MO Diagram of Oxygen

Introduction

Oxygen is the classic success of introductory MO theory. Its Lewis O=O drawing suggests a double bond but pairs the electrons on paper. The MO diagram also gives bond order two while placing one electron in each of two degenerate π orbitals, predicting the observed paramagnetism.

Core explanation

Each O atom has six valence electrons, so O₂ has twelve. For O₂, use the common ordering σ2s, σ2s , σ2p, the two π2p bonding orbitals, then the two π2p antibonding orbitals, followed by σ2p . Fill from lower to higher energy. Four electrons occupy σ2s²σ2s ². Two fill σ2p², four fill (π2p)⁴, and the final two enter separate equal-energy π2p orbitals with parallel spins under Hund's rule.

Bonding valence electrons total eight: two in σ2s, two in σ2p and four in π2p. Antibonding valence electrons total four: two in σ2s and two in π2p . The bond order is (8 − 4)/2 = 2. This aligns with an overall O=O double-bond description. The two π electrons are unpaired, so O₂ is paramagnetic in its ground state.

The Lewis picture and MO picture are not simply “one false, one true.” Lewis structure is useful for connectivity and electron bookkeeping, while MO theory captures a magnetic observable the simplest paired Lewis drawing misses. A student should not force a Lewis diagram to carry all electronic information or conclude that O₂ lacks a bond because some antibonding orbitals are occupied.

If one π electron is removed to form O₂⁺, antibonding count falls and order rises to 2.5, with one unpaired electron remaining. If one electron is added to form O₂⁻, antibonding count rises and order falls to 1.5, again with one unpaired electron in the simple model. If two are added to O₂²⁻, the π pair becomes fully occupied and the simple order falls to one with all spins paired. These comparisons are about MO occupancy; actual species' stability and chemistry require context.

Oxygen's magnetic behaviour can be demonstrated experimentally. The MO diagram explains why the ground-state molecule responds to a magnetic field. It does not by itself calculate the full magnetic susceptibility, reaction rate or bond dissociation enthalpy exactly.

Step-by-step reasoning

1. Count twelve O₂ valence electrons. 2. Select the O₂/F₂ order with σ2p below π2p. 3. Fill lower levels and put final two electrons singly in π2p pair. 4. Count eight bonding and four antibonding electrons for order two. 5. Count two unpaired electrons for paramagnetism.

Visual explanation

Draw the O₂ valence ladder with σ2p below π2p. Put ↑ in each of the two highest occupied π2p boxes, not ↑↓ in one. Circle those two single arrows and place a small magnet icon beside “paramagnetic.”

Real-world analogy

Two people in separate equal rooms remain unpaired, even though both are on the same floor. Hund filling of equal-energy π orbitals similarly leaves one electron in each, which matters to magnetism. The analogy is only a mnemonic for quantum spin rules.

Real-world example

Liquid oxygen is visibly attracted into a strong magnetic field in demonstrations. Its two unpaired ground-state electrons, predicted by MO filling, explain this behaviour more directly than a paired-electron Lewis O=O diagram.

Why?

Why is O₂ still bonded if it occupies antibonding orbitals? It has more bonding than antibonding electrons: eight versus four. The net difference produces a positive bond order of two.

Common misconception

“Two π electrons pair in one orbital because they belong to one molecule.” The two π orbitals are degenerate; Hund's rule places one electron in each before pairing, leaving two unpaired spins.

Worked example

Derive O₂'s two key predictions. Place twelve valence electrons as σ2s²σ2s ²σ2p²π2p⁴π2p ², with the last two distributed one per degenerate π orbital. Nb = 2 + 2 + 4 = 8, Na = 2 + 2 = 4, so order = 2. The separate π single occupancies yield two unpaired electrons, so the molecule is paramagnetic. Both conclusions require the filled diagram, not merely the formula O₂.

Quick check

1. How many unpaired electrons does ground-state O₂ have in the standard MO diagram? Answer: Two.

Exam focus

Use O₂'s correct 2p ordering and Hund filling. Show bonding and antibonding counts and report magnetism separately. Do not claim that antibonding occupation prevents any bond.

Advanced insight

The two unpaired electrons are associated with oxygen's triplet ground state. Excited singlet oxygen has a different spin arrangement and reactivity, illustrating that one molecular formula can have different electronic states.

Summary

O₂ has twelve valence electrons, bond order two and two unpaired π2p electrons in the basic MO model. This explains its paramagnetism while retaining a net O–O bond.

Practice questions

1. How many valence electrons are in O₂? Answer: Twelve. 2. What are Nb and Na in the standard valence diagram? Answer: Eight bonding and four antibonding electrons. 3. What is O₂'s calculated bond order? Answer: (8 − 4)/2 = 2. 4. What happens to order when one antibonding electron is removed to form O₂⁺? Answer: It rises to 2.5 in the simple model.