The oxidation state of Cr is $[C r (N H_{3})_{4} C l_{2}]$ is?

+ 3

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+ 2

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+ 1

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0

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Solution

The correct option is B $+ 2$
$[C r {({N H}_{3})}_{4} {C l}_{2}]$
We know that $C 1 = - 1$ ${N H}_{3} = 0$
So onidation state of $C r - - -$
$\begin{matrix} x + 4 \times 0 + 2 (- 1) = 0 x = + 2 \end{matrix}$
Corret Answer $- B - -$

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