Shapes of molecules

1. Overview

Molecules have definite three-dimensional shapes, and a molecule's shape helps to decide its polarity, its intermolecular forces and how it reacts. This section uses VSEPR theory (valence shell electron pair repulsion) to explain the shapes and bond angles of seven key molecules, from linear CO₂ to octahedral SF₆. The same method predicts the shape of any similar molecule or ion: count the bonding regions and lone pairs around the central atom, then let them spread out as far as possible.

Key Definitions

• VSEPR theory: valence shell electron pair repulsion theory, the model in which the electron pairs in the outer shell of a central atom repel one another and take up positions as far apart as possible, so that the repulsion is at a minimum; this fixes the shape of the molecule and its bond angles.
• Bonding pair: a pair of electrons shared between two atoms in a covalent bond, including a dative (coordinate) bond.
• Lone pair: a pair of electrons in the outer shell of an atom that is not used in bonding.
• Bonding region: a covalent bond between the central atom and one other atom; in VSEPR theory a single, double or triple bond each counts as one region.
• Bond angle: the angle between two bonds that meet at the same atom.
• Shape of a molecule: the three-dimensional arrangement of the atoms around the central atom; lone pairs affect the shape but are not included in its name.
• Order of repulsion: lone pair-lone pair repulsion is greater than lone pair-bonding pair repulsion, which is greater than bonding pair-bonding pair repulsion.
• Expanded octet: more than eight electrons in the outer shell of a central atom, which is possible for elements in Period 3 and later periods, as in PF₅ (10 electrons) and SF₆ (12 electrons).

Content

The VSEPR model

Every shape on this page is fixed by the central atom, the atom that the others are bonded to, and by the electron pairs in its outer shell.

• A bonding pair is shared between the central atom and another atom.
• A lone pair belongs to the central atom alone and is not used in bonding.

All these pairs are regions of negative charge, so they repel one another. They take up positions as far apart as possible, which keeps the repulsion to a minimum. The positions of the pairs fix the positions of the atoms, and so the bond angles.

Two rules complete the model.

1. A double or triple bond counts as one bonding region. All its electrons lie between the same two nuclei, so they point in one direction. Carbon in CO₂ has two regions around it, not four.
2. Lone pairs repel more strongly than bonding pairs. A bonding pair is pulled out between two nuclei. A lone pair is attracted by only one nucleus, so it stays closer to the central atom and spreads over a wider angle. The order of repulsion is:

$$\text{lone pair-lone pair} > \text{lone pair-bonding pair} > \text{bonding pair-bonding pair}$$

The shape of a molecule describes where its atoms are. Lone pairs change the shape, but they are not part of its name.

Shapes with no lone pairs on the central atom

When every pair around the central atom is a bonding pair, the number of bonding regions alone fixes the shape.

Bonding regions	Shape	Bond angle(s)	Example
2	linear	180°	CO₂
3	trigonal planar	120°	BF₃
4	tetrahedral	109.5°	CH₄
5	trigonal bipyramidal	120° and 90°	PF₅
6	octahedral	90°	SF₆

• CO₂. Carbon uses all four outer electrons in two C=O double bonds: two bonding regions and no lone pair. Two regions are furthest apart on opposite sides of the carbon atom, at 180°.
• BF₃. Boron has three outer electrons and forms three B-F bonds: three bonding pairs, no lone pair. They spread out in a flat triangle, 120° apart, so all four atoms lie in one plane: the molecule is planar. Boron has only six electrons in its outer shell.
• CH₄. Carbon has four bonding pairs. In a flat square they would be only 90° apart. In three dimensions they get further apart, pointing to the corners of a tetrahedron, 109.5° apart.
• PF₅. Phosphorus uses all five outer electrons to form five P-F bonds. Three F atoms form a triangle around the phosphorus atom (the equatorial positions), 120° apart. The other two sit above and below the triangle (the axial positions), at 90° to each equatorial bond and at 180° to each other.
• SF₆. Sulfur uses all six outer electrons to form six S-F bonds. Four F atoms form a square around the sulfur atom, and the other two sit above and below it. Every bond is at 90° to its four neighbours and at 180° to the bond opposite it.

