Animated Solution for Chemistry - s and p-Block Elements: The complete hydrolysis of $\text{ICl}$, ${\text{ClF}}_3$ and ${\text{BrF}}_5$, respectively, gives

The hydrolysis of interhalogen compounds is one of those beautiful topics in inorganic chemistry where a single, elegant rule unlocks the entire puzzle. You don't need to memorize a dozen different reactions; you just need to understand the underlying principle of electronegativity and oxidation states.

Imagine an interhalogen compound as a partnership between two different halogens, let's call them $\text{X}$ and ${\text{X}}^{\prime }$. In this partnership, $\text{X}$ is the larger, less electronegative atom (the central atom), and ${\text{X}}^{\prime }$ is the smaller, more electronegative atom.

When water (${\text{H}}_2\text{O}$) enters the scene, it breaks this partnership. The more electronegative halogen (${\text{X}}^{\prime }$) is greedy for electrons, so it grabs the electropositive hydrogen from water to form a hydrohalic acid (${\text{HX}}^{\prime }$).

The larger central atom ($\text{X}$), on the other hand, bonds with the remaining oxygen and hydrogen to form an oxoacid. The golden rule here is that this is not a redox reaction! The oxidation state of the central atom remains exactly the same before and after the hydrolysis.

Let's apply our master rule to the first compound, $\text{ICl}$.

Between iodine and chlorine, iodine is the larger and less electronegative element. Therefore, iodine acts as the central atom. Since chlorine is more electronegative, it pulls the shared electron pair towards itself, giving iodine an oxidation state of $+1$.

Upon hydrolysis, chlorine forms hydrochloric acid ($\text{HCl}$). Iodine, maintaining its $+1$ oxidation state, forms hypoiodous acid ($\text{HIO}$).

In an aqueous solution, this acid dissociates to give the hypoiodite ion (${\text{IO}}^{-}$).

Next up is ${\text{ClF}}_3$. Fluorine is the undisputed king of electronegativity, so chlorine is forced to be the central atom. With three fluorines attached, chlorine sits at an oxidation state of $+3$.

Following our rule, the three fluorine atoms will form three molecules of hydrogen fluoride ($\text{HF}$). Chlorine must form an oxoacid where it retains its $+3$ state. That acid is chlorous acid (${\text{HClO}}_2$).

Chlorous acid then dissociates in water to yield the chlorite ion (${\text{ClO}}_2^{-}$).

Finally, we have ${\text{BrF}}_5$. Here, bromine is the central atom, surrounded by five highly electronegative fluorine atoms. This pushes bromine to a high oxidation state of $+5$.

When water attacks, the five fluorines convert into five molecules of $\text{HF}$. Bromine, staying true to its $+5$ state, forms bromic acid (${\text{HBrO}}_3$).

Bromic acid is a strong acid that completely dissociates to give the bromate ion (${\text{BrO}}_3^{-}$).

Putting all our findings together, the complete hydrolysis of $\text{ICl}$, ${\text{ClF}}_3$, and ${\text{BrF}}_5$ yields the ions ${\text{IO}}^{-}$, ${\text{ClO}}_2^{-}$, and ${\text{BrO}}_3^{-}$, respectively.

This perfectly matches option (A). By simply tracking the oxidation states, we turned a potentially complex memorization task into a logical, step-by-step deduction!