Molecular Orbital Theory (MOT) — LCAO, bonding - antibonding orbitals: Worked examples

Before we start, let us nail down two pieces of notation and one tool, so no symbol is used before it is earned.

To fill MOs we use the Aufbau + Pauli + Hund rules exactly as for atoms:

Aufbau: fill the lowest-energy MO first (climb the energy ladder from the bottom).
Pauli: at most 2 electrons per MO, and they must have opposite spins (↑↓).
Hund: when two MOs have the same energy (we call them degenerate), put one electron in each (↑ ↑) before pairing up.

We also need the two possible energy ladders (orbital orders). The parent note proved why they differ (s–p mixing); here we just keep both handy:

The only difference between the two ladders is where $σ_{2 p_{z}}$ sits relative to the $π_{2 p}$ pair. Everything else is identical.

Below, Figure 1 draws both ladders side by side. Teal rungs are the bonding orbitals (no star); orange rungs are the antibonding (starred) ones. Notice the plum arrow: it points at the one rung that swaps between the two ladders — the $σ_{2 p_{z}}$ level. When you fill a molecule, you literally place electrons on these rungs from the bottom up, two per rung.

Figure — Molecular Orbital Theory (MOT) — LCAO, bonding - antibonding orbitals

The scenario matrix

Every MOT question is really one of these case-classes. We will hit each cell at least once.

#	Case class	What makes it tricky	Example that covers it
C1	Simplest bond (2 e⁻)	Warm-up, no ladder needed	H₂
C2	Zero bond order (molecule doesn't exist)	Bonding fully cancelled	He₂
C3	Charged species (ion) — add/remove e⁻	Must adjust electron count first	H₂⁺, O₂⁺
C4	Ladder-A molecule, paramagnetic	s–p mixing flips the order	B₂
C5	Ladder-A molecule, diamagnetic, max bond order	Triple bond, all paired	N₂
C6	Ladder-B molecule, paramagnetic (the star result)	Degenerate π* half-filled	O₂
C7	Ladder-B molecule, diamagnetic	π* fully filled	F₂
C8	Heteronuclear (two different atoms)	Unequal atoms, still countable	NO
C9	Comparison / limiting behaviour	Trend as bond order changes	O₂, O₂⁺, O₂⁻, O₂²⁻ series

The two quantities we always extract:

Bond order (B.O.) $= \frac{N _{b} - N _{a}}{2}$ — where $N_{b}$ = electrons in bonding MOs, $N_{a}$ = electrons in antibonding MOs. Higher B.O. → shorter, stronger bond.
Magnetism: any unpaired electron ⇒ paramagnetic (attracted by a magnet). All paired ⇒ diamagnetic (weakly repelled). See Paramagnetism and Diamagnetism.

Worked examples

C1 — The simplest bond: H₂

Worked example Hydrogen molecule, H₂

Find the bond order and magnetic nature of H₂.

Forecast: Two H atoms, one electron each → 2 electrons total. Guess the bond order and whether it's magnetic before reading on.

Step 1 — Count electrons. Each H has 1 electron, so H₂ has $1 + 1 = 2$ electrons. Why this step? Everything downstream (filling, counting, B.O.) depends on the total electron count — get it first.

Step 2 — Fill the ladder. H uses only the 1s shell. Lowest MO is $σ_{1 s}$ . Place both electrons there (Pauli allows 2, opposite spins): configuration $σ_{1 s}^{2}$ . Why this step? Aufbau: fill lowest energy first. Both fit in the bonding orbital.

Step 3 — Count $N_{b}$ and $N_{a}$ . $σ_{1 s}$ is bonding (no star), so $N_{b} = 2$ . No antibonding electrons: $N_{a} = 0$ . Why this step? B.O. needs the split between bonding and antibonding populations.

Step 4 — Bond order. $B.O. = \frac{N _{b} - N _{a}}{2} = \frac{2 - 0}{2} = 1.$ Why this step? Divide by 2 because one bond = one electron pair.

Step 5 — Magnetism. Both electrons paired (↑↓) ⇒ diamagnetic.

Verify: B.O. = 1 is a single bond — exactly what we know real H₂ has. It's a stable gas, and it's diamagnetic. ✅

C2 — Zero bond order: He₂

Worked example Helium "molecule", He₂

Does He₂ exist? Give the bond order.

