ChemistryISC Class 11 · Chemical Bonding and Molecular Structure

Molecular Orbital Theory: Bond Order, Stability and Magnetic Behaviour

Updated 17 October 2026 · Fact-checked

In molecular orbital theory atomic orbitals combine to form bonding and antibonding molecular orbitals. Bond order = (N_b − N_a)/2. A larger bond order means a stronger, shorter bond. A molecule with unpaired electrons is paramagnetic; one with all electrons paired is diamagnetic.

Understand Molecular Orbital Theory, Bond Order and Magnetism

Idea. When atomic orbitals overlap, they form an equal number of molecular orbitals: bonding (lower energy, electron density between the nuclei) and antibonding (higher energy, marked *, with a node between the nuclei). Electrons fill the molecular orbitals in order of increasing energy following the Aufbau, Pauli and Hund rules.

Energy order. For molecules up to N₂ (Li₂ to N₂, including B₂, C₂ and N₂): σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2p_y) < σ2p_z < (π*2pₓ = π*2p_y) < σ*2p_z. For O₂ and F₂ the σ2p_z is lower than π2p: σ2p_z < π2p.

Bond order. BO = (N_b − N_a)/2 where N_b and N_a are the numbers of electrons in bonding and antibonding orbitals. BO = 0 means the molecule does not exist (He₂, Ne₂); BO 1, 2, 3 correspond to single, double, triple bonds. A higher BO means higher bond dissociation enthalpy and shorter bond length.

Magnetic nature. Unpaired electrons in the MO diagram make the species paramagnetic (attracted to a magnetic field); all electrons paired makes it diamagnetic. The spin-only moment is μ = √(n(n + 2)) BM.

Key results. O₂ (16 electrons): ... σ2p_z² π2p⁴ π*2p² with two unpaired electrons in π*; BO 2, paramagnetic — something valence bond theory fails to predict. N₂ (14 electrons): BO 3, diamagnetic. Species: O₂⁺ (BO 2.5), O₂ (2), O₂⁻ (1.5), O₂²⁻ (1); bond length increases in the order O₂⁺ < O₂ < O₂⁻ < O₂²⁻.

Molecular Orbital Theory, Bond Order and Magnetism: formulas and key facts

Bond order: $B O = (N_{b} - N_{a}) /2$; BO = 0: species does not exist.
MO order (up to N₂): $σ 1 s < σ * 1 s < σ 2 s < σ * 2 s < π 2 p < σ 2 p_{z} < π * 2 p < σ * 2 p_{z}$; For O₂ and F₂, σ2p_z is below π2p.
Magnetic moment: $μ = \sqrt (n (n + 2)) B M$; n = number of unpaired electrons.
Trend: BO up ⇒ bond length down, bond enthalpy up; Compare isoelectronic or related species.

Molecular Orbital Theory, Bond Order and Magnetism in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Why liquid oxygen sticks to a magnet

Suppose a demonstration shows liquid oxygen being held between the poles of a strong magnet.

Explain this with the MO configuration of O₂ and find the bond order and magnetic moment.

Try it first, then show the solutionSolution

O₂ has 16 electrons: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2pₓ² π2p_y² π*2pₓ¹ π*2p_y¹.
N_b = 10 and N_a = 6, so BO = (10 − 6)/2 = 2.
Two unpaired electrons in π* orbitals: paramagnetic, μ = √(2 × 4) = 2.83 BM.

Answer: BO = 2; paramagnetic with 2 unpaired electrons (2.83 BM).

Memory hook: The unpaired electrons in the antibonding π* orbitals give the attraction to the magnet.

Tricks to remember Molecular Orbital Theory, Bond Order and Magnetism

Short memory aids. Say each one aloud once, then use it on a question the same day.

Count from the top
Fill 14 electrons for N₂-type order; for O₂ and F₂ swap σ2p_z and π2p.
Use it when: Writing MO configurations quickly.
Add to antibonding, BO falls
O₂⁻ and O₂²⁻ add electrons to π*, lowering the bond order.
Use it when: Species of O₂.

