JEE Mains Chemistry · Chemical Bonding and Molecular Structure

Hybridisation and Sigma and Pi Bonds

Every bonded pair of atoms shares one σ bond and any extra bonds are π; the central atom mixes one hybrid orbital for each σ bond and lone pair, so the steric number names the hybridisation.

All Chemical Bonding and Molecular Structure notes JEE Mains Chemistry strategy

Why this matters

Twenty-seven PYQs, nineteen of them multiple choice, and two from 2026. Eight count σ and π bonds, name the hybridisation of each carbon in a chain or test how orbitals overlap; eleven name or count the hybridisation of a central atom; eight match hybridisations to shapes, complexes such as [PtCl₄]²⁻ included. Three ideas cover the page.

Concept 1 of 3: Counting σ and π bonds

Two atoms joined by any bond share exactly one σ bond, formed by head-on overlap along the line between them. Extra bonds come from sideways overlap of p orbitals: one π bond in a double bond, two in a triple bond. So σ bonds count the connections and π bonds count the extra lines.

Definition

σ bonds = number of bonded pairs of atoms. In an open chain this is (number of atoms − 1); each ring adds one.
π bonds: 1 per double bond, 2 per triple bond.
Write out every hydrogen before counting; the name fixes the structure.
Carbon hybridisation: four σ bonds → $sp^3$ ; one double bond → $sp^2$ ; a triple bond or two double bonds → $sp$ . In allene $\mathrm{CH_2{=}C{=}CH_2}$ the middle carbon is $sp$ .
Overlap can be positive (in phase, bonding), negative (out of phase) or zero (orbitals whose orientation makes them cancel, as s with a $p_x$ approaching along z).

Counting bonds in an open chain

n_\sigma = N_{\text{atoms}} - 1 \ (+1 \text{ per ring}) \qquad n_\pi = n_{\text{double}} + 2\,n_{\text{triple}}

Worked example

Count the σ and π bonds in but-1-en-3-yne,

\mathrm{CH_2{=}CH{-}C{\equiv}CH}

, and give the hybridisation of each carbon.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q48Moderate

Example 1 · Chemical Bonding and Molecular Structure · Hybridisation and Sigma and Pi Bonds

The sum of sigma

(\sigma)

and

pi(\pi)

bonds in Hex-1,3-dien-5-yne is

\_\_\_\_

Every C–H bond is a σ bond

The hydrogens are easy to forget because the name does not list them. Draw the full structure: in a six-carbon chain the C–H bonds are often more than half of the σ total.

Give σ and π in the order asked

Options often list the same pair twice, as '13 and 3' and '3 and 13'. The question says which comes first; match that order.

Drill 7 more on counting σ and π bonds

Concept 2 of 3: Hybridisation from the steric number

The central atom needs one hybrid orbital for each σ bond and each lone pair. π bonds use leftover p orbitals, so they do not count. So the steric number tells you how many orbitals were mixed, and that names the hybridisation.

Definition

Steric number = σ bonds + lone pairs on the central atom (or $\tfrac{1}{2}(V + M - c + a)$ , as on the VSEPR page).
2 → $sp$ ; 3 → $sp^2$ ; 4 → $sp^3$ ; 5 → $sp^3d$ ; 6 → $sp^3d^2$ ; 7 → $sp^3d^3$ .
Double bonds count once: $\mathrm{SO_2}$ (2 σ + 1 lone pair) is $sp^2$ ; $\mathrm{XeOF_4}$ (5 σ + 1 lone pair) is $sp^3d^2$ .
In a network solid count the real neighbours: Si in $\mathrm{SiO_2}$ bonds to four O, so it is $sp^3$ .
Nitrogen species: $\mathrm{NO_2^-}$ $sp^2$ , $\mathrm{NO_2^+}$ $sp$ , $\mathrm{N_3^-}$ (central N) $sp$ , $\mathrm{NH_4^+}$ $sp^3$ .

