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Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 7 Alternating Current. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 7 Important Extra Questions Alternating Current

Alternating Current Important Extra Questions Very Short Answer Type

Question 1.
The instantaneous current flowing from an ac source is l = 5 sin 314 t. What is the rms value of current?
Answer:
The rms value of current is \(\frac{5}{\sqrt{2}}\).

Question 2.
The instantaneous emf of an ac source is given by E = 300 sin 314 t. What is the rms value of emf?
Answer:
The rms value of voltage is \(\frac{300}{\sqrt{2}}\)

Question 3.
Give the phase difference between the applied ac voltage and the current in an LCR circuit at resonance.
Answer:
The applied ac voltage and the current in an LCR circuit at resonance are in phase.
Hence phase difference = 0.

Question 4.
What is the phase difference between the voltage across the inductor and the capacitor in an LCR circuit?
Answer:
The phase difference is 180°.

Question 5.
What is the power factor of an LCR series circuit at resonance?
Answer:
The power factor is one.

Question 6.
In India, the domestic power supply is at 220 V, 50 Hz, while in the USA it is 110 V, 50 Hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply.
Answer:
Advantage: less power loses
Disadvantage: more fatal.

Question 7.
Define the term ‘wattles current’. (CBSE Delhi 2011)
Answer:
It is the current at which no power is consumed.

Question 8.
In a series LCR circuit, V_L = V_C ≠ V_R. What is the value of the power factor? (CBSE AI 2015)
Answer:
One.

Question 9.
Define capacitor reactance. Write its SI units. (CBSE Delhi 2015)
Answer:
It is the opposition offered to the flow of current by a capacitor. It is measured in ohm.

Question 10.
Define quality factor in series LCR circuit. What is its SI unit? (CBSE Delhi 2016)
Answer:
The quality factor is defined as the ratio of the voltage developed across the capacitor or inductor to the applied voltage. It does not have any unit.

Question 11.
A choke and a bulb are in series to a dc source. The bulb shines brightly. How does its brightness change when an iron core is inserted inside the choke coil?
Answer:
There is no change in the final brightness as the inductive reactance is zero for dc.

Question 12.
A solenoid with an iron core and a bulb are connected to a dc source. How does the brightness of the bulb change, when the iron core is removed from the solenoid?
Answer:
There Is no change In the finaL brightness as the inductive reactance is zero for dc.

Question 13.
In a serles LCR circult, the voltage across an inductor, capacitor and a resistor are 20 V, 20 V and 40 V, respectively. What Is the phase difference between the applied voltage and the current In the circuit?
Answer:
WhenV_L = V_C, then the circuit is in series resonance, therefore both current and voltage are in phase.

Question 14.
Why Is there no power consumption In an Ideal inductor connected to an ac source?
Answer:
This is because current and voltage across an ideal inductor are out of phase by 900.
Hence P = V_RMS I_RMS cos 90° = O

Question 15.
Can a choke be replaced by a capacitor of suitable capacitance?
Answer:
Yes, because even then the power consumed will be zero.

Question 16.
Find the inductance of the Inductor that would have a reactance of 50 ohm when used with an ac source of frequency 25/π kHz.
Answer:
Using X_L = 2 πf L or L = \(\frac{X_{L}}{2 \pi f}\), therefore
L = \(\frac{50 \times \pi}{2 \pi \times 25}\) = 1 H

Question 17.
The figure gIven below shows the variation of an alternating emf with time. What Is the average value of the emf for the shaded part of the graph?

Answer:
Average or mean value of ac over half cycle or in time T/2 is
E_m = 2E_o/π = 0.637 E_o = 0.637 × 314
or Em = 200 V.

Question 18.
What is the power dissipated in an ac circuit in which the voltage and current are given by V = 230 sin(ωt + π/2) and l = 10 sin ωt.
Answer:
Since the phase difference between the voltage and current is π/2, therefore power consumed = V_rms l_rms cos π/2 = 0

Question 19.
When a lamp is connected to an alternating voltage supply, it lights with the same brightness as when compared to a 12 V dc battery. What Is the peak value of alternating voltage?
Answer:
The peak value is V = 12 × \(\sqrt{2}\) = 16.97 V

Question 20.
Why is the use of a.c. voltage preferred over d.c. voltage? Give two reasons. (CBSEAI 2014)
Answer:

1. Can be increased or decreased easily.
2. Can easiLy be converted into dc.

Question 21.
Can the Instantaneous power output of an ac source ever be negative? Can the average power output be negative?
Answer:
Yes, No.

Alternating Current Important Extra Questions Short Answer Type

Question 1.
State the phase relationship between the current flowing and the voltage applied in an ac circuit for (i) a pure resistor (ii) a pure inductor.
Answer:

1. Electric current and voltage applied in a pure resistor are in same phase, i.e. Φ = 0°
2. Applied voltage leads electric current flowing through pure-inductor in an ac circuit by phase angle of π/2.

Question 2.
A light bulb is in turn connected in a series (a) across an LR circuit, (b) across an RC circuit, with an ac source. Explain, giving the necessary mathematical formula, the effect on the brightness of the bulb in case (a) and (b), when the frequency of the ac source is increased. (CBSE 2019C)
Answer:
(a) The current in LR circuit is given by
l = \(\frac{V}{\sqrt{R^{2}+\omega^{2} L^{2}}}\)

When the frequency of ac source ω increases, l decreases, and hence brightness decreases.

(b) The current in RC circuit is given by
l = \(\frac{V}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}\)

When the frequency of ac source ω increases, l increases, and hence brightness increases.

Question 3.
An air-core solenoid is connected to an ac source and a bulb. If an iron core is inserted in the solenoid, how does the brightness of the bulb change? Give reasons for your answer.
Answer:
Insertion of an iron core in the solenoid increases its inductance. This in turn increases the value of inductive reactance. This decreases the current and hence the brightness of the bulb.

Question 4.
A bulb and a capacitor are connected in series to an ac source of variable frequency. How will the brightness of the bulb change on increasing the frequency of the ac source? Give reason.
Answer:
When the frequency of the ac is increased, it will decrease the impedance of the circuit as Z = \(\sqrt{R^{2}+(1 / 2 \pi f C)^{2}}\). As a result, the current and hence the brightness of the bulb will increase.

Question 5.
An ideal inductor is in turn put across 220 V, 50 Hz, and 220 V, 100 Hz supplies. Will the current flowing through it in the two cases be the same or different?
Answer:
The current through the inductor is given by l = \(\frac{V}{X_{L}}=\frac{V}{2 \pi f L}\). The current is inversely proportional to the frequency of applied ac.

Since the frequency is different therefore the current will also be different.

Question 6.
State the condition under which the phenomenon of resonance occurs in a series LCR circuit, Plot a graph showing the variation of current with a frequency of ac source in a series LCR circuit.
Answer:
The phenomenon occurs when the inductive reactance becomes equal to the capacitive reactance., i.e. X_L – X_C
⇒ ω L = \(\frac{1}{ωC}\)
⇒ ω = \(\frac{1}{\sqrt{L C}}\)

The graph is as shown below.

Question 7.
Give two advantages and two disadvantages of ac over dc.
Answer:
Advantages of ac:
(a) The generation and transmission of ac are more economical than dc.
(b) The alternating voltage may be easily stepped up or down as per need by using suitable transformers.

Disadvantages of ac:
(a) It is more fatal than dc.
(b) It cannot be used for electrolysis.

Question 8.
In a series, LCR circuit connected to an ac source of variable frequency and voltage v = v_m sin ωt, draw a plot showing the variation of current (l) with angular frequency (ω) for two different values of resistance R₁ and R₂ (R₁ > R₂). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define the Q-factor of the circuit and give its significance. (CBSE Delhi 2013C)
Answer:
The plot is as shown.

Resonance occurs in an LCR circuit when
X_L = X_C.
The smaller the value of R sharper is the resonance. Therefore the curve will be sharper for R₂. It determines the sharpness of the resonance. The larger the value of Qsharper is the resonance.

Question 9.
You are given three circuit elements X, Y, and Z. When the element X is connected across an a.c. source of a given voltage, the current and the voltage are in the same phase. When the element Y is connected in series with X across the source, voltage is ahead of the current in phase by π/2. But the current is ahead of the voltage in phase by π/2 when Z is connected in series with X across the source. Identify the circuit elements X, Y, and Z. When all the three elements are connected in series across the same source, determine the impedance of the circuit. Draw a plot of the current versus the frequency of the applied source and mention the significance of this plot. (CBSE AI 2015)
Answer:
X-Resistor, Y-Inductor, Z-Capacitor For expression of the impedance of LCR circuit see X is a resistor
Y is a capacitor
Z is an inductor

Consider the impedance triangle
Z = \(\sqrt{R^{2}+\left(X_{t}-X_{C}\right)^{2}}\)

The plot is as shown.

Significance, at ω = ω₀ (resonance frequency) current, is maximum.

Question 10.
Given three elements X, Y, and Z to be connected across an ac source. With the only X connected across the ac source, voltage and current are found to be in the same phase. With only element Y in the circuit, the voltage lags behind the current in phase by π/2, while with the element Z in the circuit, the voltage leads the current in phase by π/2
(a) Identify the elements X, Y, and Z.
Answer:
X is resistor
Y is a capacitor
Z is an inductor

Consider the impedance triangle
Z = \(\sqrt{R^{2}+\left(X_{t}-X_{C}\right)^{2}}\)

(b) When all these elements are connected in series across the same source,
(i) determine the power factor,
Answer:
Power factor = cos Φ = R/Z
= \(\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\)

(ii) find out the condition when the circuit is in resonant state. (CBSE AI 2019)
Answer:
Circuit will be in resonance when X_L – X_C = 0

Question 11.
A coil with an air core and an electric bulb are connected in series across a 220 V 50 Hz ac source. The bulb glows with some brightness. How will the glow of the bulb be affected by introducing a capacitor in series with the circuit? Justify your answer.
Answer:
The impedance of an LR circuit is given by the expression
Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and

the impedance of a series LCR circuit is given by the expression
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\).

Now current flowing through the two circuits is given by V
l = \(\frac{V}{Z}\).

Since Z decreases when a capacitor is connected to an LR circuit therefore there is an increase in current through the circuit. This increases the brightness of the bulb.

Question 12.
In the given circuit, inductor L and resistor R have identical resistance. Two similar electric lamps B1 and B2 are connected as shown. When switch S is closed,
(i) which one of the lamps lights earlier,
(ii) will the lamps be equally bright after some time? Justify your answer.

Answer:
(i) Lamp B₂ connected with the resistor will light up first. This is because the current through the inductor will grow before attaining maximum value.
(ii) When the current through the inductor becomes maximum, after some time, both the lamps will be equally bright.

Question 13.
Figure (a), (b), and (c) show three ac circuits in which equal currents are flowing. If the frequency of emf be increased, how will the current be affected in these circuits? Give the reason for your answer.

Answer:
There will be no change in the current in figure (b) as the resistance of the resistor does not depend upon the frequency of the applied ac.

The reactance of the inductor in figure (a) is given by X_L = 2πf L. An increase in frequency increases the value of inductive reactance. This decreases the current through the circuit.

The reactance of the capacitor in figure (c) is given by X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). An increase in frequency decreases the value of capacitive reactance. This increases the current through the circuit.

Question 14.
An alternating voltage of frequency f is applied across a series LCR circuit. Let f_r be the resonance frequency for the circuit. Will the current in the circuit lag, lead, or remain in phase with the applied voltage when (i) f > f_r (ii) f < f_r Explain your answer in each case.
Answer:
(i) When f > f_r, then the circuit behaves as an inductive circuit. Thus emf leads current. This is because the inductive reactance is given by the expression X_L = 2πfL. At high-frequency X_L will be more.

(ii) When f < f_r> then the circuit behaves as a capacitive circuit. Thus emf lags current. This is because the capacitive reactance is given by the expression X_C = \(\frac{1}{2 \pi f C}\). This value is more at low 2πfC frequency.

Question 15.
For a series LCR circuit, connected to a sinusoidal ac voltage source, identify the graph that corresponds to ω > \(\frac{1}{\sqrt{L C}}\). Give reason.

Answer:
When ω > \(\frac{1}{\sqrt{L C}}\), the circuit behaves as an inductive circuit. In an inductive circuit, emf leads current. This is depicted in the graph (a).

Question 16.
Draw the graphs showing the variations of (a) inductive reactance and (b) capacitive reactance, with the frequency of applied voltage in the ac circuit. How do the values of (a) inductive, and (b) capacitive reactance change, when the frequency of applied voltage is tripled?
Answer:
The graphs are as shown below.

(a) The inductive reactance is given by the expression X_L = 2πfL. Therefore if the frequency is tripled then the value of X_L also gets tripled.
(b) The capacitive reactance is given by the expression X_C = \(\frac{1}{2 \pi f C}\). Therefore if the frequency is tripled then the value of X_C becomes one-third of its previous value.

Question 17.
Can the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the ac source? Justify your answer.
Answer:
Yes, the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the ac source. It is because the applied voltage is equal to the algebraic sum (as obtained by the use of a phasor diagram) of V_R, V_L, and V_C, i.e.
V = l\(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

whereas V_L = l X_L and V_C = l X_C.
Thus, it is self-evident that V_L or V_C may be greater than V.

Question 18.
Mention the factors on which the resonant frequency of a series LCR circuit depends. Plot a graph showing the variation of the impedance of a series LCR circuit with the frequency of the applied ac source.
Answer:
In a series LCR circuit, the resonant frequency depends on the value of inductance L and capacitance C present in the circuit.
The graph showing the variation of impedance Z of a series LCR circuit with the frequency f of the applied ac source is shown below.

Question 19.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? (NCERT)
Answer:
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitive reactance (1/ω_C), and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

Question 20.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.

The switch is closed and after some time, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases (c) is unchanged as the iron rod is inserted. Give your answer with reasons. (NCERT)
Answer:
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron rod thereby increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

Question 21.
Draw the effective equivalent circuit of the circuit shown in the figure at very high frequencies and find the effective impedance. (NCERT Exemplar)

Answer:
At high frequency, the capacitive reactance is small while the inductive reactance is large. Therefore the capacitive reactance can be neglected while no current will flow through the inductors. Therefore the equivalent reactance of the circuit is Z = R₁ + R₃ and hence the circuit becomes

Question 22.
Study the circuits (a) and (b) shown in the figure and answer the following questions.

(a) Under which conditions would the rms currents in the two circuits be the same?
Answer:
This will happen when the impedance of both the circuits is the same, i.e. R. This is possible when circuit (b) is in resonance.

(b) Can the rms current in circuit (b) be larger than that in (a)? (NCERT Exemplar)
Answer:
No, because in circuit (b)
l_rms = \(\frac{V_{\text {rms }}}{Z}=\frac{V_{\text {rms }}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\), Z cannot be less than R.

Question 23.
How does the sign of the phase angle Φ, by which the supply voltage leads the current in an LCR series circuit, change as the supply frequency is gradually increased from very low to very high values? (NCERT Exemplar)
Answer:
The phase angle for an LCR circuit is given by the expression
tan Φ = \(\frac{R}{Z}=\frac{X_{L}-X_{C}}{R}=\frac{2 \pi f L-1 / 2 \pi f C}{R}\)

At low frequencies X_L < X_C and at high frequencies X_L > X_C Therefore Φ changes from negative to zero and to positive; zero at the resonant frequency.

Question 24.
A device ‘X’ is connected to an ac source. The variation of voltage, current, and power in one complete cycle is shown in the figure.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’. (NCERT Exemplar)

Answer:
(a) A
(b) Zero
(c) L or C or LC

Question 25.
Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? (NCERT Exemplar)
Answer:
An ac current changes direction with the source frequency and the charge flow would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule’s heating effect is such property and hence it is used to define misvalue of ac.

Question 26.
A sinusoidal voltage of peak value 10 V is applied to a series LCR circuit In which resistance, capacitance, and inductance have values of 10 0, 1 μF, and 1 H, respectively.
Find (i) the peak voltage across the inductor at resonance
(ii) the quality factor of the circuit. (CBSE Sample Paper 2018-19)
Answer:
Given V_m = 10V, R = 1o Ω, L = 1 H, C = 1 μF,
V_L = ?, Q = ?