Molecular Orbital Theory, Bond Order and Magnetism: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Bond order of nitrogen · 1 mark · MCQBoard level

The bond order of the N₂ molecule is:
(a) 1
(b) 2
(c) 2.5
(d) 3

Try it first, then show the solutionSolution

N₂ has 10 bonding and 4 antibonding electrons: (10 − 4)/2 = 3.

Answer: (d) 3

Example 2: Bond order and magnetism of O₂ · 3 marks · Short answerBoard level

Write the molecular orbital configuration of O₂, and find its bond order and magnetic nature.

Try it first, then show the solutionSolution

16 electrons: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2pₓ² π2p_y² π*2pₓ¹ π*2p_y¹. (1 mark)
N_b = 10, N_a = 6, BO = 2. (1 mark)
Two unpaired electrons, so paramagnetic. (1 mark)

Answer: BO = 2; paramagnetic.

Example 3: Species of oxygen · 4 marks · Short answerHarder board level

Find the bond order and the number of unpaired electrons in O₂⁺, O₂⁻ and O₂²⁻, and arrange the four species O₂⁺, O₂, O₂⁻, O₂²⁻ in order of increasing bond length.

Try it first, then show the solutionSolution

O₂⁺ (15 e): N_a = 5, BO = (10 − 5)/2 = 2.5; 1 unpaired electron. (1 mark)
O₂⁻ (17 e): N_a = 7, BO = 1.5; 1 unpaired electron. (1 mark)
O₂²⁻ (18 e): N_a = 8, BO = 1; no unpaired electrons. (1 mark)
Bond length increases as bond order falls: O₂⁺ < O₂ < O₂⁻ < O₂²⁻. (1 mark)

Answer: O₂⁺ 2.5 (1 unpaired); O₂⁻ 1.5 (1 unpaired); O₂²⁻ 1 (0 unpaired); increasing length: O₂⁺ < O₂ < O₂⁻ < O₂²⁻.

Example 4: B₂ and C₂ · 4 marks · Short answerHarder board level

Write the MO configurations of B₂ (10 electrons) and C₂ (12 electrons) and find the bond order and magnetic nature of each.

Try it first, then show the solutionSolution

B₂: σ1s² σ*1s² σ2s² σ*2s² π2pₓ¹ π2p_y¹ (the π orbitals fill singly by Hund's rule). (1 mark)
B₂: N_b = 6, N_a = 4, BO = (6 − 4)/2 = 1; two unpaired electrons, paramagnetic. (1 mark)
C₂: σ1s² σ*1s² σ2s² σ*2s² π2pₓ² π2p_y². (1 mark)
C₂: N_b = 8, N_a = 4, BO = (8 − 4)/2 = 2 (both bonds π); all electrons paired, diamagnetic. (1 mark)

Answer: B₂: BO 1, paramagnetic. C₂: BO 2, diamagnetic.

Example 5: The nitrogen family · 5 marks · Long answerHOTS / competency

Write the MO configurations of N₂, N₂⁺ and N₂⁻ (14, 13 and 15 electrons). Find the bond order and the number of unpaired electrons in each, and arrange them in order of decreasing stability.

Try it first, then show the solutionSolution

N₂: ...π2pₓ² π2p_y² σ2p_z²; N_b = 10, N_a = 4; BO 3; no unpaired electrons. (1.5 marks)
N₂⁺ (13 e): ...π2p⁴ σ2p_z¹; N_b = 9, N_a = 4; BO 2.5; 1 unpaired electron. (1.5 marks)
N₂⁻ (15 e): ...π2p⁴ σ2p_z² π*2pₓ¹; N_b = 10, N_a = 5; BO 2.5; 1 unpaired electron. (1.5 marks)
Stability follows bond order: N₂ > N₂⁺ ≈ N₂⁻ (the latter two have the same bond order). (0.5 mark)

Answer: N₂: BO 3, 0 unpaired. N₂⁺: 2.5, 1 unpaired. N₂⁻: 2.5, 1 unpaired. Stability: N₂ greatest.

Where students lose marks in Molecular Orbital Theory, Bond Order and Magnetism

The slips examiners see most often, and what to write instead.

