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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 notesMHT-CET Chemistry strategy

Why this matters

The single biggest subtopic of this chapter — 21 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

Intuition

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 lp-lp>lp-bp>bp-bp\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. CH4\text{CH}_4, SiCl4\text{SiCl}_4, PCl5\text{PCl}_5, SF6\text{SF}_6).
  • Any lone pair distorts the shape, so a molecule with lone pairs does not show the regular parent geometry (e.g. SF4\text{SF}_4 see-saw, XeF4\text{XeF}_4 square planar).

Worked example

Of CH4\text{CH}_4 and SiCl4\text{SiCl}_4, do they have the same geometry, and how many lone pairs sit on the central atom of each?
  1. In CH4\text{CH}_4 carbon has 4 valence electrons, all used to bond four H atoms: 4 bond pairs, 0 lone pairs.
  2. In SiCl4\text{SiCl}_4 silicon has 4 valence electrons, all used to bond four Cl atoms: 4 bond pairs, 0 lone pairs.
  3. Both are AX4\text{AX}_4 with no lone pairs, so both take the same regular tetrahedral shape.
Answer:Same geometry (tetrahedral), with no lone pair on the central atom in either.
Practice this conceptself-check · 5 quick reps

Try it yourself

Among SF4\text{SF}_4, BrF5\text{BrF}_5, XeF4\text{XeF}_4 and PCl5\text{PCl}_5, which shows its regular parent geometry, and why?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Which repels more strongly: a lone pair or a bond pair?
  2. 2.
    Write the VSEPR repulsion order.
  3. 3.
    A molecule shows its regular geometry only when the central atom has how many lone pairs?
  4. 4.
    Do CH₄ and SiCl₄ have the same shape?
  5. 5.
    Does XeF₄ show a regular octahedral shape?

From the bank · past-year question

Example 1Chemical Bonding and Molecular StructureEASY
Which of the following molecules has a regular geometry as expected?

[Shift || · 2025]

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. XeF4\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 — SiCl4\text{SiCl}_4, not SF4\text{SF}_4/BrF5\text{BrF}_5/XeF4\text{XeF}_4. A lone pair always distorts, so any lone-pair molecule is disqualified.
Drill 2 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

Intuition

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) =V−(bp)2= \dfrac{V - \text{(bp)}}{2}, where VV is the number of valence electrons on the central atom (electrons left after bonding, paired up).
  • Total pairs =bp+lp= \text{bp} + \text{lp} — this sets the electron geometry (2 linear, 3 trigonal, 4 tetrahedral, 5 trigonal bipyramidal, 6 octahedral).
  • Worked counts: NH3\text{NH}_3 (N: V=5V=5, bp=3=3, lp=1=1); H2O\text{H}_2\text{O} (O: V=6V=6, bp=2=2, lp=2=2); BF3\text{BF}_3 (B: V=3V=3, bp=3=3, lp=0=0); SF6\text{SF}_6 (S: V=6V=6, bp=6=6, lp=0=0); BrF3\text{BrF}_3 (Br: V=7V=7, bp=3=3, lp=2=2).

Lone pairs on the central atom

lp=V−bp2\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 BrF3\text{BrF}_3?
  1. Central atom Br has V=7V = 7 valence electrons.
  2. It bonds three F atoms, so bp =3= 3, using 3 of the 7 electrons.
  3. Remaining electrons =7−3=4= 7 - 3 = 4, which pair up: lp =4/2=2= 4/2 = 2.
Answer:2 lone pairs (so BrF3\text{BrF}_3 is sp3dsp^3d, T-shaped).
Practice this conceptself-check · 5 quick reps

Try it yourself

Which of SCl2\text{SCl}_2, PCl3\text{PCl}_3, ClF3\text{ClF}_3 and XeF4\text{XeF}_4 has the fewest lone pairs on the central atom?

