Question

A brown colored mixture of two gases is obtained by the reduction of 6N nitric acid with metallic copper. This mixture on cooling condenses to a blue liquid which on freezing (-30 $^{o}C$ ) gives a blue solid. The correct choice for blue liquid or solid is:
A. It is referred to as the anhydride of nitrous acid.
B. It is an acidic oxide and hence dissolves in alkalies producing nitrites.
C. It can also be prepared by the action of 50% $HN{{O}_{3}}$ on arsenious oxide and then cooling to 250K.
D. All of the above

Answer 1

Reduction is nothing but the addition of hydrogens, gaining of electrons or removing of oxygen from a chemical. Generally copper is going to do reduction of chemicals. The chemical is going to accept electrons from copper when it reacts.

Complete step by step answer:
- In the question it is given that a brown colored mixture of two gases is obtained by the reduction of 6N nitric acid with metallic copper.
- The above statement can be represented as a chemical reaction as follows.

& 2Cu+6HN{{O}_{3}}\to 2Cu{{(N{{O}_{3}})}_{2}}+NO+N{{O}_{2}}+3{{H}_{2}}O \\\ & NO+N{{O}_{2}}\overset{-{{30}^{o}}C}{\longleftrightarrow}{{N}_{2}}{{O}_{3}} \\\ \end{aligned}$$ \- Generally nitric acid reacts with copper and forms copper nitrate, nitrogen monoxide and nitrogen dioxide and 3 water molecules as the products. \- The formed nitrogen monoxide and nitrogen dioxide on freezing at $-{{30}^{o}}C$ forms a blue color solid having a molecular formula of ${{N}_{2}}{{O}_{3}}$ . \- We came to what product is formed after freezing, now we have to check all the given options. \- Coming to option A, It is referred to as the anhydride of nitrous acid. Means ${{N}_{2}}{{O}_{3}}$ is anhydride of nitrous acid. $$2HN{{O}_{2}}\to {{N}_{2}}{{O}_{3}}+{{H}_{2}}O$$ \- From the above chemical equation we can say clearly that after removing a water molecule from nitrous acid ${{N}_{2}}{{O}_{3}}$ is going to form. So, option A is correct. \- Coming to option B, It is an acidic oxide and hence dissolves in alkalies producing nitrites. $${{N}_{2}}{{O}_{3}}+2KOH\to 2KN{{O}_{2}}+{{H}_{2}}O$$ \- Yes it is correct ${{N}_{2}}{{O}_{3}}$ is an acidic oxide and dissolves in alkali (see above reaction) and forms potassium nitrite. So, option B is also correct. \- Coming to option C, It can also be prepared by the action of 50% $HN{{O}_{3}}$ on arsenious oxide and then cooling to 250K. $$2HN{{O}_{3}}+A{{s}_{2}}{{O}_{3}}+2{{H}_{2}}O\to NO+N{{O}_{2}}\xrightarrow{250K}{{N}_{2}}{{O}_{3}}$$ \- From the above equation option C is also correct. **So, the correct answer is “Option D”.** **Note:** The oxides of nitrogen formed (Nitrogen monoxide and nitrogen dioxide) when nitric acid reacts with copper are going to react (condenses) at lower temperature (250K) and form a blue colored dinitrogen trioxide as the product.