Q= ω_r L/R = (10³ × 1)/10 = 100

Question 27.
Draw a labeled diagram of a step-down transformer. State its working principle Write one main cause of energy loss in this device and the method used for reducing it.
Answer:
The labeled diagram is as shown.

It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil.

One main cause of loss of energy is heat produced due to the production of eddy currents. This can be reduced by laminating the iron core.

Alternating Current Important Extra Questions Long Answer Type

Question 1.
Prove mathematically that the average power over a complete cycle of alternating current through an Ideal inductor is zero.
Answer:
Let the instantaneous value of voltage and current in the ac circuit containing a pure inductor are
V = V_m sin ωt and
l = l_m sin (ωt – π/2) = – l_m cos ωt
where π/2 is the phase angle by which voltage Leads currently when ac flows through an inductor. Suppose the voltage and current remain constant for a small-time dt. Therefore, the electrical energy consumed in the small-time dt is
dW = V l dt

The total electrical energy consumed in one time period of ac is given by

Therefore, the total electrical energy consumed in an ac circuit by a pure inductor is W = 0

Now average power is defined as the ratio of the total electrical energy consumed over the entire cycle to the time period of the cycle, therefore
P_av = \(\frac{W}{T}\) = 0

Hence, the average power consumed in an ac circuit by a pure inductor is P_av = 0
Thus a pure inductor does not consume any power when ac flows through it. Whatever energy is used in building up current is returned back during the decay of current.

Question 2.
Draw the phasor diagram of a series LCR connected across an ac source V= V_o sin ωt. Hence, derive the expression for the impedance of the circuit. Obtain the conditions for the phase angle under which the current is
(i) maximum and
(ii) minimum. (CBSE AI 2019)
Answer:
The voltages across the various elements are drawn as shown in the figure below.

From the diagram, we observe that the vector sum of the voltage amplitudes V_R, V_L, and V_C equals a phasor whose length is the maximum applied voltage V_m, where the phasor Vm makes an angle φ with the current phasor lm. Since the voltage phasors, V_L and V_C are in opposite direction, therefore, a difference phasor (V_L – V_C) is drawn which is perpendicular to the phasor V_R. Adding vectorially we have


where X_L = ω L and X_C = 1 / ω C, therefore, we can express the maximum current as
l_m = \(\frac{V_{\mathrm{m}}}{\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}}\)

The impedance Z of the circuit is defined as Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{C}\right)^{2}}\)

For maximum lm, Z should be minimum (Z = R) or X_C = X_L = 0 and Φ = 0

For (l_m)min Φ → 90° (|X_C – X_L| >> R) Z → ∞

Question 3.
(a) The graphs (i) and (ii) represent the variation of the opposition offered by the circuit element to the flow of alternating current with a frequency of the applied emf. identify the circuit element corresponding to each graph.


(b) Write the expression for the impedance offered by the series combinations of the above two elements connected across the ac source. Which will be ahead in phase in this circuit, voltage or current? (CBSE AI 2011C)
Answer:
(a) In figure (i) the opposition to the flow of current does not depend upon frequency, the circuit element is a resistor.

In figure (ii) the opposition increases with frequency, the current element is an inductor.

(b) When the resistor R and the inductance L are connected in series across an ac source, the impedance Z of the circuit is given by
Z = \(\sqrt{R^{2}+X_{L}^{2}}\), where X_L is the inductive reactance.

In an L – R circuit, the voltage is ahead of the current.

Question 4.
An Inductor L of inductance XL is connected in series with bulb B and an ac source. How would the brightness of the bulb change when
(a) the number of turns In the Inductor Is reduced,
(b) an Iron rod Is Inserted Into the Inductor and
(c) a capacitor of reactance X_C = X_L
Is Inserted in series In the circuit. Justify your answer In each case. (CBSE Delhi 2015)
Answer:
(a) When the number of turns of the inductor reduced it decreases the inductance of the inductor as (L ∝ n²) where n is the number of turns. This in turn decreases the inductive reactance X_L which increases the current in the circuit and hence the brightness of the bulb decreases.
(b) When an iron rod is inserted in the inductor, it increases the inductive reactance, which in turn decreases the current and hence the brightness of the bulb.
(c) When X_L = X_C, the circuit acts as a resistive circuit, i.e. the impedance becomes minimum and maximum current flows. This makes the bulb glow more brightly.

Question 5.
An inductor L of reactance X_L is connected in series with a bulb B to an ac source as shown in the figure below. Briefly explain how the brightness of the bulb changes, when
(a) number of turns of the inductor Is reduced? and
(b) a capacitor of reactance X_C = X_L is included in series in the same circuit?

Answer:
If R is the resistance of the bulb, then the total impedance of the circuit is Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and the corresponding current is l = V l Z.
(a) If the number of turns of the inductor is reduced then its inductance L and consequently there is a decrease in the reactance. This leads to a decrease in the impedance of the circuit. As a result the current flowing through the circuit increases. This increases the brightness of the bulb.

(b) When a capacitor of reactance X_C = X_L is included in the circuit, then the new impedance becomes Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\) = R. Thus the impedance has its minimum value. This increases the current through the circuit, which results in an increase in the brightness of the bulb.

Question 6.
A capacitor, ‘C’ a variable resistor ‘R’, and a bulb ‘B’ are connected in series to the ac mains in the circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (a) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same (b) the resistance R is increased keeping the same capacitance? (CBSE Delhi 2014)

Answer:
(a) As the dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V = Q/C). As a result, the potential drop across the bulb will increase (since both are connected in series). So, its brightness will increase.
(b) As the resistance (R) is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease (since both are connected in series). So, its brightness will decrease.

Question 7.
What is meant by impedance? Give its unit. Using a phasors diagram or otherwise derive the expression for the Impedance of an ac circuit containing L, C, and R in series. Find the expression for the resonant frequency.
Answer:
It is the opposition offered by LR or CR or LCR circuit to the flow of ac. It is measured in ohm.

For derivation of Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

The impedance Z of the circuit is defined as
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\)

The graph is shown.

The current through an LCR circuit is given by the expression l_v = \(\frac{E_{v}}{Z}\)

where Z is the impedance. Substituting the value of Z in the above equation we have
l_v = \(\frac{E_{v}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\)

Because the impedance depends on the frequency of the source, we see that the current in the LCR circuit will depend upon frequency. The current, therefore, reaches its maximum value when the impedance is minimum. This happens when X_L = X_C, corresponding to Z = R. The frequency ω at which this occurs is called resonant angular frequency. To find ω0, we use the condition X_L = X_C, from which we get

But ω_o = 2 πf_o, where f_o is called resonance frequency. Therefore, the above equation becomes f_o = \(\frac{1}{2 \pi} \frac{1}{\sqrt{L C}}\)

Question 8.
Explain the term ‘inductive reactance’. Show graphically the variation of an inductive reactance with the frequency of the applied alternating voltage.
An ac voltage E = E_o sin ωt is applied across a pure inductor of inductance L. Show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of π/2.
Answer:
It is the opposition offered to the flow of current by a pure inductor.

Consider an ac circuit consisting of a pure Inductor connected to the terminaLs of an ac source. Let the Instantaneous value of the ac source be
V = V_m sin ωt ….(1)

Let V_L be the instantaneous voltage drop across the inductor, then Kirchoff’s loop rule when applied to the circuit gives V + V_L = 0
or
V – L\(\frac{di}{dt}\) = 0 …(2)

since V_L = – L\(\frac{di}{dt}\)
Using equations (1) and (2) we have

Integrating the above equation we have

where \(\frac{V_{m}}{\omega L}\) = l_m , ω_L = 2πfL has the dimensions of resistance and is called the inductive reactance of the circuit. Using the trigonometric identity cos ωt = – sin (ωt – π/2) in equation (6) we have
l_L =l_m sin (ωt – π/2) …(7)
Comparing equation (1) with equation (7) we see that the current lags voltage by π/2 radian or 90°.

Question 9.
Explain the term ‘capacitive reactance’. Show graphically the variation of a capacitive reactance with the frequency of the applied alternating voltage.
An ac voltage E=E_o sin cot is applied across a pure capacitor of capacitance C. Show mathematically that the current flowing through it leads the applied voltage by a phase angle of π/2.
Answer:
It is the opposition offered to the flow of current by a pure capacitor.
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = V_m sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – V_C = 0
V = V_C = Vm Sin ωt …(2)

But from the definition of capacitance, V_C = Q / C or Q= V_CC. Substituting for V_C we have
Q = C V_m sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CV_m sin ωt = ω C V_m cos ωt …(4)

But dQ/dt = lc, therefore the above equation becomes,
l_C = ω C V_m cos ωt = l_m cos ωt …(5)
where
l_m = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCV_m = l_m.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
l_C = l_m sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

The graph of variation of X_C with f is as shown.

Question 10.
Derive an expression for the impedance of a series LCR circuit connected to an ac supply of variable frequency. Plot a graph showing a variation of current with the frequency of the applied voltage. Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning mechanism of a radio or a TV set. (CBSE Delhi 2011)
Answer:
The voltages across the various elements are drawn as shown in the figure below.

From the diagram, we observe that the vector sum of the voltage amplitudes V_R, V_L, and V_C equals a phasor whose length is the maximum applied voltage V_m, where the phasor V_m makes an angle Φ with the current phasor lm. Since the voltage phasors, V_L and V_C are in opposite directions, a different phasor (V_L – V_C) is drawn which is perpendicular to the phasor V_R.

Adding vectorially, we have
V_m = \(\sqrt{V_{R}^{2}+\left(V_{L}-V_{c}\right)^{2}}\)
= \(\sqrt{\left(l_{m} R\right)^{2}+\left(l_{m} X_{L}-l_{m} X_{c}\right)^{2}}\) …(1)
or
V_m = lm\(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\) …(2)

where X_L = ω L and X_C = 1/ω C, therefore we can express the maximum current as
l_m = \(\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\) …(3)

∴ l_m = \(\frac{V_{m}}{Z}\)

The impedance Z of the circuit is defined as
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\)

The graph is shown.

The graph is as shown.

The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna act as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum and that particular radio station is tuned in.

Question 11.
A device ‘X’ is connected to an ac source V = V_o sin ωt. The variation of voltage, current, and power in one cycle is shown in the following graph.

(a) Identify the device ‘X’.
Answer:
Capacitor

(b) Which of the curves A, B, and C represent the voltage, current, and power consumed in the circuit? Justify your answer.
Answer:
A: Power, B: Voltage, and C: Current.

(c) How does its impedance vary with the frequency of the ac source? Show graphically.
Answer:
For a capacitor, Impedance is given by X_C = 1 /ω_C = 1 /2πf_C. This is shown graphically as

(d) Obtain an expression for the current in the circuit and its phase relation with ac voltage. (CBSE AI 2017)
Answer:
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = V_o sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – V_C = 0
or
V = V_C = V_o sin ωt …(2)

But from the definition of capacitance,
V_C = Q / C
or
Q= V_C C.

Substituting for Vc we have
Q = CV_o sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\) CV_o sin ω t = ω C V_o cos ω t …(4)

But dQ/dt = i_c, therefore the above equation becomes,
i_c = ω C V_o cos ω t = l_o cos ω t …(5)

Here the term 1/ωc has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCV_m = l_m. Using the trigonometric identity cos ω t = sin (ω t + π/2) equation (5) can be written as
i_c = l_o sin (ω t + π/2) …(6)

Comparing this expression with equation (1) we see that the current is 90° (π/2 ) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

Question 12.
A device X is connected across an ac source of voltage V = V_o sin ωt. The current through X is given as l = l_o sin (ωt + π/2).
(a) Identify the device X and write the expression for Its reactance.
(b) Draw graphs showing the variation of voltage and current with time over one cycle of ac, for X.
(c) How does the reactance of device X vary with the frequency of the ac? Show this variation graphically.
(d) Draw the phasor diagram for the device X. (CBSE AI 2018, Delhi 2018)
Answer:
(a) X: capacitor
Reactance X_C = 1 / ω_C = 1 / 2πf_C

(b) The graphs area as shown.

(c) Reactance of the capacitor varies in inverse proportion to the frequency. Therefore, the graph is as shown.

(d) The phasor diagram is as shown.

Question 13.
A resistor of 200 Ω and a capacitor of 15.0 μF is connected In series to a 220 V, 50 Hz ac source, (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. (NCERT, CBSE Delhi 2004)
Answer:
Given R = 200 Ω, C= 15.0 μF = 15.0 × 10^-6 F, V = 220 V, f = 50 Hz
(a) In order to calculate the current, we need the impedance of the

Therefore the current in the circuit is
l = \(\frac{V}{Z}=\frac{220}{291.5}\) = 0.7555 A

(b) Since the current is the same throughout the circuit we have
V_R = I_R = 0.755 × 200 = 151 V
V_C = lX_C = 0.755 × 212.3 = 160.3 V

The algebraic sum of the two voltages VR and Vc is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learned in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem;
V = \(\sqrt{V_{R}^{2}+V_{C}^{2}}\) = 220 V

Thus if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 14.
Explain why the reactance offered by an inductor increases with the increasing frequency of an alternating voltage.
(NCERT Exemplar)
Answer:
An inductor opposes the flow of current through it by developing a back emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e. if the frequency is higher. The reactance of an Inductor, therefore, is proportional to the frequency, being given by ω_L.

Question 15.
(a) Prove that an Ideal capacitor in an ac circuit does not dissipate power.
Answer:
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = V_m sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – V_C = 0
V = V_C = V_m Sin ωt …(2)

But from the definition of capacitance, V_C = Q / C or Q= V_CC. Substituting for Vc we have
Q = C V_m sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CV_m sin ωt = ω C V_m cos ωt …(4)

But dQ/dt = lc, therefore the above equation becomes,
l_C = ω C V_m cos ωt = l_m cos ωt …(5)
where
l_m = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCV_m = l_m.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
l_C = l_m sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

The graph of variation of X_C with f is as shown.

Power in capacitive circuit P = E_v l_v cos Φ. In pure captor 0 = 90° phase difference between voltage and current is π/2.
So power consumed in pure capdtor P_av = E_v l_v cos 90° = 0.

(b) An ideal inductor of 200 mH, capacitor of 400 μF, and a resistor of 10 Ω are connected In series to an ac source of 50 V and variable frequency. Calculate
(i)the angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of effective current and
(ii) the value of Q factor in the circuit. (CBSE AI 2017C)
Answer:

Question 16.
(a) What is the principle of transformer?
(b) Explain how laminating the core of a transformer helps to reduce eddy current losses in it.
(c) Why the primary and secondary coils of a transformer are preferably wound on the same core?
Or
Show that In the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the Inductor Is constant In time. (CBSE Sample Paper 2018-19)
Answer:
(a) It works on the principle of mutual induction.
(b) In order to reduce the eddy current loss, the resistance of the core should be Increased. In a transformer, the core Is made up of thin sheets of steel, each lamination being insulated from others by a thin layer of varnish. As the laminations are thin, they will have relatively high resistance.
(c) This is done to maximize the sharing of magnetic flux and also so that magnetic flux per turn should become the same in both the primary and secondary coils.
Or
Let at any instant t, q be the charge on the capacitor and ‘l’ be the current through the inductor
q (t) = q₀ cos ωt
i (t) = – q₀ ω sin ωt
Energy stored in the capacitor at time t is
U_E = \(\frac{1}{2} \frac{q^{2}}{c}=\frac{1}{2} \frac{q_{0}^{2}}{C}\) cos² ωt

Energy stored in the inductor at time t is

This sum is constant in time as q₀ and C, both are time-independent.

Question 17.
Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency. (NCERT Exemplar)
Answer:
A capacitor does not allow the flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of the supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency; it is given by 1 /ω_C.

