Visual walkthrough — MO diagrams of H₂, He₂, N₂, O₂, F₂, NO, CO — bond order, magnetism

We assume only this: an atom has electrons (tiny charged specks) living in fuzzy clouds called orbitals (regions where an electron is likely to be found). Everything else we earn on the way. If you want the rulebook behind seating electrons, that is Hund's Rule and Pauli Exclusion; the wave-combining idea is Molecular Orbital Theory (LCAO basics).

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Step 1 — Two atoms, two clouds, nothing shared yet

WHAT. Picture two hydrogen atoms far apart. Each has one electron cloud — a round ball centred on its nucleus. We call each cloud a wavefunction and write it $\psi$ (Greek "psi", just a name-tag for "the shape of this cloud"). The left atom's cloud is ${\psi }_A$, the right atom's is ${\psi }_B$.

WHY start here. A bond is nothing more than what happens to these two clouds when the atoms drift close enough to touch. So the honest starting point is: two separate clouds, each unbothered.

PICTURE. Two identical navy balls, each with its nucleus dot. Nothing overlaps.

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Step 2 — Waves can add two ways: the birth of two orbitals

WHAT. Slide the atoms together so the clouds overlap. A cloud is a wave, and waves add. There are exactly two ways to add two waves:

• in phase — crests line up, they reinforce → cloud piles up between the nuclei;
• out of phase — a crest meets a trough, they cancel → a gap (a node) appears between the nuclei.

$\underset{\text{new cloud, glue}}{\underbrace{{\psi }_{\text{bond}}}}=\underset{\text{left}}{\underbrace{{\psi }_A}}+\underset{\text{right}}{\underbrace{{\psi }_B}}\underset{\text{new cloud, anti-glue}}{\underbrace{{\psi }_{\text{anti}}}}=\underset{\text{left}}{\underbrace{{\psi }_A}}-\underset{\text{right}}{\underbrace{{\psi }_B}}$

The $+$ means "add crests" (in phase); the $-$ means "subtract" (out of phase). Two inputs, two outputs — that is why one pair of atomic orbitals always makes exactly two molecular orbitals.

WHY this tool (adding waves) and not something else. We could ask "where does the electron go?" many ways, but electrons obey wave rules, and the one operation a wave always allows is superposition — adding. The $+$ and $-$ combinations are the only two independent ones you can build from ${\psi }_A$ and ${\psi }_B$, so they exhaust the possibilities. No case is missed.

PICTURE. Top: two crests merging into a fat middle bulge (bonding). Bottom: crest-plus-trough leaving a white slit down the middle (the node) — anti-bonding.

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Step 3 — Why "between the nuclei" means low energy (the energy ladder)

WHAT. Electron density piled between two positive nuclei glues them: each nucleus is pulled toward the shared cloud. That is a stable, comfortable, low-energy arrangement → the bonding orbital, drawn below the atomic level. The anti-bonding cloud, with its central gap, leaves the nuclei staring at each other with nothing between → they repel → high energy → drawn above, and we brand it with an asterisk: ${\sigma }^{\ast }$.

WHY we now switch to an energy picture. Counting bonds later needs heights: which orbital is lower, which is higher. So we redraw the two clouds as two rungs on an energy ladder — left rung = atom A's orbital, right rung = atom B's, middle = the two new MOs.

PICTURE. A ladder diagram: two atomic rungs at the same height on the sides; a lower middle rung ($\sigma$, bonding) and a higher middle rung (${\sigma }^{\ast }$, antibonding). Arrows show one rung sinks, one rises by the same amount.

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Step 4 — Seat the electrons: three rules, lowest first

WHAT. Now pour in the electrons the molecule actually has. Three seating rules (full story in Hund's Rule and Pauli Exclusion):

1. Aufbau — fill the lowest empty rung first.
2. Pauli — a rung holds at most two electrons, and they must have opposite spins (drawn ↑↓).
3. Hund — if two rungs are at the same height (degenerate), put one electron in each, spins parallel (↑ ↑), before doubling any up.

WHY these three. Aufbau keeps the total energy as low as possible; Pauli is a hard law (no two electrons share all properties); Hund minimises electron–electron repulsion by spreading them out. Together they fix a unique filling — no guesswork.

PICTURE. H₂: 2 electrons drop into the bonding rung as ↑↓. He₂: the next 2 are forced up into the ${\sigma }^{\ast }$ rung — anti-glue exactly cancels the glue.

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Step 5 — Count the net bonds: the bond-order formula, earned

WHAT. A "bond" is really a pair of electrons glued between nuclei. So:

• each electron in a bonding rung adds $+\frac{1}{2}$ of a bond,
• each electron in an antibonding rung undoes $\frac{1}{2}$ of a bond (it cancels a bonding partner).

Let $N_b$ = electrons in bonding rungs, $N_a$ = electrons in antibonding rungs. Add up the contributions:

$\text{Bond Order}=\frac{\stackrel{{\text{glue e}}^{-}}{\overbrace{N_b}}-\stackrel{{\text{anti-glue e}}^{-}}{\overbrace{N_a}}}{\underset{{\text{2 e}}^{-}=\text{1 bond}}{\underbrace{2}}}$

The $-N_a$ says anti-glue subtracts; dividing by 2 converts "electrons" into "pairs = bonds".

WHY divide by 2. Because a bond is a pair, not a single electron. Two glue electrons = one bond; the $/2$ is just the unit conversion.

PICTURE. H₂: $\frac{2-0}{2}=1$ (one bond). He₂: $\frac{2-2}{2}=0$ — glue and anti-glue tie, no net bond, so He₂ doesn't exist. This degenerate zero-bond case is a full case, not a footnote.

