Predict the shape of the ammonia molecule, $\text{NH}_3$, using VSEPR theory, and explain why its bond angle ($107^{\circ}$) is smaller than the ideal tetrahedral angle of $109.5^{\circ}$.

Predict the shape of the ammonia molecule, NH₃, using VSEPR theory, and explain why its bond angle (107°) is s

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\text{NH}_3

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Concept understanding — VSEPR Theory

VSEPR Theory: Why Molecules Have the Shapes They Do

Imagine you're in a crowded room. Everyone wants their personal space. If you're standing with a few friends, you'll naturally spread out so no one is too close to anyone else. That's exactly what happens inside a molecule.

The Core Intuition

Electron pairs are negatively charged. They repel each other. In a molecule, the electron pairs around a central atom will arrange themselves as far apart as possible — just like those people in the room. This simple idea is the entire foundation of VSEPR (pronounced "ves-per") Theory.

Note

VSEPR stands for Valence Shell Electron Pair Repulsion. The name tells you exactly what it's about: the repulsion between electron pairs in the valence shell.

The Precise Statement

VSEPR Theory states that the geometry around a central atom is determined by minimizing the repulsion between all electron pairs (both bonding and lone pairs) in its valence shell.

Two key points to hold onto:

All electron pairs repel — whether they are shared (bonding pairs) or unshared (lone pairs).
Lone pairs repel more strongly than bonding pairs. A lone pair is "fatter" — it's only attracted to one nucleus, so it spreads out more and pushes harder on its neighbours.

How to Predict Shape in 3 Steps

Step 1: Count the total electron pairs around the central atom.

Add the number of atoms bonded to the central atom plus the number of lone pairs on it. This gives you the steric number.

Step 2: Arrange those pairs as far apart as possible.

This gives you the electron-pair geometry — the shape if you pretend all pairs are identical.

Step 3: Replace lone pairs with "invisible" space.

The actual molecular geometry is the shape formed by the atoms alone, ignoring lone pairs.

The Common Geometries at a Glance

Steric Number	Electron-Pair Geometry	Lone Pairs	Molecular Geometry	Example	Bond Angle
2	Linear	0	Linear	$\text{CO}_2$	180°
3	Trigonal planar	0	Trigonal planar	$\text{BF}_3$	120°
3	Trigonal planar	1	Bent	$\text{SO}_2$	~119°
4	Tetrahedral	0	Tetrahedral	$\text{CH}_4$	109.5°
4	Tetrahedral	1	Trigonal pyramidal	$\text{NH}_3$	~107°
4	Tetrahedral	2	Bent	$\text{H}_2\text{O}$	~104.5°
5	Trigonal bipyramidal	0	Trigonal bipyramidal	$\text{PCl}_5$	90°, 120°
6	Octahedral	0	Octahedral	$\text{SF}_6$	90°

Watch out

A common mistake: thinking that $\text{NH}_3$ is tetrahedral. It has tetrahedral electron-pair geometry, but because one position is a lone pair, the molecular shape is trigonal pyramidal. The bond angle is 107°, not 109.5°.

Why Lone Pairs Squeeze Bond Angles

Take water ( $\text{H}_2\text{O}$ ). The central oxygen has 4 electron pairs: 2 bonding (to H atoms) and 2 lone pairs. The ideal tetrahedral angle is 109.5°. But the two lone pairs push harder on the bonding pairs, compressing the H–O–H angle to about 104.5°.

In ammonia ( $\text{NH}_3$ ), there's only one lone pair, so the compression is less — the H–N–H angle is about 107°. …

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Nitrogen in $\text{NH}_3$ has five valence electrons: three are used to form N–H bonds, and the remaining two form a single lone pair on nitrogen. This gives a total steric number of 4 (3 bonding pairs + 1 lone pair), so the underlying electron-pair geometry is tetrahedral, the same base geometry as $\text{CH}_4$ . However, the molecular shape is described only by the positions of the atoms (the three hydrogens), which gives a trigonal pyramidal shape — three N–H bonds visible, with the lone pair occupying the fourth tetrahedral position but not counted in describing the shape. The bond angle is not the ideal tetrahedral $109.5^{\circ}$ , though, because VSEPR theory ranks repulsions as lone pair–bond pair $>$ bond pair–bond pair: the lone pair, held only by the nitrogen nucleus, occupies a broader region …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 20 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is mismatch regarding shape?$
(a) XeF4 — Octahedral
(b) PCl5 — Trigonal bipyramidal
(c) SO2 — Bent
(d) CH4 — Tetrahedral
›Reveal solutionSolution
Check each pair against VSEPR theory. XeF4 has 4 bond pairs + 2 lone pairs on Xe; the electron-pair arrangement is octahedral, but the MOLECULAR SHAPE (what the atoms actually trace out) is square planar, not octahedral.

