2.2.2 (g,h) Shapes of Molecules and Ions

Syllabus

(g) the shapes of, and bond angles in, molecules and ions with up to six electron pairs (including lone pairs) surrounding the central atom as predicted by electron pair repulsion, including the relative repulsive strengths of bonded pairs and lone pairs of electrons

{Learners should be able to draw 3-D diagrams to illustrate shapes of molecules and ions.}

(h) electron pair repulsion to explain the following shapes of molecules and ions: linear, non-linear, trigonal planar, pyramidal, tetrahedral and octahedral.

{Learners are expected to know that lone pairs repel more than bonded pairs and the bond angles for common examples of each shape including CH₄ (109.5°), NH₃ (107°) and H₂O (104.5°)}

What does this mean?

Chemistry uses a theory usually referred to as VSEPR.

That is Valence Shell Electron-Pair Repulsion theory.

Where the valence shell is the outside shell.

There are a few fairly simple rules.

All pairs of electrons repel each other but lone pairs repel more than bond pairs.
Double and treble bonds count as a single pair of electrons.
Electron pairs will move as far apart as possible to minimise repulsions.

The IB refers to any bond or lone-pair as a charge centre - one of its few good points!

You know you want to hear the song...

2 charge centres.

The furthest apart two charge centres can be is when the bond angle is 180^o.

Beryllium compounds are generally linear as Be atoms have two unpaired electrons and no lone pairs.

So they form two single bonds that can repel each other to the full 180^o.

The example shows BeH₂- but might easily be BeCl_2,BeBr_2,BeI_2.

Generally you would expect these molecules to be Ionic as they are Metal-Non-metal but very small 2+ ions tend to polarise bonds so much that they end up covalent.

Carbon Dioxide is also linear.

This time there are two double bonds and no lone pairs.

But double bonds don't behave any differently to single bonds in VSEPR. So, this is still just two charge centres.

And the angle and the shape is the same.

There are lone pairs on the Oxygen atoms but we are only concerned with lone pairs if they are on the central atom

Both the molecules above are frequently required on exam papers the molecule below is not.

But it still has 2 charge centres even if one is a bond and the other a treble bond - so is still linear.

Slightly more likely to be asked is Propyne.

Both Carbon 1 and Carbon 2 has a single bond and a treble bond and no lone pairs.

So the shape is linear about both Carbon 1 & 2.

Although it's tetrahedral about Carbon 3 (more later)

Three charge-centres

Three bonds, no lone pairs

The furthest apart three charge centres can be is 120^o - we call this Trigonal Planar - join the outside atoms and you make a flat (planar) triangle.

The easiest way to achieve this is to be a group 3 element and form three single bonds without lone pairs.

Examples would include AlCl₃, AlBr₃, All₃, AlH₃, BCl₃, BBr₃, Bl₃, BH₃

Generally you would expect these molecules to be Ionic as they are Metal-Non-metal but very small 3+ ions tend to polarise bonds so much that they end up covalent.

Any molecule that includes a C=O bond and two single bonds still has three charge centres and no lone pairs.

So it will still take up a Trigonal Planar shape because double bonds don't count as anything different to single bonds.

Two bonds, one lone pairs

eg GeF₂ or GeCl₂ etc

These molecules are based on the same shape.

But because lone pairs repel about 2.5^o more than bond pairs the bond angle is around 117.5^o

The same is true of Sulphur Dioxide.

Here there are two double bonds and a single lone-pair.

And also NOF, where there's a lone pair and one single bond and a double bond (that counts as a single for VSEPR)

All 3 molecules would be referred to as Non-linear, V-shaped, or just Bent.

Four Charge-Centres

Four bond-pairs, no lone-pairs

With four bonds and no lone-pairs to consider the bond-pairs will repel as far apart as physically possible.

Though it may not seem obvious, the furthest apart that 4 bonds can be in three dimensions is 109.5^o.

If I could give you a good explanation as to why, I would.

But I can't.

So it just is and you just have to learn the angle.

Molecules like CCl₄, CCl₄, SiH₄ etc that take up this shape are said to be tetrahedral.

Tetra means four, but it is not talking about the four bonds - it is talking about the four faces of the shape created when you join the four outlying atoms.

This may not seem important now but Octahedral molecules don't have 8 bonds.

They have 6.

But when you join up the outlying atoms they produce an eight-sided shape - an Octahedron.

Drawing a two-dimensional representation of a tetrahedral molecule is impossible but you can use the wedge and dot notation.

