MHT-CET Chemistry · Chemical Bonding and Molecular Structure

VSEPR Theory and Molecular Geometry

Count the electron pairs around the central atom — bond pairs plus lone pairs — and they spread out to keep as far apart as possible; the arrangement of the bond pairs is the molecule's shape, and lone pairs push the bonds closer to distort the ideal angles.

All Chemical Bonding and Molecular Structure notes MHT-CET Chemistry strategy

Why this matters

The single biggest subtopic of this chapter — 20 PYQs, and the most reliable shape-questions in MHT-CET Chemistry. They cluster four ways: count the lone pairs on the central atom (most, fewest, equal pair, or zero), name the shape of a given molecule or AXnEm type, recall a specific bond angle, and spot which molecule keeps its regular (undistorted) geometry. Every one of them reduces to the same two-step drill: count bond pairs and lone pairs, then read the shape off the master table — so with one table memorised a student should never drop a mark here.

Concept 1 of 4: The VSEPR premise: electron pairs repel and spread out

Electron pairs around the central atom are all negatively charged, so they repel each other and settle into the arrangement that keeps them as far apart as possible. That arrangement fixes the geometry. A lone pair takes up more room than a bond pair, so lone pairs squeeze the bonds together and distort the ideal angles.

Definition

The Valence Shell Electron Pair Repulsion (VSEPR) theory:

Electron pairs (bond pairs and lone pairs) around the central atom arrange themselves to minimise repulsion — i.e. to stay as far apart as possible.
The repulsion order is $\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$ (lone-pair–lone-pair is strongest, bond-pair–bond-pair weakest).
A molecule has its regular (expected) geometry only when the central atom has no lone pairs — then the electron-pair geometry and the molecular shape coincide (e.g. $\text{CH}_4$ , $\text{SiCl}_4$ , $\text{PCl}_5$ , $\text{SF}_6$ ).
Any lone pair distorts the shape, so a molecule with lone pairs does not show the regular parent geometry (e.g. $\text{SF}_4$ see-saw, $\text{XeF}_4$ square planar).

Worked example

\text{CH}_4

and

\text{SiCl}_4

, do they have the same geometry, and how many lone pairs sit on the central atom of each?

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q77Moderate

Example 1 · Chemical Bonding and Molecular Structure · VSEPR Theory and Molecular Geometry

Which of the following molecules has a regular geometry as expected?

Lone pairs count toward the electron geometry but not the described shape

The parent (electron-pair) geometry counts every pair, but the reported shape names only where the atoms sit.

\text{XeF}_4

has an octahedral electron geometry, yet its shape is square planar — the 2 lone pairs occupy positions but aren't drawn as part of the shape. Always answer with the atom-only shape unless the question asks for the parent geometry.

'Regular geometry as expected' means zero lone pairs

When the bank asks which molecule has its regular or expected geometry, it wants the one with no lone pairs on the central atom —

\text{SiCl}_4

, not

\text{SF}_4

\text{BrF}_5

\text{XeF}_4

. A lone pair always distorts, so any lone-pair molecule is disqualified.

Drill 1 more on the vsepr premise: electron pairs repel and spread out

Concept 2 of 4: Counting bond pairs and lone pairs on the central atom

Everything in VSEPR starts here. Take the central atom's valence electrons, use one for each bond it forms, and whatever is left over pairs up into lone pairs. Bond pairs plus lone pairs is the total number of electron domains that decides the shape.

Definition

How to count electron pairs on the central atom:

Bond pairs (bp) = the number of atoms bonded to the central atom (for single bonds).
Lone pairs (lp) $= \dfrac{V - \text{(bp)}}{2}$ , where $V$ is the number of valence electrons on the central atom (electrons left after bonding, paired up).
Total pairs $= \text{bp} + \text{lp}$ — this sets the electron geometry (2 linear, 3 trigonal, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral).
Worked counts: $\text{NH}_3$ (N: $V=5$ , bp $=3$ , lp $=1$ ); $\text{H}_2\text{O}$ (O: $V=6$ , bp $=2$ , lp $=2$ ); $\text{BF}_3$ (B: $V=3$ , bp $=3$ , lp $=0$ ); $\text{SF}_6$ (S: $V=6$ , bp $=6$ , lp $=0$ ); $\text{BrF}_3$ (Br: $V=7$ , bp $=3$ , lp $=2$ ).

