Hybridization — Concepts, Formulas & Examples

Hybridisation explains the shapes and bonding of molecules by mixing atomic orbitals to form new hybrid orbitals. CBSE Class 11 and NEET test shape prediction frequently. This is a core chemical bonding concept and one of the most scoring topics once you learn the counting shortcut.

Here is the central idea: isolated atoms have s, p and d orbitals with different shapes and energies. But when an atom forms bonds, it rearranges those orbitals into equivalent hybrids that point in specific directions. Carbon does not use one 2s and three 2p orbitals separately — it mixes them into four identical $sp^3$ hybrids, each pointing to a corner of a tetrahedron. This mixing is hybridisation, and it is the bridge between atomic orbital theory and the shapes we actually observe.

Core Concepts

What hybridisation means

Mixing of atomic orbitals on the same atom to form new hybrid orbitals of equal energy, each suited to form a sigma bond or hold a lone pair. The number of hybrid orbitals equals the number of atomic orbitals mixed. The new orbitals have shapes intermediate between the parent orbitals — fatter on one side than the other, optimised for head-on overlap.

Key rules:

Only orbitals on the central atom hybridise
The number of hybrid orbitals = number of sigma bonds + number of lone pairs on the central atom
Unhybridised p (or d) orbitals are available for pi bonding

sp3 hybridisation

One s and three p orbitals mix to give four sp3 hybrids. They point to the corners of a tetrahedron, 109.5° apart.

Molecule	Bonded pairs	Lone pairs	Shape	Bond angle
$\text{CH}_4$	4	0	Tetrahedral	109.5°
$\text{NH}_3$	3	1	Trigonal pyramidal	107°
$\text{H}_2\text{O}$	2	2	Bent (V-shape)	104.5°

Notice the pattern: same hybridisation ( $sp^3$ ), but increasing lone pairs compress the bond angle because lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion.

sp2 hybridisation

One s and two p orbitals mix to give three sp2 hybrids plus one unhybridised p orbital. The three hybrids lie in a plane at 120°. The unhybridised p orbital is perpendicular to this plane and available for pi bonding.

Molecule	Bonded pairs	Lone pairs	Shape	Bond angle
$\text{BF}_3$	3	0	Trigonal planar	120°
$\text{C}_2\text{H}_4$ (ethene)	3 (on each C)	0	Planar around each C	~120°
$\text{SO}_2$	2	1	Bent	~119°

In ethene, each carbon is $sp^2$ . The unhybridised p orbitals on the two carbons overlap sideways to form the pi bond — the second bond in a C=C double bond. The sigma bond is from $sp^2$ - $sp^2$ head-on overlap.

sp hybridisation

One s and one p orbital mix to give two sp hybrids plus two unhybridised p orbitals. Linear geometry, 180°.

Molecule	Bonded pairs	Shape	Bond angle
$\text{BeCl}_2$	2	Linear	180°
$\text{C}_2\text{H}_2$ (ethyne)	2 (on each C)	Linear	180°
$\text{CO}_2$	2	Linear	180°

In ethyne, each carbon is $sp$ . Two unhybridised p orbitals on each carbon form two pi bonds — giving the C $\equiv$ C triple bond (one sigma + two pi).

Extended hybridisations (sp3d and sp3d2)

For elements in period 3 and beyond that have d orbitals available:

Hybridisation	Shape (no lone pairs)	Example	Bond angle
$sp^3d$	Trigonal bipyramidal	$\text{PCl}_5$	90° and 120°
$sp^3d^2$	Octahedral	$\text{SF}_6$	90°

With lone pairs, the shape changes:

$sp^3d$ with 1 lone pair → seesaw ( $\text{SF}_4$ )
$sp^3d$ with 2 lone pairs → T-shape ( $\text{ClF}_3$ )
$sp^3d$ with 3 lone pairs → linear ( $\text{XeF}_2$ )
$sp^3d^2$ with 1 lone pair → square pyramidal ( $\text{BrF}_5$ )
$sp^3d^2$ with 2 lone pairs → square planar ( $\text{XeF}_4$ )

The counting shortcut

This is the most practical trick for exams:

H = \text{(sigma bonds)} + \text{(lone pairs on central atom)}

$H = 2 \to sp$ , $H = 3 \to sp^2$ , $H = 4 \to sp^3$ , $H = 5 \to sp^3d$ , $H = 6 \to sp^3d^2$

This shortcut works for 95% of exam questions. Count the sigma bonds (each single bond is one sigma, each double bond has one sigma, each triple bond has one sigma) and add the lone pairs on the central atom. The total gives the hybridisation directly.

Shapes from hybridisation — the complete table

$H$ value	Hybridisation	Electron pair geometry	0 LP	1 LP	2 LP	3 LP
2	$sp$	Linear	Linear	—	—	—
3	$sp^2$	Trigonal planar	Trigonal planar	Bent	—	—
4	$sp^3$	Tetrahedral	Tetrahedral	Pyramidal	Bent	—
5	$sp^3d$	Trigonal bipyramidal	TBP	Seesaw	T-shape	Linear
6	$sp^3d^2$	Octahedral	Octahedral	Square pyramidal	Square planar	—

LP = lone pair. This table is your one-page answer sheet for all shape questions.

