Sainavle

Question 1. If the bond energies of H-H, Br-Br, and H-Br are 433, 192 and 364 kJ mol^-1 respectively, the ΔH° for the reaction \(\mathrm{H}_{2(g)}+\mathrm{Br}_{2(g)} \rightarrow 2 \mathrm{HBr}_{(g)}\) is

1. -261 kJ
2. +103kJ
3. +261 kJ
4. -103 kJ

Answer: 4. -103 kJ

⇒ \(\begin{array}{ccc}
\mathrm{H}-\mathrm{H}+\mathrm{Br}-\mathrm{Br} & \rightarrow & 2 \mathrm{H}-\mathrm{Br} \\
433+192 & & 2 \times 364 \\
=625 & & =728
\end{array}\)

Net energy released = 728 – 625 = 103 kJ

i.e. ΔH° = – 103 kJ

Question 2. For irreversible expansion of an ideal gas under isothermal conditions, the correct option is

1. \(\Delta U \neq 0, \Delta S_{\text {total }}=0\)
2. \(\Delta U=0 \quad \Delta S_{\text {total }}=0\)
3. \(\Delta U \neq 0, \Delta S_{\text {total }} \neq 0\)
4. \(\Delta U=0, \Delta S_{\text {total }} \neq 0\)

Answer: 4. \(\Delta U=0, \Delta S_{\text {total }} \neq 0\)

For reversible and irreversible expansion for an ideal gas under isothermal conditions, ΔU = 0, but ΔS_total i.e., \Delta \(S_{\text {sys }}+\Delta S_{\text {Surr }}\), is not zero for an irreversible process.

Read And Learn More Class 11 Chemistry Solutions

Question 3. For the reaction, \(2 \mathrm{Cl}_{(g)} \rightarrow \mathrm{Cl}_{2(g)}\), the correct option is

1. \(\Delta_r H>0\) and \(\Delta_r S>0\)
2. \(\Delta_r H>0\) and \(\Delta_r S<0\)
3. \(\Delta_r H<0\) and \(\Delta_r S>0\)
4. \(\Delta_r H<0\) and \(\Delta_r S<0\)

Answer: 4. \(\Delta_r H<0\) and \(\Delta_r S<0\)

In the reaction, \(2 \mathrm{Cl}_{(g)} \rightarrow \mathrm{Cl}_{2(g)}\), the randomness decreases as 2 moles of \(\mathrm{Cl}_{(\mathrm{g})}\) are converted to 1 mole of \(\mathrm{Cl}_{2(g)}\), thus, \(\Delta_r S<0\).

This is an exothermic reaction, thus, \(\Delta_r H<0\).

Question 4. In which case change in entropy is negative?

1. \(2 \mathrm{H}_{(g)} \rightarrow \mathrm{H}_{2(g)}\)
2. Evaporation of water
3. Expansion of a gas at a constant temperature
4. Sublimation of solid to gas

Answer: 1. \(2 \mathrm{H}_{(g)} \rightarrow \mathrm{H}_{2(g)}\)

If Δn_g < 0 then ΔS < 0

PSEB Class 11 Chemistry Thermodynamics MCQs

Question 5. For a given reaction, ΔH = 35.5 kJ mol^-1 and ΔS = 83.6 J K^-1 mol^-1. The reaction is spontaneous at (Assume that ΔH and ΔS do not vary with temperature.)

1. T > 425 K
2. All temperatures
3. T > 298 K
4. T < 425 K

Answer: 1. T > 425 K

For a spontaneous reaction, ΔG<0 i.e., ΔH – TΔS < 0

T > \(\frac{\Delta H}{\Delta S} ; T>\left(\frac{35.5 \times 1000}{83.6}=424.6\right) \approx 425 \mathrm{~K}\)

∴ T>425K

Question 6. For a sample of perfect gas when its pressure is changed isothermally from p_i to p_j, the entropy change is given by

1. \(\Delta S=n R \ln \left(\frac{p_f}{p_i}\right)\)
2. \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)
3. \(\Delta S=n R T \ln \left(\frac{p_f}{p_i}\right)\)
4. \(\Delta S=R T \ln \left(\frac{p_i}{p_f}\right)\)