Question 18.
(a) With the help of a labeled diagram, describe briefly the underlying principle and working of a step-up transformer.
(b) Write any two sources of energy loss in a transformer.
(c) A step-up transformer converts a low input voltage into a high output voltage. Does it violate the law of conservation of energy? Explain. (CBSE Delhi 2011)
Answer:
(a) The diagram is as shown.

It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil. In a step-up transformer, the number of turns in the secondary is more than that in the primary.

The ac source causes an alternating current in the primary, which sets up an alternating flux in the core; this induces an emf in each winding of the secondary, in accordance with Faraday’s law. The induced emf in the secondary gives rise to an alternating current in the secondary, and this delivers energy to the device to which the secondary is connected.

For step up : N_s> > N_p
\(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) and \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\)

(b) Two major sources of energy loss in this device are:

1. Heat produced due to production of eddy currents.
2. Copper loss: heat produced in the copper coils

(c) No, a transformer does not change the power. Thus if a voltage increases current gets decreased such that the energy and power remain the same.

Question 19.
(a) Explain with the help of a labeled diagram, principle, and working of a transformer. Deduce the expression for its working formula.
Answer:
For diagram as shown.

• Principle: A transformer works on the basis of mutual induction.
• Working: In a 100% efficient transformer

ε_s l_s = ε_p l_p where l and l_p, are the secondary and primary currents, therefore we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) ….(1)

Now a 1oo % efficient transformer
we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}} \frac{\frac{d \phi}{d t}}{\frac{d \phi}{d t}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}\) …(2)

Therefore form (1) and (2) we have
\(\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) = k

Efficiency of transformer = \(\frac{V_{s} I_{s}}{V_{p} I_{p}}\).

(b) Name any four causes of energy loss in an actual transformer. (CBSEAI 2013C)
Answer:

1. Copper loss
2. Iron loss
3. Flux loss
4. Hysteresis loss

Question 20.
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of the variation of (i) Magnetic flux and alternating emf versus time generated by a loop of wire rotating in a magnetic field,
Answer:
A schematic sketch is as shown.

Working principle: Electromagnetic induction.

The graphs are as shown.

(b) Why is a choke coil needed in the use of fluorescent tubes with ac mains? (CBSE Delhi 2014)
Answer:
It is used to regulate current with minimum loss of energy.

Question 21.
(a) Draw a labeled diagram of the AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.
Answer:
It is based on the principle of electromagnetic induction. When a coil is rotated about an axis perpendicular to the direction of the uniform magnetic field, an induced emf is produced across it.

Working: The working of the ac generator can be understood with the help of the various positions of the armature as shown in the figure below.


Suppose at time t = 0, the plane of the loop is perpendicular to B. As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0° and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero.

As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to the maximum value and then again becomes zero. But this time it reverses Its direction. For angles 90° and 270° maximum magnetic flux are linked with the coil hence the emf is a maximum. Thus the output of the ac generator varies sinusoidally with time. The Induced emf does not depend upon the shape of the Loop but depends only upon the area of the loop.

The emf generated Is given by the expression ε = nBAω Sin ωt, where ω is the speed of rotation of the coil

(b) A circular coil of cross-sectional area 200 cm² and 20 turns is rotated about the vertical diameter with an angular speed of 50 rad s^-1 in a uniform magnetic field of magnitude 3.O × 10^-2 T. Calculate the maximum value of the current in the coil. (CBSE Delhi 2017)
Answer:
Given A = 200 cm², ω = 50 rad s^-1, n = 20
S = 3 × 10^-2 T, ε_o = ?,

Using the relation ε0 = nBAω
= 20 × 3 × 10^-2 × 200 × 10^-4 × 50 = 0.6 V

Question 22.
Draw a labeled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of the number of turns and currents in the two coils.
A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. (CBSE Delhi 2017)
Answer:
(a) The diagram is as shown.

It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil. In a step-up transformer, the number of turns in the secondary is more than that in the primary.

The ac source causes an alternating current in the primary, which sets up an alternating flux in the core; this induces an emf in each winding of the secondary, in accordance with Faraday’s law. The induced emf in the secondary gives rise to an alternating current in the secondary, and this delivers energy to the device to which the secondary is connected.

For step up : N_s > N_p
\(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) and \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\).

Principle – It works on the principle of electromagnetic induction. When the current in one circuit changes, an induced current is set up in the neighboring circuit.

(b) N_s = ?, N_p = 3000
E_p = 2200 V, E_s = 220 V

Since, E_s/E_p = N_s/N_p
220/2200 = N_s/3000
N_s = 300

Question 23.
(a) State the principle of working of a transformer.
Answer:
It works on the principle of mutual inductance i.e. whenever the magnetic flux Linked in a coil changes an induced emf Is produced in the neighboring coil.

(b) Define the efficiency of a transformer.
Answer:
It is the ratio of the output power to the input power.

(c) State any two factors that reduce the efficiency of a transformer.
Answer:
Copper Loss, flux Loss

(d) Calculate the current drawn by the primary of a 90% efficient transformer which steps down 220 V to 22 V If the output resistance Is 440 Ω. (CBSFAI 2018 C)
Answer:
Given η = 90%, V_p = 22OV, V_s = 22V, R_o = 440 W
Now l_s = V_s/R = 22/440 = 0.05 A

Also η = \(\frac{P_{0}}{P_{1}}=\frac{V_{s} l_{s}}{V_{p} l_{p}}\) , therefore we have

\(\frac{90}{100}=\frac{22 \times 0.05}{220 \times l_{\mathrm{p}}}\)
or
l_p = 0.0056 A

Question 24.
Draw an arrangement for winding of primary and secondary coil In a transformer with two coils on a separate limb of the core.
State the underlying principle of a transformer. Deduce the expression for the ratio of secondary voltage to the primary voltage In terms of the ratio of the number of turns of the primary and secondary winding. For an ideal transformer, obtain the ratio of primary and secondary currents In terms of the ratio al the voltages in the secondary and primary voltages. Write any two reasons for the energy losses which occur In actual transformers, (CBSE Dei hi 2016C)
Answer:
Principle: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which Links the secondary and Induces an emf in it.

Induced emf, or voltage, in the secondary, with ‘N_s‘ number of turns,
ε_s = – N_s \(\frac{d \phi}{d t}\)

Back emf In the primary with Np turns,
ε_p = – N_p \(\frac{d \phi}{d t}\)

Since ε_s = V_s and ε_p = V_p
Therefore we have
\(\frac{V_{p}}{V_{s}}=\frac{N_{p}}{N_{s}}\)

For an ideal transformer,
Input power = output power
Therefore
l_pV_p = l_sV_s
Or
\(\frac{V_{p}}{V_{s}}=\frac{l_{s}}{l_{p}}\)

(a) Heat energy Loss.
(b) Humming effect Loss.

Question 25.
A pure inductor is connected across an ac source. Show mathematically that the current in it lags behind the applied emf by a phase angle of π/2. What is its inductive reactance? Draw a graph showing the variation of an inductive reactance with the frequency of the ac source.
Answer:
Consider an ac circuit consisting of a pure Inductor connected to the terminaLs of an ac source. Let the Instantaneous value of the ac source be
V = V_m sin ωt ….(1)

Let V_L be the instantaneous voltage drop across the inductor, then Kirchoff’s loop rule when applied to the circuit gives V + V_L = 0
or
V – L\(\frac{di}{dt}\) = 0 …(2)

since V_L = – L\(\frac{di}{dt}\)
Using equations (1) and (2) we have

Integrating the above equation we have

where \(\frac{V_{m}}{\omega L}\) = l_m , ω_L = 2πfL has the dimensions of resistance and is called the inductive reactance of the circuit. Using the trigonometric identity cos ωt = – sin (ωt – π/2) in equation (6) we have
l_L =l_m sin (ωt – π/2) …(7)
Comparing equation (1) with equation (7) we see that the current lags voltage by π/2 radian or 90°.

Question 26.
An alternating emf Is applied across a capacitor. Show mathematically that the current leads the emf by a phase angle π/2, What Is its capacitive reactance? Draw a graph showing the variation of a capacitive reactance with the frequency of the ac source.

Answer:
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = V_m sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – V_C = 0
V = V_C = V_m Sin ωt …(2)

But from the definition of capacitance, V_C = Q / C or Q= V_CC. Substituting for V_C we have
Q = C V_m sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CV_m sin ωt = ω C V_m cos ωt …(4)

But dQ/dt = l_C, therefore the above equation becomes,
l_C = ω C V_m cos ωt = l_m cos ωt …(5)
where
lm = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
l_C = l_m sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

The graph of variation of X_C with f is as shown.

Question 27.
(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with the frequency of the ac source.
Answer:
The voltages across the various elements are drawn as shown in the figure below.

From the diagram, we observe that the vector sum of the voltage amplitudes V_R, V_L, and V_C equals a phasor whose length is the maximum applied voltage Vm, where the phasor Vm makes an angle Φ with the current phasor lm. Since the voltage phasors, V_Land V_C is in opposite directions, a different phasor (V_L – V_C) is drawn which is perpendicular to the phasor V_R.

Adding vectorially, we have
V_m = \(\sqrt{V_{R}^{2}+\left(V_{L}-V_{c}\right)^{2}}\)
= \(\sqrt{\left(l_{m} R\right)^{2}+\left(l_{m} X_{L}-l_{m} X_{c}\right)^{2}}\) …(1)
or
V_m = l_m\(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\) …(2)

where X_L = ω L and X_C = 1/ω C, therefore we can express the maximum current as
l_m = \(\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\) …(3)

∴ l_m = \(\frac{V_{m}}{Z}\)

The impedance Z of the circuit is defined as
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\)

The graph is shown.

(b) What is the phase difference between the voltages across the inductor and the capacitor at resonance in the LCR circuit?
Answer:
The phase difference between the voltage across the inductor and the capacitor at resonance is 180°.

(c) When an inductor is connected to a 200V dc voltage, a current of 1A flows through it. When the same inductor is connected to a 200V, 50 Hz ac source, only 0.5A current flows. Explain why? Also, calculate the self-inductance of the inductor. (CBSE Delhi 2019)
Answer:
Inductor will offer an additional impedance to ac due to its self-inductance.

Now R = \(\frac{V_{\text {rms }}}{l_{\text {rms }}}=\frac{200}{1}\) = 200 Ω

The impedance of the inductor
Z = \(\frac{V_{\mathrm{rms}}}{l_{\mathrm{rms}}}=\frac{200}{0.5}\) = 400 Ω

Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\) , therefore
X_L² = Z² – R²
X_L = \(\left(\sqrt{(400)^{2}-(200)^{2}}\right)\) = 346.4 Ω

Hence inductance (L) is
L = \(\frac{X_{L}}{\omega}=\frac{346.4}{2 \times 3.14 \times 50}\) = 6.2 H

Question 28.
(a) Obtain the expression for the average power dissipated in a series LCR circuit driven by an ac source of voltage V = V_m sin ωt supplying the current i = i_m sin (ωt + Φ).
Answer:
Average Power dissipated in a series LCR circuit
Given V = V_m sin tot
And l = i_m sin (ωt + Φ)
Instantaneous Power
P = Vi

(b) Define the terms:
(i) Wattless current, and
Answer:
Wattles’s current: If the average power consumed due to the flow of current in a circuit is zero, the current is said to be wattless.

(ii) Q – a factor of LCR circuit.
Answer:
Q-factor of an LCR Circuit: Q-factor of the LCR circuit is the ratio of the potential difference across inductance (or capacitance) at resonance to the applied voltage.

Question 29.
Distinguish between the terms reactance and impedance of an ac circuit. Prove that an ideal capacitor connected to an ac source does not dissipate power.

Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = V_m sin ωt …(1)

Let V_C be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – V_C = 0
V = V_C = V_m Sin ωt …(2)

But from the definition of capacitance, V_C = Q / C or Q= V_CC. Substituting for Vc we have
Q = C V_m sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CV_m sin ωt = ω C V_m cos ωt …(4)

But dQ/dt = lc, therefore the above equation becomes,
l_C = ω C V_m cos ωt = l_m cos ωt …(5)
where
lm = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
l_C = l_m sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, current leads emf by 90°.

The graph of variation of Xc with f is as shown.

Question 30.
Define the term root mean square (rms) value of ac. Derive the relation between rms value and the peak value of ac.
Answer:
It is that value of steady current (dc) which when passed through a resistor in a given time produces the same heat as is produced by the ac. when passed through the same resistor for the same time. It is abbreviated as rms value of current. It is denoted by l_rms.

Consider an ac source. Let the instantaneous value of current be represented by the equation
i = l_m sin ωt

Let this current pass through a resistor of resistance R. Therefore in a small-time dt, the amount of heat produced in the resistor is
dH = i² R dt

The total amount of heat produced in one complete cycle of ac is given by

Let l_rms be the virtual value of ac, then the amount of heat produced in the same resistor of resistance R, in the same time T is
H = l_rms²RT …(2)

Therefore from equations (1) and (2) we have
l_rms = \(\frac{l_{m}}{\sqrt{2}}\) = 0.707 lm

Question 31.
(a) Prove that the current flowing through an Ideal Inductor connected across a.c. source lags the voltage In phase by nil.
Answer:
(a) Consider an ac circuit consisting of a pure inductor connected to the terminals of an ac source. Let the instantaneous value of the ac source be
V = V_m sin ω t …(1)

Let V_L be the instantaneous voltage drop across the inductor, then Kirchoff’s loop rule when applied to the circuit gives V + V_L = 0
or
V – L\(\frac{di}{dt}\) = 0 …(2)

since V_L = – L \(\frac{di}{dt}\)

Using equations (1) and (2) we have
L\(\frac{di}{dt}\) = V_m sin ω t …(3)
or
di = \(\frac{V_{m}}{L}\) sin ω t dt …(4)

Integrating the above equation we have

C = 0, over a period of cycle

\(\frac{V_{\mathrm{m}}}{\omega L}\) = l_m, ω_L = 2πfL has the dimensions of resistance and is called the inductive reactance of the circuit. Using the trigonometric identity
cos ω t = – sin (ωt – π/2) in equation (6) we have
l_L = l_m sin (ωt – π/2) …(7)

Comparing equation (1) with equation (7) clearly shows that the current lags voltage by π/2 radian or 90°.

(b) An inductor of self-inductance 100 mH and a bulb are connected in series with a.c. source of rms voltage 10 V, 50 Hz. It is found that the effective voltage of the circuit leads the current in phase by π/4. Calculate the resistance of the bulb used and average power dissipated in the circuit, if a current of 1 A flows in the circuit. (CBSE Delhi 2017C)
Answer:
(b) V= 10 V, f = 50 Hz, l = 1.0 A, Φ = π/4, R = ?, L = 100mH = 0.1 H
Now Z = V/l = 10/1.0 = 10 ohm

Also Z = \(\sqrt{R^{2}+X_{L}^{2}}\) = 10
Now cos Φ = \(\frac{R}{Z}\)
or
Therefore, R = cos Φ × Z
Or
R = cos π/4 × 10 = 7 ohm

Power dissipated
P_av = l_rms V_rms cos Φ = \(\frac{V_{m}^{2}}{Z}\) × cos Φ
0r
P_av = \(\frac{(10)^{2}}{10}\) × 0.707 = 7.07 W

Question 32.
(a) Draw graphs showing the variations of inductive reactance and capacitive reactance with the frequency of the applied ac source.
(b) Draw the phasor diagram for a series RC circuit connected to an ac source.
(c) An alternating voltage of 220 V is applied across a device X, a current of 0.25 A flows, which lag behind the applied voltage In phase by \(\frac{π}{2}\) radian. If the same voltage is applied across another device Y, the same current flows but now it is in phase with the applied voltage.
(i) Name the devices X and Y.
(ii) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y. (CBSEAI 2018C)
Or
(a) State the principle of working of a transformer.
(b) Define the efficiency of a transformer.
(c) State any two factors that reduce the efficiency of a transformer.
(d) Calculate the current drawn by the primary of a 90% efficient transformer which steps down 220 V to 22 V, if the output resistance is 440 Ω.
Answer:
(a) The two graphs are as shown.

