrons pf the carbon atom are involved in bond formation, in the absence of free electrons, diamond is bad conductor of electricity. It is also a poor conductor of heat.

Uses of Diamond.

1. Due to its hardness, diamond is used for cutting marble, granite and glass.
2. It is used as an abrasive and for polishing hard surfaces.
3. It is used in making special surgical knives.
4. Dies made from diamond are used for drawing wires from the metals.
5. Diamonds when properly cut and polished are used as precious stones or gems.

Question 26.
Classify fallowing oxides as neutral, acidic, basic or amphoteric :
CO, B₂O3,SiO2, CO₂, Al₂O₃, PbO₂, Tl₂O3.
Answer:

Question 27.
In some of the reactions, thallium resembles aluminium, whereas in others, it resembles with group I metals. Support this statement by giving some evidences.
Answer:
Aluminium (Al) generally exhibits + 3 oxidation state in its compounds. Thallium (Tl), the last member of the group-13, is expected to show oxidation states of + 3 and +1 (due to inert pair effect). Thus, the metal resembles aluminium in exhibiting + 3 oxidation state. Both form trichlorides TlCl₃ and AlCl₃ At the same time, it also resembles alkali metals of group 1 in exhibiting + 1 oxidation state (e.g., both thallium and sodium form monochlorides TlCl and NaCl).
Tl shows both the oxidation state +1 and +3 due to inert pair effect. Tl forms basic oxide like group I elements. TlO₂ is strongly basic.

Question 28.
When metal X is treated with sodium hydroxide, a white precipitate (A) is obtained, which is soluble in excess of NaOH to give soluble complex (B). Compound (A) is soluble in dilute HCl to form compound (C). The compound (A) when heated strongly gives (D), which is used to extract the metal. Identify (X), (A), (B), (C) and (D). Write suitable equations to support their identiti(P.I.S.A. Based)
Answer:
The data suggests that the metal ‘X’ is aluminium. The reactions in which aluminium participates leading to the ‘ formation of compounds (A), (B), (C) and (D) are given :
(1) Aluminium (X) reacts with NaOH upon heating to form a white precipitate of Al(OH)₃ e., compound (A) which dissolves in excess of NaOH to form soluble complex ‘B’

(2) The compound (A) dissolves in dil HC₁ to form aluminium chloride (C)

(3) Upon heating Al(OH)₃ is converted into alumina (D)

Al₂O₃ is used in the extraction of the Al metal.

Question 29.
What do you understand by (a) inert pair effect (b) allotropy (c) catenation ?
Answer:
(a) Inert pair effect: The pair of electron in the valence shell does not take part in bond formation is called inert pair effect.
(b)Allotropy: It is the property of the element by which an element can exist in two or more forms which have same chemical properties but different physical properties due to their structures.
(c)Catenation: The property to form chains or rings not only with single bonds but also with multiple bonds with itself is called catenation.
For example, carbon forms chains with (C-C) single bonds and also with multiple bonds (C = C or C = C).

Question 30.
A certain salt X gives the following results :
(1) Its aqueous solution is alkaline to litmus.
(2) It swells up to a glassy material Y on strong heating.
(3) When cone. H₂SO₄ is added to a hot solution of X, white crystals of an acid Z separate out.
Write equations for all the above reactions and identify X, Y and Z.
Answer:
The data suggests that the salt ‘X’ is Borax (Na₂B₄O₇.10H₂O)
(1) The aqueous solution of ‘borax is of basic nature and turns red litmus blue.

(2) Borax swells in size upon strong heating and loses molecules of water of crystallisation to form
solid (Y)

(3) Upon reacting with cone. H₂SO₄, borax forms boric acid (H₃BO₃). When crystallised from the solution, it is in the form of white crystals (Z)

Question 31.
Write balanced equations for the following :
(1) BF₃ + LiH→
(2) B₂H₆ + H₂O—>
(3)NaH + B₂H₆—>
(4) H₃BO₃—>
(5) Al + NaOH—>
(6) B₂H₆ + NH_3→
Answer:

Question 32.
Give one method for industrial preparation and one for laboratory preparation of CO and CO₂
Answer:

Carbon Monoxide (CO)
Preparation of carbon monoxide

• Carbon monoxide is a constituent of water gas also called synthesis gas (CO + H₂) and is formed by passing steam over red hot coke.
  
• Carbon monoxide can be separated by liquefaction.If air is passed over red hot coke instead of’steam producer gas is formed which is a mixture of CO and N₂ From producer gas, CO can be removed by liquefaction
  
• Carbon monoxide can be,prepared by passing carbon dioxide through red hot charcoal.
  
  From the mixture, CO₂ can be removed by passing the mixture into water under pressure.
• Oxides of certain metals upon heating with powdered coke are reduced to carbon monoxide
  
• Laboratory preparation. In the laboratory carbon monoxide can be prepared by the dehydration of formic acid with concentrated H₂SO₄
  
  

Carbon Dioxide (CO₂)
Preparation of Carbon dioxide

• Carbon dioxide can be prepared by the following methods.
  C + O₂ →CO₂
  CH₄ + 2O₂ → CO₂ + 2H₂O
• It can also be obtained by the thermal decomposition of certain carbonates of non-alkali metals.
  
• It can be prepared by the action of dilute acids on certain carbonates and bicarbonates of metals.
  Na₂CO₃ + 2HCl (dil.) → 2NaCl + H₂O + CO₂
  NaHCO₃ + H₂SO₄(dil.)→ NaHS0₄ + H₂O + CO₂
  
• In the laboratory, carbon dioxide is prepared by the action of dilute hydrochloric acid on calcium carbonate.
  CaCO₃ + 2HCl (dil)→ CaCl₂ + H₂O + CO₂
  
• On commerical scale, carbon dioxide is obtained as a by-product in the decomposition of lime stone to form lime and in the manufacture of ethyl alcohol as a result of fermentation of glucose (present in cane sugar).
  

Question 33.
An aqueous solution of borax is :
(a)neutral
(b) amphoteric
(c) basic 
(d) acide
Answer:
(c) is the correct answer.

Question 34.
Boric acid is polymeric due to :
(a) its acidic nature
(b) the presence of hydrogen bonds
(c) its monobasic nature
(d) its geometry.
Answer:
(b) The acid is polymeric due to hydrogen bonding.

Question 35.
The type of hybridisation of boron in diborane is :
(a) sp
(b) sp¹
(c) sp³  
(d) dsp²
Answer:
(c) Boron is sp³ hybridised in diborane.

Question 36.
Thermodynamically the most stable form of carbon is
(a) diamond
(b) graphite 
(c) fullerenes
(d) Coal
Answer:
(b) graphite is thermally the most stable form.

Question 37.
Elements of group 14 :
(a) exhibit oxidation state of +4 only
(b) exhibit oxidation state of +2 and +4
(c) form M²‘ and M⁴⁺ ions
(d) form M²⁺ and M⁴⁺
Answer:
both (b) and (d) are correct answers. The variable oxidation states are due to inert pair effect.

Question 38.
If the starting material for the manufacture of silicones is RSiCl₃, write the structure of the product formed.
Answer:
Silicones     
Silicones are the synthetic organosilicone polymers containing Si—O—Si linkages. These are represented by the general formula (R₂SiO)„ in which R may be alkyl (methyl or ethyl or phenyl groups). Since the general empirical formula is similar to a ketone (R₂CO), the name silicone has been given to these polymers.

Structure of Silicones
Silicones are of two types. These may be either linear polymers or cross-linked in nature. Both of them are formed by the action of SiCl₄ on Grignard reagents followed by hydrolysis leading to polymerisation.

NCERT Solutions for Class 11 Chemistry Chapter 11(A) Some p-Block Elements (Group 15 Elements)

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 11(A) Some p-Block Elements (Group 15 Elements)

Question 1.
Discuss the general characteristics of group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
Answer:
For answer, consult Section 11.3..

Question 2.
Why is the reactivity of nitrogen different from that of phosphorus ?
Answer:
Molecular nitrogen exists as a diatomic molecule (N₂) in which the two nitrogen atoms are linked to each other by triple bond (N=N). It is a gas at room temperature. Multiple bonding is not possible in case of phosphorus due to its large size. It exists as P₄ molecule (solid) in which P atoms are linked to
one another by single covalent bonds. Because of greater bond dissociation enthalpy (946 kJ mol-1) of N=N bond, molecular nitrogen is very less reactive as compared to molecular phosphorus.

Question 3.
Discuss the trends in chemical reactivity of group 15 elements.
Answer:
For answer, consult chemical properties of nitrogen family (Section 11.4).

Question 4.
Why does NH3 form hydrogen bonding while PH₃ does not ?
Answer:
The N—H bond in ammonia is quite polar on account of the electronegativity difference of N (3-0) and H (2 .1). On the contrary, P—FI bond in phosphine is almost non-polar because both P and H atoms have almost same electronegativity (21). Due to polarity, intermolecular hydrogen bonding is present in the molecules of ammonia but not in those of phosphine.

Question 5.
How is nitrogen prepared in the laboratory ? Write the chemical equations of the reactions involved.
Answer:
For answer, consult Section 7.6.

Question 6.
How is ammonia manufactured industrially ?
Answer:
For answer, consult Section 7.7.

Question 7.
Illustrate how copper metal gives different products on reaction with HNO₃.
Answer:
For answer, consult properties of HNO₃ (Section 7.8).

Question 8.
Give the resonating structures of NO₂ and N₂O5
Answer:

Question 9.
The HNH angle value is higher than those of HPH, HAsH and HSbH angles ; why ?
Answer:

The central atom (E) in all the hydrides is sp3 hybridised. However, its electronegativity decreases and atomic size increases down the group. As a result, there is a gradual decrease in the force of repulsion in the shared electron pairs around the central atom. This leads to decrease in the bond angle. For more details, consult text part Section 7.4.

Question 10.
Why does R₃P=0 exist but R₃N=0 does not (R is an alkyl group) ?
Answer:
Nitrogen does not have vacant d-orbitals on its valence’shell. Therefore, it cannot extend its covalency to five and dn-pn bonding is not possible. As a result, the molecules of R₃N=O does not exist. However, phosphorus and rest of the members of the group 15 have vacant rf-orbitals in the valence shell which can be involved in dn-pn bonding. Under the circumstances, R3P=O molecule can exist.

Question 11.
Explain why is NH₃ basic while PH₃ is feebly basic in nature.
Answer:
Both nh₃ and ph₃ are Lewis bases due to the presence of lone electron pair on the central atom. However, NH₃ is more basic than PH₃. The atomic size of nitrogen (Atomic radius = 70 pm) is less than that of phosphorus (Atomic radius = 110 pm) As a result, electron density on the nitrogen atom is more than on phosphorus. This means that electron releasing tendency of ammonia is also more and is therefore, a stronger base than phosphine.

Question 12.
Nitrogen exists as diatomic molecule (N₂) while phosphorus as tetra-atomic molecule (P₄). Why ?
Answer:
For answer, consult Section 7.3 (Physical properties of nitrogen family).

Question 13.
Write the main difference between the properties of white and red phosphorus.
Answer:
For answer, consult Section 7-11.

Question 14.
Why does nitrogen show catenation properties less than phosphorus ?
Answer:
The valence shell electronic configuration of N is 2s²2p³. In order to complete the octet, the two nitrogen atoms share three electron pairs in the valence p-sub-shell and get linked by triple bond (N=N). Thus molecular nitrogen exists as discrete diatomic species and there is no scope of any self linking or catenation involving a number of nitrogen atoms. However, in case of phosphorus, multiple bonding is not feasible due to Comparatively large atomic size of the element. For details, consult 7.3. Molecular phosphorus exists as tetra-atomic molecule (P₄) in white phosphorus. These tetrahedrons are further linked by covalent bonds to form red variety which is in polymeric form. Thus, catenation in nitrogen is less than in phosphorus.

