In which of the following pairs is there the greatest difference in the oxidation number of the underlined elements?

N - - O_{2}

and

N_{2} - - - O_{4}

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S - - O_{3}^{2 -}

and

S - - O_{4}^{2 -}

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S^{2 -} - -- -

and

S - - O_{3}^{2 -}

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S^{2 -} - -- -

and

S - - O_{4}^{2 -}

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Solution

The correct option is D $S^{2 -} - -- -$ and $S - - O_{4}^{2 -}$
In $S^{2 -} - -- -$ and $S - - O_{4}^{2 -}$ the difference in the oxidation number of the underlined elements is greatest.
(A) $N - - O_{2}$ and $N_{2} - - - O_{4}$
The oxidation numbers of the underlined elements are +4 and +4 respectively. The difference in the oxidation number is 0.
(B) $S - - O_{3}^{2 -}$ and $S - - O_{4}^{2 -}$
The oxidation numbers of the underlined elements are +4 and +6 respectively. The difference in the oxidation number is 2.
(C) $S^{2 -} - -- -$ and $S - - O_{3}^{2 -}$
The oxidation numbers of the underlined elements are -2 and +4 respectively. The difference in the oxidation number is 6.
(D) $S^{2 -} - -- -$ and $S - - O_{4}^{2 -}$
The oxidation numbers of the underlined elements are -2 and +6 respectively. The difference in the oxidation number is 8.

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