Electrochemistry Multiple Choice Questions and Answers

Question 1. The following limiting molar conductivities are given:

• \(\lambda_{m\left(\mathrm{H}_2 \mathrm{SO}_4\right)}^{\circ}=x \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)
• \(\lambda_{m\left(\mathrm{~K}_2 \mathrm{SO}_4\right)}^{\circ}=y \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
• \(\lambda_{m\left(\mathrm{CH}_3 \mathrm{COOK}\right)}^{\circ}=z \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)
• \(\lambda_m^{\circ}\) (in S \(\mathrm{cm}^2 \mathrm{~mol}^{-1}\)) for \(\mathrm{CH}_3 \mathrm{COOH}\) will be

1. \(x-y+2 z\)
2. \(x+y-z\)
3. \(x-y+z\)
4. \(\frac{(x-y)}{2}+z\)

Answer: 4. \(\frac{(x-y)}{2}+z\)

According to Kohlrausch’s law, \(\lambda_m^{\circ}\) for \(\mathrm{CH}_3 \mathrm{COOH}=\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}+\lambda_{\mathrm{H}^{+}}^{\circ}\)

⇒ \(\lambda^{\circ}\) for \(\mathrm{H}_2 \mathrm{SO}_4=2 \lambda_{\mathrm{H}^{+}}^{\circ}+\lambda_{\mathrm{SO}_4^{2-}}^{\circ}=x \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)……(1)

⇒ \(\lambda^{\circ}\) for \(\mathrm{K}_2 \mathrm{SO}_4=2 \lambda_{\mathrm{K}^{+}}^{\circ}+\lambda_{\mathrm{SO}_4^{2-}}^{\circ}=y \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)…..(2)

⇒ \(\lambda^{\circ}\) for \(\mathrm{CH}_3 \mathrm{COOK}=\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}+\lambda_{\mathrm{K}^{+}}^{\circ}=z \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)…(3)

On adding equation (1) and 2 x (3) and subtracting (2), we get

⇒ \(2 \lambda_{\mathrm{H}^{+}}^{\circ}+\lambda_{\mathrm{SO}_4^{2-}}^{\circ}+2 \lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}+2 \lambda_{\mathrm{K}^{+}}^{\circ}-2 \lambda_{\mathrm{K}^{+}}^{\circ}-\lambda_{\mathrm{SO}_4^{2-}}^{\circ}=x+2 z-y\)

⇒ \(2 \lambda_{\mathrm{H}^{+}}^{\circ}+2 \lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}=x+2 z-y\)

⇒ \(\lambda_{\mathrm{H}^{+}}^{\circ}+\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^{\circ}=\frac{(x-y)}{2}+z\)

Read And Learn More Class 12 Chemistry MCQs

Question 2. The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 x 10^-3 S cm^-1 at 298 K is

1. 2.88 S cm²/mol
2. 11.52 S cm²/mol
3. 0.086 S cm²/mol
4. 28.8 S cm²/mol

Answer: 2. 11.52 S cm²/mol

⇒ \(\Lambda_m=\frac{\kappa \times 1000}{\mathrm{Molarity}(M)}\)

= \(\frac{5.76 \times 10^{-3} \mathrm{Scm}^{-1} \times 1000}{0.5 \mathrm{~mol} \mathrm{~cm}^{-3}}=11.52 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

PSEB Class 12 Chemistry Electrochemistry MCQs

Question 3. At 25 °C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm^-1 cm² mol^-1 and at infinite dilution its molar conductance is 238 ohm^-1 cm² mol^-1. The degree of ionization of ammonium hydroxide at the same concentration and temperature is

1. 4.008%
2. 40.800%
3. 2.080%
4. 20.800%

Answer: 1. 4.008%

Degree of dissociation \((\alpha)=\frac{\text { Molar conductivity at conc. } C\left(\Lambda_m^c\right)}{\text { Molar conductivity at infinite dilution }\left(\Lambda_m^{\infty}\right)}\)

α= \(\frac{9.54 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}{238 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}=0.04008=4.008 \%\)

Question 4. Limiting molar conductivity of (i.e., \(\Lambda_{m(\mathrm{NH}_4 \mathrm{OH})}^{\circ}\)) is equal to

1. \(\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaCl})}^{\circ}-\Lambda_{m(\mathrm{NaOH})}^{\circ}\)
2. \(\Lambda_{m(\mathrm{NaOH})}^{\circ}+\Lambda_{m(\mathrm{NaCl})}^{\circ}-\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}\)
3. \(\Lambda_{m\left(\mathrm{NH}_4 \mathrm{OH}\right)}^{\circ}+\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}-\Lambda_{m(\mathrm{HCl})}^{\circ}\)
4. \(\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaOH})}^{\circ}-\Lambda_{m(\mathrm{NaCl})}^{\circ}\)

Answer: 4. \(\Lambda_{m\left(\mathrm{NH}_4 \mathrm{Cl}\right)}^{\circ}+\Lambda_{m(\mathrm{NaOH})}^{\circ}-\Lambda_{m(\mathrm{NaCl})}^{\circ}\)

Question 5. Molar conductivities (∧°_m) at infinite dilution of NaCl, HCl, and CH₃COONa are 126.4, 425.9, and 91.0 S cm² mol^-1 respectively. (∧°_m) for CH₃COOH will be

1. 425.5 S cm² mol^-1
2. 180.5 S cm² mol^-1
3. 290.8 S cm² mol^-1
4. 390.5 S cm² mol^-1

Answer: 4. 390.5 S cm² mol^-1

⇒ \(\Lambda_{\mathrm{NaCl}}^{\circ}=126.4 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

⇒  \(\Lambda_{\mathrm{HCl}}=425.9 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

⇒  \(\Lambda^{\circ}{ }_{\mathrm{CH}_3 \mathrm{COONa}}=91.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

⇒  \(\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\circ}=\Lambda_{\mathrm{CH}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}\)

=91.0+425.9-126.4

= \(390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)

Electrochemistry Multiple Choice Questions PSEB Class 12

Question 45. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

1. Increase in ionic mobility of ions
2. 100% Ionisation of electrolyte at normal dilution
3. Increase in both i.e., Number of ions and ionic mobility of ions
4. Increase in number of ions.

Answer: 1. Increase in ionic mobility of ions

Strong electrolytes are completely ionized at all concentrations. On increasing dilution, the no. of ions remains the same but the ionic mobility increases, and the equivalent conductance increases.

Question 6. Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al₂(SO₄)₃? Given that \(\Lambda_{\mathrm{Al}^{3+}}^{\circ} \text { and } \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\) are the equivalent conductances at infinite dilution of the respective ions.

1. \(2 \Lambda_{\mathrm{Al}^{3+}}^{\circ}+3 \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\)
2. \(\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\)
3. \(\left(\Lambda_{\mathrm{Al}^{3+}}^0+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\right) \times 6\)
4. \(\frac{1}{3} \Lambda_{\mathrm{Al}^{3+}}^{\circ}+\frac{1}{2} \Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\)

Answer: 2. \(\Lambda_{\mathrm{Al}^{3+}}^{\circ}+\Lambda_{\mathrm{SO}_4^{2-}}^{\circ}\)

At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards the molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.

Hence, \(\Lambda_{\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3}=\Lambda_{\mathrm{Al}^{3+}}^0+\Lambda_{\mathrm{SO}_4^{2-}}^0\)

Question 7. The equivalent conductance of the M/32 solution of a weak monobasic acid is 8.0 mho cm² and at infinite dilution is 400 mho cm². The dissociation constant of this acid is

1. 1.25 x 10^-6
2. 6.25 x 10^-4
3. 1.25 x 10^-4
4. 1.25 x 10^-5

Answer: 4. 1.25 x 10^-5

Given, \(\Lambda=8 \mathrm{mho}^2, \Lambda^{\infty}=400 \mathrm{mho} \mathrm{cm}^2\)

Degree of dissociation, \(\alpha=\frac{\Lambda}{\Lambda^{\infty}} \Rightarrow \alpha=\frac{8}{400}=2 \times 10^{-2}\)

Dissociation constant, \(\mathrm{K}=\mathrm{Ca}^2\)

Given, C=M/32

∴ K = \(\frac{1}{32} \times 2 \times 10^{-2} \times 2 \times 10^{-2}=1.25 \times 10^{-5}\)

Electrochemistry Multiple Choice Questions PSEB Class 12

Question 8. Kohlrauschs law states that at

1. Infinite dilution, each ion makes a definite contribution to the conductance of an electrolyte whatever the nature of the other ion of the electrolyte
2. In infinite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte, whatever the nature of the other ion of the electrolyte
3. In finite dilution, each ion makes a definite contribution to the equivalent conductance of an electrolyte, whatever the nature of the other ion of the electrolyte
4. Infinite dilution each ion makes a definite contribution to the equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.