Question 15.
Give one disproportionation reaction of phosphorus acid (H₃PO₃).
Answer:
Upon heating to about 573 K, phosphorus acid undergoes disproportionation as follows :

Question 16.
Can PC₅ act as oxidising as well as reducing agent ? Justify.
Answer:
In general, the molecules of a substance can behave as a reducing agent if the central atom is in a position to increase its oxidation number. Similarly, they can act as an oxidising agent if the central atom is in a position to decrease its oxidation number. Now, the maximum oxidation state or oxidation number of phosphorus (P) is +5. It cannot increase the same but at the sametime can decrease its oxidation number. In PCl₅, oxidation number of P is already +5. It therefore, cannot act as a reducing agent. However it can behave as an oxidising agent in certain reactions in which its oxidation number decreases. For example,

Question 17.
Why are pentahalides more covalent than trihalides in the members of the nitrogen family ?
Answer:
The electronic configuration of the elements of nitrogen family (group 15) is ns²p³. Because of the inert pair effect, the valence 5-electrons cannot be released easily for the bond formation. This means that the elements can form trivalent cation (E³⁺) by releasing valence p-electrons while it is difficult to form pentavalent cation (E³⁺). Under the circumstances, if all the five valence electrons are to be involved in the bond formation, the compounds showing pentavalency or +5 oxidation state must be of covalent nature. This is particularly the case in the higher members (Sb, Bi) of the family where the inert pair effect is quite prominent.

Question 18.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements ?
Answer:
Down the group, the atomic size of the element (E) increases and the bond length of the corresponding E—H bond also increases. This adversely affects the bond dissociation enthalpy. This means that amongst the trihydrides of the members of nitrogen family, the bond dissociation enthalpy of Bi—H bond is the least. Therefore, BiH₃ is the strongest reducing agent among the hydrides of group 15 elements.

Question 19.
Why is N₂less reactive at room temperature ?
Answer:
In the nitrogen molecule (N₂), two atoms of nitrogen are linked by triple bond (N = N). Due to small atomic size of the element (atomic radius = 70 pm), the bond dissociation enthalpy is very high (946 kJ mol-1). This means that it is quite difficult to cleave or break the triple bond at room temperature. As a result, N₂ is less reactive at room temperature.

Question 20.
Mention the conditions required for the maximum yield of ammonia.
Answer:
In Haber’s process, ammonia is formed by the following reaction.

According to Le-chatelier’s principle, the favourable conditions for the maximum yield of ammonia are :
Low temperature : But optimum temperature of 700 K is necessary to keep the forward reaction in progress.
High pressure : Pressure to the extent of about 200 atm is required.
Catalyst & promoter : Ip order to achieve the early attainment of equilibrium, iron oxide acts as catalyst. Along with that; K₂O, Al₂O₃ or Mo metal may act as the promoter to increase the efficiency of the catalyst.

Question 21.
How does ammonia react with blue solution having Cu²⁺ ions ?
Answer:
When ammonia gas is passed through blue solution containing Cu²⁺ ions, ammonium hydroxide formed reacts with Cu²⁺ ions to give a soluble complex with deep blue colour.


Question 22.
How does ammonia react with blue solution having Cu²⁺ions ?
Answer:
When ammonia gas is passed through blue solution containing Cu²⁺ ions, ammonium hydroxide formed reacts with Cu²⁺ ions to give a soluble complex with deep blue colour.

Question 23.
What is bond angle in  PH⁺⁴ ion higher than in PH₃
Answer:
In both PH₃ and PH⁺⁴ ion, the phosphorus atom is sp3 hybridised. However, in PH₃ the central atom has a pyramidal structure due to the presence of lone electron pair on the phosphorus atom.

Because of lone pair : shared pair repulsion which is more than that of shared pair : shared pair repulsion, the bond angle in PH₃ is nearly 93-6°. In PH⁺⁴ ion, there is no lone electron pair on the phosphorus atom. It has a tetrahedralstructure with bond angle of 109°-28′. Thus, the bond angle in PH⁺⁴ ion is higher than in PH₃.

Question 24.
What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO⁺⁴?
Answer:
A mixture of sodium hypophosphite and phosphine gas is formed upon heating the reaction mixture in an inert atmosphere.

Question 25.
What happens when PCl₅ is heated ?
Answer:
Upon heating, PCl₅ dissociates to give molecules of PCl₅ and Cl₂. Actually, the two P—Cl (a) bonds with more bond length break away from the molecule leaving three P— C(e) bonds attached to the central P atom since these are more firmly linked

Question 26.
Write the balanced equation for the hydrolytic reaction of PCl₅ in heavy water.
Answer:
With heavy water (D₂O) ; phosphorus pentachloride (PCl₅) reacts as follows :

We hope the NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements, help you. If you have any query regarding. NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 10 s-Block Elements (Alkali and Alkaline Earth Metals)

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements (Alkali and Alkaline Earth Metals).

Question 1.
What are the common physical and chemical features of alkali metals ?
Answer:

Property	Li	Na	K	Rb	Cs	Fr
Atomic number	3	11	19	37	55	87
Atomic mass (g mol-1)	6-94	22-99	39-10	85-47	132-91	(223)
Atomic (Metallic) radius (pm)	152	186	227	248	265	—
Ionic radius (pm)	76	102	138	152	167	—
Ionization enthalpy (ΔiHi) in kJ mol-1	520	496	419	403	376	—
Ionisation enthalpy (Δi-Hj) in kJ mol-1	7298	4562	3051	2633	2230	—
Hydration enthalpy (Δ HHydration ) in kJ mol	-506	-406	-330	-310	-276	—
Electronegativity (Pauling scale)	10	0-9	0-8	0-8	0-7	—
Density/g cm-3 (at 293 K)	0-53	0-97	0-86	1-53	1-90	—
Melting point/K	453-5	370-8	336-2	312-0	301-5	—
Boiling point/K E° (V) at 298 K for	1620	1154-4	1038-5	961-0	978-0	—
M+ (aq) + e–—— > M (s)	-3-03	-2-71	-2-93	-2-93	-2-92	—

Now let us try to account for the general trends in the physical properties with the help of the available data.
1. Atomic radii. Alkali metals have maximum value of the atomic radii (metallic radii) in their respective periods and these values further increase down the group. Although the last element in each period e., the noble gas element has more value of the atomic radius, but it is van der Waals radius and not atomic or metallic radius as in alkali metals. Therefore, it is not proper to make a comparison in their values.

Explanation. We know that the atomic radius is the distance of the outermost electron present in the valence shell of an atom from the center of its nucleus. Being the first element in the respective periods, alkali metal atoms have maximum atomic .radii since their atoms have only one electron in the valence shell. As a result, the magnitude of the force of attraction with the nucleus is the minimum. Down the group, the atomic radius increases mainly because of the gradual increase in the number of electron shells.

In addition to this, the magnitude of the screening effect also increases which decreases the attraction between the valence 5-electron with the nucleus of the atom. No doubt, the nuclear charge also increases which is likely to increase the attraction of the electrons with the nucleus resulting in the decreased atomic size. However, its magnitude is very small compared to the screening effect. The net result is what we actually observe i.e., atomic sizes of the elements increase down the group.

2. Ionic radii. Alkali metal atoms form monovalent cations by the loss of valence s (ns¹) The cationic radius is less as compared to that of the atom. As the values indicate, the radii increase down the group.
Explanation. A monovalent cation is formed by the loss of the only electron present in the valence shell of the atom. As a result, the remaining electrons are pulled closer to the nucleus of the atom causing a decrease in the cationic size. We all know that the size of the ion (cation or anion) is linked to that the corresponding atom. Since the atomic radii increase down the group; so are the ionic radii.

3. Ionisation enthalpies. The ionisation enthalpy as all of us know, is the minimum amount of energy needed to remove the most loosly bound electron from a neutral isolated atom in the gaseous state. Its unit is kJ mol^-1. It can also be expressed as ionisation potential with electron volt (eV) per atom as the unit
(1 eV per atom = 96-472 kJ mol ).
Alkali metals have the lowest ionisation enthalpies in their respective periods and these further decrease down the group. The first (Δ_iH₁) and second (Δ_iH₂) ionisation energies .

Explanation. The low ionisation enthalpies of the alkali metals may be attributed to very large atomic sizes as a result of which the valence x-electron (ns¹) can be readily removed. These values decrease down the group because of decrease in the magnitude of the force of attraction with the nucleus on account of increased atomic radii and magnitude of screening effect. However, there is a large difference in Δ_iH₁ and Δ_iH₂ values of a particular element. For example, A,H] value of sodium is 496 kJ mol^-1 while its Δ_i H₂ value is 4562 kJ mol¹. It is mainly because of the reason that with the release of the first electron, the monovalent cation (M⁺) is highly symmetrical and has nearest noble gas configuration. Therefore, the loss of second electron is very difficult comparatively. The difference in the Δ_i H₁ and Δ_i H₂ values can be, thus, justified.

4. Electronegativity of an element is the relative electron attracting tendency of its atom for a shared pair in a bond. Alkali metals have low electronegativities which means that their electron attracting tendencies are low. These values tend to decrease down the group.

Explanation. As the alkali metal atoms have ns¹ electronic configuration, this means that they have electron releasing rather than electron accepting tendency. Thus, they have low values of electronegativities as is evident from the data listed in the Table 10.2. Since the atomic sizes increase down the group, there is gradual decrease in capacity of the atoms to hold their valence electrons. Therefore, the electronegativities decrease down the group.

5. Oxidation states and electropositive character. All the members of the alkali metal family are strongly electropositive (tendency to form positive ion) and show + 1 oxidation states in their compounds. The electropositive character further increases down the group.
Explanation. Alkali metal atoms have strong urge to lose the valence electron and to form monovalent cation due to their low ionisation energies. These are therefore, highly electropositive in nature. This character i.e., electron releasing tendency increases further down the group as the ionisation energy decreases.

6. Metallic character. Group 1 elements are typical metals and are so soft that most of them can be easily cut with a knife. The metallic character further increases down the group.
Explanation. The metallic character of the element is related to the electron releasing tendency of its atom to form positive ion. The strength of the metallic bond depends upon magnitude of the attractive forces present in the kernels and valence electrons in the electron sea. Smaller the size of the kernel and more is the number of valence electrons, stronger will be the metallic bond and greater will be the hardness of the metal. In alkali metals, the kernels are of large size (it depends upon the atomic size) and there is only one valence electron. Therefore, metallic bond is rather weak and as a result, alkali metals are quite soft. They can be cut even with a knife. Lithium, the first element is the hardest since the kernel is the smallest in size.

7. Melting and boiling points. Alkali metals have low melting and boiling points which further decrease down the group.
Explanation. As the atoms belonging to the alkali metals have large size, their binding energies in the crystal lattice are quite low. As a result, they have low melting points. These decrease down the group because of the increase in atomic size. The explanation for the trends in boiling points is also similar.

8. Density . Alkali metals are very light. The first three members of the family are even lighter than water. The density increases downwards.
Explanation. Since the alkali metal atoms have large size, these are not so closely packed in space and have, therefore, low densities. The densities are expected to decrease further down the group because the atomic sizes increase. But the trend is the reverse which means that the densities increase downwards as listed in Table 10.2. This is because of the reason that the atomic masses of the elements increase and the effect of increase in atomic mass is more compared to increase in atomic size. Therefore, the density (mass/volume) increases downwards. However, there is an exception i.e., the density of potassium is less than that of sodium. This may be probably due to the reason that there is an abnormal increase in atomic size and atomic volume as we move from sodium to potassium.

9. Hydration enthalpies. Hydration enthalpy is the energy which is released when the ion of a particular element takes up molecules of water and gets hydrated. In general, smaller the size of the ion, more is its tendency to get hydrated or more is the hydration enthalpy. In the alkali metals, the hydration enthalpies of the monovalent cations (M⁺) are low due to their large size. These further decrease down the group since the ionic size increases.
Explanation : The hydration involves attraction between the ion and the surrounding molecules of water. Thus, smaller the size of the ion, more will be the magnitude of the charge on it and more will be its capacity to hydrate. Among the alkali metals, Li⁺ ion has the maximum hydration enthalpy. Therefore, the salts of lithium are mostly hydrated in nature e.g., LiCl.2H₂O.