(b) (The current leads the voltage by an angle 0 where 0 < θ < \(\frac{π}{2}\).) The required phasor diagram is as shown.
Here θ = tan^-1 \(\frac{1}{ωCR}\)

(c) In device X:
Current lags behind the voltage by \(\frac{π}{2}\)
∴ X is an inductor.

In device Y:
Current is in phase with the applied voltage
∴ X is a resistor.

We are given that 0.25 = \(\frac{220}{X_{L}}\)
or
X_L = \(\frac{220}{0.25}\) Ω = 880 Ω

Also 0.25 = 0.25 = \(\frac{220}{X_{R}}\)

∴ X_R = \(\frac{220}{0.25}\) Ω = 880 Ω

For the series combination of X and Y, Equivalent impedance
= \(\sqrt{X_{L}^{2}+X_{R}^{2}}\) = (880\(\sqrt{2}\) Ω

∴ Current flowing = \(\frac{220}{880 \sqrt{2}}\) A = 0.177 A
Or
(a) A transformer works on the principle of mutual induction. (Alternatively, an emf is induced in the secondary coil when the magnetic flux linked with it changes with time due to changing magnetic flux linked with the primary coil).

(b) The efficiency of a transformer equals the ratio of the output power to the input power.
Efficiency = \(\frac{\text { output power }}{\text { input power }}\)
or
Efficiency = \(\frac{V_{S} I_{S}}{V_{P} I_{p}}\)
(c)

• Eddy current Losses
• route heat Losses
• hysteresis Losses
• magnetic flux leakage Losses

We have

Question 33.
(a) Draw a schematic diagram of a step-up transformer. Explain its working principle. Assuming the transformer to be 100% efficient, obtain the relation for
(i) the current in the secondary in terms of the current in the primary, and
(ii) the number of turns in the primary and secondary windings.
(b) Mention two Important energy losses in actual transformers and state how these can be minimized. (CBSE Delhi 2011C)
Answer:
(a) For diagram as shown.

• Principle: A transformer works on the basis of mutual induction.
• Working: In a 100% efficient transformer

ε_s l_s = ε_p l_p where l and l_p are the secondary and primary currents, therefore we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) ….(1)

Now a 1oo % efficient transformer
we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}} \frac{\frac{d \phi}{d t}}{\frac{d \phi}{d t}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}\) …(2)

Therefore form (1) and (2) we have
\(\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) = k

Efficiency of transformer = \(\frac{V_{s} I_{s}}{V_{p} I_{p}}\)

(b) The two important energy losses in actual transformers are:

1. Iron losses in the core of the transformer.
2. Loss of energy in the primary and secondary due to Joule heating. The iron losses due to eddy currents are minimized by laminating the iron core.

Question 34.
(a) Draw the diagram of a device that is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.
Answer:
The labeled diagram is as shown.

It works on the principle of mutual induction, i.e. whenever magnetic flux linked with a coil changes, an emf is induced in the neighboring coil.

The possible causes of energy losses in transformers are:

• Flux leakage
• Copper loss
• Eddy currents
• Hysteresis loss

(b) A small town with a demand of 1200kW of electric power at 220 V is situated 20 km away from an electric plant generating power at 440V. The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets the power from the line through a 4000 – 220V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat. (CBSE Delhi 2019)
Answer:
Total resistance of the line = length × resistance per unit length = 40 km × 0.5 = 20 Ω

Current flowing in the line l = P/V
= 1200 × 10³ /4000 = 300 A

Power loss = l²R = (300)² × 20 = 1800 kW

Question 35.
(a) State the principle of an ac generator and explain its working with the help of a labeled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ‘co’ in a magnetic field B, directed perpendicular to the axis of rotation.
Answer:
For principle and diagram


Suppose at time t = 0, the plane of the loop is perpendicular to B. As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0° and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero. As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to the maximum value and then again becomes zero. But this time it reverses Its direction. For angles 90° and 270° maximum magnetic flux are linked with the coil hence the emf is a maximum. Thus the output of the ac generator varies sinusoidally with time. The Induced emf does not depend upon the shape of the Loop but depends only upon the area of the loop.

The emf generated Is given by the expression ε = nBAω Sin ωt, where ω is the speed of rotation of the coiL.

(b) An airplane is flying horizontally from west to east with a velocity of 900 km h^-1. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10^-4 T and the angle of dip is 30°. (CBSEAI 2018, Delhi 2018)
Answer:
Potential difference developed between the ends of the wings
ε = BLv

Given Velocity V = 900 km h^-1 = 250 m s^-1
Wing span (L₁) = 20 m

Vertical component of Earth’s magnetic field
B_v = BH tan δ = 5 × 10-4 (tan 30°) T

∴ Potential difference
= 5 × 10^-4 (tan 30°) × 20 × 250 = 1.44 V

Question 36.
Explain, with the help of a diagram, the principle and working of an ac generator. Write the expression for the emf generated in the coil in terms of its speed of rotation.
Answer:
It is based on the principle of electromagnetic induction. When a coil is rotated about an axis perpendicular to the direction of a uniform magnetic field, an induced emf is produced across it.

Working: The working of the ac generator can be understood with the help of the various positions of the armature as shown in figure below.


Suppose at time t = 0, the plane of the loop is perpendicular to B. As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0° and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero. As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to a maximum value and then again becomes zero. But this time it reverses Its direction. For angles 90° and 270° maximum magnetic flux are linked with the coil hence the emf is a maximum. Thus the output of the ac generator varies sinusoidally with time. The Induced emf does not depend upon the shape of the Loop but depends only upon the area of the loop.

The emf generated Is given by the expression ε = nBAω Sin ωt, where ω is the speed of rotation of the coiL.

Numerical Problems:
Formulae for solving numerical problems

• Capacitive reactance X_C = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
• Inductive reactance X_L = ω_L = 2πfL
• When ac flows through an LR circuit then,
  Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and tan Φ = \(\frac{X_{L}}{R}=\frac{\omega L}{R}\)
• When ac flows through a CR circuit then,
  Z = \(\sqrt{R^{2}+X_{C}^{2}}\) and tan Φ = \(\frac{X_{c}}{R}=\frac{1}{\omega C R}\)
• For a senes LCR circuit driven by voltage V = V_m sin ωt, the current is given by
  l = l_m sin (ωt + Φ) where
  l_m = \(\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\)
• The impedance of this circuit is given by
  Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
  \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}\) = k.

Question 1.
The figure shows how the reactance of an Inductor varies with frequency. (a) Calculate the value of the Inductance of the Inductor using Information given In the graph. (b) If this Inductor is connected In senes to a resistor of 8 ohms, find what would be the impedance at 300 Hz.

Answer:
(a) We know that X_L = 2πfL or L = \(\frac{X_{L}}{2 \pi f}\).
Now slope of the graph is
\(\frac{X_{L}}{f}=\frac{8-6}{400-300}=\frac{2}{100}\) = 0.02

Therefore L is L = \(\frac{X_{L}}{2 \pi f}=\frac{0.02}{2 \times 3.14}\) = 0.0032 H

(b) NowR = 8 ohm, f = 300 Hz, Z = ?
Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\) . Therefore we have
Z = \(\sqrt{R^{2}+X_{L}^{2}}\) = \(\sqrt{(8)^{2}+(6)^{2}}\) = 10 Ω

Question 2.
A 25.0 μF capacitor, a 0.10-henry Inductor, and a 25.0-ohm resistor are connected in series with an ac source whose emf is E= 310 s. In 314 t (i) what is the frequency of the .mf? (ii) Calculate (a) the reactance of the circuit (b) the Impedance of the circuit and (C) the current in the circuit.
Answer:
Given C = 25.0 μF, L = 0.10 henry, R = 25.0 ohm, E_o = 310V,
Comparing with the equation E = E_o sin ωt,
we have
(i) ω = 314 or f = 50 Hz

Question 3.
A sinusoidal voltage V = 200 sin 314 t Is applied to a resistor of 10 ohms. Calculate (i) rms value of current (ii) rms value of voltage and (iii) power dissipated as heat in watt.
Answer:
V_o = 200 V, ω = 314 rads^-1 , V_rms = ?, l_rms = ?, P = ?

Question 4.
Find the inductance of the inductor used in series with a bulb of resistance 10 ohms connected to an ac source of 80 V, 50 Hz. The power factor of the circuit is 0.5. Also, calculate the power dissipation in the circuit.
Answer:
Given R = 10 ohm, V = 80 Hz, f = 50 Hz, cos Φ = 0.5 , L = ?, P = ?
Using the formula
cos Φ = \(\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}\)
or
0.25(R² + X²) = R²
Or
O.25X_L² = O.75R²
or
X_L = 17.32 ohm

Now using X_L = 2πfL we have
L = \(\frac{X_{L}}{2 \pi f}=\frac{17.32}{2 \times 3.14 \times 50}\) = 0.055 H

Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\)
= \(\sqrt{(10)^{2}+(17.32)^{2}}\) = 20 Ω

Now P_av = l_rms V_rms cos Φ = \(\frac{V_{m}^{2}}{Z}\) × cos Φ
or
Pav = \(\frac{(80)^{2}}{2 \times 20}\) × 0.5 = 160 W

Question 5.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and Is in phase with the applied voltage. When the same voltage is applied across a device Y, the same current again flows through it, but it leads the voltage by π/2. If element ‘X’ is a pure resistor of 100 ohms,
(a) name the circuit element ‘Y’ and
Answer:
The element Y is a capacitor.

(b) calculate the rms value of current, if rms value of voltage is 141 V.
Answer:
The value of Xc is obtained as below

X_C = \(\frac{V}{I}=\frac{220}{0.5}\) = 440 ohm

Therefore impedance of the circuit
Z = \(\sqrt{R^{2}+X_{C}^{2}}=\sqrt{(100)^{2}+(440)^{2}}\) = 451.2 ohm

Therefore rms value of current V 141
l = \(\frac{V_{r m s}}{Z}=\frac{141}{451.2}\) = 0.3125 A

Question 6.
When an alternating voltage of 220 V is applied across a device X, a current of 0. 5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across a device V, the same current again flows through it, but it lags the voltage by π/2.
(a) Name the devices X and Y.
Answer:
The element X is a resistor and Y is an inductor.

(b) Calculate the current flowing through the circuit when the same voltage is applied across the series combination of the two devices X and Y.
Answer:
Now both R and XL are the same and are given by
R = X_L = \(\frac{220}{0.5}\) = 440 ohm

Hence impedance of the circuit
Z = \(\sqrt{R^{2}+X_{L}^{2}}\)
= \(\sqrt{(440)^{2}+(440)^{2}}\)
= 622.2 ohm

Therefore current flowing through the circuit is
l = \(\frac{V}{Z}\) = \(\frac{220}{622.2}\) = 0.353 A

Question 7.
A 15.0 μF capacitor Is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) In the circuit. If the frequency Is doubled, what happens to the capacitive reactance and the current?
Answer:
Given C= 15.0 μF= 15 × 10^-6 F, V= 220 V,
f = 50 HZ, X_C = ? l_m = ?

The capacitive reactance is

Now l_rms = \(\frac{V_{r m s}}{X_{c}}=\frac{220}{212}\) = 1.04 A

Peak value of current
l_m = \(\sqrt{2}\) × lrms = 1.4.1 × 1.04 = 1.47 A

This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 8.
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Answer:
Given V_m = 283 V, f = 50 Hz, R = 3 Ω
L = 25.48 mH, and C = 796 μF.
(a) To find the impedance of the circuit, we first calculate XL and Xc
X_L = 2πfL = 2 × 3.14 × 50 × 25.48 × 10^-3 = 8 Ω
X_C = \(\frac{1}{2πfC}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4 Ω

Therefore impedance of the circuit is
Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{C}\right)^{2}}\)
= \(\sqrt{3^{2}+(8-4)^{2}}\)
= 5 Ω

(b) Phase difference
Φ = tan^-1\(\frac{X_{L}-X_{c}}{R}\) = tan-1 \(\frac{8-4}{3}\) = 53.1°
Since Φ is positive therefore voltage leads current by the above phase.

(c) The power dissipated in the circuit is
P = \(\frac{V_{\mathrm{rms}}^{2}}{Z}=\frac{V_{\mathrm{m}}^{2}}{2 \mathrm{Z}}=\frac{(283)^{2}}{2 \times 5}\) = 8008.9 W
(d) power factor = cos Φ = cos 53.1° = 0.6

Question 9.
A capacitor and a resistor are connected in series with an ac source. If the potential difference across the C, R is 120 V and 90 V respectively, and if rms current of the circuit is 3 A, calculate the (i) impedance and (ii) power factor of the circuit.
Answer:
Given V_C = 120 V, V_R = 90 V, f = 3 A. and R = 90/3 = 30 ohm ,

Effective voltage in the circuit
V = \(\sqrt{V_{C}^{2}+V_{R}^{2}}=\sqrt{(120)^{2}+(90)^{2}}\)= 150 V .
(i) Therefore impedance of the circuit
Z = \(\frac{V}{l}=\frac{150}{3}\) = 50 Ω.

(ii) Now power factor of the circuit is
cos Φ = \(\frac{R}{Z}=\frac{30}{50}\) = 0.6

Question 10.
An inductor 200 mH, a capacitor C, and a resistor 10 ohm are connected in series with 100 V, 50 Hz ac source. If the current and the voltage are in phase with each other, calculate the capacitance of the capacitor.
Answer:
When current and voltage are in phase then X_L = X_C

Question 11.
An ac voltage of 100V, 50 Hz is connected across a 20-ohm resistor and 2 mH inductors in series.
Calculate (i) impedance of the circuit, and
(ii) rms current in the circuit. (NCERT)
Answer:
Given, V = 100 V, f = 50 Hz, R = 20 ohm, L = 2 mH = 2 × 10^-3 H, Z = ?, lrms = ?
Using the relation
(a) Z = \(\sqrt{(R)^{2}+\left(X_{L}\right)^{2}}=\sqrt{(R)^{2}+(2 \pi f L)^{2}}\)
Z = \(\sqrt{(20)^{2}+\left(2 \times 3.14 \times 50 \times 2 \times 10^{-3}\right)^{2}}\)
Z = 20 ohm
(b) lrms= V/Z= 100/20 = 5 A

Question 12.
A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the. bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. (NCERT)
Answer:
(a) We are given P = 100 W and V = 220 V.
The resistance of the bulb is
R = \(\frac{V^{2}}{P}=\frac{(220)^{2}}{100}\) = 484 Ω

(b) The peak voltage of the source is
Vm = \(\sqrt{2}\)V = \(\sqrt{2}\) × 220 = 311V

(c) Since P = I V, therefore l = P/V= 100/220 = 0.45 A

Question 13.
A pure Inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. (NCERT)
Answer:
The inductive reactance,
X_L = 2 πf L = 2 × 3.14 × 50 × 25 × 10^-3 Ω
= 7.85 ohm

The rms current in the circuit is
l = \(\frac{V}{X_{L}}=\frac{220}{7.85}\) = 28 A

Question 14.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle? (NCERT)
Answer :
Given u_rms = 220 V, f = 50 Hz, l_rms = ?, P = ?
(a) lrms = \(\frac{V_{\mathrm{rms}}}{R}=\frac{220}{100}\) = 2.2 A

(b) Power consumed over one cycle of ac P = V_rms × l_rms = 220 × 2.2 = 484 W

Question 15.
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current? (NCERT)
Answer:
(a) V_m = 300V, V_rms =?

