JEE Mains Chemistry · Chemical Bonding and Molecular Structure

VSEPR Theory and Lone Pairs

VSEPR theory counts the electron pairs on the central atom, bond pairs plus lone pairs, and places them as far apart as possible; lone pairs take the positions where they meet the fewest bond pairs at 90°.

All Chemical Bonding and Molecular Structure notes JEE Mains Chemistry strategy

Why this matters

Sixteen PYQs, eleven of them multiple choice, and two from 2026. Ten count the lone pairs on a central atom, most often xenon or a halogen; six ask where the lone pairs sit in a trigonal bipyramid or an octahedron and what shape results. Two ideas cover the page.

Concept 1 of 2: Counting lone pairs on the central atom

The central atom brings its valence electrons. Each bond to a hydrogen or halogen takes one of them; each double bond to an oxygen takes two; a negative charge adds one and a positive charge removes one. Whatever is left pairs up as lone pairs.

Definition

Steric number (SN) = σ bonds + lone pairs on the central atom.
Shortcut for centres bonded to H, halogens and O: $SN = \tfrac{1}{2}(V + M - c + a)$ , where $V$ is the central atom's valence electrons, $M$ the number of H or halogen atoms bonded to it, $c$ a positive charge, $a$ a negative charge. Oxygen atoms add nothing.
Lone pairs on the centre = SN − number of atoms bonded to it.
Xenon ( $V = 8$ ): $\mathrm{XeF_2}$ 3, $\mathrm{XeF_4}$ 2, $\mathrm{XeF_6}$ 1, $\mathrm{XeO_3}$ 1, $\mathrm{XeOF_4}$ 1, $\mathrm{XeO_2F_2}$ 1, $\mathrm{XeO_3F_2}$ 0.
Halogen centres: $\mathrm{IF_7}$ 0, $\mathrm{IF_5}$ 1, $\mathrm{BrF_3}$ 2, $\mathrm{ICl_4^-}$ 2, $\mathrm{I_3^-}$ 3.

Steric number and lone pairs

SN = \tfrac{1}{2}(V + M - c + a) \qquad \text{lone pairs} = SN - (\text{atoms bonded to the centre})

Worked example

Find the number of lone pairs on the central atom in

\mathrm{IF_5}

\mathrm{SO_3^{2-}}

and

\mathrm{ICl_2^-}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 8 Apr 2023 · Q145Moderate

Example 1 · Chemical Bonding and Molecular Structure · VSEPR Theory and Lone Pairs

The number of species from the following carrying a single lone pair on central atom Xenon is_____

XeF_{5}^{+},XeO_{3},XeO_{2}{\text{ }F}_{2},XeF_{5}^{-},XeO_{3}{\text{ }F}_{2},XeOF_{4}

XeF_{4}

A double bond to oxygen uses two electrons

Xe in

\mathrm{XeO_3}

uses 6 of its 8 electrons on three Xe=O bonds and keeps one lone pair. Counting each Xe=O as one electron leaves 5 electrons, which cannot pair: a sure sign of a slip.

Know what the question calls a bond pair

Some questions count π pairs as bond pairs. On that count

\mathrm{SO_2}

has 4 bond pairs and 1 lone pair, not 2 and 1. Read the options: if no option fits the σ-only count, the π pairs are included.

Drill 9 more on counting lone pairs on the central atom

Concept 2 of 2: Where lone pairs sit: equatorial and trans

In a trigonal bipyramid an axial position has three neighbours at 90°, but an equatorial one has only two. A lone pair repels more than a bond pair, so it takes the roomier equatorial site. In an octahedron all sites are alike, and two lone pairs keep as far apart as they can, opposite each other.

