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

Jul 30, 2016

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

#### Explanation:

For $C l$, and $O$, there are $7$, and $6$ valence electrons respectively associated with the neutral atoms.

For hypochlorite ion, $C l - {O}^{-}$, 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 $C l - O$ 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 $C l$, 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, $C l {O}_{3}^{-}$. We have $7 + 6 + 6 + 6 + 1$ valence electrons, 26 electrons, and 13 electron pairs to distribute. A Lewis stucture of ${\left(O =\right)}_{2} \ddot{C} l \left(- {O}^{-}\right)$ 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, $C l {O}_{4}^{-}$, where chlorine has its max Group 7 oxidation number of $+ V I I$, Here, we have $7 + 4 \times 6 + 1 = 32$ 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...