Shapes of Molecules

Why is a water molecule bent while carbon dioxide is straight? Both have a central atom bonded to two others, but water's oxygen also carries two lone pairs that push the bonds together. This note teaches valence shell electron pair repulsion (VSEPR) theory, which predicts the shape of any simple molecule or ion from a count of electron pairs. You must know seven named shapes with their bond angles, and be able to predict the shapes of unfamiliar molecules and ions by analogy: a reliable two- or three-mark question in Paper 1 and Paper 2, and the starting point for deciding whether a molecule is polar.

The idea behind VSEPR

The electrons in the outer shell of the central atom are arranged in pairs: bonding pairs and lone pairs. Every pair is a region of negative charge, and like charges repel. So the pairs move as far apart from each other as possible, and the bonded atoms end up wherever their bonding pairs point.

Why are lone pairs more repulsive? A bonding pair is pulled out between two nuclei, so its charge is spread along the bond. A lone pair is held by only one nucleus, so it sits closer to the central atom and is more concentrated, and it spreads out over a wider angle. It therefore pushes the bonding pairs closer together. Each lone pair reduces the angle between the remaining bonds by roughly $2.5^\circ$.

The shape of a molecule describes the positions of the atoms only. Lone pairs influence the shape, but they are not part of its name. That is why ammonia, with four pairs arranged tetrahedrally, is called pyramidal (the atoms form a pyramid), not tetrahedral.

The seven shapes you must know

Look at the methane, ammonia and water series. All three have four pairs of electrons around the central atom, arranged roughly tetrahedrally. The angles shrink as bonding pairs are replaced by more repulsive lone pairs:

molecule	bonding pairs	lone pairs	angle
$\ce{CH4}$	4	0	$109.5^\circ$
$\ce{NH3}$	3	1	$107^\circ$
$\ce{H2O}$	2	2	$104.5^\circ$

In the trigonal bipyramid of $\ce{PF5}$, the five bonds are not all equivalent. Three equatorial bonds lie in a plane at $120^\circ$ to each other, and two axial bonds point straight up and down, at $90^\circ$ to the plane. In the octahedron of $\ce{SF6}$, all six bonds are equivalent and every adjacent pair is at $90^\circ$.

Predicting the shape of any molecule or ion

You will often be asked about a molecule or ion you have not met, "analogous" to the seven above. The method is always the same: count electron pairs around the central atom.

Why does oxygen add nothing? A single-bonded H or halogen makes one bonding pair from one of the central atom's electrons plus one of its own, so we add its one electron. A double-bonded O makes one bonding region from two of the central atom's electrons, which already count as one pair when you halve the total. Either way, halving the total gives the number of regions.

Here are common molecules and ions predicted this way.

species	electron count	regions	bonded atoms	lone pairs	shape	angle
$\ce{NH4+}$	$5 + 4 - 1 = 8$	4	4	0	tetrahedral	$109.5^\circ$
$\ce{H3O+}$	$6 + 3 - 1 = 8$	4	3	1	pyramidal	$107^\circ$
$\ce{NH2-}$	$5 + 2 + 1 = 8$	4	2	2	non-linear	$104.5^\circ$
$\ce{SiCl4}$	$4 + 4 = 8$	4	4	0	tetrahedral	$109.5^\circ$
$\ce{BCl3}$	$3 + 3 = 6$	3	3	0	trigonal planar	$120^\circ$
$\ce{SO3}$	$6 + 0 = 6$	3	3	0	trigonal planar	$120^\circ$
$\ce{SO2}$	$6 + 0 = 6$	3	2	1	non-linear	about $118^\circ$ (less than $120^\circ$)
$\ce{CO3^2-}$	$4 + 0 + 2 = 6$	3	3	0	trigonal planar	$120^\circ$
$\ce{NO3-}$	$5 + 0 + 1 = 6$	3	3	0	trigonal planar	$120^\circ$
$\ce{SO4^2-}$	$6 + 0 + 2 = 8$	4	4	0	tetrahedral	$109.5^\circ$
$\ce{PCl5}$	$5 + 5 = 10$	5	5	0	trigonal bipyramidal	$120^\circ$ and $90^\circ$
$\ce{PCl6-}$	$5 + 6 + 1 = 12$	6	6	0	octahedral	$90^\circ$
$\ce{BeCl2}$ (gas)	$2 + 2 = 4$	2	2	0	linear	$180^\circ$

The shortcut only covers multiple bonds to O and S. For a molecule such as $\ce{HCN}$, reason directly: carbon forms one single bond (to H) and one triple bond (to N), uses all four of its electrons and has no lone pair: two regions, linear.

