Q. The oxidation number of nitrogen in $(N_{2} H_{5})^{+}$ is

2703 212 Redox Reactions Report Error

A

$- 2$

32%

B

$+ 2$

26%

C

$+ 3$

21%

D

$- 3$

21%

Solution:

$(N_{2} H_{5})^{+} \Rightarrow 2 x + 5 = + 1 \Rightarrow 2 x = - 4 \Rightarrow$ or, $x = - 2$