Oxidation number of $C r$ in ${({N H}_{3})}_{3} C r {(O_{2})}_{2}$ is:

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Solution

Let $x$ be the oxidation number of $C r$ in ${({N H}_{3})}_{3} C r {(O_{2})}_{2}$ .
Since, the overall charge on the complex is $0$ , the sum of oxidation states of all elements in it should be equal to $0$ .
Therefore, $3 (0) + x + 2 (- 2) = 0$
Hence, $x = + 4$
Thus, the oxidation number of Cr in ${({N H}_{3})}_{3} C r {(O_{2})}_{2}$ is $+ 4$ .

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