The amphoteric oxide among V₂ O₃, V₂ O₄ and V₂ O₅, upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

Solution

Chemistry ma’am · 4 steps

Find the amphoteric oxide first. An acid-base reaction never changes the metal's oxidation state.

1. Trend in acidity. For the oxides of one metal, a higher oxidation state makes the oxide more covalent and more acidic. So vanadium's oxides shift from basic towards acidic as the oxidation state rises.

   Oxide	Oxidation state of V	Nature
   V₂O₃	+3	basic
   V₂O₄	+4	less basic, dissolves in acids to give VO²⁺ salts
   V₂O₅	+5	amphoteric, though mainly acidic

2. Reaction with alkali. V₂O₅ reacts with alkali to give the vanadate ion: V₂O₅ + 6OH⁻ → 2VO₄³⁻ + 3H₂O. This is an acid-base reaction, so no electrons are transferred.
3. Check the anion. In VO₄³⁻, take V as x: x + 4(-2) = -3, so x = +5. This is the same as in V₂O₅.
4. So. Vanadium is in the +5 oxidation state in the oxide anion, which is option (b).

The idea to keep. Oxides of the same metal grow more acidic as its oxidation state rises. When an oxide dissolves in acid or alkali, the metal keeps the oxidation state it had in the oxide.

Answer from the NTA final answer key. Solution by PrepWiser.