How do you calculate the formal charge of ##Cl## in ##ClO^-## and ##ClO_3^-##?

In both examples, the chlorine atom is neutral, and the charge is presumed to reside on oxygen.

For ##Cl##, and ##O##, there are ##7##, and ##6## valence electrons respectively associated with the neutral atoms.

For hypochlorite ion, ##Cl-O^-##, we have to distribute ##7+6+1## electrons in the Lewis structure. There are thus ##7## electron pairs. One of these electron pairs is conceived to form the ##Cl-O## bond, and so around each chlorine and each oxygen atom there are 3 lone pairs of electrons. Because the bonding pair of electron is shared, i.e. one electron is claimed by ##Cl##, and one by ##O##, this means that the chlorine atom owns 7 valence electrons, and is thus formally neutral, and the oxygen atom also owns 7 valence electrons, and thus has a FORMAL negative charge.

That is oxygen, ##Z=8##, has 7 valence electrons, and 2 inner core electrons, and thus 9 electrons in total. Given this electronic formalism, the oxygen centre is formally negative, and our Lewis structure certainly represents this.

And now for chlorate, ##ClO_3^-##. We have ##7+6+6+6+1## valence electrons, 26 electrons, and 13 electron pairs to distribute. A Lewis stucture of ##(O=)_2Cl:(-O^-)## is reasonable, and I think, correctly accounts for the charge. Chlorine is neutral, and the singly bound oxygen has a negative charge. Of course, this charge is distributed to the other oxygen centres by resonance.

For completeness, we should consider perchlorate, ##ClO_4^-##, where chlorine has its max Group 7 oxidation number of ##VII##, Here, we have ##7+4xx6+1=32## valence electrons, and a Lewis structure of ##Cl(=O)_3O^-##.