Using Vrms = \(\frac{V_{\mathrm{m}}}{\sqrt{2}}=\frac{300}{\sqrt{2}}\) = 212.16 V

(b) Using
l_m = l_rms × V₂ = 10 × \(\sqrt{2}\) = 14.14 A

Question 16.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. (NCERT)
Answer:
Given L = 44 mH = 44 × 10^-3 H, V_rms = 220 V, f = 50Hz,l_rms = ?
Using the relation
l_rms = \(\frac{V_{\text {rms }}}{x_{L}}=\frac{V_{\text {rms }}}{2 \pi f L}=\frac{220}{2 \times 3.14 \times 50 \times 44 \times 10^{-3}}\)
= 15.9 A

Question 17.
A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply, (a) What is the maximum current in the coil? (b) What Is the time lag between the voltage maximum and the current maximum? (NCERT)
Answer:
Given L = 0.5 H, R = 100 Ω, V_rms = 240 V,
f= 50 Hz, ω = 2πf= 2 × 3.14 × 50 = 314 s^-1 l_m = ?
(a) Now
Z = \(\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(100)^{2}+(0.5 \times 314)^{2}}\)
Z = 186.1 Ω

Therefore
l_m = \(\frac{V_{m}}{Z}=\frac{\sqrt{2} \times V_{\mathrm{rms}}}{Z}=\frac{\sqrt{2} \times 240}{186.1}\)
l_m = 1.82 A

(b) Let Φ be the phase angle by which current lags emf, then
tan Φ = \(\frac{X_{L}}{R}=\frac{0.5 \times 314}{100}\) = 1.571
Φ = 57.5° or Φ = 57.5 × π/180 rad

Therefore timw lag
t = \(\frac{\phi T}{2 \pi}=\frac{57.5 \times \pi}{180 \times 2 \pi \times 50}\) = 3.194 × 10^-3 s

Question 18.
A coil of 0.01-henry Inductance and 1-ohm resistance is connected to 200 volts, 50 Hz ac supply. Find the Impedance of the circuit and time lag between max alternating voltage and current. (NCERT Exemplar)
Answer:
Given L = 0.01 henry, R = 1 ohm,
V_rms = 200V, f = 50Hz, Z = ?, t = ?
We know that
X_L = 2πfL = 2 × 3.14 × 50 × 0.01 = 3.14 ohm
Now
Z = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}=\sqrt{(3.14)^{2}+(1)^{2}}\) = 3.3 Ω
Also tan Φ = \(\frac{X_{L}}{R}=\frac{3.14}{1}\)= 3.14
Or
Φ = tan^-1 (3.14)
Or
Φ = 72° or 72 × π/180 rad
Now time lag Δt is
Δt = \(\frac{\phi}{\omega}=\frac{72 \times \pi}{180 \times 2 \pi \times 50}=\frac{1}{250}\) s

Question 19.
An electrical device draws 2 kW power from AC mains (voltage 223 V (rms) = V50,000 V). The current differs (lags) in phase by Φ (tan Φ = – 3/4) as compared to voltage.
Find (i)R,
(ii) X_C – X_L
(iii) lm
Another device has twice the values for R, X_C, X_L. How are the answers affected? (NCERT Exemplar)
Answer:
Given P = 2 kW = 2000 W, V_rms = 223
= \(\sqrt{50,000}\) V, tan Φ = – 3/4, R = ?,X_C – X_L = ?, l_m = ?

(i)
Solving for R we have
R = 20 ohm

(ii) Also X_C – X_L = (-3/4) R = – 15 Ω
Hence lrms =\(\frac{V}{Z}=\frac{223}{25}\) = 9 A
(iii) l_m = \(\sqrt{2}\) × lrms = 1.414 × 9 = 12.6 A

If R, X_C, X_L, are all doubled, tan Φ does not change. Therefore as Z is doubled, the current becomes halved.

Question 20.
An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 Ω. Find
(a) the frequency of the source.
(b) the rms current through the resistor. (CBSEAI 2012)
Answer:
R = 50 Ω, f = ?, l_rms = ?, V_m = 140 V
Comparing with V = V_m sin 2πft,
we have
2πft = 314 t
or
f = 50 Hz
lrms = \(\frac{V_{r m s}}{R}=\frac{140}{50 \sqrt{2}}\) = 1.98 A

Question 21.
The figure shows a series LCR circuit with L = 5.0 H, C = 80 μF, R = 40 Ω connected to a variable frequency 240 V source. Calculate

(a) The angular frequency of the source drives the circuit at resonance.
(b) The current at the resonating frequency.
(C) The rms potential drops across the capacitor at resonance. (CBSE Delhi 2012)
Answer:
Given L= 5.0 H, C= 80 μF , R = 40 Ω, Vrms = 240 V ω_o = ?, l_rms = ?, V_C = ?
(a)Angutar frequency at resonance
ω0 = \(\sqrt{\frac{1}{L C}}=\sqrt{\frac{1}{5 \times 80 \times 10^{-6}}}\) = 50 s^-1

(b)At resonance Z = R = 4O Ω
Now lrms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{240}{40}\) = 6 A

(C) V_C = l_rms × X_C = 6 (50 × 80 × 10^-6) = 1500 V

Question 22.
The figure shows a series LCR circuit connected to a variable frequency 200 V source with L = 50 mH, C = 80 μF and R = 40 Ω.
Determine
(a) the source frequency which derives the circuit in resonance;
(b) the quality factor (Q) of the circuit. (CBSEAI 2014C)

Answer:
(a) Given V = 200 V, L = 50 mH, C = 80 μF, R = 40 Ω
Source frequency at resonance

Question 23.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device Y, the same current again flows through it, but it leads the voltage by π/2 rattan.
(a) Name the devices X and Y.
(b) Calculate the current flowing In the circuit when the same voltage Is applied across the series combination of the two devices X and Y.
Answer:
(a) The two devices are resistor and capacitor, respectively.

(b) Given l = 0.5 A, V = 220 V, X_L = ?, R = ? , l = ?
Now X_C and R are equal and are given by
X_C = R = \(\frac{220}{0.5}\) = 440 Ω

Hence impedance of the circuit
Z = \(\sqrt{R^{2}+X_{C}^{2}}=\sqrt{2(440)^{2}}\) = 622.25 Ω

Therefore current through the combination is
l = \(\frac{V}{Z}=\frac{220}{622.25}\) = 0.354 A

Question 24.
In the following circuit, calculate (a) the capacitance of the capacitor, if the power factor of the circuit is unity, (b) the Q-factor of this circuit. What is the significance of the Q-factor in a.c. circuit? Given the angular frequency of the a.c. source to be 100 s^-1. Calculate the average power dissipated in the circuit. (CBSE Delhi 2017C)

Answer:
(a) when the power factor is unity the circuit is in resonance, therefore, we have

Quality factor determines the sharpness of tuning at resonance. More the Q factor more the sharpness.

Question 25.
The figure below shows how the reactance of a capacitor varies with frequency.

(a) Use the Information on the graph to calculate the value of capacity of the capacitor.
(b) An inductor of inductance L has the same reactance as the capacitor at 100 Hz. Find the value of L.
(c) Using the same axes, draw a graph of reactance against frequency for the inductor.
(d) If this capacitor and inductor were connected in series to a resistor of 10 Ω, what would be the impedance of the combination at 300 Hz?
Answer:
(a) From the graph we find that for a frequency of 100 Hz the capacitive reactance is 6 ohm, therefore

(b) Now given X_L = X_C at 100 Hz, therefore X_L = 6 ohm

(c) The X_L – f graph is as shown below.

(d) For a frequency of 300 Hz, X_L = 18 ohm and X_C = 2 ohm. If resistance is 10 ohm then the impedance of the combination is

Question 26.
The given graphs (i) and (ii) represent the variation of the opposition offered by the circuit element to the flow of alternating current, with the frequency of the applied emf. Identify the circuit element corresponding to each graph.

A circuit is set up by connecting L= 100 mH, C = 5 μf and R =100 Ω in series. An alternating emf of (150 \(\sqrt{2}\)) volt, (500/π) Hz is applied across this series combination. Calculate the impedance of the circuit. What is the average power dissipated in (a) the resistor, (b) the capacitor, (c) the inductor, and (d) the complete circuit?
Answer:
(a) Resistor and (b) Inductor
Given L = 100 mH = 0.1 H, C = 5μF = 5 × 10^-6F, R=100 Ω,

Average power dissipated in
(a) the resistor P = l²R = (1.5)² × 100 = 225 W
(b) power in inductor is zero.
(c) power in capacitor is zero.
(d) the complete circuit is same as that in a resistor.

Question 27.
The output voltage of an ideal transformer, connected to a 240 V ac mains, is 24 V. When this transformer is used to light a bulb with rating 24 V, 24 W, calculate the current in the primary coil of the circuit.
Answer:

Question 28.
An ac generator consists of a coil of 50 turns and an area of 2.5 m² rotating at an angular speed of 60 rads^-1 in a uniform magnetic field B = 0.2 tesla, between the two fixed pole pieces. The resistance of the circuit including that of the coil is 500 ohm (a) Calculate the maximum current drawn from the generator. (b) What is the flux when the current is zero and (c) Would the generator work if the coil were stationary and instead the pole pieces rotated together with the same speed as above? Give a reason for your answer.
Answer:
Given n = 50, A = 2.5 m², ω = 60 rads^-1, B = 0.2 T, R = 500 ohm, l₀ = ?
(a) The maximum induced emf produced in the coil is
ε₀ = nBAω = 50 × 0.2 × 2.5 × 60 = 1500 V

Therefore maximum current in the circuit is
l₀ = \(\frac{\varepsilon_{0}}{R}=\frac{1500}{500}\) = 3 A

(b) The flux in this case is maximum and is given by Φ = nBA = 50 × 0.2 × 2.5 = 25 Wb

(c) Yes, for a generator to work there should be relative motion between the magnet and the coil.

Question 29.
The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is also 100. The input voltage and the powers are 220 V and 1100 W respectively. Calculate:
(a) number of turns in the secondary
(b) the current In the primary
(C) the voltage across the secondary
(d) the current in the secondary
(e) power in the secondary
Answer:
Here, N = 100, Transformation ratio = 100
V_p = 220 V, P_p = 1100W
(a) Number of turns in the secondary
N_s = N_p × Transformation ratio
= 100 × 100 = 10,000

(b) Current in the primary
l_p = \(\frac{P_{p}}{V_{p}}=\frac{1100}{220}\) = 5 A

(C) Voltage across the secondary V_s = V_p × transformation ratio
= 220 × 1oo = 22000 V

(d) Current in the secondary
l_s = \(\frac{V_{p} l_{p}}{V_{s}}=\frac{220 \times 5}{22000}\) = 0.05 A

(e) Power in the secondary = power in the primary (ideal transformer) = 1100 W

Question 30.
Calculate the current drawn by the primary of a transformer which steps down 200 V to 20 V to operate a device of resistance 20 Q. Assume the efficiency of the transformer to be 80 %.
Answer:

Question 31.
The primary of a transformer has 200 turns and the secondary has 1000 turns. If the power output from the transformer at 1000 V is 9 kW, calculate
(a) The primary voltage and
(b) The heat loss in the primary coil if the resistance of the primary is 0.2 Ω and the efficiency of the transformer is 90%.
Answer:

Question 32.
A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary In order to get output power at 230 V? (NCERT)
Answer:

Question 33.
A small town with a demand of 800 kW of electric power at 220 V situated 15 km away from an electric plant generating power at 440 V The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets power from the line through a 4000 – 220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming their negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant. (NCERT)
Answer:
Electric power required P = 800 kW = 800 × 10³ W
Voltage required V0 = 220 V

Distance of the power station
d = 15km = 15 × 10³m

Resistance per kilometre = 0.5 ohm per km
Total resistance R = 0.5 × 2 × 15 = 15 ohm

Input voltage V_i = 440 V

Primary voltage of transformer ε_p = 4000 V
Secondary voltage of transformer ε_s = 220 V
(a) Rms value of current in the two wire line

Therefore power loss along the line
= l^2_rmsR = (200)² × 15 = 600 kW

(b) Assuming negligible loss due to leakage Total power supply = power demand of town + power loss along the line
P = 800 kW + 600 kW = 1400 kW

(c) Voltage drop on the line = 200 × 15 = 3000 V
Hence voltage at the generation station
V= 4000 + 3000 = 7000 V

Therefore the step-up transformer at the plant is 440 V – 7000 V.

Question 34.
A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used. (NCERT Exemplar)
Answer:
Given P_s = 60 W, l_s = 0.54 A
We know that P=V I, therefore we have V_s = P_s/l_s = 60 /0.54 = 110 V

Therefore the transformer is a step-down transformer, whose transformation ratio is 1/2
Therefore
l_p = 1/2 × l_s = 1/2 × 0.54 = 0.27 A

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 7 The p-Block Elements.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 7 Important Extra Questions The p-Block Elements

The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why? (CBSE Delhi 2012)
Answer:
PCl₄^– is not likely to exist because lone pair of PCl₃ can be donated to Cl⁺ and not to Cl^–.

Question 2.
Of PH₃ and H₂S which is more acidic and why? (CBSE Delhi 2012)
Answer:
H₂S is more acidic because of weak S-H bond. PH₃ behaves as a Lewis base because of the presence of lone pair of electrons on P.

Question 3.
Which is a stronger reducing agent, SbH₃ or BiH₃ and why? (CBSE Al 2012)
Answer:
BiH₃ is a stronger reducing agent than SbH₃. This is because BiH₃ is less stable than SbH₃ because of larger size of Bi than Sb.

Question 4.
What is the covalency of nitrogen in N₂O₅? (CBSE Delhi 2013)
Answer:
Four.

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

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

Question 6.
What is the basicity of H₃PO₃ and why? (CBSE AI 2013)
Answer:
H₃PO₃ contains two P-OH bonds and therefore can give 2H⁺ ions. Its basicity is two.

Question 7.
Why does ammonia act as a Lewis base? (CBSE 2014)
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation. Therefore,it acts as a Lewis base.

Question 8.
What is the oxidation number of phosphorus in H₃PO₂ molecule? (CBSE Delhi 2010)
Answer:
+1

Question 9.
Out of white phosphorus and red phosphorus, which one is more reactive and why? (CBSE 2015)
Answer:
White phosphorus is more reactive because of angular strain in the P₄ molecules where the angles are only 60°.

Question 10.
What is the basicity of H₃PO₄? (CBSE Delhi 2015)
Answer:
Basicity of H₃PO₄ is three because it has three ionisable hydrogen atoms.

Question 11.
On heating Pb(NO₃)₂ a brown gas is evolved which undergoes dimerisation on cooling. Identify the gas. (CBSE 2016)
Answer:
NO₂.

2NO₂ → N₂O₄

Question 12.
Solid PCl₅ is ionic in nature. Give reason. (CBSE AI 2016)
Answer:
PCl₅ is ionic in the solid state because it exists as [PCl₄]⁺ [PCl₆]^– in which the cation is tetrahedral and anion is octahedral.

Question 13.
Write the structural difference between white P and red P. (CBSE Delhi 2014)
Answer:
White P consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ∠PPP = 60°. Red P also consists of P₄ tetrahedra units but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bonds.

Question 14.
What is the basicity of H₃PO₂ acid and why? (CBSE AI 2012)
Answer:
H₃PO₂ has the structure:

It has only one ionisable hydrogen and therefore, its basicity is one.

Question 15.
Name the promoter used In Haber’s process. (CBSE Sample Paper 2017-18)
Answer:
Molybdenum.

Question 16.
NF₃ is an exothermic compound whereas NCl₃ is not. Explain. (CBSE AI 2011, 2012)
Answer:
NF₃ is an exothermic compound while NCl₃ is not because:

• The bond dissociation enthalpy of F₂ is lower than that of Cl₂.
• Size of F is small as compared to Cl and therefore F forms stronger bonds with nitrogen releasing large amount of energy.

Therefore, overall energy is released during the formation of NF₃ and energy is absorbed during the formation of NCl₃.

Question 17.
N-N single bond is weaker than P-P single bond. (CBSE Delhi 2014)
Answer:
N-N single bond is weaker than P-P single bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

Question 18.
What happens when orthophosphorous acid is heated? (CBSE Sample Paper 2017-18)
Answer:
Phosphoric acid and phosphine are formed.

Question 19.
Write the order of thermal stability of the hydrides of group 16 elements. (CBSE Delhi 2017)
Answer:
The thermal stability of the hydrides of group 16 elements decreases from H₂O to H₂Te as:
H₂O > H₂S > H₂Se > H₂Te

Question 20.
The two O – O bond lengths in ozone molecule are equal. Why? (CBSE AI 2013, CBSE Delhi 2014)
Answer:
Ozone is a resonance hybrid of two structures and therefore, the two O-O bond lengths are equal.