Go through each option:

XeF4: Xe has 8 valence electrons; 4 are used in Xe-F bonds, leaving 2 lone pairs. Total electron domains = 6, so the electron-pair geometry is octahedral -- but the two lone pairs occupy opposite (axial) positions to minimise repulsion, leaving the 4 fluorine atoms in a plane. The resulting MOLECULAR shape is square planar, not octahedral. Labelling XeF4's shape as 'Octahedral' is a mismatch.

PCl5: 5 bond pairs, no lone pairs on P -- trigonal bipyramidal. Correct match. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Match the correct pair: Square planner$
(a) XeF4
(b) Stock notation
(c) Lassaigne's test
(d) Size of orbital
(e) Exothermic reaction
›Reveal solutionSolution

XeF4 has a square planar shape — its central Xe atom has 6 electron domains (4 bond pairs + 2 lone pairs) arranged octahedrally, with the two lone pairs on opposite (axial) sides.

In XeF4, the central xenon atom has 8 valence electrons: 4 are used to form 4 Xe-F sigma bonds, leaving 2 lone pairs. This gives 6 electron domains total, which arrange themselves octahedrally to minimize repulsion. The two lone pairs occupy positions opposite each other (axial positions) to minimize lone pair-lone pair repulsion, leaving the four fluo …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is not a linear molecule ?$
(a) CO2
(b) BeCl2
(c) C2H2
(d) H2S
›Reveal solutionSolution

H2S is not linear; it is bent.

CO2 (sp, O=C=O), BeCl2 (sp, Cl–Be–Cl) and C2H2 (sp, H–C≡C–H) are all linear (180°). H2S has a central sulphur with two bond pairs and two lone pairs, so its shape …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following has the largest bond angle ?$
(a) H2O
(b) CO2
(c) NH3
(d) CH4
›Reveal solutionSolution

CO2 has the largest bond angle (180°).

Bond angles: H2O ≈ 104.5° (two lone pairs), NH3 ≈ 107° (one lone pair), CH4 = 109.5° (tetrahedral), CO2 = 180° (linear, sp carbon, no …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. What is the shape of the molecule NH3?$
(a) Square pyramidal
(b) V-shape
(c) Trigonal pyramidal
(d) Tetrahedral
›Reveal solutionSolution

NH3 has 4 electron domains (3 bond pairs + 1 lone pair) arranged tetrahedrally, but since molecular SHAPE only counts the positions of ATOMS (not lone pairs), the visible shape is trigonal pyramidal.

Step 1 — Electron domains on N: Nitrogen has 5 valence electrons. In NH3, 3 of these electrons pair up with 3 hydrogen atoms (3 sigma bonds), and the remaining 2 electrons form 1 lone pair. Total electron domains = 3 bonding + 1 lone pair = 4.

Step 2 — Electron geometry vs molecular shape: 4 electron domains arrange themselves tetrahedrally (electron-pair geometry = tetrahedral) to minimise repulsion. However, molecular SHAPE is described only by the positions of the atoms, ignoring the lone pair's position (even though the lone pair does occupy space and affect bond angles).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The repulsive interaction of electron pairs decrease in the order (Here lp = lone pair, bp = bond pair):$
(a) bp – bp > lp – bp > lp – lp
(b) lp – lp > lp – bp > bp – bp
(c) lp – lp > bp – bp > lp – bp
(d) bp – bp > lp – lp > lp – bp
›Reveal solutionSolution

VSEPR theory ranks repulsions as lone pair-lone pair strongest, then lone pair-bond pair, then bond pair-bond pair weakest.

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, a lone pair (lp) is held by only one nucleus and so occupies more space around the central atom than a bond pair (bp), which is shared (and pulled) by two nuclei and is more confined.

Because lone pairs spread out more, they repel neighbouring electron pairs more strongly. This gives the order of repulsive interaction strength:

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following molecules have trigonal bipyramidal shape?$
(a) [Ni(CN)4]2-
(b) SF6
(c) [CrF6]3-
(d) PCl5
›Reveal solutionSolution
PCl5 (sp3d, 5 bond pairs, 0 lone pairs) is trigonal bipyramidal; the others are square planar/octahedral.

