How do you calculate the formal charge on atoms of an ion?

May 19, 2016

With care!

Explanation:

Most of the time it is fairly straightforward to assign the formal charge on molecule or a radical ion, by considering the formal charge of the individual atoms.

In the given example, we considered the neutral ammonia molecule, versus the ammonium cation, $N {H}_{4}^{+}$. Here I will consider the oxygen molecules, ${O}_{2}$ versus the ozone, ${O}_{3}$, molecule.

Now both species are neutral gases, and our Lewis structures should reflect this, nevertheless, in the ozone molecule there is formal charge separation.

For $O$, $Z = 8$, there are $6$ valence electrons; the other $2$ electrons are inner core and do not participate in bonding. For the ${O}_{2} \text{ molecule}$, there are $12$ valence electrons, i.e. $6$ electron pairs to distribute over $2$ $O$ atoms, and a $O = O$ molecule results.

Why is each oxygen atom neutral here? Each oxygen atom has $2$ lone pairs of electrons, and shares the electrons involved in the double bond. Thus each $O$ atom claims $4$ electrons from the lone pairs (lone pairs devolve to the parent atom) plus it is conceived to share $\frac{1}{2}$ the electrons that form each covalent bond. Thus each oxygen atom is associated with $2 + 4 + 1 + 1$ electrons. These $8$ electrons electrostatically balance the $8$ positive charges in the $O$ nucleus, and so each $O$ is depicted as neutral.

Now contrast this with the neutral ozone molecule, ${O}_{3}$, with $18$ valence electrons. The typical VESPER structure of ozone is as bent $O = {O}^{+} - {O}^{-}$ molecule, with formal charge separation.

Going from right to left in the $O = {O}^{+} - {O}^{-}$ structure, there are $8$, $7$, and $9$ electrons on each individual oxygen atom, and the resultant charges are $0 , + 1 , - 1$ as required. Do you see the formalism?

If you re happy with this, try formal charge assignment with the neutral sulfuric acid molecule. We can represent this as either ${\left(H O -\right)}_{2} S {\left(= O\right)}_{2}$ or better, ${\left(H O -\right)}_{2} {S}^{2 +} {\left(- {O}^{-}\right)}_{2}$. How?