10. Colouration to the flame. The members of the alkali metals, particularly the chlorides of these metals when heated on a platinum wire impart characteristic colours to the flame of the burner. For example,

Explanation. As the alkali metals have low ionisation enthalpies, the energy of the flame causes the excitation of the electrons to higher energy state away from the nucleus. When these excited electrons jump back to the normal or ground states, the emitted radiations fall in the visible region of light. Therefore, the alkali metals impart colour to the flame. Among these metals, lithium has the minimum size and requires maximum energy for the electronic excitation.

Therefore, for the same excitation energy available (from the temperature of the flame) the energy level to which the electron in lithium atom jumps will be lower as compared to sodium. This also means that when the electron returns to the ground state, the energy released or the frequency of the radiations emitted will be minimum (E = hv). In other words, the frequencies of the radiations emitted will follow the order Li < Na < K < Rb < Cs. This is responsible for the colours of flame imparted by different metal atoms as their salts.

11. Photoelectric effect. Alkali metals (except Li) exhibit photoelectric effect. It may be defined as the phenomenon of the ejection of electrons from the surface of a metal when electrons of certain frequency strike against the metal surface. In nut-shell, the electrons are emitted from the metal surface under the influence of the striking photons.

Explanation. The cause of the photoelectric effect is the low ionisation enthalpies of the alkali metals. The striking photons of light have sufficient energy to knock out the electrons from the metal surface. Since lithium has comparatively high ionisation enthalpy due to its very small atomic size, the striking photons do not have sufficient energy to overcome the force of attraction of the electrons with the nucleus. The element, therefore, does not exhibit photoelectric effect in the visible region of light where the energy of striking photons is not sufficient to cause photoelectric effect.

Chemical Properties of alkali metals:
Alkali metals are highly reactive chemically because of their low ionisation enthalpies and enthalpy of atomisation. Some of the important chemical properties of the members of the family are discussed.

1. Reaction with air. When freshly cut, alkali metals have luster but their surfaces get tarnished when exposed to air due to the formation of a layer of oxide, hydroxide and carbonate.
4M    +    O₂      →    2M_zO
M₂O   +   H₂O    →   2MOH
2MOH  +   CO₂    → M₂CO₃ + H₂O
The alkali metals can not be placed in air. Similarly, these are also not placed in water due to strong affinity. These are normally kept in chemically inert solvents such as kerosene.

2. Reaction with oxygen. Alkali metals combine with oxygen upon heating to form different oxides depending upon their nature. Lithium forms a normal oxide (Li₂O), sodium forms peroxide (Na₂O₂) while potassium and rest of the metals form superoxides (MO₂ where M = K, Rb and Cs) upon heating in oxygen.

Reactivity with oxygen increases from Li to Cs

Explanation. The stabilities of these oxides are linked with the relative sizes of the cations and anions involved and also upon the charges present on them. In general :
A smaller cation can stabilise a smaller anion while a larger cation can stabilise a larger anion.
The size of the Li⁺ ion is the smallest and has a strong positive field around it. It can combine only with a small anion such as oxide ion (O^2-) with a strong negative field around it. Na⁺ ion due to bigger size has a weaker positive field around it and, therefore, can stabilise peroxide ion (O₂^2-) which has also a weaker negative field around it. Thus, sodium forms peroxide (Na₂O₂). The remaining cations (K⁺, Rb⁺ and Cs⁺) are still bigger in size and the magnitude of the positive field around them decreases in the same order. They can stabilise only an anion with a very weak negative field around it i.e., superoxide ion (Of).
The valence bond structures of oxide, peroxide and superoxide ions are given below :

 

The superoxide ion has a three electron bond and due to the presence of unpaired electron, it is paramagnetic in nature. On the other hand, both oxide and peroxide ions have no unpaired electrons. These are, therefore, diamagnetic in nature. The magnetic nature of these ions can also be explained on the basis of molecular orbital theory. For details, consult chemical bonding and molecular structure

3. Reaction with water (Formation of hydroxides). Alkali metals, their oxides, peroxides and even superoxides dissolve in water to form hydroxides which are soluble and are called alkalies (water soluble hydroxides are known as alkalies). For example,

The reactions of metals with water are so highly exothermic that the hydrogen gas evolved catches fire accompanied by explosion. Therefore, alkali metals are not kept in contact with water. These are kept under kerosene.

Question 2.
Discuss the general characteristics and gradation in properties of alkaline earth metals.
Answer:
The important physical properties of the members of the family are listed in the Table

Property	Be	Mg	Ca	Sr	Ba	Ra (Radioactive)
Atomic number	4	12	20	38	56	88
Atomic mass	901	2 4-31	40-08	87-62	137-33	226-03
Atomic radius/pm (Metallic)	112	160	197	215	222
Ionic radius/pm	31	72	100	118	135	148
Ionisation enthalpy (ΔiH1)	899	737	590	549	503	509
(kJ mof1)(ΔiH2)	1757	1450	1146	1064	965	979
Hydration enthalpy
in (kJ mof1)	-2494	-1921	-1577	-1443	-1305
Electronegativity
(Pauling scale)	1-5	1-2	1-00	1-0	0-9	0-9
Density/g mol-1 (at 293 K)	1-85	1-74	1-55	2-63	3-62	5-5
Melting point/K	1560	924	1424	1062	1002	973
Boiling point/K,	2745	1363	1767	1655	2078	1973
E° (Y) at 298 K for
M2+ (aq) + 2e– A—> M(s)	-1-97	-2-36	-2-84	-2-89	-2-92	-2-92

Based upon the above data, let us discuss the trends in the important physical properties of these elements.
1. Atomic radii. The atomic radii of the alkaline earth metals are fairly large but less than the corresponding alkali metals present in the same period. They increase down the group.
Explanation. The atoms of these elements have only two valence electrons and the magnitude of the force of attraction with the nucleus is quite small. Therefore, these elements have sufficient large atomic sizes as well atomic radii. They tend to increase down the group because of gradual increase in the number of electron shells and magnitude of screening or shielding effect.

2. Ionic radii. All the elements form divalent cation (M²⁺) by losing the valence electrons. Therefore, the ionic radius is less than the radius of the corresponding atom. These increase down the group as these are linked to the atomic radii of the elements.

3. Ionisation enthalpies. The members of the alkaline earth family have, in general, low ionisation enthalpies due to fairly large atomic sizes. As the atomic sizes increase down the group, the ionisation enthalpies are expected to decrease in the same manner. This is quite evident from the data listed in the Table

The ionisation enthalpies of radium (both AjHj and AjHf) are slightly higher than those of barium while these are expected to be less. This is probably due to the reason that it comes after the lanthanoid series off-block elements which undergo lanthanoid contraction. Therefore, radium undergoes some decrease in atomic size resulting in an increase of ionisation enthalpies.

A comparison of the ionisation enthalpies of the elements present in group 1 and group 2 belonging to the same period indicates that the Δ_iH₁, values of group 2 elements are higher while the Δ_iH₂ values are lower. The Δ_i H [ and Δ_i H₂ both Na and Mg are given to support it.

Explanation. The (A,-Hi) value of Mg is higher than that of Na because of smaller size and more symmetrical electronic configuration. But after losing one electron, Na⁺ ion acquires the configuration of noble gas neon (1 ,v²2.r2//’) while Mg⁺ ion has still one electron in the valence shell (li²25²2/>⁶3s¹). Due to greater symmetrical electronic configuration, Δ_iH₂ value of sodium is higher than that of magnesium while its A,-Hi value is less.

4. Hydration enthalpies. Just like alkali metals, the divalent cations of the alkaline earth metals have also tendency to get hydrated. The negative hydration enthalpies are more due to the smaller size of cations as compared to the cations of the alkali metals present in the same period. For example, Li⁺ (A_hydH = – 506 kJ mol ¹) ; Be²⁺ (A_hydH = – 2494 kJ mol ). The hydration enthalpies decrease down the group since the catonic size increases
Be²⁺ > Mg²⁺ > Ca²⁺ > Sr²⁺ > Ba²⁺
The salts of alkaline earth metals generally exist as hydrated salts. For example, MgCl₂ exists as MgGl₂. 6H₂O while CaCl₂ as CaCl₂.6H₂O. The corresponding salts of alkali metals NaCl and KCl are anhydrous since their cations are not hydrated so extensively.

5.electronegativity. The electronegativity values of alkaline earth metals are quite close to those of alkali metals, though slightly more. These values decrease from beryllium to radium indicating increased tendency to form ionic compounds. The high electronegativity value of beryllium (1-5) indicates that the element tends to form covalent compounds.

6. Oxidation states and electropositive character. All the members of the family exhibit + 2 oxidation states in their compounds and form divalent cations (M²⁺). This appears to be rather unacceptable because the enthalpy needed to form monovalent cation (M⁺) in all these elements is less than what is required to form a divalent cation (M²⁺). But ionisation enthalpy alone is not the sole criteria for the formation of the cations. A number of other factors are also involved. These are briefly discussed as follows :

• The divalent cation has the configuration of the nearest noble gas element while the monovalent cation is yet to achieve it.
• In the solid state, divalent cation results in a stronger lattice than monovalent cation because of its smaller ionic radius and greater magnitude of positive charge as compared to monovalent cation. A stronger lattice releases more lattice enthalpy in its formation than a weaker lattice.
• In aqueous solution, the divalent cation has a greater tendency to get hydrated because of its smaller size and stronger positive field as compared to monovalent cation. Thus, the hydration enthalpy which is released is more in case of divalent cation as compared to monovalent cation.
• The tendency to get hydrated is maximum in Be²⁺ ions due to their smallest size and is least in Ba²⁺ Their ionic mobilities follow the reverse order i. e., Be²⁺ < Mg²⁺ < Ca²⁺ < Sr²⁺ < Ba²⁺.

used upon the above discussion, one can conclude that divalent cations are formed in preference of monovalent cations when alkaline earth metals participate in the formation of chemical compounds. This can be further supported by considering the enthalpy changes which occur in the formation of MgCl and MgCl₂ in aqueous solution.

Thus, in aqueous solution, MgCl₂ is formed in perference to MgCl. The alkaline earth metals are electropositive in nature as they form positive (divalent) cations and this character increases down the group. However, the first element beryllium (Be) is the least electropositive due to its very small size and high ionisation energy. The components of beryllium are mostly covalent in nature.

7. Metallic character. Alkaline earth metals have stronger metallic bonds as compared to the alkali metals present in the same period. For example, magnesium is harder and denser than sodium.

Explanation. The stronger metallic bonds in alkaline earth metals as compared to alkali metals may be attributed to the smaller kernel size and more number of valence electrons (two) compared to alkali metals (there is only one valence electron). The stronger bonding also results in more close packed arrangement.

8. Density.  Alkaline earth methls are denser than the alkali metals present in the same period because these are more closely packed due to smaller size and.stronger metallic bonds. However, the trend in the density is not uniform. It initially increases from Be to Ca and then decreases from Ca to Ba. The irregular trends in densities are due to the difference in the crystal structures of these metals.

9. Melting and boiling points. The melting and boiling points of these metals are higher than those of the alkali metals present in the same period. This is quite expected also since they have more packed crystal lattice as compared to the alkali metals. But the data as given in the Table 10.4 indicates an irregular trend in these values. This is because of the reason that the atoms adopt different crystal structures in these metals.

• Chemical Properties of Alkaline Earth Metals

Alkaline earth metals are quite reactive chemically due to their low ionisation enthalpies and strongly electropositive nature. The reactivity further increases down the group since the ionisation energy tends to decrease. However, these are less reactive than the corresponding alkali metals which have smaller ionisation enthalpies and stronger electropositive nature. This will be quite apparent from the following discussion concerning these elements.