Steric number to hybridisation

SN = n_\sigma + n_{\text{lp}}: \ 2 \to sp,\ 3 \to sp^2,\ 4 \to sp^3,\ 5 \to sp^3d,\ 6 \to sp^3d^2,\ 7 \to sp^3d^3

Worked example

How many of these have an

sp^3

central atom:

\mathrm{H_2O}

\mathrm{BF_3}

\mathrm{NH_4^+}

\mathrm{SO_4^{2-}}

\mathrm{CO_2}

\mathrm{PCl_3}

\mathrm{SO_3}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q46Moderate

Example 2 · Chemical Bonding and Molecular Structure · Hybridisation and Sigma and Pi Bonds

Consider the following species

{BrF}_{5},{XeF}_{5}^{-},{BF}_{4}^{-},{ICl}_{4}^{-},{XeF}_{4},{SF}_{4},{NH}_{4}^{+},{ClF}_{3},{XeF}_{2},{ICl}_{2}^{-}

Number of species having sp

\ ^{3}\text{ }d

hybridized central atom is

\_\_\_\_

π bonds do not add hybrid orbitals

\mathrm{SO_3}

has three S=O double bonds but only three σ bonds, so it is

sp^2

, not

sp^3d^2

. Count σ bonds and lone pairs only.

Five atoms around the centre can still be sp³d²

\mathrm{BrF_5}

has five bonds and one lone pair, a steric number of 6:

sp^3d^2

, not

sp^3d

. Only lone-pair-free

\mathrm{PF_5}

and

\mathrm{PCl_5}

are

sp^3d

with five bonds.

Drill 10 more on hybridisation from the steric number

Concept 3 of 3: Hybridisation, orientation and complexes

Each hybridisation points its orbitals in a fixed pattern, so it fixes the arrangement of pairs. Transition-metal complexes add two more patterns: inner d orbitals can join in (d²sp³, dsp²) when strong ligands pair up the metal's electrons.

Definition

Main-group hybridisations use the outer d orbitals: $sp^3d$ , $sp^3d^2$ , $sp^3d^3$ .
Complexes with strong-field ligands ( $\mathrm{CN^-}$ , $\mathrm{NH_3}$ on $\mathrm{Co^{3+}}$ ) pair the d electrons and use inner (n−1)d orbitals: $d^2sp^3$ octahedral, $dsp^2$ square planar.
Weak-field ligands ( $\mathrm{F^-}$ , $\mathrm{Cl^-}$ ) leave the d electrons unpaired: $sp^3$ tetrahedral or $sp^3d^2$ octahedral (outer orbital).
$\mathrm{Ni(CO)_4}$ : Ni(0) is $3d^{10}$ , so it uses 4s and 4p: $sp^3$ , tetrahedral.

Hybridisation	Orientation	Main-group examples	Complex examples
$sp$	Linear, 180°	$\mathrm{BeCl_2}$ , $\mathrm{CO_2}$ , $\mathrm{NO_2^+}$	$\mathrm{[Ag(NH_3)_2]^+}$
$sp^2$	Trigonal planar, 120°	$\mathrm{BF_3}$ , $\mathrm{SO_2}$ , $\mathrm{NO_2^-}$	Rare in complexes
$sp^3$	Tetrahedral, 109.5°	$\mathrm{CH_4}$ , $\mathrm{NH_4^+}$ , $\mathrm{XeO_3}$	$\mathrm{Ni(CO)_4}$ , $\mathrm{[NiCl_4]^{2-}}$
$dsp^2$	Square planar, 90°	None	$\mathrm{[PtCl_4]^{2-}}$ , $\mathrm{[Ni(CN)_4]^{2-}}$
$sp^3d$	Trigonal bipyramidal	$\mathrm{PCl_5}$ , $\mathrm{SF_4}$ , $\mathrm{ClF_3}$ , $\mathrm{XeF_2}$	$\mathrm{Fe(CO)_5}$ is often written $dsp^3$
$sp^3d^2$	Octahedral, 90°	$\mathrm{SF_6}$ , $\mathrm{BrF_5}$ , $\mathrm{XeF_4}$	$\mathrm{[CoF_6]^{3-}}$ (outer orbital)
$d^2sp^3$	Octahedral, 90°	None	$\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Fe(CN)_6]^{3-}}$ (inner orbital) [Co(NH₃)₆]³⁺ is d²sp³, not sp³d²: a stated match of it with SF₆ is false.
$sp^3d^3$	Pentagonal bipyramidal	$\mathrm{IF_7}$ , $\mathrm{XeF_6}$ (distorted)	None

The hybridisation fixes the arrangement of pairs; the shape then depends on how many are lone pairs.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 3 Apr 2025 · Q43Moderate

Example 3 · Chemical Bonding and Molecular Structure · Hybridisation and Sigma and Pi Bonds

