VSEPR Theory: Molecular Shapes, AXE Notation & Geometry — Class 11 Chemistry Notes (JEE & NEET)

01

What is VSEPR Theory? — Complete Theory

Lewis structures (§ 4.1) fixed how many electrons an atom shares; VSEPR (Sidgwick–Powell 1940, refined by Nyholm and Gillespie 1957) answers where in space they sit. Its logic is brutally simple: electron pairs around a central atom are all negatively charged, repel one another, and therefore adopt the arrangement that maximises their mutual separation. Molecular shape is a consequence of crowd control.

Postulates. (1) The shape depends on the total number of valence-shell electron pairs — bonding and lone — around the central atom. (2) Pairs arrange to minimise repulsion and maximise distance. (3) Crucially, the pairs are not equal citizens: a lone pair is held by one nucleus only, spreads wider, and repels harder. The repulsion hierarchy: lp–lp > lp–bp > bp–bp. (4) A multiple bond counts as one super electron-pair for geometry — a double bond occupies one direction, but its extra electron density pushes neighbours a little further.

Repulsion orderlp–lp > lp–bp > bp–bp Countbond pairs (X) + lone pairs (E) Multiple bond= one super pair

The AXE notation — three letters that name every molecule. A = central atom; X_n = number of atoms bonded to it; E_m = number of lone pairs on it. CH₄ is AX₄, NH₃ is AX₃E, H₂O is AX₂E₂. The sum n + m — the steric number — fixes the electron-pair geometry; the lone pairs then decide how much of that geometry survives into the visible molecular shape. Seven steric numbers, seven geometries:

Table 1 — The master table: AXE class → geometry → shape (the whole topic on one screen)
AXE	Steric no.	Electron-pair geometry	Shape	Bond angle	Example
AX₂	2	Linear	Linear	180°	BeCl₂, CO₂
AX₃	3	Trigonal planar	Trigonal planar	120°	BF₃
AX₂E	3	Trigonal planar	Bent (V-shape)	< 120° (SO₂ 119.5°)	SO₂, O₃
AX₄	4	Tetrahedral	Tetrahedral	109.5°	CH₄
AX₃E	4	Tetrahedral	Trigonal pyramidal	107°	NH₃
AX₂E₂	4	Tetrahedral	Bent	104.5°	H₂O
AX₅	5	Trigonal bipyramidal	Trigonal bipyramidal	120°, 90°	PCl₅
AX₄E	5	TBP	See-saw	distorted	SF₄
AX₃E₂	5	TBP	T-shaped	≈ 90°	ClF₃
AX₂E₃	5	TBP	Linear	180°	XeF₂, I₃⁻
AX₆	6	Octahedral	Octahedral	90°	SF₆
AX₅E	6	Octahedral	Square pyramidal	≈ 90°	BrF₅, IF₅
AX₄E₂	6	Octahedral	Square planar	90°	XeF₄

Where lone pairs sit in a trigonal bipyramid — the positioning rule. In an AX₅ TBP the five positions split into 3 equatorial (120° apart, two neighbours each) and 2 axial (90° to three neighbours each). Lone pairs, being the greediest repellers, claim equatorial seats first — fewer close neighbours means less lp–bp suffering. That single rule generates SF₄’s see-saw (one equatorial lone pair), ClF₃’s T-shape (two equatorial), and XeF₂’s linear (three equatorial positions filled — the two axial X–F bonds survive 180° apart). In an octahedron all six positions are equivalent, so lone pairs simply go trans (opposite) to each other — XeF₄’s square planar.

Angle deviations — the ladder. Perfect geometries carry perfect angles only when all pairs are identical. Each lone pair squeezes the remaining bond-pair angles: CH₄ (no lone pairs) 109.5° → NH₃ (one) 107° → H₂O (two) 104.5°. The same logic bends SO₂ (AX₂E) just below 120°, distorts SF₄’s angles, and explains why water’s H–O–H is the most squeezed tetrahedron in the syllabus. The builder below draws each class and states its deviation reason.

02

Visualising the Shapes & Predicting Any Molecule

Ek gallery, ek predictor — the four canonical tetrahedral-family shapes first, then any AXE class on demand.