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Lone pairs on central atom of H₂O?
  2. 2.
    Lone pairs on central atom of NH₃?
  3. 3.
    Lone pairs on central atom of BF₃?
  4. 4.
    Lone pairs on central atom of SF₆?
  5. 5.
    Which has 2 lone pairs on the central atom: NH₃, H₂O, SF₄ or SO₂?

From the bank · past-year question

Example 2Chemical Bonding and Molecular StructureMODERATE
Identify a molecule having highest number of lone pair of electrons in valence shell of central atom.

[Q96 · 20 April Shift I · 2025]

Count lone pairs on the central atom only

For IF, the question asks for lone pairs on the central iodine: I has V=7V=7, one electron goes into the I–F bond, leaving 6/2=36/2 = 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; BF3\text{BF}_3, SF6\text{SF}_6 and PCl5\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)

Intuition

Once you have the bond-pair and lone-pair counts, the shape is a straight table lookup. Write the molecule as AXnEm\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 (AX2\text{AX}_2…AX6\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: NH3\text{NH}_3 (AX3E\text{AX}_3\text{E}, pyramidal), H2O\text{H}_2\text{O} (AX2E2\text{AX}_2\text{E}_2, bent), SF4\text{SF}_4/TeF4\text{TeF}_4 (AX4E\text{AX}_4\text{E}, see-saw), XeF4\text{XeF}_4 (AX4E2\text{AX}_4\text{E}_2, square planar), BrF5\text{BrF}_5 (AX5E\text{AX}_5\text{E}, square pyramidal).
Type (AXnEm)Bond pairs / Lone pairsShapeIdeal bond angleExample
AX2\text{AX}_22 / 0Linear180∘180^\circBeCl2\text{BeCl}_2, C2H2\text{C}_2\text{H}_2
AX3\text{AX}_33 / 0Trigonal planar120∘120^\circBF3\text{BF}_3
AX2E\text{AX}_2\text{E}2 / 1Bent (angular)about 119.5∘119.5^\circSO2\text{SO}_2
AX4\text{AX}_44 / 0Tetrahedral109.5∘109.5^\circCH4\text{CH}_4, SiCl4\text{SiCl}_4, NH4+\text{NH}_4^{+}Q
AX3E\text{AX}_3\text{E}3 / 1Trigonal pyramidal107∘107^\circNH3\text{NH}_3
AX2E2\text{AX}_2\text{E}_22 / 2Bent (angular)104.5∘104.5^\circH2O\text{H}_2\text{O}, SCl2\text{SCl}_2Q
AX5\text{AX}_55 / 0Trigonal bipyramidal120∘120^\circ and 90∘90^\circPCl5\text{PCl}_5Q
AX4E\text{AX}_4\text{E}4 / 1See-saw90∘90^\circ, 120∘120^\circSF4\text{SF}_4, TeF4\text{TeF}_4Q
AB4E\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.
AX3E2\text{AX}_3\text{E}_23 / 2T-shapedabout 90∘90^\circClF3\text{ClF}_3, BrF3\text{BrF}_3, ICl3\text{ICl}_3
AX2E3\text{AX}_2\text{E}_32 / 3Linear180∘180^\circXeF2\text{XeF}_2
AX6\text{AX}_66 / 0Octahedral90∘90^\circSF6\text{SF}_6
AX5E\text{AX}_5\text{E}5 / 1Square pyramidalabout 90∘90^\circBrF5\text{BrF}_5, IF5\text{IF}_5Q
AX4E2\text{AX}_4\text{E}_24 / 2Square planar90∘90^\circXeF4\text{XeF}_4Q
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

Try it yourself

Give the shape of each: CH4\text{CH}_4, C2H2\text{C}_2\text{H}_2, NH3\text{NH}_3, BF3\text{BF}_3.

Practice — Level 1 (6 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Shape of BrF₅ (AX₅E)?
  2. 2.
    Shape of an AB₄E-type molecule?
  3. 3.
    Geometry of PCl₅ (AX₅)?
  4. 4.
    Shape of XeF₄ (AX₄E₂)?
  5. 5.
    Shape of XeF₂ (AX₂E₃)?
  6. 6.
    Parent (electron-pair) geometry of TeF₄?