Question 21.
SF₆ is not easily hydrolysed. (CBSE Delhi 2005)
OR
SF₆ is kinetically inert substance. Explain. (CBSE Al 2011, CBSE Delhi 2011)
Answer:
SF₆ is chemically inert and therefore, does not get hydrolysed. Its inert nature is due to the presence of stearically protected sulphur atom which does not allow thermodynamically favourable hydrolysis reaction.

Question 22.
Which of the following compounds has a lone pair of electrons at the central atom? (CBSE Sample Paper 2011)
H₂S₂O₈, H₂S₂O₇, H₂SO₃, H₂SO₄.
Answer:
H₂SO₃

Question 23.
Ozone is thermodynamically unstable. Explain. (CBSE Sample Paper 2011)
Answer:
Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (∆H is -ve) and increase in entropy (∆S is +ve). These two factors reinforce each other resulting negative ∆G (∆G = ∆H – T∆S) for its conversion to oxygen.

Question 24.
Which neutral molecule would be isoelectronic with ClO^–? Is that molecule a Lewis base? (CBSE Delhi 2008)
Answer:
ClO^– is isoelectronic with ClF. Yes, it is a Lewis base.

Question 25.
Which compound of xenon has distorted octahedral shape? (CBSE Sample Paper 2012)
Answer:
XeF₆ has distorted octahedral shape.

Question 26.
Iodide ions can be oxidised by oxygen in acidic medium. Give chemical equation to support this. (CBSE Sample Paper 2011)
Answer:
Iodide can be oxidised by oxygen in acidic medium:
4I^–(aq) + O₂(g) + 4H⁺(oq) > 2l₂(s) + 2H₂O(l).

Question 27.
What happens when XeF6 undergoes complete hydrolysis? (CBSE Sample Paper 2017-18)
Answer:
XeO₃ is formed.
XeF₆ + 3H₂O → XeO₃ + 6HF

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

Question 29.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
What inspired N. Bartlett for carrying out reaction between Xe and PtF₆? (CBSE Delhi 2013)
Answer:
In 1962, N. Bartlett noticed that platinum hexafluoride PtF₆, is a powerful oxidising agent which combines with molecular oxygen to form ionic compound, dioxygenyl hexafluoroplatinate (V), O₂⁺ [PtF₆]^–.
O₂(S) + PtF₆(g) → O₂⁺ [PtF₆]^–
This indicates that PtF₆ has oxidised O₂ to O₂⁺. Now, oxygen and xenon have some similarities:

• The first ionisation energy of xenon gas (1170 kJ mol^-1) is fairly close to that of oxygen (1177 kJ mol^-1).
• The molecular diameter of oxygen and atomic radius of xenon are similar (4Å). These prompted Bartlett to carry out the reaction between Xe and PtF₆.

Question 2.
Calculate the number of lone pairs on central atom in the following molecule and predict the geometry. (CBSE Sample Paper 2019)
Answer:
XeF₄ has square planar structure. It involves sp³d² hybridisation of Xenon in which two positions are occupied by lone pairs.

Structure of XeF₄ (Square planar)

Question 3.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE AI 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’_n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X, e.g. ClF, ICl₃, BrF₅, IF₇, etc.

Question 4.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalides. Give reason. (CBSE 2013, CBSE Delhi 2015)
Answer:
Nitrogen belongs to second period (n = 2) and has only s and p-orbitals. It does not have d-orbitals in its valence shell and therefore, it cannot extend its octet. That is why nitrogen does not form pentahalides.

Question 5.
Why does NO₂ dimerise? Explain. (CBSE 2014, CBSE Delhi 2014)
Answer:
NO₂ contains odd number of valence electrons. It behaves as a typical molecule. In the liquid and solid state, it dimerises to form stable N₂O₄ molecule, with even number of electrons. Therefore, NO₂ is paramagnetic, while N₂O₄ is diamagnetic in which two unpaired electrons get paired.

Question 6.
Draw the structure of O₃ molecule. (CBSE Delhi 2010)
Answer:
Ozone has angular structure as:

It is resonance hybrid of the following structures:

Question 7.
Complete the following equations: (CBSE 2014)
(i) Ag + PCl₅ →
(ii) CaF₂ + H₂SO₄ →
Answer:
(i) 2Ag + PCl₅ → 2AgCl + PCl₃
(ii) CaF₂ + H₂SO₄ → CaSO₄ + 2HF

Question 8.
Write balanced chemical equations for the following processes:
(i) XeF₂ undergoes hydrolysis.
(ii) MnO₂ is heated with cone. HCl.
OR
Arrange the following in order of property indicated for each set:
(i) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character
(ii) HF, HCl, HBr, HI – decreasing bond enthalpy (CBSE Delhi 2019)
Answer:
(i) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂
(ii) MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
OR
(i) H₂O < H₂S < H₂Se < H₂Te
(ii) HF > HCI > HBr > HI

Question 9.
Explain the following giving an appropriate reason in each case.
(i) O₂ and F₂ both stabilise higher oxidation states of metals but O₂ exceeds F₂ in doing so.
(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach. (CBSE 2012)
Answer:
(i) This is because fluorine can show only -1 oxidation state, whereas oxygen can show -2 oxdiation state as well as positive oxidation states such as +2, +4 and +6.

(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach because xenon is a noble gas and its octet is complete. Therefore, according to Valence Bond theory, xenon should not combine with fluorine to form fluorides due to its stable outermost shell configuration.

Question 10.
Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₃PO₃ (CBSE 2013)
Answer:
(i)

Structure of XeOF₄

(ii)

Structure of H₃PO₃

Question 11.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’_n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X,
e.g. ClF, ICl₃, BrF₅, IF₇, etc.

Question 12.
Draw the structures of the following molecules: (CBSE 2013)
(i) N₂O₅
(ii) XeF₂
Answer:
(i)

(ii)

Structure of XeF₂ (Linear)

Question 13.
What happens when
(i) PCl₅ is heated?
(ii) H₃PO₃ is heated?
Write the reactions involved. (CBSE Delhi 2013)
Answer:
(i) On heating, PCl₅ sublimes and decomposes on strong heating:

(ii) On heating, H₃PO₃ decomposes into phosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃

Question 14.
Complete the following chemical equations:
(i) Ca₃P₂ + H₂O →
(ii) Cu + H₂SO₄(conc.) →
OR
Arrange the following in the order of property indicated against each set:
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character. (CBSE Delhi 2014)
Answer:
(i) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃
(ii) Cu + 2H₂SO₂₄;(conc.) → CuSO₄
+ SO₂ + 2H₂O
OR
(i) HI < HBr < HCl < HF
(ii) H₂O < H₂S < H₂Se < H₂Te

Question 15.
Complete the following equations:
(i) P₄ + H₂O →
(ii) XeF₄ + O₂F₂ → (CBSE 2014)
Answer:
(i) P₄ + H₂O → no reaction
(ii) XeF₄ + O₂F₂ → XeF₆ + O₆

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

Structure of XeF₂ (Linear)

(ii)

Structure of BrF₃ (T Shape)

Question 17.
(a) Draw the structure of XeF₄.
(b) What happens when CaF₂ reacts with cone. H₂SO₄? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)

(b) HF is formed.
CaF₂ + H₂SO₄ → 2HF + CaSO₄

Question 18.
(a) Draw the structure of H₂S₂O₇.
(b) What happens when carbon reacts with cone. H₂SO₄? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)

(b) Carbon dioxide is formed.
C + 2H₂SO₄(conc.) → CO₂ + 2 SO₂ + 2H₂O

Question 19.
Draw the structures of the following molecules: (CBSE 2014)
(i) XeO₂
(ii) H₂SO₄
Answer:
(i)

(ii)

Question 20.
Write the structure of the following: (HPO₃)₃ (CBSE 2016)
Answer:
(HPO₃)₃

Question 21.
Why is N₂ less reactive at room temperature? (CBSE Al 2015, H.P.S.B. 2015, 2016, Meghalaya S.B. 2017)
Answer:
In molecular nitrogen, there is a triple bond between two nitrogen atoms (NsN) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 22.
Phosphine has lower boiling point than ammonia. Give reason. (CBSE Al 2008,
CBSE Delhi 2013)
Answer:
Ammonia exists as associated molecule due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state.
This is because of low electronegativity of P than N. As a result, the boiling point of PH₃ is lower than that of NH₃.

Question 23.
Draw structures of the following:
(a) XeF₄
(b) S₂O₈^2- (CBSE AI 2019)
Answer:
(a)

(b)

Question 24.
(i) Draw the structure of phosphinic acid (H₃PO₂).
(ii) Write a chemical reaction for its use as reducing agent. (CBSE Sample Paper 2011)
Answer:
(i) Phosphinic acid, H₃PO₂.

(ii) H₃PO₂ reduces Ag⁺ ion to Ag which shows its reducing nature.
H₃PO₂ + 4AgNO₃ + 2H₂O → 4Ag + 4HNO₃ + H₃PO₄

Question 25.
Among the hydrides of Group 15 elements, which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character? (CBSE AI 2018)
Answer:
(a) PH₃
(b) NH₃
(c) NH₃
(d) BiH₃

Question 26.
The bond angles (O-N-O) are not of the same value in NO₂^– and NO₂⁺. Give reason. (CBSE Delhi 2012)
Answer:
In NO₂ there is one electron on N. Therefore, in NO₂⁺, there is no electron on N atom and there is a lone pair of electrons on N atom in NO₂^–.

Question 27.
H₂PO₂ is a stronger reducing agent than H₃PO₃. Give reason. (CBSEAI 2014, 2016)
Answer:
The reducing character of the acid is due to H atoms bonded directly to P atom. In H₃PO₂, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₂ is a stronger reducing agent than H₂PO₃.

Question 28.
Bi(V) is a stronger oxidising agent than Sb(V). Give reason. (CBSE Delhi 2014)
Answer:
On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

Question 29.
Dioxygen is a gas while sulphur is a solid at room temperature. Why?
(CBSE AI 2013, 2018)
Answer:
Oxygen exists as a stable diatomic molecule and is, therefore, a gas. On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has strong tendency to form multiple bonds with itself and forms strong O=O bonds rather than O-O bonds.

On the other hand, sulphur-sulphur double bonds (S=S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O=O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule.
Therefore, oxygen exists as a diatomic gas and sulphur exists as S₈ solid.

Question 30.
Draw structures of the following:
(a) H₂S₂O₇
(b) HClO₃ (CBSE Al 2019)
Answer:
(a) H₂S₂O₇

(b) HClO₃

Question 31.
Complete and balance the following equations:
(a) C + H₂SO₄ (cone.) →
(b) XeF₂ + PF₅ →
OR
Write balanced chemical equations involved in the following reactions:
(a) Fluorine gas reacts with water.
(b) Phosphine gas is absorbed in copper sulphate solution. (CBSE AI 2019)
Answer:
(a)

(b)

OR

(a) 2F₂ + 2H₂O → 4HF + O₂

(b)

Question 32.
Sulphuric acid has low volatility. Give chemical reactions in support of this. (CBSE Sample Paper 2011)
Answer:
Sulphuric acid has low volatility and decomposes the salts of volatile acids forming its own salts:
2MX + H₂SO₄ → 2HX + M₂SO₄
(M = metal, X = F, Cl, NO3)
e.g.,NaCl + H₂SO₄ → NaHSO₄ + HCl
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃.

Question 33.
Electron gain enthalpies of halogens are largely negative. Why? (CBSE AI 2017)
Answer:
The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns²np⁶).

Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have largely negative electron gain enthalpies.

Question 34.
Why does R₃P = 0 exist but R₃N = 0 does not (R = alkyl group). (CBSE AI 2014)
Answer:
R₃N = 0 does not exist because nitrogen cannot have covalency more than four. Moreover, R₃P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bond whereas nitrogen cannot form dπ-pπ bond.

Question 35.
Write any two oxoacids of sulphur and draw their structures. (CBSE Al 2019)
Answer:
(i) Sulphuric acid

(ii) Peroxodisulphuric acid

Question 36.
Complete and balance the following equations:
(a) NH₃ (excess) + Cl₂ →
(b) XeF₆ + 2H₂O →
OR
Write balanced chemical equations involved in the following reactions:
(a) XeF₄ reacts with SbF₅.
(b) Ag is heated with PCl₅. (CBSE AI 2019)
Answer:
(a)
8NH₃ + 3Cl₂ → 6NH₄Cl + N₂
excess
(b) XeF₆ + 2H₂O → XeO₂F₂ + 4HF

OR

(a) XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]^–
(b) 2Ag + PCl₅ → 2AgCl + PCl₃

Question 37.
Complete and balance the following equations:
(a) S + H₂SO₄ (cone.) →
(b) PCl₃ + H₂O →
Write balanced chemical equations involved in the following reactions:
(a) Chlorine gas reacts with cold and dilute NaOH
(b) Calcium phosphide Is dissolved in water. (CBSE AI 2019)
Answer:
(a) S + 2H₂SO₄(conc.) → 3SO₂ + 2H₂O
(b) PCl₃ +3H₂O → H₃PO₃ + 3HCl
OR
(a) Cl₂ + ZNaOH → NaCl + NaOCl + H₂O
(b) Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃

Question 38.
Bond enthalpy of fluorine Is lower than that of chlorine. Why? (CBSE AI 2018)
Answer:
Fluorine atoms are smalL and internuclear distance between two F atoms In F₂ molecule is also small (1 .43Å). As a result, electron-electron repulsions among the lone pairs on two fluorine atoms in F₂ molecule are large. These Llarge electron-electron repulsions weaken the F-F bond. However, the electron-electron repulsions among lone pairs in Cl₂ molecule are less because of Larger Cl atoms and larger Cl-Cl distance (1.99Å). Therefore, the bond enthalpy of fluorine is lower than that of chlorine.

The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
Give reasons for the following:
(a) Dioxygen is a gas but sulphur a solid.
(b) NO (g) released by jet aeroplanes is slowly depleting the ozone layer.
(c) Interhalogens are more reactive than pure halogens. (CBSE Al 2019)
Answer:
(a) Due to small size and high electronegativity, oxygen atom forms pπ-pπ double bond, O = O. The intermolecular forces in oxygen are weak van der Waal’s forces and therefore, oxygen exists as a gas. On the other hand, sulphur does not form stable pπ-pπ bonds and does not exist as S₂. It is linked by single bonds and forms polyatomic complex molecules having eight atoms per moledule (S₈) and has puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

(b) Nitrogen oxide (NO) gas emitted from the exhaust of jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen.
NO + O₃ → NO₂ + O₂
Since jet aeroplanes fly in the stratosphere, near ozone layer, these are responsible for the depletion of ozone layer.

(c) This is because covalent bond between dissimilar atoms in interhalogens (X- Y) is polar and weaker than between similar atoms (X-X and Y-Y). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping between similar atoms.

Question 2.
Give reasons for the following:
(i) Where R is an alkyl group, R₃P = 0 exists but R₃N = 0 does not.
(ii) PbCl₄ is more covalent than PbCl₂.
(iii) At room temperature, N₂ is much less reactive. (CBSE AI 2013)
Answer:
(i) R₃N = 0 does not exist because nitrogen cannot have covalency more than four. But R₃P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bonds whereas nitrogen cannot form dπ-pπ bond.

(ii) Pb⁴⁺ has smaller size than Pb²⁺. As a result, it can polarise the Cl^– ion to greater extent than Pb²⁺. Consequently, PbCl₄ is more covalent than PbCl₂.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 3.
Give reasons for the following:
(i) Oxygen is gas but sulphur is a solid,
(ii) O₃ acts as a powerful oxidising agent.
(iii) BiH₃ is the strongest reducing agent amongst all the hydrides of Group 15 elements. (CBSE Al 2013)
Answer:
(i) Oxygen exists as a stable diatomic molecule and is, therefore, a gas.
On the other hand, sulphur exists in solid state as S₈ molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has good tendency to form multiple bonds with itself and forms strong o = o bonds than o – o bonds. On the other hand, sulphur- sulphur double bonds (S = S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O = O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule. Therefore, oxygen exists as a diatomic gas and sulphur exists as S₈ solid.

