Question

The amphoteric oxide among $\mathrm{V}_2 \mathrm{O}_3, \mathrm{~V}_2 \mathrm{O}_4$ and $\mathrm{V}_2 \mathrm{O}_5$, upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

• A. + 4
• B. + 5
• C. + 3
• D. + 7

Hint:

Identify which vanadium oxide is amphoteric, write the product of its reaction with alkali and find the oxidation state of V in that anion.

Answer 1

The Correct Option is B

Step 1: Identify the amphoteric oxide. Oxides of a metal become more acidic as its oxidation state rises, so of $V_2O_3$ (+3), $V_2O_4$ (+4) and $V_2O_5$ (+5) it is the highest one, $V_2O_5$, that is amphoteric (it dissolves in both acids and alkalis).
Step 2: Reaction with alkali: $V_2O_5$ combines with hydroxide to give the oxide anion $VO_4^{3-}$ (orthovanadate).
Step 3: Oxidation state of V: the four O atoms contribute $-8$ and the ion carries $-3$, so V is $-3 + 8 = +5$. The oxidation state does not change in an acid-base reaction, which agrees with V being +5 in $V_2O_5$.
The required oxidation state is $+5$, option (2).