Answer: 1. Infinite dilution, each ion makes a definite contribution to the conductance of an electrolyte whatever be the nature of the other ion of the electrolyte

At infinite dilution, when dissociation is complete each ion makes a definite contribution towards the molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated and that the molar conductance of any electrolyte at infinite dilution is given by the sum of the contributions of two ions.

This is called Kohl Rausch’s law \(\Lambda_m^{\infty}=\Lambda_{+}^{\infty}+\Lambda_{-}^{\infty}\),

where, \(\Lambda_{+}^{\infty}\) and \(\Lambda_{-}^{\infty}\) are molar ionic conductance at infinite dilution for cation and anion, respectively.

Question 9. Equivalent conductances of Ba²⁺ and Cl^– ions are 127 and 76 ohm^-1 cm^-1 eq^-1 respectively. The equivalent conductance of BaCl₂ at infinite dilution is

1. 139.5
2. 101.5
3. 203
4. 279

Answer: 1. 139.5

⇒ \(\lambda_{\infty}=\frac{1}{n_{+}} \lambda_{+}^{\infty}+\frac{1}{n_{-}} \lambda_{-}^{\infty}\)

So, \(\lambda_{\infty}\left(\mathrm{BaCl}_2\right)=\frac{1}{2} \times \lambda_{\mathrm{Ba}^{2+}}^{\infty}+\frac{1}{1} \times \lambda_{\mathrm{Cl}^{-}}^{\infty}\)

= \(\frac{1}{2} \times 127+76=63 \cdot 5+76=139 \cdot 5\)

Question 10. The specific conductance of a 0.1 N KCl solution at 23 °C is 0.012 ohm^-1 cm^-1. The resistance of the cell containing the solution at the same temperature was found to be 55 ohms. The cell constant will be

1. 0.918 cm^-1
2. 0.66 cm^-1
3. 1.142 cm^-1
4. 1.12 cm^-1

Answer: 2. 0.66 cm^-1

k = \(0.012 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\)

R = \(55 \mathrm{ohm} \Rightarrow C=\frac{1}{R}=\frac{1}{55} \mathrm{ohm}^{-1}\)

Cell Constant \(\left(\frac{l}{a}\right)=\frac{\text { Specific Conductance }}{\text { Conductance }}\)

= \(\frac{0.012}{1 / 55}=55 \times 0.012=0.66 \mathrm{~cm}^{-1}\)

Question 11. On heating one end of a piece of metal, the other end becomes hot because of

1. Energized electrons move to the other end
2. Minor perturbation in the energy of atoms
3. Resistance of the metal
4. Mobility of atoms in the metal.

Answer: 1. Energised electrons moving to the other end

The conductivity of heat in metals is due to the presence of free electrons, which move due to an increase in temperature.

Class 12 Chemistry Chapter Electrochemistry MCQs

Question 12. On electrolysis of dil. sulphuric acid using a platinum (Pt) electrode, the product obtained at the anode will be

1. Hydrogen gas
2. Oxygen gas
3. H₂S gas
4. SO₂ gas.

Answer: 2. Oxygen gas

During the electrolysis of dilute sulphuric acid, the following reaction takes place at the anode.

⇒ \(2 \mathrm{H}_2 \mathrm{O}_{(l)} \rightarrow \mathrm{O}_{2(g)}+4 \mathrm{H}_{(a q)}^{+}+4 e^{-} ; E_{\text {cell }}^{\circ}=+1.23 \mathrm{~V}\)

i.e., \(\mathrm{O}_{2(g)}\) will be liberated at anode.

Question 13. The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (Atomic mass of Ca = 40 g mol^-1) is

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

Answer: 1. 1

Thus, one Faraday is required to produce 20 g of calcium from molten CaCl₂.

Question 14. During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 amperes is

1. 55 minutes
2. 110 minutes
3. 220 minutes
4. 330 minutes.

Answer: 2. 110 minutes

During the electrolysis of molten sodium chloride,

According to Faraday’s first law of electrolysis, w = Z x I x t

w = \(\frac{E}{96500} \times I \times t\)

No. of moles of \(\mathrm{Cl}_2\) gas \(\times \mathrm{Mol}\). weight of \(\mathrm{Cl}_2\) gas

= \(\frac{\text { Eq. wt. of } \mathrm{Cl}_2 \text { gas } \times I \times t}{96500}\)

0.10 x 71 = \(\frac{35.5 \times 3 \times t}{96500}\)

t = \(\frac{0.10 \times 71 \times 96500}{35.5 \times 3}=6433.33 \mathrm{sec}\)

t = \(\frac{6433.33}{60} \mathrm{~min}=107.22 \mathrm{~min} \approx 110 \mathrm{~min}\)

Class 12 Chemistry Chapter Electrochemistry MCQs

Question 15. The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 x 10^-19 C)

1. 6 x 10²³
2. 6 x 10²⁰
3. 3.75 x 10²⁰
4. 7.48 x 10²³

Answer: 3. 3.75 x 10²⁰

Q = I x t

Q = 1 x 60 = 60C

Now, 1.60 x 10^-19 C ≡ 1 electron

∴ 60 C \(\equiv \frac{60}{1.6 \times 10^{-19}}=3.75 \times 10^{20}\) electrons

Question 16. When 0.1 mol MnO₄^2- is oxidized, the quantity of electricity required to completely oxidize MnO^2-₄ to MnO^–₄ is

1. 96500 C
2. 2 x 96500 C
3. 9650 C
4. 96.50 C

Answer: 3. 9650 C

The oxidation reaction is \(\stackrel{+6}{\mathrm{MnO}_4^{2-}} \longrightarrow \stackrel{+7}{\mathrm{MnO}_4^{-}}+\underset{0.1 \mathrm{~mol}}{e^{-}}\)

Q = 0.1 x F= 0.1 x 96500 C = 9650 C

Question 17. The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be

1. 5.4 g
2. 10.8 g
3. 54.0 g
4. 108.0 g

Answer: 4. 108.0 g

According to Faraday’s second law, \(\frac{W_{\mathrm{Ag}}}{E_{\mathrm{Ag}}}=\frac{W_{\mathrm{O}_2}}{E_{\mathrm{O}_2}}\) or \(\frac{W_{\mathrm{Ag}}}{108}=\frac{\frac{5600}{22400} \times 32}{8}\)

or \(\frac{W_{\mathrm{Ag}}}{108}=\frac{8}{8} \Rightarrow W_{\mathrm{Ag}}=108 \mathrm{~g}\)

Question 18. How many grams of cobalt metal will be deposited when a solution of cobalt(2) chloride is electrolyzed with a current of 10 amperes for 109 minutes? (1 Faraday = 96,500 C; Atomic mass of Co = 59 u)

1. 4.0
2. 20.0
3. 40.0
4. 0.66

Answer: 2. 20.0

w = \(\frac{I t E}{96500}\)

= \(\frac{10 \times 109 \times 60 \times 59}{96500 \times 2}=19.99=20 \mathrm{~g}\)

PSEB Class 12 Chemistry Electrochemistry Questions and Answers

Question 19. Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 x 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminum is produced? (Assume 100% current efficiency, at. mass of Al = 27 g mol^-1)

1. 8.1 x 10⁴ g
2. 2.4 x 10⁵ g
3. 1.3 x 10⁴ g
4. 9.0 x 10³ g

Answer: 1. 8.1 x 10⁴ g

Applying E = Z x 96500

⇒ \(\frac{27}{3}=Z \times 96500 \Rightarrow Z=\frac{9}{96500}\)

Now applying the formula, w = \(Z \times I \times t\)

w = \(\frac{9}{96500} \times 4 \times 10^4 \times 6 \times 60 \times 60=8.1 \times 10^4 g\)

Question 20. 4.5 g of aluminum (at. mass 27 amu) is deposited at the cathode from Al³⁺ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H⁺ ions in solution by the same quantity of electric charge will be

1. 44.8 L
2. 22.4 L
3. 11.2 L
4. 5.6 L

Answer: 4. 5.6 L

We know that 1 Faraday charge liberates 1 eq. ofsubstance.