(ii) Ozone acts as a powerful oxidising agent because it has higher energy content and decomposes readily to give atomic oxygen as:
O₃ → O₂ + O
Therefore, ozone can oxidise a number of non-metals and other compounds. For example,
PbS (s) + 4O₃ (8) → PbSO₄ (s) + 4O₂ (g)
2I^– (aq) + H₂O(l) + O₃(g) → 2OH^– (aq) + I₃₂(s) + O₂(g)

(iii) The stability of group 15 hydrides decreases from NH₃ to BiH₃. Therefore, BiH₃ is most unstable hydride among group 15 hydrides because of lowest M-H bond strength. As a result, BiH₃ is the strongest reducing agent.

Question 4.
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalphy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bonds lengths in ozone molecule are identical. (CBSE Al 2013)
Answer:
(i) Nitrogen does not have vacant d-orbitals in its valence shell. Therefore, it cannot extend its valency beyond 3. Thus, it cannot form pentahalides.

(ii) The less negative electron gain enthalpy of fluorine as compared to chlorine is due to very small size (72 pm) of fluorine atom. As result, there are strong inter- electronic repulsions in the relatively small 2p sub-shell of fluorine and therefore, the incoming electron does not feel much attraction. Thus, its electron gain enthalpy is small.

(iii) Ozone is resonance hybrid of two structures and therefore, the two oxygen-oxygen bonds are identical.

Question 5.
How would you account for the following:
(i) H₂S is more acidic than H₂O.
(ii) The N-O bond in NO₂^– is shorter than the N-O in NO₃^–.
(iii) Both O₂ and F₂ stabilise higher oxidation states but the ability of oxygen to stabilise the higher oxidation states exceeds that of fluorine. (CBSE 2011)
Answer:
(i) Since size of S is more than that of O, the distance between central atom and hydrogen atoms is more in H₂S than in H₂O. Therefore, bond dissociation enthalpy of H₂S is less and bond cleavage is more easy than in H₂O. Therefore, H₂S is more acidic than H₂O.

(ii) In the resonance structures of NO₂^–, two bonds are sharing a double bond, while in NO₃^–, three bonds are sharing a double bond.

As a result, the bond in NO₂^– will be shorter than in NO₃^–.

(iii) The larger tendency of oxygen to stabilise the higher oxidation state is because of the tendency of oxygen to form multiple bonds with metal.

Question 6.
How would you account for the following:
(i) NF₃ is an exothermic compound but NCl₃ is not.
(ii) The acidic strength of compounds increases in the order:
PH₃ < H₂S < HCl
(iii) SF₆ is kinetically inert. (CBSE 2011)
Answer:
(i) NF₃ is an exothermic compound while NCl₃ is an endothermic compound because
(a) The bond dissociation enthalpy of F₂ is lower as compared to Cl₂.
(b) Size of F is small as compared to Cl and therefore, F forms stronger bond with N releasing large amount of energy.
Therefore, overall energy is released during the formation of NF₃ and energy is absorbed during the formation of NCl₃.

(ii) The bond dissociation enthalpy decreases in the order:
P-H > S-H > Cl-H
Therefore, acidic character follows the order:
PH₃ < H₂S < HCl

(iii) SF₆ is kinetically inert because sulphur is sterically protected by six F atoms and it is a coordinatively saturated compound. Therefore, it does not undergo thermodynamically favourable hydrolysis reaction.

Question 7.
(a) Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF₄

(ii) H₂SO₄

(b) White phosphorus consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Answer:
Red phosphorus also consists of P₄ tetrahedra units, but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bond.

Question 8.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N-N single bond is weaker than P-P single bond.
(iii) Noble gases have very low boiling points. (CBSE Delhi 2014)
Answer:
(i) On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

(ii) N-N single bond is weaker than single P-P bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

(iii) Noble gases are monoatomic gases and are held together by weak vander Waals forces. Therefore, these are liquified at very low temperatures. Hence they have low boiling points.

Question 9.
Give reasons for the following: (CBSE 2014)
(i) (CH₃)₃ P = 0 exists but (CH₃)₃ N = 0 does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃.
Answer:
(i) (CH₃)₃ N = 0 does not exist because
nitrogen cannot have covalency more than four. Moreover, (CH₃)₃ P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ – pπ bonds whereas nitrogen cannot form dπ – pπ bonds.

(ii) This is due to small size of oxygen atom so that its electron cloud is distributed over a small region of space and therefore, it repels the incoming electron. Hence the electron gain enthalpy of oxygen is less negative than sulphur.

(iii) In H₃PO₃, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₃ is stronger reducing agent than H₃PO₃.

Question 10.
Assign reason for the following:
(i) H₃PO₂ is a stronger reducing agent than H₃PO₄.
(ii) Sulphur shows more tendency for catenation than oxygen.
(iii) Reducing character increases from HF to HI. (CBSE 2016)
Answer:
(i) The reducing character of the acid is due to H atoms bonded directly to P atom. In H₃PO₂, there are two P-H bonds whereas in H₃PO₃, there is one P-H bond. Therefore, H₃PO₂ is a stronger reducing agent than H₃PO₃.

(ii) Sulphur shows more tendency for catenation than oxygen because of stronger S-S bonds than O-O bonds.

(iii) Due to decrease in bond enthalpy down the group, the thermal stability decreases from HF to HI.
As a result of decrease in stability from HF to HI, the reducing character increases down the group as:
HF < HCl < HBr < HI

Question 11.
What happens when
(i) (NH₄)₂Cr₂O₇ is heated?
(ii) PCl₅ is heated?
(iii) H₃PO₃ is heated? Write the equations involved. (CBSE AI 2015, CBSE Delhi 2008, 2013, 2017)
Answer:
(i) On heating (NH₄)₂Cr₂O₇, nitrogen gas is evolved.

(ii) On heating, PCl₅ first sublimes and then decomposes on strong heating:

(iii) On heating, H₃PO₃ disproportionates to give orthophosphoric acid and phosphine.

Question 12.
(a) Suggest a quantitative method for estimation of the gas which protects us from U.V. rays of the sun.
(b) Nitrogen oxides emitted from the exhaust system of supersonic jet aeroplanes slowly deplete the concentration of ozone layer in upper atmosphere. Comment. (CBSE Sample Paper 2011)
Answer:
(a) The gas which protects us from U.V. rays of the sun is ozone. It reacts with I^– ions to give iodine as:
O₃ + 2I^– + H₂O → O₂ + l₂ + 2OH^–
l₂ liberated is titrated against sodium thiosulphate solution and amount of O₃ can be estimated.

(b) The release of nitrogen oxides (NO_x) into stratosphere by the exhaust system of supersonic jet aeroplanes deplete the concentration of O₂ because NO reacts with O₃ to give O₂.
NO (g) + O₂ (g) → NO₂ (S) + O₂ (s)
Therefore, NO is slowly depleting the concentration of ozone.

Question 13.
Give reasons for the following:
(a) Acidic character decreases from N₂O₃ to Bi₂O₃.
(b) All the P-Cl bonds in PCl₅ are not equivalent.
(c) HF is a weaker acid than HCl in an aqueous solution. (CBSE Al 2019)
Answer:
(a) On moving down the group, metallic character increases. Therefore acidic character of oxides decreases from nitrogen to bismuth.

(b) PCl₅ has trigonalbipyramidal geometry.

In this geometry, axial bonds are larger than equatorial bonds because of large repulsions at axial positions. Therefore, all P-Cl bonds in PCl₅ are not equivalent.

(c) The H-F bond has high bond dissociation enthalpy than H-Cl bond. Therefore, H-F bond is stronger and can be dissociated into H⁺ and F^– ions with difficulty. Hence, it is weaker acid than HCl.

Question 14.
Give reasons for the following:
(a) O-O single bond is weaker than S-S single bond.
(b) Tendency to show -3 oxidation state decreases from Nitrogen (N) to Bismuth (Bi).
(c) Cl₂ acts as a bleaching agent. (CBSE AI 2019)
Answer:
(a) O-O single bond is weaker than S-S single bond because oxygen has smaller size and has high inter-electronic repulsion.

(b) Tendency to show -3 oxidation state decreases because metallic character and size increase down the group.

(c) Cl₂ acts as a bleaching agent in the presence of moisture
Cl₂ + H₂O → 2HCl + [O]
Coloured matter + [O] → Colourless matter
The nascent oxygen is responsible for the bleaching action of Cl₂.

Question 15.
(a) Draw the structures of the following molecules:
(i) XeOF₄
(ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF₄

(ii) H₂SO₄

(b) White phosphorus consists of P₄ units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Red phosphorus also consists of P₄ tetrahedra units, but it has polymeric structure consisting of P₄ tetrahedra linked together by covalent bond.

Question 16.
Account for the following:
(i) PCl₅ is more covalent than PCl₃.
(ii) Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
(iii) The two O-O bond lengths in the ozone molecule are equal. (CBSE Delhi 2014)
Answer:
(i) In PCl₅, the oxidation state of P is +5 while in PCl₃, it is +3. As a result of higher oxidation state of the central atom, PCl₅ has larger polarising power and can polarise the chloride ion (Cl^–) to a greater extent than in the corresponding PCl₃. Since larger the polarisation, larger is the covalent character; PCl₅ is more covalent than PCl₃.

(ii) Iron reacts with HCl to form ferrous chloride with the evolution of hydrogen gas.
Fe + 2HCl → FeCl₂ + H₂

(iii) The two O-O bond lengths in ozone molecule are equal and are intermediate between single and double bonds. This is because of resonance between two structures (a) and (b) and actual molecule is resonance hybrid of these two structures:

Question 17.
Account for the following:
(a) Moist SO₂ decolourises KMnO₄ solution.
(b) In general, interhalogen compounds are more reactive than halogens (except fluorine).
(c) Ozone acts as a powerful oxidising agent. (CBSE Sample Paper 2019)
Answer:
Moist sulphur dioxide behaves as a reducing agent. It reduces MnO₄^– to Mn²⁺.
5 SO₂ + 2MnO₄^– + 2H₂O → 5 SO₄^2- + 4H⁺ + 2Mn²⁺

(b) X-X’ bond in inter halogens is weaker than X-X bond in halogens except F-F bond. This is due to the fact that the overlapping of orbitals of two dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(c) Due to the ease with which it liberates atoms of nascent oxygen, Ozone acts as a powerful oxidising agent.
O₃ → O₂ + O

Question 18.
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of Fe(lll) salt?
Answer:
(a) (i) Sodium chlorate and sodium chloride are formed.

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

(b) Answer the following:
(i) What is the basicity of H₃PO₃ and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points? (CBSE Delhi 2011)
Answer:
(i) H₃PO₃ is dibasic and has basicity of
two because it has two P-OH bonds which are ionisable. The third H atom is linked to P and is non-ionisable.

H₃PO₃ ⇌ HPO₃^2- + 2H⁺

(ii) Fluorine does not play the role of a central atom in interhalogen compounds because it is highly electronegative. Moreover, it has only one electron less than the octet and does not have vacant d-orbitals in its valence shell. Therefore, it can form only one bond with other halogen atoms and cannot act as central atom in interhalogen compounds.

(iii) Noble gases have very low boiling points because only weak van der Waals’ forces are present between the atoms of the noble gases in the liquid state.

Question 19.
(a) Give reasons for the following:
(i) Bond enthalpy of F₂ is lower than that of Cl₂.
(ii) PH₃ has lower boiling point than NH₃.
Answer:
(a) (i) Due to small size of F atom, there are strong repulsions between the non-bonding electrons of F atoms in the small sized F₂ molecule. Therefore, bond enthalpy of F₂ is lower than relatively larger Cl₂ molecule in which repulsions between non-bonding electrons are less.

(ii) Ammonia exists as associated molecules due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH₃, phosphine (PH₃) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of PH₃ is lower than that of NH₃.

(b) Draw the structures of the following molecules:
(i) BrF₃
(ii) (HPO₃)₃
(iii) XeF₄
Answer:
(i)

BrF₃ (T shape)

(ii)

(HPO₃)₃

(iii)

XeF₄

OR

(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
Answer:
(a) (i) Helium alongwith oxygen (helium- oxygen mixture) is used by deep sea divers in preference to nitrogen-oxygen mixture because of its low solubility in blood.

(ii) Fluorine is the most electronegative element and therefore, it shows an oxidation state of -1 only. It does not show any positive oxidation state.

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

(b) Draw the structures of the following molecules.
(i) XeF₂
(ii) H₂S₂O₈ (CBSE Delhi 2013)
Answer:
(i)

XeF₂ (Linear)

(ii)

Question 20.
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
Answer:
(a) (i) In the vapour state, sulphur exists as S2 molecules. S2 molecule has two unpaired electrons in anti-bonding molecular orbitals (π^x and π^y) and hence shows paramagnetism.

(ii) The N-N bond is weaker than P-P bond because of high interelectronic repulsions of non-bonding electrons due to small N-N bond length (109 pm).

(iii) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is negative) and increase in entropy (ΔS positive). These two factors make large negative ΔG for its conversion to oxygen.

(b) Write the name of gas released when Cu is added to
(i) dilute HNO₃ and
(ii) conc. HNO₃
Answer:
(i) With dilute HNO₃; NO (nitric oxide)
3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO + 4H₂O

(ii) With cone. HNO₃: NO₂ (nitrogen dioxide)
Cu + 4HNO₃(conc) → Cu(NO₃)₂ + 2NO₂ + 2H₂O

OR

(a) (i) Write the disproportionation reaction of H₃PO₃.
(ii) Draw the structure of XeF₄.
Answer:
(i) 4H₃PO₃ → 3H₃PO₄ + PH₃

(ii)

(b) Account for the following:
(i) Although fluorine has less negative electron gain enthalpy, yet F₂ is strong oxidising agent.
(ii) Acidic character decreases from N₂O₃ to Bi₂O₃ in group 15.
Answer:
(i) Fluorine is the strongest oxidising agent, although it has less negative electron gain enthalpy. This is because of its smaller size, large bond dissociation enthalpy of F₂ and high exothermic hydration enthalpy of small F^– ion.

(ii) On moving down the group, the metallic character increases. Therefore, the acidic character of oxides decreases down the group from N₂O₃ to Bi₂O₃

(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. (CBSE Delhi 2019)
Answer:
SO₂ decolourises pink violet colour of acidified potassium permanganate solution.
2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄

Question 21.
(a) Complete the following chemical reaction equations:
(i) P₄ + SO₂Cl₄ →
(ii) XeF₆ + H₄O →
Answer:
(i) P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂

(ii) XeF₆ + H₂O → XeOF₄ + 2HF

(b) Predict the shape and the asked angle (90° or more or less) in each of the following cases:
(i) SO₃^2- and the angle O – S – O
(ii) ClF₃ and the angle F – Cl – F
(iii) XeF₂ and the angle F – Xe – F
Answer:
(i) SO₃^2-

The O-S-O angle is more than 90°.

(ii) ClF₃

The F-Cl-F bond angle is equal to 90°.

(iii)
XeF₂

F-Xe-F bond angle is equal to 180° (greater than 90°).

OR

(a) Complete the following chemical equations:
(i) NaOH + Cl₂ →
(hot and cone.)
(ii) XeF₄ + O₂F₂ →
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
(Hot and conc.)

(ii) XeF₄ + O₂F₂ → XeF₆ + O₂

(b) Draw the structures of the following molecules:
(i) H₃PO₂
(ii) H₂S₂O₇ (CBSE 2012)
Answer:
(i) H₃PO₂

(ii) H₂S₂O₇

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

(ii)

(b) Explain the following observations:
(i) The molecules NH₃ and NF₃ have dipole moments which are of opposite directions.
(ii) All the bonds in PCl₅ molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
Answer:
(i) Both NH₃ and NF₃ have pyramidal shape with one lone pair on N atom.

The lone pair on N is in opposite direction to the N-F bond moments and therefore, it has very low dipole moment (about 0.234 D). But ammonia has high dipole moment because its lone pair is in the same direction as the N-H bond moments.

