SCIENCE · INVESTIGATE AND EXPLAIN

Use orbital occupancy to compare bond order and magnetism

Fill a supplied degenerate orbital scheme and check two independent consequences.

Learning outcome: Compare10,12,13 and14 electrons in the higher-Z ordering. Capture orbital boxes, electron sums, bond orders and unpaired counts; compare the two supplied orderings at6 electrons.

Starting recap

Review the electron and formal-charge budget, bond multiplicity, signed interaction energies and the distinction between a model and a measured property.

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Try

Twelve valence electrons in the oxygen-like ordering give8 bonding and4 antibonding electrons: bond order2 and two unpaired electrons in pi-star orbitals.

Open the experiment

Predict

Can a Lewis double-bond drawing alone explain why ground-state oxygen is paramagnetic?

Your prediction and reason

Experiment

Compare10,12,13 and14 electrons in the higher-Z ordering. Capture orbital boxes, electron sums, bond orders and unpaired counts; compare the two supplied orderings at6 electrons.

Fill a supplied molecular-orbital ordering

Worked teaching model · synthetic values · Supplied second-period homonuclear valence ordering. 2s and 2s* fill first; twofold pi sets obey Hund single occupancy before pairing. Ordering0 approximates B2–N2, ordering1 O2–F2. Formal other counts/orderings are comparisons, not claims of stable molecules or valid heteronuclear diagrams.

Valence electron count: 12 electrons p ordering: 0=pi below sigma, 1=sigma below pi: 1 choice

Reset keeps your prediction, captures and practice history.

Low-to-high energy ordering supplied by the selected model; row spacing is not an energy scale
Orbital set	Type	Spatial orbitals and spin projections
σ2s	bonding	↑↓
σ*2s	antibonding	↑↓
σ2p	bonding	↑↓
π2p	bonding	↑↓↑↓
π*2p	antibonding	↑↑
σ*2p	antibonding	–

Calculated evidence under the declared representation
Quantity	Value	Unit
Bonding electrons	8	electrons
Antibonding electrons	4	electrons
Bond order	2	ratio
Unpaired electrons	2	electrons
Electron total	12	electrons

Bond order and unpaired occupancy test different aspects of the supplied diagram.

A positive formal bond order is not a complete stability calculation. Core 1s bonding/antibonding occupancies are omitted together and cancel from this valence bond order. Degenerate pi orbitals fill singly with parallel spin projections before pairing. The ordering choice is a stipulated approximation, not a universal sequence for all molecules.

Bond order and unpaired occupancy test different aspects of the supplied diagram.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

A different representation can test a different observation

Molecular orbitals extend over a molecule rather than assigning every pair to one localized bond. Combining atomic functions produces bonding and antibonding combinations. A bonding orbital contributes positively to the formal bond-order count, while an antibonding orbital contributes negatively. The elementary bond-order expression is(Nbonding−Nantibonding)/2.

Fill the declared ordering

The workbench includes valence sigma2s and sigma-star2s, then supplied2p sigma/pi orderings. Two degenerate pi orbitals fill singly before pairing in the ground-configuration model. Each spatial orbital accepts at most two electrons with opposite spin projections. Core1s bonding/antibonding pairs are omitted together; their contributions cancel in this valence bond-order count.

Reconcile oxygen’s two observations

With12 valence electrons in the oxygen-like ordering,8 are bonding and4 antibonding. The bond order is2, consistent with a basic double-bond count, but the last two occupy separate degenerate pi-star orbitals. Two unpaired electrons explain paramagnetism, an observation that an all-paired Lewis drawing misses. The lesson does not ask for liquid-oxygen handling or a magnetic experiment.

Predict a changed electron count

Adding one electron gives8 bonding and5 antibonding, bond order1.5 and one unpaired electron. Adding a second gives bond order1 and no unpaired electrons in this supplied scheme. Removing an antibonding electron instead increases formal bond order. These comparisons distinguish electron accounting from the unsupported slogan that adding electrons always strengthens a bond.

Respect the ordering boundary

The lower-Z B2–N2 approximation places bonding pi2p below sigma2p; the O2–F2 approximation reverses that order. At6 valence electrons, the former produces two unpaired pi electrons, while the latter pairs them in sigma. Choosing the right model matters to the magnetic prediction. Arbitrary control combinations are formal comparisons, not a catalog of stable species. Heteronuclear energy offsets, excited states and full energy calculations require additional evidence beyond this bounded diagram.

Sources and evidence boundary

Conceptual sources checked 30 October 2026: Molecular orbital theory. All prose, cases, code and diagrams are original. The supplied models are not measured structures, a performed laboratory investigation or certification of independent mastery. No chemical handling, heating, pressure vessel or cryogenic experiment is requested. Retain predictions, contrasting evidence and a changed-case explanation.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Why is a positive formal bond order not a complete stability proof or a valid diagram for every heteronuclear molecule?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Audit sigma, pi and hybrid descriptions without literal orbital machinery Next: Compare dispersion contacts without a mass-only shortcut

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