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Step 6 — Second-row atoms: the rung order, and why it FLIPS

WHAT. For row-2 atoms we have $2s$ and $2p$ orbitals too. Three $p$ orbitals overlap two ways: head-on ($p_z$ along the axis → one ${\sigma }_{2p}$) and sideways (the two $p_x,p_y$ → two degenerate ${\pi }_{2p}$). So near the top the rungs are ${\pi }_{2p}$ (a pair, same height) and ${\sigma }_{2p}$ (single).

Their order depends on the atom:

B, C, N (atomic number $Z\le 7$): the $2s$ and $2p$ energies sit close, so ${\sigma }_{2s}$ and ${\sigma }_{2p}$ repel each other — this is s–p mixing (s–p mixing and orbital energy ordering). It shoves ${\sigma }_{2p}$ up above the ${\pi }_{2p}$: $\dots <{\pi }_{2p}<{\sigma }_{2p}<{\pi }_{2p}^{\ast }<{\sigma }_{2p}^{\ast }$

O, F ($Z\ge 8$): the $2s$–$2p$ gap is now wide, mixing is weak, natural order returns — ${\sigma }_{2p}$ below ${\pi }_{2p}$: $\dots <{\sigma }_{2p}<{\pi }_{2p}<{\pi }_{2p}^{\ast }<{\sigma }_{2p}^{\ast }$

WHY the flip matters. It changes which rung fills last, which changes the magnetism. Get the order wrong and you predict the wrong number of unpaired electrons.

PICTURE. Two ladders side by side: left ($Z\le 7$) with ${\sigma }_{2p}$ perched above ${\pi }_{2p}$; right ($Z\ge 8$) with ${\sigma }_{2p}$ sunk below. A curved arrow shows the mixing shove.

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Step 7 — The flagship case: O₂ and the magnet

WHAT. Oxygen has 8 electrons per atom → 16 total. Use the $Z\ge 8$ ladder (no mixing). Fill lowest-first: ${\sigma }_{1s}^2{\sigma }_{1s}^{\ast 2}{\sigma }_{2s}^2{\sigma }_{2s}^{\ast 2}{\sigma }_{2p}^2{\pi }_{2p}^4{\pi }_{2p}^{\ast 2}$ The last two electrons land in the two degenerate ${\pi }_{2p}^{\ast }$ rungs. Hund's rule forces them one each, parallel (↑ ↑) — two unpaired electrons.

Count: $N_b=10$ (the ${\sigma }_{1s},{\sigma }_{2s},{\sigma }_{2p}^2,{\pi }_{2p}^4$ bonding electrons), $N_a=6$ (${\sigma }_{1s}^{\ast }2,{\sigma }_{2s}^{\ast }2,{\pi }_{2p}^{\ast }2$). $\text{BO}=\frac{10-6}{2}=2$

WHY this is the whole point. A Lewis dot structure pairs everything and screams "diamagnetic". MO theory, by honestly obeying Hund on the degenerate ${\pi }^{\ast }$, predicts two unpaired electrons — and real liquid O₂ sticks to a magnet. That is paramagnetism (see Paramagnetism and Diamagnetism), and it is MO theory's crown jewel.

PICTURE. O₂ ladder fully filled; the two ${\pi }^{\ast }$ rungs each hold a single ↑ arrow, highlighted red, labelled "2 unpaired → paramagnetic".

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The one-picture summary

Everything above, compressed: two atomic rungs → split into bonding + antibonding → fill by Aufbau/Pauli/Hund → count $N_b-N_a$ over 2 → read magnetism off the unpaired arrows. The trend ${\text{N}}_2(3)>{\text{O}}_2(2)>{\text{F}}_2(1)$ falls out because each new electron past N₂ pours into antibonding rungs and tears the bond down (this is the Bond Order, Bond Length, Bond Energy correlation).

Recall Feynman: the whole walkthrough in plain words

Start with two fuzzy balls, one on each atom (Step 1). Push them together and the fuzz can either pile up in the middle (glue) or leave a gap in the middle (anti-glue) — two shapes, always (Step 2). Glue-in-the-middle is comfy and sits low; gap-in-the-middle is tense and sits high, so draw a ladder with a low rung and a high rung (Step 3). Now bring your electrons and seat them like people in a theatre: lowest seats first, two per seat facing opposite ways, and if two seats are the same height put one person in each before doubling up (Step 4). Count the glue-people, subtract the anti-glue-people, divide by two — that's how many bonds hold the atoms; if it hits zero, there's no molecule at all, like He₂ (Step 5). For bigger atoms there are more rungs, and a sneaky repulsion between the $s$ and $p$ sausage-orbitals flips two rungs for boron-through-nitrogen but relaxes back for oxygen and fluorine (Step 6). Oxygen's last two electrons are forced to sit alone in two same-height seats, spins matched — two lonely electrons — which is exactly why liquid oxygen jumps at a magnet, the thing Lewis dots could never explain (Step 7). Odd molecules like NO leave one lonely electron and give a half-a-bond bond order — that's fine, nature doesn't owe us whole numbers.

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Connections

• 2.3.13 MO diagrams of H₂, He₂, N₂, O₂, F₂, NO, CO — bond order, magnetism (Hinglish)
• Molecular Orbital Theory (LCAO basics)
• Hund's Rule and Pauli Exclusion
• Paramagnetism and Diamagnetism
• Bond Order, Bond Length, Bond Energy correlation
• s–p mixing and orbital energy ordering
• Isoelectronic species (CO, N₂, CN⁻, NO⁺)