Answer: 2. \(\Delta S=n R \ln \left(\frac{p_i}{p_f}\right)\)

For an ideal gas undergoing reversible expansion, when the temperature changes from T_i to T_f and pressure changes from p_i to p_f

∴ \(\Delta S=n C_p \ln \frac{T_f}{T_i}+n R \ln \frac{p_i}{p_f}\)

For an isothermal process, \(T_i=T_f\) so, In 1=0

∴ \(\Delta S=n R \ln \frac{p_i}{p_f}\)

Question 7. The correct thermodynamic conditions for the spontaneous reaction at all temperatures is

1. ΔH < 0 and ΔS > 0
2. ΔH < 0 and ΔS < 0
3. ΔH < 0 and ΔS = 0
4. ΔH > 0 and ΔS < 0

Answer: 1. ΔH < 0 and ΔS > 0; and 3. ΔH < 0 and ΔS = 0

ΔG = ΔH – TΔS

If ΔH <0 and ΔS>0

ΔG =(-ve)-T(+ve)

then at all temperatures, ΔG = -ve, spontaneous reaction

I fΔH<0 and ΔS=0

ΔG = (-ve) – T(0) = -ve at all temperatures

Question 8. Consider the following liquid-vapour equilibrium. Liquid ⇔ Vapour Which of the following relations is correct?

1. \(\frac{d \ln P}{d T^2}=\frac{-\Delta H_v}{T^2}\)
2. \(\frac{d \ln P}{d T}=\frac{\Delta H_v}{R T^2}\)
3. \(\frac{d \ln G}{d T^2}=\frac{\Delta H_v}{R T^2}\)
4. \(\frac{d \ln P}{d T}=\frac{-\Delta H_v}{R T}\)

Answer: 2. \(\frac{d \ln P}{d T}=\frac{\Delta H_v}{R T^2}\)

This is Clausius-Clapeyron equation’

Question 9. Which of the following statements is correct for the spontaneous adsorption of a gas?

1. ΔS is negative and, therefore ΔH should be highly positive.
2. ΔS is negative and therefore, ΔH should be highly negative.
3. ΔS is positive and therefore, ΔH should be negative.
4. ΔS is positive and therefore, ΔH should also be highly positive.

Answer: 2. ΔS is negative and therefore, ΔH should be highly negative.

Using Gibbs-Helmholtz equation, ΔG = ΔH- TΔS

During the adsorption of a gas, entropy decreases i.e ΔS < 0

For spontaneous adsorption, ΔG should be negative, which is possible when ΔH is highly negative.

PSEB Class 11 Chemistry Thermodynamics MCQs

Question 10. For the reaction, \(\mathrm{X}_2 \mathrm{O}_{4(l)} \longrightarrow 2 \mathrm{XO}_{2(\mathrm{~g})}\) ΔU= 2.1 kcal, ΔS = 20 cal K^-1 at 300 K Hence, ΔG is

1. 2.7 kcal
2. -2.7 kcal
3. 9.3 kcal
4. -9.3 kcal

Answer: 2. -2.7 kcal

⇒ \(\Delta H=\Delta U+\Delta n_g R T\)

Given, \(\Delta U=2.1 \mathrm{kcal}, \Delta n_g=2\)

R \(=2 \times 10^{-3} \mathrm{kcal}, T=300 \mathrm{~K}\)

∴ \(\Delta H=2.1+2 \times 2 \times 10^{-3} \times 300=3.3 \mathrm{kcal}\)

Again, \(\Delta G=\Delta H-T \Delta S\)

Given, \(\Delta S=20 \times 10^{-3} \mathrm{kcal} \mathrm{K}^{-1}\)

On putting the values of \(\Delta H\) and \(\Delta S\) in the equation, we get

Δ\(G=3.3-300 \times 20 \times 10^{-3}\)

= \(3.3-6 \times 10^3 \times 10^{-3}=-2.7 \mathrm{kcal}\)

Question 11. A reaction having equal energies of activation for forward and reverse reactions has

1. ΔH = 0
2. ΔH = ΔG = ΔS = 0
3. ΔS = 0
4. ΔG = 0

Answer: 1. ΔH = 0

ΔH = \(\left(E_a\right)_f-\left(E_a\right)_b=0\)

Question 12. In which of the following reactions, standard reaction entropy change (ΔS°) is positive and standard Gibbs energy change (ΔG°) decrease sharply with increasing temperature?

