Bonding in Homonuclear Diatomic Molecules: H₂ to Ne₂
The theory of § 4.7 now walks the periodic table: ten diatomic molecules, three that refuse to exist, two that hide unpaired electrons — and the ionisation paradox where removing one electron weakens N₂ but strengthens O₂.
The Table, Molecule by Molecule — Complete Theory
Everything here runs on § 4.7’s engine: fill MOs by aufbau + Hund, count Nb and Na, take (Nb − Na)/2. The only wrinkle is the two fill orders — Z ≥ 8 (O₂, F₂, Ne₂) keep σ2pz below π2p; Z ≤ 7 (B₂, C₂, N₂) swap them via 2s–2p mixing. Now the walkthrough, in three families:
Family 1 — the s-only quartet. H₂ (2 e⁻): σ1s², BO 1, diamagnetic — exists. He₂ (4 e⁻): σ1s² σ*1s², BO 0 — does not exist; antibonding cancels bonding exactly. Li₂ (6 e⁻): core plus σ2s², BO 1 — exists (lithium vapour really is Li₂). Be₂ (8 e⁻): BO 0 again — beryllium vapour is monatomic. And the ions that shouldn’t work but do: H₂⁺ (1 e⁻, BO 0.5) and He₂⁺ (3 e⁻, BO 0.5) — one-electron and three-electron bonds are real, just weak.
Family 2 — the mixed-order trio. B₂ (10 e⁻): after σ2s² σ*2s², the next two electrons enter the degenerate π2p pair singly (Hund) — BO 1, paramagnetic (2). C₂ (12 e⁻): π pair now full — BO 2, diamagnetic. N₂ (14 e⁻): σ2pz² completes the set — BO 3, diamagnetic, one of the strongest bonds known (946 kJ/mol). Note the swap: had we used O₂’s order, B₂ would wrongly come out diamagnetic.
Family 3 — the Z ≥ 8 trio and the ghost. O₂ (16 e⁻): BO 2, paramagnetic (2) — the two π* singles. F₂ (18 e⁻): π* full — BO 1, diamagnetic. Ne₂ (20 e⁻): σ*2pz also full — BO 0, does not exist. The scoreboard runs BO 1 → 0 → 1 → 0 → 1 → 2 → 3 → 2 → 1 → 0 across the row: chemistry’s strongest bond sits dead centre, and both ends of the row are ghosts.
H₂ · Li₂ · B₂ · C₂ · N₂ · O₂ · F₂
GhostHe₂ · Be₂ · Ne₂ (BO 0)
ParamagneticB₂ · O₂ (2 unpaired each)
| Species | Change vs neutral | Bond order | Magnetism | Stability vs neutral |
|---|---|---|---|---|
| H₂⁺ | σ1s¹ — lost bonding e⁻ | 0.5 | paramagnetic (1) | weaker, but exists |
| He₂⁺ | σ*1s¹ — 3 e⁻ total | 0.5 | paramagnetic (1) | transient existence |
| N₂⁺ | σ2pz¹ — lost bonding e⁻ | 2.5 | paramagnetic (1) | weaker than N₂ |
| O₂⁺ | π*¹ — lost antibonding e⁻ | 2.5 | paramagnetic (1) | stronger than O₂ |
| O₂⁻ | π*³ — gained antibonding e⁻ | 1.5 | paramagnetic (1) | weaker than O₂ |
| O₂²⁻ | π*⁴ — π* filled | 1 | diamagnetic | weakest of the family |
Visualising the Paradox & Browsing the Scoreboard
Ek paradox, ek scoreboard — see exactly which electron each ionisation removes, then browse all sixteen species.
Diatomic Scoreboard
| Species | e⁻ | BO | Magnetism |
|---|
Click any row — the mini MO diagram, configuration and notes update. Filter chips isolate paramagnetic species, the BO-0 ghosts, or just the ions.
Solved Examples (Step-by-Step)
Config → count → verdict. Jo chain yahan chalti hai, wahi scoreboard me live chalti hai.
The nitrogen cation’s papers
Write the MO configuration of N₂⁺, calculate its bond order and predict its magnetic behaviour and stability relative to N₂.
- Fill13 e⁻, mixed order:
σ1s² σ*1s² σ2s² σ*2s² π2px² π2py² σ2pz¹ - Count
Nb = 9, Na = 4→BO = (9 − 4)/2 = 2.5 - MagnetismOne unpaired σ2pz electron → paramagnetic (N₂ was diamagnetic).
- StabilityBO 2.5 < 3 → less stable than N₂ — the electron left a bonding shelf (verify on the scoreboard).
BO 2.5 · paramagnetic · weaker than N₂
Same operation, opposite effects
Explain why removing an electron from N₂ decreases the bond order while removing one from O₂ increases it.
- N₂Highest occupied = σ2pz (bonding) in the mixed order → BO 3 → 2.5. Weaker.
