Why is ordinary oxygen attracted to a magnetic field when a simple O=O Lewis structure makes its electrons look paired? I like beginning H3 Chemistry molecular orbital theory with that question, because it reveals what a model is for. A Lewis structure captures useful bonding information, but it cannot show every feature of the molecule’s electronic state. A molecular orbital diagram offers a different, testable account: O₂ contains two unpaired electrons in separate π* antibonding orbitals, making it paramagnetic.
This is a practical H3 Chemistry tutor Hougang topic because the Singapore–Cambridge 2027 H3 Chemistry 9813 syllabus asks students to construct and interpret molecular orbital diagrams for simple homonuclear diatomic molecules, including O₂ and F₂. Students also identify HOMO and LUMO levels, but a correctly filled diagram must come before those labels. Here I will teach the electron-counting and magnetism question in full, then connect it to bond order.
SEAB 2027 H3 Chemistry 9813 syllabus · OpenStax molecular orbital theory and O₂ example · HOMO and LUMO foundations
Go to: Why Lewis structures are insufficient · Orbital order · Fill the O₂ diagram · Bond order · Paramagnetism · Compare F₂ · Practice · FAQs
The O₂ Puzzle: A Good Model Must Match an Observation
The usual O=O drawing is valuable. It tells us that the two oxygen atoms are connected by a double bond in a standard valence picture. But when we represent the remaining electrons as pairs, we have not explained the experimentally observed magnetic behaviour. Oxygen is paramagnetic: in an applied magnetic field, its unpaired-electron spins produce a magnetic response.
Rather than discarding Lewis structures, chemists use a model that can answer the new question. In molecular orbital theory, atomic orbitals combine into orbitals spread over the molecule. Electrons fill these molecular orbitals according to their energies and spin rules. Some orbitals stabilise bonding; others are antibonding.
The lesson is not that the first model was useless. The lesson is that a model earns our trust by the observations it explains, and we must know where its limits lie.
Know the Orbital Energy Order Before Filling Electrons
For the usual simplified O₂/F₂ valence molecular orbital diagram, the levels in increasing energy are σ(2s), σ*(2s), σ(2p), π(2p) pair, π*(2p) pair, σ*(2p). The asterisk means antibonding. Each sigma orbital can hold two electrons of opposite spin. Each set of two degenerate pi orbitals can hold four electrons in total.
The important practical rule is that the two equal-energy π* orbitals are filled according to Hund’s rule: place one electron in each before pairing them. That single decision reveals the magnetism of O₂.
Be aware that the relative ordering of certain 2p sigma and pi orbitals differs in lighter diatomic molecules because of orbital mixing. For O₂ and F₂, use the appropriate energy ordering shown here; do not apply one universal memorised ladder to every molecule.
Step-by-Step: Filling the Oxygen O₂ Molecular Orbital Diagram
Each oxygen atom contributes six valence electrons (2s²2p⁴), so O₂ has 12 valence electrons. Counting only the valence-shell MOs keeps this problem focused; the fully occupied core 1s contributions cancel in the bonding calculation.
- First 2 electrons: fill σ(2s)², a bonding orbital.
- Next 2: fill σ*(2s)², an antibonding orbital.
- Next 2: fill σ(2p)², a bonding orbital.
- Next 4: fill the two π(2p) bonding orbitals completely, giving π(2p)⁴.
- Final 2: place one in each of the degenerate π*(2p) antibonding orbitals, giving π*(2p)² with two unpaired electrons.
The compact valence configuration is σ(2s)² σ*(2s)² σ(2p)² π(2p)⁴ π*(2p)². The superscript two on π*(2p)² describes the total electrons in the pair of degenerate orbitals; it does not mean the two electrons must be paired together inside a single orbital.
If the student draws both arrows in the first π* box and leaves the second empty, the diagram is wrong for the ground state. It violates Hund’s rule and destroys the correct prediction of paramagnetism.
Calculate O₂ Bond Order Without Guessing
The simplified molecular orbital bond-order expression is bond order = (number of bonding electrons − number of antibonding electrons) / 2. Count these only after filling every orbital.
- Bonding electrons: σ(2s)² contributes 2, σ(2p)² contributes 2 and π(2p)⁴ contributes 4. Total = 8.
- Antibonding electrons: σ*(2s)² contributes 2 and π*(2p)² contributes 2. Total = 4.
- Bond order = (8 − 4) / 2 = 2.
That predicted bond order is consistent with a double bond in O₂. But the MO model explains something extra: the two unpaired electrons. The most useful exam response mentions both conclusions and connects each to its respective evidence in the diagram.
Why O₂ Is Paramagnetic
An unpaired electron carries a magnetic moment. A molecule with one or more unpaired electrons is generally paramagnetic, meaning that it exhibits a response attracted towards stronger magnetic fields in the familiar educational context. A molecule in which every electron is paired is diamagnetic; that is a different, typically much weaker magnetic response.
The two π* electrons in O₂ occupy separate equal-energy orbitals with parallel spins in the ground-state diagram. They do not cancel as an electron pair in the same orbital would. Oxygen is consequently paramagnetic. This is a classic case where a measured physical property helps chemists assess and prefer one bonding description over another.
