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QUESTION

# 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^-.