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(D) The correct option is $D$.We calculate the oxidation state $(x)$ for the metal in each compound:$(i)$ In $K_2CrO_4$: $2(+1) + x + 4(-2) = 0 \impli

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$Cr$

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(D) The correct option is $D$.We calculate the oxidation state $(x)$ for the metal in each compound:$(i)$ In $K_2CrO_4$: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$.The oxidation state of $Cr$ is $+6$.$(ii)$ In $NbCl_5$: $x + 5(-1) = 0 \implies x - 5 = 0 \implies x = +5$.The oxidation state of $Nb$ is $+5$.$(iii)$ In $MnO_2$: $x + 2(-2) = 0 \implies x - 4 = 0 \implies x = +4$.The oxidation state of $Mn$ is $+4$.Comparing the values $(+6, +5, +4)$,$Cr$ has the highest oxidation state.

The metal with the highest oxidation state present in $K_2CrO_4$,$NbCl_5$,and $MnO_2$ is

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