Formation of Covalent Bonds

Lessons

Senior 4 · Chemistry

Formation of Covalent Bonds

5 lessons

Lesson 4 of 5

VSEPR Theory and the Shapes of Molecules

25 min

Why is water bent?

Carbon dioxide, O=C=O, is a straight molecule. Water, H–O–H, has the same pattern of three atoms, yet it is bent, with an angle of 104.5° between its bonds. That bend is why water molecules have positive and negative ends, why water dissolves salt and sugar, and why ice floats. Chemists predict shapes like this with a simple model based on one idea: electron pairs repel each other. You can check every prediction with a model kit, and you should, because these are three-dimensional objects.

Definition

VSEPR theory (valence shell electron pair repulsion): the pairs of electrons in the outer shell of the central atom, bonding pairs and lone pairs, repel one another and arrange themselves as far apart as possible. The positions of the atoms then give the shape of the molecule.

Bond angle: the angle between two bonds that meet at the same atom.

The rules

Count the electron pairs (regions of electron density) around the central atom. A double or triple bond counts as one region, because all its electrons point the same way.
The regions arrange themselves as far apart as possible: 2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral.
Lone pairs are held closer to the central atom and spread out more than bonding pairs, so they repel more strongly. The order of repulsion is $\text{lone pair–lone pair} > \text{lone pair–bond pair} > \text{bond pair–bond pair}$ Each lone pair squeezes the bonding pairs closer together, by about 2.5° in the tetrahedral family.
Name the shape from the positions of the atoms only; lone pairs are invisible in the name.

To find the number of pairs: take the central atom's outer electrons, add one for each atom bonded to it by a single bond, add one for each negative charge (subtract one for each positive charge), and divide by two. The number of bonding pairs is the number of atoms bonded; the rest are lone pairs.

Table 4.1: Shapes from electron pairs

Pairs	Bonding	Lone	Shape	Bond angle	Example
2	2	0	linear	180°	$\mathrm{BeCl_2}$ , $\mathrm{CO_2}$
3	3	0	trigonal planar	120°	$\mathrm{BF_3}$
3	2	1	bent	about 119°	$\mathrm{SO_2}$
4	4	0	tetrahedral	109.5°	$\mathrm{CH_4}$ , $\mathrm{NH_4^{+}}$
4	3	1	trigonal pyramidal	107°	$\mathrm{NH_3}$
4	2	2	bent (V-shaped)	104.5°	$\mathrm{H_2O}$
5	5	0	trigonal bipyramidal	90° and 120°	$\mathrm{PCl_5}$
5	4	1	see-saw	about 87° and 102°	$\mathrm{SF_4}$
5	3	2	T-shaped	about 87.5°	$\mathrm{ClF_3}$
6	6	0	octahedral	90°	$\mathrm{SF_6}$
6	5	1	square pyramidal	about 85°	$\mathrm{BrF_5}$
6	4	2	square planar	90°	$\mathrm{XeF_4}$

In the trigonal bipyramid, lone pairs go in the equatorial positions, where they have more room (only two neighbours at 90° instead of three). In the octahedron, two lone pairs go opposite each other.

Worked example 1: methane, ammonia and water

Use VSEPR to explain why the bond angles are $\mathrm{CH_4}$ 109.5°, $\mathrm{NH_3}$ 107° and $\mathrm{H_2O}$ 104.5°.

Solution.

C: $\frac{4 + 4}{2} = 4$ pairs, all bonding → tetrahedral, 109.5°.
N: $\frac{5 + 3}{2} = 4$ pairs: 3 bonding, 1 lone. The four pairs are still roughly tetrahedral, but the lone pair repels the bonding pairs more strongly, squeezing them to 107°. Shape (atoms only): trigonal pyramidal.
O: $\frac{6 + 2}{2} = 4$ pairs: 2 bonding, 2 lone. Two lone pairs squeeze the bonds further, to 104.5°. Shape: bent.

Worked example 2: the ions in solid phosphorus pentachloride

In the solid state, phosphorus pentachloride exists as $[\mathrm{PCl_4}]^{+}[\mathrm{PCl_6}]^{-}$ . Predict the shape and bond angles of each ion.

Solution.

$\mathrm{PCl_4^{+}}$ : P has 5 outer electrons, +4 from the Cl atoms, −1 for the positive charge = 8 electrons, 4 pairs, all bonding → tetrahedral, 109.5°.
$\mathrm{PCl_6^{-}}$ : 5 + 6 + 1 for the negative charge = 12 electrons, 6 pairs, all bonding → octahedral, 90°.

Answer: a tetrahedral cation and an octahedral anion, while the gaseous $\mathrm{PCl_5}$ molecule is trigonal bipyramidal.

Worked example 3: chlorine trifluoride and the amide ion

Predict the shapes of $\mathrm{ClF_3}$ and $\mathrm{NH_2^{-}}$ .

