Molecular Geometry and VSEPR Theory

Master molecular geometry and vsepr theory with conceptual clarity, solved examples, and exam strategies.

CBSE JEE-MAIN NEET 10 min read

What Is Molecular Geometry?

When atoms bond together, they don’t arrange themselves randomly. The shape of a molecule — the three-dimensional arrangement of its atoms — determines everything from its physical properties (boiling point, polarity) to how it interacts biologically (enzyme-substrate fit, drug-receptor binding).

VSEPR (Valence Shell Electron Pair Repulsion) theory is the simplest model that lets us predict these shapes from just the Lewis structure. The core idea: electron pairs around a central atom repel each other and arrange themselves to be as far apart as possible.

The VSEPR Logic — Step by Step

Step 1: Draw the Lewis structure to count electron pairs.

Step 2: Count the steric number (total electron domains around the central atom):

Steric Number=bonding pairs+lone pairs\text{Steric Number} = \text{bonding pairs} + \text{lone pairs}

Note: A double bond or triple bond counts as one domain (one region of electron density).

Step 3: Identify the electron geometry (based on steric number, ignoring whether domains are bonded or lone pairs).

Step 4: Identify the molecular geometry (shape of only the atoms — some domains may be invisible lone pairs that still affect angles).

The key insight: Lone pairs repel more strongly than bonding pairs (because they’re held closer to the nucleus and occupy more space). This causes bonding angles to be compressed from their ideal values.

Fundamental Shapes

Steric #Electron GeometryBonding/Lone PairsMolecular ShapeBond AngleExample
2Linear2 BP, 0 LPLinear180°CO₂, BeCl₂
3Trigonal planar3 BP, 0 LPTrigonal planar120°BF₃, SO₃
3Trigonal planar2 BP, 1 LPBent/V-shaped<120°SO₂, SnCl₂
4Tetrahedral4 BP, 0 LPTetrahedral109.5°CH₄, SiF₄
4Tetrahedral3 BP, 1 LPTrigonal pyramidal107°NH₃
4Tetrahedral2 BP, 2 LPBent/V-shaped104.5°H₂O
5Trigonal bipyramidal5 BP, 0 LPTrigonal bipyramidal90°, 120°PCl₅
5Trigonal bipyramidal4 BP, 1 LPSee-sawSF₄
6Octahedral6 BP, 0 LPOctahedral90°SF₆
6Octahedral5 BP, 1 LPSquare pyramidalBrF₅
6Octahedral4 BP, 2 LPSquare planar90°XeF₄, [PtCl₄]²⁻

Understanding Lone Pair Repulsion

The bond angle in NH₃ (107°) is less than the tetrahedral angle (109.5°) because one lone pair on nitrogen repels the three bonding pairs, compressing them.

In H₂O (104.5°), two lone pairs compress the two bonding pairs even further.

The order of repulsion strength: Lone pair–Lone pair > Lone pair–Bonding pair > Bonding pair–Bonding pair

This is why bond angles decrease in the series: CH₄ (109.5°) → NH₃ (107°) → H₂O (104.5°)

To remember that lone pairs compress angles: think of lone pairs as “fat” — they take more room, squeezing the bonding pairs together. This mental model is crude but never fails.

Key Molecules in Detail

CO₂ — Linear (180°)

Carbon has 2 double bonds to oxygen → 2 electron domains → linear. Each C=O bond is polar, but because they point in exactly opposite directions, the bond dipoles cancel → CO₂ is non-polar overall despite having polar bonds.

H₂O — Bent (104.5°)

Oxygen has 2 bonding pairs (to 2 H atoms) and 2 lone pairs → steric number 4 → based on tetrahedral electron geometry. But the molecular shape (just the atoms) is bent. Both OH bonds are polar and they don’t cancel (the molecule is V-shaped, not linear) → H₂O is polar.

NH₃ — Trigonal Pyramidal (107°)

Nitrogen has 3 bonding pairs and 1 lone pair → tetrahedral electron geometry → trigonal pyramidal molecular shape. The lone pair points upward, 3 NH bonds point downward. NH₃ is polar and can act as a Lewis base (donating the lone pair).

PCl₅ — Trigonal Bipyramidal

Phosphorus has 5 bonding pairs (expanded octet, possible because P is in Period 3 and has access to 3d orbitals). The two axial positions and three equatorial positions are NOT equivalent — the three equatorial bonds are longer and slightly weaker than the two axial bonds. Lone pairs, when present in this geometry (SF₄, ClF₃), occupy equatorial positions to minimise repulsion.

XeF₄ — Square Planar

Xenon has 4 bonding pairs and 2 lone pairs → steric number 6 → octahedral electron geometry. The two lone pairs occupy opposite axial positions (to maximise their separation) → the 4 F atoms are in the equatorial plane → square planar molecule.

XeF₄ being square planar (not tetrahedral) is a JEE-favourite trap. Students assume 4 bonded pairs = tetrahedral, forgetting the 2 lone pairs. Always count lone pairs first.

Hybridisation Connection

VSEPR geometry directly corresponds to hybridisation:

Steric NumberHybridisationGeometry
2spLinear
3sp²Trigonal planar
4sp³Tetrahedral
5sp³dTrigonal bipyramidal
6sp³d²Octahedral

Hybridisation = mixing of atomic orbitals to form equivalent hybrid orbitals. sp means 1 s + 1 p orbital mixed → 2 equivalent sp orbitals at 180°. sp³ means 1 s + 3 p orbitals → 4 equivalent sp³ orbitals at 109.5°.

