When $K_{2} C r_{2} O_{7}$ is converted into $K_{2} C r O_{4},$ the change in oxidation no. of Cr is:

0.0

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Solution

The correct option is A 0.0
$K_{2} C r_{2} O_{7} \to K_{2} C r O_{4}$
Let, the oxidation state of $C r$ in $K_{2} C r_{2} O_{7}$ be $x$ .
$\Rightarrow 2 + 2 x + 7 \times (- 2) = 0$
$∴ x = + 6$
Let, the oxidation state of $C r$ in $K_{2} C r O_{4}$ be $y$ .
$\Rightarrow 2 + y + 4 \times (- 2) = 0$
$∴ y = + 6$
So, no change in oxidation number of $C r$ .

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