Lecture 10-Hybrid

Lecture 10: Hybrid Orbital Theory

Overview of Hybrid Orbital Theory

• Explores how atomic orbitals combine to form new hybrid orbitals.

• Fundamental to understanding bonding in molecules.

Orbitals and Bonding (p. 98)

• Electrons occupy atomic orbitals; bonding orbitals result from overlapping valence orbitals of bonding atoms.

• Bonding orbitals position electrons between atomic nuclei, resulting in lower energy (more stability).

• Diatomic molecules' bonding explained by ordinary atomic orbital overlap.

Bonding in Diatomic Molecules

Example: Hydrogen (H)

• Electron configuration: H: 1s¹

• Bond formation: H∙ + •H → H-H

Example: Fluorine (F)

• Electron configuration: F: 1s² 2s² 2p⁵

• Bond Formation: Two F atoms share electrons.

Sigma Bonds (σ) (p. 99)

• Electron density resides on the internuclear axis.

• Forms via end-to-end overlap of orbitals (e.g., pz + pz, 1s + 1s).

• Sigma bonds allow rotation without breaking the bond.

Bonding in F₂ (p. 101)

• Electron configuration: F: 1s² 2s² 2p⁵

• Bonds formed: Single bond electrons located in a sigma bonding orbital.

• All single bonds are sigma bonds.

Bonding in O₂ (p. 101)

Electron Configuration

• O: 1s² 2s² 2p⁴

• Bond Formation: O + O → O=O

• Contains one sigma bond (like F₂) and one pi bond formed using 2px or 2py orbitals.

Pi Bonds (π) (p. 99)

• Electron density above and below the internuclear axis.

• Formed from side-to-side overlap of parallel p orbitals (px + px, py + py).

• Cannot be rotated without breaking the bond; necessary for multiple bonds.

Bonding in N₂ (p. 101)

• N: 1s² 2s² 2p³

• Bonds formed: One sigma bond and two pi bonds, making the triple bond.

• Second pi bond formed from the overlapping 2py orbitals.

Bonding in More Complicated Molecules

Example: Methane (CH₄)

• Electron count: 4 + 4(1) = 8 e⁻.

• Geometry: Tetrahedral structure with 109° bond angles, not achievable with 2s and 2p orbitals alone.

• Hybridization required to form the correct angles.

Hybridization and Bonding Geometry

Bonding in Molecules with Tetrahedral Geometry

• Example: CH₄ exhibits sp³ hybridization, bond angles of 109°.

Bonding in Ethane (C₂H₆) (p. 101)

• Electron count: 2(4) + 6(1) = 14 e-.

• Each carbon atom uses sp³ hybrid orbitals for bonding, maintaining tetrahedral geometry.

Bonding in Ammonia (NH₃) (p. 101)

• Electron count: 5 + 3(1) = 8 e-.

• Central nitrogen atom exhibits sp³ hybridization; all N-H bonds are sigma bonds.

Bonding in PCl₅ (p. 102)

• Central phosphorus atom utilizes dsp³ hybridization for bonding; single bonds are sigma bonds.

Bonding in SF₆ (p. 102)

• Central sulfur atom exhibits d²sp³ hybridization for an octahedral arrangement of six electron pairs.

Clicker Questions on Hybridization

For XeF₂ and XeF₄

• Expected hybridizations:

  • XeF₂: dsp³

  • XeF₄: d²sp³

Summary of Hybrid Orbital Theory (p. 98, 99)

• Hybrid orbitals form sigma bonds and house lone pairs.

• Unhybridized p orbitals form pi bonds.

• Bond compositions:

  • Single bond = 1 sigma bond.

  • Double bond = 1 sigma + 1 pi bond.

  • Triple bond = 1 sigma + 2 pi bonds.

Delocalized Pi Electrons

• Example: Benzene (C₆H₆) demonstrates delocalized pi electrons leading to equivalent bond strengths in resonance structures.

• Each carbon in benzene exhibits sp² hybridization and bond angles of 120°.

Delocalized Electrons in NO₃⁻

• Central nitrogen atom is sp² hybridized; all nitrogen-oxygen bonds are equivalent.