CBSE Class 11 Chemistry HOTs Redox Reactions Set 02

Higher Order Thinking Questions

Question 1. In the reaction 2FeCl₃ + H₂S → 2FeCl₂ + 2HCl + S:
(a) FeCl₃ acts as an oxidising agent
(b) Both H₂S and FeCl₃ are oxidised
(c) FeCl₃ is oxidised while H₂S is reduced
(d) H₂S acts as an oxidising agent
Answer:
Ans. (a) Fe goes from +3 to +2 (reduced), so FeCl₃ is the oxidising agent. S goes from −2 to 0 (oxidised), so H₂S is the reducing agent.
In simple words: FeCl₃ gains an electron, so it oxidises H₂S.

Teacher's Note:
1. Say which element changes and by how much.

Question 2. When KMnO₄ reacts with acidified FeSO₄:
(a) Only FeSO₄ is oxidised
(b) Only KMnO₄ is oxidised
(c) FeSO₄ is oxidised and KMnO₄ is reduced
(d) None of the above
Answer:
Ans. (c) Fe²⁺ is oxidised to Fe³⁺ and Mn (+7) is reduced to Mn²⁺ (+2). Oxidation and reduction occur together.
In simple words: Iron rises from +2 to +3 and manganese falls from +7 to +2.

Teacher's Note:
1. Redox reactions always have both changes.

Question 3. Which of the following is a redox reaction?
(a) H₂SO₄ with NaOH
(b) Formation of O₃ from O₂ in the atmosphere by lightning
(c) Formation of nitrogen oxides from nitrogen and oxygen by lightning
(d) Evaporation of H₂O
Answer:
Ans. (c) N₂ + O₂ → 2NO: nitrogen goes from 0 to +2 and oxygen from 0 to −2, so it is a redox reaction.
H₂SO₄ with NaOH is a neutralisation (no change), O₂ → O₃ is 0 to 0 and evaporation is a physical change.
In simple words: Only the nitrogen–oxygen reaction changes oxidation numbers.

Teacher's Note:
1. Check whether any element changes oxidation number.

Question 4. Which of the following reactions do not involve oxidation or reduction? (more than one may be correct)
(a) 2Rb + 2H₂O → 2RbOH + H₂
(b) 2CuI₂ → 2CuI + I₂
(c) NH₄Cl + NaOH → NaCl + NH₃ + H₂O
(d) 4KCN + Fe(CN)₂ → K₄[Fe(CN)₆]
Answer:
Ans. (c) and (d)
(a) Rb goes from 0 to +1 and H from +1 to 0: redox. (b) Cu goes from +2 to +1 and I from −1 to 0: redox.
(c) is an acid–base reaction and (d) is the formation of a complex; no oxidation numbers change in either.
In simple words: Reactions (c) and (d) leave every oxidation number unchanged.

Teacher's Note:
1. Do the oxidation-number check for each option before choosing.

Question 5. Which of the following can act both as an oxidising agent and as a reducing agent? (more than one may be correct)
(a) HNO₂
(b) H₂O₂
(c) H₂S
(d) SO₂
Answer:
Ans. (a), (b) and (d)
N in HNO₂ is +3 (between −3 and +5), O in H₂O₂ is −1 (between −2 and 0) and S in SO₂ is +4 (between −2 and +6), so each can act both ways.
S in H₂S is −2, its lowest state, so H₂S can only act as a reducing agent.
In simple words: Intermediate oxidation state means both roles are possible.

Teacher's Note:
1. H₂S is the odd one out because sulphur is already at its minimum.

Question 6. Assertion (A): MnO₂ can act as an oxidising agent as well as a reducing agent.
Reason (R): The oxidation state of Mn in MnO₂ lies between its highest and lowest oxidation states.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) Mn in MnO₂ is +4, between +7 (highest) and +2 (lowest, in the common compounds). So it can be reduced (to Mn²⁺) or oxidised (to MnO₄²⁻ and MnO₄⁻). Hence R explains A.
In simple words: An intermediate state lets MnO₂ go up or down.

Teacher's Note:
1. Link Reason to Assertion with the words 'because it is intermediate'.

