In which of the following is the difference i | vedclass

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(D) $1$. For option $A$: In $\underline{N}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{N}_2H_4$,$2x + 4(+1) = 0 \implies x = -2$. Difference =

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$\underline{S}^{2-}$ and $\underline{S}O_4^{2-}$

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(D) $1$. For option $A$: In $\underline{N}O_2$,$x + 2(-2) = 0 \implies x = +4$. In $\underline{N}_2H_4$,$2x + 4(+1) = 0 \implies x = -2$. Difference = $|4 - (-2)| = 6$.$2$. For option $B$: In $\underline{S}O_3^{2-}$,$x + 3(-2) = -2 \implies x = +4$. In $\underline{S}O_4^{2-}$,$x + 4(-2) = -2 \implies x = +6$. Difference = $|6 - 4| = 2$.$3$. For option $C$: In $\underline{S}^{2-}$,$x = -2$. In $\underline{S}O_3^{2-}$,$x = +4$. Difference = $|4 - (-2)| = 6$.$4$. For option $D$: In $\underline{S}^{2-}$,$x = -2$. In $\underline{S}O_4^{2-}$,$x = +6$. Difference = $|6 - (-2)| = 8$.$5$. Comparing the differences,the greatest difference is $8$ in option $D$.

In which of the following is the difference in the oxidation numbers of the underlined elements the greatest?

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