Question

Pure ${{N}_{2}}$ can be obtained from:
A. $N{{H}_{3}}+NaN{{O}_{2}}$
B. $N{{H}_{4}}Cl+NaN{{O}_{2}}$
C. ${{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}$
D. ${{N}_{2}}O+Ca$

Answer 1

Nitrogen gas is colorless and odorless. It is non-flammable in nature. There are various processes through which nitrogen gas can be produced. But in this question, we have to find out that pure nitrogen, without any impurities is produced through which process.

Complete step-by-step answer:
Let us look into the given options:
A. The given reaction is $N{{H}_{3}}+NaN{{O}_{2}}$. When ammonia reacts with sodium nitrite, nitrogen gas is not obtained. Reaction for the same is mentioned below
$N{{H}_{3}}+NaN{{O}_{2}}\to N{{H}_{3}}N{{O}_{2}}+Na$
Thus, this option is wrong as there is no nitrogen gas produced in this reaction.

B. The given reaction in option B is $N{{H}_{4}}Cl+NaN{{O}_{2}}$. This is a laboratory method for the preparation of nitrogen. The reaction for the same is mentioned below.
$N{{H}_{4}}Cl+NaN{{O}_{2}}\to {{N}_{2}}+2{{H}_{2}}O+NaCl$
The nitrogen produced is a pure gas, it contains only a minimum amount of Nitrous oxide and nitric acid as impurities, but these can be easily removed by passing the gas through aqueous solution of sulfuric acid containing potassium dichromate. Thus, the nitrogen gas produced is in its pure form. Thus, this option is correct.

C. The given option is ammonium dichromate. Thermal decomposition of it yields nitrogen, but not in that pure form.
${{\left( N{{H}_{4}} \right)}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{heat}{{N}_{2}}+4{{H}_{2}}O+C{{r}_{2}}{{O}_{3}}$. Hence, this option is incorrect.

D. Similarly, Option D does not produce pure nitrogen gas, as Nitrous oxide is not much reactive at room temperature. Hence, this option is also incorrect.

Therefore, the correct answer is Option B.

Note: It should be noted that a very pure form of nitrogen can be obtained by the thermal decomposition of sodium or barium azide. The reactions for the same are mentioned below,
$Ba{{\left( {{N}_{3}} \right)}_{2}}\xrightarrow{\Delta }Ba+3{{N}_{2}} \\\ 2Na{{N}_{3}}\xrightarrow{\Delta }2Na+3{{N}_{2}} \\\$