This is the Faraday law.

equation weight of AI = 27/3 =9

Number of equation of AI = \(\frac{\text { wt. of } \mathrm{Al}}{\text { eq. wt. }}=\frac{4.5}{9}=0.5\)

Number of Faradays required = 0.5

A number of equations of H₂ produced = 0.5 eq.

The volume occupied by 1 equation of H₂ = 22.4/2 = 11.2 L

Volume occupied by 0.5 eq. of H₂ = 11.2 x 0.5 = 5.6 L at STP

Question 21. In electrolysis of NaCl when the Pt electrode is taken then H₂ is liberated at the cathode while with H₂ cathode it forms sodium amalgam. The reason for this is

1. Hg is more inert than Pt
2. More voltage is required to reduce H⁺ at Hg than at Pt
3. Na is dissolved in Hg while it does not dissolve in Pt
4. The cone, of H⁺ ions is larger when Pt electrode is taken.

Answer: 2. More voltage is required to reduce H+ at Hg than at Pt

When sodium chloride is dissolved in water, it ionizes as \(\mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \text {. }\)

Water also dissociates as: \(\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}\)

During the passing of electric current through this solution using a platinum electrode, Nan and Hn ions move toward the cathode. However, only H⁺ ions are discharged more readily than Na⁺ ions because of their low discharge potential (in the electromotive series hydrogen is lower than sodium). These H⁺ ions gain electrons and change into neutral atoms.

At cathode: \(\mathrm{H}^{+}+e^{-} \longrightarrow \mathrm{H}, \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2\)

Cl^– and OH^– ions move toward the anode. Cl^– ions lose electrons and change into neutral atoms.

At anode, \(\mathrm{Cl}^{-}-e^{-} \longrightarrow \mathrm{Cl}, \mathrm{Cl}+\mathrm{Cl} \longrightarrow \mathrm{Cl}_2\)

If mercury is used as the cathode, H⁺ ions are not discharged at the mercury cathode because mercury has a high hydrogen overvoltage. Na⁺ ions are discharged at the cathode in preference to H⁺ ions, yielding sodium, which dissolves in mercury to form sodium amalgam.

At cathode: \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\)

PSEB Class 12 Chemistry Electrochemistry Questions and Answers

Question 22. A 5-ampere current is passed through a solution of zinc sulfate for 40 minutes. The amount of zinc deposited at the cathode is

1. 0.4065 g
2. 65.04 g
3. 40.65 g
4. 4.065 g

Answer: 4. 4.065 g

Current (I) = 5 ampere and time (t) = 40 minutes = 2400 seconds.

Amount of electricity passed (Q) = I x t = 5 x 2400 = 12000C

Now, \(\mathrm{Zn}^{2+}+2 e^{-} \longrightarrow \mathrm{Zn}(1 \text { mole }=65.39 \mathrm{~g})\)

Since, two charges (i.e., 2 x 96500 C) deposit 65.39 g of zinc, therefore 12000 C will deposit.

= \(\frac{65.39 \times 12000}{2 \times 96500}=4.065 \mathrm{~g} \text { of zinc }\)

Question 23. Sodium is made by the electrolysis of a molten mixture of about 40% NaCl and 60% CaCl₂ because

1. Ca⁺⁺ can reduce NaCl to Na
2. Ca⁺⁺ can displace Na from NaCl
3. CaCl₂ helps in the conduction of electricity
4. This mixture has a lower melting point than NaCl.

Answer: 4. This mixture has a lower melting point than NaCl.

Sodium is obtained by electrolytic reduction of its chloride. The melting point of chloride in sodium is high so in order to lower its melting point, calcium chloride is added to it.

Question 24. When CuSO₄ is electrolyzed using platinum electrodes,

1. Copper is liberated at the cathode, sulfur at the anode
2. Copper is liberated at the cathode, oxygen at the anode
3. Sulfur is liberated at the cathode, oxygen at the anode
4. Oxygen is liberated at the cathode, and copper at the anode.

Answer: 2. Copper is liberated at the cathode, oxygen at anode

⇒ \(\mathrm{CuSO}_4 \rightleftharpoons \mathrm{Cu}^{2+}+\mathrm{SO}_4^{2-}\)

⇒ \(\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-}\)

At cathode: \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}\)

At anode: \(4 \mathrm{OH}^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+4 e^{-}\)

PSEB Class 12 Chemistry Electrochemistry Questions and Answers

Question 25. On electrolysis of dilute sulphuric acid using platinum electrodes, the product obtained at the anode will be

1. Hydrogen
2. Oxygen
3. Hydrogen sulphide
4. Sulfur dioxide.

Answer: 2. Oxygen

During the electrolysis of dilute sulphuric acid, the product obtained at the anode will be oxygen.

At anode: \(4 \mathrm{OH}^{-} \rightleftharpoons 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{O}_2+4 e^{-}\)

Question 26. A device that converts energy of the combustion of fuels like hydrogen and methane, directly into electrical energy is known as

1. Dynamo
2. Ni-Cd cell
3. Fuel cell
4. Electrolytic cell.

Answer: 3. Fuel cell

Question 27. The efficiency of a fuel cell is given by

1. ΔG/ΔS
2. ΔG/ΔH
3. ΔS/ΔG
4. ΔH/ΔG

Answer: 2. ΔG/ΔH

The thermal efficiency, η of a fuel conversion device is the amount of useful energy produced relative to the change in enthalpy, ΔH between the product and feed streams.

= \(\frac{\text { useful energy }}{\Delta H}\)

In an ideal case of an electrochemical converter, such as a fuel cell, the change in Gibbs free energy, ΔG of the reaction is available as useful electric energy at that temperature of the conversion.

Hence, \(\eta_{\text {ideal }}=\frac{\Delta G}{\Delta H}\)

Question 28. Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because

1. Zinc is lighter than iron
2. Zinc has a lower melting point than iron
3. Zinc has a lower negative electrode potential than iron
4. Zinc has a higher negative electrode potential than iron.

Answer: 4. Zinc has a higher negative electrode potential than iron.

Reduction potential values of \(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\) and \(E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.44 \mathrm{~V}\)

Thus, due to the higher negative electrode potential value of zinc than iron, iron cannot be coated on zinc.

Question 29. The most convenient method to protect the bottom of the ship made of iron is

1. Coating it with red lead oxide
2. White tin plating
3. Connecting it with Mg block
4. Connecting it with Pb block.

Answer: 2. White tin plating

The most convenient method to protect the bottom of the ship made of iron is white tin plating preventing the build-up of barnacles.

Question 30. To protect iron against corrosion, the most durable metal plating on it is

1. Copper plating
2. Zinc plating
3. Nickel plating
4. Tinplating.

Answer: 2. Zinc plating

Question 1. Among the following, which one is a wrong statement?

1. PH₅ and BiCl₅ do not exist.
2. pπ-dπ bonds are present in SO₂.
3. SeF₄ and CH₄ have the same shape.
4. \(\mathrm{I}_3^{+}\) has bent geometry.

Answer: 3. SeF₄ and CH₄ have the same shape.

Question 2. The outer orbitals of C in an ethene molecule can be considered to be hybridized to give three equivalent sp² orbitals. The total number of sigma (σ) and pi (π) bonds in ethene molecule is

1. 3 sigma (σ) and 2 pi (π) bonds
2. 4 sigma (σ) and 1 pi (π) bonds
3. 5 sigma (σ) and 1 pi (π) bonds
4. 1 sigma (σ) and 2 pi (π) bonds.

Answer: 3. 5 sigma (σ) and 1 pi (π) bonds

Question 3. In which of the following pairs both species have sp³ hybridization?