1. \(\mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}\)
2. \(\mathrm{CO}_{(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}\)
3. \(\mathrm{Mg}_{(s)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{MgO}_{(s)}\)
4. \(\frac{1}{2} \mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \frac{1}{2} \mathrm{CO}_{2(g)}\)

Answer: 1. \(\mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}\)

⇒ \(\mathrm{C}_{\text {(graphite) }}+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}\)

⇒ \(\Delta n_g=1-\frac{1}{2}=\frac{1}{2}\)

As the amount of gaseous substance is increasing in the product side thus, \(\Delta S\) is positive for this reaction.

And we know that \(\Delta G=\Delta H-T \Delta S\)

As \(\Delta S\) is positive, thus increase in temperature will make the term \((-T \Delta S)\) more negative and \(\Delta G\) will decrease.

Question 13. The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is

1. 10.52 cal/(mol K)
2. 21.04 cal/(mol K)
3. 5.260 cal/(mol K)
4. 0.526 cal/(mol K)

Answer: 3. 5.260 cal/(mol K)

⇒ \(\Delta H_{\text {fus }}=1.435 \mathrm{kcal} / \mathrm{mol}\)

⇒ \(\Delta S_{f u s}=\frac{\Delta H_{f u s}}{T_{f u s}}=\frac{1.435 \times 10^3}{273}=5.26 \mathrm{cal} /(\mathrm{mol} \mathrm{K})\)

Question 14. If the enthalpy change for the transition of liquid water to steam is 30 kJ mol^-1 at 27°C, the entropy change for the process would be

1. 10 J mol^-1 K^-1
2. 1.0 J mol^-1 K^-1
3. 0.1 J mol^-1 K^-1
4. 100 J mol^-1 K^-1

Answer: 4. 100 J mol^-1 K^-1

We know that ΔG = ΔH – TΔS

0 = ΔH – TΔS ∴ (ΔG = 0 as transition of \(\mathrm{H}_2 \mathrm{O}_{(t)} \rightleftharpoons \mathrm{H}_2 \mathrm{O}_{(v)}\) is at equilibrium)

∴ \(\Delta S=\frac{\Delta H}{T}=\frac{30 \times 10^3}{300}=100 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\)

Thermodynamics Multiple Choice Questions PSEB Class 11

Question 15. Standard entropies of X₂, Y₂ and XY₃ are 60, 40 and 50 J K^-1 mol^-1 respectively. For the reaction \(1 / 2 X_2+3 / 2 Y_2 \rightleftharpoons X Y_3, \Delta H=-30 \mathrm{~kJ}\), to be at equilibrium, the temperature should be

1. 750 K
2. 1000 K
3. 1250 K
4. 500 K

Answer: 1. 750 K

Given reaction is \(\frac{1}{2} X_2+\frac{3}{2} Y_2 \rightleftharpoons X Y_3\)

We know, \(\Delta S^{\circ}=\Sigma S_{\text {products }}^{\circ}-\Sigma S_{\text {reactants }}^{\circ}\)

= \(50-\left(\frac{1}{2}(60)+\frac{3}{2}(40)\right)\)

= \(50-(30+60)=-40 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

At equilibrium \(\Delta G^{\circ}=0\) \(\Delta H^{\circ}=T \Delta S^{\circ}\)

∴ T = \(\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{-30 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{-40 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}}=750 \mathrm{~K}\)

Question 16. For the vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol^-1 and 108.8 J K^-1 mol^-1 respectively. The temperature when Gibbs’ energy change (ΔG) for this transformation will be zero, is

1. 273.4 K
2. 393.4 K
3. 373.4 K
4. 293.4 K

Answer: 3.  373.4 K

According to Gibbs equation, ΔG = ΔH- TΔS

when ΔG=0, ΔH= TΔS

Given, ΔH = 40.63 kJ mol^-1 = 40.63 x 10³ J mol^-1

ΔS = 108.8 J K^-1 mol^-1

∴ T = \(\frac{\Delta H}{\Delta S}=\frac{40.63 \times 10^3}{108.8}=373.43 \mathrm{~K}\)

