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

1 Answer
May 19, 2016

Answer:

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, #NH_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 " 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 #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 #(HO-)_2S(=O)_2# or better, #(HO-)_2S^(2+)(-O^-)_2#. How?