1. \(\mathrm{SiF}_4, \mathrm{BeH}_2\)
2. \(\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{NF}_3, \mathrm{BF}_3\)
4. \(\mathrm{H}_2 \mathrm{~S}, \mathrm{BF}_3\)

Read And Learn More Class 11 Chemistry Solutions

Answer: 2. \(\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}\)

NF₃ and H₂O are sp³-hybridised.

PSEB Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs

Question 4. Which one of the following pairs is isostructural (i.e., having the same shape and hybridization)?

1. \(\left[\mathrm{BCl}_3\right]\) and \(\left.\mathrm{BrCl}_3\right]\)
2. \(\left[\mathrm{NH}_3\right]\) and \(\left.\mathrm{NO}_3^{-}\right]\)
3. \(\left[\mathrm{NF}_3\right]\) and \(\left.\mathrm{BF}_3\right]\)
4. \(\left[\mathrm{BF}_4^{-}\right]\) and \(\left.\mathrm{NH}_4^{+}\right]\)

Answer: 4. \(\left[\mathrm{BF}_4^{-}\right]\) and \(\left.\mathrm{NH}_4^{+}\right]\)

• \(\mathrm{BCl}_3 \Rightarrow s p^2\), trigonal planar
• \(\mathrm{BrCl}_3 \Rightarrow s p^3 d\), T-shaped
• \(\mathrm{NH}_3 \Rightarrow s p^3\) pyramidal
• \(\mathrm{NO}_3^{-} \Rightarrow s p^2\) trigonal planar
• \(\mathrm{NF}_3 \Rightarrow s p^3\), pyramidal
• \(\mathrm{BF}_3 \Rightarrow s p^2\), trigonal planar
• \(\mathrm{BF}_4^{-} \Rightarrow s p^3\), tetrahedral
• \(\mathrm{NH}_4^{+} \Rightarrow s p^3\), tetrahedral

Question 5. Which of the two ions from the list given below that have the geometry that is explained by the same hybridization of orbitals, \(\mathrm{NO}_2^{-}, \mathrm{NO}_3^{-}, \mathrm{NH}_2^{-}, \mathrm{NH}_4^{+},\), SCN^–?

1. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NO}_3^{-}\)
2. \(\mathrm{NH}_4^{+}\) and \(\mathrm{NO}_3^{-}\)
3. \(\mathrm{SCN}^{-}\) and \(\mathrm{NH}_2^{-}\)
4. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NH}_2^{-}\)

Answer: 1. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NO}_3^{-}\)

⇒ \(\begin{array}{cc}
\text { Ions } & \text { Hybridisation } \\
\mathrm{NO}_2^{-} & s p^2 \\
\mathrm{NO}_3^{-} & s p^2 \\
\mathrm{NH}_2^{-} & s p^3 \\
\mathrm{NH}_4^{+} & s p^3 \\
\mathrm{SCN}^{-} & s p
\end{array}\)

PSEB Class 11 Chemistry Chemical Bonding and Molecular Structure MCQs

Question 6. In which of the following pairs of molecules/ions, do the central atoms have sp² hybridization?

1. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NH}_3\)
2. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)
3. \(\mathrm{NH}_2\) and \(\mathrm{H}_2 \mathrm{O}\)
4. \(\mathrm{BF}_3\) and \(\mathrm{NH}_2^{-}\)

Answer: 2. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)

The hybridisation of calculated as

H = \(\frac{1}{2}\left[\begin{array}{r}
\left(\begin{array}{l}
\text { No. of electrons } \\
\text { in valence shell } \\
\text { of atom }
\end{array}\right)+\left(\begin{array}{l}
\text { No. of monovalent } \\
\text { atoms around } \\
\text { central atom }
\end{array}\right) \\
-\left(\begin{array}{l}
\text { Charge on } \\
\text { cation }
\end{array}\right)+\left(\begin{array}{l}
\text { Charge on } \\
\text { anion }
\end{array}\right)
\end{array}\right]\)

∴ For \(\mathrm{BF}_3, H=\frac{1}{2}[(3)+(3)-(0)+(0)]\)

= \(3 \Rightarrow s p^2\) hybridisation

For \(\mathrm{NO}_2^{-}, H=\frac{1}{2}[(5)+(0)-(0)+(1)]\)

= \(3 \Rightarrow s p^2\) hybridisation.

Question 7. In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three?

1. \(\mathrm{SF}_4\)
2. \(\mathrm{I}_3\)
3. \(\mathrm{SbCl}_5^{2-}\)
4. \(\mathrm{PCl}_5\)

Answer: 3. \(\mathrm{SbCl}_5^{2-}\)

Hybridization of the central atom can be calculated as

H = \(\frac{1}{2}\left[\begin{array}{c}
\left(\begin{array}{l}
\text { No. of valence } \\
\text { electrons in the } \\
\text { central atom }
\end{array}\right)+\left(\begin{array}{l}
\text { No. of monovalent } \\
\text { atoms around } \\
\text { central atom }
\end{array}\right) \\
-\left(\begin{array}{l}
\text { Charge on } \\
\text { cation }
\end{array}\right)+\left(\begin{array}{l}
\text { Charge on } \\
\text { anion }
\end{array}\right)
\end{array}\right]\)

Applying this formula we find that all the given species except [SbCl₅]^2- have central atoms with sp³d (corresponding to H = 5) hybridization. In [SbCl₅]^2-, Sb is sp³d² hybridized.

Question 8. In which of the following molecules the central atom does not have sp³ hybridization?

1. \(\mathrm{CH}_4\)
2. \(\mathrm{SF}_4\)
3. \(\mathrm{BF}_4\)
4. \(\mathrm{NH}_4^{+}\)

Answer: 2. \(\mathrm{SF}_4\)

For neutral molecules,

No. of electron pairs = No. of atoms bonded to it + 1/2[Gp. number of the central atoms – Valency of the central atom]

∴ For \(\mathrm{CH}_4\), number of \(e^{-}\) pairs = \(4+\frac{1}{2}[4-4]=4\left(s p^3\right.\) hybridisation)

For \(\mathrm{SF}_4\), number of \(e^{-}\) pairs \(=4+\frac{1}{2}[6-4]=5\left(s p^3 d\right.\) hybridisation)

For ions,

No. of electron pairs = No. of atoms bonded to it +1/2 [Gp. no. of central atom – Valency of central atom ± No. of electrons equals to the units of charge]

∴ For \(\mathrm{BF}_4^{-}\), number of \(e^{-}\) pairs = \(4+\frac{1}{2}[3-4+1]\)

= \(4(s p^3\) hybridisation)

∴ For \(\mathrm{NH}_4^{+}\), number of \(e^{-}\) pairs = \(4+\frac{1}{2}[5-4-1]\)

= \(4\left(s p^3 \text { hybridisation }\right)\)

Question 9. Some of the properties of the two species, NO₃^– and H₃O⁺ are described below. Which one of them is correct?

1. Dissimilar in hybridization for the central atom with different structures.
2. Isostructural with the same hybridization for the central atom.
3. Isostructural with different hybridization for the central atom.
4. Similar in hybridization for the central gap atom with different structures.

Answer: 1. Dissimilar in hybridization for the central atom with different structures.

No. of electron pairs at the central atom = No. of atoms bonded to it + 1/2[Group number of the central atom – Valency of the central atom ± No. of electrons equals the units of charge]

No. of electron pairs at the central atom in \(\mathrm{NO}_3^{-}\) = 3 + 1/2(5-6+1) = 3 (sp² hybridization).

No. of electron pairs at the central atom in \(\mathrm{H}_3 \mathrm{O}^{+}=3+\frac{1}{2}[6-3-1]=4\) (sp³ hybridization)

Question 10. In which of the following molecules/ions BF₃, NO^–₂, NH^–_2, and H₂O, the central atom is sp² hybridized?

