Question

# The anhydride of Nitrous Acid$\left({\mathrm{HNO}}_{2}\right)$ is

A

${\mathrm{N}}_{2}{\mathrm{O}}_{3}$

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B

${\mathrm{N}}_{2}\mathrm{O}$

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C

$\mathrm{NO}$

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D

${\mathrm{N}}_{2}{\mathrm{O}}_{4}$

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Solution

## 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$2\left(\mathrm{p}\right)+3\left(-2\right)=0\phantom{\rule{0ex}{0ex}}2\mathrm{p}-6=0\phantom{\rule{0ex}{0ex}}\mathrm{p}=+3$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$2\left(\mathrm{q}\right)+\left(-2\right)=0\phantom{\rule{0ex}{0ex}}2\mathrm{q}-2=0\phantom{\rule{0ex}{0ex}}2\mathrm{q}=2\phantom{\rule{0ex}{0ex}}\mathrm{q}=+1$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$\mathrm{r}+\left(-2\right)=0\phantom{\rule{0ex}{0ex}}\mathrm{r}-2=0\phantom{\rule{0ex}{0ex}}\mathrm{r}=+2$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$2\left(\mathrm{s}\right)+4\left(-2\right)=0\phantom{\rule{0ex}{0ex}}2\mathrm{s}-8=0\phantom{\rule{0ex}{0ex}}2\mathrm{s}=8\phantom{\rule{0ex}{0ex}}\mathrm{s}=+4$The oxidation number of N in ${\mathrm{N}}_{2}{\mathrm{O}}_{4}$ is ‘+4’Hence, Option(A) is correct.

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