Question 17. The values of ΔH and ΔS for the reaction, \(\mathrm{C}_{\text {(graphite) }}+\mathrm{CO}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}_{(g)}\) are 170 kJ and 170 J K^-1, respectively. This reaction will be spontaneous at

1. 910 K
2. 1110 K
3. 510 K
4. 710 K

Answer: 2. 1110 K

For the reaction to be spontaneous, ΔG = -ve

Given : ΔH = 170 kJ = 170 x 10³ J, ΔS = 170 J K^-1

Applying, ΔG = ΔH – TΔS, the value of ΔG = -ve only when TΔS > ΔH, which is possible only when T = 1110 K

∴ ΔG = 170 x 10³ – (1110 x 170) = -18700J

Thus, the reaction is spontaneous at T = 1110 K

Question 18. For the gas phase reaction, \(\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}\) which of the following conditions are correct?

1. ΔH < 0 and ΔS < 0
2. ΔH > 0 and ΔS < 0
3. ΔH = 0 and ΔS < 0
4. ΔH > 0 and ΔS > 0

Answer: 4. ΔH > 0 and ΔS > 0

Gas phase reaction, \(\begin{aligned}
& \mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(\mathrm{~g})} \\
& \Delta H=\Delta E+\Delta n_g R T
\end{aligned}\)

Δn_g = Change in number of moles of products and reactants species.

Since Δn_g = +ve, hence ΔH = +ve also one inole of PCl₅ is dissociated into two moles of PCl₃ and Cl₂ in the same phase.

Therefore, ΔS = S_Product– S_reactants

ΔS = +ve.

Thermodynamics Multiple Choice Questions PSEB Class 11

Question 19. Identify the correct statement for change of Gibbs energy for a system (ΔG_system) at constant temperature and pressure.

1. If ΔG_system < 0, the process is not spontaneous.
2. If AΔG_system > 0, the process is spontaneous.
3. If ΔG_system = 0, the system has attained equilibrium.
4. If ΔG_system = 0, the system is still moving in a particular direction.

Answer: 3. If ΔG_system = 0, the system has attained equilibrium.

The criteria for spontaneity of a process in terms of ΔG is as follows:

If ΔG is negative, the process is spontaneous’

If ΔG is positive, the process does not occur in the forward direction. It may occur in the backward direction

If ΔG is zero, the system is in equilibrium

Question 20. The enthalpy and entropy change for the reaction: \(\mathrm{Br}_{2(l)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{BrCl}_{(\mathrm{g})}\) are 30 kJ mol^-1 and 105 J K^-1 mol^-1 respectively. The temperature at which the reaction will be in equilibrium is

1. 300 K
2. 285.7 K
3. 273 K
4. 450 K

Answer: 2. 285.7 K

⇒ \(\mathrm{Br}_{2(t)}+\mathrm{Cl}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{BrCl}_{(\mathrm{g})}\)

Δ\(H=30 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta S=105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

Δ\(\mathrm{S}=\frac{\Delta H}{T} \text { i.e. } 105=\frac{30}{T} \times 1000\)

∴ T = \(\frac{30 \times 1000}{105}=285.7 \mathrm{~K}\)

Question 21. Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?

1. Exothermic and increasing disorder
2. Exothermic and decreasing disorder
3. Endothermic and increasing disorder
4. Endothermic and decreasing disorder

Answer: 1. Exothermic and increasing disorder

For spontaneous reaction, ΔH= -ve, ΔS = +ve

Spontaneity depends upon both critical minimum energy and maximum randomness disorderness.

Question 22. A reaction occurs spontaneously if

1. TΔS < ΔH and both ΔH and ΔS are +ve
2. TΔS > ΔH and ΔH is +ve and ΔS is -ve
3. TΔS > ΔH and both ΔH and ΔS are +ve
4. TΔS = ΔH and both ΔT and ΔS are +ve.