1. Reaction with oxygen (Formation of oxides and peroxides) with the exception barium and radium which form peroxides (MO₂), rest of the metals form normal oxides (MO) on heating with excess of oxygen.

Beryllium is quite unreactive and reacts only above 600°C when the powder form of the metal is heated with oxygen. Magnesium burns brightly on heating in oxygen and the reactivity with oxygen increases down the group. The enthalpies of formation of these oxides are quite high and these are quite stable in nature.

2. Reaction with water (Formation of hydroxides). The members of alkaline earth metals are less reactive towards water as compared to the corresponding alkali metals because these drfe’less electropositive in nature. Beryllium, the first member of the family has no action while magnesium reacts with boiling water or steam to form magnesium hydroxide (also called milk of magnesia). The rest of the members react even with cold water and their reactivity increases down the group

The hydroxides are also formed, when the metal oxides are dissolved in water BeO

3. Reaction with hydrogen (Formation of hydrides). The members of the family, except beryllium, combine with hydrogen directly upon heating to form metal hydrides

The hydride of beryllium (BeH₂) can be prepared indirectly by the reduction of BeCl₂ with LiΔ_iH₄ dissolved in anhydrous ether

BeH₂ and MgH₂ are covalent in nature while the hydrides of the other metals have ionic structure. The ionic hydrides such as CaH₂ (also called hydrolith) react with water to evolve hydrogen gas

Structure of BeH₂. It exists as a polymeric solid (BeH₂)„ in which each H atom is involved in a three centred bond and exists as a bridge

4. Reaction with halogens (Formation of halides). The members of the family combine directly with halogens at appropriate temperatures to form corresponding halides i. e., MX₂. The halides can also be formed by the action of halogen acid (HX) on oxides, hydroxides and carbonates of these metals.

Berlyllium chloride can be obtained by passing vapour of chorine over BeO and carbon at 1073 K

5. Reaction with carbon (Formation of carbides). Beryllium carbide is formed by heating BeO with coke (C) at about 1900 – 2000°C temperature when a brick red coloured ionic carbide Be₂C is formed. It reacts with water to evolve methane gas. Be₂C is quite often called Methanide.

The other elements of group-2 generally form ionic carbides with the formula MC₂ (M = Mg, Ca, Sr and Ba) by heating the metal directly with carbon (coke) in an electric arc furnace. These can also be formed by heating the metal oxides with coke.

Out of these carbides, calcium carbide evolves acetylene gas on reacting with water. It is therefore, quite often called acetylide.

6. Reaction with nitrogen (Formation of nitrides). Upon heating in an atmosphere of nitrogen (N₂), the members of alkaline earth family form nitrides with the formula M₃N₂. For example,

These metal nitrides are crystalline solids. They react with water forming corresponding hydroxides and liberating ammonia gas.

Mg₃N₂ + 6H₂O  →  3Mg(OH)₂ + 2NH₃

Ca₃N₂+ 6H₂0    →   3Ca(OH)₂ + 2NH₃

7. Reaction with sulphuric acid (Formation of sulphates). Sulphates of the members of group 2 elements can be formed by the action of dilute H₂SO₄
M+ H₂SO₄  →MSO₄ + H₂

Alternatively, these can be obtained by reacting dilute H₂SO₄ with oxides, hydroxides or carbonates of these elements.

MO  +  H₂SO₄ → MSO₄ + H₂O
M(OH)₂+ H₂SO₄ → MSO₄ +2H₂O
MCO₃ +H₂SO₄ →  MSO₄  +   CO₂ + H₂O

8. Formation of carbonates. Carbonates of metals can be formed as white precipitates by passing vapours of CO₂ in limited amount through aqueous solutions of the metal hydroxides.

These can also be formed by adding aqueous solution of sodium carbonate to the aqueous salt solution e.g., metal chlorides.

9. Complex formation. Generally, the members of the family do not form complexes. However, smaller ions like Be^{2 +} and Mg²⁺ form complexes with the electron donor species. For example, the stable complexes of Be are : [BeF₃]^–, [BeF₄]^2- and [Be(H₂O)₄]²⁺.Chlorophyll the complex compound of Mg plays an important role in photosynthesis. Ca, Sr and Ba form complexes only with strong complexing agents like acetylacetone and EDTA etc.

10. Formation of organo-metallic compounds. Both beryllium and magnesium form a number of organo-metallic compounds containing M—C bonds with certain organic compounds. For example, magnesium reacts with alkyl halide in the presence of dry ether to give Grignard reagent.

BeCl₂ reacts with Grignard reagent to form dialkyl beryllium.

Dialkyls or diaryls of Ca, Sr and Ba can also be formed indirectly in the same way.

11. Reducing nature. Just like alkali metals, the members of the family also act as reducing agents. However, their reducing strengths are comparatively less since their ionisation enthalpies are higher and they have less negative reduction potentials (E°) as compared to the alkali metals. The alkaline earth metals behave as a reducing agents by losing their valence electrons to form divalent cations (M²⁺). The reducing character increases down the group due to decrease in the magnitude of ionisation enthalpies and also due to increase in the values of negative reduction potentials of the elements.

The first element beryllium (Be) is expected to behave as a very weak reducing agent because of its small atomic radius and high values of ionisation enthalpies (Δ_iH₁ and Δ_iH₂). However, it behaves as a reducing agent due to large enthalpy of hydration as well as sublimation (also called enthalpy of atomisation).

12. Solution in liquid ammonium. Like alkali metals, the metals belonging to alkaline earth family (group 2) dissolve in liquid ammonia to give deep blue solution which is conducting in nature.

Question 3.
Why are alkali metals not found in nature ?
Answer:
Alkali metals are highly reactive due to low ionisation enthalpy and strong electro-positive character. They do not occur in free or native state and are always combined with other elements. As a result, alkali metals are not generally found in nature.

Question 4.
Find the oxidation state of sodium in Na₂O₂
Answer:

Question 5.
Explain why is sodium less reactive than potassium.
Answer:
This is mainly due to higher ionisation enthalpy (Δ_iH₁ = 496 kJ mol^-1) of sodium as compared to potassium (A,H- = 419 kJ mol^-1). As a result, the potassium is more electropositive and stronger reducing agent than sodium. It also reacts with water more violently than sodium. Thus, sodium is less reactive than potassium.

Question 6.
Compare the alkali metals and alkaline earth metals with respect to
(1) ionisation enthalpy
(2) basicity of oxides
(3)solubility of hydroxides.
Answer:
(1) Ionisation enthalpy. The ionisation enthalpy of alkaline earth metals (group 2) is more as compared to alkali metals (group 1) present in the same period due to smaller size and more symmetrical configuration. For example,

(2) Basicity of oxides. Oxides of alkali metals are stronger bases as compared to those of alkaline earth metals present in the same period. This is quite evident from the fact that when Na₂O is dissolved in water, NaOH formed is a stronger base than when MgO is dissolved in water to form Mg(OH)₂.
Na₂O + H₂O  → 2NaOH ; MgO + H₂O→ Mg(OH)₂
This is on account of lesser ionisation enthalpy of alkhli metals as compared to alkaline earth metals. Therefore, NaOH can release OH’ ions in solution more readily than Mg(OH)₂.

(3) Solubility of hydroxides. Alkali metal hydroxides are more soluble in water as compared to the hydroxides of alkaline earth metals present in the same period. This is on account of higher lattice enthalpy of the hydroxides of alkaline earth family as compared to those of alkali metals.

Question 7.
In what ways does lithium show similarity to magnesium in its chemical behaviour ?
Answer:
Similarity in Lithium and Magnesium (Diagonal Relationship)

In the discussion of group-1 elements we have seen that lithium, the first member of the family differs from the rest of the members (congeners) in most of the characteristics. However, it has a close resemblance with magnesium, the second element of alkaline earth metals (group 2) which it is linked diagonally. In other words, lithium and magnesium exhibit diagonal resemblance or diagonal relationships. In fact the first three elements of second period (Li, Be, B) show diagonal similarity with the elements (Mg, Al, Si) of third period placed on the right hand side.

It may be noted that this relationship is noticed only in these pairs of elements and not anywhere else in the periodic table.Cause of Diagonal Relationship. The cause of diagonal relationship is due to opposing trends in the main periodic properties of the elements along a period and down the group. For example, both atomic and ionic radii decrease along a period and increase down the group. Ionisation enthalpy, electron gain enthalpy and electronegativity follow opposing trends i.e., they increase along a period and decrease down the group. On moving diagonally, these opposing trends tend to cancel mutually and as a result, the elements listed above show diagonal relationalship.

Question 8.
Explain why alkali and alkaline earth metals cannot be obtained by chemical reduction.
Answer:
The metals belonging to both these families are very strong reducing agents. It is, therefore not possible to reduce their oxides by reacting with common reducing agents like carbon (coke), zinc etc. These are normally isolated by carrying out the electrolysis of the salts of these metals in the molten state.

Question 9.
Why are potassium and cesium, rather than lithium used in photoelectric cells ?
Answer:
The ionisation enthalpy of lithium is quite high. The photons of light are not in a position to eject electrons from the surface of lithium metal. Therefore, photoelectric effect is not noticed. However, both potassium and cesium have comparatively low ionisation enthalpies. This means that the electrons can quite easily be ejected from the surface of these metals when photons of certain minimum frequency (threshold frequency v°) strike against their surface.

Question 10.
When an alkali metal is dissolved in liquid ammonia, the solution can acquire different colours. Explain the reason for this type of colour change.
Answer:
Solubility in liquid ammonia. Alkali metals dissolve in liquefied ammonia to give highly conducting solution with blue colour.

Explanation. Alkali metals due to their low ionisation energies, ionise in ammonia solution to form ammoniated cations and ammoniated electrons.

The blue colour of the solution is attributed to the fact that when light falls on the ammoniated electrons, they absorb energy corresponding to red colour and the transmitted light has a blue colour. The electrical conductivity of the solution is because of ammoniated cations as well as ammoniated electrons.

• In concentrated solution (3 M solution) the colour changes from blue to bronze which is copper-coloured. The blue solution is paramagnetic while the concentrated solution is diamagnetic in nature.
• The presence of free ammoniated electrons makes the blue solution reducing in nature.
• When dry ammonia gas is passed over heated metals, amides are formed and hydrogen gas is evolved.
  

Alkali metal amides are powerful reducing agents due to the presence of amide (NH₂~) ions.

Question 11.
Beryllium and magnesium do not give colour to the flame while other alkaline earth metals do so. Explain.
Answer:
Colouration to the flame. With the exception of beryllium and magnesium, the rest of the elements impart characteristic colours to the flame. For example,

Explanation. The reason for imparting characteristic colours to the flame when volatile chlorides of these metals are heated on the tip of a platinum wire is the same as given under alkali metals. The heat energy provided by the flame is sufficient to excite the electrons to higher energy states. When these electrons jump back to the normal or ground states, the emitted radiations fall in the visible region of light resulting in characteristic colours. Since elements differ in their ionisation enthalpies, they are excited to different extent and also differ in the energy that is emitted as light radiations. This leads to different flame colours.

Beryllium and magnesium, the first two members of the family fail to impart any colours to the flame because of smaller size and high ionisation enthalpies. The energy of the flame is not sufficient to excite the electrons in these atoms.

Question 12.
Discuss the various reactions which occur in Solvay process.
Answer:
Details of the process : The plant used for the manufacturing process is shown

1. Ammoniation Tower. A strong solution of brine (30% NaCl solution) is introduced from the top of the tower made up of iron. A mixture of ammonia and carbon dioxide which are formed in the ammonia recovery tower is led from a side into the tower. As a result, brine gets saturated with ammonia. Soluble impurities of some calcium and magnesium salts like CaCl₂ and MgCl₂ associated with sodium chloride are precipitated as carbonates by ammonium carbonate which is formed in the reaction.
2NH₃ + CO₂ + H₂O → (NH₄)₂CO₃ CaCl₂ + (NH₄)₂CO₃

The solution is then passed through filters in order to remove these precipitates.

