VSEPR Theory — Predicting Molecular Shapes
VSEPR — Valence Shell Electron Pair Repulsion — answers a question a Lewis structure alone cannot: what does the molecule actually look like in three dimensions? The whole theory rests on one sentence. Electron pairs in the valence shell of the central atom repel one another and arrange themselves as far apart as possible. Everything else, including every shape name you have to memorise, follows from counting.
Step 1 — count electron domains
An electron domain (also called a region of electron density, or a steric number) is anything occupying space around the central atom:
- each single bond = 1 domain
- each double bond = 1 domain (not two)
- each triple bond = 1 domain
- each lone pair on the central atom = 1 domain
For a central atom bonded only to monovalent atoms (H, F, Cl, Br, I), there is a quick formula for the steric number:
where V = valence electrons of the central atom, M = number of monovalent atoms attached, c = positive charge on the ion, a = negative charge on the ion. For oxo-species (SO₃²⁻, ClO₃⁻) draw the Lewis structure instead and count directly.
Step 2 — read the shape off the table
The domains fix the electron geometry. The molecular shape is what you see when you look only at the atoms, ignoring the lone pairs — the lone pairs are still there, still pushing, but they are invisible in the shape name.
| Domains (SN) | Electron geometry | Lone pairs | Molecular shape | Approx. bond angle | Example |
|---|---|---|---|---|---|
| 2 | Linear | 0 | Linear | 180° | BeCl₂, CO₂ |
| 3 | Trigonal planar | 0 | Trigonal planar | 120° | BF₃, SO₃ |
| 3 | Trigonal planar | 1 | Bent (V-shaped) | just under 120° (≈119° in SO₂) | SO₂, SnCl₂ |
| 4 | Tetrahedral | 0 | Tetrahedral | 109.5° | CH₄, NH₄⁺, SO₄²⁻ |
| 4 | Tetrahedral | 1 | Trigonal pyramidal | ≈107° | NH₃, PCl₃, SO₃²⁻ |
| 4 | Tetrahedral | 2 | Bent (V-shaped) | ≈104.5° | H₂O, OF₂ |
| 5 | Trigonal bipyramidal | 0 | Trigonal bipyramidal | 120° equatorial, 90° axial–equatorial | PCl₅, PF₅ |
| 5 | Trigonal bipyramidal | 1 | See-saw | ≈102° equatorial, ≈173° axial–axial | SF₄ |
| 5 | Trigonal bipyramidal | 2 | T-shaped | just under 90° (≈87.5°) | ClF₃, BrF₃ |
| 5 | Trigonal bipyramidal | 3 | Linear | 180° | XeF₂, I₃⁻ |
| 6 | Octahedral | 0 | Octahedral | 90° | SF₆, PF₆⁻ |
| 6 | Octahedral | 1 | Square pyramidal | just under 90° (≈85°) | BrF₅, IF₅ |
| 6 | Octahedral | 2 | Square planar | 90° | XeF₄, ICl₄⁻ |
| 7 | Pentagonal bipyramidal | 0 | Pentagonal bipyramidal | 72° equatorial, 90° axial–equatorial | IF₇ |
Experimental bond angles vary a little between sources and with the substituents, so treat every "≈" value as the accepted approximate figure rather than an exact constant.
Step 3 — apply the repulsion order
Not all domains push equally. A lone pair is held by only one nucleus, so it spreads out wider and takes up more angular room than a bonding pair, which is pulled tight between two nuclei. The order is:
This single rule explains the most-asked comparison in the chapter — all three molecules below have four domains, so all three start from a tetrahedron, and the angle shrinks as lone pairs are added:
CH₄ — 4 bonding, 0 lone pairs → 109.5°
NH₃ — 3 bonding, 1 lone pair → ≈107°
H₂O — 2 bonding, 2 lone pairs → ≈104.5°
Each lone pair squeezes the bond angle by roughly 2–3°.
Worked example 1 — SF₄, and why the lone pair goes equatorial
SN = ½[6 + 4] = 5 → trigonal bipyramidal electron geometry, 4 bonding pairs and 1 lone pair → see-saw.
