• A limitation of the model of covalent bonding is that when drawing Lewis structures for molecules, it is sometimes possible to come up with more than one structure while still obeying the?octet rule
• This leads to the problem of deciding which structure is appropriate and is consistent with other information such as spectroscopic data on bond lengths and electron density
• One approach to determining which is the preferred structure is to determine the?formal charge (FC)?of all the atoms present in the molecule
• It is a kind of electronic book keeping involving the bonding, non-bonding and valence electrons
• Formal charge is described as the charge assigned to an atom in a molecule, assuming that all the electrons in the bonds are shared equally between atoms, regardless of differences in?electronegativity
• The formula for calculating FC is

FC= (number of valence electrons) - ?(number of bonding electrons) - (number of non-bonding electrons)

or

FC= V - ?B - N

• The Lewis structure which is preferred is the one which:
  • the?difference?in FC of the atoms is?closest to zero
  • has negative charges located on the most electronegative atoms
• The process of drawing a Lewis structure has been covered previously, but here is a reminder of how to draw the Lewis structure of tetrachloromethane, CCl₄

Steps in drawing the Lewis structure for CCl₄

• To work our the formal charge of the C and Cl atoms in the structure simply apply the FC formula:

FC for carbon = (4) - ?(8) - 0 = 0

FC for chlorine = (7) - ?(2) - 6 = 0

• Notice that formal charge is calculated for one of each type of atom and does not count the total number of atoms in the molecule

Answer

• • Boron is a group 13 element, so has 3 valence electrons. Hydrogen has one valence electron and the charge on the ion is -1, so there are 8 electrons in the diagram. The Lewis structure is therefore:

Lewis structure of BH₄^-

• • The number of bonded electrons is 8 and the number of non-bonded electrons is zero. So the formal charge on B is:

FC (B) = (3) - ?(8) - 0 = -1

• It is possible to draw three resonance structures for sulfur dioxide, SO₂:

The three resonance structures of sulfur dioxide

• The first structure is an illustration of the expansion of the octet as the sulfur has 10 electrons around it
• Formal charge can be used to decide which of the Lewis structures is preferred
• The FC on the first structure is as follows:

FC on sulfur = (6) - ?(8) -(2) = 0

FC on oxygen = (6) - ?(4) -(4) = 0

Difference in FC = ΔFC = FC_max- FC_min?= 0

• The FC on the second (and third) structures is as follows:

FC on sulfur = (6) - ?(6) -(2) = +1

FC on left side oxygen = (6) - ?(2) -(6) = -1

FC on right side oxygen = (6) - ?(4) -(4) = 0

Difference in FC = ΔFC = FC_max- FC_min?= 2

Answer

Structure I

FC on carbon = (4) - ?(8) -(0) = 0

FC on oxygen = (6) - ?(4) -(4) = 0

Difference in FC = ΔFC = FC_max- FC_min?= 0

Structure II

FC on carbon = (4) - ?(8) -(0) = 0

FC on left oxygen = (6) - ?(6) -(2) = +1

FC on right oxygen = (6) - ?(2) -(6) = -1

Difference in FC = ΔFC = FC_max- FC_min?= 2

Structure I is the preferred structure as the difference is zero