Molecular Orbital Theory Diagrams

Why This Matters

Molecular orbital (MO) theory is one of the most powerful tools you'll encounter in General Chemistry II because it explains why molecules behave the way they do. While Lewis structures give you a quick sketch, MO diagrams reveal the electronic architecture that determines whether a molecule will form at all, how strong its bonds are, and whether it's attracted to a magnetic field.

You're being tested on your ability to construct and interpret MO diagrams, calculate bond order, predict paramagnetism vs. diamagnetism, and explain how orbital mixing affects energy levels. Exam questions love to ask you to compare molecules like $O_2$ and $N_2$, or explain why $CO$ is unusually stable. Don't just memorize electron configurations. Know what each orbital filling means for molecular properties.

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Foundational Concepts: Bonding vs. Antibonding

Before diving into specific molecules, you need to understand the core principle: when atomic orbitals combine, they form bonding orbitals (lower energy, stabilizing) and antibonding orbitals (higher energy, destabilizing). The balance between these determines everything about the molecule.

Bonding and Antibonding Orbitals

• Bonding orbitals (σ, π) form from constructive interference of atomic orbitals. Electron density concentrates between the nuclei, which is what holds the atoms together.
• Antibonding orbitals (σ*, π*) form from destructive interference. There's a node between the nuclei where electron density drops to zero, which actively weakens the bond.
• Bond order tells you the net bonding strength:

$\text{Bond Order} = \frac{(\text{bonding electrons}) - (\text{antibonding electrons})}{2}$

A higher bond order means a stronger, shorter bond. A bond order of zero means the molecule won't form.

MO Energy Level Diagrams

Filling MO diagrams follows the same rules you already know from atomic electron configurations:

1. Aufbau principle: fill the lowest-energy orbitals first.
2. Pauli exclusion principle: each orbital holds at most two electrons with opposite spins.
3. Hund's rule: when orbitals are degenerate (same energy), add one electron to each before pairing.

The tricky part is that orbital ordering shifts depending on the molecule. For homonuclear diatomics with $Z \leq 7$ (that's $B_2$, $C_2$, $N_2$), s-p mixing pushes the $\sigma_{2p}$ orbital higher in energy than the $\pi_{2p}$ orbitals. For $O_2$ and $F_2$, the "expected" order holds, with $\sigma_{2p}$ below $\pi_{2p}$. Getting this wrong will mess up your electron configuration and your magnetism prediction.

Unpaired electrons in the completed diagram mean the molecule is paramagnetic (attracted to a magnetic field). All electrons paired means diamagnetic.

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Homonuclear Diatomics: Symmetric Orbital Combinations

When two identical atoms combine, their atomic orbitals have equal energies, creating perfectly symmetric MO diagrams. The key challenge is knowing when orbital ordering changes.

Homonuclear Diatomic Molecules ($H_2$, $N_2$, $O_2$, $F_2$)

These are the molecules you'll diagram most often. Because the atoms are identical, bonding and antibonding orbitals form from equivalent atomic orbital contributions on each side.

• $H_2$ has bond order 1 (two bonding electrons, zero antibonding). It's the simplest MO diagram you can draw.
• $N_2$ has bond order 3, corresponding to a triple bond. All electrons are paired, so it's diamagnetic.
• $O_2$ has bond order 2, corresponding to a double bond. Its MO diagram shows two unpaired electrons in the degenerate $\pi^*_{2p}$ orbitals, making it paramagnetic. This is a fact that Lewis structures completely miss, and it's one of MO theory's biggest wins.
• $F_2$ has bond order 1. All electrons are paired (diamagnetic), and the bond is relatively weak because many antibonding orbitals are occupied.

Orbital Ordering in Homonuclear Diatomics

This is the detail that trips people up most on exams:

• For $B_2$, $C_2$, $N_2$ ($Z \leq 7$): s-p mixing is significant. The 2s and 2p orbitals are close enough in energy that they interact, pushing $\sigma_{2p}$ above $\pi_{2p}$. So $\pi_{2p}$ fills first.
• For $O_2$, $F_2$ ($Z > 7$): the energy gap between 2s and 2p is larger, so s-p mixing is minimal. $\sigma_{2p}$ sits below $\pi_{2p}$, which is the "expected" order.

Why does this matter? Consider $B_2$: with the reversed ordering, its two highest-energy electrons go into the two degenerate $\pi_{2p}$ orbitals (one each, by Hund's rule), making $B_2$ paramagnetic. With the wrong ordering, you'd predict it's diamagnetic.

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Heteronuclear Diatomics: Unequal Contributions

When two different atoms bond, their atomic orbitals sit at different energies because of electronegativity differences. This asymmetry shifts electron density and creates polarity.

Heteronuclear Diatomic Molecules ($CO$, $NO$)

In a heteronuclear MO diagram, the more electronegative atom's atomic orbitals are drawn lower in energy. That atom contributes more to the bonding molecular orbitals, meaning bonding electrons spend more time near it. The less electronegative atom contributes more to the antibonding orbitals.

