Which hybrid orbitals are used by carbon atoms in the following molecules? (a) CH₃–CH₃ (b) CH₃–CH=CH₂ (c) CH₃-

Exercises · 4.30

Q. Which hybrid orbitals are used by carbon atoms in the following molecules?

(a)

CH_3\text{–}CH_3

(b)

CH_3\text{–}CH=CH_2

(c)

CH_3\text{-}CH_2\text{-}OH

(d)

CH_3\text{-}CHO

(e)

CH_3COOH

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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

s

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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Count the number of sigma bonds and lone pairs around each carbon to determine its steric number: 4 → $sp^3$ , 3 → $sp^2$ , 2 → $sp$ . All five molecules contain only $sp^3$ and $sp^2$ carbons.

The key to identifying hybridization lies in understanding what orbitals a carbon atom needs to form its bonds. Carbon forms covalent bonds by overlapping its hybrid orbitals with orbitals from other atoms. The type of hybridization depends on the steric number: the total count of sigma bonds plus lone pairs around the atom.

Steric number 4 (four regions of electron density) → $sp^3$ hybridization → tetrahedral geometry
Steric number 3 (three regions) → $sp^2$ hybridization → trigonal planar geometry
Steric number 2 (two regions) → $sp$ hybridization → linear geometry

Remember: only sigma bonds and lone pairs count for the steric number. Pi bonds (the second and third bonds in double and triple bonds) use unhybridized $p$ orbitals and don't affect hybridization.

Let me analyze each molecule carbon by carbon:

(a) $\ce{CH3-CH3}$ (Ethane)

Left carbon ( $\ce{CH3}$ ): Forms four single bonds (three $\ce{C-H}$ and one $\ce{C-C}$ ). Each single bond is a sigma bond.

Steric number = 4 → $sp^3$ hybridized
Right carbon ( $\ce{CH3}$ ): Identical situation — four sigma bonds.

Steric number = 4 → $sp^3$ hybridized

Both carbons are $sp^3$ .

(b) $\ce{CH3-CH=CH2}$ (Propene)

First carbon ( $\ce{CH3}$ ): Four single bonds (three $\ce{C-H}$ , one $\ce{C-C}$ ).

Steric number = 4 → $sp^3$ hybridized
Second carbon ( $\ce{CH=}$ ): One single bond to $\ce{CH3}$ , one double bond to $\ce{CH2}$ , one bond to $\ce{H}$ .

The double bond consists of one sigma + one pi bond. Count only the sigma: three sigma bonds total.

Steric number = 3 → $sp^2$ hybridized
Third carbon ( $\ce{=CH2}$ ): One double bond to the middle carbon (counts as one sigma), two $\ce{C-H}$ bonds.

Steric number = 3 → $sp^2$ hybridized

Carbons: $sp^3$ , $sp^2$ , $sp^2$ .

(c) $\ce{CH3-CH2-OH}$ (Ethanol)

First carbon ( $\ce{CH3}$ ): Three $\ce{C-H}$ bonds, one $\ce{C-C}$ bond.

Steric number = 4 → $sp^3$ hybridized
Second carbon ( $\ce{CH2}$ ): Two $\ce{C-H}$ bonds, one $\ce{C-C}$ bond, one $\ce{C-O}$ bond.

Steric number = 4 → $sp^3$ hybridized

Both carbons are $sp^3$ .

Note

The oxygen in $\ce{-OH}$ is also $sp^3$ hybridized (two bonds + two lone pairs), but the question asks only about carbon atoms.

(d) $\ce{CH3-CHO}$ (Acetaldehyde)

First carbon ( $\ce{CH3}$ ): Three $\ce{C-H}$ bonds, one $\ce{C-C}$ bond.

Steric number = 4 → $sp^3$ hybridized
Second carbon (aldehyde $\ce{CHO}$ ): One $\ce{C-H}$ bond, one single bond to $\ce{CH3}$ , one double bond to oxygen ( $\ce{C=O}$ ).

The carbonyl double bond = one sigma + one pi. Count three sigma bonds.

Steric number = 3 → $sp^2$ hybridized

Carbons: $sp^3$ , $sp^2$ .

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Q.Which hybrid orbitals are used by carbon atoms in the following molecules?

(a) CHX3−CHX3\ce{CH3-CH3}CHX3​−CHX3​ (Ethane)

(b) CHX3−CH=CHX2\ce{CH3-CH=CH2}CHX3​−CH=CHX2​ (Propene)

(c) CHX3−CHX2−OH\ce{CH3-CH2-OH}CHX3​−CHX2​−OH (Ethanol)

(d) CHX3−CHO\ce{CH3-CHO}CHX3​−CHO (Acetaldehyde)

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(a) $\ce{CH3-CH3}$ (Ethane)

(b) $\ce{CH3-CH=CH2}$ (Propene)

(c) $\ce{CH3-CH2-OH}$ (Ethanol)

(d) $\ce{CH3-CHO}$ (Acetaldehyde)

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