3.7.3 sp³ Hybridization

This topic explains sp³ hybridization — the mixing of one s orbital and three p orbitals to form four new hybrid orbitals. It covers the definition, examples, and bond angles.

Definition

sp³ hybridization is the mixing of one s and three p orbitals to form four sp³ hybrid orbitals of equal energy and shape.

Key Features

• Four sp³ hybrid orbitals are formed.
• They are arranged in a tetrahedral geometry at 109.5°.
• Each has 25% s-character and 75% p-character.
• They form sigma bonds.

Examples

• Carbon undergoes sp³ hybridization.
• Four sp³ orbitals form sigma bonds with four H atoms.
• Shape: Tetrahedral.
• Bond angle: 109.5°.

• Nitrogen undergoes sp³ hybridization.
• Three sp³ orbitals form sigma bonds with three H atoms.
• One sp³ orbital contains a lone pair.
• Shape: Trigonal pyramidal.
• Bond angle: 107°.

• Oxygen undergoes sp³ hybridization.
• Two sp³ orbitals form sigma bonds with two H atoms.
• Two sp³ orbitals contain lone pairs.
• Shape: Bent (V-shaped).
• Bond angle: 104.5°.

Comparison Table

Main Idea

In sp³ hybridization, one s orbital and three p orbitals mix to form four sp³ hybrid orbitals. These orbitals are arranged in a tetrahedral geometry at 109.5° to each other. Each sp³ hybrid orbital has 25% s-character and 75% p-character. The sp³ hybrid orbitals form sigma bonds. This hybridization is common in carbon, nitrogen, oxygen, and other atoms that form four bonds or have lone pairs.

Core Concept

sp³ hybridization occurs when a central atom needs to form four bonds (or has four electron pairs, including lone pairs). The four sp³ hybrid orbitals point toward the corners of a tetrahedron (109.5° apart), giving a tetrahedral geometry.

Detailed Explanation

What is sp³ Hybridization?

• The intermixing of one s and three p orbitals having different energy and different shape.
• Produces four hybridized orbitals of the same energy and same shape.
• Each sp³ hybrid orbital has 25% s-character and 75% p-character.
• The sp³ hybrid orbitals form sigma bonds.

Example 1: Methane (CH₄)

• Ground state: 1s² 2s² 2p_x¹ 2p_y¹
• There are only two unpaired electrons in the ground state.

• Carbon promotes one of its 2s electrons into the empty 2p orbital in the excited state.
• Excited state configuration: 1s² 2s¹ 2p_x¹ 2p_y¹ 2p_z¹

• During the formation of methane molecule, the carbon atom undergoes sp³ hybridization.
• In the excited state, by mixing one 2s and three 2p orbitals, four half-filled sp³ hybrid orbitals are formed.
• These are oriented in tetrahedral symmetry in space around the carbon atom.

• Each of these sp³ hybrid orbitals forms a sigma bond with the s orbital of a hydrogen atom.
• Thus, carbon forms four σ sp³-s bonds with four hydrogen atoms.
• Methane molecule is tetrahedral in shape with 109.5° bond angle.

Example 2: Ethane (C₂H₆)

• Just as in methane molecules, each carbon atom undergoes sp³ hybridization in the excited state.
• This gives four sp³ hybrid orbitals in tetrahedral geometry.
• The two carbon atoms form a sp³-sp³ bond with each other due to overlapping of sp³ hybrid orbitals along the inter-nuclear axis.
• The six carbon-hydrogen bonds are formed from overlaps between the remaining sp³ orbitals on the two carbons and the 1s orbitals of hydrogen atoms.
• All of these are sigma bonds.

Example 3: Ammonia (NH₃)

• Ground state: 1s² 2s² 2p_x¹ 2p_y¹ 2p_z¹
• There are three unpaired electrons in the 2p sublevel.

• In ammonia, nitrogen atom undergoes sp³ hybridization of a 2s and three 2p orbitals.
• This gives four sp³ orbitals, which are arranged in tetrahedral symmetry.
• This arrangement gives more stability to the molecule due to minimization of repulsions.
• Among them, three are half-filled and one is fully filled.

