
Extra very pure ${{\text{N}}_{\text{2}}}$ can be obtained by heating:
(A) ${\text{N}}{{\text{H}}_{\text{3}}}$ with ${\text{CuO}}$
(B) ${\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}$ ${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$
(C) ${{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$
(D) ${\text{Ba}}{\left( {{{\text{N}}_{\text{3}}}} \right)_{\text{2}}}$
Answer
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Hint: Pure form of elemental dinitrogen means it is free from any impurities.
- In a decomposition reaction, a compound (reactant) is getting broken into two or smaller products with the aid of absorbing heat or light energy applied.
Complete Solution :
All the compounds given in the above options are either used in the synthesis of nitrogen or in the synthesis of nitrogen compounds. So let us find the reactions of the compounds mentioned one by one.
(A) ${\text{N}}{{\text{H}}_{\text{3}}}$ with ${\text{CuO}}$-
Ammonia $\left({{\text{N}}{{\text{H}}_{\text{3}}}}\right)$ reacts with ${\text{CuO}}$ (Copper oxide) to form a complex ion. The copper ion in the compound CuO is in +2 oxidation state, exists as ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}$. To detect this ion ammonia is made to react with copper oxide. The complex ion formed from the reaction of ammonia with copper oxide is a deep blue or violet colored compound and its name is tetraamminecopper (II) ion. The chemical equation of the reaction is,
${\text{4N}}{{\text{H}}_{\text{3}}}{\text{ + C}}{{\text{u}}^{{\text{ + 2}}}} \rightleftharpoons {\left[ {{\text{Cu}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{4}}}} \right]^{{\text{ + 2}}}}$(Deep blue)
(B) Ammonium nitrate
$\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}} \right)$, is a nitrogen compound which undergoes thermal decomposition to give a colorless, neutral gas, dinitrogen oxide or Nitrogen (II) oxide$\left( {{{\text{N}}_{\text{2}}}{\text{O}}} \right)$. The chemical reaction, decomposition of ammonium nitrate is as follows,
$\text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{\text{3}}}\xrightarrow{heat}{{\text{N}}_{\text{2}}}\text{O + 2}{{\text{H}}_{\text{2}}}\text{O}$
(C) Dinitrogen can be prepared by the thermal decomposition of ammonium dichromate,$\left( {{{\left[{{\text{N}}{{\text{H}}_{\text{4}}}}\right]}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}} \right)$
The decomposition reaction of ammonium dichromate is given as follows,
${{\left[ \text{N}{{\text{H}}_{\text{4}}} \right]}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\xrightarrow{heat}{{\text{N}}_{\text{2}}}\uparrow \text{ + 4}{{\text{H}}_{\text{2}}}\text{O + C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
But in this case, the dinitrogen is not pure, it contains impurities.
(D) Barium azide $\left[ {{\text{Ba}}{{\left( {{{\text{N}}_{\text{3}}}} \right)}_{\text{2}}}} \right]$ and Sodium azide$\left[ {{\text{Na}}{{\text{N}}_{\text{3}}}} \right]$are known to produce very pure nitrogen gas upon thermal decomposition.
The thermal decomposition reaction of barium azide is,
$\text{Ba}{{\left( {{\text{N}}_{\text{2}}} \right)}_{\text{3}}}\xrightarrow{heat}\text{Ba + 3}{{\text{N}}_{\text{2}}}$
- The decomposition of sodium azide upon heating is given as
$\text{2Na}{{\text{N}}_{\text{3}}}\xrightarrow{heat}\text{2Na + 3}{{\text{N}}_{\text{2}}}$
Here the correct option from the given options is option (D), ${\text{Ba}}{\left( {{{\text{N}}_{\text{2}}}} \right)_{\text{3}}}$
- As in this process first we are making the azide to absorb dinitrogen and ten only we will give the heat and it decomposes to the respective metal and nitrogen and here the dinitrogen produced will not have any of the impurities. Hence the extra pure dinitrogen is produced.
So, the correct answer is “Option D”.
