3.7.2 sp² Hybridization

This topic explains sp² hybridization — the mixing of one s orbital and two p orbitals to form three new hybrid orbitals. It covers the definition, examples, and bond angles.

Definition

sp² hybridization is the mixing of one s and two p orbitals to form three sp² hybrid orbitals of equal energy and shape.

Key Features

• Three sp² hybrid orbitals are formed.
• They are arranged in a trigonal planar geometry at 120°.
• Each has 33.33% s-character and 66.67% p-character.
• They form sigma bonds.
• One unhybridized p orbital forms pi bonds.

Example: BCl₃

• Ground state of B: 1s² 2s² 2p¹
• Excitation: 1s² 2s¹ 2p_x¹ 2p_y¹
• Hybridization: 2s + 2px + 2py → 3 sp² orbitals
• sp² orbitals overlap with Cl p orbitals
• Shape: Trigonal planar
• Bond angle: 120°

Example: C₂H₄ (Ethylene)

• Ground state of C: 1s² 2s² 2p_x¹ 2p_y¹
• Excitation: 1s² 2s¹ 2p_x¹ 2p_y¹ 2p_z¹
• Hybridization: 2s + 2px + 2py → 3 sp² orbitals
• sp² orbitals form sigma bonds
• Unhybridized 2p_z orbital forms pi bond
• Shape: Trigonal planar (planar molecule)
• Bond angle: 120°

Main Idea

In sp² hybridization, one s orbital and two p orbitals mix to form three sp² hybrid orbitals. These orbitals are arranged in a trigonal planar geometry at 120° to each other. Each sp² hybrid orbital has 33.33% s-character and 66.67% p-character. The sp² hybrid orbitals form sigma bonds. The remaining unhybridized p orbital forms pi bonds.

Core Concept

sp² hybridization occurs when a central atom needs to form three bonds and has no lone pairs (or sometimes one lone pair). The three sp² hybrid orbitals point toward the corners of an equilateral triangle (120° apart), giving a trigonal planar geometry.

Detailed Explanation

What is sp² Hybridization?

• The intermixing of one s and two p orbitals of different energies and different shapes.
• Produces three new hybrid orbitals of the same energy and same shape.
• Each sp² hybrid orbital has 33.33% s-character and 66.67% p-character.
• The sp² hybrid orbitals form sigma bonds.
• One unhybridized p orbital remains, which can form pi bonds.

Example 1: Boron Trichloride (BCl₃)

• Ground state: 1s² 2s² 2p¹
• There is only one unpaired electron in the ground state.

• Boron undergoes excitation by promoting one of its 2s electrons into an empty 2p orbital.
• Excited state configuration: 1s² 2s¹ 2p_x¹ 2p_y¹

• In the excited state, the boron atom undergoes sp² hybridization by mixing a 2s and two 2p orbitals.
• This gives three half-filled sp² hybrid orbitals oriented in trigonal planar symmetry.

• In BCl₃, three sp² hybridized orbitals of boron atom overlap with three p orbitals of three chlorine atoms.
• This forms three σ sp²-p bonds.
• Thus, the shape of BCl₃ is trigonal planar with bond angles equal to 120°.

Example 2: Ethylene (C₂H₄)

Electronic Configuration of Carbon: - Ground state: 1s² 2s² 2p_x¹ 2p_y¹

• 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

• Each carbon atom undergoes sp² hybridization in its excited state by mixing 2s and two 2p orbitals.
• This gives three half-filled sp² hybrid orbitals oriented in trigonal planar symmetry.
• There is also one half-filled unhybridized 2p orbital on each carbon perpendicular to the plane of sp² hybrid orbitals.

• The carbon atoms form a σ bond with each other by using sp² hybrid orbitals.
• A π bond is also formed between them due to the parallel overlapping of unhybridized 2p_z orbitals.
• Thus, there is a double bond (σ and π) between two carbon atoms.
• Each carbon atom also forms two σ bonds between sp² orbital of carbon atom and s orbitals of two hydrogen atoms.

• Ethylene molecule is planar with HCH and HCC bond angles equal to 120°.
• All the atoms are present in one plane.

Summary of sp² Hybridization

Important Points

• 1. sp² hybridization = mixing of one s and two p orbitals.
• 2. Three sp² hybrid orbitals are formed.
• 3. They are arranged in a trigonal planar geometry at 120°.
• 4. Each has 33.33% s-character and 66.67% p-character.
• 5. They form sigma bonds.
• 6. One unhybridized p orbital forms pi bonds.
• 7. Example: BCl₃ (trigonal planar), C₂H₄ (planar, double bond).
• 8. In C₂H₄, unhybridized p orbitals form pi bonds.

Remember

• - sp² hybridization: 1 s + 2 p → 3 sp² orbitals
• - Geometry: Trigonal planar
• - Bond angle: 120°
• - s-character: 33.33%
• - p-character: 66.67%
• - Bonds: Sigma (and pi with unhybridized p)
• - Examples: BCl₃, C₂H₄

Common Mistakes

• 1. Thinking sp² hybridization gives tetrahedral shape: It gives trigonal planar shape.
• 2. Forgetting the 33.33% s and 66.67% p character: Each sp² orbital has 33.33% s and 66.67% p.
• 3. **Confusing sp² with sp or sp³:** sp² = 1 s + 2 p; sp = 1 s + 1 p; sp³ = 1 s + 3 p.
• 4. Ignoring pi bonds in C₂H₄: The unhybridized p orbital forms a pi bond.

Quick Revision

• sp² hybridization = mixing of one s and two p orbitals.
• Three sp² hybrid orbitals are formed.
• They are arranged in a trigonal planar geometry at 120°.
• Each has 33.33% s-character and 66.67% p-character.
• They form sigma bonds.
• One unhybridized p orbital forms pi bonds.
• Example: BCl₃ (trigonal planar), C₂H₄ (planar, double bond).
• In C₂H₄, unhybridized p orbitals form pi bonds.

60-Second Revision

• sp² hybridization: 1 s + 2 p → 3 sp² orbitals
• Geometry: Trigonal planar
• Bond angle: 120°
• s-character: 33.33%
• p-character: 66.67%
• Bonds: Sigma (and pi with unhybridized p)
• Examples: BCl₃, C₂H₄

QUALITY VALIDATION

• Topic: 3.7.2 sp² Hybridization
• Chapter: 3 (Chemical Bonding)

• Source re-read: PASS
• Source scope followed: PASS
• Knowledge level matched: PASS
• Important concepts covered: PASS
• Definitions checked: PASS
• Equations checked: N/A
• Derivations checked: N/A
• Examples checked: PASS
• Unsupported knowledge added: NO
• Unnecessary repetition: NO
• Notes format maintained: PASS
• Important source content omitted: NO

• Source limitations:
• NONE

NEXT TOPIC