1. Identify the amphoteric oxide among the given vanadium oxides

   The oxidation states of vanadium are:

   • In $\mathrm{V_2O_3}$: $2x+3(-2)=0 \Rightarrow x=+3$
   • In $\mathrm{V_2O_4}$: $2x+4(-2)=0 \Rightarrow x=+4$
   • In $\mathrm{V_2O_5}$: $2x+5(-2)=0 \Rightarrow x=+5$

   In general:

   • Lower oxidation state oxides are more basic
   • Higher oxidation state oxides are more acidic
   • Intermediate oxidation state oxides are often amphoteric

   Therefore:

   • $\mathrm{V_2O_3}$ is mainly basic
   • $\mathrm{V_2O_5}$ is acidic
   • $\mathrm{V_2O_4}$ is amphoteric

2. Reaction of amphoteric oxide with alkali

   Since $\mathrm{V_2O_4}$ is amphoteric, it reacts with alkali to form an oxoanion (oxide anion of vanadium), namely vanadate-type species.

   On oxidation/alkaline treatment, vanadium forms vanadate ion where vanadium is in the $+5$ oxidation state, such as $\mathrm{VO_4^{3-}}$ or related oxoanions.

3. Find oxidation state of V in the oxide anion

   For example, in $\mathrm{VO_4^{3-}}$: $x+4(-2)=-3$ $x-8=-3$ $x=+5$

   Hence, the oxidation state of V in the oxide anion formed is $+5$.

4. Check options

   • A: $+4$ ❌
   • B: $+5$ ✅
   • C: $+3$ ❌
   • D: $+7$ ❌

Therefore, the correct answer is B.