MHT-CET Chemistry · Chemical Bonding and Molecular Structure

Hybridization

Hybridization mixes an atom's pure s, p (and sometimes d) atomic orbitals into an equal number of identical hybrid orbitals — and the count of those hybrids, the steric number, fixes the molecule's shape and bond angle.

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

Why this matters

Seven PYQs, mostly EASY recall plus one MODERATE — the most reliable single topic in Chemical Bonding. They come in three shapes: name-the-hybridization-and-geometry (CH4 -> sp3, tetrahedral; SF4 -> sp3d, see-saw), pick-the-odd-one-out (which molecule is NOT sp3), and match hybridization to shape (sp2 -> trigonal planar). Every one of them is settled by counting the steric number = bond pairs + lone pairs on the central atom, so master that one count and you never drop a mark here.

Concept 1 of 4: Valence bond theory: sigma and pi bonds

Valence bond theory pictures a covalent bond as the overlap of two half-filled atomic orbitals. Head-on (axial) overlap gives a strong sigma bond; sideways overlap of parallel p-orbitals gives a weaker pi bond. Count them by bond multiplicity: single = one sigma, double = one sigma + one pi, triple = one sigma + two pi.

Definition

Valence bond theory (Heitler-London, Pauling):

A covalent bond forms when two half-filled atomic orbitals overlap and pair their electrons; the greater the overlap, the stronger the bond.
Sigma $\sigma$ bond — end-on (axial) overlap along the internuclear axis (s-s, s-p or p-p axial). Strong; allows free rotation.
Pi $\pi$ bond — sideways (lateral) overlap of two parallel p-orbitals, above and below the axis. Weaker than sigma; no free rotation.
Bond multiplicity: single $= 1\sigma$ ; double $= 1\sigma + 1\pi$ ; triple $= 1\sigma + 2\pi$ . The first bond between two atoms is always a sigma bond.

Bond	Sigma and pi	Example
Single bond	$1\sigma$	$\text{H}-\text{H}$ ; C-C in ethane
Double bond	$1\sigma + 1\pi$	$\text{C}=\text{C}$ in ethene; $\text{O}=\text{O}$
Triple bond	$1\sigma + 2\pi$	$\text{C}\equiv\text{C}$ in ethyne; $\text{N}\equiv\text{N}$

The first bond between two atoms is always a sigma bond; any extra bonds are pi.

Practice this concept5 quick reps

The first bond is always sigma

In any single, double or triple bond the FIRST bond is a sigma bond; only the additional bonds are pi. So a double bond is

1\sigma + 1\pi

(not

2\pi

), and a triple bond is

1\sigma + 2\pi

Sigma is stronger than pi; pi locks rotation

Axial overlap in a sigma bond is more effective than the sideways overlap of a pi bond, so a sigma bond is stronger. A pi bond also fixes the geometry — there is no free rotation about a double or triple bond.

Concept 2 of 4: What hybridization is

Pure atomic orbitals (one s, three p, ...) have different shapes and energies, so they cannot on their own explain why the four C-H bonds in methane are identical. Hybridization is the idea that the central atom blends its valence orbitals into a new set of equivalent hybrid orbitals — as many hybrids as the orbitals mixed — which then point in fixed directions and hold the bonds.

Definition

Hybridization is the intermixing of atomic orbitals of nearly equal energy on the same atom to form an equal number of new, identical hybrid orbitals:

The number of hybrid orbitals formed = number of atomic orbitals mixed. One s + three p give four $sp^3$ hybrids.
The hybrids are equivalent in shape and energy and point in fixed directions, which is what sets the molecular shape and bond angle.
More s-character pulls the electrons closer to the nucleus and widens the bond angle: $sp$ (50% s) $\to 180^\circ$ , $sp^2$ (33% s) $\to 120^\circ$ , $sp^3$ (25% s) $\to 109.5^\circ$ .
Each hybrid orbital forms one $\sigma$ bond or holds one lone pair; $\pi$ bonds use the leftover unhybridized p orbitals.