• $CO$ has bond order 3. It's isoelectronic with $N_2$ (both have 14 electrons), and its bond is actually stronger than the $N_2$ triple bond (1072 kJ/mol vs. 945 kJ/mol). This exceptional stability comes from the favorable energy match between carbon and oxygen orbitals.
• $NO$ has bond order 2.5. It has 15 electrons, with one unpaired electron in a $\pi^*$ orbital. That makes it paramagnetic and highly reactive as a free radical.

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Polyatomic Molecules: Geometry Meets MO Theory

For molecules with three or more atoms, MO theory combines with hybridization and VSEPR to explain geometry. The shape depends on how orbitals mix and how lone pairs arrange themselves.

Linear Triatomic Molecules ($BeH_2$, $CO_2$)

• 180° bond angles result from sp hybridization of the central atom, which creates two equivalent hybrid orbitals pointing in opposite directions.
• In $CO_2$, each C=O bond consists of one σ bond and one π bond. The π electrons are delocalized across the molecule.
• Neither $BeH_2$ nor $CO_2$ has lone pairs on the central atom, so there's nothing to distort the geometry away from linear.

Angular Triatomic Molecules ($H_2O$)

• 104.5° bond angle results from sp³ hybridization with two lone pairs on oxygen. Those lone pairs compress the H-O-H angle below the ideal tetrahedral angle of 109.5°.
• Lone pairs occupy more space than bonding pairs because they're held closer to the nucleus and spread out more. This extra repulsion pushes the bonded atoms closer together.
• The bent geometry gives water a net dipole moment, which is responsible for hydrogen bonding and many of water's unusual properties.

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Larger Molecular Geometries: Expanding the Octet

Molecules with central atoms from period 3 and beyond can accommodate more than eight electrons. These expanded octets lead to geometries that go beyond what sp, sp², and sp³ can produce.

Tetrahedral Molecules ($CH_4$)

• sp³ hybridization creates four equivalent orbitals arranged at 109.5° around carbon.
• Each C-H bond is a σ bond formed from overlap of an sp³ hybrid orbital on carbon with hydrogen's 1s orbital.
• The symmetrical tetrahedral arrangement cancels all bond dipoles, making methane nonpolar.

Octahedral Molecules ($SF_6$)

• sp³d² hybridization creates six equivalent orbitals pointing toward the vertices of an octahedron, with 90° bond angles.
• Sulfur expands its octet using 3d orbitals to form six S-F σ bonds.
• Despite each S-F bond being polar, the perfect octahedral symmetry cancels all dipoles, so $SF_6$ is nonpolar overall.

Square Planar Molecules ($XeF_4$)

$XeF_4$ has six electron domains around xenon (four bonding pairs and two lone pairs), which start in an octahedral arrangement. The two lone pairs position themselves opposite each other (in the axial positions) to minimize repulsion, leaving the four fluorine atoms in a square plane.

• The hybridization is sp³d² (same electron domain count as octahedral).
• Bond angles are 90° within the plane.
• The symmetric arrangement of fluorine atoms cancels all dipoles, making $XeF_4$ nonpolar despite having lone pairs.

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Pi Bonding and Delocalization

Pi bonds form from side-by-side overlap of unhybridized p orbitals and are weaker than sigma bonds. They're crucial for understanding double/triple bonds and the special stability of aromatic systems.

Pi-Bonding in Organic Molecules (Ethylene, Benzene)

Ethylene ($C_2H_4$) has a C=C double bond made of one σ bond (from sp² orbital overlap) and one π bond (from unhybridized p orbital overlap). The π bond locks the molecule into a planar shape and prevents rotation around the double bond.

Benzene ($C_6H_6$) takes π bonding further. Each carbon is sp² hybridized, leaving one unhybridized p orbital perpendicular to the ring. These six p orbitals overlap to form a continuous molecular orbital above and below the ring plane, and the six π electrons are delocalized across all six carbons. This delocalization is what gives benzene its aromatic stability.

Both molecules have approximately 120° bond angles due to sp² hybridization.

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Quick Reference Table

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Self-Check Questions

1. Why does $O_2$ exhibit paramagnetism while $N_2$ is diamagnetic, even though both are homonuclear diatomic molecules? Draw the MO diagrams to support your answer.

2. Calculate the bond order for $NO$. Based on this value, would you predict $NO$ to be more or less stable than $N_2$? Explain.

3. Compare the molecular geometries of $CO_2$ and $H_2O$. Both have three atoms. Why does one end up linear and the other bent?

4. If an FRQ asks you to explain why $CO$ has an unusually strong bond despite being heteronuclear, what concepts from MO theory would you use in your response?

5. How does the MO diagram for $N_2$ differ from that of $O_2$ in terms of orbital ordering, and what causes this difference?