When $K_{2} C r_{2} O_{7}$ is mixed with $H_{2} S O_{4}$ and thoroughly shaken with $H_{2} O_{2}$ in presence of ether, then a floated blue coloured complex $X$ is formed. The change in oxidation state and the percentage of $C r$ in the complex $X$ is:

+ 6, 49.4

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+ 4, 39.4

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0, 39.4

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+ 4, 59.4

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Solution

The correct option is C $0, 39.4$
When $K_{2} C r_{2} O_{7}$ reacts with $H_{2} O_{2}$ , a blue colored compound is formed which is $C r O_{5}$ .

The oxidation number of $C r$ in $C r O_{5}$ is $+ 6$ , because four of oxygen atoms are involved in peroxide linkage.

So, change in oxidation state from $K_{2} C r_{2} O_{7}$ to $C r O_{5}$ is $0$ .

The percentage of chromium in the complex is $\frac{52}{132} \times 100 = 39.4$ %.

The structure of $C r O_{5}$ is shown above.

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