Lone pairs on the central atom

\text{lp} = \frac{V - \text{bp}}{2}

Vvalence electrons of the central atom
$\text{bp}$ bond pairs = number of atoms bonded to it (single bonds)
$\text{lp}$ lone pairs left on the central atom

Worked example

How many lone pairs sit on the central atom of

\text{BrF}_3

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q96Moderate

Example 2 · Chemical Bonding and Molecular Structure · VSEPR Theory and Molecular Geometry

Identify a molecule having highest number of lone pair of electrons in valence shell of central atom.

Count lone pairs on the central atom only

For IF, the question asks for lone pairs on the central iodine: I has

V=7

, one electron goes into the I–F bond, leaving

6/2 = 3

lone pairs on I. Don't add the 3 lone pairs sitting on F — the central-atom count is 3.

BF₃ has zero lone pairs — boron is electron-deficient

Boron has only 3 valence electrons and forms 3 bonds, so nothing is left over — 0 lone pairs (an incomplete octet with 6 electrons). Students often assume every central atom carries a lone pair;

\text{BF}_3

\text{SF}_6

and

\text{PCl}_5

are common zero-lone-pair molecules.

Drill 7 more on counting bond pairs and lone pairs on the central atom

Concept 3 of 4: The master shape table (AXnEm to geometry)

Once you have the bond-pair and lone-pair counts, the shape is a straight table lookup. Write the molecule as

\text{AX}_n\text{E}_m

— A central atom, X bonded atoms, E lone pairs — and read off the name and ideal bond angle. This one table answers every 'what is the shape?' PYQ.

Definition

Notation: A is the central atom, X each bonded atom (bond pair), E each lone pair. Total pairs $=$ X count $+$ E count fixes the parent geometry; the lone pairs then decide the atom-only shape:

No lone pairs ( $\text{AX}_2$ … $\text{AX}_6$ ) give the regular parent geometries.
With lone pairs, the lone pairs take the roomiest positions and the shape is named by where the atoms end up.
Key examples the bank uses: $\text{NH}_3$ ( $\text{AX}_3\text{E}$ , pyramidal), $\text{H}_2\text{O}$ ( $\text{AX}_2\text{E}_2$ , bent), $\text{SF}_4$ / $\text{TeF}_4$ ( $\text{AX}_4\text{E}$ , see-saw), $\text{XeF}_4$ ( $\text{AX}_4\text{E}_2$ , square planar), $\text{BrF}_5$ ( $\text{AX}_5\text{E}$ , square pyramidal).

Type (AXnEm)	Bond pairs / Lone pairs	Shape	Ideal bond angle	Example
$\text{AX}_2$	2 / 0	Linear	$180^\circ$	$\text{BeCl}_2$ , $\text{C}_2\text{H}_2$
$\text{AX}_3$	3 / 0	Trigonal planar	$120^\circ$	$\text{BF}_3$
$\text{AX}_2\text{E}$	2 / 1	Bent (angular)	about $119.5^\circ$	$\text{SO}_2$
$\text{AX}_4$	4 / 0	Tetrahedral	$109.5^\circ$	$\text{CH}_4$ , $\text{SiCl}_4$ , $\text{NH}_4^{+}$ Q
$\text{AX}_3\text{E}$	3 / 1	Trigonal pyramidal	$107^\circ$	$\text{NH}_3$
$\text{AX}_2\text{E}_2$	2 / 2	Bent (angular)	$104.5^\circ$	$\text{H}_2\text{O}$ , $\text{SCl}_2$ Q
$\text{AX}_5$	5 / 0	Trigonal bipyramidal	$120^\circ$ and $90^\circ$	$\text{PCl}_5$ Q
$\text{AX}_4\text{E}$	4 / 1	See-saw	$90^\circ$ , $120^\circ$	$\text{SF}_4$ , $\text{TeF}_4$ Q $\text{AB}_4\text{E}$ has a trigonal-bipyramidal parent geometry but a see-saw shape — the bank tests both the type-to-shape and the parent-geometry versions.
$\text{AX}_3\text{E}_2$	3 / 2	T-shaped	about $90^\circ$	$\text{ClF}_3$ , $\text{BrF}_3$ , $\text{ICl}_3$
$\text{AX}_2\text{E}_3$	2 / 3	Linear	$180^\circ$	$\text{XeF}_2$
$\text{AX}_6$	6 / 0	Octahedral	$90^\circ$	$\text{SF}_6$
$\text{AX}_5\text{E}$	5 / 1	Square pyramidal	about $90^\circ$	$\text{BrF}_5$ , $\text{IF}_5$ Q
$\text{AX}_4\text{E}_2$	4 / 2	Square planar	$90^\circ$	$\text{XeF}_4$ Q

Read off the shape from the AXnEm type: count X (bonded atoms) and E (lone pairs), then look up the row.

