Q. Oxidation number of $N$ in $(N H_{4})_{2} S O_{4}$ is

1795 222 Redox Reactions Report Error

A

$- 3$

B

$- l$

C

$+ 1$

D

$- 1/3$

Solution:

$(N H_{4})_{2} S O_{4} ⇌ 2 N H_{4}^{+} + S O_{4}$
$N * H_{4}^{+}$
$x + 4 = + 1; x = 11 - 4 = - 3$