Question 7. The oxidation number of cobalt in K₃[Co(NO₂)₆] is:
(a) 0
(b) +4
(c) +3
(d) +6
Answer:
Ans. (c) The complex ion is [Co(NO₂)₆]³⁻ (three K⁺ balance it). Each NO₂⁻ ligand is −1, so x + 6(−1) = −3 and x = +3.
In simple words: Six ligands of −1 and an overall −3 charge give Co = +3.

Teacher's Note:
1. First find the charge on the complex ion, then solve for Co.

Question 8. Phosphorus has the oxidation state of +3 in:
(a) Phosphorous acid
(b) Orthophosphoric acid
(c) Hypophosphorous acid
(d) Metaphosphoric acid
Answer:
Ans. (a) Phosphorous acid H₃PO₃: 3(+1) + x + 3(−2) = 0, so x = +3.
H₃PO₄ and HPO₃ have P at +5, and H₃PO₂ has P at +1.
In simple words: Count H as +1 and O as −2 in each acid.

Teacher's Note:
1. Remember: H₃PO₂ (+1), H₃PO₃ (+3), H₃PO₄ (+5).

Question 9. When Cl₂ reacts with cold aqueous NaOH, the oxidation number of chlorine changes from 0 to: (more than one may be correct)
(a) −1
(b) +1
(c) −2
(d) +2
Answer:
Ans. (a) and (b) Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O. Chlorine is oxidised to ClO⁻ (+1) and reduced to Cl⁻ (−1).
In simple words: A disproportionation reaction gives one higher and one lower oxidation state.

Teacher's Note:
1. Chlorine is the only element here that is both oxidised and reduced.

Question 10. Assertion (A): KO₂ is a superoxide.
Reason (R): The oxidation state of oxygen in KO₂ is −2.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (c) KO₂ is a superoxide, so A is true. In a superoxide O is −1/2, not −2 (the usual oxide value), so R is false.
In simple words: A is true and R is false.

Teacher's Note:
1. Superoxide: −1/2. Peroxide: −1. Normal oxide: −2.

Question 11. The oxidation number of chlorine in NOClO₄ is ______.
Answer:
Ans. +7 NOClO₄ consists of NO⁺ and ClO₄⁻. In ClO₄⁻, x + 4(−2) = −1, so x = +7.
In simple words: Split the compound into its two ions first.

Teacher's Note:
1. ClO₄⁻ (perchlorate) has Cl at +7.

Question 12. Assertion (A): F₂ does not undergo disproportionation reactions.
Reason (R): Fluorine shows only 0 and −1 oxidation states.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) A fluorine atom is the most electronegative, so it has no positive oxidation state. With only 0 and −1 available, it cannot be both oxidised and reduced from 0. So R is the correct explanation of A.
In simple words: Disproportionation needs an intermediate state; fluorine has none.

Teacher's Note:
1. Compare chlorine, which shows −1, 0, +1, +3, +5 and +7.

Question 13. Identify the correct statements about the reaction P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻ (more than one may be correct)
(a) Phosphorus is undergoing reduction only
(b) Phosphorus is undergoing oxidation only
(c) Phosphorus is undergoing oxidation as well as reduction
(d) Hydrogen is undergoing neither oxidation nor reduction
Answer:
Ans. (c) and (d) P goes from 0 to −3 in PH₃ (reduction) and from 0 to +1 in H₂PO₂⁻ (oxidation). H stays at +1 throughout.
In simple words: White phosphorus is both oxidised and reduced, so this is a disproportionation reaction.

Teacher's Note:
1. Check H separately; it does not change.

Question 14. Assertion (A): White phosphorus reacts with aqueous NaOH to form PH₃ and NaH₂PO₂. It is a disproportionation reaction.
Reason (R): In a disproportionation reaction the same substance is oxidised as well as reduced simultaneously.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Answer:
Ans. (a) P₄ (0) is reduced to PH₃ (−3) and oxidised to NaH₂PO₂ (+1). R is the definition of disproportionation, so it explains A.
In simple words: Same substance, two opposite changes.

Teacher's Note:
1. Write the oxidation numbers before and after to support the assertion.