In which compounds does vanadium have an oxidation number of +4?

N H_{4} V O_{2}

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K_{4} [V (C N)_{6}]

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V O S O_{4}

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C r O_{2} C l_{2}

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Solution

The correct option is B $V O S O_{4}$
Oxidation state of V in $N H_{4} V O_{2}$ is +3.
Oxidation state of V in $K_{4} [V (C N)_{6}]$ is +2.
Oxidation state of Cr in $C r O_{2} C l_{2}$ is +6.
And
$V O S O_{4} = V O^{2 +} + S O_{4}^{2 -}$ ;
Let the oxidation number of V be X. So, X x(-2) = + 2
$∴ X = + 4$

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