1. \(\mathrm{NH}_2^{-}\) and \(\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{NO}_2^{-}\) and \(\mathrm{H}_2 \mathrm{O}\)
3. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)
4. \(\mathrm{NO}_2^{-}\) and \(\mathrm{NH}_2^{-}\)

Answer: 3. \(\mathrm{BF}_3\) and \(\mathrm{NO}_2^{-}\)

∴ \(\mathrm{BF}_3 \rightarrow s p^2, \mathrm{NO}_2^{-} \rightarrow s p^2, \mathrm{NH}_2^{-} \rightarrow s p^3, \mathrm{H}_2 \mathrm{O} \rightarrow s p^3\)

Question 11. Among the following, the pair in which the two species are not isostructural is

1. \(\mathrm{SiF}_4\) and \(\mathrm{SF}_4\)
2. \(\mathrm{IO}_3^{-}\) and \(\mathrm{XeO}_3\)
3. \(\mathrm{BH}_4^{-}\) and \(\mathrm{NH}_4^{+}\)
4. \(\mathrm{PF}_6^{-}\) and \(\mathrm{SF}_6\)

Answer: 1. \(\mathrm{SiF}_4\) and \(\mathrm{SF}_4\)

SiF₄ has a symmetrical tetrahedral shape which is due to sp³ hybridisation of the centrai silicon atom. SF₄ has distorted tetrahedral or see-saw geometry which arises due to sp³d hybridisation of a central sulfur atom and due to the presence of one lone pair of electrons in one of the equatorial hybrid orbitals.

Question 12. Which of the following two are isostructural?

1. \(\mathrm{XeF}_2, \mathrm{IF}_2^{-}\)
2. \(\mathrm{NH}_3, \mathrm{BF}_3\)
3. \(\mathrm{CO}_3^{2-}, \mathrm{SO}_3^{2-}\)
4. \(\mathrm{PCl}_5, \mathrm{ICl}_5\)

Answer: 1. \(\mathrm{XeF}_2, \mathrm{IF}_2^{-}\)

Compounds having the same shape with the same hybridization are known as isostructural.

∴ \(\mathrm{XeF}_2, \mathrm{IF}_2^{-}\) + both are sp³d hybridised linear molecules

Question 13. The bond length between hybridized carbon atoms and other carbon atoms is minimal in

1. Propene
2. Propyne
3. Propane
4. Butane.

Answer: 2. Propyne

The C-C bond length = 1.54 Å, C = C bond length = 1.34 Å, and C: C bond length = 1.20 Å.

Since propyne has a triple bond, therefore it has a minimum bond length.

Question 14. Which of the following has sp²-hybridisation?

1. \(\mathrm{BeCl}_2\)
2. \(\mathrm{C}_2 \mathrm{H}_2\)
3. \(\mathrm{C}_2 \mathrm{H}_6\)
4. \(\mathrm{C}_2 \mathrm{H}_4\)

Answer: 4. \(\mathrm{C}_2 \mathrm{H}_4\)

BeCl₂ and C₂H₂ have sp-hybridisation and C₂H₆ has sp³-hybridisation. C₂H₄ has sp² hybridisation.

Chemical Bonding and Molecular Structure MCQs PSEB Class 11

Question 15. When the hybridization state of the carbon atom changes from sp³ to sp² and finally to sp, the angle between the hybridized orbitals

1. Decreases gradually
2. Decreases considerably
3. Is not affected
4. Increases progressively.

Answer: 4. Increases progressively.

The angle increases progressively, sp³ (109° 28′), sp² (120°), sp (180°).

Question 16. Which one of the following has the shortest carbon-carbon bond length?

1. Benzene
2. Ethene
3. Ethyne
4. Ethane

Answer: 3. Ethyne

There is a triple bond in the ethyne molecule (H – C ≡ C – H) and due to this triple bond, the carbon-carbon bond distance is the shortest in ethyne.

Question 17. Which structure is linear?

1. \(\mathrm{SO}_2\)
2. \(\mathrm{CO}_2\)
3. \(\mathrm{CO}_3^{2-}\)
4. \(\mathrm{SO}_4^{2-}\)

Answer: 2. \(\mathrm{CO}_2\)

The CO₂ molecule is sp-hybridized and thus, it is linear while \(\mathrm{CO}_3^{2-}\) is planar (sp² -hybridized), SO₂ is an angular molecule with sp² hybridization and \(\mathrm{SO}_4^{2-}\) is tetrahedral (sp³-hybridised).

Question 18. A sp³ hybrid orbital contains

1. 1/4 s-character
2. 1/2 s-character
3. 1/3 s-character
4. 2/3 s-character.

Answer: 1. 1/4 s-character

sp³ orbital has 1/4(25%) s-character.

Question 19. The complex ion [Co(NH₃)₆]³⁺ is formed by sp³d² hybridization. Hence the ion should possess

1. Octahedral geometry
2. Tetrahedral geometry
3. Square planar geometry
4. Tetragonal geometry.

Answer: 1. Octahedral geometry

According to VSEPR theory, a molecule with 6 bond pairs must be octahedral.

Question 20. Which of the following molecules does not have a linear arrangement of atoms?

1. \(\mathrm{H}_2 \mathrm{~S}\)
2. \(\mathrm{C}_2 \mathrm{H}_2\)
3. \(\mathrm{BeH}_2\)
4. \(\mathrm{CO}_2\)

Answer: 1. \(\mathrm{H}_2 \mathrm{~S}\)

For the linear arrangement of atoms, the hybridization is sp (bond angle = 180°).

Only H₂S has sp³-hybridisation and hence it has an angular shape while C₂H₂, BeH_2, and CO₂ all involve sp-hybridization and hence, have a linear arrangement of atoms.

Question 21. In which one of the following molecules the central atom can be said to adopt sp² hybridization?

1. \(\mathrm{BeF}_2\)
2. \(\mathrm{BF}_3\)
3. \(\mathrm{C}_2 \mathrm{H}_2\)
4. \(\mathrm{NH}_3\)

Answer: 2. \(\mathrm{BF}_3\)

BF₃ involves sp²-hybridisation.

Question 22. Equilateral shape has

1. sp hybridization
2. sp² hybridisation
3. sp³ hybridisation
4. dsp³ hybridisation

Answer: 2. sp² hybridization

Equilateral or triangular planar shape involves sp² hybridization, for example, BCl₃.

Question 23. The correct order of energies of molecular orbitals of N₂ molecules is

1. \(\sigma 1 s<\sigma^* 1 s< \sigma 2 s<\sigma^* 2 s<\sigma 2 p_z<\sigma^* 2 p_z\) \(<\left(\pi 2 p_x=\pi 2 p_y\right) <\left(\pi^* 2 p_x=\pi^* 2 p_y\right)\)
2. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x\right.\left.=\pi 2 p_y\right)\) \(<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma 2 p_z<\sigma^* 2 p_z\)
3. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x=\pi 2 p_y\right)\) \(<\sigma 2 p_z<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)
4. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\sigma 2 p_z<\left(\pi 2 p_x=\pi 2 p_y\right)\) \(<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)

Answer: 3. \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x=\pi 2 p_y\right)\) \(<\sigma 2 p_z<\left(\pi^* 2 p_x=\pi^* 2 p_y\right)<\sigma^* 2 p_z\)

For molecules Li₂, Be₂, Br₂, C_2, and N₂ the order of energies of various molecular orbitals is \(\sigma 1 s<\sigma^* 1 s<\sigma 2 s<\sigma^* 2 s<\left(\pi 2 p_x=\pi 2 p_y\right)<\sigma 2 p_z<\left(\pi^* 2 p_x\right. \left.=\pi^* 2 p_y\right) <\sigma^* 2 p_z\)

Question 24. Consider the following species: CN⁺, CN^–, NO and CN. Which one of these will have the highest bond order?

1. NO
2. CN^–
3. CN⁺
4. CN

Answer: 2. CN^–

NO(15): \((\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\sigma 2 p_z\right)^2, \left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2,\left(\pi^* 2 p_x\right)^1=\left(\pi^* 2 p_y\right)^0\)

B.O. = \(\frac{10-5}{2}=2.5\)

∴ \(\mathrm{CN}^{-}(14):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2,\left(\sigma 2 p_z\right)^2\)

B.O. = \(\frac{10-4}{2}=3\)

CN(13): \((\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2,\left(\sigma 2 p_z\right)^1\)

B.O. = \(\frac{9-4}{2}=2.5\)

∴ \(\mathrm{CN}^{+}(12):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\pi 2 p_x\right)^2=\left(\pi 2 p_y\right)^2\)

B.O. \(=\frac{8-4}{2}=2\)

Hence, CN^– has the highest bond order.

