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Multiple Choice Questions

One and More than One Correct

1. Reduction of the metal centre in aqueous permanganate ion involves [JEE 2011 P-II]

(a) 3 electrons in neutral medium

(b) 5 electrons in neutral medium

(d) 5 electrons in acidic medium

Answer: (a, d)

Explanation: In acidic medium,

$$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \longrightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$$

In neutral medium,

$$\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \longrightarrow \text{MnO}_2 + 4\text{OH}^-$$

Hence, number of electron loose in acidic and neutral medium 5 and 3 electrons respectively.

2. The equilibrium $2\text{Cu}^+ \rightleftharpoons \text{Cu}^{2+} + \text{Cu}^\circ$ in aqueous medium at $25^\circ\text{C}$ shifts towards the left in the presence of [JEE 2011 P-II]

(a) $\text{NO}_3^-$

(b) $\text{Cl}^-$

(d) $\text{CN}^-$

Answer: (b, c, d)

Explanation: $\text{Cu}^{2+}$ ions will react with $\text{CN}^-$ and $\text{SCN}^-$ forming $[\text{Cu}(\text{CN})_4]^{3-}$ and $[\text{Cu}(\text{SCN})_4]^{3-}$ leading the reaction in the backward direction.

$$\text{Cu}^{2+} + 2\text{CN}^- \longrightarrow \text{Cu}(\text{CN})_2$$

$$2\text{Cu}(\text{CN})_2 \longrightarrow 2\text{CuCN} + (\text{CN})_2$$

$$\text{CuCN} + 3\text{CN}^- \longrightarrow [\text{Cu}(\text{CN})_4]^{3-}$$

$$\text{Cu}^{2+} + 4\text{SCN}^- \longrightarrow [\text{Cu}(\text{SCN})_4]^{3-}$$

$\text{Cu}^{2+}$ also combines with $\text{Cu}$ which reacts with $\text{CuCl}_2$ to produce $\text{CuCl}$ pushing the reaction in the backward direction.

$$\text{Cu} + \text{CuCl}_2 \longrightarrow 2\text{CuCl} \downarrow$$

3. For the reaction: $\text{I}^- + \text{ClO}_3^- + \text{H}_2\text{SO}_4 \longrightarrow \text{Cl}^- + \text{HSO}_4^- + \text{I}_2 + \text{H}_2\text{O}$ The correct statement(s) in the balanced equation is/are [JEE Adv. 2014 P-I]

(a) Stoichiometric coefficient of $\text{HSO}_4^-$ is 6

(b) Iodide is oxidised

(d) $\text{H}_2\text{O}$ is one of the products.

Answer: (a, b, d)

Explanation: Balanced chemical equation:

$$\text{ClO}_3^- + 6\text{I}^- + 6\text{H}_2\text{SO}_4 \longrightarrow 3\text{I}_2 + \text{Cl}^- + 6\text{HSO}_4^- + 3\text{H}_2\text{O}$$

Stoichiometric coefficient of $\text{HSO}_4^-$ is 6, iodide ($\text{I}^-$) is oxidised to $\text{I}_2$, and $\text{H}_2\text{O}$ is one of the products.

4. From the following statements regarding $\text{H}_2\text{O}_2$, choose the incorrect statement [JEE Main 2015]

(a) It can act only as an oxidising agent

(b) It decomposes on exposure to light

(d) It has to be kept away from dust

Answer: (a)

Explanation: It can act as an oxidising as well as reducing agent.

5. Which of the following reactions is an example of redox reaction? [JEE Main 2017]

(a) $\text{XeF}_6 + 2\text{H}_2\text{O} \longrightarrow \text{XeO}_2\text{F}_2 + 4\text{HF}$

(b) $\text{XeF}_4 + \text{O}_2\text{F}_2 \longrightarrow \text{XeF}_6 + \text{O}_2$

(d) $\text{XeF}_6 + \text{H}_2\text{O} \longrightarrow \text{XeOF}_4 + 2\text{HF}$

Answer: (b)

Explanation: In the reaction

$$\overset{+4}{\text{Xe}}\text{F}_4 + \overset{+1}{\text{O}_2}\text{F}_2 \longrightarrow \overset{+6}{\text{Xe}}\text{F}_6 + \overset{0}{\text{O}_2}$$

Xenon undergoes oxidation while oxygen undergoes reduction.