Mistake: Wrong MO order for O₂ and F₂
Using the N₂ order for O₂.
Do this: For O₂ and F₂ the σ2p_z is below π2p.
Usually costs: 1 to 2 marks.
Mistake: Pairing π* electrons
Writing π*2p⁴ with paired electrons for O₂.
Do this: Hund's rule: the two π* electrons occupy separate orbitals with parallel spins.
Usually costs: 2 marks.
Mistake: Counting inner shell electrons wrongly in BO
Leaving out σ1s and σ*1s.
Do this: They cancel, but count them or state that they cancel.
Usually costs: 1 mark.

Exam tips

Write the electron count and the MO configuration before the bond order.
State the bond order formula.
Link magnetism to the number of unpaired electrons.
Compare bond lengths through bond orders.

Practice questions on Molecular Orbital Theory, Bond Order and Magnetism

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following molecules is paramagnetic?
(a) N₂
(b) C₂
(c) O₂
(d) F₂
Hint
Look for unpaired electrons in the MO diagram.
Answer
(c) O₂
Q2 · 2 marks · Short answerBoard level
Why does the He₂ molecule not exist? Give the bond order.
Hint
He₂ has 4 electrons: σ1s² σ*1s².
Answer
N_b = 2, N_a = 2, so BO = 0; the stabilisation of the bonding electrons is cancelled by the antibonding pair, so no stable molecule forms.
Q3 · 3 marks · Short answerHarder board level
Find the bond order and magnetic nature of Li₂ and Be₂.
Hint
Li₂ has 6 electrons; Be₂ has 8.
Answer
Li₂: σ1s² σ*1s² σ2s²; BO = (4 − 2)/2 = 1; diamagnetic. Be₂: σ1s² σ*1s² σ2s² σ*2s²; BO = 0, so it does not exist as a stable molecule.
Q4 · 3 marks · Short answerHarder board level
Find the bond order of F₂ and Ne₂ using the MO order for O₂ and F₂.
Hint
F₂ has 18 electrons; Ne₂ has 20.
Answer
F₂: N_b = 10, N_a = 8, BO = 1 (a single bond, diamagnetic). Ne₂: N_b = 10, N_a = 10, BO = 0, so it does not exist.
Q5 · 3 marks · Short answerHarder board level
Find the bond order of CO (14 electrons), CN⁻ (14 electrons) and NO (15 electrons), and the magnetic nature of NO.
Hint
CO and CN⁻ are isoelectronic with N₂.
Answer
CO and CN⁻: BO = 3, diamagnetic. NO (15 e): N_b = 10, N_a = 5, BO = 2.5; one unpaired electron in π*, so NO is paramagnetic.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose liquid oxygen is observed to be attracted to a magnet.
(i) Give the MO configuration of O₂. (1 mark)
(ii) State the number of unpaired electrons and the spin-only magnetic moment. (1 mark)
(iii) Explain why valence bond theory (with two double bonds and all electrons paired) fails here. (1 mark)
(iv) What would happen to the bond length if an electron were removed from O₂? (1 mark)
Hint
Removing an electron from π* raises the bond order.
Answer
(i) σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p⁴ π*2p². (ii) 2 unpaired electrons; μ = √8 = 2.83 BM. (iii) Valence bond theory pairs all electrons and predicts diamagnetism; MO theory correctly puts two electrons singly in the degenerate π* orbitals. (iv) The bond order rises to 2.5 in O₂⁺, so the bond becomes shorter.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Bond order formula?Show answer
(N_b − N_a)/2.
2.Bond order of O₂?Show answer
2.
3.Is N₂ paramagnetic?Show answer
No, diamagnetic.
4.What does BO = 0 mean?Show answer
The molecule does not exist.

Molecular Orbital Theory, Bond Order and Magnetism: frequently asked questions

Why do antibonding orbitals lie higher in energy?

They have a node between the nuclei, so electrons in them are not shared in the bonding region and repel the nuclei apart.

Why is O₂ paramagnetic?

Its two highest-energy electrons occupy two degenerate π* orbitals singly with parallel spins.

Is MO theory better than valence bond theory?

It explains magnetic behaviour and the existence of species like O₂⁺ and H₂⁺ more directly, while valence bond theory is simpler for shapes.