Formal Charge

Sometimes the five-step recipe above yields MORE THAN ONE plausible Lewis structure for the same molecule. Consider carbon dioxide (CO₂): the skeletal structure is O-C-O; total valence electrons = [1×4 (C)] + [2×6 (O)] = 16; three single bonds use 4 electrons, leaving 12 to distribute; placing six lone pairs (three per oxygen) satisfies both oxygens' octets but leaves carbon TWO electron pairs short. Two different fixes are possible -- move two lone pairs from ONE oxygen (giving that oxygen a triple bond to carbon and leaving the other oxygen singly bonded) or move one lone pair from EACH oxygen (giving two symmetric C=O double bonds) -- producing two different candidate Lewis structures (Fig 10.7(a)).

Formal charge is the tool used to decide which candidate structure is the better representation. The formal charge of an atom in a molecule is the electrical-charge difference between the number of valence electrons that atom has in its free, isolated state and the number of electrons formally assigned to it within the Lewis structure. The formula is:

where $N_v$ = number of valence electrons of the atom in its isolated state, $N_l$ = number of electrons present as lone pairs around the atom in the Lewis structure, and $N_b$ = number of electrons present in the bonds around the atom (bond pairs, each bond contributing 2 electrons) in the Lewis structure.

Worked for CO₂ structure 1 (two C=O double bonds, symmetric):

• Carbon: $N_v = 4$, $N_l = 0$, $N_b = 8$ (two double bonds = 4 bond pairs = 8 electrons) → FC $= 4 - 0 - 8/2 = 4 - 4 = 0$.
• Each oxygen: $N_v = 6$, $N_l = 4$ (two lone pairs), $N_b = 4$ (one double bond = 2 bond pairs) → FC $= 6 - 4 - 4/2 = 6 - 4 - 2 = 0$. So structure 1 has ZERO formal charge on every atom.

Worked for CO₂ structure 2 (one single bond, one triple bond):

• The singly-bonded oxygen: $N_v = 6$, $N_l = 6$ (three lone pairs), $N_b = 2$ (one single bond) → FC $= 6 - 6 - 2/2 = -1$.
• The triply-bonded oxygen: $N_v = 6$, $N_l = 2$ (one lone pair), $N_b = 6$ (one triple bond = 3 bond pairs) → FC $= 6 - 2 - 6/2 = 6 - 2 - 3 = +1$.
• Carbon works out to 0 as before. So structure 2 carries formal charges of +1 and −1 on the two oxygens (Fig 10.7(b)).

Selection guidelines, applied after calculating formal charges:

1. A structure in which ALL formal charges are zero is preferred over one with non-zero charges.
2. A structure with SMALLER formal charges (in magnitude) is preferred over one with larger formal charges. …