Polarity and Molecular Shape

A molecule’s polarity depends on both:

  1. Whether individual bonds are polar (electronegativity difference)
  2. Whether the bond dipoles cancel (based on molecular geometry)

Symmetric molecules with polar bonds (net dipole = 0):

  • CO₂ (linear), BF₃ (trigonal planar), CCl₄ (tetrahedral), PCl₅ (trigonal bipyramidal), SF₆ (octahedral)

Asymmetric molecules with polar bonds (net dipole ≠ 0):

  • H₂O (bent), NH₃ (trigonal pyramidal), SO₂ (bent), CHCl₃ (tetrahedral but asymmetric)

“Is CO₂ polar?” is a JEE Main 2023 classic. Answer: CO₂ has polar C=O bonds but the molecule is linear — the two bond dipoles are equal and opposite, so they cancel. CO₂ is non-polar.

Solved Examples

Example 1 (CBSE): Predict the shape and bond angle of BF₃

B has 3 bonding pairs to F, no lone pairs. Steric number = 3.
Electron geometry = trigonal planar
Molecular shape = trigonal planar (no lone pairs to distort it)
Bond angle = 120°
Hybridisation: sp²

Example 2 (JEE Main): Compare bond angles in NH₃ and NF₃

In NH₃: N has 3 BP + 1 LP. H is less electronegative than N, so bonding pairs are closer to N — more repulsion from LP → angles ≈ 107°.

In NF₃: F is more electronegative than N, so bonding pairs are pulled away from N (toward F) — less electron density near N → bonding pairs are “closer to F,” reducing BP-BP repulsion → angle slightly less than NH₃ (≈ 102.5°).

NF₃ has a smaller bond angle than NH₃.

Example 3 (JEE Advanced): Why is PCl₅ non-polar despite having polar P–Cl bonds?

PCl₅ has trigonal bipyramidal geometry. The 5 P–Cl bonds are all polar, but the geometry is symmetric enough that all bond dipoles cancel:

  • 3 equatorial bonds at 120° cancel each other in the equatorial plane
  • 2 axial bonds at 180° cancel each other along the axial direction Net dipole = 0 → PCl₅ is non-polar.

Common Mistakes to Avoid

Mistake 1: Counting double/triple bonds as multiple domains. A C=O is ONE electron domain (one sigma bond forms the geometry; the pi bond just adds electron density in the same region). CO₂ has 2 domains, not 4.

Mistake 2: Forgetting to count lone pairs in the steric number. H₂O has steric number 4 (2 BP + 2 LP), not 2. The shape is bent, not linear.

Mistake 3: Confusing electron geometry with molecular geometry. In NH₃, electron geometry is tetrahedral (4 domains) but molecular geometry (shape of atoms only) is trigonal pyramidal. The question usually asks for molecular shape.

Mistake 4: Assuming all molecules with the same number of atoms have the same shape. H₂O (bent) and CO₂ (linear) both have 3 atoms but completely different shapes because CO₂ has no lone pairs and H₂O has two.

Practice Questions

Q1. What is the hybridisation and shape of SF₆?

SF₆: S has 6 bonding pairs (expanded octet), 0 lone pairs. Steric number = 6. Hybridisation: sp³d². Shape: octahedral, bond angle = 90°.

Q2. Explain why H₂O has a smaller bond angle than H₂S.

Both H₂O and H₂S have bent geometry with 2 BP + 2 LP. In H₂O, oxygen is more electronegative and has a smaller size — the lone pairs are more concentrated and exert stronger repulsion on bonding pairs, compressing the angle more. H₂O: ≈ 104.5°; H₂S: ≈ 92°. Also, S is larger so S–H bonds are more diffuse, less repulsion overall.

Q3. Is CCl₄ polar or non-polar? Explain.

CCl₄ is non-polar. Even though each C–Cl bond is polar (Cl is more electronegative than C), the four bonds are arranged symmetrically in a perfect tetrahedral geometry (109.5°). All four bond dipoles point outward at equal angles — they cancel completely. Net dipole moment = 0.

Q4. What is the shape and bond angle of ClF₃?

Cl has 5 valence electrons, 3 bonding pairs to F, and 2 lone pairs. Steric number = 5. Electron geometry: trigonal bipyramidal. Lone pairs occupy equatorial positions (lower repulsion). Molecular shape: T-shaped. Bond angles: approximately 87.5° (compressed from 90°/120° by lone pair repulsion).

Q5. Among BeCl₂, PCl₃, and SiCl₄ — which is/are non-polar?

BeCl₂ (linear, no lone pairs, bond dipoles cancel → non-polar) and SiCl₄ (tetrahedral, symmetric, bond dipoles cancel → non-polar). PCl₃ has a lone pair on P → trigonal pyramidal → asymmetric → polar.

FAQs

What is the difference between electron geometry and molecular geometry? Electron geometry considers ALL electron domains (bonding pairs + lone pairs) around the central atom. Molecular geometry describes only the positions of atoms. They are the same when there are no lone pairs; they differ when lone pairs are present.

Why do lone pairs occupy equatorial positions in trigonal bipyramidal geometry? Equatorial positions have fewer 90° angles to other positions (two 90° axial neighbours) while axial positions have three 90° angles (to the three equatorial groups). Lone pairs, which have greater repulsion, prefer equatorial positions to minimise close 90° repulsions.

Can VSEPR predict the shapes of molecules with multiple central atoms? VSEPR works for each central atom independently. For a molecule like ethane (CH₃CH₃), each C has 4 bonding pairs → tetrahedral geometry around each carbon. Apply VSEPR to each central atom separately.

What is a polar bond vs a polar molecule? A polar bond forms between atoms with different electronegativities — one end is δ+ and the other is δ−. A polar molecule has an overall uneven charge distribution (net dipole moment ≠ 0). You can have polar bonds in a non-polar molecule if the geometry makes all bond dipoles cancel (like CO₂).

Practice Questions