(ii) PCl₅ has trigonal bipyramidal structure in which there are three P-Cl equatorial bonds and two P-Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90° while the three equatorial bonds are being repelled by two bond pairs at 90°. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm).

(iii) In vapour state, sulphur partly exists as S₂ molecule and S₂ molecule like O₂ has two unpaired electrons in anti-bonding π* molecular orbitals. Therefore, it is paramagnetic.

OR

(a) Complete the following chemical equations:
(i) XeF₄ + SbF₅ →
(ii) Cl₂ + F₂ (excess ) →
Answer:
(i) XeF₄ + SbF₅ → XeF₄.SbF₅ → [XeF₃]⁺ [SbF₆]^–

(ii)

(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in NO₂^– and NO₂⁺. (CBSE Delhi 2012)
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) The stablity of +5 oxidation state decreases down the group because of inert pair effect. Therefore, the +5 oxidation state of Bi is less stable than that of Sb.

(iii) In NO₂, there is one electron on N while in NO₂⁺ there is no electron and in NO₂⁺, there is a lone pair of electrons on N as shown below:

NO₂⁺ molecule is linear and has bond angle of 180°. The repulsion by single electron is less as compared to repulsion by a lone pair of electrons. Therefore, bond pairs in NO₂^– are forces more closer than in NO₂.

Question 23.
(a) Draw the structure of the following molecule: H₃PO₂
Answer:
(a)

(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is nonpolar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) Though electronegativity of F is more than O, yet water forms more extensive hydrogen bonding than HF. In H₂O, each oxygen is tetrahedrally surrounded by two covalent bonds and two hydrogen bonds, (-O….H). In HF there is only one hydrogen bond (-F….H).

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

OR

(a) Draw the structures of the following molecules:
(i) N₂O₅
(ii) HClO₄
Answer:

(a)
(i)

(ii)

(b) Explain the following observations:
(i) H₂S is more acidic than H₂O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (CBSE 2012)
Answer:
(i) The size of S is more than that of O. Therefore, the distance between S and H, i.e. S-H bond length, is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in H₂S than O-H bond in water. Therefore, H₂S will be more acidic than H₂O.

(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of – 1 only. It does not show any positive oxidation state.

(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

Question 24.
Write balanced equations for the following reactions:
(a) P₄ + NaOH + H₂O → CBSE Delhi 2009)
(b) As₄ + Cl₂(excess) →
(c) P₄O₁₀ + H₂O →
(d) Ca₃P₂ + H₂O → (CBSE Delhi 2008, 2014)
(e) POCl₃ + H₂O →
(f) HgCl₂ + PH₃ → (CBSE AI 2010)
(g) Ag + PCl₅ → (CBSE AI 2014)
Answer:
(a) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂
(b) As₄ + 10Cl₄(excess) → 4AsCl₅
(c) P₄O₁₀ + 6H₄O → 4H₃PO₄
(d) Ca₃P₂ + 6H₂O → 2PH₃ + 3 Ca(OH)₂
(e) POCl₃ + 3H₂O → H₃PO₄ + 3HCl
(f) 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl
(g) 2Ag + PCl₅ → 2 AgCl + PCl₃

Question 25.
Complete the following reactions
(i) NaOH + Cl₂ →
hot and conc.
(ii) NH₃ + Cl₂ (excess) → (CBSE Delhi 2017)
(iii) NaNO₂ + HCl →
(iv) F₂(g) + H₂O(l) → (CBSE Delhi 2008)
(v) K₂CO₃ + HCl →
(vi) Cl₂ + H₂O → (CBSE Delhi 2017)
(vii) F₂ + 2Cl^– → (CBSE Delhi 2017)
Answer:
(i) 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
hot and conc.
(ii) NH₃ + 3Cl₂ (excess) → NCl₃ + 3HCl
(iii) 2NaNO₂ + 2HCl → 2NaCl + H₂O + NO₂ + NO
(iv) 2F₂(g) + 2H₂O(l) → 4H⁺(aq) + 4F^– (aq) + O₂(g)
(v) K₂CO₃ + 2HCl → 2KCl + CO₂ + H₂O
(vi) 2Cl₂ + 2H₂O → 4HCl + O₂
(vii) F₂ + 2Cl^– → 2F^– + Cl₂

Question 26.
Complete the following reactions:
(i) XeF₄ + SbF₅ → (CBSE Delhi 2012)
(ii)

(CBSE Delhi 2012, CBSE Al 2012, 2014)
(iii) XeF₂ + H₂O →
(iv) XeF₄ + H₂O →
(v) XeF₆ + H₂O → (CBSE Delhi 2012)
(vi) XeF₆ + 2H₂O → (CfiSE Delhi 2017)
(vii) XeF₆ + 3H₂O → (CBSE Delhi 2017)
Answer:
(i) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
(ii)

(iii) XeF₂ + H₂O → 2Xe + 4HF + O₂
(iv) XeF₄ + H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
(v) XeF₆ + H₂O → XeOF₄ + 2HF
(vi) XeF₆ + 2H₂O → XeO₂F₂ + 4HF
(vii) XeF₆ + 3H₂O → XeO₃ + 6HF

Question 27.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
Answer:
(a) (i) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is -ve) and increase in entropy ΔS is +ve). These two factors reinforce each other resulting in negative ΔG (ΔG = ΔH – TΔS) for its conversion to oxygen.

(ii) PCl₅ has trigonal bipyramidal structure and is not very stable. It splits up into more stable tetrahedral and octahedral structures which are stable as
PCl₅ ⇌ [PCl₄]⁺ [PCl₆]^–
Therefore, it exists as ionic.

(iii) Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

(b) Draw the structure of
(i) BrF₅
(ii) XeF₄
Answer:
(i)

Square Pyramidal

(ii)

OR

(i) Compare the oxidising action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂. Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Write the conditions to maximise the yield of H₂SO₄ by contact process.
Answer:
Conditions for maximum yield of H₂SO₄ by Contact Process:
(a) Low temperature (optimum temperature 720 K)
(b) High pressure (optimum pressure 2 bar)
(c) Presence of catalyst (V₂O₅ catalyst).

(iii) Arrange the following in the increasing order of property mentioned:
(a) H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)
(b) NH₃, PH₃, ASH₃, SbH₃, BiH₃ (Base strength) (CBSE 2016)
Answer:
(a) H₃PO₂ > H₃PO₃ > H₃PO₄
(b) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

Question 28.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
Answer:
(a) (i) In gaseous state, hydrogen halides are covalent. But in aqueous solution, they ionise and behave as acids. The acidic strength of these acids decreases in the order:
HI > HBr > HCl > HF
Thus, HF is the weakest acid and HI is the strongest acid among these hydrogen halides.

The above order of acidic strength is reverse of that expected on the basis of electronegativity. Fluorine is the most electronegative halogen, therefore, the electronegativity difference will be maximum in HF and should decrease gradually as we move towards iodine through chlorine and bromine.

Thus, HF should be most ionic in nature and consequently it should be strongest acid. Although many factors contribute towards the relative acidic strengths, the major factor is the bond dissociation energy. The bond dissociation energy decreases from HF to HI so that HF has maximum bond dissociation energy and HI has the lowest value.

Since H-I bond is weakest, it can be dissociated into H⁺ and I^– ions readily while HF can be dissociated with maximum difficulty. Thus, HI is the strongest acid while HF is the weakest acid among the hydrogen halides.

(ii) Oxygen molecule is held by weak van der Waals forces because of the small size and high electronegativity of oxygen. On the other hand, sulphur molecules do not exist as S₂ but form polyatomic molecules having eight atoms per molecule (S₈) linked by single bonds. Therefore, S atoms are strongly held together by intermolecular forces and its melting point is higher than that of oxygen. Hence, there is large difference in melting and boiling points of oxygen and sulphur.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation enthalpy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(b) Draw the structures of the following:
(i) CIF₃
(ii) XeF₄
Answer:
(i)

T – Shaped molecule

(ii)

Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
Answer:
White phosphorus is most reactive of all the allotropes because it is unstable due to angular strain on P₄ molecule with bond angle of 60°.

(ii) How are the supersonic jet aeroplanes responsible for the depletion of ozone layers?
Answer:
Nitrogen oxide emitted from exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near ozone layer, they are responsible for the depletion of ozone layer.

(iii) F₂ has lower bond dissociation enthalpy than Cl₂. Why?
Answer:
Because of small size of fluorineatoms, there are strong electron-electron repulsions between the lone pairs of electrons on F atoms. Hence, bond dissociation enthalpy of F₂ is lower than that of Cl₂.

(iv) Which noble gas is used in filling balloons for meteorological observations?
Answer:
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF₂ + PF₅ →
Answer:
XeF₂ + PF₅ → [XeF]⁺ [PF₆]^–

Question 29.
(a) Account for the following:
(i) Interhalogens are more reactive than pure halogens.
(ii) N₂ is less reactive at room temperature.
(iii) Reducing character increases from NH₃ to BiH₃.
Answer:
(i) Interhalogen compounds are more reactive than component halogens.
This is because covalent bond between dissimilar atoms in interhalogen compounds is polar and weaker than between similar atoms in halogens (except F-F). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(ii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol^-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(iii) The reducing character of hydrides of group 15 depends upon the stability of the hydride. On going down the group, the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases.

The greater unstability of a hydride, the greater is its reducing character. Since the stability of the group 15 hydrides decreases from NH₃ to BiH₃, hence reducing character increases.

(b) Draw the structures of the following:
(i) H₄P₂O₇ (Pyrophosphoric acid)
(ii) XeF₄
Answer:
(i)

(ii)

OR

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

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

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Answer:
Fluorine has lower bond dissociation enthalpy of F-F bond than Cl-Cl bond of Cl₂ and high enthalpy of hydration because of smaller size of F ion. As a result it has greater tendency to accept electron in solution and is stronger oxidising agent than chlorine.

(d) What happens when H₃PO₃ is heated?
Answer:

(e) Complete the equation:
PbS + O₃ → (CBSE 2015)
Answer:
PbS + 4O₃ → PbSO₄ + 4O₂

Question 30.
(a) Account for the following observations:
(i) SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
(ii) Chlorine water is a powerful bleaching agent.
(iii) Bi(V) is a stronger oxidising agent than Sb(V).
Answer:
(a) (f) S atom in SF₄ is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF₆ is protected by six F atoms. Thus attack by water molecules cannot take place easily.

(ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.
Cl₂ + H₂O → 2HCl + O

(iii) Due to inert pair effect Bi(V) can accept a pair of electrons to form more stable Bi (III) (+3 oxidation state of Bi is more stable than its +5 oxidation state).

(b) What happens when
(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO₂.
(ii) XeF₆ undergoes partial hydrolysis.
(Give the chemical equations involved).
Answer:
(i) Phosphorus undergoes disproportionation reaction to form phosphine gas.
P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

(ii) On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
XeF₆ + H₂O → XeOF₄ + 2HF

OR

(a) What inspired N.Bartlett for carrying out reaction between Xe and PtF₆?
Answer:
N. Bartlett first prepared a red compound O₂⁺PtF₆^–. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF₆.

(b) Arrange the following in the order of property indicated against each set:
(i) F₂, I₂, Br₂, Cl₂ (increasing bond dissociation enthalpy)
(ii) NH₃, ASH₃, SbH₃, BiH₃, PH₃ (decreasing base strength)
Answer:
(i) I₂ < F₂ < Br₂ < Cl₂
(ii) NH₃ > PH₃ > ASH₃ > SbH₃ > BiH₃

(c) Complete the following equations:
(i) Cl₂ + NaOH (cold and dilute) →
(ii) Fe³⁺ + SO₂ + H₂O →
(CBSE Sample Paper 2018)
Answer:
(i) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(ii) 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺

Question 31.
(a) Give reasons:
(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.
(ii) When Cl₂ reacts with excess of F₂, ClF₃ is formed and not FCl₃.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
Answer:
(i) In +3 oxidation state phosphorus tends to disproportionate to higher and lower oxidation states / Oxidation state of P in H₃PO₃ is +3 so it undergoes disproportionation but in H₃PO₄ it is +5 which is the highest oxidation state.

(ii) F cannot show positive oxidation state as it has highest electronegativity/ Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding due to pπ-pπ bonding whereas sulphur shows catenation. Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic and held by strong intermolecular forces.

(b) Draw the structures of the following:
(i) XeF₄
(ii) HClO₃
Answer:
(i)

(ii)

OR

(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas Intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
Answer:
(i) A = NO₂, B = N₂O₄
(ii)

(iii) NO₂ contains an odd electron. So it dimerises to give N₂O₄.

(b) Arrange the following In the decreasing order of their reducing character:
HF, HCl, HBr, HI
Answer:
HI > HBr > HCl > HF

Complete the following reaction:
XeF₄ + SbF₅ →, (CBSE 2018)
Answer:
(C) XeF₄ + SbF₅ → [XeF₃] [SbF₆]

Question 32.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF₂ undergoes hydrolysis?
Answer:
(a) 2NaOH + Cl₂ → NaCl + NaOCl + H₂O
(cold and dilute)
(b) 2XeF₂(s) + 2H₂O (l) → 2Xe (g) + 4HF(aq) + O₂(S)

(ii) Assign suitable reasons for the following:
(a) SF₆ Is inert towards hydrolysis.
(b) H₃PO₃ is diprotic.
(C) Out of noble gases only Xenon is known to form established chemical compounds.
Answer:
(a) Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are present as -OH groups, thus behaves as dibasic acid.

(c) Xe has least ionisation energy among the noble gases and hence it forms chemical compounds particularly with O₂ and F₂.

OR

(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F₂ and Cl₂.
Answer:
F₂ is stronger oxidising agent than Cl₂. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:

The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F₂ than for Cl₂.
Hence, F₂ is stronger oxidising agent than Cl₂.

(ii) Complete the following reactions :
(a) Cu + HNO₃(dilute) →
(b) Fe³⁺ + SO₂ + H₂O →
(c) XeF₄ + O₂F₂ → (CBSE 2018)
Answer:
(a) 3Cu + 8 HNO₃ (dilute) → 3Cu(NO₃)₂ + 2NO + 4H₃O
(b) 2Fe³⁺ + SO₃ + 2H₃O → 2 Fe²⁺ + SO₄^2- + 4H⁺
(c) XeF₄ + O₂F₂ → XeF₆ + O₂

Question 33.
A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified KMnO₄ (aq.) solution and reduces Fe³⁺ to Fe²⁺. Identify ‘A’ and ‘B’ and write the reactions involved. (CBSE 2019C)
Answer:
A = S₈ / Sulphur
S₈ + 8 O₂ → 8SO₂ / S + O₂ → SO₂
B = SO₂

Decolourises KMnO₄
2KMnO₄ + 5 SO₂ + 2H₄O → 2H₂SO₄ + 2MnSO₄ + K₂SO₄ / 2MnO₄^– + 5SO₂ + 2H₂O → 4H⁺ + 2Mn²⁺ + 5SO₄^2-
Reduces Fe³⁺ to Fe²⁺
2Fe³⁺ + SO₂ + 2 H₂O → 2 Fe²⁺ + SO₄^2- + 4H⁺

OR

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

(b) Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why?
Answer:
PCl^4- is not likely to exist because lone pair on P in PCl₃ can be donated to Cl⁺ and not to Cl^–. Phosphorus has 10e^– which cannot be accommodated in sp³ orbitals.

(c) Which aliotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur is thermally stable at room temperature. Its melting point is 385.8 K. All other varieties of sulphur change into this form on standing. It has low thermal and electrical conductivity.

(d) Write the formula of a compound of phosphorus which is obtained when cone. HNO₃ oxidises P₄.
Answer:
HNO₃ oxidises phosphorus to phosphoric acid. H₃PO₄ is formed
P₄ + 2OHNO₃ → 4H₃PO₄ + 2ONO₂ + 4H₂O

(e) Why does PCl₃ fume in moisture?
Answer:
PCl₃ fumes in moist air and reacts with water violently to form phosphorus acid. It hydrolyses in the presence of moisture to give fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl