Oxidation–Reduction and Oxidation Number Electronic concepts of oxidation and reduction, oxidation number, rules for assigning oxidation number,: NEET MCQs with Answers

15 free questions on Oxidation–Reduction and Oxidation Number Electronic concepts of oxidation and reduction, oxidation number, rules for assigning oxidation number, for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
Which oxidation states are NOT exhibited by Mn in the reaction given below? [Re-NEET 2024]
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
A. +6 B. +2 C. +4 D. +7 E. +3
  1. D and E only
  2. B and D only
  3. A and B only
  4. B and E only
Show answer
Answer: (D) B and E only
Explanation: Mn is +6 in MnO₄²⁻, +7 in MnO₄⁻ and +4 in MnO₂. Thus +6 (A), +4 (C) and +7 (D) are shown, while +2 (B) and +3 (E) are not shown. Hence the answer is B and E only.

Question 2

Previous-year question
How many moles of MnO₄⁻ are reduced to Mn²⁺ in acidic medium by 4.517 × 10²⁴ electrons? [NEET 2024]
  1. 7.5 moles
  2. 2.5 moles
  3. 5.0 moles
  4. 1.5 moles
Show answer
Answer: (D) 1.5 moles
Explanation: Half-reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so 1 mol MnO₄⁻ needs 5 mol electrons = 5 × 6.022 × 10²³ = 30.11 × 10²³ electrons. Moles of electrons supplied = 4.517 × 10²⁴ / 6.022 × 10²³ = 7.5 mol. Moles of MnO₄⁻ reduced = 7.5/5 = 1.5 mol.

Question 3

Previous-year question
Which of the following reactions is NOT a disproportionation reaction? [NEET 2024 (Swaimadhopur)]
  1. Cl₂(g) + 2OH⁻(aq) → ClO⁻(aq) + Cl⁻(aq) + H₂O(l)
  2. 2NO₂(g) + 2OH⁻(aq) → NO₂⁻(aq) + NO₃⁻(aq) + H₂O(l)
  3. 2H₂O₂(aq) → 2H₂O(l) + O₂(g)
  4. 2F₂(g) + 2OH⁻(aq) → 2F⁻(aq) + OF₂(g) + H₂O(l)
Show answer
Answer: (D) 2F₂(g) + 2OH⁻(aq) → 2F⁻(aq) + OF₂(g) + H₂O(l)
Explanation: Fluorine, being the most electronegative element, cannot show a positive oxidation state; in this reaction F goes from 0 to −1 in both F⁻ and OF₂, while O is oxidised from −2 to +2. Since F is only reduced, it is not a disproportionation reaction. In the reaction of Cl₂ with OH⁻, Cl goes from 0 to +1 and −1; in the reaction of NO₂ with OH⁻, N goes from +4 to +3 and +5; and in the decomposition of H₂O₂, O goes from −1 to −2 and 0. Each of these is a disproportionation.

Question 4

Previous-year question
Identify the set in which every species is capable of undergoing a disproportionation reaction. [NEET 2022 (Overseas)]
  1. Cl₂, ClO₂⁻, ClO₃⁻, S₈
  2. ClO₄⁻, ClO⁻, ClO₂⁻, F₂
  3. ClO₃⁻, ClO₄⁻, H₂O₂, ClO⁻
  4. ClO₂⁻, ClO₃⁻, ClO₄⁻, Cl₂
Show answer
Answer: (A) Cl₂, ClO₂⁻, ClO₃⁻, S₈
Explanation: A species can disproportionate only if its element is in an intermediate oxidation state, so it can be both oxidised and reduced. Cl₂ (0), ClO₂⁻ (+3), ClO₃⁻ (+5) and S₈ (0) all qualify. ClO₄⁻ (Cl +7, highest state) and F₂ (most electronegative, cannot show positive states) cannot disproportionate, so any set containing either of them is ruled out.

Question 5

Previous-year question
When the salt KHC₂O₄·H₂C₂O₄·4H₂O is used as a reducing agent, its equivalent weight is: [NEET 2021 (Paper-2)]
  1. Mol. wt/1
  2. Mol. wt/2
  3. Mol. wt/3
  4. Mol. wt/4
Show answer
Answer: (D) Mol. wt/4
Explanation: As a reducing agent each oxalate unit is oxidised to CO₂: C₂O₄²⁻ → 2CO₂ + 2e⁻ (C: +3 → +4). One formula unit contains 2 oxalate units (HC₂O₄⁻ and H₂C₂O₄), so n-factor = 2 × 2 = 4. Equivalent weight = Molecular weight/n-factor = Mol. wt/4.

Question 6

Previous-year question
The oxidation numbers of the underlined atoms in some species are given below. Identify the INCORRECT option. [NEET 2020 (COVID-19)]
  1. O in Cu₂O is −1
  2. Cl in ClO₃⁻ is +5
  3. Cr in K₂Cr₂O₇ is +6
  4. Au in HAuCl₄ is +3
Show answer
Answer: (A) O in Cu₂O is −1
Explanation: In Cu₂O, Cu is +1 and O is −2 (2(+1) + x = 0, x = −2), so stating −1 is incorrect. The others are correct: Cl in ClO₃⁻: x + 3(−2) = −1, x = +5; Cr in K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0, x = +6; Au in HAuCl₄: +1 + x + 4(−1) = 0, x = +3.