6. The oxidation states of Cr in $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$, $[\text{Cr}(\text{C}_6\text{H}_6)_2]$, and $\text{K}_2[\text{Cr}(\text{CN})_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$ respectively are: [JEE Main 2018]

(a) +3, 0 and +4

(b) +3, +4 and +6

(d) +3, 0 and +6

Answer: (d)

Explanation: $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$

$\Rightarrow x + 0 \times 6 - 1 \times 3 = 0 \implies x = +3$

$[\text{Cr}(\text{C}_6\text{H}_6)_2]$

$\Rightarrow x + 2 \times 0 = 0 \implies x = 0$

$\text{K}_2[\text{Cr}(\text{CN})_2(\text{O})_2(\text{O}_2)(\text{NH}_3)]$

$\Rightarrow 1 \times 2 + x - 1 \times 2 - 2 \times 2 - 2 \times 1 = 0 \implies x - 6 = 0 \implies x = +6$

7. In order to oxidise a mixture one mole of each of $\text{FeC}_2\text{O}_4$, $\text{Fe}_2(\text{C}_2\text{O}_4)_3$, $\text{FeSO}_4$ and $\text{Fe}_2(\text{SO}_4)_3$ in acidic medium , the number of moles of $\text{KMnO}_4$ required is: [JEE Main 2019]

(a) 3

(b) 2

(d) 1.5

Answer: (b)

Explanation: $\text{eq. } \text{KMnO}_4 = \text{eq. } [\text{FeC}_2\text{O}_4 + \text{Fe}_2(\text{C}_2\text{O}_4)_3 + \text{FeSO}_4]$

Or $n \times 5 = 1 \times 3 + 1 \times 6 + 1 \times 1$

$\therefore n \times 5 = 10 \implies n = 2$

8. The one that is extensively used as a piezoelectric material is: [JEE Main 2019]

(a) Quartz

(b) Amorphous silica

(d) Tridymite

Answer: (a)

Explanation: Quartz (Information)

9. Good reducing nature of $\text{H}_3\text{PO}_2$ attributed to the presence of: [JEE Main 2019]

(a) one P-OH bond

(b) one P-H bond

(d) two P-OH bonds

Answer: (c)

Explanation: $\text{H}_3\text{PO}_2$ is good reducing agent due to presence of two P—H bonds.

10. The chemical nature of hydrogen peroxide is:- [JEE Main 2019]

(a) Oxidising and reducing agent in acidic medium, but not in basic medium.

(b) Oxidising and reducing agent in both acidic and basic medium

(d) Oxidising agent in acidic medium, but not in basic medium.

Answer: (b)

Explanation: $\text{H}_2\text{O}_2$ act as oxidising agent and reducing agent in acidic medium as well as basic medium.

$\text{H}_2\text{O}_2$ act as oxidant:

$$\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \longrightarrow 2\text{H}_2\text{O} \quad \text{(In acidic medium)}$$

$$\text{H}_2\text{O}_2 + 2e^- \longrightarrow 2\text{OH}^- \quad \text{(In basic medium)}$$

$\text{H}_2\text{O}_2$ Act as reductant:

$$\text{H}_2\text{O}_2 \longrightarrow 2\text{H}^+ + \text{O}_2 + 2e^- \quad \text{(In acidic medium)}$$

$$\text{H}_2\text{O}_2 + 2\text{OH}^- \longrightarrow 2\text{H}_2\text{O} + \text{O}_2 + 2e^- \quad \text{(In basic medium)}$$

11. The element that usually does not show variable oxidation states is: [JEE Main 2019]

(a) V

(b) Ti

(d) Cu

Answer: (c)

Explanation: Usually Sc (Scandium) does not show variable oxidation states.

Most common oxidation states of:

(i) Sc : +3

(ii) V : +2, +3, +4, +5

(iii) Ti : +2, +3, +4

(iv) Cu : +1, +2

12. The relative stability of +1 oxidation state of group 13 elements follows the order:- [JEE Main 2019]

(a) Al < Ga < Tl < In

(b) Tl < In < Ga < Al

(d) Ga < Al < In < Tl

Answer: (c)

Explanation: Due to inert pair effect as we move down the group in 13th group lower oxidation state becomes more stable.

$$\text{Al} < \text{Ga} < \text{In} < \text{Tl}$$

13. Iodine reacts with concentrated $\text{HNO}_3$ to yield Y along with other products. The oxidation state of iodine in Y is:- [JEE Main 2019]

(a) 5

(b) 3

(d) 7

Answer: (a)

Explanation: $$\text{I}_2 + 10\text{HNO}_3 \longrightarrow 2\text{HIO}_3 + 10\text{NO}_2 + 4\text{H}_2\text{O}$$

In $\text{HIO}_3$ oxidation state of iodine is +5. Closest option is (a).

