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

1 Answer
Jul 30, 2016

Answer:

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

Explanation:

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=)_2ddotCl(-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 #(""^(-)O-)Cl(=O)_3#, again with the charge FORMALLY residing on an oxygen atom....which we CONCEIVE to be singly bound to chlorine...