Bonding Angles Explained

[AustralianIB]

I have been tutoring Chemistry for over 2 years now and one of the topics students continually struggle with is Bonding, especially with regards to bond angles and VSEPR Theory.

The key points to remember from VSEPR Theory are:

- Electrons repel each other and thus spread out to minimise repulsion

- Lone pairs further compress the bonding angle (I use 2 degrees as a rule of thumb)

Throughout this explanation, I will be referring to negative charge centres, defined as:

- A Single, Double or Triple Bond (Counts as one negative charge centre)

- A non-bonding pair

HL students can use this definition:

- sigma bond

- non-bonding pair

The hard part is working out how many bonding and non-bonding pairs there are. This comes through practice and being able to recognise molecules from experience (I'll give a few tips about this at the end)

The easy part is memorising is knowing what to do after you know how many negative charge centres and non-bonding pairs there are. Let's go through this now:

Note - when I write say 4 negative charge centres, 2 non-bonding pairs, I mean:

- 4 negative charge centres, of which 2 are non-bonding pairs

- NOT 4 negative charge centres and 2 non-bonding pairs

1 negative charge centre:

If there is only 1 negative charge centre, the shape can only be linear (180 degrees)

Examples include H2, O2 Cl2 F2 N2 etc.

2 Negative charge centres:

0 non-bonding pairs

If there are 2 negative charge centres, e.g. CO2, the charge centres will automatically repel each other to be as far apart as possible.

Hence: Linear, 180 degrees

1 non- bonding pair

This essentially means that there is only 1 bond, so linear, 180 degrees

3 Negative charge centres:

0 non-bonding pairs

3 negative charge centres trying to spread out to minimise repulsion means basically 360 degrees/3 = 120 degrees.

Shape: trigonal planar

1 non-bonding pair

Using the rule of thumb of -2, this would be 118 degrees

Shape: Bent

4 negative charge centres:

0 non-bonding pairs

Regular tetrahedron - 109.5 degrees

1 non-bonding pair

Trigonal pyramidal - 107 degrees (-2 and rounding down)

2 non-bonding pairs

V-shaped - 105 degrees

5 negative charge centres:

0 non-bonding pairs

Trigonal Bi-pyramidal - 120 degrees and 90 degrees (Note that there are 2 different angles here)

1 non-bonding pair

See-saw - 118 degrees and 90 degrees

2 non-bonding pairs

T - shaped - 90 degrees (Think T)

6 negative charge centres:

0 non-bonding pairs

Octahedral - 90 degrees

1 non-bonding pair

Square pyramid - 88 degrees

2 non-bonding pairs

Square planar - 90 degrees

Now for a little bit to help ease the hard part:

For SL students, if you see a 4 at the end, e.g. NH4, CH4 - this is tetrahedral 109.5 degrees. Why? Because 5 and 6 negative charge centres are not examinable in SL.

BF3 is trigonal planar - 120 degrees

NH3 is trigonal pyramidal - 107 degrees

H2O has 4 negative charge centres, of which 2 are non-bonding - V-shaped, 105 degrees

For HL Students, if you see a 6, it is octahedral. As 7+ negative charge centres are not examinable.

Be careful when you see 2s however, as it can be linear or V-shaped.

For HL students, be careful when you see 4, as it is not necessarily tetrahedral.

Hope this helps!

[SerUmbras]

As a point, I found an exam that asked for bond angles of XeF₄ and XeO_4, as well as PF₅ and IF₅. It's those sorts of really nasty questions that they'll throw at you in HL, but if you read the all of the above, you'll nail it. It does require knowledge of 'expanding the octet' though.

To add on to AusIB's statement;

For an atom to 'expand the octect' it needs two things; a), to be in period 3 or higher (in number, not higher up the table) and b), to have spare n.b. electron pairs.

For each spare electron pair, an atom can create 2 more bonds than it normally could - in the example above phosphorus, P, can normally only make 3 bonds, but it can "sacrifice" its n.b. electron pair to make 2 more bonds.

NOTE: In the example above, phosphorus makes either 3 OR 5 bonds, not up to 5 bonds. Each n.b. pair that is lost MUST make 2 bonds.

For other groups it's a similar deal;

Group 5 - 3 OR 5 (1 n.b. pair)

Group 6 - 2 OR 4 OR 6 (2 n.b. pairs)

Group 7 - 1 OR 3 OR 5 OR 7 (3 n.b. pairs)

Group 8 - Usually makes none with the exception of Xenon. Though I've not seen a question where it makes 2 bonds, theoretically it can make 0, 2, 4, 6 OR 8 bonds. This is usually a favourite for IB (so my teacher says)

So yeah. That's about all that is needed for bonding. Angles and names are all covered, as is some HL theory. Sweet.

GL, HF, DD

Edited August 3, 2012 by illidan101