VSEPR theory and the shapes of molecules

What is VSEPR theory?

In plain words: electron pairs are all negatively charged, so they push each other as far apart as they can round the central atom, like balloons tied together at their knots. Count the pairs and you know the arrangement. Then remember that only the atoms are “seen” in the shape, and that a lone pair, held by one nucleus only, spreads out more and pushes harder than a bond pair.

The postulates

Lewis structures cannot explain molecular shapes. Sidgwick and Powell proposed VSEPR theory in 1940, and Nyholm and Gillespie developed and refined it in 1957 (NCERT Section 4.4, pages 112–113). Its main postulates (page 113):

• The shape of a molecule depends on the number of valence-shell electron pairs (bonded or non-bonded) round the central atom.
• Electron pairs repel one another, since their clouds are negatively charged.
• They take up positions that minimise repulsion and so maximise the distance between them.
• The valence shell is treated as a sphere, with the pairs on its surface as far from one another as possible.
• A multiple bond is treated as one electron pair, a “super pair”.
• Where a molecule has resonance structures, VSEPR can be applied to any one of them.

The repulsions decrease in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Nyholm and Gillespie explained why:

While the lone pairs are localised on the central atom, each bonded pair is shared between two atoms.

NCERT Chemistry Part I (Class 11), Section 4.4, page 113

So a lone pair occupies more space round the central atom and repels more strongly (page 113).

Molecules with no lone pair on the central atom

For isolated molecules with no central lone pairs, the shape follows the electron-domain arrangement (NCERT Table 4.6, page 114; ball-and-stick models in NCERT Fig. 4.6, page 113):

VSEPR predicts the shapes of a great many molecules accurately, especially compounds of the p-block elements, even where the energy difference between possible structures is very small. NCERT adds that its theoretical basis is still a matter of discussion (page 113).

Molecules with lone pairs

With lone pairs, the arrangement of electron pairs still follows the table above, but the shape describes only the positions of the atoms. Writing A for the central atom, B for a bonded atom and E for a lone pair (NCERT Table 4.7, page 115; the AB₂E₃ row is beyond the NCERT table):

NCERT’s Table 4.7 labels the water row “AB₃E₂”, though it lists two bonding pairs; water is AB₂E₂, as Table 4.8 has it. Table 4.8 in turn gives SO₂ (AB₂E) “4” bonding pairs; it has 2 (counting each double bond as one super pair).

How lone pairs distort the angles (NCERT Table 4.8, pages 116–117):

• SO₂ (AB₂E): three pairs would give 120°, but lp–bp repulsion is greater than bp–bp, so the O–S–O angle closes slightly to 119.5°; the molecule is bent.
• NH₃ (AB₃E): four pairs would give a tetrahedron, but the lone pair pushes the N–H bonds together, to 107°.
• H₂O (AB₂E₂): two lone pairs push harder still (lp–lp > lp–bp > bp–bp), to 104.5°.
• SF₄ (AB₄E): in a trigonal bipyramid the lone pair goes equatorial: there it has two lp–bp repulsions at 90°, against three if it were axial. The shape is a see-saw (a distorted tetrahedron).
• ClF₃ (AB₃E₂): both lone pairs go equatorial. They are then 120° apart, with no lp–lp repulsion at 90° and four lp–bp repulsions at 90° (one lone pair axial and one equatorial would give an lp–lp repulsion at 90°; both axial would give six lp–bp repulsions at 90°). The shape is a T.
• XeF₂ (AB₂E₃): ½(8 + 2) = 5 pairs; the three lone pairs take all three equatorial positions, so F–Xe–F is 180°: linear.

NCERT’s reason for the T shape of ClF₃ (Table 4.8, page 117), that the equatorial structure has fewer lp–bp repulsions than the others, is not right as stated: it has four lp–bp repulsions at 90°, while the structure with one lone pair axial has only three. What makes it the most stable is that it avoids the strongest repulsion, lp–lp at 90°.

Using VSEPR step by step

1. Draw the Lewis structure, or count the electron pairs on the central atom.
2. Count bond pairs (a multiple bond counts once) and lone pairs.
3. Their total fixes the arrangement (2 linear, 3 trigonal planar, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral).
4. Name the shape from the atoms only; lone pairs often compress some angles, but symmetry and other factors also matter.

A common shortcut for the number of pairs round the central atom (beyond the NCERT text):

where V is the number of valence electrons of the central atom, M the number of monovalent atoms bonded to it (H, halogens), c the charge of a cation and a the charge of an anion. This shortcut works for the simple hydrides, halides and oxides illustrated here; it is not universal. Confirm the count from a complete Lewis structure, especially for charged, delocalised or unusual species. Each bonded atom gives one bonding domain, and each lone pair one additional domain. For NO₂ the unpaired electron is treated as a weaker extra domain, not literally an electron pair.

