JEE Main 2023 · 31 January · morning · Single correct · +4 / −1
The correct order of basicity of oxides of vanadium is :
- (a) V₂O₃ > V₂O₅ > V₂O₄
- (b) V₂O₃ > V₂O₄ > V₂O₅
- (c) V₂O₅ > V₂O₄ > V₂O₃
- (d) V₂O₄ > V₂O₃ > V₂O₅
Answer and solution
Answer (b) V₂O₃ > V₂O₄ > V₂O₅
Solution
Chemistry ma’am · 3 steps
Find vanadium's oxidation state in each oxide: the higher it climbs, the more acidic the oxide gets.
- Oxidation states. Each O is -2. In V₂O₃, 2x = 6, so V is +3. In V₂O₄ it is +4, and in V₂O₅ it is +5.
- Why charge matters. A higher positive charge makes the metal ion smaller and more polarising. It pulls electron density from oxygen and makes the V–O bond more covalent. A low-valent oxide gives O²⁻ and acts as a base. A high-valent oxide behaves like a non-metal oxide and acts as an acid.
Oxide Oxidation state of V Nature V₂O₃ +3 basic V₂O₄ +4 amphoteric, less basic V₂O₅ +5 amphoteric, mainly acidic - So. Basicity falls as the oxidation state rises: V₂O₃ > V₂O₄ > V₂O₅, which is option (b).
The idea to keep. For a metal with several oxides, the lowest oxidation state gives the most basic oxide and the highest gives the most acidic one. Think MnO basic and Mn₂O₇ acidic.
Answer from the NTA final answer key. Solution by PrepWiser.