From two orbitals to a working ladder
In the previous guide you learned the machinery: two atomic orbitals combine into one bonding orbital lower in energy and one antibonding orbital higher up, and a molecular orbital diagram is just the energy ladder you fill with electrons from the bottom. Here we put that machinery to work on the simplest, most revealing family in all of chemistry: the homonuclear diatomics, two identical atoms bonded together, marched across the second row from H2 to F2. Same recipe every time, only the electron count changes — and yet the answers swing wildly, from no bond at all to the strongest bond nature makes.
Because the two atoms are identical, the diagram is beautifully symmetric: the same atomic orbitals sit at the same energy on the left and on the right, and every molecular orbital in the middle is shared dead evenly between them. We only ever draw the valence orbitals — the deep core 1s electrons are too low in energy to mix and contribute nothing — so for the second row that means just four atomic orbitals per atom: one 2s and three 2p (call them 2p along the bond axis, plus two 2p perpendicular to it).
Those orbitals overlap in two distinct ways, and the distinction runs through this whole guide. The two 2s orbitals, and the two 2p that point head-on along the bond axis, overlap end-to-end to give cylindrical sigma orbitals; the two pairs of 2p that point sideways overlap flank-to-flank to give pi orbitals — exactly the sigma and pi bonds you met in valence-bond language, now recast as full-molecule orbitals. Crucially, the two perpendicular 2p directions are equivalent, so they give a matched pair of pi orbitals at identical energy. That degeneracy — two orbitals at exactly the same height — is the seed of the oxygen mystery we will crack at the end.
The first two rungs: H2 makes a bond, He2 cannot
Start with the smallest cases, where only the 1s orbitals are involved. In H2 the two 1s combine into a bonding sigma and an antibonding sigma-star; the molecule's two electrons both drop into the bonding sigma, opposite spins. The bond order — bonding electrons minus antibonding electrons, all over two — is (2 - 0)/2 = 1. A clean single bond, with every electron paired, so the molecule is diamagnetic, faintly pushed out of a magnetic field. This is the textbook success of the whole theory: a real, stable bond built from nothing but adding two atomic orbitals.
Now He2. Helium brings four electrons to the same little two-orbital ladder. Two fill the bonding sigma, but the next two have nowhere to go but the antibonding sigma-star. Bond order is (2 - 2)/2 = 0. The destabilizing antibonding pair exactly cancels the stabilizing bonding pair — and remember from last time that the antibonding orbital is raised slightly more than the bonding one is lowered, so the canceled state is actually a touch worse than two free atoms. So He2 simply does not form. The very same diagram that explains why hydrogen bonds explains why the next element refuses to: the answer is not the bonding electrons alone but the difference between bonding and antibonding.
The second row, and the twist of s-p mixing
From Li2 onward the 2s and 2p orbitals join the game, and now there are eight molecular orbitals to stack: from the 2s come a sigma2s and sigma-star2s; from the 2p come a sigma2p, a matched pair of pi2p, their antibonding partners pi-star2p, and a sigma-star2p. The naive ladder, lowest to highest, would put the head-on sigma2p below the sideways pi2p, since head-on overlap is stronger. For O2, F2 (and Ne2, were it to exist) that naive order is exactly right. But for the early second row — Li2, Be2, B2, C2, N2 — the levels reorder, and you have to know why or you will get the magnetism of B2 and the bond order of C2 wrong.
The cause is s-p mixing. The sigma2s and the sigma2p are both sigma-symmetric about the bond axis, so they are allowed to interact with each other — and orbitals of the same symmetry that lie close in energy always repel, pushing apart. The sigma orbital built mainly from 2s sinks a little lower; the sigma orbital built mainly from 2p is shoved a little higher — high enough, in the early second row, to climb above the pi2p pair. The pi orbitals have the wrong symmetry to take part, so they sit unmoved and become the lowest 2p-derived level. This is just the symmetry-and-energy matching rule from earlier doing its quiet work inside one molecule.
Why does the mixing fade by O2? Across the row the nuclear charge climbs, the 2s drops in energy much faster than the 2p, and the 2s-2p energy gap widens. The wider the gap, the weaker the s-p mixing (recall: mixing is strong only between orbitals close in energy), so by oxygen the sigma2p has fallen back below the pi2p and the naive order is restored. The crossover sits between N2 and O2. So there are really two standard diagrams: the s-p-mixed order (sigma2p above pi2p) for Li2 through N2, and the unmixed order (sigma2p below pi2p) for O2, F2, Ne2.
Filling order, valence MOs (lowest at bottom) Li2 -> N2 (s-p mixed) O2 -> Ne2 (unmixed) -------------------- -------------------- sigma*2p sigma*2p pi*2p pi*2p pi*2p pi*2p sigma2p <-- swap pi2p pi2p pi2p pi2p these two sigma2p sigma*2s sigma*2s sigma2s sigma2s Only the sigma2p / pi2p pair trade places.