FIG. 1 — The angle ladder made visible: every panel shares the tetrahedral electron-pair geometry, but each lone pair claims a vertex and squeezes the surviving bond angle — 109.5° → 107° → 104.5°. BF₃ (flat, AX₃) shows what happens with no lone pairs at all: perfect 120°.

Try it live

Molecular Shape Predictor

AXE class

Example

Molecular shape—

Electron-pair geometry—

Ideal angle—

LP positioning—

Dashed red lobes = lone pairs; solid lines = bonds. Angle values are idealised — real deviations follow the lp–lp > lp–bp > bp–bp order. TBP classes show the equatorial lone-pair rule in action.

03

Solved Examples (Step-by-Step)

Lewis → count → AXE → shape. Jo chain yahan chalti hai, wahi predictor me live chalti hai.

EXAMPLE 01Foundation · NEET

The full address of ammonia

Predict the shape of NH₃ using VSEPR: count the pairs, assign the AXE class, and state the geometry, shape and bond angle with the reason for deviation.

LewisN (5 valence e⁻) bonds 3 H → 3 bond pairs, 1 lone pair left on N.
AXEAX₃E — steric number 4.
Geometry/shapeElectron pairs: tetrahedral; shape (atoms only): trigonal pyramidal.
AngleOne lp–bp squeeze → 109.5° → 107° (verify on the predictor).

AX₃E · tetrahedral geometry · trigonal pyramidal · 107°

EXAMPLE 02JEE Main · TBP positioning

XeF₂: three lone pairs, and yet linear

XeF₂ has five electron pairs around xenon. Predict its shape, explaining where the three lone pairs sit and why.

CountXe: 8 valence e⁻ → 2 bond pairs (2 Xe–F) + 3 lone pairs → AX₂E₃, steric 5.
PositioningTBP: lone pairs take equatorial seats (only two 90° neighbours each vs axial’s three) → all three equatorials are lp.
ShapeThe two F atoms survive on the axial positions — linear, 180°.
Why not bentAny axial lp would suffer three 90° lp–bp repulsions — the equatorial choice minimises them.

AX₂E₃ · equatorial lone pairs · linear 180°

EXAMPLE 03JEE Main · Angle ordering

CH₄, NH₃, H₂O — one geometry, three angles

All three molecules have four electron pairs around the central atom. Arrange them in decreasing bond angle and justify with the repulsion order.

ClassesCH₄ = AX₄ (0 lp) · NH₃ = AX₃E (1 lp) · H₂O = AX₂E₂ (2 lp).
Repulsion countEach lp adds stronger lp–bp repulsion → angles compress stepwise: 109.5° → 107° → 104.5°.
OrderCH₄ > NH₃ > H₂O — the repulsion order lp–lp > lp–bp > bp–bp made measurable.
Exam formThe same question returns as “why is HOH angle less than HNH?” — answer with the repulsion hierarchy, never with electronegativity alone.

CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°)

04

Practice Questions (With Solutions)

Attempt first — options lock after one shot, exactly like the real exam. Then read the working, chahe galti ho ya na ho.

Attempted 0/4 · Correct 0

Q1NEET · Identify shape

The shape of SF₄ is:

Solution

S: 6 valence e⁻ → 4 S–F bonds + 1 lone pair → AX₄E, steric 5 → TBP electron geometry.
The lone pair takes an equatorial seat → four F atoms form a see-saw.
Option B names the electron geometry, not the shape — the § 4.4 split in action.

(C) See-saw

Q2JEE Main · Angle ordering

The correct decreasing order of bond angle is:

Solution

0 lp → 1 lp → 2 lp: each lone pair adds stronger lp–bp repulsion and compresses the angle.
109.5° > 107° > 104.5°. Options with H₂O first invert the repulsion logic.

(B) CH₄ > NH₃ > H₂O

Q3JEE Main · Ions count too

The shape of the ammonium ion NH₄⁺ is:

Solution

In NH₄⁺, N donates its lone pair to H⁺ (a dative bond) — the pair becomes a bond pair.
Now AX₄, four identical bond pairs, zero lone pairs → perfect tetrahedral, 109.5° — the angle springs back from NH₃’s 107°.

