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?

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

Tardigrade Admin · Accepted Answer

overset2+Zn overset-2S +H overset+5NO → overset+2ZN(N overset-5 O3)+ H2 S overset+6O4 + N overset +4O2 Change in O.N. of Zn Zn =0 S = 6 - (-2) = 8 N = 5 -4 = 1

Q. 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?

$+ 2, + 4, —1$

$+ 2, + 6, —2$

$0, + 4, - 2$

$0, + 8, - 1$

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

Q. 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?

+2,+4,—1

+2,+6,—2

0,+4,−2

0,+8,−1

Zn2+S−2+HN+5O→ZN+2(NO3​−5​)+H2​SO+64​+NO2​+4​ Change in O.N. of Zn Zn=0 S=6−(−2)=8 N=5−4=1

$+ 2, + 4, —1$

$+ 2, + 6, —2$

$0, + 4, - 2$

$0, + 8, - 1$

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