2. Carbonation Tower. It is also made up of iron and is fitted with a number of horizontal plates. Each plate has a hole in the center covered by a perforated cover. The brine saturated with ammonia (or ammoniated brine) is introduced from the top and the vapours of carbon dioxide from the lime kiln are introduced from the side. As the vapours rise, they come in contact with the ammoniated brine and the following reactions take place.
CO₂ + H₂O + NH₃ →NH₄HCO₃
NaCl + NH₄HCO₃ → NaHCO₃ + NH₄Cl

The temperature in the carbonation tower is between 300 – 310 K (very low) and crystals of sodium hydrogen carbonate are formed. Carbon dioxide needed for the reaction is obtained by heating lime stone (CaCO₃) in a lime kiln

Quick lime formed is dissolved in water to form Ca(OH)₂ which is led into the ammonia recovery tower as shown in the figure.

3. Filtration. The solution coming out of the cafbonation tower is passed through filters when the precipitated sodium hydrogen carbonate gets separated. The solution containing NH₄Cl and small amount of NH₄HCO₃ is taken to the ammonia recovery tower where it meets calcium hydroxide.

4. Calcination of sodium hydrogen carbonate. Sodium hydrogen carbonate formed above is heated strongly in the absence of air (calcined) in a furnace to give sodium carbonate.

5. Ammonia recovery tower. In this tower, ammonia is formed by the reaction between NH₄Cl and Ca(OH)₂ and the reaction mixture is heated with the help of steam coil.

Ammonium hydrogen carbonate present in the tower also decomposes to evolve NH₃ and CO₂

Both these gases are pumped into the ammoniation tower where they take part in the reaction.

Merits of the Ammonia Solvay process. This process is extremely useful because of the following reasons.

• Sodium carbonate (or soda ash) is almost completely pure.
• The process is also very cheap because both NH₃ and CO₂ are formed in the plant itself.
• There is hardly any pollution problem because the waste like flue gases are not evolved.

Question 13.
Potassium carbonate cannot be prepared by Solvay process. Explain.
Answer:
In Solvay Ammonia process for getting sodium carbonate, NaHCO₃ formed gets precipitated and upon heating it gives sodium carbonate. KHCO₃ is expected to be formed if the same process is used for the preparation of K₂CO₃. But as KHCO₃ is highly soluble in water, it cannot be isolated.

Question 14.
Lithium carbonate decomposes on heating while sodium carbonate does not.
Answer:
Upon heating, Li₂CO₃ decomposes to form Li₂O and CO₂. The smaller size of Li⁺ ion makes the lattice of Li₂O more stable than that of Li₂CO₃. But due to the bigger size of Na⁺ ion, the lattice of Na₂O is less stable than that of Na₂CO₃. Therefore, Na₂CO₃ fails to decompose on heating while lithium carbonate decomposes.

Question 15.
Compare the solubility and thermal stability of the following compounds of alkali metals with those of alkaline earth metals :
(a) Nitrates (b) Carbonates (c) Sulphates
Answer:
(a) Nitrates of alkali and alkaline earth metals.
Solubility. Nitrates of alkali metals are water soluble. Their solubility increases down the group because their lattice ., enthalpy decreases more rapidly than the hydration enthalpy.(Please note that decrease in lattice enthalpy favours solubility of a solid while decrease in hydration enthalpy opposes the same) Nitrates of alkaline earth follow the reverse trends i.e., their solubility decreases down the group because hydration enthalpy decreases more rapidly than the lattice enthalpy.

Thermal stability. The nitrates of alkali metals (except lithium nitrate) upon heating decompose to form corresponding nitrites and evolve oxygen.

Lithium nitrate decomposes to form lithium oxide and evolves NO₂ as well as O₂.

The alkaline earth metal nitrates decompose in the same way as lithium nitrate.

(b) Carbonates of alkali and alkaline earth metals
Solubility. The trend in the solubility of alkali metal carbonates is the same as that of nitrates i. e., it increases down the group. The trend in the solubility of the alkaline earth metal carbonates is also the same i.e., it decreases down the group

Thermal stability. Except for lithium carbonate?which decomposes upon strong heating to evolve CO2, the carbonates of rest of the alkali metals are quite stable to heat i.e., they donot decompose

In fact, Li₂O is more stable than UCO3 because due to small size of Li⁺ ion, the lattice of Li₂O is quite stable. For rest of the alkali metal carbonates, M₂O is less stable than MCO₃ due to bigger size of the metal ion. The alkaline earth metal carbonates decompose in the same way as lithium carbonate.

Due to smaller size of M²⁺ ion, the lattice of MO is more stable than that of MCO₃. However, the stability of metal carbonates increases down the group because of gradual increase in the size of the M²⁺ ion and lesser stability of MO as compared to MCO₃.

(C) Sulphates of alkali and alkaline earth metals
Solubility. The trend in the solubility of both alkali metal sulphates and alkaline earth metal sulphates is the same as in case of nitrates and carbonates.
Thermal stability. Except for Li₂SO₄ which decomposes upon heating the sulphates of other alkali metals are thermally stable i.e., they do not decompose on heating

The sulphates of alkaline earth metals also decompose in the same way.

Question 16.
Starting from sodium chloride, how will you proceed to prepare (1) sodium metal (2) sodium hydroxide (3) sodium peroxide (4) sodium carbonate.
Answer:
(1) Preparation of sodium metal. Sodium metal is formed by carrying out the electrolytic reduction of the salt in the molten state.

(2) Preparation of sodium hydroxide. Sodium hydroxide is prepared by carrying out the electrolysis of the aqueous solution of sodium chloride either in Nelson’s cell or Castner Kellner cell.

(3) Preparation of sodium peroxide. Sodium chloride is first converted to sodium by electrolytic reduction. The metal is then heated with excess of oxygen at about 573K in an atmosphere free from moisture and carbon dioxide to form sodium peroxide.

(4) Preparation of sodium carbonate. From sodium chloride, sodium carbonate is prepared by Solvay ammonia process.

Anhydrous sodium carbonate is also galled soda ash as it is present in the ashes of certain marine plants. This was the only natural source till 1790. Washing soda, Na₂CO₃. 10H₂O is very popular in laundries for washing clothes. Though a number of methods are available, the most common among them is the Solvay Ammonia Process. This is used for the manufacture of anhydrous sodium carbonate.

Theory : Carbon dioxide is passed through concentrated solution of sodium chloride or brine (about 30%) which is saturated with ammonia to form sodium hydrogen carbonate as follows :

In the presence of Na⁺ ions present in the solution, sodium hydrogen carbonate gets precipitated. The precipitate is removed by filtration and upon heating, it gives sodium carbonate.

Question 17.
What happens when : –

1. Magnesium is burnt in air
2. Quick lime is heated with silica
3. Chlorine reacts with slaked lime
4. Calcium nitrate is heated.

Answer:
(1) A mixture of magnesium oxide and magnesium nitride is formed

(2) Calcium silicate is formed

(3) Calcium oxychloride (bleaching powder) is formed

(4) Nitrogen dioxide is evolved

Question 18.
Describe two important uses of each of the following (1) caustic soda (2) sodium carbonate (3) quick lime.
Answer:

1. Uses Sodium hydroxide  Or Caustic Soda

• in the manufacture of soap, paper, rayon (artificial silk) and a large number of chemicals,
• in cotton industry for mercerizing (or making unshrinkable) cotton fabrics,
• in the refining of petroleum and also of vegetable oils,
• in the preparation of soda lime (a mixture of NaOH and CaO), ,
• as a cleansing agent for machines and metal sheets.

2. Uses : Sodium carbonate is

• for the manufacture of soap, glass, paper, caustic soda, borax etc.
• as a fusion mixture when mixed with K₂CO₃.
• for washing of clothes in laundary under the name washing soda,
• in textile industry and also in the refining of petroleum.
• as a laboratory reagent.

3. Uses of Quick Lime :

• It is used for making building materials such as mortar cement etc.
• It is useful in the manufacture of cement, glass, calcium carbide, sodium carbonate etc.
• It is used as- a basic flux in metallurgy.
• It is helpful in tanning industry to remove hairs from hides.
• Quick lime is used for drying a number of gases and alcohol.

Question 19.
Draw the structures of (1) BeCl₂ (in solid state) (2) BeCl₂ (in vapour state)’.
Answer:

Structure of BeCl₂. In solid state, beryllium chloride has a chain structure (polymer) in which the electron pairs from chlorine atoms present on neighbouring molecules are donated to electron deficient Be atom to form co-ordinate bonds as follows :

The Be atom in the above chain structure is sp³ hybridised but the Cl—Be—Cl bond angle is considerably less (98°) as compared to normal tetrahedral bond angle of (109-5°). In the vapour state, the compound exists as a dimer (Be atom is sp² hybridised) which decomposes at about 1000 K to give a monomer in which Be atom is in sp hybridisation state.

Question 20.
The hydroxides and carbonates of sodium and potassium are easily soluble in water while the corresponding compounds of magnesium and calcium are sparingly soluble. Explain.
Answer:
All the compounds are crystalline solids and their solubility in water is guided by both lattice enthalpy and hydration enthalpy. In case of sodium and potassium compounds, the magnitude of lattice enthalpy is quite small as compared to hydration enthalpy since the catonic sizes are large. Therefore, the compounds of sodium and potassium that are mentioned, readily dissolve in water. However, in case of corresponding magnesium and calcium compounds, the cations have smaller sizes and more magnitude of positive charge. This means that their lattice enthalpies are more as compared to the compounds of sodium and potassium. Therefore, the hydroxides and carbonates of these metals are only sparingly soluble in water. For more details, consult text part.

Question 21.
Describe the importance of the following :
(1) Lime stone (2) Cement (3) Plaster of Paris.
Answer:

1. Lime stone or Calcium Carbonate or Marble (CaCOs)

Calcium carbonate occurs in nature as lime stone and also as marble rocks. It can be prepared on small scale by the following methods.

• By passing carbon dioxide through slaked lime in limited amount.
  Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
  Excess of the gas should be avoided as it will lead to soluble calcium bicarbonate.

• By reacting aqueous solutions of sodium carbonate and calcium chloride
  Na₂CO₃ + CaCl₂ → CaCO₃ + 2NaCl

Properties :

• It is a white fluffy amorphous powder and is only slightly soluble in water.
• Upon heating to 1200 K, it decomposes to evolve carbon dioxide.
  
• It liberates carbon dioxide on reacting with dilute mineral acids.
  CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
  CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

(2) Cement:

Cement is one of the most important materials used in the construction of buildings, dams, bridges, roads etc. In 1824, Joseph Asplin, a mason working in Leeds (U.K.), for the first time, heated a mixture of lime stone, clay and water and allowed the mass to stand for some time when it hardened into a stone like mass. It resembled Portland rock which was an important naturally occurring building stone used in England during those days. He named it portland cement. In India, cement industry came into existence in 1914.

Cement is a greyish fine powder consisting of a mixture of various silicates and aluminates of calcium such as tricalcium silicate (3CaO.SiO₂), tricalcium aluminate (3CaO.Al₂O₃) and dicalcium silicate (2CaO.SiO₂). On mixing with water, it sets into a hard mass with a good strength.

(3) Plaster of Paris : (CaSO₄.1/2H₂O) or (2CaSO4.H₂O)   
Preparation : Plaster of paris as mentioned above is prepared by the partial dehydration of gypsum at 390 K

Temperature should not be allowed to rise beyond 390 K in order to check the formation of completely anhydrous CaSO₄.
Properties :

1. It is a white amorphous powder.
2. When mixed with three times its weight of water, it forms a plastic mass which sets into a hard mass within ten to fifteen minutes. This is known as setting of plaster of paris. The setting is most probably because of rehydration back to gypsum.
   
   During setting, the mass undergoes a slight expansion (about 1 %) in volume. This helps plaster of paris to take the shape of any mould in which it is put.

