$V O_{2}$ is an amphoteric oxide and in acidic medium it forms:

V O^{2 +}

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V O_{2}^{+}

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V O^{3 +}

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V O_{2}^{2 +}

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Solution

The correct option is A $V O^{2 +}$

$V O_{2}$ is an amphoteric in nature and it reacts with alkalies as well as with acids to give $V_{4} O_{9}^{2 -}$ and $V O^{2 +}$ respectively.
Hence option A is correct.

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V O^{2 +}

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M n^{2 +}

{V O}_{4}^{3 -} + {F e}^{2 +} ⟶ {F e}^{3 +} + {V O}^{2 +}

V O_{2}

{V O}_{2}^{2 +}

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V^{2 +} + V O^{2 +} + 2 H^{+} \to 2 V^{3 +} + H_{2} O; E^{o} = 0.616 V

V^{3 +} + A g^{+} + H_{2} O \to V O^{2 +} + 2 H^{+} + A g; E^{o} = 0.439 V

E_{V^{3 +} / V^{2 +}}^{o}

E_{A g^{+} / A g}^{o} = 0.799 V

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