Answer: 3. TΔS > ΔH and both ΔH and ΔS are +ve

ΔG = ΔH-TΔS

ΔG = -ve for spontaneous reaction

When ΔS = +ve, ΔH= +ve and TΔS > ΔH ⇒ ΔG = -ve

Question 23. The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are -382.64 kJ mol^-1 and -145.6 J mol^-1, respectively. Standard Gibbs’ energy change for the same reaction at 298 K is

1. -221.1 kJ mol^-1
2. -339.3 kJ mol^-1
3. -439.3 kJ mol^-1
4. -523.2 kJ mol^-1

Answer: 2. -339.3 kJ mol^-1

Δ\(G=\Delta H-T \Delta S\)

= \(-382.64-298\left(\frac{-145.6}{1000}\right)\)

= \(-382.64+43.38=-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Question 24. Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is

1. \(\Delta S_{\text {system }}+\Delta S_{\text {surroundings }}>0\)
2. \(\Delta S_{\text {system }}-\Delta S_{\text {surroundings }}>0\)
3. \(\Delta S_{\text {system }}>0\) only
4. \(\Delta S_{\text {surroundings }}>0\) only.

Answer: 1. \(\Delta S_{\text {system }}+\Delta S_{\text {surroundings }}>0\)

For spontaneous process, \(\Delta S_{\text {total }}>0\).

∴ \(\Delta S_{\text {sys }}+\Delta S_{\text {surt }}>0\)

PSEB Class 11 Chemistry Chapter 6 MCQs with Answers

Question 25. What is the entropy change (in J K^-1 mol^-1) when one mole of ice is converted into water at 0°C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol^-1 at 0°C.)

1. 20.13
2. 2.013
3. 2.198
4. 21.98

Answer: 4. 21.98

S = \(\frac{q_{\text {rev }}}{T}=\frac{6000}{273}=21.978 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

Question 26. The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm^-3, respectively. If the standard free energy difference (ΔG°) is equal to 1895 J mol^-1, the pressure at which graphite will be transformed into diamond at 298 K is

1. 9.92 x 10⁸ Pa
2. 9.92 x 10⁷ Pa
3. 9.92 x 10⁶ Pa
4. 9.92 x 10⁵ Pa

Answer: 1. 9.92 x 10⁸ Pa

ΔG° = – PΔV = Work done

ΔV \(=\left(\frac{12}{3.31}-\frac{12}{2.25}\right) \times 10^{-3} \mathrm{~L}=-1.71 \times 10^{-3} \mathrm{~L} \)

Δ\(G^{\circ}=\text { Work done }=-\left(-1.71 \times 10^{-3}\right) \times P \times 101.3 \mathrm{~J}\)

P = \(\frac{1895}{1.71 \times 10^{-3} \times 101.3}=10.93 \times 10^3 \mathrm{~atm}\)

=11.08 x 10⁸ Pa= 9.92 x 10⁸ pa (1 atm = 101325 pa)

Question 27. The unit of entropy is

1. J K^-1 mol^-1
2. J mol^-1
3. J^-1 K^-1 mol^-1
4. J K mol^-1

Answer: 1. J K^-1 mol^-1

Entropy change \((\Delta S)\) is given by \(\Delta S=\frac{q_{r e v}}{T}\)

∴ Unit of entropy \(=\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\)

Question 28. 2 moles of ideal gas at 27°C temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change. (R = 2 cal/mol K)

1. 92.1
2. 0
3. 4
4. 9.2

Answer: 4. 9.2

The change of entropy dS = \(\frac{q_{\text {rev }}}{T}\)

From the first law of thermodynamics, \(d q=d U+P d V=C_V d T+P d V\)

⇒ \(\frac{d q}{T}=C_V \frac{d T}{T}+\frac{P}{T} d V\)

⇒  \(\frac{d q}{T}=C_V \frac{d T}{T}+\frac{R d V}{V}\) (For 1mole of a gas \(\frac{P}{T}=\frac{R}{V}\))

dS = \(C_V \frac{d T}{T}+R \frac{d V}{V}\)

⇒ \(\Delta S=C_V \ln \frac{T_2}{T_1}+R \ln \frac{V_2}{V_1}\) for one mole of ideal gas

Here \(T_2=T_1=27^{\circ} \mathrm{C}=300 \mathrm{~K}\)

ln \(\frac{T_2}{T_1}=0\)

∴ \(\Delta S=R \ln \frac{V_2}{V_1}=2 \ln \frac{20}{2}=2 \ln 10=4.605\)