PF₅ (10 outer electrons on P) and SF₆ (12 on S) have an expanded octet. Elements from Period 3 onwards can do this. A Period 2 atom holds at most eight outer electrons, so nitrogen forms NF₃ but no NF₅, while phosphorus forms both PF₃ and PF₅.

Shapes with lone pairs on the central atom

• NH₃. Nitrogen has five outer electrons: three form N-H bonds and two are a lone pair. The four pairs point roughly to the corners of a tetrahedron, but only three hold atoms, so the shape is pyramidal (also called trigonal pyramidal). The lone pair repels the bonding pairs more strongly than they repel each other, so it pushes the N-H bonds closer together: 107°, not 109.5°.
• H₂O. Oxygen has six outer electrons: two form O-H bonds and four are two lone pairs. Again the four pairs point roughly to the corners of a tetrahedron, but only two hold atoms, so the shape is non-linear (also called bent or V-shaped). Two lone pairs squeeze the bonds closer still: 104.5°.

Bonding pairs	Lone pairs	Shape	Bond angle	Example
4	0	tetrahedral	109.5°	CH₄
3	1	pyramidal	107°	NH₃
2	2	non-linear	104.5°	H₂O

All three have four pairs around the central atom. In these three molecules each lone pair closes the bond angle by about 2.5°, a handy way to remember the values.

Three regions with one lone pair. In SO₂, sulfur forms two S=O double bonds and keeps one lone pair. The three regions lie in a flat triangle, with atoms in only two positions, so SO₂ is non-linear. The extra repulsion from the lone pair makes the O-S-O angle a little less than 120°: write a value such as 119°, not 120°.

Dative bonds. A dative (coordinate) bond counts as a bonding pair like any other. When NH₃ gains an H⁺ ion, its lone pair forms a dative bond to the H⁺. NH₄⁺ has four identical N-H bonds and no lone pair, so it is tetrahedral and the angle opens from 107° to 109.5°. In Al₂Cl₆ the dative bond points the other way: each aluminium atom accepts a lone pair from a chlorine atom of the other AlCl₃ unit. With its three ordinary covalent bonds, that gives aluminium four bonding pairs, arranged roughly tetrahedrally, whereas a single AlCl₃ molecule is trigonal planar, like BF₃.

Predicting the shape of any molecule or ion

1. Find the central atom and its number of outer electrons: 3 for B and Al, 4 for C and Si, 5 for N and P, 6 for O and S, 7 for the halogens.
2. Adjust for a charge: add one electron for each negative charge and take away one for each positive charge.
3. Take away the electrons the central atom uses in bonds: one for each single bond, two for each double bond and three for each triple bond. An outer H or halogen atom forms one single bond.
4. Halve what is left to get the number of lone pairs.
5. Add the lone pairs to the number of bonded atoms to get the number of regions. The regions fix the arrangement, the bonded atoms fix the shape, and the lone pairs adjust the angle.

For example, in H₃O⁺ oxygen has 6 − 1 = 5 electrons. Three O-H bonds use 3, leaving 2: one lone pair. Three bonded atoms and one lone pair make H₃O⁺ pyramidal, 107°, like NH₃.

Step 2 works when every outer atom is H or a halogen. In other ions the extra electrons usually belong to outer atoms. In an oxoanion, an ion in which oxygen atoms surround the central atom, they belong to outer oxygen atoms. For these ions, draw the dot-and-cross diagram and count the lone pairs on the central atom from it instead. In the carbonate ion, CO₃²⁻, carbon forms one C=O bond and two C-O⁻ bonds and has no lone pair, so the ion is trigonal planar, 120°.

Worked example 1: explaining the shape of OF₂

Question: In oxygen difluoride, OF₂, the oxygen atom is joined to each fluorine atom by a single bond. What shape is the molecule, and what is the F-O-F bond angle? Explain your answer using VSEPR theory.

Step 1: count the pairs. Oxygen has 6 outer electrons and uses 2 in the O-F bonds. The 4 left over are 2 lone pairs, so oxygen has 2 bonding pairs and 2 lone pairs.

Step 2: shape. Four pairs repel towards the corners of a tetrahedron, as in water. Only two positions hold atoms, so the molecule is non-linear.

