Question

Anhydride of nitrous acid is:
A. \[{{\text{N}}_{\text{2}}}{\text{O}}\]
B. \[{\text{NO}}\]
C. \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]
D. \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]

Answer 1

Hint:Calculate the oxidation number of nitrogen atoms in nitrous acid and in the given oxides of nitrogen. Then find out the oxide of nitrogen that has the same oxidation number of nitrogen as that in nitrous acid.

Complete answer:
You can write the chemical symbol of nitrous acid as \[{\text{HN}}{{\text{O}}_{\text{2}}}\] .
Let X be the oxidation number of nitrogen atoms in nitrous acid. The oxidation numbers of hydrogen and oxygen are +1 and -2 respectively.
In a neutral molecule, the sum of the oxidation numbers of all the elements is zero.
Calculate the oxidation number of nitrogen atoms in nitric acid.
\[
\Rightarrow {\text{1 + X + 2}}\left( { - 2} \right) = 0 \\
\Rightarrow 1 + {\text{X}} - 4 = 0 \\
\Rightarrow {\text{X}} - 3 = 0 \\
\Rightarrow {\text{X}} = + 3
 \]
Hence, the oxidation number of nitrogen atoms in nitrous acid is +3.
Let X be the oxidation number of nitrogen atoms in \[{{\text{N}}_{\text{2}}}{\text{O}}\] . The oxidation number of oxygen is -2
Calculate the oxidation number of nitrogen atoms in \[{{\text{N}}_{\text{2}}}{\text{O}}\] .
\[
\Rightarrow {\text{2X + 2}}\left( { - 2} \right) = 0 \\
\Rightarrow 2{\text{X}} - 4 = 0 \\
\Rightarrow {\text{2X}} - 4 = 0 \\
\Rightarrow {\text{X}} = + 2
 \]
Hence, the oxidation number of nitrogen atoms in \[{{\text{N}}_{\text{2}}}{\text{O}}\] is +2.
Let X be the oxidation number of nitrogen atoms in \[{\text{NO}}\] . The oxidation number of oxygen is -2
In a neutral molecule, the sum of the oxidation numbers of all the elements is zero.
Calculate the oxidation number of nitrogen atoms in \[{\text{NO}}\] .
\[
\Rightarrow {\text{X + }}\left( { - 2} \right) = 0 \\
\Rightarrow {\text{X}} - 2 = 0 \\
\Rightarrow {\text{X}} = + 2
 \]
Hence, the oxidation number of nitrogen atoms in \[{\text{NO}}\] is +2.
Let X be the oxidation number of nitrogen atom in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] . The oxidation number of oxygen is -2
In a neutral molecule, the sum of the oxidation numbers of all the elements is zero.
Calculate the oxidation number of nitrogen atom in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] .
\[
\Rightarrow {\text{2X + 3}}\left( { - 2} \right) = 0 \\
\Rightarrow {\text{2X}} - 6 = 0 \\
\Rightarrow {\text{2X}} = + 6 \\
\Rightarrow {\text{X}} = + 3
 \]
Hence, the oxidation number of nitrogen atom in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] is +3.
\[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
Let X be the oxidation number of nitrogen atom in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] . The oxidation number of oxygen is -2
In a neutral molecule, the sum of the oxidation numbers of all the elements is zero.
Calculate the oxidation number of nitrogen atom in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] .
\[
\Rightarrow {\text{2X + 4}}\left( { - 2} \right) = 0 \\
\Rightarrow {\text{2X}} - 8 = 0 \\
\Rightarrow {\text{2X}} = + 8 \\
\Rightarrow {\text{X}} = + 4
 \]
Hence, the oxidation number of nitrogen atom in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\] is +4.

The oxidation number of nitrogen is same in nitrous acid and in \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]
Hence, the anhydride of nitrous acid is \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\]

Hence, the correct option is the option C.

Note:
When one molecule of \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}}\] reacts with one molecule of water, it forms two molecules of nitrous acid.
\[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{3}}} + {\text{ }}{{\text{H}}_2}{\text{O }} \to {\text{ 2 HN}}{{\text{O}}_{\text{2}}}\]