ChemistryCBSE Class 11 · Chemical Bonding and Molecular Structure

MO Theory: Oxygen, Nitrogen and Related Diatomic Species Class 11

Q: Why is a higher bond order associated with a shorter bond?

More net bonding electrons pull the nuclei closer.

Q: What does the star in σ* mean?

It marks an antibonding orbital.

Q: Can bond order be a fraction?

Yes, e.g. 2.5 for O₂⁺ and NO.

Q: Why do Be₂ and Ne₂ not exist?

Their bonding and antibonding electrons are equal in number, so the bond order is zero.

Q: What is the difference between MO and valence bond descriptions?

VB treats electrons as localised in bonds between two atoms; MO treats them as delocalised over the whole molecule and explains paramagnetism.

Updated 17 October 2026 · Fact-checked

In molecular orbital theory bond order = (N_b − N_a)/2, where N_b and N_a are the electrons in bonding and antibonding orbitals; a larger bond order means a shorter, stronger bond, and unpaired electrons make the species paramagnetic. For O₂ the filling order is σ2s, σ*2s, σ2pz, π2px = π2py, π*2px = π*2py, σ*2pz, giving bond order 2 and two unpaired electrons; for N₂ and lighter molecules the π2p orbitals lie below σ2pz.

Understand MO Theory: Oxygen, Nitrogen and Related Diatomic Species

Principle. Atomic orbitals combine to give molecular orbitals: a bonding MO lower in energy and an antibonding MO (marked *) higher in energy. Electrons fill MOs by the Aufbau, Pauli and Hund rules.
$ $Bond order = \frac{1}{2} (N_{b} - N_{a}) .$ $
Bond order 0 means no stable molecule; higher bond order means higher bond enthalpy and shorter bond length. A species with unpaired electrons is paramagnetic; with all electrons paired it is diamagnetic.

Orbital order.

For O₂, F₂ (and Ne₂): $σ 1 s < σ^{*} 1 s < σ 2 s < σ^{*} 2 s < σ 2 p_{z} < (π 2 p_{x} = π 2 p_{y}) < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$ .
For B₂, C₂, N₂ (s–p mixing): $σ 2 s < σ^{*} 2 s < (π 2 p_{x} = π 2 p_{y}) < σ 2 p_{z} < (π^{*} 2 p_{x} = π^{*} 2 p_{y}) < σ^{*} 2 p_{z}$ .

Oxygen and its ions (valence MO count of 12 electrons for O₂: $N_{b} = 8$ , $N_{a} = 4$ ): O₂⁺ has 11 valence electrons (BO 2.5, one unpaired), O₂ has 12 (BO 2, two unpaired in $π^{*}$ ), O₂⁻ has 13 (BO 1.5, one unpaired), O₂²⁻ has 14 (BO 1, diamagnetic). Bond length increases in the order O₂⁺ < O₂ < O₂⁻ < O₂²⁻. Adding an electron to $π^{*}$ lowers the bond order; removing an electron from $π^{*}$ raises it.

Nitrogen group. N₂ (10 valence electrons): $N_{b} = 8$ , $N_{a} = 2$ , BO 3, diamagnetic. Removing an electron from N₂ takes it from a bonding orbital ( $σ 2 p_{z}$ ): N₂⁺ has BO 2.5.

Heteronuclear diatomics (similar atoms). NO (11 valence electrons) is treated like O₂⁺: BO 2.5, one unpaired electron in $π^{*}$ . CO, CN⁻, NO⁺ and N₂ are isoelectronic (10 valence electrons): BO 3, diamagnetic. CN has 9 valence electrons (BO 2.5, one unpaired).

Reading the data. Compare species by (i) electron count, (ii) bond order, (iii) magnetic behaviour, and (iv) bond length. A lower ionisation energy for O₂ than for N₂ arises because the electron is removed from an antibonding orbital in O₂ (raising the bond order) but from a bonding orbital in N₂.

MO Theory: Oxygen, Nitrogen and Related Diatomic Species: formulas and key facts

Bond order: $B O = \frac{1}{2} (N_{b} - N_{a})$; Bonding minus antibonding.
O₂ valence configuration: $σ 2 s^{2} σ^{*} 2 s^{2} σ 2 p_{z}^{2} π 2 p^{4} π^{*} 2 p^{2}$; BO = 2.
N₂ valence configuration: $σ 2 s^{2} σ^{*} 2 s^{2} π 2 p^{4} σ 2 p_{z}^{2}$; BO = 3.
Magnetism: $unpaired electrons > 0 \Rightarrow paramagnetic$; Else diamagnetic.
Bond length trend: $B O ↑\Rightarrow length ↓$; Higher order, shorter bond.
Isoelectronic species: $CO, CN^{-}, NO^{+}, N_{2} : B O = 3$; 10 valence electrons.

How to solve MO Theory: Oxygen, Nitrogen and Related Diatomic Species questions

Count the valence electrons, fill the MOs in the correct order for the element pair, and read off BO and unpaired electrons.

