Question

The anhydride of Nitrous Acid$\left({\mathrm{HNO}}_2\right)$ is

• A. ${\mathrm{N}}_2{\mathrm{O}}_3$
• B. ${\mathrm{N}}_2\mathrm{O}$
• C. $\mathrm{NO}$
• D. ${\mathrm{N}}_2{\mathrm{O}}_4$

Answer 1

The correct option is A

${\mathrm{N}}_2{\mathrm{O}}_3$

The oxidation number of nitrogen in Nitrous Acid$\left({\mathrm{HNO}}_2\right)$ is ‘+3’.

The oxidation number for nitrogen atom in anhydride of Nitrous Acid$\left({\mathrm{HNO}}_2\right)$ must be ‘+3’.

Explanation of correct option-

Let the oxidation number of ‘N’ in ${\mathrm{N}}_2{\mathrm{O}}_3$ is ‘p’.

Then

$$\begin{gathered} 2\left(\mathrm{p}\right)+3(-2)=0 \\ 2\mathrm{p}-6=0 \\ \mathrm{p}=+3 \end{gathered}$$

The oxidation number of ‘N’ in ${\mathrm{N}}_2{\mathrm{O}}_3$ is ‘+3’.

Explanation of incorrect options-

Let the oxidation number of N in ${\mathrm{N}}_2\mathrm{O}$ is ‘q’.

Then

$$\begin{gathered} 2\left(\mathrm{q}\right)+(-2)=0 \\ 2\mathrm{q}-2=0 \\ 2\mathrm{q}=2 \\ \mathrm{q}=+1 \end{gathered}$$

The oxidation number of N in ${\mathrm{N}}_2\mathrm{O}$ is ‘+1’.

Let the oxidation number of N in $\mathrm{NO}$ is ‘r’.

Then

$$\begin{gathered} \mathrm{r}+(-2)=0 \\ \mathrm{r}-2=0 \\ \mathrm{r}=+2 \end{gathered}$$

The oxidation number of N in $\mathrm{NO}$ is ‘+2’.

Let the oxidation number of N in ${\mathrm{N}}_2{\mathrm{O}}_4$ is ‘s’.

Then

$$\begin{gathered} 2\left(\mathrm{s}\right)+4(-2)=0 \\ 2\mathrm{s}-8=0 \\ 2\mathrm{s}=8 \\ \mathrm{s}=+4 \end{gathered}$$

The oxidation number of N in ${\mathrm{N}}_2{\mathrm{O}}_4$ is ‘+4’

Hence, Option(A) is correct.