Step 3: angle. The two lone pairs repel more strongly than the bonding pairs, so they push the O-F bonds closer together than 109.5°.

Answer: OF₂ is non-linear, with an F-O-F angle of about 104.5°. A full explanation: "The oxygen atom has two bonding pairs and two lone pairs. The four electron pairs repel and take up positions as far apart as possible. Lone pairs repel more strongly than bonding pairs, so the bond angle is reduced from 109.5° to about 104.5°."

Worked example 2: two ions and a molecule with double bonds

Question: Predict the shape of, and the bond angle in, (a) the PH₄⁺ ion, (b) the PF₆⁻ ion and (c) a molecule of SO₃, in which sulfur forms three S=O double bonds.

Step	PH₄⁺	PF₆⁻	SO₃
Outer electrons of the central atom	P: 5	P: 5	S: 6
Adjust for the charge	5 − 1 = 4	5 + 1 = 6	no charge: 6
Electrons used in bonds	4 single bonds: 4	6 single bonds: 6	3 double bonds: 6
Left over, and lone pairs	0, so 0	0, so 0	0, so 0
Regions (bonded atoms + lone pairs)	4	6	3

In PF₆⁻, the negative charge gives phosphorus the sixth electron it needs for a sixth bond. Without it, phosphorus would have five electrons for six bonds, which is impossible.

Answer: (a) PH₄⁺ is tetrahedral, 109.5°, like NH₄⁺. (b) PF₆⁻ is octahedral, 90°, like SF₆. (c) SO₃ is trigonal planar, 120°, like BF₃: its three double bonds are three regions, not six.

Worked example 3: bond angles around several atoms

Question: Cyanoethanoic acid, N≡C-CH₂-COOH, contains a C≡N triple bond, a C=O bond and an O-H group. Predict (a) the C-C≡N bond angle at the nitrile carbon, (b) the C-C-C bond angle at the CH₂ carbon, (c) the O=C-O bond angle in the COOH group and (d) the C-O-H bond angle. Then put the four angles in order from smallest to largest.

Treat each atom in turn as a central atom.

Atom	Bonding regions	Lone pairs	Shape around the atom	Predicted angle
C of C≡N	2 (C≡N, C-C)	0	linear	(a) 180°
C of CH₂	4 single bonds	0	tetrahedral	(b) 109.5°
C of COOH	3 (C=O, C-O, C-C)	0	trigonal planar	(c) 120°
O of OH	2 (O-C, O-H)	2	non-linear	(d) 104.5°

The nitrile carbon uses three outer electrons in C≡N and one in C-C, so it has no lone pair, and the triple bond counts once: two regions. The carboxyl carbon uses two in C=O and one each in C-O and C-C: three regions, no lone pair.

Answer: (a) 180°, (b) 109.5°, (c) 120°, (d) 104.5°. From smallest to largest: d, b, c, a.

Which angle to write in an exam

VSEPR theory gives the correct shape for every species on this page. When the central atom has lone pairs, though, the angle is only an estimate. Measured angles can differ by a few degrees, and by more when the central atom is from Period 3 onwards. VSEPR cannot predict these differences, so unless a question gives you data, write the angle of the matching example:

• three bonding pairs and one lone pair: 107°, as in NH₃;
• two bonding pairs and two lone pairs: 104.5°, as in H₂O;
• two bonding regions and one lone pair: a value just below 120°, such as 119°, as in SO₂.

Drawing shapes in three dimensions

• A plain line is a bond in the plane of the paper.
• A solid wedge is a bond pointing towards you, wide end at the outer atom.
• A hashed wedge (dashed) is a bond pointing away from you, behind the paper.

Shape	How to draw it
linear, trigonal planar	plain lines only, 180° or 120° apart
tetrahedral	two plain lines forming a V, plus one solid and one hashed wedge; label 109.5° between the plain bonds
pyramidal, non-linear	as tetrahedral, but with a lone pair (two dots) in place of one of the plain bonds (pyramidal) or two lone pairs in place of the two wedges (non-linear)
trigonal bipyramidal	axial bonds as plain lines up and down; equatorial bonds as one plain line, one solid and one hashed wedge
octahedral	axial bonds as plain lines up and down; the square as two solid and two hashed wedges

Draw each angle as an arc between two bonds that meet at the central atom, not between a bond and an atom.

