Question

$N{H}_{2}OH\to N_{2}O$

In the above conversion, the equivalent weight of $N{H}_{2}OH$ will be:

• A. $\frac{M}{4}$
• B. $\frac{M}{2}$
• C. $\frac{M}{5}$
• D. $\frac{M}{6}$

Answer 1

Answer: (B)

$\frac{M}{2}$

Oxidation state of $N$ in $N{H}_{2}OH$,

$x+3-2=0$

$x=-1$

Therefore, the charge on nitrogen is $-1$

Oxidation state of $N$ in $N_{2}O$

$2x-2=0$

$x=+1$

Since oxidation state of $N$ changes from $-1$ to $+1$, the equivalent weight of $N{H}_{2}OH$ is $\frac{M}{2}$.