Explanation

(a)
F_{2} on reaction with NaOH gives different products under different conditions.
(i) $F_{2} + \ dil,\ cold\ NaOH$
$\ \ 2F_{2} + 2NaOH(cold)(dil) \longrightarrow 2NaF + H_{2}O + OF_{2}\$
_{oxygen diflouride}
(ii)F_{2}+hot ,conc.NaOH
4F_{2}+NaOH (hot)(dil) ⟶4NaF+2 H_{2}O+O_{2}