Question

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?

• A. $+2, +4, —1$
• B. $+2, +6, —2$
• C. $0, +4, -2$
• D. $0, +8, -1$

Answer 1

Answer: (D)

$\overset{2+}{Zn} \overset{-2}{S} +H \overset{+5}{N}O \to \overset{+2}{ZN}(N \overset{-5 }{O_{3}})+ H_{2} S \overset{+6}{O}_{4} + N \overset {+4}{O_{2}}$
Change in O.N. of Zn
$Zn =0$
$S = 6 - (-2) = 8$
$N = 5 -4 = 1$