Uses :

• Plaster of paris is commonly used in making moulds for pottery, ceramics etc.
• It is also used in surgical bandages for setting broken bones of the body.
• It is used for making statues, models, decorative materials and black board chalks.

Question 22.
Why are lithium salts commonly hydrated while those of other alkali metal ions are usually anhydrous ?
Answer:
In the lithium salts, the lithium ion (Li⁺) due to very small size gets readily hydrated on coming in contact with moisture (or water). Therefore, lithium salts are commonly hydrated. But the other alkali metal ions are comparatively big in size. They have therefore, lesser tendency to get hydrated. These salts are usually anhydrous.

Question 23.
Why is LiF almost insoluble in water while LiCl is soluble not only in water but also in acetone ?
Answer:
The low solubility of LiF in water is due to its very high lattice enthalpy (F~ ion is very small in size). On the other hand, in lithium chloride (LiCl), the lattice enthalpy is comparatively very small due to comparatively large size of CL ion. This means that the magnitude of hydration enthalpy is quite large. Therefore, lithium chloride dissolves in water and also in acetone due to dipolar attraction. (Acetone is polar in nature

Question 24.
Explain significance of sodium, potassium, magnesium and calcium as biological fluids.
Answer:

• Sodium and Potassium

Living beings need nearly 27 elements out of which about 15 are metals. Alkali metals (Na, K) and alkaline earth metals (Ca, Mg) are needed in major quantities. Minor quantities of other metals like Fe, Co, Cu, Zn, Mo, Ni, Al etc. are also needed in different organisms.

Both sodium and potassium in the form of ions are common and essential constituents of biological fluids. In a person weighing 70 kg, sodium and potassium are present to the extent of 90 g and 170 g respectively along with 5 g of iron and 0-06 g of copper.

Although both Na⁺ and K⁺ are chemically similar, they have different biological functions to perform. The Na⁺ ions are present in blood plasma and in the interstitial fluid which surrounds the cells. These ions take part in the transmission of nerve signals, regulate the flow of water across cell membranes and help in the transport of different sugars and amino acids into cells. The K⁺ ions present in abundance in the cell fluids, activate a variety of enzymes and also promote oxidation of glucose into ATP which is a source of energy. Both sodium and potassium are also responsible for the transmission of nerve signals.

The Na⁺ and K⁺ ions differ in concentration on the opposite sides of the cell membranes. For example, in blood plasma, the concentrations of Na⁺ and K⁺ ions are 143 milli mol/L and 5 milli mole/L respectively. However, within the red blood cells, the respective concentrations of these ions are 10 milli mol/L and 105 milli mol/L. This ratio is known as concentration gradient or ionic gradient. In order to maintain the concentration gradient in the cells, work has to be done. For this, a sodium potassium pump operates across the cell membranes and consumes nearly one third of the ATP used by a resting animal.

• Magnesium and Calcium

We have discussed earlier the biological importance of sodium and potassium ions. The ions of magnesium and calcium are also found in large proportions in biological fluids and they perform a variety of biological functions. An adult body contains about 25 g of magnesium and 200 g of calcium as compared to 5 g of iron and 0 06 g of copper.

Mg²⁺ ions are the constituent of chlorophyll, the green colouring pigment in plants. It absorbs photons from light and carries photosynthesis in plants. Ca²⁺ ions are the major constituent of our bones in the form as apatite Ca₃(PO4)2. The deficiency of these ions occurs particularly in old age. Similarly, Ca²⁺ ions are a constituent of the enamel of our teeth as fluorapatite [3Ca₃(P0₄)2 CaF₂] They also play an important role in muscle contraction. These functions are related to the energy storage in biological systems in the form of pyrophosphate linkage. Both Mg²⁺ and Ca²⁺ act as catalysts in the formation of these linkages. The release of energy due to hydrolysis of pyrophosphate is controlled by Ca²⁺ ions. Calcium occurs mainly in milk. Vitamin D helps in the deposition of calcium in bones.

Question 25.
What happens when   

1. Sodium metal is dropped in water
2. Sodium metal is heated in free supply of air
3. Sodium peroxide dissolves in water ?

Answer:
(1) Sodium hydroxide is forced. The metal catches fire. Actually, the hydrogen evolved is highly combustible and it catches fire

(2) Sodium peroxide is formed
2Na + O₂ → Na₂O₂

(3) Oxygen gas is evolved
2Na₂O₂ + 2H₂O —> 4NaOH + O₂

Question 26.
(a) Why is LiF least soluble in water among the fluorides of alkali metals ?
(b) Justify the given order of mobilities of the alkali metal cations in aqueous solution :
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺
(c) Lithium is the only alkali metal which forms a nitride directly. Explain.
(d) E° for M²⁺(aq) —> M(s) (where M = Ca, Sr or Ba) is nearly constant. Discuss.
Answer:
(a) Lithium fluoride (LiF) is of covalent nature because of the high polarising power of Li⁺ ion due to its very small size and high effective nuclear charge. It distorts the electron cloud of the F^– ion to the maximum as compared to the cations of other alkali metals. It is therefore, least soluble in water. On the other hand, the fluorides of other alkali metals are generally ionic and are water soluble.
(b) This is attributed to the hydration of the cation in water. As a result, size of the cation increases and its mobility decreases. Due to the smallest size, Li⁺ ion is hydrated to the maximum and exists as Li⁺ (aq) and has least mobility. Cs⁺ion due to least hydration exists as Cs⁺ (aq) has maximum mobility.
(c) Lithium is a very strong reducing agent. As a result, it directly exists as Cs⁺ (aq) combines with nitrogen to form its nitride (Li₃N)

(d) The overall magnitude of reduction potential (E°) depends upon three factors. These are

1. sublimation enthalpy
2. ionisation enthalpy and
3.  hydration enthalpy.

In case of the metals listed, the overall magnitude of E° values remain almost the same. Therefore, these metals have almost same reducing strength

Question 27.
State as to why :
(a)A solution of Na₂CO3 is alkaline.
(b)Alkali metals are prepared by the eletrolysis of their fused chlorides.
(c)Sodium is found to be more useful than potassium.
Answer:
(a)Sodium carbonate being a salt of strong base (NaOH) and weak acid (H₂CO₃), forms an alkaline solution upon hydrolysis

(b)Since alkali metals are highly reactive chemically, they react with water to evolve hydrogen gas. These are therefore, prepared by the electrolysis of their fused chlorides.
(c) Sodium is relatively more abundant than potassium. At the same time, it is also less reactive and its reactions with other substances can be better controlled.

Question 28.
Write the balanced equations for the reactions between :
(1) Na₂O₂ and water
(2) KO₂ and water
(3) Na₂O and CO₂
aNS:

1. 2Na₂O₂ + 2H₂O → 4NaOH + O₂
   
2. 2KO₂ + 2H₂O → 2KOH + H₂O₂ + O₂
   
3. Na₂O + CO₂ → Na₂CO₃.

Question 29.
How would you explain the following :       

1. BeO is insoluble in water while BeSO₄ is soluble.
2. Ba(OH)₂ is soluble in water while BaSO₄ is almost insoluble.
3. Lil is more soluble than KI in ethanol.
4. NaHCO₃ is known in the solid state but Ca(HCO₃)₂ is not isolated in the solid state.

Answer:

1. The lattice enthalpy of BeO is higher as compared to BeSO₄ because the size of O^2- ion is very small while SO₄^2- ion has bigger size. Since high lattice enthalpy opposes the solubility of a substance in water therefore, BeO is almost insoluble while BeSO₄ is soluble in water.
2. The size of the SO₄^2- ion in BaSO₄ is quite big as compared to that of OH^– ion in Ba(OH)₂. The SO₄^2- ion has masked the Ba²⁺ ion in BaSO₄ to a large extent with the result that the cation has a very little tendency to get hydrated. On the other hand, the OH^– ion due to smaller size masks the
3. Ba²⁺ ions to lesser extent which means that hydration energy released when Ba(OH)₂ dissolves in water is quite large. Therefore, Ba(OH)₂ readily dissolves in water while BaSO₄ is almost insoluble.
4.  Lil is mainly covalent while KI has ionic nature. In fact, the size of Li⁺ ion is smaller than that of K⁺ ion and it polarises the electron cloud of I^– ion to a greater extent. With the result, LiI dissolves in ethanol or ethyl alcohol (organic solvent) whereas KI is almost insoluble.
5. NaHCO₃ is less soluble in water than Ca(HCO₃)₂. Therefore, it can be precipitated from the solution while it is difficult to precipitate Ca(HCO₃)₂. Actually, Ca²⁺ ion has a greater tendency to get
   hydrated than Na⁺ ion due to its small size and more magnitude of positive charge. Therefore Ca(HCO₃)₂ is more soluble in water than NaHCO₃.

Question 30.
Which of the following alkali metals is having the least melting point ?
(a) Na (b) K (c) Rb (d)
Answer:
(d) Cesium (Cs) has the least m.p. due to its maximum size the least lattice enthalpy.

Question 31.
Which of the following alkali metals gives hydrated salts ?
(a) Li (b) Na (c) K (d)
Answer:
(a) The salts of lithium are generally hydrated because the size of the Li⁺ ion is very small. It has maximum hydration enthalpy.

Question 32.
Thermally the most stable alkaline earth metal carbonate is :
(a) MgCO₃ (b) CaCO₃ (c) SrCO₃ (d) BaCO₃.
Answer:
(d) Barium carbonate (BaCO₃) is thermally the most stable. For more details, consult section 10.14.

We hope the NCERT Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements (Alkali and Alkaline Earth Metals), help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry Chapter 10 The s-Block Elements (Alkali and Alkaline Earth Metals), drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

Question 1.
Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer:
Position of Hydrogen in the Periodic Table

In the periodic table, hydrogen has been placed at the top of the alkali metals in group 1 although it is not a member of the group. However, its position is not properly justified simply because of its electronic configuration (Is¹). It can be placed along with the alkali metals as they too have ns¹ configuration. However, it can also be placed along with halogens in group 17 since just like halogens, it also requires any one electron to have the configuration of the nearest noble gas element helium. Let us compare its characteristics with the members of both the families

Resemblance with Alkali metals
Hydrogen and alkali metals resemble in the following respects.
Electronic configuration. Hydrogen has one electron in its valence shell like the alkali metals. For example,
Element                                                 H                   Li                         Na
Atomic No.                                            1                     3                           11
Electronic configuration                      Is¹               1s²2s’               ls²2s²2p⁶3s¹

Question 2.
Write the names of the three isotopes of hydrogen. What is the mass ratio of these isotopes ?
Answer:
The three isotopes of hydrogen are : protium (\( _{ 1 }^{ 1 }{ H }\)) deuterium( \( _{ 1 }^{ 2 }{ H }\)or D) and tritium ( \( _{ 1 }^{ 3 }{ H }\) or T). Mass ratio of the three isotopes is :

Question 3.
Why does hydrogen occur in diatomic form rather than in monoatomic form under normal conditions ?
Answer:
In monoatomic form, hydrogen atom has only one electron in K-shell (Is¹) while in diatomic form, the K-shell is complete (Is²). This means that in diatomic form, hydrogen (H₂) has acquired the configuration of noble gas helium. It is therefore, quite stable.

Question 4.
How can the production of dihydrogen from ‘coal gasification’ be increased ?
Answer:
In coal gasification reaction, steam vapours are passed through coke to form mixture of CO and H₂ also called syn gas.

The production of dihydrogen can be increased by reacting CO(g) present in syn gas with steam in the presence of iron chromate catalyst With the removal of CO(g) or CO₂(g), the reaction shifts in the forward direction. This is known as water gas shift reaction. As a result, the production of dihydrogen will increase

Question 5.
Describe the bulk preparation of dihydrogen by electrolytic method. What is the role of an electrolyte in the process ?
Answer:
Dihydrogen is formed mainly from water by carrying its electrolysis in the presence of suitable electrolytes i.e., small amount of an acid or alkali. Actually, water as such is a poor conductor of electricity. Addition of small amount of electrolyte increases its conductivity.

