Formal Charge (College Board AP® Chemistry): Revision Note

What is the formal charge of boron in the BH₄^- ion?

Answer:

• Boron is in group 13, so it has 3 valence electrons

• Hydrogen is in group 1, so it has 1 valence electron

• The negative charge means BH₄^- has gained one electron, so the total number of valence electrons is 3 + (4 × 1) + 1 = 8

• By following the steps to draw a Lewis diagram, the structure should look like this:

• The number of bonding electrons in B is 8 and the number of nonbonding electrons in B is 0

FC = V - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$B - N

FC on boron = 3 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(8) - 0 = -1

Two resonance structures for carbon dioxide, CO₂, are shown below:

Calculate the formal charge on every atom in each structure and deduce which is the preferred structure.

Answer:

Structure I

FC = V - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$B - N

FC on carbon = 4 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(8) - 0 = 0

FC on each oxygen = 6 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(4) - 4 = 0

Structure II

FC = V - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$B - N

FC on carbon = 4 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(8) - 0 = 0

FC on the triple-bonded oxygen = 6 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(6) - 2 = +1

FC on the single-bonded oxygen = 6 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(2) - 6 = -1

• In both structures, the formal charges add up to 0, because CO₂ is a neutral molecule

• Structure I is the preferred structure because every atom has a formal charge of 0

• Structure II separates the charge into +1 and −1, and puts a positive formal charge on oxygen, a very electronegative atom, so it is less favorable

Two possible Lewis diagrams for the formula HCN are shown below. The atoms are joined in a different order, so these are two different molecules (isomers), not resonance structures

Use formal charge to deduce which arrangement of atoms is more likely.

Answer:

Structure I

FC = V - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$B - N

FC on hydrogen = 1 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(2) - 0 = 0

FC on carbon = 4 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(8) - 0 = 0

FC on nitrogen = 5 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(6) - 2 = 0

Structure II

FC = V - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$B - N

FC on hydrogen = 1 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(2) - 0 = 0

FC on carbon = 4 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(6) - 2 = -1

FC on nitrogen = 5 - $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$(8) - 0 = +1

• In both structures, the formal charges add up to 0, because the molecule is neutral

• Structure I (H–C≡N) is more likely because every atom has a formal charge of 0

• Structure II has charges of +1 and −1, and its negative formal charge is on carbon, which is less electronegative than nitrogen

• Formal charge belongs to each atom. It is not the overall charge of the ion, and it is not the number of valence electrons

• To justify the best diagram, calculate the formal charge on every atom in each diagram and compare them. Do not argue from the octet alone or from "charges on the outside"

• Lewis diagram and Lewis structure mean the same thing