Practice this conceptself-check · 6 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q92Easy

Example 3 · Chemical Bonding and Molecular Structure · VSEPR Theory and Molecular Geometry

Identify the correct geometry for the following molecules: CH

_4

, C

_2

_2

, NH

_3

, BF

_3

H₂O is bent, not linear

Water is

\text{AX}_2\text{E}_2

: the 2 lone pairs on oxygen push the two O–H bonds down to about

104.5^\circ

, giving a bent shape — not the

180^\circ

linear shape you might expect from just 'two bonds'. Its shape-twin in the bank is

\text{SCl}_2

, also bent.

SF₄ is not tetrahedral — it has a lone pair

\text{SF}_4

has 4 bonded atoms but S carries 1 lone pair (

\text{AX}_4\text{E}

), so 5 electron domains give a see-saw shape, not tetrahedral. Only the zero-lone-pair

\text{AX}_4

molecules (

\text{CH}_4

\text{SiCl}_4

\text{NH}_4^{+}

) are tetrahedral.

Parent geometry versus molecular shape

For

\text{TeF}_4

(

\text{AX}_4\text{E}

) the parent geometry is trigonal bipyramidal (5 domains) but the molecular shape is see-saw. If the question says 'geometry', answer the parent trigonal bipyramidal; if it says 'shape', answer see-saw. Read the wording.

Drill 7 more on the master shape table (axnem to geometry)

Concept 4 of 4: Bond angles and how lone pairs shrink them

Start from the ideal angle for the electron geometry, then knock it down a little for every lone pair — because a lone pair pushes harder than a bond pair, it squeezes the bond angles smaller. The classic run

\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}

is the same tetrahedral parent with 0, 1 and 2 lone pairs.

Definition

Bond angle depends on the electron geometry and the number of lone pairs:

Ideal angles by parent geometry: linear $180^\circ$ , trigonal planar $120^\circ$ , tetrahedral $109.5^\circ$ , octahedral $90^\circ$ .
Each lone pair pushes the bonds closer, shrinking the angle below the ideal.
The signature tetrahedral-family sequence: $\text{CH}_4\ (109.5^\circ) > \text{NH}_3\ (107^\circ) > \text{H}_2\text{O}\ (104.5^\circ)$ — same parent, more lone pairs, smaller angle.
$\text{BF}_3$ keeps its full $120^\circ$ (no lone pairs); $\text{SO}_2$ is bent at about $119.5^\circ$ (one lone pair barely dents the $120^\circ$ parent).

Molecule	Bond pairs / Lone pairs	Bond angle	Note
$\text{CH}_4$	4 / 0	$109.5^\circ$	Ideal tetrahedral — no lone pair to distort.
$\text{NH}_3$	3 / 1	$107^\circ$	One lone pair shrinks $109.5^\circ$ a little.
$\text{H}_2\text{O}$	2 / 2	$104.5^\circ$	Two lone pairs shrink it further.
$\text{BF}_3$	3 / 0	$120^\circ$	Trigonal planar, no lone pair — full angle.Q
$\text{SO}_2$	2 / 1	about $119.5^\circ$	Bent; one lone pair barely dents the $120^\circ$ parent.Q SO $_2$ is the O–S–O $119.5^\circ$ the bank tests — not $109.5^\circ$ or $180^\circ$ ; its parent is trigonal, not tetrahedral.

Take the ideal angle for the parent geometry, then subtract for each lone pair.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q51Easy

Example 4 · Chemical Bonding and Molecular Structure · VSEPR Theory and Molecular Geometry

Identify the angle O-S-O in SO

_2

molecule.

SO₂ is 119.5°, not 109.5°

\text{SO}_2

has a trigonal (not tetrahedral) parent — 2 bond pairs and 1 lone pair around S — so its O–S–O angle is about

119.5^\circ

, close to the

120^\circ

trigonal ideal. The

107.5^\circ

109^\circ

distractors are tetrahedral-family angles that don't apply here.

More lone pairs, smaller angle

Because a lone pair repels harder than a bond pair, adding lone pairs to the same parent geometry always shrinks the bond angle:

\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}

. Don't quote

109.5^\circ

for all three — only the zero-lone-pair member keeps the ideal.

