Question

$HCl$ is prepared by heating $NaCl$ with conc. $H_{2}S{O}_4$ but $HI$ cannot be prepared by treating $NaI$ with conc. $H_{2}S{O}_4$. Identify the reason.

• A. $H_{2}S{O}_4$ is oxidised by $HI$ and not oxidised by $HCl$.
• B. $HI$ is oxidised by $H_{2}S{O}_4$ and $HCl$ is not oxidised by $H_{2}S{O}_4$.
• C. $HI$ dissociates on treatment with $H_{2}S{O}_4$ and $HCl$ does not dissociate.
• D. The reaction of $H_{2}S{O}_4$ with $HI$ is reversible whereas with $HCl$ is irreversible.

Answer 1

Answer: (B)

$HI$ is oxidised by $H_{2}S{O}_4$ and $HCl$ is not oxidised by $H_{2}S{O}_4$.

$H_{2}S{O}_4$ acts as oxidizing agent and oxidises $HI$ but $HCl$ is not oxidised by $H_{2}S{O}_4$.

Hence, $HCl$ is prepared by heating $NaCl$ with conc. $H_{2}S{O}_4$ but $HI$ cannot be prepared by treating $NaI$ with conc. $H_{2}S{O}_4$.