Molecular Orbital Theory, Intermolecular Forces, and Bond Energetics
Lewis structures and VSEPR theory are excellent starting points for understanding molecular geometry, but they often fail to explain complex behaviors like magnetism, excited states, or why certain bonds are stronger than others. To truly master chemical bonding for university-level exams, you must bridge the gap between simple diagrams and quantum mechanical reality.
This guide dives into Molecular Orbital (MO) Theory, the energetics of bond formation, and the intermolecular forces that dictate the physical state of matter.
Molecular Orbital Theory (MOT) and Bond Order
While Valence Bond Theory (VSEPR) treats bonds as localized overlaps between atomic orbitals, Molecular Orbital Theory assumes that electrons are delocalized over the entire molecule. When atomic orbitals combine, they form new molecular orbitals (MOs).
The LCAO Method: Constructive vs. Destructive Interference
According to the Linear Combination of Atomic Orbitals (LCAO) method, orbitals combine in two primary ways:
- Bonding Orbitals (Constructive Interference): Wave functions combine "in-phase." Electron density concentrates between the nuclei, shielding them from mutual repulsion and lowering the system's energy.
- Antibonding Orbitals (Destructive Interference): Wave functions combine "out-of-phase." A node (zero electron density) forms between the nuclei, causing repulsion and raising the system's energy. These are denoted with an asterisk (e.g., $\sigma^*, \pi^*$).
Consider two $p$-orbitals perpendicular to the internuclear axis. If they combine in-phase (side-on overlap), they produce a bonding $\pi$-orbital.
- The resulting orbital has $\pi$ symmetry (lobes above and below the axis).
- It is lower in energy than the original atomic orbitals.
- Crucial Concept: Removing an electron from this bonding orbital decreases the bond order, weakening the bond. Conversely, removing an electron from an antibonding orbital would increase stability.
Calculating Bond Order and Stability
MO theory quantifies stability using Bond Order. A bond order of 0 implies the molecule is too unstable to exist.
$$ \text{Bond Order} = \frac{1}{2} (\text{Number of Bonding } e^- - \text{Number of Antibonding } e^-) $$Practice Problem: The Instability of He$_2^+$
Why is the HeH$^+$ bond stronger than the He$_2^+$ bond?
To solve this, we compare their bond orders:
- HeH$^+$: 2 electrons total. Both occupy the $\sigma$ bonding orbital.
$$ \text{B.O.} = \frac{1}{2}(2 - 0) = 1 $$ Result: A stable single bond. - He$_2^+$: 3 electrons total (He has 2, He$^+$ has 1). Two electrons fill the $\sigma$ bonding orbital, but the third electron must go into the destabilizing $\sigma^*$ antibonding orbital.
$$ \text{B.O.} = \frac{1}{2}(2 - 1) = 0.5 $$ Result: The antibonding electron partially cancels the bonding interaction, making the bond significantly weaker.
Comparing Models: MO Theory vs. Lewis/VSEPR
Students are often asked to compare these theories. Here is the distinction:
- Lewis/VSEPR: Best for predicting geometry (shape) and bond connectivity. It relies on minimizing electron repulsion but ignores energy levels.
- MO Theory: Best for predicting energetics, magnetism (paramagnetism vs. diamagnetism), and spectroscopy. It captures electron delocalization that Lewis structures miss.
A Closer Look at Bond Energies and Energetics
Bond energy is the energy required to break a bond (endothermic). Conversely, energy is released when bonds form (exothermic). Understanding the structure of a molecule is the first step to calculating its total energetic profile.
Counting $\sigma$ and $\pi$ Bonds
Before calculating enthalpies, you must accurately identify the bond types. Remember:
- Single Bond: 1 $\sigma$ bond.
- Double Bond: 1 $\sigma$ bond + 1 $\pi$ bond.
- Triple Bond: 1 $\sigma$ bond + 2 $\pi$ bonds.
Question: How many $\sigma$ and $\pi$ bonds are in 1-butene ($CH_2=CH-CH_2-CH_3$)?
Step-by-Step Breakdown:
- Count C-H bonds: All are single $\sigma$ bonds.
Total = 2 + 1 + 2 + 3 = 8 $\sigma$ bonds. - Count C-C bonds:
- C=C double bond contains 1 $\sigma$ + 1 $\pi$.