Key Equations

Lone pairs on the central atom ($V$ = outer electrons of the central atom; $z$ = charge on the ion with its sign, 0 for a molecule; use the formula for an ion only when every outer atom is H or a halogen, and read other ions from a dot-and-cross diagram; $b$ = electrons the central atom uses in bonds: 1 per single, 2 per double and 3 per triple bond; all are counts, with no unit): $$\text{lone pairs} = \frac{V - z - b}{2}$$

For PF₆⁻: $V = 5$, $z = -1$ and $b = 6$, so the number of lone pairs is $(5 + 1 - 6)/2 = 0$.

Number of regions (a double or triple bond counts as one region; a count, no unit): $$\text{regions} = \text{bonded atoms} + \text{lone pairs}$$

Rule of thumb for four regions only (approximate): $$\text{bond angle} \approx 109.5^\circ - 2.5^\circ \times (\text{number of lone pairs})$$

This gives 109.5° for CH₄, 107° for NH₃ and 104.5° for H₂O.

Common Mistakes to Avoid

• Mistake: counting only the bonding pairs. Fix: count the lone pairs on the central atom too. OF₂ has two bonding pairs and two lone pairs, so it is non-linear (about 104.5°), not linear (180°).
• Mistake: naming the arrangement of the electron pairs as the shape. Fix: name the shape from the atoms only. The four pairs in NH₃ point towards the corners of a tetrahedron, but NH₃ is pyramidal; H₂O is non-linear, not tetrahedral.
• Mistake: incomplete or wrong shape names. Fix: write "trigonal bipyramidal" in full for PF₅, not "trigonal pyramidal" (the shape of NH₃) or "bipyramidal". Write "octahedral" for SF₆, not "hexagonal" or "octagonal".
• Mistake: giving only one angle for SF₆ or PF₅. Fix: in SF₆, neighbouring bonds are at 90° and opposite bonds at 180°. PF₅ has 120° (equatorial to equatorial), 90° (axial to equatorial) and 180° (axial to axial).
• Mistake: counting a double bond as two regions. Fix: CO₂ has two regions, so it is linear (180°), not tetrahedral; SO₃ has three, so it is trigonal planar (120°).
• Mistake: handling an ion's charge wrongly. Fix: for PF₆⁻, phosphorus has 5 + 1 = 6 electrons for six bonds, so the ion is octahedral. For an oxoanion such as CO₃²⁻, adding the charge to carbon would wrongly give it a lone pair and a pyramidal shape; the dot-and-cross diagram shows no lone pair, so it is trigonal planar.
• Mistake: reading the shape from a flat drawing, or writing "the atoms repel". Fix: reason from the electron pairs. CH₄ drawn flat seems to have 90° angles, but its four bonding pairs repel in three dimensions to 109.5°. Write "the electron pairs repel" and "lone pairs repel more strongly than bonding pairs".
• Mistake: labelling an angle between a bond and an atom, or a 3D drawing that could not have the stated shape. Fix: draw the arc between two bonds at the central atom. For CH₄, use two plain bonds, one solid wedge and one hashed wedge, and label 109.5° between the plain bonds.

Exam Tips

• State a shape by its full name and give the angle with a degree sign, for example "pyramidal, 107°".
• Explain a shape in three linked points: the numbers of bonding pairs and lone pairs around the central atom; the pairs repel and take up positions as far apart as possible; lone pairs repel more strongly than bonding pairs, so the angle is smaller than with no lone pairs.
• Predict an unfamiliar shape by showing your count of outer electrons, bonds and lone pairs, then giving the shape and angle of the example with the same pattern of pairs.
• Compare two bond angles by stating the numbers of bonding pairs and lone pairs on each central atom: NH₃ (three bonding pairs, one lone pair, 107°) has a larger angle than H₂O (two bonding pairs, two lone pairs, 104.5°).
• Draw a 3D shape with plain lines and wedges, show any lone pairs on the central atom, and label at least one bond angle. If the molecule has several central atoms, give the angle at each one separately, as in Worked example 3.