Solution. $\mathrm{ClF_3}$ : $\frac{7 + 3}{2} = 5$ pairs, 3 bonding and 2 lone. Five pairs form a trigonal bipyramid; the two lone pairs take equatorial positions, leaving the three F atoms in a T shape, with F–Cl–F angles a little under 90°. $\mathrm{NH_2^{-}}$ : $\frac{5 + 2 + 1}{2} = 4$ pairs, 2 bonding and 2 lone → bent, about 104.5°, like water.

Warning

Common mistakes. (1) Naming the shape from all the electron pairs: $\mathrm{NH_3}$ has four pairs but its shape is trigonal pyramidal, not tetrahedral. (2) Forgetting the charge on an ion when counting electrons. (3) Counting a double bond as two pairs for the shape: $\mathrm{CO_2}$ has two regions, so it is linear. (4) Drawing a 3D shape flat: use wedges and dashes, and give the angle.

Tip

Set out every shape answer in four lines: pairs around the central atom → bonding and lone pairs → shape (atoms only) → angle, and why it is reduced ("lone pairs repel more than bonding pairs"). This earns all the marks in most exam questions.

Professor Dave Explains: VSEPR Theory and Molecular Geometry (7 min). Watch how the number of electron domains sets the arrangement, and how lone pairs change the shape's name and shrink the angles.

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Practice

State the main idea of VSEPR theory and the order of repulsion between lone pairs and bonding pairs.
Predict the shape and bond angle of: $\mathrm{SiH_4}$ , $\mathrm{PH_3}$ , $\mathrm{H_2S}$ , $\mathrm{BCl_3}$ , $\mathrm{BeF_2}$ .
Draw $\mathrm{NH_3}$ and $\mathrm{NH_4^{+}}$ with wedges and dashes, and explain why their bond angles differ.
Predict the shapes of $\mathrm{SF_4}$ and $\mathrm{XeF_4}$ . Explain where the lone pairs go in each.
Predict the shape of the $\mathrm{H_3O^{+}}$ ion and estimate its bond angle.
Carbon dioxide is linear but sulfur dioxide is bent. Explain the difference.
Predict the shape of $\mathrm{ICl_4^{-}}$ .

Show answersHide answers

Electron pairs around the central atom repel and get as far apart as possible; lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
$\mathrm{SiH_4}$ tetrahedral 109.5°; $\mathrm{PH_3}$ trigonal pyramidal (about 93.5° in fact, below 107°); $\mathrm{H_2S}$ bent (about 92° in fact); $\mathrm{BCl_3}$ trigonal planar 120°; $\mathrm{BeF_2}$ linear 180°. (VSEPR predicts angles a little under 109.5° for $\mathrm{PH_3}$ and $\mathrm{H_2S}$ ; the real values are smaller still.)
$\mathrm{NH_3}$ : 3 bonding + 1 lone pair, 107°. $\mathrm{NH_4^{+}}$ : 4 bonding pairs, 109.5°. The lone pair in $\mathrm{NH_3}$ repels more strongly and closes the angle; in $\mathrm{NH_4^{+}}$ it has become a bonding pair.
$\mathrm{SF_4}$ : $\frac{6 + 4}{2} = 5$ pairs, 4 bonding + 1 lone; the lone pair is equatorial → see-saw. $\mathrm{XeF_4}$ : $\frac{8 + 4}{2} = 6$ pairs, 4 bonding + 2 lone; the lone pairs are opposite each other → square planar.
$\frac{6 + 3 - 1}{2} = 4$ pairs, 3 bonding + 1 lone → trigonal pyramidal, about 107°.
In $\mathrm{CO_2}$ carbon has two double bonds and no lone pairs: 2 regions, linear. In $\mathrm{SO_2}$ sulfur has two bonding regions and one lone pair: 3 regions, bent (about 119°).
$\frac{7 + 4 + 1}{2} = 6$ pairs, 4 bonding + 2 lone → square planar.

Quick check

What is the shape of an ammonia molecule, NH₃?

Key points

Electron pairs around the central atom repel and get as far apart as possible; a multiple bond counts as one region.
2 pairs linear 180°; 3 trigonal planar 120°; 4 tetrahedral 109.5°; 5 trigonal bipyramidal 90° and 120°; 6 octahedral 90°.
Lone pairs repel more (LP–LP > LP–BP > BP–BP): $\mathrm{NH_3}$ is trigonal pyramidal (107°), $\mathrm{H_2O}$ bent (104.5°).
Name the shape from the atoms only; count the charge of an ion when you count electrons.

Sources and credits

Image: Molymods.jpg by Sonia, via Wikimedia Commons, licensed under CC BY 3.0.
Bond angles are standard experimental values.
Video: VSEPR Theory and Molecular Geometry, Professor Dave Explains, embedded from YouTube.
The shape drawings, table, worked examples and Practice items were written and drawn by Ganzaa.

1Why is water bent?

2The rules

Why is water bent?

The rules