Which of the following halogen forms maximum number of oxoacids?

The formation of oxoacids by halogens depends on the availability of d-orbitals to expand the octet, allowing multiple oxidation states.

Fluorine cannot expand its octet and forms only one oxoacid: hypofluorous acid $\mathrm{HOF}$ (oxidation state $+1$).

Chlorine, with available d-orbitals, forms four stable oxoacids: hypochlorous $\mathrm{HClO}$ ($+1$), chlorous $\mathrm{HClO_2}$ ($+3$), chloric $\mathrm{HClO_3}$ ($+5$), and perchloric acid $\mathrm{HClO_4}$ ($+7$).

Bromine forms hypobromous $\mathrm{HBrO}$ ($+1$), bromic $\mathrm{HBrO_3}$ ($+5$), and perbromic acid $\mathrm{HBrO_4}$ ($+7$), omitting the unstable bromous acid $\mathrm{HBrO_2}$ ($+3$).

Iodine forms hypoiodous $\mathrm{HIO}$ ($+1$), iodic $\mathrm{HIO_3}$ ($+5$), and periodic acid $\mathrm{HIO_4}$ or $\mathrm{H_5IO_6}$ ($+7$), excluding the unstable iodous acid $\mathrm{HIO_2}$ ($+3$).

Chlorine forms the maximum number of stable oxoacids.