(B) Tetrahedral

Q4NEET · Same formula test

CO₂ is linear while SO₂ is bent. The correct explanation is:

Solution

C: AX₂ → linear 180°; AX₂E → trigonal-planar geometry with one invisible lp → bent ≈ 119.5°.
Strength, size and bond type never decide VSEPR shape — the lone-pair count does.

(C) Lone pair on S

05

Key Rules & Takeaways

Shape card

Eight lines that solve this topic

Repulsion order: lp–lp > lp–bp > bp–bpThe engine of every deviation — lone pairs are greedy repellers.

AXE: A + X_n (bonds) + E_m (lone pairs)Steric number n + m fixes the electron-pair geometry.

Geometry counts pairs · shape counts atomsThey coincide only for AX_n; every E bends the story.

2 linear · 3 trigonal planar · 4 tetrahedral5 trigonal bipyramidal · 6 octahedral — the seven geometries.

TBP: lone pairs take equatorial seatsFewer 90° neighbours — generates see-saw, T-shape, linear.

Octahedral: lone pairs go transAX₅E square pyramidal · AX₄E₂ square planar (XeF₄).

Angle ladder: 109.5° → 107° → 104.5°CH₄ → NH₃ → H₂O — one lone pair per step of compression.

Multiple bond = one super pairCO₂’s two double bonds count as two directions, not four.

Anchor shapes: BeCl₂/CO₂ linear 180° · BF₃ 120° · CH₄ 109.5° · PCl₅ TBP 120°/90° · SF₆ octahedral 90° · Deviations: SO₂ 119.5° · NH₃ 107° · H₂O 104.5°

Count pairs, then let them flee each other — every shape in the table is minimal-repulsion arithmetic, no memorisation needed beyond the seven geometries.
Geometry and shape are different questions — pairs vs atoms; NH₃’s tetrahedral geometry hiding a pyramidal shape is the most-tested distinction.
Lone pairs have seating preferences — equatorial in a TBP, trans in an octahedron; these two rules generate SF₄, ClF₃, XeF₂, BrF₅ and XeF₄.
Angles carry the physics — the 109.5/107/104.5 ladder is the repulsion order measured in degrees; quote lp–bp squeeze, never hand-wave “lone pair effect”.

06

FAQs

What are the main postulates of VSEPR theory?

VSEPR states that the shape of a molecule depends on the number of valence-shell electron pairs around its central atom, which arrange themselves to minimise repulsion and maximise distance. The repulsion order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, so lone pairs compress the angles between bonding pairs. A multiple bond is treated as one super electron pair for geometry.

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

Electron-pair geometry is the arrangement of all electron pairs — bonding and lone — around the central atom, while molecular shape describes only the positions of the atoms. NH3 has tetrahedral electron-pair geometry but trigonal pyramidal shape, because the fourth vertex of the tetrahedron is occupied by a lone pair that is invisible to the shape. Every AXnEm molecule with m greater than 0 shows this difference.

Why is the bond angle of NH3 (107°) smaller than that of CH4 (109.5°)?

CH4 has four identical bond pairs, giving the ideal tetrahedral angle of 109.5°. In NH3 one vertex is a lone pair, and a lone pair occupies more space and repels bond pairs more strongly than bond pairs repel each other. That extra repulsion squeezes the H–N–H angle down to 107°. In H2O two lone pairs squeeze further, giving 104.5° — the ladder 109.5° → 107° → 104.5° is the repulsion order made visible.

Why is BeCl2 linear while H2O is bent, despite both having the formula AB2?

The formula alone cannot decide shape — the electron pairs can. BeCl2 (AX2) has two bond pairs and no lone pairs on beryllium, so the pairs sit 180° apart: linear. H2O (AX2E2) has two bond pairs and two lone pairs; the four pairs adopt a tetrahedral arrangement, and the two invisible lone pairs leave the two O–H bonds bent at 104.5°. Same AB2 formula, different AXE class, different shape.