A trigonal bipyramid has two distinct positions. If the lone pair sat in an axial site it would face three neighbours at 90°. In an equatorial site it faces only two at 90° (the other two are at 120°, which is far less severe). Fewer close contacts means less repulsion, so lone pairs always take equatorial positions in a trigonal bipyramid. The same reasoning makes ClF₃ T-shaped (2 equatorial lone pairs) and XeF₂ linear (3 equatorial lone pairs).
Worked example 2 — XeF₄ and ICl₄⁻
XeF₄: SN = ½[8 + 4] = 6 → octahedral, 4 bonding, 2 lone pairs. In an octahedron every position is equivalent, so the two lone pairs simply get as far apart as possible — 180°, i.e. trans, one above and one below. The four F atoms are left in a plane: square planar, angles 90°.
ICl₄⁻: SN = ½[7 + 4 + 1] = 6 → the same analysis, also square planar. The extra electron from the negative charge is what makes an iodine compound match a xenon one.
Worked example 3 — a case where you must draw the Lewis structure
SO₃²⁻. Total valence electrons = 6 (S) + 3 × 6 (O) + 2 (charge) = 26, i.e. 13 pairs. Three S–O bonding pairs plus three lone pairs on each oxygen accounts for 3 + 9 = 12 pairs, leaving one lone pair on sulphur.
So SN = 3 bonding + 1 lone = 4 → tetrahedral electron geometry, trigonal pyramidal shape — the same shape as NH₃.
Contrast SO₃ (neutral): no lone pair on sulphur, SN = 3, so it is trigonal planar at 120°. One negative charge, and the shape changes completely.
Two refinements worth knowing
Electronegativity of the substituent. A more electronegative atom pulls the bonding pair away from the central atom, so that pair occupies less room near the centre and the angle closes slightly. NH₃ is about 107°, but NF₃ is about 102° for exactly this reason.
Multiple bonds are fatter. A double bond holds more electron density than a single bond and pushes harder. In COCl₂ the Cl–C–Cl angle is a little under 120° while the O=C–Cl angles open a little above it.
Where VSEPR stops working — state this honestly
VSEPR is a model, not a law, and examiners at higher levels test its limits:
- It gives shapes, not accurate angles. It correctly predicts H₂O and H₂S are both bent, but cannot explain why H₂S (≈92°) is so much more compressed than H₂O (≈104.5°) — that needs an argument about the reduced s–p mixing in heavier central atoms.
- Some lone pairs are stereochemically inactive. [SeCl₆]²⁻ and [SbCl₆]³⁻ have a lone pair on the central atom yet are regular octahedra, which VSEPR does not predict.
- It does not apply to transition-metal complexes, where d-electrons and crystal field effects decide the geometry instead.
Common mistakes that cost marks
- Counting a double bond as two domains. CO₂ has 2 domains and is linear, not bent.
- Giving the electron geometry when the shape was asked. NH₃ is tetrahedral in electron geometry but its shape is trigonal pyramidal. Write which one you mean.
- Counting lone pairs on the outer atoms. Only lone pairs on the central atom affect the shape.
- Putting the lone pair axial in a trigonal bipyramid. It is always equatorial — this decides see-saw vs T-shaped vs linear.
- Quoting exact angles. Say "slightly less than 109.5°" unless you know the experimental value; inventing a precise number is worse than an inequality.
- Confusing VSEPR with hybridisation. They agree for SN 2–4 (sp, sp², sp³) but they are separate arguments. VSEPR counts repulsions; hybridisation describes orbitals.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11 | Predict shape and bond angle; explain why NH₃ > H₂O in bond angle |
| JEE/NEET | Shape + hybridisation of interhalogens and noble-gas compounds; dipole moment from shape |
| IIT-JAM / CUET-PG | Shapes of oxo-anions; stereochemically inactive lone pairs |
| GATE / CSIR-NET | Point-group assignment, which begins with getting the shape right |
Every VSEPR question starts with valence electrons. The steric-number formula needs V, the valence-electron count of the central atom, and the group number that gives it. The interactive periodic table puts the group, block and electron configuration of all 118 elements one tap away, so you never guess V again.
Open the Interactive Periodic Table →Chemical bonding is the Class 11 chapter that every later chapter leans on. ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online classes across India — details at abcchemistry.in.