Forecast: He has 2 electrons, so He₂ has 4. Guess whether that fourth electron kills the bond.

Step 1 — Count electrons. $2 + 2 = 4$ electrons.

Step 2 — Fill. $σ_{1 s}$ takes 2 (full), then the next rung $σ_{1 s}^{*}$ (the starred, antibonding one) takes 2: config $σ_{1 s}^{2} σ_{1 s}^{* 2}$ . Why this step? Aufbau: once the bonding $σ_{1 s}$ is full, electrons must climb to the antibonding $σ_{1 s}^{*}$ .

Step 3 — Count. $N_{b} = 2$ (in $σ_{1 s}$ ), $N_{a} = 2$ (in $σ_{1 s}^{*}$ ).

Step 4 — Bond order. $B.O. = \frac{2 - 2}{2} = 0.$ Why this step? The antibonding pair actively cancels the bonding pair — this is why starred electrons subtract.

Verify: B.O. = 0 means no net glue → He₂ does not exist as a stable molecule. This is MOT's famous win over Valence Bond Theory, which struggles to say why. ✅

C3 — Charged species: H₂⁺ and O₂⁺

Worked example The ions H₂⁺ and O₂⁺

Find bond order and magnetism of (a) H₂⁺ and (b) O₂⁺.

Forecast: A "+" means we removed one electron. Does removing an electron always weaken a bond? Guess.

(a) H₂⁺

Step 1 — Adjust electron count. Neutral H₂ has 2; the "+" removes one → 1 electron. Why this step? For ions you must fix the electron count before filling, or every later step is wrong.

Step 2 — Fill. One electron in the lowest MO: $σ_{1 s}^{1}$ .

Step 3 — Count & B.O. $N_{b} = 1$ , $N_{a} = 0$ : $B.O. = \frac{1 - 0}{2} = 0.5.$ Why this step? Half a bond is allowed — MOT lets bond orders be fractional.

Step 4 — Magnetism. One unpaired electron ⇒ paramagnetic.

Verify: H₂⁺ is a real, observed species with a weak half-bond — B.O. 0.5 < 1, so it's weaker than H₂. ✅

(b) O₂⁺

Step 1 — Adjust count. Neutral O₂ has 16; the "+" removes one → 15 electrons. O is $Z = 8$ , so use Ladder B.

Step 2 — Fill 15 electrons on Ladder B. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 1}$ Count as we place: $2 + 2 + 2 + 2 + 2 + 2 = 12$ fill the first six rungs, then $+ 2 = 14$ on the two π rungs, then the 15th goes singly into one $π^{*}$ . Why this step? We removed the electron from the highest occupied MO of neutral O₂ — a $π^{*}$ (antibonding) electron.

Step 3 — Count (valence rungs only). Bonding (unstarred): $σ_{2 s} (2) + σ_{2 p_{z}} (2) + π_{2 p} (4) = 8$ , and the cancelled inner pair adds $σ_{1 s} (2)$ to both sides equally, so tracking it explicitly: $N_{b} = 2 + 8 = 10$ . Antibonding (starred): $σ_{1 s}^{*} (2) + σ_{2 s}^{*} (2) + π^{*} (1) = 5$ , so $N_{a} = 5$ . Why this step? Apply the universal counting rule — unstarred into $N_{b}$ , starred into $N_{a}$ . Here $2 + 2 + 2 + 4 = 10$ for bonding.

Step 4 — B.O. $B.O. = \frac{10 - 5}{2} = 2.5.$ Why this step? We removed an antibonding electron, so bond order went up (from 2 in O₂ to 2.5). Removing antibonding density strengthens the bond!

Step 5 — Magnetism. One unpaired electron in $π^{*}$ ⇒ paramagnetic.

Verify: O₂⁺ (dioxygenyl) has a shorter, stronger bond than O₂ — consistent with B.O. rising from 2 to 2.5. ✅

C4 — Ladder-A, paramagnetic: B₂

Worked example Diboron, B₂

Find bond order and magnetism of B₂. (Boron: $Z = 5$ .)

Forecast: Common trap — students expect B₂ diamagnetic. Guess before the s–p mixing bites.

Step 1 — Count electrons. Each B has 5 → $5 + 5 = 10$ electrons.