Matrix Match/Column Match

14. The standard reduction potential data at $25^\circ\text{C}$ is given below: [JEE Adv. 2013 P-II]

$$E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\text{ V}; \quad E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V}$$

$$E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}; \quad E^\circ(\text{Cu}^+/\text{Cu}) = +0.52\text{ V}$$

$$E^\circ[\text{O}_2(\text{g}) + 4\text{H}^+ + 4e^- \longrightarrow 2\text{H}_2\text{O}] = +1.23\text{ V}$$

$$E^\circ[\text{O}_2(\text{g}) + 2\text{H}_2\text{O} + 4e^- \longrightarrow 4\text{OH}^-] = +0.40\text{ V}$$

$$E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74\text{ V}; \quad E^\circ(\text{Cr}^{2+}/\text{Cr}) = -0.91\text{ V}$$

Match of the redox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists:

Column I	Column II
(A) $E^\circ(\text{Fe}^{3+}/\text{Fe})$	1. $-0.18\text{ V}$
(B) $E^\circ(4\text{H}_2\text{O} \rightleftharpoons 4\text{H}^+ + 4\text{OH}^-)$	2. $-0.4\text{ V}$
(C) $E^\circ(\text{Cu}^{2+} + \text{Cu} \longrightarrow 2\text{Cu}^+)$	3. $-0.04\text{ V}$
(D) $E^\circ(\text{Cr}^{3+}/\text{Cr}^{2+})$	4. $-0.83\text{ V}$

(a) A$\rightarrow$ 3; B$\rightarrow$ 4; C$\rightarrow$ 1; D$\rightarrow$ 2

(b) A$\rightarrow$ 1; B$\rightarrow$ 2; C$\rightarrow$ 3; D$\rightarrow$ 4

(d) A$\rightarrow$ 3; B$\rightarrow$ 4; C$\rightarrow$ 2; D$\rightarrow$ 1

Answer: (a)

Explanation: A$\rightarrow$ 3; B$\rightarrow$ 4; C$\rightarrow$ 1; D$\rightarrow$ 2

(A) $\Delta G^\circ(\text{Fe}^{3+}/\text{Fe}) = \Delta G^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) + \Delta G^\circ(\text{Fe}^{2+}/\text{Fe})$

$$\implies -3FE^\circ(\text{Fe}^{3+}/\text{Fe}) = -1FE^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) + (-2FE^\circ(\text{Fe}^{2+}/\text{Fe}))$$

$$\implies E^\circ(\text{Fe}^{3+}/\text{Fe}) = -0.04\text{ V}$$

(B) $$\text{O}_2(\text{g}) + 2\text{H}_2\text{O} + 4e^- \longrightarrow 4\text{OH}^- \quad E^\circ = 0.40\text{ V}$$

$$2\text{H}_2\text{O} \longrightarrow \text{O}_2(\text{g}) + 4\text{H}^+ + 4e^- \quad E^\circ = -1.23\text{ V}$$

$$4\text{H}_2\text{O} \rightleftharpoons 4\text{H}^+ + 4\text{OH}^- \quad E^\circ = 0.40 - 1.23 = -0.83\text{ V}$$

(C) $\Delta G^\circ(\text{Cu}^{2+}/\text{Cu}^+) = \Delta G^\circ(\text{Cu}^{2+}/\text{Cu}) + \Delta G^\circ(\text{Cu}/\text{Cu}^+)$

$$\implies -2FE^\circ(\text{Cu}^{2+}/\text{Cu}) = -1FE^\circ(\text{Cu}^{2+}/\text{Cu}^+) + (-1FE^\circ(\text{Cu}^+/\text{Cu}))$$

$$\implies E^\circ(\text{Cu}^{2+}/\text{Cu}^+) = -0.18\text{ V}$$

(D) $\Delta G^\circ(\text{Cr}^{3+}/\text{Cr}) = \Delta G^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) + \Delta G^\circ(\text{Cr}^{2+}/\text{Cr})$

$$\implies -3FE^\circ(\text{Cr}^{3+}/\text{Cr}) = -1FE^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) + (-2FE^\circ(\text{Cr}^{2+}/\text{Cr}))$$

$$\implies E^\circ(\text{Cr}^{3+}/\text{Cr}^{2+}) = -0.4\text{ V}$$

Integer and Subjective

15. Among the following the number of elements showing only one non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti. [JEE 2010 P-II]

Answer: 2

Explanation: Na, F show only one non-zero oxidation state. Answer is no. of elements is 2.