1Count the total valence electrons (add for negative charge, subtract for positive).
2Choose the energy order: O₂/F₂ type or B₂–N₂ type.
3Fill the MOs with Aufbau, Pauli and Hund rules.
4Count N_b and N_a and find the bond order.
5Count unpaired electrons for magnetism.
6Compare bond lengths and stabilities by bond order.

MO Theory: Oxygen, Nitrogen and Related Diatomic Species 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.

Liquid oxygen between magnet poles

Suppose a small amount of liquid oxygen is poured between the poles of a strong magnet and stays there.

Use MO theory to explain.

Try it first, then show the solutionSolution

O₂ has 12 valence electrons; the last two occupy the degenerate π*2px and π*2py orbitals singly with parallel spins (Hund).
Two unpaired electrons make O₂ paramagnetic, so it is attracted to the magnet.

Answer: O₂ is paramagnetic (2 unpaired electrons).

Memory hook: Valence-bond theory cannot explain this.

Poisonous CO

Suppose a student wonders why carbon monoxide has a very strong bond.

Find its bond order by MO theory.

Try it first, then show the solutionSolution

CO has 10 valence electrons, isoelectronic with N₂: N_b = 8, N_a = 2.
BO = (8 − 2)/2 = 3.

Answer: Bond order 3.

Memory hook: CO has a triple bond like N₂.

Tricks to remember MO Theory: Oxygen, Nitrogen and Related Diatomic Species

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

O₂ ladder
O₂⁺ 2.5, O₂ 2, O₂⁻ 1.5, O₂²⁻ 1.
Use it when: Bond orders of oxygen species.
Ten valence electrons
N₂, CO, CN⁻, NO⁺: BO 3.
Use it when: Isoelectronic.
Antibonding electrons hurt
Adding to π* lowers BO; removing raises it.
Use it when: Ion questions.
Paramagnetic check
Any unpaired electron means paramagnetic.
Use it when: Magnetism.

MO Theory: Oxygen, Nitrogen and Related Diatomic Species: solved examples

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

Example 1: Oxygen molecule · 2 marks · Very short answerBoard level

Write the valence molecular orbital configuration of O₂, find its bond order and say whether it is paramagnetic.

Try it first, then show the solutionSolution

σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹ (N_b = 8, N_a = 4). (1)
BO = (8 − 4)/2 = 2; two unpaired electrons in π* make it paramagnetic. (1)

Answer: BO = 2; paramagnetic.

Example 2: Oxygen ions · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

O₂⁺ (11 e): N_b = 8, N_a = 3, BO = 2.5, 1 unpaired. (1)
O₂⁻ (13 e): N_a = 5, BO = 1.5, 1 unpaired; O₂²⁻ (14 e): N_a = 6, BO = 1, 0 unpaired. (1)
Higher BO means shorter bond: O₂⁺ < O₂ < O₂⁻ < O₂²⁻. (1)

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

Example 3: Nitric oxide and relatives · 3 marks · Short answerHarder board level

Using the MO order of homonuclear diatomics, find the bond order and magnetic behaviour of NO, NO⁺ and NO⁻ (valence electrons 11, 10 and 12).

Try it first, then show the solutionSolution

NO (11 e): the extra electron goes into π*: N_b = 8, N_a = 3, BO = 2.5, one unpaired electron, paramagnetic. (1)
NO⁺ (10 e): N_a = 2, BO = 3, diamagnetic. (1)
NO⁻ (12 e): N_a = 4, BO = 2, two unpaired electrons, paramagnetic (like O₂). (1)

Answer: NO 2.5 (para); NO⁺ 3 (dia); NO⁻ 2 (para).

Example 4: Isoelectronic species and ionisation · 4 marks · Long answerHarder board level

(a) Name four species isoelectronic with N₂ and give their bond order. (b) Find the bond order of N₂⁺ and O₂⁺ and say how ionisation changes the bond length in each case. (c) Explain why the ionisation energy of O₂ is lower than that of N₂.

Try it first, then show the solutionSolution

(a) CO, CN⁻, NO⁺ and C₂²⁻ (10 valence electrons): BO = 3 each. (1)
(b) N₂⁺: an electron is removed from the bonding σ2p_z, N_b = 7, BO = 2.5, so the bond lengthens. (1)
O₂⁺: an electron is removed from the antibonding π*, N_a = 3, BO = 2.5, so the bond shortens. (1)
(c) O₂ loses a high-energy antibonding electron, which is easier than removing a bonding electron from N₂. (1)

Answer: (a) CO, CN⁻, NO⁺, C₂²⁻ (BO 3); (b) N₂⁺ 2.5 (longer), O₂⁺ 2.5 (shorter); (c) antibonding electron.

