JEE Main 2022 · 27 July · morning · Single correct · +4 / −1
Given below are two statements.
Statement I : Iron (III) catalyst, acidified K₂ Cr₂ O₇ and neutral KMnO₄ have the ability to oxidise I⁻ to I₂ independently.
Statement II : Manganate ion is paramagnetic in nature and involves p π-p π bonding.
In the light of the above statements, choose the correct answer from the options given below.
- (a) Both Statement I and Statement II are true.
- (b) Both Statement I and Statement II are false.
- (c) Statement I is true but Statement II is false.
- (d) Statement I is false but Statement II is true.
Answer and solution
Answer (b) Both Statement I and Statement II are false.
Solution
Chemistry ma’am · 3 steps
Check each oxidant's product: neutral KMnO₄ takes iodide all the way to iodate, not to I₂.
- Statement I. Fe³⁺ and acidified K₂Cr₂O₇ do give I₂. Neutral or faintly alkaline KMnO₄ is stronger here and oxidises I⁻ (oxidation state -1) all the way to iodate, IO₃⁻ (oxidation state +5). Since not all three give I₂, Statement I is false.
2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻
- Statement II. In manganate, MnO₄²⁻, Mn is +6, so it is 3d¹ with one unpaired electron. That makes it paramagnetic, and this half is correct. But its π bonds form when filled oxygen p orbitals overlap with empty Mn d orbitals. That is pπ–dπ bonding, not pπ–pπ, so Statement II is false.
- So. Both statements are false, which is option (b).
The idea to keep. Neutral or alkaline KMnO₄ turns I⁻ into IO₃⁻, while acidified oxidants stop at I₂. In MnO₄⁻ and MnO₄²⁻, the π bonds are always pπ–dπ.
Answer from the NTA final answer key. Solution by PrepWiser.