∴ ΔS=4.605 cal/mol K

Entropy change for 2 moles of gas = 2 x 4.605 cal/K = 9.2 cal/K

PSEB Class 11 Chemistry Chapter 6 MCQs with Answers

Question 29. \(\mathrm{PbO}_2 \rightarrow \mathrm{PbO}; \Delta G_{298}<0\) \(\mathrm{SnO}_2 \rightarrow \mathrm{SnO}; \Delta G_{298}>0\) The most probable oxidation state of Pb and Sn will be

1. Pb⁴⁺, Sn⁴⁺
2. Pb⁴⁺, Sn²⁺
3. Pb²⁺, Sn²⁺
4. Pb²⁺, Sn⁴⁺

Answer: 4. Pb²⁺, Sn⁴⁺

The sign and magnitude of Gibbs free energy is a criterion of spontaneity for a process.

When ΔG > 0 or +ve, it means G_products >G_reactants

as ΔG = G_products – G_reactants

the reaction will not take place spontaneously, i.e. the reaction should be spontaneous in the reverse direction.

ΔG < 0 or -ve, the reaction or change occurs spontaneously.

Question 30. Cell reaction is spontaneous when

1. ΔG° is negative
2. ΔG° is positive
3. ΔE°_red is positive
4. ΔE°_red is negative.

Answer: 1. ΔG° is negative

For a cell reaction to be spontaneous, ΔG° should be negative. As ΔG° = -nFE°_cell, the value will be -ve only when E°_cell is +ve.

Question 31. Identify the correct statement regarding entropy.

1. At absolute zero temperature, the entropy of all crystalline substances is taken to be zero.
2. At absolute zero temperature, the entropy of a perfectly crystalline substance is +ve.
3. At absolute zero temperature, the entropy of a perfectly crystalline substance is taken to be zero.
4. At 0°C, the entropy of a perfectly crystalline substance is taken to be zero.

Answer: 3. At absolute zero temperature, the entropy of a perfectly crystalline substance is taken to be zero.

The entropy of a substance increases with an increase in temperature. However, at absolute zero, the entropy of a perfectly crystalline substance is taken as zero, which is also called as third law of thermodynamics.

Question 32. Following reaction occurring in an automobile \(2 \mathrm{C}_8 \mathrm{H}_{18(\mathrm{~g})}+25 \mathrm{O}_{2(\mathrm{~g})} \rightarrow 16 \mathrm{CO}_{2(\mathrm{~g})}+18 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}\) The sign of ΔH, ΔS and ΔG would be

1. -,+,+
2. +,+,-
3. +,-,+
4. -,+,-

Answer: 4. -,+,-

1. The given reaction is a combustion reaction, therefore ΔH is less than 0. Hence, ΔH is negative.
2. Since there is an increase in the number of moles of gaseous products, therefore ΔS is positive.
3. Since the reaction is spontaneous, therefore ΔG is negative.

Question 1. How many electrons can fit in the orbital for which n = 3 and l = 1?

1. 2
2. 6
3. 10
4. 14

Answer: 1. 2

For n=3 and l = 1, the subshell is 3p and a particular 3p orbital can accommodate only 2 electrons.

Question 2. Which of the following pairs of d-orbitals will have electron density along the axes?

1. \(d_{z^2}, d_{x z}\)
2. \(d_{x z}, d_{y z}\)
3. \(d_{z^2}, d_{x^2-y^2}\)
4. \(d_{x y} d_{x^2-y^2}\)

Answer: 3. \(d_{z^2}, d_{x^2-y^2}\)

∴ \(d_{x^2-y^2}\) and \(d_z^2\) orbitals have electron density along the axes while d_xy, d_yz, and dxz orbitals have electron density in between the axes.

Read And Learn More Class 11 Chemistry Solutions

Question 3. Two electrons occupying the same orbital are distinguished by

1. Azimuthal quantum number
2. Spin quantum number
3. Principal quantum number
4. Magnetic quantum number.

Answer: 2. Spin quantum number

For the two electrons occupying the same orbital values of n, I and m_l are the same but m_s is different, _-31, i.e., \(\frac{1}{2}\) and –\(\frac{1}{2}\)

Question 4. Which is the correct order of increasing energy of the listed orbitals in the atom of titanium? (Atomic number Z = 22)

1. 4s 3s 3p 3d
2. 3s 3p 3d 4s
3. 3s 3p 4s 3d
4. 3s 4s 3p 3d

Answer; 3. 3s 3p 4s 3d

Ti(22) : ls² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d²

∴ The order of increasing energy is 3s,3p,4s,3d.

