What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.

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Concept understanding — Orbital Hybridization Theory

Orbital Hybridization Theory – From Intuition to Precision

The Problem That Started It All

Imagine you are looking at a methane molecule, $\text{CH}_4$ . Carbon has four valence electrons: two in the $2s$ orbital and two in the $2p$ orbitals. If carbon used its pure atomic orbitals to bond, you would expect two bonds from the $2s$ (identical, but one direction) and two from the $2p$ (at $90^\circ$ to each other). That would give you three different bond types and bond angles of $90^\circ$ and something else.

But experiment says methane is perfectly tetrahedral: all four bonds are identical in length, strength, and energy, and the bond angle is $109.5^\circ$ , not $90^\circ$ . Something is fundamentally wrong with the "pure orbital" picture.

This is the puzzle that hybridization theory solves.

The Core Intuition

Think of atomic orbitals as shapes that an electron can occupy. The $s$ orbital is a sphere. The $p$ orbitals are dumbbells along the $x$ , $y$ , and $z$ axes. When an atom forms bonds, it wants to mix these shapes together to create new, hybrid shapes that point in directions that maximise bond strength and minimise repulsion.

It is like mixing primary colours to get new colours. You don't have to use red, blue, and yellow separately — you can blend them to get green, orange, or purple. Similarly, an atom can blend its $s$ and $p$ orbitals to get new hybrid orbitals that are better suited for bonding.

The key insight: hybridization is a mathematical mixing of atomic orbitals on the same atom to produce an equal number of new, equivalent hybrid orbitals. The number of hybrid orbitals formed always equals the number of atomic orbitals mixed.

The Precise Statement

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The theory rests on three pillars:

Conservation of orbitals: Mixing $n$ atomic orbitals gives exactly $n$ hybrid orbitals. No orbitals are created or destroyed.
Energy averaging: The hybrid orbitals have energies that are intermediate between the original $s$ and $p$ energies.
Directionality: Hybrid orbitals point in specific directions to minimise electron pair repulsion, which directly determines molecular shape.

The Three Common Hybridizations

Hybridization	Orbitals Mixed	Number of Hybrids	Geometry	Bond Angle	Example
$sp$	one $s$ + one $p$	2	Linear	$180^\circ$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	3	Trigonal planar	$120^\circ$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	4	Tetrahedral	$109.5^\circ$	$\text{CH}_4$

Note

The superscript in $sp^2$ or $sp^3$ tells you how many $p$ orbitals were mixed. $sp^3$ means one $s$ and three $p$ orbitals were blended. It does not mean there are three $s$ orbitals — there is only one $s$ orbital per shell.

How It Works: The Methane Example

Carbon in its ground state has the configuration $1s^2 2s^2 2p_x^1 2p_y^1$ . Only two unpaired electrons — it should form only two bonds. But we know carbon forms four bonds.

Step 1: Promotion. One electron from the $2s$ orbital is promoted (excited) to the empty $2p_z$ orbital. This costs a small amount of energy, but it is more than compensated by the energy released when four strong bonds form instead of two.

Step 2: Hybridization. The one $2s$ orbital and three $2p$ orbitals mix to form four equivalent $sp^3$ hybrid orbitals. Each hybrid has 25% $s$ character and 75% $p$ character.

Step 3: Bonding. Each $sp^3$ hybrid overlaps with the $1s$ orbital of a hydrogen atom, forming four identical $\sigma$ bonds. The hybrids point to the corners of a tetrahedron, giving the $109.5^\circ$ angle. …

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Hybridisation is the mathematical mixing of atomic orbitals to form new, equivalent hybrid orbitals that better explain molecular geometry and bonding. The three common types — $sp$ (linear), $sp^2$ (trigonal planar), and $sp^3$ (tetrahedral) — each have distinct shapes and bond angles.

The Concept Behind Hybridisation

Valence bond theory faced a puzzle in the early 20th century. Consider carbon: its ground-state electronic configuration is $1s^2 2s^2 2p^2$ , with only two unpaired electrons in the $2p$ orbitals. Yet carbon forms four equivalent bonds in methane ( $\text{CH}_4$ ), not two. Even if we promote an electron from $2s$ to $2p$ to get four unpaired electrons, we'd expect three bonds from $p$ orbitals (mutually perpendicular) and one from an $s$ orbital (spherical) — but experiment shows all four C–H bonds in methane are identical, pointing toward the corners of a tetrahedron at $109.5°$ .