From the bank · past-year question

Example 3Chemical Bonding and Molecular StructureEASY
Identify the correct geometry for the following molecules: CH4_4, C2_2H2_2, NH3_3, BF3_3

[Q92 · 9th May Shift 2 · 2023]

H₂O is bent, not linear

Water is AX2E2\text{AX}_2\text{E}_2: the 2 lone pairs on oxygen push the two O–H bonds down to about 104.5∘104.5^\circ, giving a bent shape — not the 180∘180^\circ linear shape you might expect from just 'two bonds'. Its shape-twin in the bank is SCl2\text{SCl}_2, also bent.

SF₄ is not tetrahedral — it has a lone pair

SF4\text{SF}_4 has 4 bonded atoms but S carries 1 lone pair (AX4E\text{AX}_4\text{E}), so 5 electron domains give a see-saw shape, not tetrahedral. Only the zero-lone-pair AX4\text{AX}_4 molecules (CH4\text{CH}_4, SiCl4\text{SiCl}_4, NH4+\text{NH}_4^{+}) are tetrahedral.

Parent geometry versus molecular shape

For TeF4\text{TeF}_4 (AX4E\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

Intuition

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 CH4>NH3>H2O\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∘180^\circ, trigonal planar 120∘120^\circ, tetrahedral 109.5∘109.5^\circ, octahedral 90∘90^\circ.
  • Each lone pair pushes the bonds closer, shrinking the angle below the ideal.
  • The signature tetrahedral-family sequence: CH4 (109.5∘)>NH3 (107∘)>H2O (104.5∘)\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.
  • BF3\text{BF}_3 keeps its full 120∘120^\circ (no lone pairs); SO2\text{SO}_2 is bent at about 119.5∘119.5^\circ (one lone pair barely dents the 120∘120^\circ parent).
MoleculeBond pairs / Lone pairsBond angleNote
CH4\text{CH}_44 / 0109.5∘109.5^\circIdeal tetrahedral — no lone pair to distort.
NH3\text{NH}_33 / 1107∘107^\circOne lone pair shrinks 109.5∘109.5^\circ a little.
H2O\text{H}_2\text{O}2 / 2104.5∘104.5^\circTwo lone pairs shrink it further.
BF3\text{BF}_33 / 0120∘120^\circTrigonal planar, no lone pair — full angle.Q
SO2\text{SO}_22 / 1about 119.5∘119.5^\circBent; one lone pair barely dents the 120∘120^\circ parent.Q
SO2_2 is the O–S–O 119.5∘119.5^\circ the bank tests — not 109.5∘109.5^\circ or 180∘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

Try it yourself

Arrange CH4\text{CH}_4, NH3\text{NH}_3 and H2O\text{H}_2\text{O} in decreasing order of bond angle and explain.

Practice — Level 1 (5 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    F–B–F bond angle in BF₃?
  2. 2.
    O–S–O bond angle in SO₂?
  3. 3.
    H–C–H bond angle in CH₄?
  4. 4.
    H–N–H bond angle in NH₃?
  5. 5.
    H–O–H bond angle in H₂O?

From the bank · past-year question

Example 4Chemical Bonding and Molecular StructureEASY
Identify the angle O-S-O in SO2_2 molecule.