- Two C-C single bonds contribute 2 $\sigma$.
- Total:
$\sigma$ bonds = 8 (C-H) + 3 (C-C) = 11 $\sigma$ bonds.
$\pi$ bonds = 1 $\pi$ bond.
Potential Energy Curves and Equilibrium
Why do atoms stop approaching each other at a specific distance? This is governed by the potential energy curve.
- Attraction: As atoms approach, potential energy decreases (stabilizes).
- Repulsion: If they get too close, internuclear repulsion causes energy to spike.
- Equilibrium Separation ($r_e$): The specific distance where potential energy is at its minimum. Here, attractive and repulsive forces are perfectly balanced. This distance is the bond length.
Resonance, Delocalization, and Stability
Resonance describes molecules where a single Lewis structure is insufficient. The electrons are delocalized across a network of atoms, significantly increasing stability.
Benzene: The Classic $\pi$-System
Benzene (C$_6$H$_6$) is the archetype of resonance stabilization. It has 6 carbons, each contributing one $p$-orbital electron to the $\pi$-system (Total = 6 $\pi$-electrons).
In the MO model of benzene:
- The 6 atomic $p$-orbitals combine to form 6 molecular orbitals.
- These split into 3 low-energy bonding MOs and 3 high-energy antibonding MOs.
- Following the Aufbau principle, the 6 $\pi$-electrons fill the lowest available energy levels.
- Conclusion: All 6 $\pi$-electrons reside in bonding molecular orbitals. This complete filling of bonding shells is what provides benzene its exceptional aromatic stability.
Nitrogen Gas ($N_2$): Optimization and Orbitals
Nitrogen is inert because of its incredibly strong triple bond ($N \equiv N$).
- Valence Bond View: Each Nitrogen is $sp$ hybridized. The $\sigma$ bond forms from $sp-sp$ overlap. The two $\pi$ bonds form from the side-by-side overlap of the remaining unhybridized $p$-orbitals ($p_x-p_x$ and $p_y-p_y$).
- Orbital Composition: The triple bond consists of one $\sigma$ and two $\pi$ orbitals.
Intermolecular Forces (IMFs) and Physical Properties
While bonds hold atoms together within a molecule, Intermolecular Forces (IMFs) act between molecules. These forces determine boiling points, vaporization energy, and solubility.
Hierarchy of IMFs (Weakest to Strongest)
- London Dispersion Forces (LDF): Present in all molecules. Depends on polarizability (molecular size/mass).
- Dipole-Dipole: Interactions between permanent dipoles in polar molecules.
- Hydrogen Bonding: Strong interaction between H bonded to N, O, or F and a lone pair on another electronegative atom.
- Ion-Dipole: Interaction between an ion and a polar molecule (e.g., salt dissolving in water).
Analyzing Vaporization Energies
The energy required to vaporize a liquid ($\Delta H_{vap}$) is a direct measure of IMF strength.
Given the liquids $N_2$, $Cl_2$, $HCl$, and $CH_3OH$, and relative energy values [1.0, 2.9, 3.6, 6.2], which value belongs to HCl?
Logic:
- Analyze Forces:
- $N_2$: Small nonpolar molecule. Weak LDF. (Lowest Energy)
- $Cl_2$: Larger nonpolar molecule. Stronger LDF than $N_2$.
- $HCl$: Polar molecule. Dipole-dipole forces (stronger than simple LDF).
- $CH_3OH$ (Methanol): Contains O-H group. Hydrogen bonding. (Highest Energy)
- Rank: $N_2 < Cl_2 < HCl < CH_3OH$
- Match Values: $1.0 < 2.9 < 3.6 < 6.2$
- Answer: HCl corresponds to 3.6.
Key Takeaways
- MO Theory explains electron delocalization and magnetic properties better than Lewis theory. Bond order dictates stability: $B.O. = 0.5 \times (bonding - antibonding)$.
- Counting Bonds: A triple bond always consists of 1 sigma and 2 pi bonds.
- Energetics: The equilibrium bond distance ($r_e$) is where potential energy is minimized.
- IMFs: Physical properties like vaporization energy are determined by the strength of intermolecular forces, ranked: Dispersion < Dipole-Dipole < Hydrogen Bonding.