The oxidation state of $M o$ in its oxido-complex species $[M o_{2} O_{4} (C_{2} H_{4})_{2} (H_{2} O)_{2}]^{2 -}$ is:

+ 2

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+ 3

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+ 4

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+ 5

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Solution

The correct option is B $+ 3$
We know that $H_{2} O$ and $C_{2} H_{4}$ are neutral ligands. So, they carry zero charge. $O$ has $- 2$ charge.

Let $x$ be the oxidation state of $M o$ in $[M o_{2} O_{4} (C_{2} H_{4})_{2} (H_{2} O)_{2}]^{2 -}$ .
So, by substituting the values, we get
$2 x - 8 = - 2$
$\Rightarrow x = + 3$ .

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M o

[M o_{2} O_{4} (C_{2} H_{4})_{2} (H_{2} O)_{2}]^{2 -}

[M o_{2} O_{4} (C_{2} H_{4})_{2} (H_{2} O_{2})]^{2 -}]

_{2}

_{4}

_{2}

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_{2}

_{2}

_{2}

^{2 -}

\frac{- 1}{2}

N a_{2} M o - -- - O_{4}

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