Reading about how to draw Lewis structures, all I know is that formal charge is the charge assigned to an atom in a molecule assuming that electrons are shared equally, regardless of differing electronegativity between atoms. It is just a concept used to arrive at the correct Lewis structure and does not represent actual charges on atoms. But what I don't understand is, why does using this concept lead us to the correct structure?

For example, right now I am looking at the example of the sulfate ion. I get to this point before formal charges need to be calculated:

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All the atoms have an octet around them (though I'm aware sulfur can have more than eight electrons around it), but it is not the correct structure. Why does considering the formal charge give the right structure, if as stated above it doesn't reflect any actual charges? It also says on Wikipedia "Formal charge is a test to determine the efficiency of electron distribution of a molecule. This is significant when drawing structures.", but I am not sure what it means by this.

  • $\begingroup$ It's mostly for electron bookeeping, especially useful when drawing resonance structure. $\endgroup$ – L to the V Feb 28 '16 at 10:31

Formal charges are helpful for two reasons:

  • at lower levels, you use them to show your instructor that you can properly count electrons and ‘keep the books’. You would need to realise that each oxygen has an additional electron and therefore a formal negative charge while sulfur is lacking two electrons and therefore has a formal positive charge. Adding up the formal charges should give the charge of the molecule.

    Since all lone pairs are typically drawn at this level, lacking formal charges can easily be rederived.

  • at higher levels, lone pairs are typically not drawn unless they are important for some reason (e.g. if they take part in a resonance mechanism discussed at this very moment). Therefore, the only information on the number of electrons around an atom we can extract from the formula is the formal charge.

    If at higher levels, you draw a tetravalent sulfur atom without formal charges, we must assume that it is actually a formally neutral sulfur and that there is an additional lone pair. This, of course, makes a big difference; compare the electronic states of sulfur in $\ce{SO4^2-}$ and $\ce{SF4}$.

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