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Which of the following options represents the oxidation state of $\ce{Co}$ and $\ce{Cr}$ in the given complex?

$$\ce{[Co(NH3)4(NO2)2][Cr(NH3)3(NO2)3]}$$

(A) $2, 3$
(B) $3, 2$
(C) $3, 3$
(D) $2, 2$

I am unable to judge the net charge on complex when it breaks into ions and therefore I am having trouble with this question. I assumed that some unknown charge with opposite signs must be on the complex ions.

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Conventionally, first complex ion is positively charged and the latter is negatively charged. Suppose thess charges are $p+$ and $q-$, respectively. Since two ion ratio is 1:1, $|p+|= |q-|$.

Suppose oxidation number of $\ce{Co}$ is $n$ and that of $\ce{Cr}$ is $m$. Hence, for cation complex (Note that the net charges of $\ce{NH3}$ and $\ce{NO2-}$ is $0$ and $-1$, respectively): $$n+0\times4+(-1)\times2=+p = n-2$$ and for anion complex: $$m+0\times3+(-1)\times3=-q=m-3$$ Since, $|p+|= |q-|$, $$m-3= +p \; \text{and } n-2=-p$$ Or, $m+n=5$. The only answer agree with this is (A) or (B):

If (A) is correct, then, $\ce{Co^2+}$ and $\ce{Cr^3+}$. Substitute these values in original two equation: $+2+0\times4+(-1)\times2=+0 $ and $+3+0\times4+(-1)\times3=+0 $. This can't be.

If (B) is correct, then, $\ce{Co^3+}$ and $\ce{Cr^2+}$. Substitute these values in original two equation: $+3+0\times4+(-1)\times2=+1 $ and $+2+0\times4+(-1)\times3=-1 $. This is correct.

Therefore, your answer is (B), and the oxidation numbers of $\ce{Co}$ and $\ce{Cr}$ ared $+3$ and $+2$, respectively.

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