Step 2 — Pick the ladder. $Z = 5 \leq 7$ ⇒ Ladder A ( $π_{2 p}$ below $σ_{2 p_{z}}$ ). Why this step? Choosing the wrong ladder is the #1 mistake here — it decides whether the last two electrons pair or split.

Step 3 — Fill 10 electrons on Ladder A. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p_{x}}^{1} π_{2 p_{y}}^{1}$ Count: $2 + 2 + 2 + 2 = 8$ fill the first four rungs, then 2 electrons left. The next rung is the degenerate $π_{2 p_{x}} = π_{2 p_{y}}$ pair. Why this step? Because they're degenerate, Hund's rule puts one electron in each (↑ ↑), not both in one.

Step 4 — Count (valence rungs). Bonding (unstarred): $σ_{2 s} (2) + π_{2 p} (2) = 4$ , so $N_{b} = 4$ . Antibonding (starred): $σ_{2 s}^{*} (2) = 2$ , so $N_{a} = 2$ . (The cancelled $σ_{1 s}^{2} σ_{1 s}^{* 2}$ inner pair contributes equally to both and is dropped by our rule.)

Step 5 — B.O. $B.O. = \frac{4 - 2}{2} = 1.$

Step 6 — Magnetism. Two unpaired electrons (one in each π) ⇒ paramagnetic.

Verify: B₂ is experimentally paramagnetic — only Ladder A predicts this. Had we used Ladder B, both electrons would pair in $σ_{2 p_{z}}$ and wrongly give diamagnetic. ✅

Figure 2 shows this filling on Ladder A. Follow the electrons (drawn as ↑/↓ arrows) climbing rung by rung; the plum callout marks the two lone ↑ ↑ electrons sitting one-each in the degenerate $π$ rung — that split is exactly what makes B₂ paramagnetic.

C5 — Ladder-A, diamagnetic, max bond order: N₂

Worked example Dinitrogen, N₂

Find bond order and magnetism of N₂. (Nitrogen: $Z = 7$ .)

Forecast: N₂ is famously inert. What bond order would explain "hard to break"?

Step 1 — Count. $7 + 7 = 14$ electrons.

Step 2 — Ladder. $Z = 7 \leq 7$ ⇒ Ladder A.

Step 3 — Fill 14 on Ladder A. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}$ Count: $2 + 2 + 2 + 2 = 8$ , then the π pair fills ( $+ 4 = 12$ ), then $σ_{2 p_{z}}$ fills ( $+ 2 = 14$ ). ✔ Why this step? On Ladder A the π rung comes before $σ_{2 p_{z}}$ , so π fills first.

Step 4 — Count (valence rungs). Bonding (unstarred): $σ_{2 s} (2) + π_{2 p} (4) + σ_{2 p_{z}} (2) = 8$ , so $N_{b} = 8$ . Antibonding (starred): $σ_{2 s}^{*} (2) = 2$ , so $N_{a} = 2$ .

Step 5 — B.O. $B.O. = \frac{8 - 2}{2} = 3.$

Step 6 — Magnetism. Every electron paired ⇒ diamagnetic.

Verify: B.O. = 3 is a triple bond — the strongest common bond, explaining why N₂ is so unreactive. Diamagnetic, correct. ✅

C6 — Ladder-B, paramagnetic (the star result): O₂

Worked example Dioxygen, O₂

Find bond order and magnetism of O₂. (Oxygen: $Z = 8$ .)

Forecast: MOT's crown jewel. Liquid O₂ sticks to a magnet — so which is it, para or dia?

Step 1 — Count. $8 + 8 = 16$ electrons.

Step 2 — Ladder. $Z = 8 \geq 8$ ⇒ Ladder B ( $σ_{2 p_{z}}$ below π).

Step 3 — Fill 16 on Ladder B. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 1} π_{2 p_{y}}^{* 1}$ Count: $2 + 2 + 2 + 2 = 8$ , then $σ_{2 p_{z}}$ ( $+ 2 = 10$ ), then the π pair ( $+ 4 = 14$ ), then 2 left. The next rung is the degenerate $π^{*}$ pair → Hund splits them (↑ ↑). Why this step? This is the whole payoff — the last two electrons land singly in the two $π^{*}$ orbitals.

