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(D) To find the change in oxidation state,we calculate the oxidation number of the central atom in both the reactant and the product.$A$: In $C_{2}O_{

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$MnO_{4}^{-} \rightarrow Mn^{2+}$

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(D) To find the change in oxidation state,we calculate the oxidation number of the central atom in both the reactant and the product.$A$: In $C_{2}O_{4}^{2-}$,$2x + 4(-2) = -2$ $\Rightarrow 2x = 6$ $\Rightarrow x = +3$. In $CO_{2}$,$x + 2(-2) = 0 \Rightarrow x = +4$. Change = $|4 - 3| = 1$.$B$: In $CrO_{4}^{2-}$,$x + 4(-2) = -2 \Rightarrow x = +6$. In $Cr^{3+}$,$x = +3$. Change = $|3 - 6| = 3$.$C$: In $Cr_{2}O_{7}^{2-}$,$2x + 7(-2) = -2$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$. In $Cr^{3+}$,$x = +3$. Change per $Cr$ atom = $|3 - 6| = 3$.$D$: In $MnO_{4}^{-}$,$x + 4(-2) = -1 \Rightarrow x = +7$. In $Mn^{2+}$,$x = +2$. Change = $|2 - 7| = 5$.Thus,the process with a change of $5$ is $MnO_{4}^{-} \rightarrow Mn^{2+}$.

Identify the process in which the change in the oxidation state is $5$:

Similar Questions

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
$(a)$ $K\underline {I_3}$
$(b)$ $H_2\underline {S_4}O_6$
$(c)$ $\underline {Fe_3}O_4$
$(d)$ $\underline {C}H_3\underline {C}H_2OH$
$(e)$ $\underline {C}H_3\underline {C}OOH$

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