Question

The mixed anhydride of nitrogen is:
(A) ${{N}_{2}}{{O}_{2}}(NO)$
(B) ${{N}_{2}}{{O}_{4}}(2N{{O}_{2}})$
(C) ${{N}_{2}}{{O}_{5}}$
(D) ${{N}_{2}}{{O}_{2}}$

Answer 1

Hint: All non-metals form metal oxides with oxygen, which reacts with water that will form acids or with bases that form salts. Most nonmetals form oxoacids from their acidic oxides. Oxides such as sulfur trioxide ( $S{{O}_{3}}$ ) and dinitrogen pentoxide ( ${{N}_{2}}{{O}_{5}}$ ) are similar to common oxidation numbers, which are known as acidic anhydrides. These anhydrides form oxoacids while reacting with water with no change of their oxidation number.

Complete step by step solution:
Nitrogen forms oxides with each of its oxidation numbers from +1 to +5.

Name	Formula	Oxidation number	Description
Dinitrogen oxide	${{N}_{2}}O$	+1	Colourless gas
Nitrogen monoxide	$NO$	+2	Colourless gas
Dinitrogen trioxide	${{N}_{2}}{{O}_{3}}$	+3	Blue solid
Nitrogen dioxide	$N{{O}_{2}}$	+4	Brown gas
Dinitrogen tetroxide	${{N}_{2}}{{O}_{4}}$	+4	Colourless liquid

Nitrous oxide is formed when ammonium nitrate is heated.
$N{{H}_{4}}N{{O}_{3}}\overset{\Delta }{\mathop{\to }}\,{{N}_{2}}O+2{{H}_{2}}O$
Nitrogen monoxide is formed while sodium nitrate is reacted with ferrous sulfate in an acidic medium.
$2NaN{{O}_{2}}+2FeS{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2NaHS{{O}_{4}}+2{{H}_{2}}O+2NO$
When a mixture of equal parts of nitric oxide and nitrogen dioxide is cooled then dinitrogen trioxide will form.
$2NO+{{N}_{2}}{{O}_{4}}\overset{250K}{\mathop{\to }}\,2{{N}_{2}}{{O}_{3}}$
Nitrogen dioxide is prepared commercially by heating the mixture of lead nitrate.
$2Pb{{(N{{O}_{3}})}_{2}}+heat\to 2PbO+4N{{O}_{2}}+{{O}_{2}}$
Nitrogen dioxide a mixture of anhydride means the compound forms a mixture of two acidic anhydrides. Two oxides of nitrogen dioxide react with water to form nitrogen-containing mixed acids.
$2N{{O}_{2}}+{{H}_{2}}O\to HN{{O}_{3}}+HN{{O}_{2}}$

Hence, the mixed anhydride of nitrogen is ${{N}_{2}}{{O}_{4}}(2N{{O}_{2}})$, the correct answer is option (B).

Note: Nitrogen dioxide reacts with water may be the nitrogen both oxidized and reduced in a reaction. This reaction ${{N}^{+4}}$ is reduced to ${{N}^{+2}}$ (in NO) and oxidized to ${{N}^{+5}}$ in nitric acid. This reaction is called a disproportionation reaction, in which the same element is both oxidized and reduced.