Worked example

Explain why carbon in methane forms four identical bonds at 109.5 degrees, and name its hybridization.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q74Easy

Example 2 · Chemical Bonding and Molecular Structure · Hybridization

Which hybridisation and geometry does

\text{CH}_4

exhibit?

Hybrids formed = orbitals mixed, not bonds made

The number of hybrid orbitals equals the number of atomic orbitals that intermix, not the number of bonds. Water's oxygen is

sp^3

(four hybrids) even though it makes only two bonds — the other two hybrids hold lone pairs.

Concept 3 of 4: The steric-number master table

You do not need to memorise a shape for every molecule — you need one count. Add the bond pairs and lone pairs on the central atom to get the steric number, and the steric number maps straight to the hybridization, the geometry and the bond angle.

Definition

The steric number (SN) = (number of $\sigma$ -bonded atoms) + (number of lone pairs) on the central atom. It fixes everything:

SN 2 $\to sp$ , linear, $180^\circ$ .
SN 3 $\to sp^2$ , trigonal planar, $120^\circ$ .
SN 4 $\to sp^3$ , tetrahedral, $109.5^\circ$ .
SN 5 $\to sp^3d$ , trigonal bipyramidal, $90^\circ$ and $120^\circ$ .
SN 6 $\to sp^3d^2$ , octahedral, $90^\circ$ .

Lone pairs count toward SN (so they set the hybridization) but the shape name you report is the arrangement of the atoms only.

Steric number	Hybridization	Geometry	Bond angle	Example
2	$sp$	Linear	$180^\circ$	$BeCl_2$ , $C_2H_2$ Q Acetylene $C_2H_2$ has $sp$ carbons (two atoms + one triple bond that counts as one $\sigma$ ) — the bank's classic $sp$ example.
3	$sp^2$	Trigonal planar	$120^\circ$	$BF_3$ , $C_2H_4$ Q Trigonal planar geometry means $sp^2$ — the answer to 'which hybridisation gives trigonal geometry'.
4	$sp^3$	Tetrahedral	$109.5^\circ$	$CH_4$ , $NH_3$ , $H_2O$
5	$sp^3d$	Trigonal bipyramidal	$90^\circ,\,120^\circ$	$PCl_5$ , $SF_4$ Q $SF_4$ is $sp^3d$ (4 bond pairs + 1 lone pair = SN 5); the lone pair distorts it to a see-saw shape but the hybridization stays $sp^3d$ .
6	$sp^3d^2$	Octahedral	$90^\circ$	$SF_6$ , $XeF_4$ Q $XeF_4$ is $sp^3d^2$ (4 bond pairs + 2 lone pairs = SN 6), square planar — NOT $sp^3$ .

Count the steric number first; the row it lands in gives the hybridization, geometry and angle.

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q87Easy

Example 3 · Chemical Bonding and Molecular Structure · Hybridization

Which of the following types of hybridisation result in trigonal geometry?

Geometry name reports atoms only; SN includes lone pairs

SF_4

has steric number 5, so it is

sp^3d

— but its shape is called see-saw, not trigonal bipyramidal, because one of the five positions is a lone pair. The hybridization follows the steric number; the shape name follows only the bonded atoms.

d-orbitals only appear from steric number 5

sp^3d

and

sp^3d^2

need d-orbitals and only occur for central atoms with an expanded octet (period 3 and below). If the steric number is 4 or less, the answer is

sp

sp^2

sp^3

— never a d-form.

Drill 3 more on the steric-number master table

Concept 4 of 4: Determining a central atom's hybridization

Given a formula like

H_2O

XeF_4

, you find the hybridization by drawing (or picturing) the central atom, counting how many atoms are bonded to it, and adding the lone pairs it carries. That sum, the steric number, reads straight off the master table — the whole trick is remembering to count the lone pairs, which the distractors bank on you forgetting.

Definition

To find the hybridization of the central atom:

Count the $\sigma$ -bonded atoms attached to it (a double or triple bond still counts as one $\sigma$ bond, i.e. one bonded atom).
Add the lone pairs on the central atom: lone pairs $= \dfrac{V - (\text{bonding electrons used})}{2}$ , or just read them from the Lewis structure.
Steric number = bonded atoms + lone pairs; then map SN to hybridization: 2 $\to sp$ , 3 $\to sp^2$ , 4 $\to sp^3$ , 5 $\to sp^3d$ , 6 $\to sp^3d^2$ .

