When ZnS is boiled with strong nitric acid, the products are zinc nitrate, sulphuric acid and nitrogen dioxide. What are the changes in the oxidation numbers of Zn, S and N ?

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$0, + 4, - 2$

$0, 8, 1$

$+ 2, + 6, - 2$

$+ 2, + 4, - 1$

answer is D.

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Detailed Solution

$\overset{+ 2 - 2}{ZnS} + \overset{+ 5}{HNO} \overset{}{\to} \overset{+ 2}{Zn} {(\overset{+ 5}{N} O_{3})}_{2} + H_{2} \overset{+ 6}{S} O_{4} + \overset{+ 4}{N} O_{2}$

Change in O.N. of Zn = 0

S = 6 - (-2) = 8

N = 5 - 4 = 1

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