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(A) To determine the highest oxidation state,we calculate the oxidation number for each metal:$1$. $MnO_4^-$: Let $x$ be the oxidation state of $Mn$.

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$MnO_4^-$ and $CrO_2Cl_2$

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(A) To determine the highest oxidation state,we calculate the oxidation number for each metal:$1$. $MnO_4^-$: Let $x$ be the oxidation state of $Mn$. $x + 4(-2) = -1 \implies x = +7$. $CrO_2Cl_2$: Let $y$ be the oxidation state of $Cr$. $y + 2(-2) + 2(-1) = 0 \implies y = +6$. Both $Mn$ $(+7)$ and $Cr$ $(+6)$ are in their highest possible oxidation states.$2$. $[NiCl_4]^{2-}$: $Ni$ is $+2$; $[CoCl_4]^{2-}$: $Co$ is $+2$.$3$. $[Fe(CN)_6]^{3-}$: $Fe$ is $+3$; $[Cu(CN)_4]^{2-}$: $Cu$ is $+2$.$4$. $[FeCl_4]^-$: $Fe$ is $+3$; $Co_2O_3$: $Co$ is $+3$.Thus,the pair $MnO_4^-$ and $CrO_2Cl_2$ contains metals in their highest oxidation states.

The pair of compounds having metals in their highest oxidation state is

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