Linus Pauling resolved this by proposing hybridisation: the idea that atomic orbitals on the same atom can mix mathematically to produce a new set of hybrid orbitals. These hybrids are equivalent in energy, have specific directional character, and their geometry matches observed molecular shapes. The number and type of atomic orbitals mixed equals the number of hybrid orbitals formed.

Think of it as redistributing electron density: instead of using "pure" $s$ and $p$ orbitals with their own shapes, the atom creates custom orbitals optimized for bonding in particular directions.

The Three Principal Hybridisation Schemes

1. $sp$ Hybridisation (Linear geometry)

One $s$ orbital mixes with one $p$ orbital to form two equivalent $sp$ hybrid orbitals.

Formation:

\text{1 } s + \text{1 } p \longrightarrow \text{2 } sp \text{ hybrids}

Shape and orientation:

Each $sp$ hybrid has two lobes — one large lobe pointing in one direction and a smaller lobe on the opposite side. The two $sp$ hybrids orient themselves $180°$ apart along a straight line to minimize repulsion, giving linear geometry.

Characteristics:

Bond angle: $180°$
Each hybrid has 50% $s$ character and 50% $p$ character
Two unhybridised $p$ orbitals remain perpendicular to the bonding axis (available for $\pi$ bonding)

Example: Beryllium in $\text{BeH}_2$ or carbon in acetylene ( $\text{HC≡CH}$ ). In acetylene, each carbon uses two $sp$ hybrids for $\sigma$ bonds (one C–H, one C–C), while the two remaining $p$ orbitals form two $\pi$ bonds in the triple bond.

2. $sp^2$ Hybridisation (Trigonal planar geometry)

One $s$ orbital mixes with two $p$ orbitals to form three equivalent $sp^2$ hybrid orbitals.

Formation:

\text{1 } s + \text{2 } p \longrightarrow \text{3 } sp^2 \text{ hybrids}

Shape and orientation:

Each $sp^2$ hybrid orbital has a large lobe and a small lobe, similar to $sp$ but with different proportions. The three hybrids lie in the same plane, oriented $120°$ apart in a trigonal planar arrangement.

Characteristics:

Bond angle: $120°$
Each hybrid has 33.3% $s$ character and 66.7% $p$ character
One unhybridised $p$ orbital remains perpendicular to the plane (available for $\pi$ bonding)

Example: Boron in $\text{BF}_3$ or carbon in ethene ( $\text{H}_2\text{C=CH}_2$ ). In ethene, each carbon uses three $sp^2$ hybrids for $\sigma$ bonds (two C–H, one C–C), and the remaining $p$ orbital forms the $\pi$ bond of the double bond.

3. $sp^3$ Hybridisation (Tetrahedral geometry)

One $s$ orbital mixes with three $p$ orbitals to form four equivalent $sp^3$ hybrid orbitals.

Formation:

\text{1 } s + \text{3 } p \longrightarrow \text{4 } sp^3 \text{ hybrids}

Shape and orientation:

Each $sp^3$ hybrid has one large lobe and one small lobe. The four hybrids point toward the corners of a regular tetrahedron, maximizing separation. …

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Q.What is meant by hybridisation of atomic orbitals? Describe the shapes of spspsp, sp2sp^2sp2, sp3sp^3sp3 hybrid orbitals.

The Concept Behind Hybridisation

The Three Principal Hybridisation Schemes

1. spspsp Hybridisation (Linear geometry)

2. sp2sp^2sp2 Hybridisation (Trigonal planar geometry)

3. sp3sp^3sp3 Hybridisation (Tetrahedral geometry)

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Q.What is meant by hybridisation of atomic orbitals? Describe the shapes of $sp$ , $sp^2$ , $sp^3$ hybrid orbitals.

1. $sp$ Hybridisation (Linear geometry)

2. $sp^2$ Hybridisation (Trigonal planar geometry)

3. $sp^3$ Hybridisation (Tetrahedral geometry)

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