§ 4.7NCERT Class 11 · Chemistry · Chapter 4

Molecular Orbital Theory: Bond Order & Magnetism

Valence bond theory drew neat electron pairs — then liquid oxygen stuck to a magnet and the picture broke. MOT delocalises electrons across the whole molecule, counts them in bonding and antibonding shelves, and predicted what VBT never could: O₂ has two unpaired electrons.

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 14 min read

0unpaired e⁻

O₂’s lone electrons in π* — MOT’s vindication
VBT: all paired (wrong) · MOT: two unpaired (right)
Bond order 2 · paramagnetic — verified by liquid O₂

What is Molecular Orbital Theory? — Complete Theory

Valence bond theory (§ 4.5) treats a bond as two electrons shared between two atoms — and predicts, correctly, that every electron in O₂ should be paired. Experiment disagrees: liquid oxygen is paramagnetic — it clings to a magnet, which demands unpaired electrons. In 1932 Hund and Mulliken rebuilt the picture: electrons in a molecule occupy molecular orbitals (MOs) that spread over the entire molecule, not between two specific atoms. That single shift explains everything VBT could not.

LCAO — where MOs come from. Molecular orbitals are built by the linear combination of atomic orbitals: take two atomic wave functions ψ_A and ψ_B and either add or subtract them. Addition (in-phase) gives a bonding MO — electron density increases between the nuclei, energy lower than the parent AOs, the molecule held together. Subtraction (out-of-phase) gives an antibonding MO — a node between the nuclei, density pushed outside, energy higher; electrons here destabilise the bond. Antibonding orbitals carry the star (σ*).

σ and π MOs. Head-on overlap of orbitals along the bond axis gives σ MOs (cylindrically symmetric — from s–s, s–p or p_z–p_z overlap); sidewise overlap of p_x/p_y orbitals gives π MOs, always produced as degenerate pairs (π_x and π_y of equal energy, capacity 2 + 2 = 4). Every AO combination yields one bonding and one antibonding partner — nothing is lost, energy is just re-shelved.

Bonding MOψ = ψ_A + ψ_B · lower E Antibonding MOψ = ψ_A − ψ_B · node inside Bond order(N_b − N_a)/2

The two fill orders. Electrons enter MOs by aufbau + Hund, exactly as in atoms — but the 2p-region order depends on the atom. For O₂, F₂, Ne₂ (Z ≥ 8), σ2p_z sits below the π2p pair. For B₂, C₂, N₂ (Z ≤ 7), 2s–2p mixing pushes σ2p_z above the π pair. Both orders are correct — for their own atoms. The builder below uses the right scheme for each species automatically.

Table 1 — The two energy ladders, level by level (bottom → top)
Level	O₂, F₂, Ne₂ (Z ≥ 8)	B₂, C₂, N₂ (Z ≤ 7)
1	σ1s	σ1s
2	σ*1s	σ*1s
3	σ2s	σ2s
4	σ*2s	σ*2s
5	σ2p_z	π2p_x = π2p_y
6	π2p_x = π2p_y	σ2p_z
7	π2p_x = π2p_y	π2p_x = π2p_y
8	σ*2p_z	σ*2p_z

Bond order — the payoff formula. Count electrons in bonding MOs (N_b) and antibonding MOs (N_a):

Bond orderBO = (N_b − N_a)/2 StabilityBO > 0 → exists · BO ↑ = stronger LengthBO ↑ = bond length ↓

Bond order 1, 2, 3 mean single, double, triple bonds — and fractional orders are legal: H₂⁺ carries 0.5 (a half-bond), O₂⁻ carries 1.5. Zero bond order means bonding and antibonding populations cancel — the molecule does not exist. Magnetism falls out of the same diagram for free: count unpaired electrons; any → paramagnetic, none → diamagnetic.

Table 2 — The diatomic scoreboard (results worth memorising)
Species	Total e⁻	Bond order	Magnetism	Verdict
H₂	2	1	diamagnetic	exists
He₂	4	0	diamagnetic	does not exist
Li₂	6	1	diamagnetic	exists
Be₂	8	0	diamagnetic	does not exist
B₂	10	1	paramagnetic (2)	exists
C₂	12	2	diamagnetic	exists
N₂	14	3	diamagnetic	exists — very strong
O₂	16	2	paramagnetic (2)	exists
F₂	18	1	diamagnetic	exists
Ne₂	20	0	diamagnetic	does not exist

Visualising the O₂ Diagram & Building Any Molecule

Ek canonical diagram, ek builder — the O₂ picture that made MOT famous first, then fill MOs for fifteen species yourself.