Note: Azide is an anion formed by three nitrogen atoms bonded linearly. As it is a negatively charged ion it combines with metals such as sodium mostly, but other azides are also known and have important applications.
- In a decomposition reaction, a compound (reactant) is getting broken into two or smaller products with the aid of absorbing heat or light energy applied.
Complete Solution :
All the compounds given in the above options are either used in the synthesis of nitrogen or in the synthesis of nitrogen compounds. So let us find the reactions of the compounds mentioned one by one.
(A) ${\text{N}}{{\text{H}}_{\text{3}}}$ with ${\text{CuO}}$-
Ammonia $\left({{\text{N}}{{\text{H}}_{\text{3}}}}\right)$ reacts with ${\text{CuO}}$ (Copper oxide) to form a complex ion. The copper ion in the compound CuO is in +2 oxidation state, exists as ${\text{C}}{{\text{u}}^{{\text{ + 2}}}}$. To detect this ion ammonia is made to react with copper oxide. The complex ion formed from the reaction of ammonia with copper oxide is a deep blue or violet colored compound and its name is tetraamminecopper (II) ion. The chemical equation of the reaction is,
${\text{4N}}{{\text{H}}_{\text{3}}}{\text{ + C}}{{\text{u}}^{{\text{ + 2}}}} \rightleftharpoons {\left[ {{\text{Cu}}{{\left( {{\text{N}}{{\text{H}}_{\text{3}}}} \right)}_{\text{4}}}} \right]^{{\text{ + 2}}}}$(Deep blue)
(B) Ammonium nitrate
$\left( {{\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}} \right)$, is a nitrogen compound which undergoes thermal decomposition to give a colorless, neutral gas, dinitrogen oxide or Nitrogen (II) oxide$\left( {{{\text{N}}_{\text{2}}}{\text{O}}} \right)$. The chemical reaction, decomposition of ammonium nitrate is as follows,
$\text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{\text{3}}}\xrightarrow{heat}{{\text{N}}_{\text{2}}}\text{O + 2}{{\text{H}}_{\text{2}}}\text{O}$
(C) Dinitrogen can be prepared by the thermal decomposition of ammonium dichromate,$\left( {{{\left[{{\text{N}}{{\text{H}}_{\text{4}}}}\right]}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}} \right)$
The decomposition reaction of ammonium dichromate is given as follows,
${{\left[ \text{N}{{\text{H}}_{\text{4}}} \right]}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\xrightarrow{heat}{{\text{N}}_{\text{2}}}\uparrow \text{ + 4}{{\text{H}}_{\text{2}}}\text{O + C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{3}}}$
But in this case, the dinitrogen is not pure, it contains impurities.
(D) Barium azide $\left[ {{\text{Ba}}{{\left( {{{\text{N}}_{\text{3}}}} \right)}_{\text{2}}}} \right]$ and Sodium azide$\left[ {{\text{Na}}{{\text{N}}_{\text{3}}}} \right]$are known to produce very pure nitrogen gas upon thermal decomposition.
The thermal decomposition reaction of barium azide is,
$\text{Ba}{{\left( {{\text{N}}_{\text{2}}} \right)}_{\text{3}}}\xrightarrow{heat}\text{Ba + 3}{{\text{N}}_{\text{2}}}$
- The decomposition of sodium azide upon heating is given as
$\text{2Na}{{\text{N}}_{\text{3}}}\xrightarrow{heat}\text{2Na + 3}{{\text{N}}_{\text{2}}}$
Here the correct option from the given options is option (D), ${\text{Ba}}{\left( {{{\text{N}}_{\text{2}}}} \right)_{\text{3}}}$
- As in this process first we are making the azide to absorb dinitrogen and ten only we will give the heat and it decomposes to the respective metal and nitrogen and here the dinitrogen produced will not have any of the impurities. Hence the extra pure dinitrogen is produced.
So, the correct answer is “Option D”.
Note: Azide is an anion formed by three nitrogen atoms bonded linearly. As it is a negatively charged ion it combines with metals such as sodium mostly, but other azides are also known and have important applications.
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