Using Lewis structures with Cl=O double bonds and no charge separation, the total number of lone pairs on all atoms (Cl and O) in the four molecules $\mathrm{HOCl}$, $\mathrm{HClO}{}_2$, $\mathrm{HClO}{}_3$ and $\mathrm{HClO}{}_4$ taken together is

Count lone pairs on Cl from 7 valence electrons less the electrons used in bonds, and 2 on each terminal O and 2 on O of OH. HOCl: Cl 3 + O 2 = 5; HClO2: Cl 2 + OH 2 + O 2 = 6; HClO3: Cl 1 + OH 2 + 4 = 7; HClO4: Cl 0 + OH 2 + 6 = 8. The total is 26.