FIG. 1 — The canonical O₂ diagram: 16 electrons, bonding shelf (N_b = 10) against antibonding shelf (N_a = 6), bond order 2. The circled pair — two single ↑ arrows in the degenerate π* orbitals — is the paramagnetism liquid oxygen demonstrates. Atomic 2p boxes at the sides follow Hund’s rule within each atom.

Try it live

MO Builder

Species

Bond order2

Bonding e⁻ (Nb)10

Antibonding e⁻ (Na)6

Magnetismparamagnetic (2)

Filling follows aufbau + Hund — degenerate π pairs take single electrons before pairing. Scheme note: B₂/C₂/N₂ (Z ≤ 7) use the s–p-mixed order (π 2p below σ 2p); the diagram marks it when active. Electrons in antibonding levels show dashed red shelves.

Solved Examples (Step-by-Step)

Count → fill → subtract → verdict. Jo chain yahan chalti hai, wahi builder me live chalti hai.

EXAMPLE 01Foundation · NEET

The classic: bond order and magnetism of O₂

Write the MOT configuration of O₂ (16 electrons), calculate its bond order, and predict its magnetic behaviour.

Fillσ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹ — last two electrons enter the degenerate π* pair singly (Hund).
CountN_b = 2+2+2+4 = 10 · N_a = 2+2+2 = 6
Bond order(10 − 6)/2 = 2 — a double bond.
MagnetismTwo unpaired π* electrons → paramagnetic ✓ (matches liquid-O₂ experiment).

BO = 2 · paramagnetic (2 unpaired e⁻)

EXAMPLE 02JEE Main · Ion ladder

O₂⁺, O₂, O₂⁻, O₂²⁻ — stability and bond length orders

Arrange O₂⁺, O₂, O₂⁻ and O₂²⁻ in order of (a) decreasing stability and (b) increasing bond length, with bond orders.

BaseO₂: BO 2. All changes happen in the π* antibonding level.
Remove oneO₂⁺: π* has 1 e⁻ → BO = (10 − 5)/2 = 2.5 — removing an antibonding electron strengthens the bond.
Add one, twoO₂⁻: π*³ → BO 1.5 · O₂²⁻: π*⁴ → BO 1.
OrdersStability: O₂⁺ > O₂ > O₂⁻ > O₂²⁻ · Bond length: exact reverse — O₂²⁻ > O₂⁻ > O₂ > O₂⁺.

BO 2.5 → 2 → 1.5 → 1 · length order exactly reversed

EXAMPLE 03JEE Main · Existence logic

Why He₂ does not exist — but He₂⁺ does

Using MOT, explain why the He₂ molecule is not observed, while the He₂⁺ ion can exist transiently.

He₂4 electrons: σ1s² σ*1s² → N_b = 2, N_a = 2 → BO = 0.
MeaningZero bond order — antibonding destabilisation exactly cancels bonding: no net stabilisation, so He₂ never forms.
He₂⁺3 electrons: σ1s² σ*1s¹ → BO = (2 − 1)/2 = 0.5 — a net half-bond: exists transiently in discharge tubes.
PatternSame logic kills Be₂ (8 e⁻, BO 0) and Ne₂ (20 e⁻, BO 0) — Table 2’s three “does not exist” rows.

BO 0 kills He₂ · BO 0.5 lets He₂⁺ live

Practice Questions (With Solutions)

Attempt first — options lock after one shot, exactly like the real exam. Then read the working, chahe galti ho ya na ho.

Attempted 0/4 · Correct 0

Q1NEET · Superoxide ion

The bond order of the superoxide ion O₂⁻ is:

Solution

O₂⁻ has 17 electrons: the extra electron joins a π* orbital → N_a = 7, N_b = 10.
BO = (10 − 7)/2 = 1.5 — fractional orders are routine in MOT (verify on the builder).

(B) 1.5

Q2JEE Main · Magnetism

Which of the following species is paramagnetic?

Solution

B₂ (Z ≤ 7 scheme): after σ2s/σ*2s, the last two electrons enter the degenerate π2p pair singly — 2 unpaired → paramagnetic.
N₂, O₂²⁻ and F₂ are all fully paired — diamagnetic. The s–p mixing swap is what makes B₂ the odd one out.

