$V_{2} O_{5}$ reacts with alkalies as well as acids to give:

V O_{4}^{3 -}

and

V O^{2 +}

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V O^{2 +}

and

V O_{4}^{+}

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V O_{2}^{+}

and

V O^{2 +}

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V O_{4}^{3 -}

and

V O_{4}^{+}

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Solution

The correct option is A $V O_{4}^{3 -}$ and $V O^{2 +}$
$V_{2} O_{5}$ is, amphoteric in nature. But mainly it is acidic. Still, it gives $v V O_{4}^{3 -}$ as well as $V O^{2 +}$ salts. In vanadium, there will be a gradual change from the basic $V_{2} O_{3}$ to less basic $V_{2} O_{4}$ and to the amphoteric $V_{2} O_{5}$ . $V_{2} O_{4}$ dissolves in acids to give $V O^{2 +}$ salts. Similarly, $V_{2} O_{5}$ reacts with alkalies as well as acids to give $V O_{4}^{3 -}$ and $V O^{2 +}$ respectively.

The correct option is A.

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Similar questions

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

V O_{2}

V O_{2}

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

{V O}^{2 +}

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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