Identify the process in which the change in t | vedclass

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

Column-$I$	Column-$II$
$A$. Positively charged ion	$1$. $+7$
$B$. Oxidation number of all neutral atoms	$2$. $-1$
$C$. Oxidation number of hydrogen $(H^+)$	$3$. $+1$
$D$. Oxidation number of fluorine in $NaF$	$4$. $0$
$E$. Negatively charged ion	$5$. Cation
	$6$. Anion

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

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