Shapes with lone pairs on five or six pairs

The syllabus names only lone-pair-free examples for five and six pairs, but analogous questions occasionally use species such as $\ce{XeF4}$, $\ce{SF4}$ or $\ce{ClF3}$. Treat these as unfamiliar applications of the same rules. The key point is where the lone pairs go: lone pairs take the positions that keep them furthest apart from each other and from bonds.

pairs	bonding	lone	shape	example
6	5	1	square pyramidal	$\ce{BrF5}$
6	4	2	square planar ($90^\circ$; the two lone pairs are opposite each other)	$\ce{XeF4}$
5	2	3	linear ($180^\circ$; the three lone pairs are equatorial)	$\ce{XeF2}$, $\ce{I3-}$

Shape and polarity

A molecule's shape decides whether it is polar overall. Each polar bond has a dipole, pointing towards its more electronegative atom. If the molecule is symmetrical, the dipoles cancel and the molecule has no overall dipole moment; if it is unsymmetrical, they do not cancel and the molecule is polar.

molecule	polar bonds?	shape	overall dipole?
$\ce{CO2}$	yes (C=O)	linear, symmetrical	no
$\ce{H2O}$	yes (O–H)	non-linear	yes
$\ce{BF3}$	yes	trigonal planar, symmetrical	no
$\ce{NH3}$	yes	pyramidal	yes
$\ce{CCl4}$	yes	tetrahedral, symmetrical	no
$\ce{CHCl3}$	yes	tetrahedral but unsymmetrical (different atoms)	yes

You will use this in the next note to decide which intermolecular forces a substance has.

Worked examples

Practice

Answers

1. Electron pairs (bonding and lone) in the outer shell of the central atom repel each other and arrange themselves as far apart as possible to minimise repulsion. Lone pairs repel more strongly than bonding pairs (lone–lone > lone–bond > bond–bond). Multiple bonds count as a single region.
2. (a) Tetrahedral, $109.5^\circ$ (4 bonding pairs). (b) Trigonal planar, $120^\circ$ (3 bonding pairs). (c) Non-linear, about $104.5^\circ$ by analogy with water (2 bonding pairs, 2 lone pairs). (d) Octahedral, $90^\circ$.
3. In $\ce{BF3}$, boron has three outer electrons, all used in bonds: three bonding pairs and no lone pairs, which are as far apart as possible in a plane at $120^\circ$. In $\ce{NF3}$, nitrogen has five outer electrons: three bonding pairs and one lone pair. Four pairs are arranged tetrahedrally and the lone pair pushes the bonds closer, giving a pyramidal shape (about $107^\circ$).
4. Nitrogen: $5 + 2 + 1 = 8$ electrons, so 4 pairs: 2 bonding pairs and 2 lone pairs. Pairs repel to be as far apart as possible; lone pairs repel more strongly, so the shape is non-linear with an angle of about $104.5^\circ$, like water.
5. Three F atoms in a triangular plane around P (equatorial, $120^\circ$ apart) and two F atoms directly above and below (axial). Axial-to-equatorial angle $90^\circ$; equatorial-to-equatorial angle $120^\circ$.
6. $\ce{CO3^2-}$: $4 + 0 + 2 = 6$, 3 regions, no lone pairs: trigonal planar, $120^\circ$. $\ce{SO4^2-}$: $6 + 0 + 2 = 8$, 4 regions, no lone pairs: tetrahedral, $109.5^\circ$.
7. Both have polar bonds (O is more electronegative than C or S). $\ce{CO2}$ is linear (two regions, no lone pairs on C), so the two bond dipoles are equal and opposite and cancel. $\ce{SO2}$ has a lone pair on sulfur, so it is non-linear; the two dipoles do not cancel, giving an overall dipole.
8. In $\ce{NH3}$: 3 bonding pairs and 1 lone pair, pyramidal, $107^\circ$. In the product, the lone pair on N has become a bonding pair (the coordinate bond), so N has 4 bonding pairs: tetrahedral, about $109.5^\circ$. In $\ce{BF3}$: 3 bonding pairs, trigonal planar, $120^\circ$. In the product B has 4 bonding pairs: tetrahedral, about $109.5^\circ$. Both angles move to the tetrahedral value.
9. Iodine: $7 + 4 + 1 = 12$ electrons, 6 pairs: 4 bonding and 2 lone pairs. The six pairs point to the corners of an octahedron; the lone pairs repel most strongly so they are opposite each other. The four Cl atoms lie in a plane: square planar, $90^\circ$.
10. Central iodine: $7 + 2$ (two bonded I atoms, one electron each) $+ 1$ (charge) $= 10$ electrons, 5 pairs: 2 bonding and 3 lone pairs. Five pairs adopt a trigonal bipyramidal arrangement. The lone pairs occupy the three equatorial positions, where they are $120^\circ$ apart from each other; in the axial positions they would be at $90^\circ$ to three other pairs, giving more repulsion. The two bonded atoms are therefore axial, at $180^\circ$: a linear ion.