Sigma and pi bonds

Sigma bond: head-on overlap of hybrid orbitals. Stronger, allows free rotation. Every single bond is one sigma bond.
Pi bond: sideways overlap of unhybridised p orbitals. Weaker, restricts rotation. A double bond = 1 sigma + 1 pi. A triple bond = 1 sigma + 2 pi.

In a molecule: count each single bond as 1 $\sigma$ , each double bond as 1 $\sigma$ + 1 $\pi$ , each triple bond as 1 $\sigma$ + 2 $\pi$ .

Example: $\text{CH}_3\text{CHO}$ (acetaldehyde) has 6 single bonds + 1 double bond = 7 $\sigma$ + 1 $\pi$ .

Worked Examples

Both are $sp^3$ hybridised. $\text{CH}_4$ has four bonded pairs and no lone pairs — perfect tetrahedron with 109.5°. $\text{NH}_3$ has three bonded pairs plus one lone pair, which repels more strongly than a bonding pair, compressing the H-N-H angle to 107°. The lone pair ‘pushes’ the bonding pairs closer together.

Each carbon in $\text{C}_2\text{H}_4$ is $sp^2$ hybridised. Two of the three $sp^2$ orbitals overlap with hydrogen 1s orbitals (C-H sigma bonds). The third overlaps with the $sp^2$ orbital of the other carbon (C-C sigma bond). The remaining unhybridised p orbital on each carbon overlaps sideways to form the C=C pi bond. The pi bond prevents rotation — ethene is planar.

In $\text{CH}_4$ : C has 4 sigma bonds, 0 lone pairs → $H = 4$ → $sp^3$ → tetrahedral.

In $\text{C}_2\text{H}_4$ : each C has 3 sigma bonds (2 C-H + 1 C-C), 0 lone pairs → $H = 3$ → $sp^2$ → planar around each C.

In $\text{C}_2\text{H}_2$ : each C has 2 sigma bonds (1 C-H + 1 C-C), 0 lone pairs → $H = 2$ → $sp$ → linear.

Carbon goes from $sp^3$ to $sp^2$ to $sp$ as bond order increases from single to double to triple.

Xe in $\text{XeF}_2$ : 2 sigma bonds + 3 lone pairs = $H = 5$ → $sp^3d$ hybridised. Electron pair geometry is trigonal bipyramidal. The three lone pairs occupy equatorial positions (more space), and the two F atoms occupy axial positions. Result: linear shape with 180° bond angle. Molecular shape ≠ electron pair geometry when lone pairs are present.

$\text{CH}_3\text{COOH}$ : The molecule has: 3 C-H bonds (3 $\sigma$ ), 1 C-C bond (1 $\sigma$ ), 1 C=O bond (1 $\sigma$ + 1 $\pi$ ), 1 C-O bond (1 $\sigma$ ), 1 O-H bond (1 $\sigma$ ). Total: 7 sigma bonds and 1 pi bond. The carbon in $\text{CH}_3$ is $sp^3$ ; the carbon in COOH is $sp^2$ .

S in $\text{SO}_3$ : 3 sigma bonds (each S=O double bond has one sigma), 0 lone pairs on S → $H = 3$ → $sp^2$ hybridised → trigonal planar, 120°. Each S=O bond also has a pi component from the unhybridised p orbital.

Common Mistakes

Saying lone pairs do not affect shape. They affect molecular shape (but not electron pair geometry). $\text{NH}_3$ and $\text{CH}_4$ have the same electron pair geometry (tetrahedral) but different molecular shapes (pyramidal vs tetrahedral) because of the lone pair.

Confusing hybridisation and bond order. Hybridisation is about orbital mixing and shape prediction. Bond order is about the number of shared electron pairs between two specific atoms. An $sp^2$ carbon can have both single and double bonds.

Writing that all $sp^3$ molecules are tetrahedral. Only those without lone pairs are tetrahedral. With one lone pair → pyramidal ( $\text{NH}_3$ ). With two lone pairs → bent ( $\text{H}_2\text{O}$ ). Always account for lone pairs when naming the molecular shape.

Counting pi bonds when determining hybridisation. The shortcut counts only sigma bonds and lone pairs. Pi bonds use unhybridised orbitals and do not participate in hybridisation. Including pi bonds gives a wrong $H$ value.

Assuming that period 2 elements can have expanded octets. Carbon, nitrogen and oxygen cannot be $sp^3d$ or $sp^3d^2$ hybridised because they have no accessible d orbitals. Extended hybridisation is only for period 3 and beyond (Si, P, S, Cl, etc.).

