Question

When concentrated sulfuric acid was added to an unknown salt present in a test tube, a brown gas A was evolved. This gas intensified when copper turning was added to this test tube. On cooling, the gas A changed to a colorless solid B. Write the structures of A and B.

Answer 1

From the above conditions, we have to find the compound A and B. To find these compounds, initially we have to check which compound, when mixed with concentrated sulfuric acid, gives a brown color gas. When copper turnings are added with concentrated sulfuric acid itself, it produces a brown color.

Complete step by step solution:

The general reaction from the above information is given below:

${{\text{H}}_2}{\text{S}}{{\text{O}}_4} + {\text{X}} \to {\text{A}}\xrightarrow{{{\text{Cu}}}}{\text{Y}} \to {\text{B}}$

Brown gas produced when the sulfuric acid is added to an unknown salt is an indication of nitrogen dioxide gas. We have seen in several organic reactions that the nitrating mixture is the combination of concentrated sulfuric acid and nitric acid. It produces a brown fuming gas. Thus the compound A is ${\text{N}}{{\text{O}}_2}$ gas.

When copper turnings are added to the test tube, it is converted to copper nitrate. The gas formed is ${\text{NO}}$ and some water molecules are also formed. This nitrogen oxide reacts with atmospheric oxygen and gets converted to nitrogen dioxide again.

When it is cooled, nitrogen dioxide gas changes to colorless ${{\text{N}}_2}{{\text{O}}_4}$. Thus the compound B is ${{\text{N}}_2}{{\text{O}}_4}$.

So we can say that the compounds A and B are ${\text{N}}{{\text{O}}_2}$ and ${{\text{N}}_2}{{\text{O}}_4}$.

The structures of A and B are given below:

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Note:**

Copper nitrate has a blue crystalline solid structure. It occurs in different hydrate forms in which trihydrate and hemipentahydrate are common.

After cooling, the compound produced is diamagnetic while ${\text{N}}{{\text{O}}_2}$ is paramagnetic.