Reading bond order, strength, and magnetism off the page
With the ladder fixed, filling it is pure bookkeeping — Aufbau from the bottom, two opposite-spin electrons per orbital, and Hund's rule to spread electrons singly across the degenerate pi pair before pairing them. Then you read three things straight off. Bond order, as before, is (bonding minus antibonding) over two. Bond strength tracks bond order closely: higher bond order means a shorter, stiffer, harder-to-break bond. And magnetism is decided by a single yes-or-no question — are there any unpaired electrons? If yes, the molecule is paramagnetic, drawn into a magnetic field; if no, it is diamagnetic, gently repelled.
Walk the row and watch the bond order climb then fall. Li2 has two valence electrons filling sigma2s: bond order 1, a weak single bond (lithium vapor really does contain Li2). Be2, with four, also fills sigma-star2s, giving bond order 0 — like He2, essentially no molecule. B2 has six: after the 2s pair and its antibonding pair cancel, the last two electrons enter the degenerate pi2p pair, one in each by Hund's rule. Bond order 1 — and, surprisingly, two unpaired electrons, so B2 is paramagnetic. That single fact is the proof that the s-p-mixed order is real, because the naive order would have put both electrons paired into sigma2p and predicted, wrongly, a diamagnetic B2.
Press on. C2 has eight electrons; both pi2p fill completely, leaving sigma2p empty, for bond order 2 — a curious double bond made of two pi bonds and no net sigma bond, and diamagnetic. N2 has ten: it fills both pi2p and the sigma2p, reaching bond order (8 - 2)/2 = 3, the celebrated triple bond, all electrons paired, diamagnetic. That triple bond is one of the strongest in chemistry — about 945 kilojoules per mole — which is exactly why N2 is so inert and why fixing nitrogen industrially is so hard. Nitrogen is the high-water mark; from here on, each new electron must go into an antibonding orbital, and the bond order starts coming back down.
Oxygen's two unpaired electrons — the riddle solved
Here is the failure that launched the whole subject. Draw O2 as a Lewis structure and you get a tidy double bond, O=O, with every electron neatly paired — which predicts a diamagnetic molecule. Yet pour liquid oxygen between the poles of a magnet and it clings there, visibly attracted: O2 is paramagnetic. Lewis dots, and even the simple valence-bond double bond, are flatly wrong about one of the most abundant molecules on Earth. This is the riddle the rung promised, and molecular orbital theory dissolves it on the first try.
- Oxygen sits past the crossover, so use the unmixed order: from the bottom, sigma2s, sigma-star2s, then sigma2p, then the degenerate pi2p pair, then the degenerate pi-star2p pair, and sigma-star2p on top.
- Pour in O2's twelve valence electrons: four fill sigma2s and sigma-star2s (canceling), two fill sigma2p, four fill the pi2p pair. Ten placed, two to go.
- The last two electrons must enter the degenerate pi-star2p pair. By Hund's rule they go in singly, one in each, with parallel spins — two unpaired electrons, sitting in antibonding orbitals.
- Count: eight bonding minus four antibonding, over two, gives bond order 2 (matching the O=O double bond) — but with two unpaired electrons, so O2 is paramagnetic. Both facts fall out of one diagram.
That is the whole trick, and it is worth savoring why it works. The double bond comes out right because eight-minus-four-over-two really is two — so MO theory agrees with Lewis on bond order. But where Lewis has to cram all the electrons into pairs, the MO diagram has a degenerate pi-star pair, and Hund's rule forces the last two electrons to occupy them singly with parallel spins. The unpaired spins are exactly what a magnet feels. The delocalized, energy-resolved picture captures something the localized dot picture structurally cannot: that you can have a definite bond order and still have unpaired electrons. The honest caveat is that the simple LCAO diagram is still a model — it gets bond order, oxygen's paramagnetism, and the level ordering right, but precise energies need heavier computation.
Closing the row, and what carries forward
Two steps remain. F2 has fourteen valence electrons: everything fills except the top sigma-star2p, and the count gives (8 - 6)/2 = 1, a weak single bond, all paired, diamagnetic — fluorine's bond is feeble, which is part of why F2 is so ferociously reactive. Ne2, with sixteen, fills the last antibonding level too: bond order 0, no molecule, the noble gas keeping to itself. So the whole second-row story is a single arc: bond order climbs 1, 0, 1, 2, 3 to the peak at N2, then falls 2, 1, 0 — and bond strength rises and falls in lockstep, because what you are really tracking is bonding-minus-antibonding electrons, one orbital at a time.
Step back and notice what you can now do that Lewis structures never could. You can predict half-integer bond orders, you can predict magnetism, and you can predict how bond strength shifts when you add or remove a single electron. The price is honesty about what the diagram is: these orderings are real and experimentally confirmed for the homonuclear diatomics, but the homonuclear diatomic MO diagram is the easy case — both atoms are identical, so the orbitals match perfectly. The moment the two atoms differ, the picture tilts.
That tilt is exactly the next guide. When the two atoms are different — CO, NO, HF — their atomic orbitals start at different energies, the molecular orbitals lean toward the more electronegative atom, and a half-filled level can leave a molecule like NO with an odd electron and a fractional bond order. The skill is the same skill you just built here; only the symmetric ladder becomes a lopsided one. Get the homonuclear diagram into your fingertips, and the heteronuclear case is a short step away.