Think of a ball and stick model of methane.

Lay it on a table and one bond must point directly up.

Another could be arranged so that it was parallel to the table-edge.

This would leave one pointing to the rear of the table and one pointing towards you.

We draw the upwards and parallel bonds are straight lines.

The bond pointing towards you can be a wedge shape, the one pointing away can be the dotted line.

Practise this - it's important!

Three bond-pairs, one lone-pair

With three bonds and a lone-pair to consider the pairs will repel as far apart as physically possible but the lone pair will repel about 2.5^o more.

This leaves a bond angle of 107^o.

It is no longer a tetrahedron

It is a triangular based pyramid.

So we say the molecule is trigonal pyramidal or just pyramidal.

Examiners really like Ammonia as their example - a wedge and dot diagram would look like this.

Two bond-pairs, two lone-pair

With two bonds and two lone-pairs to consider the pairs will repel as far apart as physically possible but the lone pairs will repel about 5^o more.

This leaves a bond angle of 104.5^o.

It is no longer a tetrahedron or a trigonal pyramid

It is bent or V-shaped --- but with a different bond angle to the bent SO₂ molecule.

Examiners really like water as their example.

You don't need wedges or dots to draw two bonds!

So, all molecules with four charge centres are based on the tetrahedral shape.

Every lone-pair that replaces a bond-pair changes the angle by about 2.5^o.

This means you only really need to remember the tetrahedral bond angle rather than trying to learn all the bond angles.

You still have to remember the names of the shapes though.

Six Charge-Centres

As always the charge centres repel to minimise repulsion.

The OCR are only interested in molecules with six bonds so we don't need to worry about the effects of lone-pairs.

Logically the furthest apart the bonds can be is 90^o.

This is the octahedral shape because joining up the outer atoms makes an eight-sided figure.

The usual wedge and dot representation of an octahedral molecule is shown on the right.

So how do we predict the shape of an unfamiliar molecule?

What group of the Periodic Table is the central atom in? This is how many valence (outershell) electrons it has available.
How many bonds has it formed? This is how many electrons are used up in the bonding.
How many electrons are left? Half this number and you know how many lone-pairs there are.
Is the molecule charged? Add an extra electron for every negative charge, take one away for every positive charge.
Apply the rules above.

eg. SiCl₄

Silicon is in Group 4 - 4 valence electrons.
There are 4 single bonds - 4 electrons used in bonding
There are no electrons left over - no lone-pairs
There is no charge
4 bonds and no lone-pairs is tetrahedral

eg PCl₃

Phosphorus is in Group 5 - 5 valence electrons.
There are 3 single bonds - 3 electrons used in bonding
There are two electrons left over - one lone-pair
There is no charge
3 bonds and 1 lone-pair is pyramidal

eg SCl₂

Sulphur is in Group 6 - 6 valence electrons.
There are 2 single bonds - 2 electrons used in bonding
There are four electrons left over - two lone-pairs
There is no charge
2 bonds and 2 lone-pairs is bent

eg NH₄+

Nitrogen is in Group 5 - 5 valence electrons.
There are enough electrons for 5 bonds but only 4 are needed
There aren't enough electrons left over for a lone-pair
There is a 1+ charge so remove the spare electron
4 bonds and no lone-pairs is tetrahedral

Video

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Exam-style Questions

(1) Draw the shape of the PCl5 molecule and of the PCl4 + ion. State the value(s) of the bond angles.

PCl5 PCl4+

Bond angle(s) ...................................... Bond angle(s) ..........................................(4)

(a) Draw diagrams to show the shapes of the following molecules and in each case show the value of the bond angle on the diagram.

BeCl2 BF3

CCl4 SF6

(8)

(b) Explain why the shape of NF3 is not the same as the shape of BF3.

.......................................................................................................................................................................

.......................................................................................................................................................................(3)

(Total 11 marks)

(3) Predict the shape of the AlH₄^- ion. Explain why it has this shape.

Shape .......................................................................................................................................................................

Explanation .......................................................................................................................................................................

.....................……………………….............................................................................………………………………………………

.....................………………………….........................................................................……………………………………………….(3)

Answers

(1) 4

PCl5 shown as trigonal bipyramid PCl4+ shown as tetrahedral

[Look for: ONE solid linear Cl-P-Cl bond] NO solid linear Cl-P-Cl bonds]

Bond Angle(s) 90° and 120° (1) Bond angle(s) 109 or 109.5° (1)

(2)

(a)