Chemical Bonding and Molecular Structure MCQs PSEB Class 11

Question 25. Which one of the following pairs of species have the same bond order?

1. \(\mathrm{O}_2, \mathrm{NO}^{+}\)
2. \(\mathrm{CN}^{-}, \mathrm{CO}\)
3. \(\mathrm{N}_2, \mathrm{O}_2^{-}\)
4. \(\mathrm{CO}, \mathrm{NO}\)

Answer: 2. \(\mathrm{CN}^{-}, \mathrm{CO}\)

Molecular orbital electronic configurations and bond order values are \(\mathrm{O}_2(16): \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2,\pi^* 2 p_x^1=\pi^* 2 p_y^1\)

B.O. = \(\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(10-6)=2\)

∴ \(\mathrm{NO}^{+}(14): \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

∴ \(\mathrm{CN}^{-}(14): \sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

CO(14): \(1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

∴ \(N_2\) (14): \(\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\)

B.O. = \(\frac{1}{2}(10-4)=3\)

∴ \(\mathrm{O}_2^{-}(17)=\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2,\pi^* 2 p_x^2=\pi^* 2 p_y^1\)

B.O. = \(\frac{1}{2}(10-7)=1.5\)

NO(15): \(\sigma 1 s^2, \sigma^* 1 s^2, \sigma 2 s^2, \sigma^* 2 s^2, \sigma 2 p_z^2, \pi 2 p_x^2=\pi 2 p_y^2, \pi^* 2 p_x^1\)

B.O. = \(\frac{1}{2}(10-5)=2.5\)

Question 26. Which of the following is paramagnetic?

1. \(\mathrm{CN}^{-}\)
2. \(\mathrm{NO}^{+}\)
3. \(\mathrm{CO}\)
4. \(\mathrm{O}_2{ }^{-}\)

Answer: 4. \(\mathrm{O}_2{ }^{-}\)

∴ \(\mathrm{O}_2^{-}(17)\) superoxide has one unpaired electron. \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^1\)

Question 27. The pair of species that has the same bond order in the following is

1. \(\mathrm{CO}, \mathrm{NO}^{+}\)
2. \(\mathrm{NO}^{-}, \mathrm{CN}^{-}\)
3. \(\mathrm{O}_2, \mathrm{~N}_2\)
4. \(\mathrm{O}_2, \mathrm{~B}_2\)

Answer: 1. \(\mathrm{CO}, \mathrm{NO}^{+}\)

CO = 6 + 8 = 14electrons

NO⁺ = 7 + B – 1 = 14 electrons

Electronic configuration of \(\mathrm{NO}^{+}:\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2\)

Electronic configuration of \(\mathrm{CO}:\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2\)

So, both have bond order = \(\frac{10-4}{2}=3\)

PSEB Class 11 Chemistry Chapter 4 MCQs with Answers

Question 28. In which of the following ionization processes does the bond energy increase and the magnetic behavior change from paramagnetic to diamagnetic?

1. \(\mathrm{O}_2 \rightarrow \mathrm{O}_2^{+}\)
2. \(\mathrm{C}_2 \rightarrow \mathrm{C}_2^{+}\)
3. \(\mathrm{NO} \rightarrow \mathrm{NO}^{+}\)
4. \(\mathrm{N}_2 \rightarrow \mathrm{N}_2^{+}\)

Answer: 3. \(\mathrm{NO} \rightarrow \mathrm{NO}^{+}\)

Molecular orbital configuration of \(\mathrm{O}_2(16): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1 \pi^* 2 p_y{ }^1\)

⇒ Paramagnetic-Bond order =\(\frac{10-6}{2}=2\)

∴ \(\mathrm{O}_2^{+}(15): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1\)

⇒ Paramagnetic- Bond order = \(\frac{10-5}{2}=2.5\)

∴ \(\mathrm{C}_2(12): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2\)

⇒ Diamagnetic ; Bond order \(=\frac{8-4}{2}=2\)

∴ \(\mathrm{C}_2^{+}(11): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y{ }^1\)

⇒ Paramagnetic; Bond order = \(\frac{7-4}{2}=1.5\)

∴ \(\mathrm{NO}(15): \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1\)

⇒ Paramagnetic;Bond order \(=\frac{10-5}{2}=2.5\)

∴ \(\mathrm{NO}^{+}(14): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2\)

⇒ Diamagnetic; Bond order = \(\frac{10-4}{2}=3\)

∴ \(\mathrm{~N}_2(14): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^2 \)

⇒ Diamagnetic; Bond order = \(\frac{10-4}{2}=3\)

∴ \(\mathrm{~N}_2^{+}(13): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2 \pi 2 p_y^2 \sigma 2 p_z^1\)

⇒ Paramagnetic Bond order = \(\frac{9-4}{2.5}=2.5\)

Thus from NO → NO⁺, bond order increases i.e., bond energy increases, and magnetic behaviour changes from paramagnetic to diamagnetic

Question 29. The pair of species with the same bond order is

1. \(\mathrm{O}_2^{2-}, \mathrm{B}_2\)
2. \(\mathrm{O}_2^{+}, \mathrm{NO}^{+}\)
3. \(\mathrm{NO}, \mathrm{CO}\)
4. \(\mathrm{N}_2, \mathrm{O}_2\)

Answer: 1. \(\mathrm{O}_2^{2-}, \mathrm{B}_2\)

⇒ \(\begin{array}{ll}
\mathrm{O}_2^{2-} \rightarrow 1.0 & \mathrm{~B}_2 \rightarrow 1.0 \\
\mathrm{O}_2^{+} \rightarrow 2.5 & \mathrm{NO}^{+} \rightarrow 3.0 \\
\mathrm{NO} \rightarrow 2.5 & \mathrm{CO} \rightarrow 3.0 \\
\mathrm{~N}_2 \rightarrow 3.0 & \mathrm{O}_2 \rightarrow 2.0
\end{array}\)

PSEB Class 11 Chemistry Chapter 4 MCQs with Answers

Question 30. During the change of O₂ to O^–₂ ion, the electron adds on which one of the following orbitals?

1. π* orbital
2. π orbital
3. σ* orbital
4. σ orbital

Answer: 1. π* orbital

Electronic configuration of O₂(16) \(\sigma(1 s)^2 \sigma^*(1 s)^2 \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_z\right)^2 \pi\left(2 p_x\right)^2=\pi\left(2 p_y\right)^2 \pi^*\left(2 p_x\right)^1=\pi^*\left(2 p_y\right)^1\)

Question 31. Which of the following is isoelectronic?

1. \(\mathrm{CO}_2, \mathrm{NO}_2\)
2. \(\mathrm{NO}_2^{-}, \mathrm{CO}_2\)
3. \(\mathrm{CN}^{-}, \mathrm{CO}\)
4. \(\mathrm{SO}_2, \mathrm{CO}_2\)

Answer: 3. \(\mathrm{CN}^{-}, \mathrm{CO}\)

In CO, the number of electrons = 6 + 8 = 14 Molecular orbital electronic configuration of

CO: \((\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\sigma 2 p_z\right)^2\)

CN^– have also gets (6 + 7 + 1) 14 electrons and the configuration is similar to that of CO.

CN^– and CO are isoelectronic species.

Question 32. Which species does not exhibit paramagnetism?

1. \(\mathrm{N}_2^{+}\)
2. \(\mathrm{O}_2^{-}\)
3. \(\mathrm{CO}\)
4. \(\mathrm{NO}\)

Answer: 3. \(\mathrm{CO}\)

In ‘CO’ (14 electrons), there is no unpaired electron in its molecular orbital. Therefore, this does not exhibit paramagnetism.

Question 33. The number of anti-bonding electron pairs in O₂^2- molecular ion on the basis of the molecular orbital theory is (Atomic number of O is 8.)

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

Answer: 4. 4

∴ \(\mathrm{O}_2^{2-}(18):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^* 2 s\right)^2,\left(\sigma 2 p_z\right)^2,\left(\pi 2 p_x\right)^2, \left(\pi 2 p_y\right)^2,\left(\pi^{\star} 2 p_x\right)^2,\left(\pi^{\star} 2 p_y\right)^2\)

Thus the number of electrons in O₂^2- ion is 8(4 pairs).