Shape is decided by the number of electron pairs (VSEPR) around the central atom:

$[Ni(CN)_4]^{2-}$ : Ni is $dsp^2$ hybridized, 4 bond pairs → square planar.

$SF_6$ : S is $sp^3d^2$ hybridized, 6 bond pairs → octahedral.

$[CrF_6]^{3-}$ : Cr is $d^2sp^3/sp^3d^2$ hybridized, 6 bond pairs → octahedral. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Shape of molecule of CH4 is:$
(a) Trigonal
(b) Bent structure
(c) Tetrahedral
(d) Linear
›Reveal solutionSolution

CH4 is tetrahedral because carbon is sp3 hybridised with 4 bond pairs and no lone pairs.

Carbon in methane has 4 valence electrons, each forming a single bond with one hydrogen atom (4 bond pairs, 0 lone pairs on carbon). To minimise repulsion between these 4 electron pairs (VSEPR theory), carbon undergoes sp3 hybridisation, and the four sp3 orbitals point towards the corners of a regu …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following is linear?$
(a) CO2
(b) NO2
(c) SO2
(d) ClO2
›Reveal solutionSolution

CO2 (O=C=O) is linear; NO2, SO2, ClO2 are all bent due to a lone pair/odd electron on the central atom.

In CO2, the central carbon forms two σ (and two π) bonds to oxygen with no lone pairs, so by VSEPR the two bonding domains arrange at 180° — a linear molecule. In NO2, SO2, and ClO2, the central atom (N, S, Cl respectivel …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following molecules has a linear shape?$
(a) BeCl2
(b) BCl3
(c) CCl4
(d) PCl5
›Reveal solutionSolution
BeCl2's central Be atom has only 2 electron domains (2 bond pairs, no lone pairs), so VSEPR predicts a linear geometry.

Applying VSEPR to each option's central atom:

BeCl2: Be has 2 valence electrons, forms 2 bonds to Cl, no lone pairs -> sp hybridisation -> linear, bond angle 180 degrees.

BCl3: B has 3 bond pairs, no lone pairs -> sp2 hybridisation -> trigonal planar (120 degrees). …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The shape of IF5 molecule is __________.$
(a) Square pyramidal
(b) Trigonal bipyramidal
(c) Octahedral
(d) Square planar
›Reveal solutionSolution

IF5 has 5 bond pairs and 1 lone pair on iodine; the underlying electron geometry is octahedral, but with a lone pair occupying one vertex, the actual molecular shape is square pyramidal.

Iodine in IF5 has 7 valence electrons; it forms 5 sigma bonds to the 5 fluorine atoms (using 5 electrons) and retains 1 lone pair (the remaining 2 electrons). By VSEPR (Valence Shell Electron Pair Repulsion) theory:

Total electron domains around I = 5 bond pairs + 1 lone pair = 6 domains.

6 electron domains arrange themselves to minimise repulsion in an OCTAHEDRAL electron-pair geometry. However, the MOLECULAR SHAPE describes only the positions of the atoms, not the lone pair. With one of the six octahedral positions occupied by a lone pair (which pushes the four equatorial F atoms slightly away from it and pulls the axial F atom in), the five fluorine atoms end up arranged as a square pyramid: four F atoms forming a square base around iodine, with the fifth F atom at the apex, and the lone pair taking the remaining position opposite that ap …

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CBSE 2024Set ANNUAL1 mark
$NH_3$
›Reveal solutionSolution

More lone pairs on the central atom compress the bond angle more, because lone pair-bond pair repulsion exceeds bond pair-bond pair repulsion.

$CH_4$ has 4 bond pairs and 0 lone pairs on carbon — a perfect tetrahedral angle of 109.5°.

$NH_3$ has 3 bond pairs and 1 lone pair on nitrogen; the extra lone pair-bond pair repulsion compresses the H-N-H angle to about 107°.

…

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Q.Predict the shape of the ammonia molecule, NH3\text{NH}_3NH3​, using VSEPR theory, and explain why its bond angle (107∘107^{\circ}107∘) is smaller than the ideal tetrahedral angle of 109.5∘109.5^{\circ}109.5∘.

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Q.Predict the shape of the ammonia molecule, $\text{NH}_3$ , using VSEPR theory, and explain why its bond angle ( $107^{\circ}$ ) is smaller than the ideal tetrahedral angle of $109.5^{\circ}$ .

Unlock everything free for 14 days