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Faraday's First Law

VO 2 2+ and ...

Question

${V O}_{2}^{2 +}$ and ${V O}^{2 +}$ are known as vanadyl ion. The oxidation number of vanadium in each ions are, respectively :

+ 6, - 4

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- 6, - 4

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- 6, + 4

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+ 6, + 4

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Solution

The correct option is B $+ 6, + 4$
The oxidation number of $V$ is $+ 6$ in ${V O}_{2}^{2 +}$ and $+ 4$ in ${V O}^{2 +}$ .

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