• Three sp³ orbitals of nitrogen atom overlap with s orbital of three hydrogen atoms by using three half-filled sp³ hybrid orbitals.
• There is also a lone pair on nitrogen atom belonging to the fully filled sp³ hybrid orbital.
• It occupies more space than the bond pairs.
• However, the H−N−H bond angle is not equal to the normal tetrahedral angle (109.5°).
• The reported bond angle is 107°.
• The observed decrease in the bond angle is due to the repulsion caused by the lone pair over the bond pairs.
• That is why ammonia molecule is trigonal pyramidal in shape.

Example 4: Water (H₂O)

• Ground state: 1s² 2s² 2p_x² 2p_y¹ 2p_z¹
• There are two unpaired electrons in oxygen atom.

• The experimental bond angle reported is 104.5°.
• To account for this, sp³ hybridization before the bond formation was proposed.
• During the formation of water molecule, the oxygen atom undergoes sp³ hybridization by mixing a 2s and three 2p orbitals.
• This furnishes four sp³ hybrid orbitals oriented in tetrahedral geometry.
• Among them, two are half-filled and the remaining two are completely filled.

• The oxygen atom forms two sp³-s bonds with hydrogen atoms by using half-filled hybrid orbitals.
• Bond angle between H−O−H in water molecule is 104.5°.
• It is again due to repulsions caused by two lone pairs on the bond pairs.
• Thus, water molecule gets angular (V-shaped) shape.

Summary of sp³ Hybridization

Comparison of sp³ Hybridized Molecules

Important Points

• 1. sp³ hybridization = mixing of one s and three p orbitals.
• 2. Four sp³ hybrid orbitals are formed.
• 3. They are arranged in a tetrahedral geometry at 109.5°.
• 4. Each has 25% s-character and 75% p-character.
• 5. They form sigma bonds.
• 6. Example: CH₄ (tetrahedral), NH₃ (trigonal pyramidal), H₂O (bent).
• 7. Lone pairs reduce bond angles.
• 8. More lone pairs → smaller bond angle.

Remember

• - sp³ hybridization: 1 s + 3 p → 4 sp³ orbitals
• - Geometry: Tetrahedral
• - Bond angle: 109.5°
• - s-character: 25%
• - p-character: 75%
• - Bonds: Sigma
• - Examples: CH₄ (109.5°), NH₃ (107°), H₂O (104.5°)

Common Mistakes

• 1. Thinking sp³ hybridization always gives tetrahedral shape: It gives tetrahedral only when there are no lone pairs (CH₄). With lone pairs, it gives trigonal pyramidal (NH₃) or bent (H₂O).
• 2. Forgetting the 25% s and 75% p character: Each sp³ orbital has 25% s and 75% p.
• 3. **Confusing sp³ with sp² or sp:** sp³ = 1 s + 3 p; sp² = 1 s + 2 p; sp = 1 s + 1 p.
• 4. Ignoring the effect of lone pairs: Lone pairs reduce bond angles.
• 5. Thinking NH₃ is tetrahedral: NH₃ is trigonal pyramidal because of the lone pair.

Quick Revision

• sp³ hybridization = mixing of one s and three p orbitals.
• Four sp³ hybrid orbitals are formed.
• They are arranged in a tetrahedral geometry at 109.5°.
• Each has 25% s-character and 75% p-character.
• They form sigma bonds.
• Example: CH₄ (tetrahedral), NH₃ (trigonal pyramidal), H₂O (bent).
• Lone pairs reduce bond angles.
• More lone pairs → smaller bond angle.

60-Second Revision

• sp³ hybridization: 1 s + 3 p → 4 sp³ orbitals
• Geometry: Tetrahedral
• Bond angle: 109.5°
• s-character: 25%
• p-character: 75%
• Bonds: Sigma
• Examples: CH₄ (109.5°), NH₃ (107°), H₂O (104.5°)

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• Topic: 3.7.3 sp³ Hybridization
• Chapter: 3 (Chemical Bonding)

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