Question 34. Which of the following species is paramagnetic?

1. \(\mathrm{CO}\)
2. \(\mathrm{CN}^{-}\)
3. \(\mathrm{O}_2^{2-}\)
4. \(\mathrm{NO}\)

Answer: 4. \(\mathrm{NO}\)

As per their molecular orbital electronic configurations \(\mathrm{CO}, \mathrm{CN}^{-}\) and \(\mathrm{O}_2^{2-}\) are diamagnetic and NO is paramagnetic.

Question 35. Which of the following is an incorrect statement?

1. The bond orders of \(\mathrm{O}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{-}\) and \(\mathrm{O}_2^{2-}\) are 2.5, 2, 1.5 and 1, respectively.
2. C₂ molecule has four electrons in its two degenerate Tt molecular orbitals
3. \(\mathrm{H}_2^{+}\) ion has one electron.
4. \(\mathrm{O}_2^{+}\) ion is diamagnetic.

Answer: 4. \(\mathrm{O}_2^{+}\) ion is diamagnetic.

∴ \(\mathrm{O}_2^{+}: \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1\)

Due to the presence of one unpaired electron, \(\mathrm{O}_2^{+}\) is paramagnetic in nature.

PSEB Class 11 Chemistry Chapter 4 MCQs

Question 36. Identify a molecule that does not exist.

1. \(\mathrm{He}_2\)
2. \(\mathrm{Li}_2\)
3. \(\mathrm{C}_2\)
4. \(\mathrm{O}_2\)

Answer: 1. \(\mathrm{He}_2\)

He₂ does not exist as it has zero bond order \(\mathrm{He}_2: \sigma 1 s^2, \sigma^* 1 s^2\)

Bond order = \(\frac{1}{2}\left(N_b-N_a\right)=\frac{1}{2}(2-2)=0\)

Question 37. Which of the following diatomic molecular species has only π bonds according to Molecular Orbital Theory?

1. Be₂
2. O₂
3. N₂
4. C₂

Answer: 4. C₂

∴ \(\mathrm{Be}_2(8): K K \sigma(2 s)^2 \sigma^*(2 s)^2\)

• \(\mathrm{O}_2(16): K K \sigma(2 s)^2 \sigma^*(2 s)^2 \sigma\left(2 p_z\right)^2 \pi\left(2 p_x\right)^2 \pi\left(2 p_y\right)^2 \pi^*\left(2 p_x\right)^1 \pi^*\left(2 p_y\right)^1\)
• \(\mathrm{~N}_2(14): K K \sigma(2 s)^2 \sigma^*(2 s)^2 \pi\left(2 p_x\right)^2 \pi\left(2 p_y\right)^2 \sigma\left(2 p_z\right)^2\)
• \(\mathrm{C}_2(12): K K \sigma(2 s)^2 \sigma^*(2 s)^2 \pi\left(2 p_x\right)^2 \pi\left(2 p_y\right)^2\)

Therefore, C₂ contains 2 π bonds as, it has 4 electrons in two pi-molecular orbitals.

Question 38. Which of the following is paramagnetic?

1. N₂
2. H₂
3. Li₂
4. O₂

Answer: 4. O₂

∴ \(\mathrm{N}_2(14): K K \sigma 2 s^2 \sigma^* 2 s^2 \pi 2 p_x^2=\pi 2 p_y^2 \sigma 2 p_z^2\); Diamagnetic

• \(\mathrm{H}_2(2): \sigma l s^2\); Diamagnetic
• \(\mathrm{Li}_2(6): \sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2\); Diamagnetic
• \(\mathrm{O}_2(16): \sigma 1 s^2 \sigma^{\star} 1 s^2 \sigma 2 s^2 \sigma^{\star} 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2\)
• \(\pi^* 2 p_x^1=\pi^* 2 p_y^1 ;\) Paramagnetic

Question 39. Decreasing order of stability of \(\mathrm{O}_2, \mathrm{O}_2^{-}, \mathrm{O}_2^{+}\) and \(\mathrm{O}_2^{2-}\)

1. \(\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{O}_2^{+}\)
2. \(\mathrm{O}_2>\mathrm{O}_2^{+}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{-}\)
3. \(\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}>\mathrm{O}_2^{+}>\mathrm{O}_2\)
4. \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

Answer: 4. \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

∴ \(\mathrm{O}_2(16): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1=\pi^* 2 p_y^1\)

Bond order \(=\frac{1}{2}(8-4)=2\)

∴ \(\mathrm{O}_2^{2-}(18): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y^2\)

Bond order = \(\frac{1}{2}(8-6)=1\)

∴ \(\mathrm{O}_2^{-}(17): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^2=\pi^* 2 p_y{ }^1\)

Bond order \(=\frac{1}{2}(8-5)=1.5\)

∴ \(\mathrm{O}_2^{+}(15): K K \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2=\pi 2 p_y^2 \pi^* 2 p_x^1\)

Bond order \(=\frac{1}{2}(8-3)=2.5\)

As, bond order ∝ stability

The decreasing order of stability is \(\mathrm{O}_2^{+}>\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{O}_2^{2-}\)

PSEB Class 11 Chemistry Chapter 4 MCQs

Question 40. The correct bond order in the following species is

1. \(\mathrm{O}_2^{+}<\mathrm{O}_2^{-}<\mathrm{O}_2^{2+}\)
2. \(\mathrm{O}_2^{-}<\mathrm{O}_2^{+}<\mathrm{O}_2^{2+}\)
3. \(\mathrm{O}_2^{2+}<\mathrm{O}_2^{+}<\mathrm{O}_2^{-}\)
4. \(\mathrm{O}_2^{2+}<\mathrm{O}_2^{-}<\mathrm{O}_2^{+}\)

Answer: 2. \(\mathrm{O}_2^{-}<\mathrm{O}_2^{+}<\mathrm{O}_2^{2+}\)

∴ \(\mathrm{O}_2^{-}<\mathrm{O}_2^{+}<\mathrm{O}_2^{2+}\)

B.O.: 1.5 = 2.5 3.0

Question 41. Bond order of 1.5 is shown by

1. \(\mathrm{O}_2^{+}\)
2. \(\mathrm{O}_2^{-}\)
3. \(\mathrm{O}_2^{2-}\)
4. \(\mathrm{O}_2\)

Answer: 2. \(\mathrm{O}_2^{-}\)

Configuration of \(\mathrm{O}_2(16)\): \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2 \sigma 2 p_z^2 \pi 2 p_x^2 \pi 2 p_y^2 \pi^* 2 p_x^1 \pi^* 2 p_y^1\)

Bond order \(=\frac{\begin{array}{l}\text { No. of } e^{-} \text {in } \\ \text { bonding M.O. }-\end{array} \begin{array}{c}\text { No. of } e^{–} \text {in } \\ \text { antibonding M.O. }\end{array}}{2}\)

Bond order of \(\mathrm{O}_2^{+}=\frac{10-5}{2}=2.5\)

Bond order of \(\mathrm{O}_2^{-}=\frac{10-7}{2}=1.5\)

Bond order of \(\mathrm{O}_2^{2-}=\frac{10-8}{2}=1.0\)

Bond order of \(\mathrm{O}_2=\frac{10-6}{2}=2\)

Question 42. Which of the following has the minimum bond length?

1. \(\mathrm{O}_2^{+}\)
2. \(\mathrm{O}_2^{-}\)
3. \(\mathrm{O}_2^{2-}\)
4. \(\mathrm{O}_2\)

Answer: 1. \(\mathrm{O}_2^{+}\)

Electronic configuration \(\mathrm{O}_2: \mathrm{KK}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_y\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\pi^* 2 p_x\right)^1\left(\pi^* 2 p_y\right)^1\)

Bond order = \(\frac{1}{2}(8-4)=2\)

⇒ \(\mathrm{O}_2^{+}\): Bond order = \(\frac{1}{2}(8-3)=2 \frac{1}{2}\)

⇒\(\mathrm{O}_2^{-}\): Bond order = \(\frac{1}{2}(8-5)=1 \frac{1}{2}\)

⇒ \(\mathrm{O}_2^{2-}: \text { Bond order }=\frac{1}{2}(8-6)=1\)

As bond order increases, bond length decreases

Question 43. The pairs of species of oxygen and their magnetic behavior are noted below. Which of the following presents the correct description?

