VSEPR Theory: How to Predict the Shape of a Molecule

Chemistry · Chemical Bonding · NEET

VSEPR theory says electron pairs around a central atom push each other as far apart as they can, and this spread decides the shape of the molecule. You just count the bond pairs and lone pairs on the central atom, then match that count to a known shape. Memory hook: "Electrons hate each other, so they run to opposite corners."
Same 4 electron pairs, different shapes (lone pairs bend it)CH4 (0 lp)CTetrahedral109.5°NH3 (1 lp)NlpTrigonal pyramidal107°H2O (2 lp)OlplpBent (V-shape)104.5°
All three central atoms have 4 electron pairs (sp3), but each added lone pair (yellow) takes more space and squeezes the visible shape and bond angle: tetrahedral 109.5 degrees, to trigonal pyramidal 107 degrees, to bent 104.5 degrees.

Your doubts, answered

What is VSEPR theory in the simplest words?

VSEPR stands for Valence Shell Electron Pair Repulsion. The idea is very simple: the electron pairs in the outer shell of the central atom all carry negative charge, so they push away from each other. To get as far apart as possible, they arrange in a fixed pattern. That pattern decides the shape of the molecule. So VSEPR is just electrons trying to avoid each other.

How do I find the shape of a molecule step by step?

Step 1: Find the central atom (usually the least electronegative, and written first, like C in CH4). Step 2: Count how many atoms are bonded to it (bond pairs). Step 3: Count the lone pairs left on the central atom = (valence electrons of central atom - electrons used in bonds) / 2. Step 4: Add bond pairs + lone pairs to get total electron pairs. Step 5: Match the total to the arrangement (2=linear, 3=trigonal, 4=tetrahedral, 5=trigonal bipyramidal, 6=octahedral). Step 6: Remove the lone pair positions to see the final shape. For NEET, this 6-step method solves almost every shape question.

What is the difference between electron-pair geometry and molecular shape?

Geometry counts ALL electron pairs (bond pairs + lone pairs). Shape counts only the ATOMS you can actually see. Example: H2O has 4 electron pairs, so its geometry is tetrahedral. But it has 2 lone pairs that are invisible, so the shape you see is bent (V-shaped). Lone pairs decide geometry but they are not part of the visible shape. NEET loves this trap, so always read whether the question asks geometry or shape.

Why do lone pairs matter so much in VSEPR?

A lone pair belongs to only one atom, so it stays fat and close to the central atom. A bond pair is shared between two atoms, so it is pulled thinner and further away. Because lone pairs take more space, the repulsion order is: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. This is why lone pairs push bonded atoms closer and change both the shape and the bond angle. This exact order was asked in NEET 2016.

What does notation like AB2E3 or AB3E mean?

This is a shorthand for VSEPR. A is the central atom, B is a bonded atom (each B = one bond pair), and E is a lone pair on the central atom. So AB3E means 3 bond pairs and 1 lone pair, like NH3 (trigonal pyramidal). AB2E3 means 2 bond pairs and 3 lone pairs, like XeF2 (linear). Learn to write the AB-E code first, and the shape follows automatically.

How do I count lone pairs on the central atom quickly?

Use: lone pairs = (valence electrons of central atom - number of single bonds it makes) / 2. Add 1 electron for each negative charge and subtract 1 for each positive charge. Example ClF3: Cl has 7 valence electrons, makes 3 bonds, so (7-3)/2 = 2 lone pairs. That gives AB3E2, which is T-shaped. This shortcut was directly tested in NEET 2018 and NEET 2026.

⚠️ The NEET trap
H2O and CH4 are both sp3, so they must have the same bond angle of 109.5 degrees.
They are both sp3, but lone pairs shrink the angle. CH4 (0 lone pairs) = 109.5 degrees, NH3 (1 lone pair) = 107 degrees, H2O (2 lone pairs) = 104.5 degrees. More lone pairs means smaller angle.
🧠 Same hybridisation does NOT mean same angle. Count lone pairs first, then decide the angle.

Real NEET questions

NEET 2016 Phase 1

Consider the molecules CH4, NH3 and H2O. Which of the given statements is false?

A · The H–C–H, H–N–H and H–O–H bond angles are all greater than 90 degrees.
B · The H–O–H bond angle in H2O is larger than the H–C–H bond angle in CH4.
C · The H–O–H bond angle in H2O is smaller than the H–N–H bond angle in NH3.
D · The H–C–H bond angle in CH4 is larger than the H–N–H bond angle in NH3.
Solution: All three central atoms are sp3. As lone pairs increase, lone pair-bond pair repulsion increases and squeezes the angle: CH4 (109.5 degrees, 0 lone pairs) > NH3 (107 degrees, 1 lone pair) > H2O (104.5 degrees, 2 lone pairs). So the H–O–H angle is smaller, not larger, than H–C–H. Statement B is false.
NEET 2016 Phase 1

Predict the correct order of repulsion between electron pairs as per VSEPR theory.

A · Lone pair–lone pair > lone pair–bond pair > bond pair–bond pair
B · Lone pair–lone pair > bond pair–bond pair > lone pair–bond pair
C · Bond pair–bond pair > lone pair–bond pair > lone pair–lone pair
D · Lone pair–bond pair > bond pair–bond pair > lone pair–lone pair
Solution: A lone pair is held by only one nucleus, so it stays close and spread out, repelling strongly. A bond pair is shared between two atoms and is pulled thinner. So the repulsion order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. This single rule explains why lone pairs bend molecules and shrink angles.
NEET 2024

Match the compound with its shape. A. NH3; B. BrF5; C. XeF4; D. SF6. Shapes: I. Trigonal pyramidal; II. Square planar; III. Octahedral; IV. Square pyramidal.

A · A-II, B-IV, C-III, D-I
B · A-III, B-IV, C-I, D-II
C · A-II, B-III, C-IV, D-I
D · A-I, B-IV, C-II, D-III
Solution: Write the AB-E code for each. NH3 = AB3E (3 bonds + 1 lone pair) = trigonal pyramidal (I). BrF5 = AB5E (5 bonds + 1 lone pair) = square pyramidal (IV). XeF4 = AB4E2 (4 bonds + 2 lone pairs) = square planar (II). SF6 = AB6 (6 bonds, no lone pair) = octahedral (III). So A-I, B-IV, C-II, D-III.

Solved Chemical Bonding NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

See all 51 Chemical Bonding NEET PYQs ›
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Frequently asked

Does VSEPR theory work for every molecule?

VSEPR works very well for main-group molecules with one clear central atom, which covers almost all NEET questions. It can fail for some transition metal complexes and molecules with strong electron delocalisation, but you will not be tested on those exceptions in the shape questions.

Do we count double and triple bonds as more than one region?

No. For VSEPR shape, a double or triple bond counts as ONE electron region (one bond pair location), not two or three. For example CO2 has two double bonds but only 2 regions, so it is linear. This is a very common mistake.

Is VSEPR the same as hybridisation?

They are linked but not the same. VSEPR predicts the shape by counting electron pairs. Hybridisation (sp, sp2, sp3, sp3d, sp3d2) is the orbital explanation for the same arrangement. Total electron pairs of 2, 3, 4, 5, 6 match sp, sp2, sp3, sp3d, sp3d2. So once you count pairs, you get both the shape and the hybridisation.

Why is this important for NEET?

Chemical Bonding gives 2 to 3 questions almost every year, and VSEPR shape is one of the most repeated topics (asked in 2016, 2018, 2019, 2021, 2024, 2025 and 2026). Learning this one counting method lets you answer many shape and hybridisation questions quickly without memorising each molecule.