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(A) To determine the number of electrons transferred,we calculate the change in the oxidation state of the central atom:$1$. In $MnO_4^{-}$,the oxidat

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$MnO_4^{-} \longrightarrow Mn^{2+}$

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(A) To determine the number of electrons transferred,we calculate the change in the oxidation state of the central atom:$1$. In $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$. In $Mn^{2+}$,it is $+2$. The change is $7 - 2 = 5$ electrons.$2$. In $CrO_4^{2-}$,the oxidation state of $Cr$ is $+6$. In $Cr^{3+}$,it is $+3$. The change is $6 - 3 = 3$ electrons.$3$. In $NO_3^{-}$,the oxidation state of $N$ is $+5$. In $NH_4^{+}$,it is $-3$. The change is $5 - (-3) = 8$ electrons.$4$. In $Cr_2O_7^{2-}$,the oxidation state of $Cr$ is $+6$. In $2Cr^{3+}$,the total charge change for two $Cr$ atoms is $2 	imes (6 - 3) = 6$ electrons.Thus,the change involving the transfer of $5$ electrons is $MnO_4^{-} \longrightarrow Mn^{2+}$.

Which of the following changes involves the transfer of $5$ electrons?

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