Question 6.
Complete the following

Answer:


Question 7.
Describe the consequences of high bond enthalpy of H-H bond in terms of chemical reactivity of dihydrogen.
Answer:
The bond enthalpy or bond dissociation enthalpy of dihydrogen (H₂) is very high (AH = 435 -9 kJ mol^-1). This is on account of its small atomic size and also small bond length (74 pm) of H—H bond. Consequently, the bond cleavage is extremely difficult and molecular hydrogen or dihydrogen takes part in chemical reactions only under specific conditions.

Question 8.
What do you understand by
(1) electron deficient
(2) electron precise and
(3) electron rich compounds of hydrogen ? Provide justification with suitable examples.
Answer:
(1) Electron deficient hydrides : These hydrides have lesser number of electrons available for writing their conventional structures according to Lewis concept. One popular example of the hydrides belonging to this class is of diborane (B₂Hg). No electrons are available to account for the bonding between the two atoms of boron. We shall discuss the details about the structure of diborane in unit-11 onp-block elements. Most of the hydrides of elements of group 13 fall under this category. This electron deficient hydrides may be regarded as Lewis acids and are electron acceptors.

(2) Electron precise hydrides : These are the covalent hydrides in which the atoms have the required number of electrons which are permissible according to Lewis concept of bond formation. Hydrocarbons like methane (CH₄), ethane (C₂Hg), ethene (C₂H₄) and ethyne (C₂H₂) etc. are electron precise hydrides. We shall discuss them in unit-13.

(3) Electron rich hydrides : In these hydrides, the participating elements donot have the required number of hydrogen atoms. As a result, they have one or more lone pairs of electrons present. The elements generally belong to groups 15, 16 and  17 A few important of out of these are

They have one, two and three lone pairs of electrons present one N, O and F atoms respectively. Due to the presence of lone electron pairs, these are expected to behave as Lewis bases. However, except for NH₃, both H₂O and HF hardly exhibit basic nature of oxygen and fluorine. However, all the three hydrides which are listed, participate in intermolecular hydrogen bonding and get associated due to strong polar character.

Question 9.
What characteristics do you expect from an electron deficient hydride with respect to its structure and chemical reaction ?
Answer:
It is expected to be a Lewis acid. For example, diborane B₂H₆ (dimer of BH₃) forms a complex with LiH which gives H^– ion (Lewis base)

Similarly it also combines with trimethylamine (organic compound) and carbon monoxide which act as Lewis base to form addition compounds.

Question 10.
Do you expect the carbon hydride of the type (C_nH_2n+2) to act as Lewis acid or base ? Justify you answer.
Answer:
It is neither a Lewis acid nor a Lewis base. In all the hydrides belonging to this type, the carbon atoms have complete octet. Therefore, these hydrides behave as normal covalent hydrides also called saturated hydrocarbons or alkanes, (e.g., CH₄, C₂H₆, C₃H₈ etc.). These are also called electron precise hydrides. We shall study in details about these hydrides at a later stage in unit 13.

Question 11.
What do you understand by the term non-stoichiometric hydrides ? Do you expect these types of hydrides to be formed from alkali metals ? Justify your answer.
Answer:
Metallic or Non-stiochjometric (or Interstitial) Hydrides
Many transition and inner transition metals absorb hydrogen into the interstices of their lattices to yield metal like hydrides also called interstitial hydrides. The hydrogen is present in the atomic form in these hydrides. It is interesting to note that the transition metals of groups 3, 4 arid 5 form metallic hydrides. In group 6, chromium alone has a tendency to form hydride CrH. Then there is a gap and the metals of groups 7, 8 and 9 donot form hydrides. This is known as hydride gap. Latest studies have shown that only the hydrides of Ni, Pd, Ce and Ac are interstitial in nature which means that hydrogen atoms occupy the interstitial sites.

The rest of the hydrides have different lattices and it is not proper to call them interstitial hydrides. The elements of/-block form limited number of hydrides. The hydrides are generally non-stoichiometric and their composition vary with temperature and pressure. For example, TiH_1.73, CeH_2.7, LaH_2.8 etc. These hydrides have metallic appearance and their properties are closely related to those of the parent metal. Most of them are strong reducing agents probably due to the presence of free hydrogen atoms in metal lattice. Metallic hydrides can be used for storing hydrogen Alkali metals do not form these type of hydrides because in their crystal lattices, atoms of hydrogen, which are very small in size do not fit in.

Question 12.
How would you expect the metallic hydrides to be useful for hydrogen storage ? Explain.
Answer:
Some metals like palladium (Pd), platinum (Pt) etc. have a capacity to adsorb large volume of hydrogen on their surface forming hydrides. In fact, hydrogen dissociates on the surface of metal as H atoms which are adsorbed. In order to accommodate these atoms, the metal lattice expands and becomes rather unstable upon heating. The hydride releases hydrogen and changes back to the metallic state. The hydrogen evolved in this manner can be used as a fuel. The metals listed above (belonging to transition metals) can be used to store as well as transport hydrogen which is to be used as a fuel. Thus, metal hydrides have a very useful role in hydrogen economy.

Question 13.
How does atomic hydrogen or oxy-hydrogen torch function for cutting and welding purposes ? Explain.
Answer:
Atomic hydrogen torch.When molecular hydrogen is passed through tungsten electric arc (2000 – 3000°C), at low pressure, it dissociates to form atoms of hydrogen, known as atomic hydrogen.

Atomic hydrogen is extremely unstable (life period is about 0-3 sec). Therefore, hydrogen atoms immediately unite to form molecular hydrogen again accompanied by the liberation of a large amount of heat energy. The temperature rises to 4000 – 5000°C. This is the principle of atomic hydrogen torch which is used for welding purposes. oxy-hydrogen torch.

When hydrogen is burnt in oxygen, the reaction is highly exothermic in nature. A temperature ranging between 2800^c to 4000°C is generated. This temperature can be employed for welding purpose in the form of oxy-hydrogen torch.

Question 14.
Among NH₃, H₂O and HF, which would you expect to have highest magnitude of hydrogen bonding and Why ?
Answer:
Hydrogen bonding represents dipolar attraction. Its strength depends upon the magnitude of the polarity of the bond. The electro negativities of three non-metals F, O and N involved in these compounds are 4, 3-5 and 3 respectively. This means that H—F bond is maximum polar and as a result, highest magnitude of hydrogen bonding is HF molecules.

Question 15.
Saline hydrides are known to react violently with water producing fire. Can CO₂, a well known extinguisher, be used in this case ? Explain.
Answer:
Whenever a saline hydride (NaH or CaH₂) reacts with water, the reaction is so highly exothermic that the hydrogen evolved catches fire. For example,

NaH(s) + H₂O(aq) → NaOH(aq) + H₂(g) + heat
CaH₂(s) + 2H₂O(aq) → Ca(OH)₂(aq) + 2H₂(g) + heat

CO₂ which is normally used as fire extinguisher cannot be used in this case because it will react with the hydroxide formed in the reaction to form a carbonate. This will increase the rate of the forward reaction in which heat is evolved.

2 NaOH(aq) + CO₂(g) →Na₂CO₃(aq) + H₂O(aq).

Question 16.
Arrade the following:
(1) IH, NaH and CsH in order of increasing ionic character.
(2) Hlr-H, D—D, and F—F in order of increasing bond dissociation enthalpy.
(3) N₄₁, MgH₂ and H₂O in order of increasing reducing power.
Answer:
(1) Increasing ionic character: LiH < Nail < CsH.
Reason : The ionisation enthaiphy decreases in the order Li > Na> Cs. This influences the ionic character adversely which increases as shown
(2) Increasing bond dissociation enthalpy : F—F < H—H < D—D.
Reason : The bond dissociation enthalphy of :F—F: fluorine is very small (242-6 kJ mol^-1) due to the repulsion in the lone pairs of electrons present on the two F atoms. Out of H₂ and D₂, the bond dissociation enthalphy of H—H (435-88 kJ mol^-1) is less than that of D—D (443-35 kJ mol^-1).
(3)  Increasing reducing power : H₂O < MgH₂ < NaH.
Reason : NaH being ionic in nature is the strongest reducing agent. Both H₂O and MgH₂ are covalent in nature but bond dissociation enthalpy of H₂O is higher. Therefore, it is a weaker reducing agent than MgH₂.

Question 17.
Compare the structures of H₂O and H₂O₂
Answer:
The structure of H₂O is angular while that of H₂O₂ is non-linear just like two opposite open pages of a book. For actual structure of H₂O,
Structure of water (H₂O).

H₂O is a covalent molecule in which the two hydrogen atoms are linked to the oxygen atoms by single covalent bonds. The oxygen atom is surrounded by two shared pairs with hydrogen atoms and it has also two lone pairs of electrons. To have a minimum force of repulsion in the four electron pairs around the oxygen atom, the structure of H₂O molecule is expected to be tetrahedral.

However, the two lone pairs distort its geometry and its bond angle (104-5°) is less than the bond angle of regular tetrahedron (109-28°). As the oxygen atom is more electronegative (3-5) than the hydrogen atom (2-1), therefore, both the O—H bonds are polar in nature. Because of the unsymmetrical nature of H₂O molecule, the bond polarities do not cancel. Therefore, it is polar with dipole moment (ju) equal to 1 -84 D.

Structure of Hydrogen Peroxide
Hydrogen peroxide is a dihydroxy compound (H-O-O-H) and the 0-0 linkage is known as a peroxide linkage. It is a non­linear molecule as the two 0-H bonds are in different planes. The interplanar (dihedral) angle is 111-5° in the gaseous phase but is reduced to 9p-2° in the crystalline state because of hydrogen bonding. The molecular dimensions of hydrogen peroxide in the gas and solid phases are shown.

Question 18.
What do you understand by ‘auto-protolysis’ of water ? What is its significance ?
Answer:
Auto-protolysis of water means ‘self-ionisation’ which proceeds as follows :

Because of auto-protolysis, water behaves both as a Bronsted acid and Bronsted base i.e. it is amphoteric in nature

Question 19.
Consider the reaction of water with F₂ and suggest in terms of oxidation and reduction which species are oxidised/reduced.
Answer:

• Water (H₂O) is oxidised to O₂
  
• Fluorine (F₂) is reduced to F^– ions or HF

Question 20.
Complete the following :
(1) PbSfs) + H₂O₂(aq)→
(2) MnO₄(dq) + H₂O₂(aq) + H⁺(aq) →
(3) CaO(s) + H₂O(g)→
(4) AlCl₃(s) + H₂O(l) →
(5) Ca₃N₂(s) + H₂O(l) →
Answer:

1. PbS(s)+ 4H₂O₂(aq) →PbSO₄(.s) + 4U₂Oaq)
2. 2MnO^–₄ (aq) + 6iA⁺(aq) + 5R₂O₂(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5O₂(g)
3. CaO(s) + H₂O(g) → Ca(OH)₂(s)
4. AlCl₃(s) + 3H₂O(l) → Al(OH)₃(s) + 3HCl (l)
5. Ca₃N_2(s) + 6H₂O(l) → 3Ca(OH)₂(s) + 2NH₃(g).

Question 21.
Describe the structure of common form of ice.
Answer:
1. Ice fleets over water. In ice (solid state), the H₂O molecules are arranged tetrahedrally ini space. The oxygen atom in each H₂O molecule is linked to two hydrogen atoms by covalent bonds and at the same time its forms hydrogen bonds with the hydrogen atoms of the neighbouring H₂O molecules. This leads to a cage like structure as shown in the Figure 9.5. The structure is also porous because of the voids or empty spaces left.

Now, as ice mbits to form liquid water, the heat energy supplied tends to break some of the hydrogen bonds in H₂O molecules. As a result, the tetrahedral arrangements start collapsing and the H₂O molecules in water come closer to one another. The number of vacant spaces also start decreasing. Therefore, the density of water is more than that of ice. This means that ice always floats over water.