Drill 1 more on bond angles and how lone pairs shrink them

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Counting bond pairs and lone pairs on the central atom
Lone pairs on the central atom
$\text{lp} = \frac{V - \text{bp}}{2}$

Reference tables (2)

The master shape table (AXnEm to geometry)13 rows

Type (AXnEm)	Bond pairs / Lone pairs	Shape	Ideal bond angle	Example
$\text{AX}_2$	2 / 0	Linear	$180^\circ$	$\text{BeCl}_2$ , $\text{C}_2\text{H}_2$
$\text{AX}_3$	3 / 0	Trigonal planar	$120^\circ$	$\text{BF}_3$
$\text{AX}_2\text{E}$	2 / 1	Bent (angular)	about $119.5^\circ$	$\text{SO}_2$
$\text{AX}_4$	4 / 0	Tetrahedral	$109.5^\circ$	$\text{CH}_4$ , $\text{SiCl}_4$ , $\text{NH}_4^{+}$ Q
$\text{AX}_3\text{E}$	3 / 1	Trigonal pyramidal	$107^\circ$	$\text{NH}_3$
$\text{AX}_2\text{E}_2$	2 / 2	Bent (angular)	$104.5^\circ$	$\text{H}_2\text{O}$ , $\text{SCl}_2$ Q
$\text{AX}_5$	5 / 0	Trigonal bipyramidal	$120^\circ$ and $90^\circ$	$\text{PCl}_5$ Q
$\text{AX}_4\text{E}$	4 / 1	See-saw	$90^\circ$ , $120^\circ$	$\text{SF}_4$ , $\text{TeF}_4$ Q $\text{AB}_4\text{E}$ has a trigonal-bipyramidal parent geometry but a see-saw shape — the bank tests both the type-to-shape and the parent-geometry versions.
$\text{AX}_3\text{E}_2$	3 / 2	T-shaped	about $90^\circ$	$\text{ClF}_3$ , $\text{BrF}_3$ , $\text{ICl}_3$
$\text{AX}_2\text{E}_3$	2 / 3	Linear	$180^\circ$	$\text{XeF}_2$
$\text{AX}_6$	6 / 0	Octahedral	$90^\circ$	$\text{SF}_6$
$\text{AX}_5\text{E}$	5 / 1	Square pyramidal	about $90^\circ$	$\text{BrF}_5$ , $\text{IF}_5$ Q
$\text{AX}_4\text{E}_2$	4 / 2	Square planar	$90^\circ$	$\text{XeF}_4$ Q

Read off the shape from the AXnEm type: count X (bonded atoms) and E (lone pairs), then look up the row.

Bond angles and how lone pairs shrink them5 rows

Molecule	Bond pairs / Lone pairs	Bond angle	Note
$\text{CH}_4$	4 / 0	$109.5^\circ$	Ideal tetrahedral — no lone pair to distort.
$\text{NH}_3$	3 / 1	$107^\circ$	One lone pair shrinks $109.5^\circ$ a little.
$\text{H}_2\text{O}$	2 / 2	$104.5^\circ$	Two lone pairs shrink it further.
$\text{BF}_3$	3 / 0	$120^\circ$	Trigonal planar, no lone pair — full angle.Q
$\text{SO}_2$	2 / 1	about $119.5^\circ$	Bent; one lone pair barely dents the $120^\circ$ parent.Q SO $_2$ is the O–S–O $119.5^\circ$ the bank tests — not $109.5^\circ$ or $180^\circ$ ; its parent is trigonal, not tetrahedral.

Take the ideal angle for the parent geometry, then subtract for each lone pair.

Watch out for (9)

Lone pairs count toward the electron geometry but not the described shape
→ The VSEPR premise: electron pairs repel and spread out
'Regular geometry as expected' means zero lone pairs
→ The VSEPR premise: electron pairs repel and spread out
Count lone pairs on the central atom only
→ Counting bond pairs and lone pairs on the central atom
BF₃ has zero lone pairs — boron is electron-deficient
→ Counting bond pairs and lone pairs on the central atom
H₂O is bent, not linear
→ The master shape table (AXnEm to geometry)
SF₄ is not tetrahedral — it has a lone pair
→ The master shape table (AXnEm to geometry)
Parent geometry versus molecular shape
→ The master shape table (AXnEm to geometry)
SO₂ is 119.5°, not 109.5°
→ Bond angles and how lone pairs shrink them
More lone pairs, smaller angle
→ Bond angles and how lone pairs shrink them

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