Many molecules and ions have more than one Lewis structure that obeys the octet rule. Which one is best? A bookkeeping tool called formal charge answers that, by asking how each atom's share of electrons compares with what it brought.
Key idea
$$\text{FC} = (\text{valence electrons}) - (\text{lone-pair electrons}) - (\text{number of bonds})$$ The formal charges in a structure add up to the overall charge. The best structure has formal charges as close to zero as possible, and any negative formal charge on the more electronegative atom.
Why it works
Formal charge pretends every bond is shared perfectly evenly: each atom gets all of its lone-pair electrons and half of each bonding pair (one electron per bond). Compare that count with the atom's valence electrons in the free atom. If the atom "owns" one electron fewer than it brought, its formal charge is $+1$. Structures that need a lot of charge separation, or put negative charge on an atom that holds electrons weakly, cost energy, so they contribute less to the real molecule.
Worked example: Formal charges in nitrate
In the Lewis structure of $\mathrm{NO_3^-}$ with one N=O and two N–O bonds, find each atom's formal charge.
- N: 5 valence electrons, no lone pairs, 4 bonds: $5 - 0 - 4 = +1$.
- Double-bonded O: $6 - 4 - 2 = 0$.
- Each single-bonded O: $6 - 6 - 1 = -1$.
- Sum: $+1 + 0 - 1 - 1 = -1$, the ion's charge.
Answer: N $+1$; O $0$, $-1$, $-1$.
Worked example: Choosing the best structure
The cyanate ion, $\mathrm{OCN^-}$, has three octet structures: (a) $\mathrm{[O{=}C{=}N]^-}$, (b) $\mathrm{[O{-}C{\equiv}N]^-}$, (c) $\mathrm{[O{\equiv}C{-}N]^-}$. Which is best?
- (a): O $6 - 4 - 2 = 0$, C $4 - 0 - 4 = 0$, N $5 - 4 - 2 = -1$.
- (b): O $6 - 6 - 1 = -1$, C $0$, N $5 - 2 - 3 = 0$.
- (c): O $6 - 2 - 3 = +1$, C $0$, N $5 - 6 - 1 = -2$.
- (c) has the largest charges: worst. (a) and (b) both have one $-1$, but (b) puts it on oxygen, the more electronegative atom.
Answer: (b), $\mathrm{[O{-}C{\equiv}N]^-}$.
Common mistake
Confusing formal charge with oxidation number or with the real charge on an atom. Formal charge splits every bond evenly; oxidation numbers give all the bonding electrons to the more electronegative atom; the real charge distribution is in between. In $\mathrm{NH_4^+}$ the nitrogen has a formal charge of $+1$, but its oxidation number is $-3$, and calculations of the real charge distribution find a partial negative charge on it. Check: formal charges must add to the overall charge, and they are used only to compare structures.