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 :
Solution
<p>$$
\begin{aligned}
&\mathrm{V}_2 \mathrm{O}_5+\text { alkali } \rightarrow \mathrm{VO}_4^{3-}\\
&\text { In } \mathrm{VO}_4^{3-} \text { ion, vanadium is in }+5 \text { oxidation state }
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: d and f Block Elements · Topic: Properties of Transition Metals
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