Complete Hydrolysis Products of Interhalogen Compounds

To determine the complete hydrolysis products of the interhalogen compounds $\text{ICl}$, $\text{ClF}_3$, and $\text{BrF}_5$, we analyze the oxidation state of the central (less electronegative) halogen atom in each compound:

1. Hydrolysis of $\text{ICl}$: In $\text{ICl}$, iodine is in the $+1$ oxidation state. Hydrolysis yields hypoiodous acid ($\text{HIO}$): $\text{ICl} + \text{H}_2\text{O} \longrightarrow \text{HIO} + \text{HCl}$ The corresponding oxyanion formed by ionization of $\text{HIO}$ is the hypoiodite ion, $\text{IO}^-$.

2. Hydrolysis of $\text{ClF}_3$: In $\text{ClF}_3$, chlorine is in the $+3$ oxidation state. Hydrolysis yields chlorous acid ($\text{HClO}_2$): $\text{ClF}_3 + 2\text{H}_2\text{O} \longrightarrow \text{HClO}_2 + 3\text{HF}$ The corresponding oxyanion formed by ionization of $\text{HClO}_2$ is the chlorite ion, $\text{ClO}_2^-$.

3. Hydrolysis of $\text{BrF}_5$: In $\text{BrF}_5$, bromine is in the $+5$ oxidation state. Hydrolysis yields bromic acid ($\text{HBrO}_3$): $\text{BrF}_5 + 3\text{H}_2\text{O} \longrightarrow \text{HBrO}_3 + 5\text{HF}$ The corresponding oxyanion formed by ionization of $\text{HBrO}_3$ is the bromate ion, $\text{BrO}_3^-$.

Thus, the complete hydrolysis of $\text{ICl}$, $\text{ClF}_3$, and $\text{BrF}_5$ gives $\text{IO}^-$, $\text{ClO}_2^-$, and $\text{BrO}_3^-$, respectively.

Correct Option: (A)