Lecture Notes on Molecular Orbital Theory

Valence vs Core Orbitals

• Lithium (Li): $1s^2 2s^1$
  • Valence electrons: highest energy electrons, chemically accessible. Lewis theory focuses on valence electrons and ignores core electrons.
  • Born out of Quantum theory.
  • Example: Li₂
• Core orbitals (e.g., 1s orbitals in Li) are too low in energy to interact effectively with valence orbitals (e.g., 2s in Li).
• Core orbitals remain largely atom-centered.
  • Provide an atomic screen (shield) for the valence electrons.
• Valence electrons have approximately similar energies and can effectively form delocalized molecular orbitals (MOs).

Dihydrogen (H₂) vs Dihelium (He₂)

• Why is H₂ stable, but He₂ is not?
• Bond Order:
  • Bond Order = {{# Bonding \ e^- - # Antibonding \ e^-} \over {2}}
  • Indicator of bond strength: larger bond order indicates greater bond strength.
• H₂:
  • Bond order = $\frac{2-0}{2} = 1$
  • Making a molecule stabilizes relative to two H-atoms.
• He₂:
  • Bond order = $\frac{2-2}{2} = 0$
  • A high energy antibonding orbital ($\sigma^*$) is occupied, destabilizing the system relative to two He atoms.

Nitrogen (N₂) vs Fluorine (F₂) and Orbital Mixing

• Focusing only on valence orbitals.
• F₂: $2s^2 2p^5$
• N₂: $2s^2 2p^3$
• F₂:
  • Electronic configuration: $\sigma(2s)^2 \sigma^<em>(2s)^2 \sigma(2p)^2 \pi(2p)^4 \pi^</em>(2p)^4$
  • Bond Order = $\frac{8-6}{2} = 1$
• N₂:
  • Electronic configuration: $\sigma(2s)^2 \sigma^*(2s)^2 \pi(2p)^4 \sigma(2p)^2$
  • Bond Order = $\frac{8-2}{2} = 3$
• Orbital Mixing:
  • In N₂, mixing pushes the $\sigma<em>{2p</em>z}$ combo above the $\pi<em>{2p</em>x}$ and $\pi<em>{2p</em>y}$.
  • No mixing in F₂ due to higher effective nuclear charge (Zeff).
    • Increased energy difference between s and p orbitals.

Second Row Homodiatomics

• Effective Nuclear Charge (Zeff):
  • $Z_{eff} = Z - S$
    • Z = # of Protons
    • S = # of Shielding electrons
• As Zeff increases, orbital energy decreases.
  • Large Zeff = large energy difference = small s-p mixing.
• The size of the effect depends on the 2s-2p energy difference.
• Bond order changes across the second row diatomics.
• Examples:
  • Li₂: Bond order 1, unpaired electrons.
  • Be₂: Bond order 0
  • N₂: $\pi$ higher than $\sigma$
  • O₂ and Ne₂: Bond order 0

Molecular Spin State

• Oxygen (O₂) vs. Fluorine (F₂)
• O₂:
  • Electronic configuration: $\sigma(2s)^2 \sigma^<em>(2s)^2 \sigma(2p)^2 \pi(2p)^4 \pi^</em>(2p)^2$
  • Bond Order = $\frac{8-4}{2} = 2$ (matches Lewis structure)
  • S (Spin) = 1 (Triplet state). Paramagnetic (unpaired electrons leading to attraction to magnetic field).
• F₂:
  • Electronic configuration: $\sigma(2s)^2 \sigma^<em>(2s)^2 \sigma(2p)^2 \pi(2p)^4 \pi^</em>(2p)^4$
  • Bond Order = $\frac{8-6}{2} = 1$
  • S = 0 (Singlet state). Diamagnetic (paired electrons leading to no attraction).
• Spin State:
  • S = 0: Singlet
  • S = 1/2: Doublet
  • S = 1: Triplet
  • S = 3/2: Quartet

Heterodiatomics

• Different atoms: need to consider the relative energies of the atomic orbitals (AOs).
• Example: molecule AB
  • If A is more electronegative than B:
    • $\phi<em>1$ orbital will look more like $s</em>A$
    • $\phi<em>2$ orbital will look more like $s</em>B$
  • Molecular orbitals (MOs) more closely resemble the AOs that are closer in energy.
• Non-bonding orbitals may exist in some cases.
  • An s-orbital may be too high in energy to interact with another s-orbital, but could be close in energy to a pz orbital!

Heterodiatomics: Lithium Hydride (LiH) vs Hydrogen Fluoride (HF)

• Why does LiH have a hydridic hydrogen ($H^-$), but HF has an acidic hydrogen ($H^+$)?
• LiH: $Li^+H^-$
  • The bonding orbital is close in energy to H (1s).
  • The bonding orbital will have a lot of H character on the Li, not all electrons are on F-orbitals.
  • The 2s atomic orbital is closest to H.
  • The 2s orbital is "empty"
• HF: $H^+F^-$
  • Electrons are on F-orbitals.