Solved examples

Common mistakes to avoid

Quick revision summary

• VSEPR (Sidgwick and Powell 1940; Nyholm and Gillespie 1957): electron pairs round the central atom repel and keep as far apart as possible.
• Repulsion: lp–lp > lp–bp > bp–bp; a multiple bond counts as one super pair.
• Pairs 2, 3, 4, 5, 6: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral.
• With lone pairs: AB₂E bent (SO₂), AB₃E pyramidal (NH₃, 107°), AB₂E₂ bent (H₂O, 104.5°), AB₄E see-saw (SF₄), AB₃E₂ T-shape (ClF₃), AB₅E square pyramidal (BrF₅), AB₄E₂ square planar (XeF₄), AB₂E₃ linear (XeF₂).
• In a trigonal bipyramid, lone pairs take equatorial positions.

Questions students ask

What is the full form of VSEPR?

VSEPR stands for valence shell electron pair repulsion, and the name, usually pronounced ‘vesper’, sums up the theory: the electron pairs in the valence shell of the central atom repel one another, so the molecule takes the shape that keeps them as far apart as possible. It was proposed by Sidgwick and Powell in 1940 and developed by Nyholm and Gillespie in 1957.

What is the difference between VSEPR theory and hybridisation?

VSEPR theory predicts the shape of a molecule by counting the electron pairs round the central atom and letting them repel, while hybridisation, part of valence bond theory, explains that shape through the orbitals the atom uses to form its bonds. In the traditional school mapping, two to six electron pairs correspond to sp, sp², sp³, sp³d and sp³d², so four pairs mean a tetrahedral arrangement in VSEPR and sp³ orbitals in valence bond theory. The d-containing labels for hypervalent main-group molecules are textbook models, not evidence that d orbitals must participate substantially. Use VSEPR to find the shape and the steric number to name the hybridisation.

Why is the bond angle in NH₃ larger than in NF₃?

The H–N–H angle in NH₃ (107°) is larger than the F–N–F angle in NF₃ (about 102°) because fluorine, being far more electronegative than nitrogen, pulls the bonding pairs of NF₃ away from the nitrogen atom. With the bond pairs farther from the central atom and from one another, bond pair–bond pair repulsion is weaker, and the lone pair on nitrogen squeezes the bonds closer together. In NH₃ nitrogen is the more electronegative atom, so the bond pairs stay close to it, repel more strongly and keep the angle nearer the tetrahedral 109.5°.

What are the limitations of VSEPR theory?

VSEPR theory predicts the shapes of most p-block molecules well, but it has limits: it does not explain how bonds form or how strong they are, it tells only whether bond angles open or close, not their exact values, and it works poorly for transition-metal compounds, whose shapes depend on their d electrons. Some species also break its rules: the [TeCl₆]²⁻ ion has seven electron pairs round tellurium, one of them a lone pair, yet it is nearly regular octahedral in many salts; the precise distortion depends on the crystal environment. NCERT itself notes that the theoretical basis of VSEPR is still a matter of discussion.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Predict the shape of SF₆ and of PCl₅, and give their bond angles.

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   SF₆: ½(6 + 6) = 6 bond pairs: octahedral, 90° between neighbouring bonds (180° between opposite bonds). PCl₅: ½(5 + 5) = 5 bond pairs: trigonal bipyramidal, 120° in the equatorial plane, 90° between axial and equatorial bonds and 180° between the two axial bonds; the axial bonds are slightly longer than the equatorial ones (NCERT Section 4.6.3, page 125).

2. Q2. Why is ClF₃ T-shaped and not trigonal planar?

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   Cl has ½(7 + 3) = 5 pairs: 3 bond pairs and 2 lone pairs, a trigonal bipyramidal arrangement. The two lone pairs go equatorial, an arrangement with no lp–lp repulsion at 90° and fewer 90° lp–bp contacts than the both-axial option (one axial and one equatorial would put them 90° apart; both axial would give six lp–bp repulsions at 90° and a trigonal planar shape). This leaves the three F atoms in a T.

3. Q3. Arrange CH₄, NH₃ and H₂O in order of decreasing bond angle and explain.

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   CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°). All have four pairs; the number of lone pairs rises from 0 to 1 to 2, and lone pairs repel more strongly than bond pairs, so the bond angle shrinks.