For a second independent worked explanation, see OpenStax Chemistry 2e on molecular orbital diagrams, bond order and unpaired electrons.
O₂ Versus F₂: Two More Electrons Change the Result
The comparison with fluorine is particularly instructive. Each F atom has seven valence electrons, so F₂ has 14 valence electrons, two more than O₂. The lower MOs are filled in the same simplified sequence, but those two additional electrons pair with the electrons in the π* pair.
- For O₂: 12 valence electrons; 8 bonding and 4 antibonding; bond order 2; two unpaired electrons; paramagnetic.
- For F₂: 14 valence electrons; 8 bonding and 6 antibonding; bond order 1; all electrons paired; diamagnetic.
Two things have happened. Filling more antibonding orbitals reduces the net bond order, and completing the degenerate π* pair removes the unpaired spins. It is not true that ‘more electrons always mean a stronger bond’. Where the new electrons go is what matters.
A useful cross-check is H₂. Its two electrons occupy the bonding σ(1s) orbital, giving bond order (2−0)/2 = 1, and the electrons are paired. That confirms the counting method can be applied to a simpler diatomic molecule.
Original H3 Chemistry Practice With Answers
Question 1: A misplaced pair
A candidate puts both of O₂’s final two electrons into one π* orbital and leaves the other π* orbital empty. What rule has been neglected?
Answer: Hund’s rule. Degenerate orbitals are singly occupied with parallel spins before pairing; the correct ground-state O₂ configuration has one electron in each π* orbital.
Question 2: Counting the bond
A molecule has eight valence electrons in bonding MOs and four in antibonding MOs. Calculate its bond order.
Answer: (8 − 4)/2 = 2. State that this uses the molecular orbital electron counts, not the number of lines in a drawn Lewis formula.
Question 3: Magnetism and electron count
A diagram shows two electrons in two separate degenerate antibonding orbitals. What magnetic classification is predicted?
Answer: Paramagnetic, because the electrons are unpaired. The explanation must identify the actual spin occupancy, not only the word ‘antibonding’.
Question 4: F₂ versus O₂
Why does adding two valence electrons when moving from O₂ to F₂ decrease the bond order and remove paramagnetism?
Answer: The extra electrons occupy previously singly occupied π* antibonding orbitals. This increases the antibonding count from four to six, reducing bond order from two to one, and pairs the remaining unpaired electrons.
What I Would Correct Before an H3 Test
- Counting all electrons without distinguishing valence and core. Include core consistently if using a full diagram, or work with valence MOs consistently as done here.
- Assuming π*(2p)² is an electron pair in one orbital. Two degenerate π* orbitals share that total occupancy.
- Using bond order to decide paramagnetism. Bond order and magnetic classification are related to occupancy but answer separate questions.
- Assuming any antibonding occupancy makes a molecule unstable. Bonding and antibonding contributions must be compared; O₂ is stable despite populated π* orbitals.
- Applying the O₂/F₂ ordering indiscriminately. Some lighter diatomics have a different ordering of 2p-derived levels.
How I Would Move a Student From Memorisation to H3 Reasoning
I would first ask the student to draw the O₂ energy ladder without electrons, naming bonding and antibonding levels. Then they fill 12 arrows with the correct spin rules. They must give two independent predictions: the bond order and the magnetic behaviour. Only then would I change the species to F₂ and ask what was added and why the results differ.
This approach works because it forces an explanation of cause and effect. If a student can identify which orbitals acquired electrons, they can often infer how an observed property changes. It is the same disciplined movement from a model to a testable prediction that matters in spectroscopy, materials research and later chemical studies.
For the broader molecular-orbital framework, see Molecular Orbitals, HOMO and LUMO. For integrated application questions, use the 2027 H3 exam and stimulus-question guide. This article owns the narrower oxygen magnetism and bond-order calculation.
Families in Hougang considering specialist support can look at the eduKate JC2 H3 tutorial overview and enquire about appropriate placement, group size and venue. The existing centre details are not evidence of a separate Hougang classroom, so confirm the teaching location directly.
Frequently Asked Questions
How many unpaired electrons does O₂ have?
Two in the usual ground-state molecular orbital diagram, occupying separate degenerate π* antibonding orbitals.
What is the bond order of O₂?
Two, from eight electrons in bonding valence orbitals and four in antibonding valence orbitals: (8−4)/2 = 2.
Why is F₂ diamagnetic?
Its electrons are paired in the ground-state molecular orbital configuration. The two π* orbitals that are singly occupied in O₂ are fully occupied in F₂.
Is the O₂ molecular orbital diagram part of the H3 syllabus?
Yes. The 2027 syllabus 9813 specifies molecular orbital diagrams for simple homonuclear diatomic molecules including H₂, O₂ and F₂.
A good Chemistry student remembers that oxygen is paramagnetic. An excellent one can build the diagram, place the electrons, calculate the bond order and explain exactly why the magnetic prediction follows. The scientific reward is a habit: when a model is challenged by an observation, improve the explanation instead of ignoring the observation.
Hougang JC Chemistry tuition hub · H3 Chemistry Singapore subject guide · Ask eduKate about H3 Chemistry tutorials