Question

On addition of conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$to a chloride salt, colorless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

• A. ${\mathrm{H}}_2{\mathrm{SO}}_4$ reduces $\mathrm{HI}$to ${\mathrm{I}}_2$
• B. $\mathrm{HI}$ is of violet color
• C. $\mathrm{HI}$ gets oxidized to ${\mathrm{I}}_2$
• D. $\mathrm{HI}$ changes to ${\mathrm{HIO}}_3$

Answer 1

The correct option is C

$\mathrm{HI}$ gets oxidized to ${\mathrm{I}}_2$

Explanation for correct answer

(C) $\mathbf{HI}$ gets oxidized to ${\mathbf{I}}_{\mathbf{2}}$

1. Hydrogen Iodide is a more powerful reducing agent than sulfuric acid, It converts Sulfuric acid (${\mathrm{H}}_2{\mathrm{SO}}_4$) to ${\mathrm{SO}}_2$ and $\mathrm{HI}$ to ${\mathrm{I}}_2$.
2. When chloride salts are treated with Sulfuric acid, $\mathrm{HCl}$ gas is formed which produces a colorless gas.
3. Chemical reaction involved in chloride: $\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+\underset{SodiumChloride}{NaCl\left(s\right)}\to \underset{\mathrm{Hydrochloric}\mathrm{acid}}{\mathrm{HCl}\left(\mathrm{g}\right)}+\underset{\mathrm{Sodium}\mathrm{Hydrogen}\mathrm{Sulfate}}{{\mathrm{NaHSO}}_4\left(\mathrm{s}\right)}$
4. Chemical reaction involved in iodide: $\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}+\underset{HydrogenIodide}{HI\left(s\right)}\to \underset{\mathrm{Iodide}}{{\mathrm{I}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Sulfur}\mathrm{dioxide}}{{\mathrm{SO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)}$
5. Violet fumes are produced in the reaction as a result of formation of Iodine gas. Hence option (C) is correct.

Explanation for incorrect answer

(A) ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ reduces $\mathbf{HI}$ to ${\mathbf{I}}_{\mathbf{2}}$

1. As Hydrogen Iodide is more powerful reducing agent than Sulfuric acid, It reduces ${\mathrm{H}}_2{\mathrm{SO}}_4$ to ${\mathrm{SO}}_2$ and $\mathrm{HI}$ to ${\mathrm{I}}_2$ neither Sulfuric acid reduces.
2. Hence, option (A) is incorrect.

(B) $\mathbf{HI}$ is of Violet color

1. Hydrogen Iodide is not of Violet color but when it converts into ${\mathrm{I}}_2$ it is of violet color.
2. Hence option (B) is incorrect.

(D) $\mathbf{HI}$ changes to ${\mathbf{HIO}}_{\mathbf{3}}$

1. Hydrogen Iodide changes into ${\mathrm{I}}_2$ not ${\mathrm{HIO}}_3$.
2. Hence option (D) is incorrect.

Therefore, Violet fumes are observed is of Iodine gas and Colorless gas is of Hydrochloric acid. Hence option (C) is correct.