The amphoteric oxide among $$V_{2}O_{3}$$ and $$V_{2}O_{5}$$ , upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

We need to identify the amphoteric oxide between $$V_2O_3$$ and $$V_2O_5$$, and find the oxidation state of V in the oxide anion formed with alkali.

In vanadium oxides, the nature changes with oxidation state: $$V_2O_3$$ (V in +3 state) is a basic oxide because lower oxidation state oxides of transition metals tend to be basic, whereas $$V_2O_5$$ (V in +5 state) is an amphoteric oxide that can react with both acids and bases.

When $$V_2O_5$$ reacts with NaOH (alkali), it forms the metavanadate ion according to $$V_2O_5 + 2NaOH \rightarrow 2NaVO_3 + H_2O$$. The oxide anion formed is $$VO_3^-$$.

To calculate the oxidation state of vanadium in $$VO_3^-$$, let the oxidation state of V be $$x$$; each oxygen contributes $$-2$$ and the overall charge is $$-1$$, so $$x + 3(-2) = -1$$, which simplifies to $$x - 6 = -1$$ and thus $$x = +5$$.

The correct answer is Option D: +5.