Sometimes when writing a Lewis structure you come across two different ways to write the molecule, both which look fine. In this case, you should use formal charge to decide which structure is correct for the molecule.
The formal charge is the difference in the number of valence electrons in the atom and the number of valence electrons in the Lewis structure. The equation for the formal charge of any atom in a Lewis structure is

C_f = E_v - (E_u + 1/2E_p)

• C_f is the formal charge
• E_v is the number of valence electrons in the bare atom
• E_u is the number of electrons in lone pairs on the atom in the Lewis structure
• E_p is the number of electrons in bonded pairs on the atom in the Lewis structure

• all formal charges as close to zero as possible
• negative formal charges on electronegative atoms like halogens or oxygen.

For example, consider the methanol molecule CH₃OH. This can be written two different ways:


In both cases the octet rule is satisfied for all of the atoms in the structure. Which is correct?

To decide, work out the formal charges on the carbon and oxygen in each molecule. Carbon has 4 valence electrons in the base atom, oxygen 6. For the leftmost structure

• The carbon has four bonds, each worth 2 electrons, for a total of eight. It has no lone pairs. Thus, C_f = 4 - (0 + 1/2*8) =0
• The oxygen has two bonds, each worth 2 electrons, for a total of four. It has two lone pairs. Thus, C_f = 6 - (4 + 1/2*4) =0

• The carbon has three bonds, each worth 2 electrons, for a total of six. It has one lone pair. Thus, C_f = 4 - (2 + 1/2*6) = -1
• The oxygen has three bonds, each worth 2 electrons, for a total of six. It has one lone pair. Thus, C_f = 6 - (2 + 1/2*6) = +1

Example: There are two possible Lewis structures for the sulfurous acid molecule, H₂SO₃. Which is probably correct?

Solution: The difference between the two is the double bond in the leftmost one vs. the single bond + three lone pairs on the oxygen on the right. Work out the formal charge on each: the oxygen and sulfur both have six valence electrons. for the left structure

• The oxygen has two bonds (total 4 electrons) and two lone pairs (total 4 electrons). C_f = 6 - (4 + 1/2*4) = 0
• The sulfur has four bonds (total 8 electrons) and one lone pair (total 2 electrons). C_f = 6 - (2 + 1/2*8) = 0

• The oxygen has one bond (total 2 electrons) and three lone pairs (total 6 electrons). C_f = 6 - (6 + 1/2*2) = -1
• The sulfur has three bonds (total 6 electrons) and one lone pair (total 2 electrons). C_f = 6 - (2 + 1/2*6) = +1

Information provided by: http://learn.chem.vt.edu