Step 4 — Count (valence rungs, one consistent method). Bonding (unstarred): $σ_{2 s} (2) + σ_{2 p_{z}} (2) + π_{2 p} (4) = 8$ ; adding the cancelled inner $σ_{1 s} (2)$ that our rule drops, the full bonding total is $N_{b} = 10$ . Antibonding (starred): $σ_{2 s}^{*} (2) + π^{*} (2) = 4$ ; adding the cancelled inner $σ_{1 s}^{*} (2)$ , the full antibonding total is $N_{a} = 6$ . Either way $N_{b} - N_{a} = 10 - 6 = 8 - 4 = 4$ — the inner pair cancels, exactly as promised. Why this step? Because the inner $σ_{1 s}$ / $σ_{1 s}^{*}$ pair adds equally to $N_{b}$ and $N_{a}$ , you may include it or drop it — the difference $N_{b} - N_{a}$ is unchanged. We show both to prove they agree.

Step 5 — B.O. $B.O. = \frac{10 - 6}{2} = \frac{8 - 4}{2} = 2.$

Step 6 — Magnetism. Two unpaired electrons (one per $π^{*}$ ) ⇒ paramagnetic.

Verify: Double bond (B.O. 2) and two unpaired electrons — exactly matching experiment. VBT's O=O structure wrongly predicts diamagnetic; MOT nails it. ✅

C7 — Ladder-B, diamagnetic: F₂

Worked example Difluorine, F₂

Find bond order and magnetism of F₂. (Fluorine: $Z = 9$ .)

Forecast: Two more electrons than O₂. What do they do to the half-filled $π^{*}$ ?

Step 1 — Count. $9 + 9 = 18$ electrons.

Step 2 — Ladder. $Z = 9 \geq 8$ ⇒ Ladder B.

Step 3 — Fill 18 on Ladder B. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 2} π_{2 p_{y}}^{* 2}$ Count: $2 \times 9 = 18$ . The two extra electrons (vs O₂) now fully fill the $π^{*}$ pair. Why this step? Once each degenerate π* has 1 electron, the next two pair them up (Pauli), leaving no unpaired spins.

Step 4 — Count (valence rungs). Bonding (unstarred): $σ_{2 s} (2) + σ_{2 p_{z}} (2) + π_{2 p} (4) = 8$ ; with the dropped inner $σ_{1 s} (2)$ , $N_{b} = 10$ . Antibonding (starred): $σ_{2 s}^{*} (2) + π^{*} (4) = 6$ ; with the dropped inner $σ_{1 s}^{*} (2)$ , $N_{a} = 8$ .

Step 5 — B.O. $B.O. = \frac{10 - 8}{2} = 1.$

Step 6 — Magnetism. All paired ⇒ diamagnetic.

Verify: Single bond (B.O. 1) — weaker than O₂'s double bond, matching F₂'s longer, weaker bond. Diamagnetic, correct. ✅

C8 — Heteronuclear: NO

Worked example Nitric oxide, NO

Find bond order and magnetism of NO. (N: $Z = 7$ , O: $Z = 8$ .)

Forecast: Two different atoms and an odd total. Odd number of electrons — can it possibly be diamagnetic?

Step 1 — Count. $7 + 8 = 15$ electrons (odd!). Why this step? An odd total guarantees at least one unpaired electron → paramagnetic. Spotting this early is a free sanity check.

Step 2 — Pick the ladder. We use Ladder B for NO. Why this step? The rule " $Z \leq 7$ → Ladder A, $Z \geq 8$ → Ladder B" is set by whether s–p mixing is strong. Mixing weakens once an oxygen (with its lower, more contracted 2s) is in the molecule, because O pulls the $σ_{2 p_{z}}$ down. Since NO contains an oxygen ( $Z = 8$ ), its diagram behaves like the O₂/F₂ family — Ladder B — so $σ_{2 p_{z}}$ sits below the π pair, and we fill exactly as we did for O₂⁺ (also 15 electrons).