Match the LIST-I with LIST-II

	LIST-I (Molecules/ion)		LIST-II (Hybridisation of central atom)
A.	${PF}_{5}$	I	${dsp}^{2}$
B.	${SF}_{6}$	II.	${sp}^{3}\text{ }d$
C.	$Ni(CO)_{4}$	III.	${sp}^{3}{\text{ }d}^{2}$
D.	$\left\lbrack {PtCl}_{4} \right\rbrack^{2 -}$	IV.	${sp}^{3}$

Choose the correct answer from the options given below:

sp³d with a lone pair gives unequal bonds

\mathrm{PF_5}

\mathrm{XeF_4}

\mathrm{SF_4}

and

\mathrm{XeF_2}

, only

\mathrm{SF_4}

sp^3d

, carries a lone pair and has two bond lengths.

\mathrm{XeF_2}

sp^3d

with lone pairs, but its two bonds are equal.

Square planar means dsp², not sp³

Four ligands do not always mean tetrahedral.

\mathrm{[PtCl_4]^{2-}}

uses one inner d orbital and is square planar;

\mathrm{[NiCl_4]^{2-}}

sp^3

and tetrahedral.

Drill 7 more on hybridisation, orientation and complexes

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 (2)

Counting σ and π bonds
Counting bonds in an open chain
$n_\sigma = N_{\text{atoms}} - 1 \ (+1 \text{ per ring}) \qquad n_\pi = n_{\text{double}} + 2\,n_{\text{triple}}$
Hybridisation from the steric number
Steric number to hybridisation
$SN = n_\sigma + n_{\text{lp}}: \ 2 \to sp,\ 3 \to sp^2,\ 4 \to sp^3,\ 5 \to sp^3d,\ 6 \to sp^3d^2,\ 7 \to sp^3d^3$

Reference tables (1)

Hybridisation, orientation and complexes8 rows

Hybridisation	Orientation	Main-group examples	Complex examples
$sp$	Linear, 180°	$\mathrm{BeCl_2}$ , $\mathrm{CO_2}$ , $\mathrm{NO_2^+}$	$\mathrm{[Ag(NH_3)_2]^+}$
$sp^2$	Trigonal planar, 120°	$\mathrm{BF_3}$ , $\mathrm{SO_2}$ , $\mathrm{NO_2^-}$	Rare in complexes
$sp^3$	Tetrahedral, 109.5°	$\mathrm{CH_4}$ , $\mathrm{NH_4^+}$ , $\mathrm{XeO_3}$	$\mathrm{Ni(CO)_4}$ , $\mathrm{[NiCl_4]^{2-}}$
$dsp^2$	Square planar, 90°	None	$\mathrm{[PtCl_4]^{2-}}$ , $\mathrm{[Ni(CN)_4]^{2-}}$
$sp^3d$	Trigonal bipyramidal	$\mathrm{PCl_5}$ , $\mathrm{SF_4}$ , $\mathrm{ClF_3}$ , $\mathrm{XeF_2}$	$\mathrm{Fe(CO)_5}$ is often written $dsp^3$
$sp^3d^2$	Octahedral, 90°	$\mathrm{SF_6}$ , $\mathrm{BrF_5}$ , $\mathrm{XeF_4}$	$\mathrm{[CoF_6]^{3-}}$ (outer orbital)
$d^2sp^3$	Octahedral, 90°	None	$\mathrm{[Co(NH_3)_6]^{3+}}$ , $\mathrm{[Fe(CN)_6]^{3-}}$ (inner orbital) [Co(NH₃)₆]³⁺ is d²sp³, not sp³d²: a stated match of it with SF₆ is false.
$sp^3d^3$	Pentagonal bipyramidal	$\mathrm{IF_7}$ , $\mathrm{XeF_6}$ (distorted)	None

The hybridisation fixes the arrangement of pairs; the shape then depends on how many are lone pairs.

Watch out for (6)

Every C–H bond is a σ bond
→ Counting σ and π bonds
Give σ and π in the order asked
→ Counting σ and π bonds
π bonds do not add hybrid orbitals
→ Hybridisation from the steric number
Five atoms around the centre can still be sp³d²
→ Hybridisation from the steric number
sp³d with a lone pair gives unequal bonds
→ Hybridisation, orientation and complexes
Square planar means dsp², not sp³
→ Hybridisation, orientation and complexes

Test yourself on Chemical Bonding and Molecular Structure

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Bonding and Molecular Structure chapter test Drill all 27 questions