The complete hydrolysis of ICl, ClF$$_3$$ and BrF$$_5$$, respectively, gives

The rule for the complete (exhaustive) hydrolysis of an inter-halogen $$XY_n$$ is:
  • Each $$F$$ (or the more electronegative halogen) is replaced by $$OH$$, giving $$HF$$ in the medium.
  • The central halogen $$X$$ keeps the same oxidation number it had in $$XY_n$$ and appears as an oxy-anion $$XO_m^{\,(n-2m+1)-}$$ after loss of the acidic hydrogens.

Let us apply this idea one compound at a time.

$$\textbf{ICl}$$
Oxidation number of I: $$+1$$ (because $$\text{Cl}$$ is more electronegative).
Step-1 (replacement by $$OH$$): $$\mathrm{ICl + H_2O \;\longrightarrow\; HOI + HCl}$$
Step-2 (deprotonation in water): $$\mathrm{HOI \;\longrightarrow\; IO^- + H^+}$$
Thus the stable oxy-anion is $$\mathbf{IO^-}$$ (iodide in the +1 state), together with $$\mathrm{Cl^-}$$ from $$\mathrm{HCl}$$.

$$\textbf{ClF$$_3$$}$$
Oxidation number of Cl: $$+3$$.
Step-1: $$\mathrm{ClF_3 + 3\,H_2O \;\longrightarrow\; Cl(OH)_3 + 3\,HF}$$
Step-2: $$\mathrm{Cl(OH)_3 \;\longrightarrow\; HClO_2 + H_2O}$$ (re-arrangement to the oxy-acid having Cl in +3).
Step-3 (deprotonation): $$\mathrm{HClO_2 \;\longrightarrow\; ClO_2^- + H^+}$$
Hence the oxy-anion present after complete hydrolysis is $$\mathbf{ClO_2^-}$$.

$$\textbf{BrF$$_5$$}$$
Oxidation number of Br: $$+5$$.
Step-1: $$\mathrm{BrF_5 + 5\,H_2O \;\longrightarrow\; Br(OH)_5 + 5\,HF}$$
Step-2: $$\mathrm{Br(OH)_5 \;\longrightarrow\; HBrO_3 + 2\,H_2O}$$ (conversion to the oxy-acid with Br in +5).
Step-3 (deprotonation): $$\mathrm{HBrO_3 \;\longrightarrow\; BrO_3^- + H^+}$$
Therefore the final oxy-anion is $$\mathbf{BrO_3^-}$$.

Collecting the three results:
ICl  ➔ $$IO^-$$
ClF$$_3$$ ➔ $$ClO_2^-$$
BrF$$_5$$ ➔ $$BrO_3^-$$

The sequence matches Option A.

Answer: Option A which is: IO$$^{-}$$, ClO$$_2^{-}$$ and BrO$$_3^{-}$$