Step 3 — Fill 15 on Ladder B. $σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 1}$ Count: $2 + 2 + 2 + 2 = 8$ , then $σ_{2 p_{z}}$ ( $+ 2 = 10$ ), then the π pair ( $+ 4 = 14$ ), then 1 electron alone in a $π^{*}$ ( $+ 1 = 15$ ). ✔

Step 4 — Count (valence rungs). Bonding (unstarred): $σ_{2 s} (2) + σ_{2 p_{z}} (2) + π_{2 p} (4) = 8$ ; with the dropped inner $σ_{1 s} (2)$ , $N_{b} = 10$ . Antibonding (starred): $σ_{2 s}^{*} (2) + π^{*} (1) = 3$ ; with the dropped inner $σ_{1 s}^{*} (2)$ , $N_{a} = 5$ .

Step 5 — B.O. $B.O. = \frac{10 - 5}{2} = 2.5.$

Step 6 — Magnetism. One unpaired electron ⇒ paramagnetic (as our odd-count check foretold).

Verify: NO has bond order 2.5 and is a known paramagnetic radical. Removing that lone $π^{*}$ electron gives NO⁺ with B.O. 3 (stronger) — consistent. ✅

C9 — Limiting behaviour: the O₂ family trend

Worked example The series O₂⁺, O₂, O₂⁻, O₂²⁻

Rank these four by bond order and hence by bond strength. Each is built from O₂'s 16 electrons by adding or removing electrons in the $π^{*}$ level.

Forecast: Adding electrons to an antibonding orbital — will the bond get stronger or weaker? Guess the trend before computing.

Step 1 — Set up electron counts. All use Ladder B with the bonding total $N_{b} = 10$ fixed, and we track $N_{a}$ as the $π^{*}$ population changes:

O₂⁺: 15 e⁻ → $π^{* 1}$ → $N_{a} = 5$
O₂: 16 e⁻ → $π^{* 2}$ → $N_{a} = 6$
O₂⁻: 17 e⁻ → $π^{* 3}$ → $N_{a} = 7$
O₂²⁻: 18 e⁻ → $π^{* 4}$ → $N_{a} = 8$ Why this step? Every added electron goes into the antibonding $π^{*}$ , so each one raises $N_{a}$ and lowers B.O.

Step 2 — Compute each bond order. Using $B.O. = \frac{10 - N _{a}}{2}$ :

O₂⁺: $\frac{10 - 5}{2} = 2.5$
O₂: $\frac{10 - 6}{2} = 2.0$
O₂⁻: $\frac{10 - 7}{2} = 1.5$
O₂²⁻: $\frac{10 - 8}{2} = 1.0$

Step 3 — Rank. Bond order (and thus strength): $O_{2}^{+} > O_{2} > O_{2}^{-} > O_{2}^{2 -}$ . Bond length runs the opposite way (higher B.O. ⇒ shorter bond): $O_{2}^{+} < O_{2} < O_{2}^{-} < O_{2}^{2 -}$ .

Verify: Each extra antibonding electron drops B.O. by exactly 0.5 and lengthens/weakens the bond — a clean monotonic trend confirmed by measured bond lengths (O₂⁺ ≈ 112 pm, O₂ ≈ 121 pm, O₂⁻ ≈ 128 pm, O₂²⁻ ≈ 149 pm). ✅

Recall

Recall What does a star (*) on an orbital name mean?

It marks the antibonding MO — out-of-phase combination, node between the nuclei, higher energy. Its electrons go into $N_{a}$ and subtract in bond order. And no star? ::: Bonding MO — in-phase, density between nuclei, lower energy; electrons go into $N_{b}$ .

Recall Which ladder for which molecule, and why?

Ladder A ( $π$ below $σ_{2 p_{z}}$ ) for $Z \leq 7$ (B₂, C₂, N₂) because strong s–p mixing pushes $σ_{2 p_{z}}$ up; Ladder B ( $σ_{2 p_{z}}$ below $π$ ) for $Z \geq 8$ (O₂, F₂, and molecules containing O like NO) where mixing is weak. Which ladder gives B₂ its paramagnetism? ::: Ladder A — the two π electrons split by Hund's rule.

Recall Effect of adding an electron to an antibonding orbital

Bond order drops by 0.5, so the bond gets longer and weaker (O₂ → O₂⁻). Removing an antibonding electron does the opposite (O₂ → O₂⁺: stronger).

Recall Fast paramagnetism check

An odd total electron count always means paramagnetic (at least one unpaired). An even count may still be paramagnetic if a degenerate pair is half-filled (like O₂).