AP Chemistry: Hybridization and Molecular Orbital Theory (Basics)

AP Chemistry – Hybridization and Molecular Orbital Theory (Basics)

AP Chemistry Study Guide: Hybridization & Molecular Orbital Theory (Basics)

What This Is

Hybridization and Molecular Orbital (MO) Theory explain how atoms bond to form molecules—why some bonds are stronger, why molecules have certain shapes, and how electrons are shared. The AP exam tests your ability to predict molecular geometry (via hybridization) and bond order (via MO Theory). Real-world example: The double bond in ethylene (C₂H₄)—used to make plastics—relies on sp² hybridization to form its flat, rigid structure. Without hybridization, we wouldn’t have polyethylene (plastic bags, bottles) or understand why DNA’s base pairs stack so neatly.

Key Terms & Concepts

• Hybridization: Mixing atomic orbitals (s, p, d) to form new hybrid orbitals that explain molecular shape.

• Example: Carbon in methane (CH₄) uses sp³ hybridization to form 4 identical bonds.

• Hybrid Orbitals (sp, sp², sp³, sp³d, sp³d²):

• sp: 2 hybrid orbitals (linear, 180°), e.g., CO₂.
• sp²: 3 hybrid orbitals (trigonal planar, 120°), e.g., BF₃.
• sp³: 4 hybrid orbitals (tetrahedral, 109.5°), e.g., CH₄.
• sp³d: 5 hybrid orbitals (trigonal bipyramidal), e.g., PCl₅.

• sp³d²: 6 hybrid orbitals (octahedral), e.g., SF₆.

• Sigma (σ) Bond: Head-to-head orbital overlap; single bonds are always σ bonds.

• Pi (π) Bond: Side-to-side p-orbital overlap; found in double/triple bonds (1 π in double, 2 π in triple).

• Molecular Orbital (MO) Theory: Bonds form from the combination of atomic orbitals into molecular orbitals (bonding/antibonding).

• Bonding MO (σ, π): Lower energy, stabilizes the molecule.

• Antibonding MO (σ, π): Higher energy, destabilizes the molecule (denoted with *).

• Bond Order (BO):
  BO = ½ (# bonding e⁻ – # antibonding e⁻)

• Example: O₂ has BO = 2 (double bond), but MO Theory explains its paramagnetism (unpaired e⁻).

• Paramagnetism vs. Diamagnetism:

• Paramagnetic: Unpaired electrons (attracted to magnets), e.g., O₂.

• Diamagnetic: All electrons paired (repelled by magnets), e.g., N₂.

• HOMO & LUMO:

• HOMO (Highest Occupied MO): Where the "last" electrons sit.
• LUMO (Lowest Unoccupied MO): Next available orbital for reactions.

Step-by-Step: Predicting Hybridization & Bonding

1. Hybridization (VSEPR + Lewis Structure)

Problem: What is the hybridization of the central atom in NH₃? Steps:
1. Draw the Lewis structure → N has 3 bonds + 1 lone pair (4 electron domains).
2. Count electron domains (bonds + lone pairs) → 4 domains = sp³.
3. Check geometry → Tetrahedral electron geometry, trigonal pyramidal molecular shape.

2. Molecular Orbital (MO) Diagram (Diatomics)

Problem: Draw the MO diagram for N₂ and find its bond order.
Steps:
1. Write electron configuration → N: 1s² 2s² 2p³ (total 10 valence e⁻ for N₂).
2. Fill MO diagram (σ2s, σ2s, π2p, σ2p, π2p, σ2p).
- Order for B₂/C₂/N₂: σ2s < σ2s < π2p < σ2p < π2p < σ2p.
3. Count bonding/antibonding e⁻ → 8 bonding, 2 antibonding.
4. Calculate bond order → BO = ½(8 – 2) = 3 (triple bond).
5. Check magnetism → All electrons paired → diamagnetic.

Common Mistakes

• Mistake: Forgetting lone pairs count as electron domains for hybridization.
  Correction: Lone pairs do count! NH₃ has 4 domains (3 bonds + 1 lone pair) → sp³, not sp².

• Mistake: Mixing up σ and π bonds in double/triple bonds.
  Correction: A double bond = 1 σ + 1 π; a triple bond = 1 σ + 2 π.

• Mistake: Using the wrong MO energy order for O₂/F₂ vs. B₂/C₂/N₂.
  Correction:

• B₂/C₂/N₂: π2p below σ2p (due to s-p mixing).

• O₂/F₂: σ2p below π2p (no s-p mixing).

• Mistake: Assuming all molecules with double bonds are sp² hybridized.
  Correction: Check the central atom’s domains! CO₂ is sp (2 domains), not sp².

• Mistake: Forgetting antibonding orbitals in bond order calculations.
  Correction: BO = ½(bonding e⁻ – antibonding e⁻). Antibonding e⁻ weaken the bond!

AP Exam Insights

• FRQ Favorite: Predict hybridization, draw MO diagrams, or explain bond order/magnetism (e.g., "Why is O₂ paramagnetic?").
• Multiple-Choice Traps:
• Hybridization: Watch for lone pairs! (e.g., H₂O is sp³, not sp²).
• MO Theory: The order of π2p and σ2p flips between B₂/N₂ and O₂/F₂.
• Bond Order: A BO of 0 means the molecule doesn’t exist (e.g., He₂).
• Tricky Distinction: Hybridization explains shape; MO Theory explains bond strength/magnetism.

Quick Check Questions

1. Multiple Choice

What is the hybridization of the carbon atom in CO₂? (A) sp (B) sp² (C) sp³ (D) sp³d

Answer: (A) sp. CO₂ has 2 electron domains (2 double bonds, no lone pairs) → linear → sp.

2. Short FRQ

a) Draw the MO diagram for O₂ and label the HOMO.
b) Is O₂ paramagnetic or diamagnetic? Explain.

Answer:
a) MO diagram: σ2s² σ2s² σ2p² π2p⁴ π2p² (HOMO = π2p).
b) Paramagnetic (2 unpaired e⁻ in π2p orbitals).

Last-Minute Cram Sheet

1. Hybridization = electron domains (bonds + lone pairs).
2. sp³ = 4 domains (tetrahedral), sp² = 3 domains (trigonal planar), sp = 2 domains (linear).
3. Double bond = 1 σ + 1 π; triple bond = 1 σ + 2 π.
4. MO Theory order for B₂/C₂/N₂: σ2s < σ*2s < π2p < σ2p.
5. MO Theory order for O₂/F₂: σ2s < σ*2s < σ2p < π2p.
6. Bond order = ½(bonding e⁻ – antibonding e⁻).
7. BO = 0 → molecule doesn’t exist (e.g., He₂).
8. Paramagnetic = unpaired e⁻ (O₂); diamagnetic = all paired (N₂).
9. ⚠️ Lone pairs count for hybridization! (e.g., H₂O is sp³).
10. ⚠️ π bonds are weaker than σ bonds (break first in reactions).