# Does formal charge affect polarity?

Jun 27, 2016

It's impossible. Formal charge is an accounting method, and cannot affect an observable.

No, it can't. No accounting scheme that we make up affects something we can observe in real life.

Formal charge is determined by assuming the bonds are perfectly covalent, meaning that electrons are perfectly evenly shared.

That is, if we cleave the bond in half, exactly one electron goes to each atom.

Then, the formal charge is calculated as:

$\setminus m a t h b f \left({\text{FC" = "valence e"^(-) - "owned e}}^{-}\right)$

where:

• The number of valence electrons is based off of the typical charge of the ionized atom.
• The number of owned electrons is based off of how many valence electrons the atom has upon bond cleavage.

So in sulfate, ${\text{SO}}_{4}^{2 -}$, the major resonance structure has:

• two oxygen atoms with a formal charge of $6 - 6 = \textcolor{b l u e}{0}$
• two oxygen atoms with a formal charge of $6 - 7 = \textcolor{b l u e}{- 1}$
• one sulfur atom with a formal charge of $6 - 6 = \textcolor{b l u e}{0}$

The resonance hybrid structure is, however, completely symmetrical, despite what you perceive in the major resonance structure.

So, all dipoles cancel out perfectly and an undisturbed ${\text{SO}}_{4}^{2 -}$ ion is nonpolar.