Question 7

Previous-year question
What is the change in the oxidation number of carbon in the following reaction? CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g) [NEET 2020]
  1. 0 to +4
  2. −4 to +4
  3. 0 to −4
  4. +4 to +4
Show answer
Answer: (B) −4 to +4
Explanation: In CH₄ hydrogen is +1, so carbon is −4. In CCl₄ chlorine is −1, so carbon is +4. The carbon has therefore been oxidised through eight units, from −4 to +4.
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Question 8

Previous-year question
The reaction of aqueous KMnO₄ with H₂O₂ in acidic conditions gives: [AIPMT 2014]
  1. Mn⁴⁺ and O₂
  2. Mn²⁺ and O₂
  3. Mn²⁺ and O₃
  4. Mn⁴⁺ and MnO₂
Show answer
Answer: (B) Mn²⁺ and O₂
Explanation: In acidic medium MnO₄⁻ (Mn⁷⁺) is reduced to Mn²⁺ while H₂O₂ is oxidised to O₂.

Question 9

Previous-year question
(a) H₂O₂ + O₃ → H₂O + 2O₂ (b) H₂O₂ + Ag₂O → 2Ag + H₂O + O₂. Role of hydrogen peroxide in the above reactions is respectively: [AIPMT 2014]
  1. Oxidizing in (a) and reducing in (b)
  2. Reducing in (a) and oxidizing in (b)
  3. Reducing in (a) and (b)
  4. Oxidizing in (a) and (b)
Show answer
Answer: (C) Reducing in (a) and (b)
Explanation: In both reactions H₂O₂ is oxidised (O: −1→0, evolving O₂) while reducing O₃ and Ag₂O, so it acts as a reducing agent in both.

Question 10

Previous-year question
In which of the following compounds, nitrogen exhibits highest oxidation state? [AIPMT 2012]
  1. N₃H
  2. NH₂OH
  3. N₂H₄
  4. NH₃
Show answer
Answer: (A) N₃H
Explanation: N states: N₃H (−1/3) > NH₂OH (−1) > N₂H₄ (−2) > NH₃ (−3). The highest (least negative) is in N₃H.

Question 11

Previous-year question
When Cl₂ gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from: [AIPMT 2012]
  1. Zero to −1 and zero to +3
  2. Zero to +1 and zero to −3
  3. Zero to +1 and zero to −5
  4. Zero to −1 and zero to +5
Show answer
Answer: (D) Zero to −1 and zero to +5
Explanation: Hot conc. NaOH: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O. Cl goes 0→−1 (NaCl) and 0→+5 (NaClO₃).

Question 12

Previous-year question
How much amount of CuSO₄·5H₂O required for liberation of 2.54 g I₂ when titrated with KI? [AIPMT 2011]
  1. 2.5 g
  2. 4.99 g
  3. 2.4 g
  4. 1.2 g
Show answer
Answer: (B) 4.99 g
Explanation: 2Cu²⁺ + 4I⁻ → 2CuI + I₂. Moles I₂ = 2.54/254 = 0.01, so Cu²⁺ = 0.02 mol. Mass = 0.02 × 249.5 ≈ 4.99 g.

Question 13

Previous-year question
When S₂O₈²⁻ oxidize Fe²⁺, then product formed is: [AIPMT 2010]
  1. S₂O₆²⁻
  2. SO₄²⁻
  3. S₂O₃²⁻
  4. SO₅²⁻
Show answer
Answer: (B) SO₄²⁻
Explanation: Peroxydisulphate oxidises Fe²⁺ to Fe³⁺ and is itself reduced to sulphate as its O–O peroxide bond breaks: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺.

Question 14

Previous-year question
Correct order of oxidizing strength is: [AIPMT 2010]
  1. MnO₄⁻, VO₂⁺, Cr₂O₇²⁻
  2. Cr₂O₇²⁻, MnO₄⁻, VO₂⁺
  3. Cr₂O₇²⁻, VO₂⁺, MnO₄⁻
  4. MnO₄⁻, Cr₂O₇²⁻, VO₂⁺
Show answer
Answer: (D) MnO₄⁻, Cr₂O₇²⁻, VO₂⁺
Explanation: Standard reduction potentials: MnO₄⁻ (1.51 V) > Cr₂O₇²⁻ (1.33 V) > VO₂⁺ (1.00 V), so oxidising strength decreases in that order.

Question 15

Previous-year question
Oxidation number of P in PO₄³⁻, of S in SO₄²⁻ and that of Cr in Cr₂O₇²⁻ are respectively: [AIPMT 2009]
  1. −3, +6 and +6
  2. +5, +6 and +6
  3. +3, +6 and +5
  4. +5, +3 and +6
Show answer
Answer: (B) +5, +6 and +6
Explanation: PO₄³⁻: P = +5; SO₄²⁻: S = +6; Cr₂O₇²⁻: Cr = +6.

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