The oxidation number of chromium in $K_{2} C r O_{4}$ is:

+ 2

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+ 3

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+ 6

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+ 8

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Solution

The correct option is A $+ 6$
In $K_{2} C r O_{4}$ ,
Oxidation no. of $K = + 1.$
Oxidation no. of $O = - 2.$
Oxidation no. of $C r$ be $x$ .
$2 (1) + x + 4 (- 2) = 0, 2 + x - 8 = 0, x - 6 = 0, x = 6.$
$∴$ Oxidation no. of $C h r o m i u m$ in $K_{2} C r O_{4}$ is $6$ .

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When $K_{2} C r_{2} O_{7}$ is converted to $K_{2} C r O_{4}$ the change in the oxidation state of chromium is

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