Explain hybridisation of carbon in ethene ethyne and ethane

Question

Explain the hybridisation of carbon in ethane ( $\text{C}_2\text{H}_6$ ), ethene ( $\text{C}_2\text{H}_4$ ), and ethyne ( $\text{C}_2\text{H}_2$ ). State the geometry, bond angles, and bond lengths in each case.

Solution — Step by Step

In ethane ( $\text{C}_2\text{H}_6$ ), each carbon is bonded to 3 H atoms and 1 C atom — total 4 bonds, no lone pairs on carbon.

Hybridisation: $sp^3$ — the 2s orbital mixes with all three 2p orbitals to form four equivalent $sp^3$ hybrid orbitals.

Each $sp^3$ orbital overlaps with one H (sigma bond), and the two carbons form a C–C sigma bond by $sp^3$ – $sp^3$ overlap.

Geometry: Tetrahedral around each carbon
Bond angle: H–C–H = 109.5°
C–C bond length: 154 pm (single bond, longest)
Bond type: Only sigma ( $\sigma$ ) bonds

In ethene ( $\text{C}_2\text{H}_4$ ), each carbon forms a double bond with the other carbon and two sigma bonds with H atoms — 3 groups around each carbon.

Hybridisation: $sp^2$ — the 2s mixes with only TWO of the three 2p orbitals to form three $sp^2$ hybrid orbitals. One p orbital remains unhybridised.

The three $sp^2$ orbitals form:

2 sigma bonds with H atoms
1 sigma bond with the other carbon ( $sp^2$ – $sp^2$ overlap)

The two unhybridised p orbitals (one on each carbon, perpendicular to the molecular plane) overlap laterally to form a pi ( $\pi$ ) bond.

Geometry: Trigonal planar around each carbon
Bond angle: H–C–H ≈ 120°
C=C bond length: 134 pm (shorter than single bond — pi bond pulls carbons closer)
Bond type: One $\sigma$ + one $\pi$ in the C=C bond

In ethyne ( $\text{C}_2\text{H}_2$ ), each carbon forms a triple bond with the other carbon and one sigma bond with H — 2 groups around each carbon.

Hybridisation: $sp$ — the 2s mixes with only ONE 2p orbital to form two $sp$ hybrid orbitals. Two p orbitals remain unhybridised.

The two $sp$ orbitals form:

1 sigma bond with H
1 sigma bond with the other carbon

The four unhybridised p orbitals (two per carbon, perpendicular to each other) form two pi bonds — creating the triple bond.

Geometry: Linear around each carbon
Bond angle: H–C–C = 180°
C≡C bond length: 120 pm (shortest — two pi bonds pull carbons very close)
Bond type: One $\sigma$ + two $\pi$ in the C≡C bond

Property	Ethane ( $\text{C}_2\text{H}_6$ )	Ethene ( $\text{C}_2\text{H}_4$ )	Ethyne ( $\text{C}_2\text{H}_2$ )
Hybridisation	$sp^3$	$sp^2$	$sp$
Geometry	Tetrahedral	Trigonal planar	Linear
Bond angle	109.5°	120°	180°
C–C bond	Single (154 pm)	Double (134 pm)	Triple (120 pm)
Pi bonds	0	1	2
s-character	25%	33%	50%

Why This Works

As we go from $sp^3$ to $sp^2$ to $sp$ , the percentage of s-character in the hybrid orbital increases: 25% → 33% → 50%. Greater s-character means the electrons are held closer to the nucleus (s orbitals are closer to nucleus than p). This results in:

Shorter, stronger bonds — C≡C is shorter and stronger than C=C, which is shorter and stronger than C–C.
Greater acidity of terminal H — In ethyne, the terminal C–H bond has 50% s-character, making the carbon very electronegative. The H is slightly more acidic (can be removed by strong bases). This is why ethyne reacts with sodium (forms sodium acetylide) but ethane doesn’t.
Bond angles increase — More p-character in the hybrid = more p-like (90° preferred angle). More s-character pushes angles toward 180°.

Alternative Method — Count Groups, Assign Hybridisation

Quick rule: count the number of groups around the carbon (bonds + lone pairs on carbon):

4 groups → $sp^3$
3 groups → $sp^2$
2 groups → $sp$

For ethane C: 4 bonds, 0 lone pairs = 4 groups → $sp^3$ . For ethene C: 2 single + 1 double = 3 groups → $sp^2$ . For ethyne C: 1 single + 1 triple = 2 groups → $sp$ .

Common Mistake

Saying ethene has 4 bonds per carbon and should be $sp^3$ . Counting individual bond lines: C in ethene has 2 C–H bonds and 1 C=C bond. Count the double bond as ONE group (not two). So 3 groups → $sp^2$ . Similarly in ethyne: 1 C–H and 1 C≡C = 2 groups → $sp$ . Always count groups (sigma bonds + lone pairs), not electron pairs in pi bonds.

Question

Solution — Step by Step

Why This Works

Alternative Method — Count Groups, Assign Hybridisation

Common Mistake

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