JEE Main 2023 · 29 January · evening · Single correct · +4 / −1
The set of correct statements is :
(i) Manganese exhibits +7 oxidation state in its oxide.
(ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides.
(iii) Sc shows +4 oxidation state which is oxidizing in nature.
(iv) Cr shows oxidising nature in +6 oxidation state.
- (a) (ii), (iii) and (iv)
- (b) (ii) and (iii)
- (c) (i) and (iii)
- (d) (i), (ii) and (iv)
Answer and solution
Answer (d) (i), (ii) and (iv)
Solution
Chemistry ma’am · 3 steps
Check each metal's top state against its valence electrons: Sc has no fourth electron to give.
- Highest state rule. A transition metal reaches its highest oxidation state by losing all its (n-1)d and ns electrons. Oxygen is very good at stabilising these high states, which is why they show up in oxides.
Statement Valence electrons Check Verdict (i) Mn, +7 in oxide 3d⁵ 4s², 7 electrons Mn₂O₇ exists, Mn is +7 Correct (ii) Ru, Os, +8 in oxides 4d⁷ 5s¹ and 5d⁶ 6s², 8 electrons each RuO₄ and OsO₄ exist, metal is +8 Correct (iii) Sc, +4 and oxidising 3d¹ 4s², only 3 electrons Sc shows only +3. A fourth electron would have to come from the stable argon core Wrong (iv) Cr oxidising at +6 3d⁵ 4s¹, 6 electrons Cr₂O₇²⁻ is a strong oxidant and goes to stable Cr³⁺ Correct - Why Cr(VI) oxidises. Across the 3d series, the highest state gets less stable as you move right. So Cr in +6 readily accepts electrons and falls to +3, which makes dichromate a common oxidising agent.
- So. Statements (i), (ii) and (iv) are correct and (iii) is wrong, so the answer is (d).
The idea to keep. A transition metal's highest oxidation state equals its (n-1)d plus ns electrons, and oxides bring it out. Sc stops at +3, Mn reaches +7, and Ru and Os reach +8.
Answer from the NTA final answer key. Solution by PrepWiser.