Examples

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CO₃^–²

If we try joining all the atoms with single bonds, we get both O and C in non-normal valences. The O with only one bond looks like Li or F (which has three lone pairs that we don't bother to show here yet). Li would make for a +6 formal charge so we'll eliminate that one quickly as unreasonable. If the O is considered to be like F, it has a formal charge -1.

The C has only a valence of 3 meaning it could look like B or like N. If we consider C to be isoelectronic with N (with a lone pair that is not drawn in yet) then the C would have a formal charge of -1 .

In this model the total of all the formal charges would add up to -4 but the actual species we're trying to model has an overall charge of only –2, i.e., this model has too many electrons in it.

Here, we consider the possibility that the C looks like B and hence has a formal charge of +1.

This gives us a valid Lewis dot structure since the sum of the formal charges does add up to the total charge. This means that once we put in our lone pairs, we will have the correct number of electrons. However, there is a better Lewis structure that we can draw, which has less formal charges.

If we make a double bond the situation improves. Now, the C has a valence of 4 which is its normal valence and hence it has a formal charge of zero. One O (double bond) also has a normal valence (of 2) and a formal charge of zero. The two single-bonded oxygens still look like F and have a formal charge of -1 each. The formal charges add up to the overall charge on the species and we've lowered the total number of charges as much as we can. If we tried to create even more double bonds, we would create a hypervalent C which is not possible since C is an n = 2 atom and has no available d-orbitals.

This is a good model which we'll see later can be improved by using the concept of resonance.

In this final diagram, we've added the lone pairs as follows:

	C is in its normal valence and has no lone pairs.
	O (double bonded) is in its normal valence has two lone pairs.
	O (single bonded) look like F and thus have three lone pairs and a formal charge of -1.

Now check electron sums

    3O: 18
     C:  4
Charge:  2
   sum: 24 or 12 pairs.

If we count the electron pairs in our drawing we find 12 pairs exactly matching the number of valence electrons.

O₂

Here, we've joined the O atoms with a single bond. They both look like F and would have a formal charge of -1 each for a total charge of -2. Obviously, wrong.

Now, with a double bond, both O atoms are in their normal valence state with zero formal charge. This one looks good.

In this final diagram, we've added the lone pairs as follows:

O (double bonded) is in its normal valence has two lone pairs.

e^- sums

2O: 12 ==> 6 pairs, exactly as drawn in our Lewis dot structure.

AlCl₃

	Here, we've joined the atoms with single bonds. The Cl atoms are in their normal valence state and so are neutral. The Al is also in it's normal valence state. This one was easy.
	Now, we've filled in the lone pairs on all the atoms according to their appearance in the table of normal and hypervalences.
e^- sums	3Cl: 21 Al: 3 sum: 24 => 12 pairs	we see 12 pairs of electrons in our Lewis dot structure exactly as predicted.

IF₃

Here, we've joined the I and F atoms with single bonds. The F atoms are in their normal valence and have a formal charge of 0 each. The I is in it's first hypervalent state and also has formal charge of 0.

In this final diagram, we've added the lone pairs as follows:

	F is in its normal valence has three lone pairs.
	I is in its first hypervalence state, i.e., one of the three lone pairs from the normal valence state is split to create two extra bonding sites for a total valence of three and leaving only two lone pairs.

e^- sums

 3F: 21
  I:  7
sum: 28 => 14 pairs

We see 14 pairs drawn on the diagram

XeF₆

Here, we've joined the F atoms to the Xe with a single bond. The F atoms are all in their normal valence state and would have a formal charge of 0. The Xe is in its third hypervalence state (check the table of normal and hypervalences) and also has a formal charge of 0.

In this final diagram, we've added the lone pairs as follows:

	F is in its normal valence has three lone pairs but we've ignored them since they contribute nothing to our understanding of the structure. This is an acceptable abbreviated form in Lewis Dot Structures
	Xe is in its third hypervalence state, i.e., three of the four lone pairs from the normal valence state is split to create two extra bonding sites each for a total valence of 6 and leaving only one lone pair.

e^- sums

  6F: 42
  Xe:  8
 sum: 50 => 25 pairs

If we remember to count the the three lone pairs on each F which we didn't draw in for simplicity sake, we find that we have indeed drawn 25 pairs of electrons.