- O₂Highest occupied = π*2p (antibonding) → BO 2 → 2.5. Stronger.
- RuleElectrons leaving bonding shelves weaken; electrons leaving antibonding shelves strengthen. The shelf decides, not the operation.
Shelf decides — FIG. 1 shows both extractions
O₂²⁻ and F₂: separated at birth
Show that O₂²⁻ and F₂ are isoelectronic, and compare their bond orders and magnetic behaviour.
- CountsO₂²⁻: 16 + 2 = 18 e⁻ · F₂: 9 + 9 = 18 e⁻ — isoelectronic ✓
- Same fillingBoth:
… σ2pz² π*2px² π*2py²— π* full. - VerdictsBoth BO 1, both diamagnetic — isoelectronic species share MOT verdicts (check both on the scoreboard).
18 e⁻ twins — BO 1, diamagnetic, both
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
Which of the following species is paramagnetic?
Solution
- N₂⁺ (13 e⁻): one electron in σ2pz → paramagnetic.
- N₂ (all paired), O₂²⁻ (π* full) and F₂ (all paired) are diamagnetic. The scoreboard’s paramagnetic filter isolates the family instantly.
(C) N₂⁺
When N₂ is ionised to N₂⁺, the bond order changes from 3 to 2.5 because the electron is removed from:
Solution
- Z ≤ 7 order puts σ2pz above the π pair — the highest occupied orbital, and the first to ionise.
- Removing from a bonding level drops BO. (O₂’s ionisation removes from π* — opposite effect, the FIG. 1 pairing.)
(B) Bonding σ2pz
Which of the following molecules is NOT expected to exist on the basis of MOT?
Solution
- Be₂ (8 e⁻): σ2s² σ*2s² → BO 0 — no net bonding.
- Li₂ (1), B₂ (1) and C₂ (2) all carry positive bond orders. He₂ and Ne₂ are the other BO-0 ghosts.
(B) Be₂
Which statement about N₂⁺ and O₂⁺ is correct?
Solution
- Both reach BO 2.5 — but by opposite shelves: N₂⁺ lost a bonding electron (weaker than N₂), O₂⁺ lost an antibonding electron (stronger than O₂).
- Both ions are paramagnetic (one unpaired each) — option D fails too.
(C) BO 2.5, opposite stabilities
Key Formulas & Takeaways
Eight lines that solve this topic
Scoreboard: BO 1→0→1→0→1→2→3→2→1→0 across H₂…Ne₂ · Isoelectronic pairs: O₂²⁻ ≡ F₂ (18 e⁻) · N₂ ≡ CO (14 e⁻) · Para ions: H₂⁺, He₂⁺, N₂⁺, O₂⁺, O₂⁻ — one unpaired each
- Three families, one engine — s-only quartet, mixed-order trio, Z ≥ 8 trio; every verdict is the same (Nb − Na)/2 arithmetic.
- BO 0 is a death certificate — He₂, Be₂, Ne₂ cancel themselves out; no experiment will ever find them.
- The paradox is shelf logic — ionisation weakens N₂⁺ (bonding loss) and strengthens O₂⁺ (antibonding loss); quote the shelf, not the operation.
- Isoelectronic species share verdicts — O₂²⁻ ≡ F₂ and N₂ ≡ CO run identical MO arithmetic; match counts before comparing.
FAQs
How does MOT explain that H2+ exists but He2 does not?
H2+ has one electron in the bonding σ1s orbital: bond order (1 − 0)/2 = 0.5, so a net half-bond stabilises it. He2 would place two electrons in σ1s and two in σ*1s: bond order (2 − 2)/2 = 0, so antibonding cancels bonding exactly and no stable molecule forms.
Why is N2+ less stable than N2 while O2+ is more stable than O2?
In N2 (Z ≤ 7 order) the highest occupied orbital is the bonding σ2pz, so removing an electron on ionisation drops the bond order from 3 to 2.5. In O2 the highest occupied orbitals are the antibonding π*2p pair, so removing an electron raises the bond order from 2 to 2.5. Ionisation weakens N2 but strengthens O2 — the same operation, opposite effects.
Why is B2 paramagnetic while C2 is diamagnetic?
B2 has 10 electrons: after the σ2s and σ*2s levels, its last two electrons occupy the two degenerate π2p orbitals singly with parallel spins (Hund's rule) — two unpaired electrons, paramagnetic. C2 has 12 electrons: both π2p orbitals are fully paired — diamagnetic. Both use the mixed order where π2p lies below σ2pz.
How do the oxygen ions compare in bond order and magnetism?
O2 (16 electrons) has bond order 2 with two unpaired π* electrons — paramagnetic. O2+ (15) loses one antibonding electron: bond order 2.5, one unpaired — paramagnetic. O2− (17) gains one: bond order 1.5, one unpaired — paramagnetic. O2²⁻ (18): bond order 1, all paired — diamagnetic.
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