1. \(\mathrm{O}_2^{-}, \mathrm{O}_2^{2-}\) – Both diamagnetic
2. \(\mathrm{O}^{+}, \mathrm{O}_2^{2-}\) – Both paramagnetic
3. \(\mathrm{O}_2^{+}, \mathrm{O}_2\) – Both paramagnetic
4. \(\mathrm{O}, \mathrm{O}_2^{2-}\) – Both paramagnetic

Answer: 3. \(\mathrm{O}_2^{+}, \mathrm{O}_2\) – Both paramagnetic

O⁺₂ and O₂ are paramagnetic in nature as they contain one and two unpaired electrons respectively.

Question 44. Which one of the following species does not exist under normal conditions?

1. \(\mathrm{Be}_2^{+}\)
2. \(\mathrm{Be}_2\)
3. \(\mathrm{B}_2\)
4. \(\mathrm{Li}_2\)

Answer: 2. \(\mathrm{Be}_2\)

Be₂ does not exist.

Be₂ has an electronic configuration of: \(\sigma 1 s^2 \sigma^* 1 s^2 \sigma 2 s^2 \sigma^* 2 s^2\)

∴ Bond order = \(\frac{4-4}{2}=0\) Thus, \(\mathrm{Be}_2\) does not exist.

Question 45. According to MO theory which of the lists ranks the nitrogen species in terms of increasing bond order?

1. \(\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}<\mathrm{N}_2\)
2. \(\mathrm{N}_2<\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}\)
3. \(\mathrm{N}_2^{-}<\mathrm{N}_2^{2-}<\mathrm{N}_2\)
4. \(\mathrm{N}_2^{-}<\mathrm{N}_2<\mathrm{N}_2^{2-}\)

Answer: 1. \(\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}<\mathrm{N}_2\)

According to MOT, the molecular orbital electronic configuration of

∴ \(\mathrm{N}_2(14):(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\sigma 2 p_z\right)^2\)

B.O = \(\frac{10-4}{2}=3\)

∴ \(\mathrm{~N}_2^{-}(15):(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\)

∴ \(\left(\sigma 2 p_z\right)^2\left(\pi^* 2 p_x\right)^1\)

B.O = \(\frac{10-5}{2}=2.5\)

∴ \(\mathrm{~N}_2^{2-}(16):(\sigma 1 s)^2\left(\sigma^* 1 s\right)^2(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\)

B.O. = \(\frac{10-6}{2}=2\)

Hence, bond order increases as: \(\mathrm{N}_2^{2-}<\mathrm{N}_2^{-}<\mathrm{N}_2\)

Question 46. Right order of dissociation energy N₂ and \(\mathrm{N}_2^{+}\) is

1. \(\mathrm{N}_2>\mathrm{N}_2^{+}\)
2. \(\mathrm{N}_2=\mathrm{N}_2^{+}\)
3. \(\mathrm{N}_2^{+}>\mathrm{N}_2\)
4. None.

Answer: 1. \(\mathrm{N}_2>\mathrm{N}_2^{+}\)

∴ \(\mathrm{N}_2(14):(\sigma 1 s)^2,\left(\sigma^* 1 s\right)^2,(\sigma 2 s)^2,\left(\sigma^{\star} 2 s\right)^2\) \(\left(\pi 2 p_x\right)^2,\left(\pi 2 p_y\right)^2,\left(\sigma 2 p_z\right)^2\)

In \(N_2\), bond order \(=\frac{N_b-N_a}{2}=\frac{10-4}{2}=3\)

In \(\mathrm{N}_2^{+}\), bond order \(=\frac{9-4}{2}=2 \cdot 5\)

As the bond order in \(\mathrm{N}_2\) is more than \(\mathrm{N}_2^{+}\) so the dissociation energy of \(\mathrm{N}_2\) is higher than \(\mathrm{N}_2^{+}\)

Question 47. N₂ and O₂ are converted into monocations, \(\mathrm{N}_2^{+}\) and \(\mathrm{O}_2^{+}\) respectively. Which is wrong?

1. In \(\mathrm{O}_2^{+}\), paramagnetism decreases.
2. \(\mathrm{N}_2^{+}\) becomes diamagnetic.
3. In \(\mathrm{N}_2^{+}\), the N-N bond weakens.
4. In \(\mathrm{O}_2^{+}\), the O-O bond order increases.

Answer: 2. \(\mathrm{N}_2^{+}\) becomes diamagnetic.

Diamagnetism is caused due to the absence of unpaired electrons. But in \(\mathrm{N}_2^{+}\) there is an unpaired electron So, it is paramagnetic.

Question 48. N₂ and O₂ are converted into monoanionic N^–₂ and O^–₂ respectively, which of the following statements is wrong?

1. In \(\mathrm{O}_2^{-}\) bond length increases.
2. \(\mathrm{N}_2^{-}\) becomes diamagnetic.
3. In \(\mathrm{N}_2^{-}\), N-N bond weakens.
4. In \(\mathrm{O}_2^{-}\), the O-O bond order decreases.

Answer: 2. \(\mathrm{N}_2^{-}\) becomes diamagnetic.

∴ \(\mathrm{N}_2^{-}\) becomes paramagnetic due to one unpaired electron in \(\pi^* 2 p_x\), orbital.

Question 49. The ground state electronic configuration of valence shell electrons in nitrogen molecule (N₂) is written as \(K K, \sigma 2 s^2, \sigma^* 2 s^2, \pi 2 p_x^2=\pi 2 p_y^2, \sigma 2 p_z^2\). Hence the bond order in nitrogen molecules is

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

Answer: 2. 3

Number of electrons in bonding orbitals, \(\mathrm{N}_b\) = 10 and number of electrons in antibonding orbitals, \(\mathrm{N}_a\) = 4

Therefore bond order = 1/2(\(\mathrm{N}_b\) – \(\mathrm{N}_a\)) = 1/2(10 – 4) = 3

Question 50. Which of the following molecules has the highest bond order?

1. \(\mathrm{O}_2^{-}\)
2. \(\mathrm{O}_2\)
3. \(\mathrm{O}_2^{+}\)
4. \(\mathrm{O}_2^{2-}\)

Answer: 3. \(\mathrm{O}_2^{+}\)

The bond order of \(\mathrm{O}_2^{+}=2.5, \mathrm{O}_2^{2-}=1\), \(\mathrm{O}_2^{-}\) = 1 .5 and that of O₂ = 2

Question 51. Which one of the following compounds shows the presence of an intramolecular hydrogen bond?

1. H₂O₂
2. HCN
3. Cellulose
4. Concentrated acetic acid

Answer: 3. Cellulose

H₂O₂, HCN and cons. CH₃COOH forms intermolecular hydrogen bonding while cellulose has intramolecular hydrogen bonding.

Question 51. In X – H….Y, X, and Y both are electronegative elements. Then

1. Electron density on X will increase and on H will decrease
2. In both electron density will increase
3. In both electron density will decrease
4. On X electron density will decrease and on H increases.

Answer: 1. Electron density on X will increase and on H will decrease

∴ \({ }^{\delta-} X-\mathrm{H}^{\delta+} \ldots . . . Y\), the electrons of the covalent bon<l are shifted towards the more electronegative atom’ This partially positively charged H-atom forms hydrogen bond with the other more electronegative atom.

Question 52. Strongest hydrogen bond is shown by

1. Water
2. Ammonia
3. Hydrogen fluoride
4. Hydrogen sulfide.

Answer: 3. Hydrogen fluoride

H – F shows the strongest H-bonds because fluorine is the most electronegative.

Question 53. Which one shows maximum hydrogen bonding?

1. \(\mathrm{H}_2 \mathrm{O}\)
2. \(\mathrm{H}_2 \mathrm{Se}\)
3. \(\mathrm{H}_2 \mathrm{~S}\)
4. \(\mathrm{HF}\)

Answer: 1. \(\mathrm{H}_2 \mathrm{O}\)

H₂O shelves maximum H-bonding because each H₂O molecule is linked to four H₂O molecules through H-bonds.

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