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

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Step 0 — What is the thing we are combining?

Before any adding or subtracting, we need one honest picture of a single atom's orbital.

• WHAT. An atomic orbital is described by a wavefunction, written $\psi$ (Greek letter "psi", say "sigh"). Think of $\psi$ as a height you can read off at every point in space — like the surface of a still pond that has been nudged. Where $\psi$ is tall and positive, the wave crests up; where $\psi$ is negative, it dips down; where $\psi =0$, the water is flat.
• WHY a wave at all. Electrons are not tiny balls sitting still; quantum mechanics says each electron is a standing wave. The only thing we can measure is ${\psi }^2$ — the electron density, i.e. how likely you are to find the electron there. Density can never be negative, and squaring a number (positive or negative) always gives something $\ge 0$, which is exactly why we square.
• PICTURE. Below, one hydrogen $1s$ orbital: a single smooth positive bump centred on the nucleus. There is no sign-change anywhere — a $1s$ wave is positive everywhere. Remember its shape; everything else is two copies of this.

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Step 1 — Bring two atoms close: line up their waves

• WHAT. Put two identical atoms, A and B, a short distance apart on a line. Call that line the internuclear axis (the straight line through both nuclei). Each atom carries its own $1s$ wave: ${\psi }_A$ centred on A, ${\psi }_B$ centred on B.
• WHY. A bond only happens when the two waves physically reach into each other's territory — this shared middle region is where all the action is. Far apart, the waves never touch and nothing new forms.
• PICTURE. Two identical positive bumps, their tails overlapping in the middle. Notice the overlap region (shaded): here both ${\psi }_A$ and ${\psi }_B$ are nonzero at the same time. That coexistence is what lets us add or subtract them.

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Step 2 — The in-phase combination: add the waves

• WHAT. Line the two waves up crest-to-crest (both positive in the middle) and add them point by point: ${\psi }_b=c(\underset{\text{wave on A}}{\underbrace{{\psi }_A}}+\underset{\text{wave on B}}{\underbrace{{\psi }_B}})$ Here $c$ is just a number (a normalisation constant) that rescales the total so the electron still counts as exactly one electron. The $+$ is the whole story: same sign meeting same sign.
• WHY "in phase". Two water waves whose crests coincide reinforce — the combined crest is taller. This is constructive interference. Nature offers this combination for free, so we must include it.
• PICTURE. In the middle, where both waves are positive, the sum bulges higher than either wave alone. The result is one fat hump piled up between the nuclei. No sign change anywhere → no node.

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Step 3 — Square it: why the middle glows

• WHAT. Density is ${\psi }^2$, so square the bonding wave. Squaring a sum uses $(x+y{)}^2=x^2+y^2+2xy$: ${\psi }_b^2=c^2(\underset{\text{density from A}}{\underbrace{{\psi }_A^2}}+\underset{\text{density from B}}{\underbrace{{\psi }_B^2}}+\underset{\text{the bonus}}{\underbrace{2{\psi }_A{\psi }_B}})$
• WHY this matters. If atoms simply sat side by side without interacting, you'd expect just ${\psi }_A^2+{\psi }_B^2$ — each atom's own density. The extra $+2{\psi }_A{\psi }_B$ is pure interference profit. In the middle, both ${\psi }_A$ and ${\psi }_B$ are positive, so $2{\psi }_A{\psi }_B>0$: extra density appears exactly where the two nuclei both pull.
• PICTURE. The density curve for ${\psi }_b^2$ shows a bright pile between the nuclei — highlighted in coral. Electrons sitting there feel attraction from both positive nuclei at once → the energy drops. That is the glue.

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Step 4 — The out-of-phase combination: subtract the waves

• WHAT. Now flip one wave upside down (make ${\psi }_B$ negative on its half) and add — i.e. subtract: ${\psi }_a=c(\underset{\text{positive}}{\underbrace{{\psi }_A}}-\underset{\text{now flipped}}{\underbrace{{\psi }_B}})$
• WHY. This is the out-of-phase pairing: a crest meets a trough. Where one wave says "up +1" the other says "down −1", and $+1+(-1)=0$. That's destructive interference. It is the other combination nature hands us — and since two atomic orbitals must always yield two molecular orbitals, we are required to keep it.
• PICTURE. Between the nuclei the two opposite-signed waves exactly cancel. Right at the midpoint the sum crosses zero — a brand-new node appears that wasn't in either atom.