Definition

Steric number 5 (trigonal bipyramid): lone pairs go equatorial, where each meets only two bond pairs at 90° instead of three.
One lone pair gives a see-saw ( $\mathrm{SF_4}$ ); two give a T-shape ( $\mathrm{ClF_3}$ , $\mathrm{BrF_3}$ ); three give a linear molecule ( $\mathrm{XeF_2}$ , $\mathrm{I_3^-}$ ), 180°.
Steric number 6 (octahedron): one lone pair gives a square pyramid ( $\mathrm{BrF_5}$ ); two lone pairs sit trans (180° apart) and give a square planar shape ( $\mathrm{XeF_4}$ , $\mathrm{ICl_4^-}$ , $\mathrm{BrF_4^-}$ ).
In $\mathrm{XeO_2F_2}$ (SN 5, one lone pair) the lone pair and the two O atoms are equatorial and the two F atoms axial, so F–Xe–F is near 180° and O–Xe–O about 105°.

Species	Bond pairs, lone pairs	Lone pairs sit	Shape
$\mathrm{SF_4}$ , $\mathrm{SeF_4}$	4, 1	Equatorial	See-saw
$\mathrm{ClF_3}$ , $\mathrm{BrF_3}$	3, 2	Both equatorial	T-shaped (bent T), about 87.5°
$\mathrm{XeF_2}$ , $\mathrm{I_3^-}$ , $\mathrm{ICl_2^-}$	2, 3	All three equatorial	Linear, 180°
$\mathrm{XeO_2F_2}$	4, 1	Equatorial, with the two O	See-saw, F atoms axial
$\mathrm{BrF_5}$ , $\mathrm{IF_5}$	5, 1	Any one octahedral site	Square pyramidal
$\mathrm{XeF_4}$ , $\mathrm{BrF_4^-}$	4, 2	Trans, opposite each other	Square planar, 90°
$\mathrm{BrF_2^+}$	2, 2	Two corners of a tetrahedron	Bent

Five pairs: lone pairs equatorial. Six pairs: two lone pairs trans.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 June 2022 · Q123Moderate

Example 2 · Chemical Bonding and Molecular Structure · VSEPR Theory and Lone Pairs

In the structure of

SF_{4}

, the lone pair of electrons on

S

is in.

Lone pairs are never axial in a trigonal bipyramid

An axial lone pair would meet three bond pairs at 90°. The stable structure puts every lone pair equatorial, so a statement that axial lone pairs minimise repulsion in

\mathrm{ClF_3}

is false.

Three lone pairs make a straight molecule

\mathrm{XeF_2}

and

\mathrm{I_3^-}

have five electron pairs, yet they are linear. The three lone pairs fill the equator and the two atoms sit on the axis, 180° apart.

Drill 5 more on where lone pairs sit: equatorial and trans

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 lone pairs on the central atom
Steric number and lone pairs
$SN = \tfrac{1}{2}(V + M - c + a) \qquad \text{lone pairs} = SN - (\text{atoms bonded to the centre})$

Reference tables (1)

Where lone pairs sit: equatorial and trans7 rows

Species	Bond pairs, lone pairs	Lone pairs sit	Shape
$\mathrm{SF_4}$ , $\mathrm{SeF_4}$	4, 1	Equatorial	See-saw
$\mathrm{ClF_3}$ , $\mathrm{BrF_3}$	3, 2	Both equatorial	T-shaped (bent T), about 87.5°
$\mathrm{XeF_2}$ , $\mathrm{I_3^-}$ , $\mathrm{ICl_2^-}$	2, 3	All three equatorial	Linear, 180°
$\mathrm{XeO_2F_2}$	4, 1	Equatorial, with the two O	See-saw, F atoms axial
$\mathrm{BrF_5}$ , $\mathrm{IF_5}$	5, 1	Any one octahedral site	Square pyramidal
$\mathrm{XeF_4}$ , $\mathrm{BrF_4^-}$	4, 2	Trans, opposite each other	Square planar, 90°
$\mathrm{BrF_2^+}$	2, 2	Two corners of a tetrahedron	Bent

Five pairs: lone pairs equatorial. Six pairs: two lone pairs trans.

Watch out for (4)

A double bond to oxygen uses two electrons
→ Counting lone pairs on the central atom
Know what the question calls a bond pair
→ Counting lone pairs on the central atom
Lone pairs are never axial in a trigonal bipyramid
→ Where lone pairs sit: equatorial and trans
Three lone pairs make a straight molecule
→ Where lone pairs sit: equatorial and trans

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