The pair of compounds in which the metals are in their highest oxidation state is:

M n O_{2}, F e C l_{3}

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M n O_{4}^{-}, C r O_{2} C l_{2}

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M n C l_{2}, C r C l_{3}

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[N i C l_{4}]^{2 -}, [C o C l_{4}]^{-}

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Solution

The correct option is C $M n O_{4}^{-}, C r O_{2} C l_{2}$
The charges on oxygen and chlorine atoms are $- 2$ and $- 1$ respectively. Therefore, the oxidation states of the central atoms in the given compounds are calculated as follows:

$M n O_{2}, F e C l_{3} \to + 4, + 3$

$M n O_{4}^{-}, C r O_{2} C l_{2} \to + 7, + 6$
$M n C l_{2}, C r C l_{3} \to + 2, + 3$

$[N i C l_{4}]^{- 2}, [C o C l_{4}]^{-} \to + 2, + 3$

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