Vidaara.orgClass 11 · Chemistry
CodeVID-C11-08-T1-01
Assignment — Oxidation Number & Redox Concepts
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
Reduction in terms of electrons is:
- A.loss of electrons
- B.gain of electrons
- C.loss of protons
- D.no change
2.
The oxidation number of oxygen in OF2 is:
- A.-2
- B.-1
- C.+2
- D.0
3.
The oxidation number of nitrogen in HNO3 is:
- A.+3
- B.+4
- C.+5
- D.-3
4.
An oxidising agent is a substance that:
- A.loses electrons
- B.gains electrons
- C.is oxidised
- D.donates protons only
5.
Reaction 2KClO3 → 2KCl + 3O2 is classified as:
- A.combination
- B.decomposition
- C.displacement
- D.comproportionation
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Define oxidation and reduction in terms of electron transfer and state the OIL RIG rule.
7.
Find the oxidation number of Cr in K2Cr2O7.
8.
Distinguish between an oxidising agent and a reducing agent with one example each.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Identify the oxidising and reducing agents in MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O.
10.
Explain disproportionation with the example of H2O2 decomposition, citing oxidation numbers.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State the rules for assigning oxidation numbers and use them to find the oxidation states of all elements in K2MnO4 and in S2O32-.
Answer Key
Section A — Multiple Choice Questions
- (B) gain of electrons
- (C) +2
- (C) +5
- (B) gains electrons
- (B) decomposition
Section B — Short Answer (2 marks)
- Oxidation is loss of electrons and reduction is gain of electrons. OIL RIG = Oxidation Is Loss, Reduction Is Gain.
- 2(+1) + 2x + 7(-2) = 0 gives 2x = +12, so Cr = +6.
- An oxidising agent gains electrons and is reduced (e.g. KMnO₄); a reducing agent loses electrons and is oxidised (e.g. Zn).
Section C — Short Answer (3 marks)
- Mn goes +4 to +2 (reduced) so MnO₂ is the oxidising agent; Cl goes -1 to 0 (oxidised) so HCl is the reducing agent.
- In 2H₂O₂ → 2H₂O + O₂, oxygen at -1 is both reduced to -2 (in H₂O) and oxidised to 0 (in O₂); the same element changes in both directions, which defines disproportionation.
Section D — Long Answer (5 marks)
- Rules: free elements 0; ions equal their charge; O = -2 (peroxide -1), H = +1 (hydride -1); Group 1 = +1; sum = charge. K₂MnO₄: K = +1, O = -2, so 2(+1) + x + 4(-2) = 0 gives Mn = +6. In S₂O₃²⁻ (thiosulphate): 2x + 3(-2) = -2 gives 2x = +4, so the average oxidation number of S = +2 (structurally the two S atoms differ, but the average is +2).
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