14.1 Further Aspects of Covalent Bonding & Structure

Formal Charge

You should recognise a Lewis Structure from Topic 4.

It's not impossible for some molecules to have several different Lewis Structures without breaking the Octet Rule.

How would we know which to believe?

We use Formal Charge (FC) and pick the structure that comes out closest to zero.

You really should be able to draw a molecule of Sulphuric Acid.

If we apply FC to it we get:

So, this is probably the correct structure.

The usual Lewis Structure for SO₂ gives the S atom 10 electrons instead of 8.

Use FC to show that it is reasonable.

The SCN^- ion has more than one possible Lewis Structure. which is best?

Use FC to show that these Lewis Structures are reasonable.

We've already seen that having more than 8 electrons around one atom doesn't necessarily make the Lewis Structure wrong.

Use FC to show which Lewis Structure is more correct

In the case of structure (b) we have two different FCs for the O atoms.

We then say ∆FC = FC_max - FC_min

If ∆FC for the O atoms in (b) is closer to zero than FC of the O atoms in (a) then it is better.

We don't really need to know why expanded octets happen.

The usual explanation involves unused an partly unused d-orbitals.

It's a good idea to know which elements are likely to expand their octets though.

Some elements form compounds that are electron deficient - don't have a full octet.

Assign Formal Charges to Aluminium Chloride

We usually find Aluminium Chloride as a dimer - two molecules joined together as below.

Assign FC to the Aluminium Atoms and the two types of Cl atoms in the dimer.

Is this structure better?

Remember: ∆FC = FC_max - FC_min

eg Beryllium Chloride

Not all electron deficient molecules can dimerise.

Assign FC to BeCl₂

Five domains

If an atom can expand its octet then it can have more than 4 electron domains.

This leads to new shapes of molecules.

With 5 domains we can have 5 bonds; 4 bonds and 1 lone pair; or 3 bonds and 2 lone pairs.

All are based on the trigonal bipyramidal model

You should learn the names and bond angles for each.

Enjoy!

We could also have two bonds and three lone pairs.

It is still based on trigonal bipyramidal but would appear linear in shape.

Six domains

With 6 domains we can have 6 bonds; 5 bonds and 1 lone pair; or 4 bonds and 2 lone pairs.

All are based on the octahedral model

You should learn the names and bond angles for each.

Enjoy Again!

We could also have three bonds and three lone pairs.

But this isn't on the IB syllabus!!!!!

Draw a 3-D representation using wedges and dots for the following compounds.

You should include bond angles

1. BrF₅

2. PCl₅

3. SCl₆

4. SBr₄

5. XeCl₄

6. BrCl₃

There are three reasonable structures for the Nitrate ion.

Or

Each Lewis Structure is effectively the same, so why would one happen and not the other.

The idea of RESONANCE is that the 3 structures flip from one to the other because there's no difference in their energies.

We usually show one resonance form of the ion as:

Where we think of the double bond being delocalised (spread) across the three N-O bonds.

This gives each bond a bond order of 1.33.

It also spreads the 2- charges across the three N atoms, giving each a charge of -2/3

QUESTION: Draw three resonance forms of the CO₃^2- ion and the delocalised form.

QUESTION: Draw two resonance forms of the NO₂^- ion and the delocalised form.

QUESTION: Draw two resonance forms of the C₃H₅⁺ ion and the delocalised form.

.

QUESTION: Draw two resonance forms of Ozone, O₃ (one O will have a + charge, another a -) and the delocalised form.