Read And Learn More Class 11 Chemistry Solutions

PSEB Class 11 Chemistry Chapter 2 Structure of Atom MCQs

Question 5. The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following?

1. d-electrons in Fe (Z = 26)
2. p-electrons in Ne (Z = 10)
3. s-electrons in Mg (Z = 12)
4. p-electrons in Cl (Z= 17)

Answer: 4. p-electrons in Cl (Z= 17)

Number of d-electrons in Fe²⁺ = 6

Number of p-electrons in Cl = 11

Question 6. The angular momentum of an electron in the ‘d’ orbital is equal to

1. 2√3 h
2. h
3. √6 h
4. √2 h

Answer: 3. √6 h

Angular momentum = \(\sqrt{l(l+1)} \hbar\)

For d-orbital, l = 2

Angular momentum = \(\sqrt{2(2+1)} \hbar=\sqrt{6} \hbar\)

Question 7. What is the maximum number of orbitals that can be identified with the following quantum numbers? n = 3, l= 1, m_l = 0

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Only one orbital, 3p_z has the following set of quantum numbers, n = 3, l = 1, and m_l = 0.

Question 8. What is the maximum number of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1 and m = -1

1. 4
2. 2
3. 10
4. 6

Answer: 2. 2

The orbital associated with n = 3, l = 1 is 3p. One orbital (with m= -1) of 3p-subshell can accommodate a maximum of 2 electrons.

Question 9. The outer electronic configuration of Gd (Atomic number 64) is

1. \(4 f^5 5 d^4 6 s^1\)
2. \(4 f^7 5 d^1 6 s^2\)
3. \(4 f^3 5 d^5 6 s^2\)
4. \(4 f^4 5 d^5 6 s^1\)

Answer: 2. \(4 f^7 5 d^1 6 s^2\)

The electronic configuration of ₆₄Gd is [Xe]4f⁷ 5d¹ 6s²

Question 10. The maximum number of electrons in a subshell with l = 3 and n = 4 is

1. 14
2. 16
3. 10
4. 12

Answer: 1. 14

l = 3 and n= 4 represents 4f. So, total number of electrons in a subshell = 2(2l + 1) = 2(2 x 3 + 1) = 14 eiectrons.

Hence, the f-subshell can contain a maximum of 14 electrons.

Structure of Atom Multiple Choice Questions PSEB Class 11

Question 11. The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is

1. 5, 1,1,+1/2
2. 6, 0, 0,+1/2
3. 5, 0, 0, +1/2
4. 5,1, 0, +1/2

Answer: 3. 5, 0, 0, +1/2

Rb(37): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹

For 5s, n =5,l=0,m=0, s = +1/2 or -1/2

Question 12. The orbital angular momentum of a p-electron is given as

1. \(\frac{h}{\sqrt{2} \pi}\)
2. \(\sqrt{3} \frac{h}{2 \pi}\)
3. \(\sqrt{\frac{3}{2}} \frac{h}{\pi}\)
4. \(\sqrt{6} \frac{h}{2 \pi}\)

Answer: 1. \(\frac{h}{\sqrt{2} \pi}\)

Orbital angular momentum (m) = \(\sqrt{l(l+1)} \frac{h}{2 \pi}\)

For p-electrons; l = 1

Thus, m = \(\sqrt{1(1+1)} \frac{h}{2 \pi}=\frac{\sqrt{2} h}{2 \pi}=\frac{h}{\sqrt{2} \pi}\)

Question 13. The total number of atomic orbitals in the fourth energy level of an atom is

1. 8
2. 16
3. 32
4. 4

Answer: 2. 16

The total number of atomic orbitals in any energy level is given by n².