[Q51 · 16th May Shift 1 · 2023]

SO₂ is 119.5°, not 109.5°

SO2\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∘119.5^\circ, close to the 120∘120^\circ trigonal ideal. The 107.5∘107.5^\circ/109∘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: CH4>NH3>H2O\text{CH}_4 > \text{NH}_3 > \text{H}_2\text{O}. Don't quote 109.5∘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)

Reference tables (2)

The master shape table (AXnEm to geometry)13 rows
Type (AXnEm)Bond pairs / Lone pairsShapeIdeal bond angleExample
AX2\text{AX}_22 / 0Linear180∘180^\circBeCl2\text{BeCl}_2, C2H2\text{C}_2\text{H}_2
AX3\text{AX}_33 / 0Trigonal planar120∘120^\circBF3\text{BF}_3
AX2E\text{AX}_2\text{E}2 / 1Bent (angular)about 119.5∘119.5^\circSO2\text{SO}_2
AX4\text{AX}_44 / 0Tetrahedral109.5∘109.5^\circCH4\text{CH}_4, SiCl4\text{SiCl}_4, NH4+\text{NH}_4^{+}Q
AX3E\text{AX}_3\text{E}3 / 1Trigonal pyramidal107∘107^\circNH3\text{NH}_3
AX2E2\text{AX}_2\text{E}_22 / 2Bent (angular)104.5∘104.5^\circH2O\text{H}_2\text{O}, SCl2\text{SCl}_2Q
AX5\text{AX}_55 / 0Trigonal bipyramidal120∘120^\circ and 90∘90^\circPCl5\text{PCl}_5Q
AX4E\text{AX}_4\text{E}4 / 1See-saw90∘90^\circ, 120∘120^\circSF4\text{SF}_4, TeF4\text{TeF}_4Q
AB4E\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.
AX3E2\text{AX}_3\text{E}_23 / 2T-shapedabout 90∘90^\circClF3\text{ClF}_3, BrF3\text{BrF}_3, ICl3\text{ICl}_3
AX2E3\text{AX}_2\text{E}_32 / 3Linear180∘180^\circXeF2\text{XeF}_2
AX6\text{AX}_66 / 0Octahedral90∘90^\circSF6\text{SF}_6
AX5E\text{AX}_5\text{E}5 / 1Square pyramidalabout 90∘90^\circBrF5\text{BrF}_5, IF5\text{IF}_5Q
AX4E2\text{AX}_4\text{E}_24 / 2Square planar90∘90^\circXeF4\text{XeF}_4Q
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
MoleculeBond pairs / Lone pairsBond angleNote
CH4\text{CH}_44 / 0109.5∘109.5^\circIdeal tetrahedral — no lone pair to distort.
NH3\text{NH}_33 / 1107∘107^\circOne lone pair shrinks 109.5∘109.5^\circ a little.
H2O\text{H}_2\text{O}2 / 2104.5∘104.5^\circTwo lone pairs shrink it further.
BF3\text{BF}_33 / 0120∘120^\circTrigonal planar, no lone pair — full angle.Q
SO2\text{SO}_22 / 1about 119.5∘119.5^\circBent; one lone pair barely dents the 120∘120^\circ parent.Q
SO2_2 is the O–S–O 119.5∘119.5^\circ the bank tests — not 109.5∘109.5^\circ or 180∘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)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Chemical Bonding and Molecular StructureEASY
Identify the correct statement regarding geometry and lone pair of electrons present in CH4CH_{4} and SiCl4SiCl_{4}.

[Q63 · 22 April Shift I · 2025]

Example 2Chemical Bonding and Molecular StructureEASY
Which from following molecules does not have lone pair of electrons in valence shell of central atom?

[Q65 · 4th May Shift 1 · 2023]

Example 3Chemical Bonding and Molecular StructureEASY
What is the shape of AB4E\text{AB}_{4}\text{E} type of molecule according to VSEPR? [E → Lone pair]

[Q90 · 12th May Shift 1 · 2024]

Example 4Chemical Bonding and Molecular StructureEASY
What is bond angle F-B-F in BF3\text{BF}_3?

[Q70 · 9th May Shift 1 · 2023]

Example 5Chemical Bonding and Molecular StructureMODERATE
Which of the following molecules has a regular geometry as expected?

[Q77 · 19 April Shift II · 2025]

Drill every past-year question on this subtopic

21 questions from the bank — paginated, with cart and Word-export support.

Drill the 21 questions

Related notes