The halogen which oxidizes water to ${\mathrm{O}}_2$ exothermally is

Among the halogens electronegativity decreases down the group. So electronegativity  order is $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$.

The atom or molecule whose electronegativity is greater that ${\mathrm{O}}_2$, that atom or  molecule have power to oxidize water to ${\mathrm{O}}_2$ .

Among the halogens electronegativity of fluorine is greater than ${\mathrm{O}}_2$. Therefore it can be able to oxidize water to ${\mathrm{O}}_2$  and ${\mathrm{O}}_3$ 

$2{\mathrm{F}}_2+2{\mathrm{H}}_2\mathrm{O}\to 4\mathrm{HF}+{\mathrm{O}}_2$         

$3{\mathrm{F}}_2+3{\mathrm{H}}_2\mathrm{O}\to 4\mathrm{HF}+{\mathrm{O}}_3$

Hence the correct option is (A).