2. Water has maximum density at 4°C. At 273 K (or 0°C), both ice and water coexist. As the temperature is increased, the heat energy supplied will further break the tetrahedral arrangements because of decrease in hydrogen bonding. Therefore, the density of water is expected to increase. But the rise in temperature is also expected to increase the average kinetic energy of the H₂O molecules leading to increase in volume (or “decrease in density). But this effect is negligible upto 4°C and the density of water, therefore, increases. However, if the temperature is increased beyond 4°C, then the effect of the increase in kinetic energy will be more as compared to the effect of increase in density. This is likely to increase the volume of water and decrease its density. Thus, we conclude that upto 4°C, the density of water increases and then decreases after this temperature. In other words, water has maximum density and minimum volume at 4°C.

This property of water is extremely helpful for the animals living under sea water. In severe cold, the surface of the sea almost freezes. But below the surface, there is water at a temperature of about 4°C. The aquatic animals can safely live in water at this temperature.

Question 22.
What causes temporary and permanent hardness of water ?
Answer:
1. Temporary hardness is due to the presence of hydrogen carbonates of calcium and magnesium dissolved in water. It is called temporary because it can be easily removed by simply boiling hard water.

2. Permanent hardness is due to the presence of chlorides and sulphates of calcium and magnesium dissolved in water. It is called permanent hardness as it cannot be removed on boiling hard water

Question 23.
Discuss the principle and method of softening of hard water by synthetic ion-exchange method.
Answer:
Softening of hard water. Softening of water means the removal of the ions from water which make it hard.

Question 24.
Write chemical reactions to show amphoteric nature of water.
Answer:
water covers nearly three-fourth of the earth’s surface in the form of snow over mountains as liquid in the rivers, lakes, springs and oceans. Next to oxygen, water is the most important for human life. One can live without food for a number of days but not without water. It is also used as a solvent for many substances and takes part in a variety of chemical reactions. Water is also a source of heavy water (D₂O) which is extremely essential for nuclear reactors to control the speed of neutrons.

Question 25.
Hydrogen peroxide can act both as an oxidising as well as reducing agent. Explain.
Answer:
The Change in oxidation number accompanying the decomposition of H₂O₂ is as follows :

Since oxygen atom in H₂O₂ can undergo an increase as well as decrease in oxidation number therefore, it can act both as reducing as well as oxidising agent. This is supported by the following reactions.
As reducing agent:

In this reaction, H₂O₂ acts as reducing agent and has reduced Ag₂O K to metallic Ag.
As oxidising agent:

In this reaction, H₂O₂ has oxidised PbS to PbSO₄.

Question 26.
What is meant by demineralised water ? How is it obtained ?
Answer:
Water free from cations (Ca²⁺, Mg²⁺ etc.) and anions (Cl^–, SO^–₄, HCO₃ etc.) responsible for hardness is known as de-ionised or demineralised water. It is formed by passing water repeatedly through cation and anion exchangers.

Now let us try to investigate as to why hard water does not form lather with soap readily. Actually soaps are water soluble sodium (or potassium salts) of higher fatty acids with formula NaX [where X may represent C₁₇H₄₅COO^– (palmitate), C₁₇H₄₃COO^– (oleate), or C₁₇H₃₅COO^–(stereate) ions]. When soap is added to the hard water, the Na⁺ ions are replaced by Ca²⁺ and Mg²⁺ ions present in hard water to form the corresponding calcium and magnesium salts of the acids which get precipitated.

Thus, a lot of soap is wasted due to formation of curdy white precipitate and hard water is not useful in lauhdary. In addition, it cannot be employed in boilers for raising steam. Over a period of a time, the salts present in hard water get deposited inside the walls of the boiler in the form of hard scale. The boiler scale does not allow the flow of heat since it is a poor conductor. Sometimes, it also damages the boiler in case it cracks. Therefore, removal of these salts from water is necessary. This is called softening of water.

Question 27.
Is dimineralised or distilled water always useful for drinking purposes ? If not, how can it be made
useful ?
Answer:
No, dimineralised water or distilled water is not always useful for drinking purposes. It is usually tasteless. Moreover,some ions such as Na⁺, K⁺ and Mg²⁺ etc. are essential to the body. In order to make dimineralised water more useful, some useful salts of sodium and potassium etc. must be dissolved in it.

Question 28.
What properties of water make it useful as a solvent ? What type of compounds can it dissolve and hydrolyse ?
Answer:
Water is a useful, rather excellent solvent due to the following properties :

1. It has high enthalpy of vaporisation and heat capacity.
2. It is a liquid over a wide range of temperature (0° to 100°C)
3. It is polar in nature and high dielectric constant (78-39).
   It can dissolve polar substances and also some organic compounds due to hydrogen bonding.Water can hydrolyse oxides, halides, phosphides, nitrides etc.

Question 29.
Describe be the usefulness of water in biosphere and biological systems.
Answer:

• Water is a liquid with freezing point of 273-2 K and boiling point of 373-2 K.
• Water has the maximum density at 277 K (4°C) e. 1 gm cm^-3.
• In water, the molecules are hydrogen bonded and the hydrogen bonding influences all the physical properties such as state, heat of fusion, heat of vaporisation, melting point and boiling point etc. For example, H₂O is the hydride of the first member of group 16 e. oxygen. The rest of the members of the group are sulphur, selenium, tellurium and polonium. The hydrides of these elements do not show any hydrogen bonding and their physical properties are quite different from H₂O. For example, H₂O is a liquid while H₂S is a gas at room temperature.
• Water is of polar nature. As a result, most of the inorganic substances and also many organic substances having polar bonds in their molecules are water soluble because df inter-molecular hydrogen bonding. In addition to this, water has high heat of vaporisation and it exists in the liquid form over a wide range of temperature (0 to 100°C). Therefore, water is regarded as an excellent solvent and is quite often knoy/n as universal solvent.
• Because of polar nature,’‘there are strong intermolecular forces of attraction and hydrogen bonding present in the H₂O molecules. As a results water has higher specific heat, thermal conductivity, surface tension, dipole moment as compared to other liquids. ‘These properties enable water to play a major role in biosphere. It is responsible for maintaining the body temperature of living beings and also for moderating climate.

Question 30.
Knowing the properties of H₂O and D₂O, do you think that D₂O can be used for drinking purposes ?
Answer:
Heavy water (D₂O) is quite injurious to living beings, plants and animals since it slows down the rates of reactions which occur in them. It fails to supporfelife and has no utility in biosphere.

Question 31.
Hydrolysis is different from hydration. Elaborate.
Answer:
Hydrolysis is a chemical reaction in which a substance reacts with water under neutral, acidic or alkaline conditions. For example, aluminium chloride is hydrolysed by water as follows:
AlCl₃ + 3H₂O—>Al(OH)₃ + 3HCl
Hydration on the other hand is the property of a chemical compound to take up molecules of water of crystallisation and get hydrated. For example, anhydrous copper sulphate (CuSO₄) which is white in colour takes up five molecules of H₂O to form hydrated copper sulphate (CuSO₄ 5H₂O) which has blue colour.

Question 32.
Saline hydrides are generally used to remove traces of water from organic coin pounds. Explain.
Answer:
Saline hydrides (e.g., NaH, CaH₂ etc.) absorb water and react with it as follows :
NaH(s) + H₂O(l) → NaOH (ag) + H₂(g)
CaU₂(s) + 2H₂O(l) →Ca(OH)₂(aq) + H₂(g)
The hydrpgen escapes leaving behind metal hydroxides. Thus, these hydrides can be used to remove traces of water from the organic compounds.

Question 33.
What do you expect the nature of hydrides if formed by the elements of atomic numbers 15, 19, 23 and 44 with dihydrogem?
Answer:

• The elefpent with atomic number (Z) = 15 is P and the corresponding hydride is phosphine [PH₃]. It is covalent in nature.
• The element with atomic number (Z) = 19 is K and the corresponding hydride is potassium hydride (K⁺H“). It is ionic in nature.
• The element with atomic number (Z) = 23 is V(vanadium). It is a transition metal. It forms an interstitial hydride (VH_x.g)
• The element with atomic number (Z) = 44 is Rh (Ruthenium). It is also a transition metal but does not form any hydride due to hydride gap (as it is present in group 8).Out of the hydrides listed, only the ionic hydrides react with water to evolve hydrogen gas.
  2K⁺H-(s) + 2H₂O(l)→ 2KOH(aq) + H₂(g)

Question 34.
Do you are expect different products in solution when aluminum(III) chloride and potassium chloride are treated separately with (1) normal water (2) acidified water and (3) alkaline water ?   (P.I.S.A. Based)
Answer:
Both the compounds are salts and they react in different manner with water of different nature.
(a) Aluminium (III) chloride or AlCl₃ will react with water as follows :
AlCl₃ + 3H₂O →Al(OH)₃ + 3HCl.
The reaction is known as hydrolysis.

• In normal water, Both Al(OH)₃ and HCl will be present.
• In alkaline water, HCl will be neutralised by the alkali added. Therefore, it will contain mainly Al(OH)₃ and will be alkaline in nature.
• In acidic water. In this medium, Al(OH)₃ will be neutralised by the acid added. Therefore, water will contain mainly HCl and will be acidic in nature.

(b) Potassium chloride is a salt of strong acid and strong base. It will remain as such under all the conditions and will not undergo any chemical reaction.

Question 35.
How does H₂O₂ behave as a bleaching agent ?
Answer:
Bleaching nature of hydrogen peroxide is due to oxidation

Question 36.
What do you understand by the terms :
(1) hydrogen economy
(2) hydrogenation
(3) syn gas
(4) water-gas shift reaction
(5) fuel cell ?
Answer:
1. Hydrogen Economy. The basic principle of hydrogen economy is the transportation and storage of energy in the form of liquid or gaseous dihydrogen. In India, dihydrogen was used for the first time in October 2005 for running automobiles like four- wheelers. Initially 5% dihydrogen was mixed with C.N.G. which is commonly used. The experiment is successful till to-day. Efforts are on to increase the percentage of dihydrogen gradually.

Hydrogen oxygen fuel cells can also be used to generate electricity to run electric powered cars. A fuel cell produces electricity from a chemical reaction just as in a battery and can work as long as hydrogen and oxygen are present. One such fuel­ cell made by German motors was on display at Sydney Olympics. A number of companies are working on these projects because of the limited availability and very high cost of petrol. A number of cars were launched in the year 2004 and it is predicted that by 2025, automobiles employing hydrogen as fuel will capture about 25% of the world market.

In these fuel cells, porous carbon electrodes are placed in a concentrated aqueous solution of KOH or NaOH. The temperature is maintained to about 400 K when the gases react to form water accompanied by the release of electrical energy. We will study in detail the working of fuel cells in next class under Electrochemistry .

2. Hydrogenation : Hydrogenation of vegetable oils. It is done by passing hydrogen gas through edible oils (groundnut oil, cotton seed oil etc.) in the presence of Ni at 473 K. As a result, the oils are converted into solid fats, also called vegetable ghees. Actually, oils are unsaturated due to the presence C=C bond. On hydrogenation, the bond changes into C—C bond and as a result, unsaturated oils change into saturated fats.

3.Syn gas : The mixture of CO and H₂ was previously known as water gas. These days, it is known as synthesis gas or syn gas. In addition to hydrocarbons and coke, syn gas’ can also be produced from some other materials containing carbon and hydrogen, e.g., wood scrap, saw dust, newspapers, sewage etc. The process of preparing ‘syn gas’ from coke, coal and other materials is known as coal gasification.

4. Water-gas shift reaction :In place of iron chromate, a mixture of ferric oxide (Fe₂O₃) and chromium oxide (Cr₂O₃) can be used. This reaction is known as Water-gas shift reaction. From the gaseous mixture, carbon dioxide can be removed by either passing into water under pressure or by scrubbing with sodium arsenite solution.

5. Fuel cell : It is a cell which converts chemical energy of fuel directly into electrical energy.

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