Steric number

\text{SN} = (\sigma\text{-bonded atoms}) + (\text{lone pairs on central atom})

$\text{SN}$ steric number, which fixes the hybridization
$\sigma\text{-bonded atoms}$ atoms directly bonded (each multiple bond counts once)
$\text{lone pairs}$ non-bonding electron pairs on the central atom

Worked example

Determine the hybridization of the central atom in water,

H_2O

Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q93Easy

Example 4 · Chemical Bonding and Molecular Structure · Hybridization

Identify the molecule in which central atom undergoes sp

^3

hybridisation?

Count lone pairs on the central atom, not just the atoms

H_2O

and

NH_3

each have only 2 or 3 bonded atoms, yet both are

sp^3

— because oxygen carries 2 lone pairs and nitrogen 1. If you count only the bonded atoms you would wrongly call water

sp

and ammonia

sp^2

. Always add the lone pairs.

A multiple bond is one atom, not two, in the steric count

In acetylene

C_2H_2

each carbon is bonded to one H and one C (a triple bond). The triple bond counts as one

\sigma

bond, so the steric number is 2, giving

sp

— not

sp^3

. Never count the extra

\pi

bonds toward the steric number.

Drill 1 more on determining a central atom's hybridization

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)

Determining a central atom's hybridization
Steric number
$\text{SN} = (\sigma\text{-bonded atoms}) + (\text{lone pairs on central atom})$

Reference tables (2)

Valence bond theory: sigma and pi bonds3 rows

Bond	Sigma and pi	Example
Single bond	$1\sigma$	$\text{H}-\text{H}$ ; C-C in ethane
Double bond	$1\sigma + 1\pi$	$\text{C}=\text{C}$ in ethene; $\text{O}=\text{O}$
Triple bond	$1\sigma + 2\pi$	$\text{C}\equiv\text{C}$ in ethyne; $\text{N}\equiv\text{N}$

The first bond between two atoms is always a sigma bond; any extra bonds are pi.

The steric-number master table5 rows

Steric number	Hybridization	Geometry	Bond angle	Example
2	$sp$	Linear	$180^\circ$	$BeCl_2$ , $C_2H_2$ Q Acetylene $C_2H_2$ has $sp$ carbons (two atoms + one triple bond that counts as one $\sigma$ ) — the bank's classic $sp$ example.
3	$sp^2$	Trigonal planar	$120^\circ$	$BF_3$ , $C_2H_4$ Q Trigonal planar geometry means $sp^2$ — the answer to 'which hybridisation gives trigonal geometry'.
4	$sp^3$	Tetrahedral	$109.5^\circ$	$CH_4$ , $NH_3$ , $H_2O$
5	$sp^3d$	Trigonal bipyramidal	$90^\circ,\,120^\circ$	$PCl_5$ , $SF_4$ Q $SF_4$ is $sp^3d$ (4 bond pairs + 1 lone pair = SN 5); the lone pair distorts it to a see-saw shape but the hybridization stays $sp^3d$ .
6	$sp^3d^2$	Octahedral	$90^\circ$	$SF_6$ , $XeF_4$ Q $XeF_4$ is $sp^3d^2$ (4 bond pairs + 2 lone pairs = SN 6), square planar — NOT $sp^3$ .

Count the steric number first; the row it lands in gives the hybridization, geometry and angle.

Watch out for (7)

The first bond is always sigma
→ Valence bond theory: sigma and pi bonds
Sigma is stronger than pi; pi locks rotation
→ Valence bond theory: sigma and pi bonds
Hybrids formed = orbitals mixed, not bonds made
→ What hybridization is
Geometry name reports atoms only; SN includes lone pairs
→ The steric-number master table
d-orbitals only appear from steric number 5
→ The steric-number master table
Count lone pairs on the central atom, not just the atoms
→ Determining a central atom's hybridization
A multiple bond is one atom, not two, in the steric count
→ Determining a central atom's hybridization

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