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Step 5 — Square it: why the middle goes dark

• WHAT. Square the antibonding wave with $(x-y{)}^2=x^2+y^2-2xy$: ${\psi }_a^2=c^2(\underset{\text{density from A}}{\underbrace{{\psi }_A^2}}+\underset{\text{density from B}}{\underbrace{{\psi }_B^2}}-\underset{\text{the penalty}}{\underbrace{2{\psi }_A{\psi }_B}})$
• WHY. Same two atomic densities as before, but now the interference term is $-2{\psi }_A{\psi }_B$. Since both waves are positive in the middle, this term is negative there — it removes density from the middle. The electrons get shoved out to the far sides.
• PICTURE. The density curve for ${\psi }_a^2$ dives to zero at the centre (a slate-coloured node), with two humps banished to the outsides. With no electron glue in the middle, the two bare positive nuclei repel each other → the energy rises.

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Step 6 — Stack the two results into an energy-level diagram

• WHAT. We now have two molecular orbitals from two atomic orbitals. Draw the two starting atomic levels (${\psi }_A$, ${\psi }_B$) at the same height in the middle, and split them: ${\psi }_b$ drops down (stabilised), ${\psi }_a$ rises up (destabilised).
• WHY the split is lopsided. The antibonding orbital rises slightly more than the bonding orbital falls. Reason: the node forces electrons close to the nuclei on both sides where they screen poorly and repel more, so the "away" penalty is harsher than the "between" reward. This asymmetry is why filling both cancels any net bond (and then some).
• PICTURE. The classic MO ladder: two atomic rungs on the sides feeding one low bonding rung and one high antibonding rung in the centre. Arrows show electrons dropping into the lowest available slot first — this is Aufbau/Pauli/Hund applied to the molecule.

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Step 7 — The degenerate case: when the two rungs never split (He₂)

• WHAT. Take four electrons (as in ${\text{He}}_2$). Two fill ${\psi }_b$, and — because Pauli forbids a third and fourth in the same orbital — the next two are forced up into ${\psi }_a$: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}$.
• WHY it's a special case. Now the "up" pair and the "down" pair fight each other. Since antibonding rises more than bonding falls (Step 6), a full antibonding pair doesn't just cancel a full bonding pair — it wins. Net result: no glue. $\text{Bond order}=\frac{N_b-N_a}{2}=\frac{2-2}{2}=0.$
• PICTURE. Both rungs full; the down-arrow of stabilisation is over-matched by the up-arrow of destabilisation. The molecule falls apart → ${\text{He}}_2$ does not exist. This is the degenerate/limiting case the derivation must cover, and it is exactly the case Valence Bond Theory (see Valence Bond Theory) struggles with.

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The one-picture summary

Everything above in a single frame: two identical $1s$ waves at the top; add them (left branch) → constructive → density pooled between nuclei → bonding, low energy; subtract them (right branch) → destructive → node in the middle → antibonding, high energy. Fill from the bottom, count $N_b-N_a$, divide by 2.

Recall Feynman retelling — the whole walkthrough in plain words

Two atoms show up, each humming its own wave (Step 0–1). You can line the hums up two ways. Line the crests up (add): the sound gets loud right in the middle — that loud spot is a puddle of electrons glued between the two nuclei, and because each electron is now tugged by both nuclei, everything settles into a lower, comfier energy. That's the bonding orbital, and squaring the wave shows the bonus $+2{\psi }_A{\psi }_B$ lighting up the middle (Steps 2–3). Flip one hum upside down (subtract): now crest meets trough and they cancel dead-silent in the middle — a node. Electrons get pushed to the outsides, the two naked nuclei glare at each other and repel, energy shoots up. That's the antibonding orbital, and squaring shows the penalty $-2{\psi }_A{\psi }_B$ emptying the middle (Steps 4–5). Stack the two: bonding sinks, antibonding rises even more (Step 6). Pour electrons in from the bottom. Count how many landed in the glued (bonding) seats minus the pushed-apart (antibonding) seats, halve it because a bond is a pair — that's the bond order. Two electrons → ${\text{H}}_2$, order 1, real. Four electrons → ${\text{He}}_2$, order 0, doesn't exist (Step 7). That halving-of-a-difference is Molecular Orbital Theory in one breath.

Recall

Which combination piles density between the nuclei? ::: The in-phase sum ${\psi }_A+{\psi }_B$ (bonding), because $+2{\psi }_A{\psi }_B>0$ there. What single feature defines the antibonding orbital? ::: A node (zero density) between the nuclei, from the $-2{\psi }_A{\psi }_B$ term. Why does antibonding rise more than bonding falls? ::: The node forces electrons near the nuclei where screening is poor and repulsion is stronger, so the penalty outweighs the reward — which is why ${\text{He}}_2$ has bond order 0. Bond order of ${\text{H}}_2$ and of ${\text{He}}_2$? ::: ${\text{H}}_2$: $(2-0)/2=1$. ${\text{He}}_2$: $(2-2)/2=0$.