Exam Weightage and Strategy

Hybridisation is tested in CBSE Class 11 boards (3-5 marks), NEET (1-2 questions per year) and JEE Main (1-2 questions). The most common question formats are: (a) predict the hybridisation and shape of a given molecule, (b) arrange molecules in order of bond angle, (c) count sigma and pi bonds in an organic molecule.

NEET PYQ favourites:

What is the hybridisation of carbon in $\text{C}_2\text{H}_2$ ? ( $sp$ )
Which of the following is planar? (anything with $sp^2$ and no lone pairs)
How many sigma and pi bonds in $\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2$ ?
What is the shape of $\text{XeF}_4$ ? (square planar)

Use the count rule — sigma bonds plus lone pairs on the central atom gives $H$ , which directly tells the hybridisation. Practice it on 20 molecules and it becomes automatic. This single shortcut handles 90% of hybridisation questions in any exam.

Practice Questions

Q1. Predict the hybridisation, electron pair geometry and molecular shape of $\text{ClF}_3$ .

Cl in $\text{ClF}_3$ : 3 sigma bonds + 2 lone pairs = $H = 5$ → $sp^3d$ . Electron pair geometry: trigonal bipyramidal. The 2 lone pairs occupy equatorial positions. Molecular shape: T-shaped. Bond angles: approximately 87° (slightly less than 90° due to lone pair compression).

Q2. Why is the bond angle in $\text{H}_2\text{O}$ (104.5°) less than in $\text{NH}_3$ (107°)?

Both are $sp^3$ . $\text{NH}_3$ has 1 lone pair, which compresses the bond angle from 109.5° to 107°. $\text{H}_2\text{O}$ has 2 lone pairs, providing even more repulsion, compressing the angle further to 104.5°. More lone pairs = more compression = smaller bond angle.

Q3. How many sigma and pi bonds are in $\text{HC} \equiv \text{C} - \text{CH} = \text{CH}_2$ ?

$\text{HC} \equiv \text{C}$ : 1 C-H sigma, 1 C $\equiv$ C = 1 sigma + 2 pi. $\text{C}-\text{CH}$ : 1 C-C sigma. $\text{CH}=\text{CH}_2$ : 1 C=C = 1 sigma + 1 pi, 1 C-H sigma, 2 C-H sigmas. Total: 7 sigma bonds and 3 pi bonds. The first carbon is $sp$ , the second is $sp$ , the third is $sp^2$ , the fourth is $sp^2$ .

Q4. Why is $\text{BF}_3$ a Lewis acid?

B in $\text{BF}_3$ is $sp^2$ hybridised with an empty unhybridised p orbital. This empty orbital can accept an electron pair from a Lewis base (like $\text{NH}_3$ ). When it does, the hybridisation changes to $sp^3$ and the shape changes from trigonal planar to tetrahedral. The product $\text{BF}_3 \cdot \text{NH}_3$ is a Lewis acid-base adduct.

Q5. What is the hybridisation of each carbon in $\text{CH}_3\text{COCH}_3$ (acetone)?

The two $\text{CH}_3$ carbons: each has 4 sigma bonds (3 C-H + 1 C-C), 0 lone pairs → $sp^3$ . The central carbonyl carbon: 3 sigma bonds (1 C=O counted as 1 sigma + 1 C-C + 1 C-C), 0 lone pairs → $sp^2$ . The C=O has one sigma and one pi bond, with the pi bond from the unhybridised p orbital.

FAQs

Does hybridisation actually happen in nature?

Hybridisation is a mathematical model, not a physical process. Atoms do not literally ‘mix’ orbitals in real time. The model correctly predicts shapes and bond angles, which is why we use it. Molecular orbital theory is more rigorous but harder to apply quickly in exams.

Why can carbon form four bonds when it has only two unpaired electrons?

In the ground state, carbon has the configuration $1s^2 2s^2 2p^2$ with only two unpaired p electrons. During bonding, one 2s electron is promoted to the empty 2p orbital, giving four unpaired electrons ( $2s^1 2p^3$ ). These four orbitals then hybridise. The energy cost of promotion is more than compensated by the energy released in forming four bonds instead of two.

What is the hybridisation of the oxygen in water?

$sp^3$ . Oxygen has 2 bond pairs (O-H) and 2 lone pairs → $H = 4$ → $sp^3$ . The electron pair geometry is tetrahedral, but the molecular shape is bent (V-shaped) with a bond angle of 104.5°.

Can nitrogen be $sp^3d$ hybridised?

No. Nitrogen is in period 2 and has no accessible d orbitals. It can only be $sp$ , $sp^2$ or $sp^3$ hybridised. The maximum number of bonds/lone pairs it can accommodate is four ( $sp^3$ ). Elements from period 3 onward (P, S, Cl, etc.) can use d orbitals for expanded hybridisation.

Hybridisation is the bridge between atomic orbitals and molecular shapes. Master the counting shortcut and the shape table, and this becomes one of the most reliable scoring topics in chemistry.