Example 5: Peroxide ion and oxygen · 4 marks · Long answerHOTS / competency

Suppose the O–O bond lengths are 121 pm in O₂ and 149 pm in the peroxide ion O₂²⁻. (a) Show that O₂²⁻ is isoelectronic with F₂ and find its bond order. (b) Correlate the lengths with the bond orders. (c) Which species contains more unpaired electrons: O₂ or O₂²⁻?

Try it first, then show the solutionSolution

(a) O₂²⁻ has 16 + 2 = 18 electrons, the same as F₂ (2 × 9); valence electrons 14 each. (1)
N_b = 8, N_a = 6, BO = 1. (1)
(b) O₂ (BO 2) has the shorter bond (121 pm) and O₂²⁻ (BO 1) the longer one (149 pm). (1)
(c) O₂ has two unpaired electrons; O₂²⁻ has all electrons paired. (1)

Answer: (a) isoelectronic with F₂, BO = 1; (b) BO 2 gives 121 pm, BO 1 gives 149 pm; (c) O₂.

Where students lose marks in MO Theory: Oxygen, Nitrogen and Related Diatomic Species

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

Mistake: Using the O₂-type order for N₂.
Mixing the two energy orders.
Do this: N₂ and lighter have π2p below σ2p_z.
Usually costs: The BO.
Mistake: Counting total electrons rather than valence electrons for BO.
Including the 1s electrons.
Do this: Count only the valence MOs or include σ1s, σ*1s consistently (they cancel).
Usually costs: The BO.
Mistake: Claiming a species is diamagnetic when π* is half-filled.
Ignoring Hund's rule.
Do this: Fill degenerate orbitals singly first.
Usually costs: The magnetism mark.
Mistake: Saying removing an electron always lowers BO.
Ignoring antibonding orbitals.
Do this: Removing from an antibonding orbital raises BO.
Usually costs: Concept mark.

Exam tips

Write the full MO configuration before the bond order.
State the MO energy order used for the molecule.
Give the number of unpaired electrons explicitly.
Link bond order with bond length and bond energy.
Mention isoelectronic species when relevant.

Practice questions on MO Theory: Oxygen, Nitrogen and Related Diatomic Species

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 species is paramagnetic?
(a) N₂ (b) O₂²⁻ (c) CO (d) O₂⁻
Hint
Look for unpaired electrons.
Answer
(d) O₂⁻ has one unpaired electron.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The oxygen molecule is paramagnetic although all its valence electrons can be paired in a Lewis structure.
Reason (R): In the MO diagram the last two electrons occupy the degenerate π* orbitals singly with parallel spins.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Hund's rule.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the bond order of Li₂ and of Be₂ and say which exists.
Hint
Li₂: σ2s²; Be₂: σ2s² σ*2s².
Answer
Li₂: BO = (2 − 0)/2 = 1 (exists); Be₂: BO = (2 − 2)/2 = 0 (does not exist as a stable molecule).
Q4 · 3 marks · Short answerHarder board level
Find the bond order and the number of unpaired electrons of CN⁺, CN and CN⁻ (valence electrons 8, 9 and 10; use the N₂-type MO order).
Hint
Order: σ2s, σ*2s, π2p, σ2p.
Answer
CN⁺ (8 e): π2p⁴, BO = (6 − 2)/2 = 2, diamagnetic. CN (9 e): extra electron in σ2p, BO 2.5, one unpaired. CN⁻ (10 e): BO 3, diamagnetic.
Q5 · 4 marks · Long answerHOTS / competency
Use MO theory to explain (a) why N₂ has a greater bond dissociation energy than O₂, (b) why F₂ has a lower bond order than O₂, (c) why Ne₂ does not exist, and (d) the order of bond lengths of N₂, O₂ and F₂.
Hint
Compute BO for each.
Answer
(a) BO: N₂ 3, O₂ 2. (b) F₂ (14 valence e): N_b = 8, N_a = 6, BO = 1. (c) Ne₂ (16 valence e): N_b = 8, N_a = 8, BO = 0. (d) Higher BO gives a shorter bond: N₂ < O₂ < F₂.

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
½(N_b − N_a).
2.BO of O₂?Show answer
2.
3.BO of O₂⁻?Show answer
1.5.
4.BO of N₂?Show answer
3.
5.Why is O₂ paramagnetic?Show answer
Two unpaired electrons in π*.
6.Isoelectronic with N₂?Show answer
CO, CN⁻, NO⁺.

MO Theory: Oxygen, Nitrogen and Related Diatomic Species: frequently asked questions

Why is a higher bond order associated with a shorter bond?

More net bonding electrons pull the nuclei closer.

What does the star in σ* mean?

It marks an antibonding orbital.

Can bond order be a fraction?

Yes, e.g. 2.5 for O₂⁺ and NO.

Why do Be₂ and Ne₂ not exist?

Their bonding and antibonding electrons are equal in number, so the bond order is zero.

What is the difference between MO and valence bond descriptions?

VB treats electrons as localised in bonds between two atoms; MO treats them as delocalised over the whole molecule and explains paramagnetism.