Q3NEET · Bond lengths

The correct order of increasing bond length for the oxygen ion series is:

Solution

Bond order: O₂⁺ 2.5 > O₂ 2 > O₂⁻ 1.5 > O₂²⁻ 1.
Bond length ∝ 1/BO → increasing length = O₂⁺ < O₂ < O₂⁻ < O₂²⁻.
Option A is the stability order wearing a length costume — the classic inversion bait.

(B) O₂⁺ < O₂ < O₂⁻ < O₂²⁻

Q4JEE Main · Highest BO

Which of the following has the highest bond order?

Solution

N₂: 14 e⁻ → N_b = 10, N_a = 4 → BO 3 — a triple bond, among chemistry’s strongest.
O₂⁺ = 2.5, O₂ = 2, O₂⁻ = 1.5 — the oxygen family never reaches 3.

Key Formulas & Takeaways

Formula card

Eight lines that solve this topic

LCAO: + → σ (bonding) · − → σ* (antibonding)In-phase addition lowers energy; subtraction creates a node between nuclei.

Order (Z ≥ 8): σ1s σ*1s σ2s σ*2s σ2p_z π2p π*2p σ*2p_zO₂, F₂, Ne₂ — σ2p below π2p.

Order (Z ≤ 7): π2p below σ2p_zB₂, C₂, N₂ — 2s–2p mixing swaps levels 5 and 6.

BO = (N_b − N_a)/2Fractional orders legal: H₂⁺ 0.5 · O₂⁻ 1.5 · O₂⁺ 2.5.

BO > 0 → exists · BO = 0 → does notHe₂, Be₂, Ne₂ carry zero bond order.

BO ↑ ⇒ strength ↑ · length ↓O₂⁺ > O₂ > O₂⁻ > O₂²⁻ in stability, reversed in length.

Unpaired e⁻ → paramagneticO₂: 2 unpaired in π* — the prediction VBT could not make.

Degenerate π pair: Hund fills singly firstCapacities: σ 2 · π pair 4 (2 + 2); π* pair likewise.

Anchor results: O₂ BO 2, paramagnetic (2) · N₂ BO 3, diamagnetic · B₂ BO 1, paramagnetic (2) · H₂⁺ BO 0.5 · He₂/Be₂/Ne₂ BO 0 · O₂²⁻ ≡ F₂ (18 e⁻, BO 1)

MOT delocalises, then counts — electrons fill molecule-wide MOs; bond order is shelf arithmetic: (N_b − N_a)/2.
Two ladders, one rule — Z ≤ 7 mixes 2s–2p and lifts σ2p above π2p; applying the wrong ladder flips B₂’s magnetism.
Magnetism is a free by-product — unpaired π* electrons made O₂ paramagnetic, the experiment that crowned MOT over VBT.
The ion ladder is one electron deep — every π* electron added or removed moves BO by 0.5 and flips the stability/length ordering.

FAQs

What is bond order in molecular orbital theory?

Bond order is half the difference between the number of electrons in bonding molecular orbitals and antibonding molecular orbitals: bond order = (Nb − Na)/2. It measures net bonding — bond order 1, 2 and 3 correspond to single, double and triple bonds. Higher bond order means greater stability, greater bond strength and shorter bond length.

Why is O2 paramagnetic?

The O2 molecule has 16 electrons. Its MOT configuration places the last two electrons singly in the degenerate π*2px and π*2py antibonding orbitals with parallel spins, following Hund's rule. These two unpaired electrons make O2 paramagnetic — liquid oxygen clings to a magnet — a fact valence bond theory could not explain and molecular orbital theory predicted naturally.

Why is the MO energy order of N2 different from that of O2?

For B2, C2 and N2 (Z ≤ 7), significant mixing between the 2s and 2p atomic orbitals pushes the σ2pz molecular orbital above the π2p orbitals. For O2 and F2 (Z ≥ 8), the 2s–2p energy gap is large, mixing is negligible, and σ2pz sits below the π2p pair. The two orders are both correct — each for its own range of atoms.

Why does the He2 molecule not exist?

He2 would have four electrons filling σ1s² and σ*1s². Bonding and antibonding populations are equal, so bond order = (2 − 2)/2 = 0. Zero bond order means no net stabilisation, so He2 does not exist as a stable molecule — though He2+, with three electrons and bond order 0.5, can exist transiently.

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