Question 14. If n = 6, the correct sequence for the filling of electrons will be

1. ns → (n – 2)f → (n – 1)d → np
2. ns → (n – 1 )d  → (n – 2)f → np
3. ns → (n – 2)f → np → (n -1)d
4. ns → np → (n – 1)d → (n – 2)f

Answer: 1. ns → (n – 2)f →(n – 1)d → np

Question 15. The maximum number of electrons in a subshell of an atom is determined by the following

1. 2l +1
2. 4l – 2
3. 2n²
4. 4l + 2

Answer: 4. 4l+ 2

For a given shell, n the number of subshells = (2l + 1)

Since each orbital can accommodate 2 electrons of opposite spin, so maximum number of electrons in a subshell = 2(2l+1) =4l +2.

Question 16. Which of the following is not a permissible arrangement of electrons in an atom?

1. n = 5, l = 3, m = 0, s = +1/2
2. n = 3, l = 2, m = -3, s = -1/2
3. n – 3, l = 2, m = -2, s = -1/2
4. n = 4, l = 0, m = 0, s = -1/2

Answer: 2. n = 3, l = 2, m = -3, s = -1/2

In an atom, for any value of ru, the values of l = 0 to (n-1).

For a given value of l, the values of m_l = -l to 0 to +l and the value of s = +1/2 or -1/2.

In option (2), l = 2 and m_l = -3

This is not possible, as values of m7 which are possible for l = 2 are -2, -1,0, +1 and +2 only

Structure of Atom Multiple Choice Questions PSEB Class 11

Question 17. The orientation of an atomic orbital is governed by

1. Principal quantum number
2. Azimuthal quantum number
3. Spin quantum number
4. Magnetic quantum number.

Answer: 4. Magnetic quantum number.

The principal quantum number represents the name, size, and energy of the shell to which the electron belongs.

The azimuthal quantum number describes the spatial distribution of electron cloud and angular momentum. The magnetic quantum number describes the orientation or distribution of the electron cloud.

The spin quantum number represents the direction of electron spin around its own axis.

Question 18. In a given atom no two electrons can have the same values for all four quantum numbers. This is called

1. Hunds Rule
2. Aufbau principle
3. Uncertainty principle
4. Pauli’s Exclusion Principle.

Answer: 4. Pauli’s Exclusion principle.

This is a Pauli’s exclusion principle

Question 19. The order of filling of electrons in the orbitals of an atom will be

1. 3d, 4s, 4p, 4d, 5s
2. 4s, 3d, 4p, 5s, 4d, 4s
3. 5s, 4p, 3d, 4d, 5s
4. 3d,4p,4s,4d,5s

Answer: 2. 4s, 3d, 4p, 5s, 4d, 4s

The higher the value of n +l) for an orbital, the higher is its energy. However, if two different types of orbitals have the same value of (n +l), the orbital with the lower value of n has lower energy

Question 20. The electronic configuration of Cu (atomic number 29) is

1. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 3 d^9\)
2. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\)
3. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 4 p^6 5 s^2 5 p^1\)
4. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 4 s^2 4 p^6 3 d^3\)

Answer: 2. \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\)

The electronic configuration of Cu⁺ is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1\)

Question 21. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are

1. 2
2. 4
3. 6
4. 8

Answer: 3. 6

n=2, l = 1

It means 2p-orbitals.

Total no. of electrons that can be accommodated in all the 2p orbitals = 6

PSEB Class 11 Chemistry Structure of Atom MCQs with Answers

Question 22. An ion has 18 electrons in the outermost shell, it is

1. Cu⁺
2. Th⁴⁺
3. Cs⁺
4. K⁺

Answer: 1. Cu⁺

Cu⁺ ion has 18 electrons in its outermost shell. The electronic configuration of Cu⁺ is \(1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10}\)

Question 23. The number of unpaired electrons in N²⁺ is/are

1. 2
2. 0
3. 1
4. 3

Answer: 3. 1

N²⁺ = 1s²2s³2p¹

∴ No. of unpaired electrons = 1

Question 24. The maximum number of electrons in a subshell is given by the expression

1. 4l – 2
2. 4l + 2
3. 2l + 2
4. 2n²

Answer: 2. 4l + 2

No. of orbitals in a subshell = 2l + l

No. of electrons = 2(2l + 1) = 4l + 2

Question 25. The number of spherical nodes in 3p orbitals is/are

1. One
2. Three
3. None
4. Two

Answer: 1. One

No. of radial nodes in 3p-orbita = n – l – 1 = 3 – 1 -1 = 1