Homonuclear Diatomic MOs

Liquid oxygen is pale blue. It also sticks to the poles of a magnet. Neither fact makes sense if you draw O=O with two neat Lewis lone-pair-free double bonds. Both facts fall out immediately once you build the molecular orbital diagram of O₂. You count electrons into it, and the diagram predicts something directly. Without any extra assumption, it shows O₂ has two unpaired electrons. These sit in degenerate antibonding orbitals. This single picture is the cleanest demonstration in introductory chemistry. It shows molecular orbital (MO) theory is not just a complicated alternative to Lewis structures. It is the theory that gets magnetism, bond order, and photoelectron spectra right. Simple bonding pictures quietly get these wrong. This article builds the MO diagrams for every homonuclear diatomic molecule and ion across Period 2. It covers H₂ through F₂, and glances at Ne₂. We combine atomic orbitals according to symmetry and energy match. We fill electrons using the same aufbau/Hund's-rule logic used for atoms. Then we read off bond order, bond length, bond energy, and magnetism directly from the filled diagram. The core machinery was assumed known beforehand. This includes linear combination of atomic orbitals (LCAO), the σ/π symmetry classification, and the bonding/antibonding split. Here the job is different. We assemble that machinery into a working periodic table of diatomic molecules. We also show why the ordering of the σ_g(2p) and π_u(2p) levels changes partway across the period. The big picture is this: two atoms bring a fixed inventory of valence atomic orbitals. These are 1s for H, and 2s and 2p for Li–F. The number of molecular orbitals that come out equals the number of atomic orbitals that went in. No orbitals are created or destroyed, only recombined. Each atomic orbital pair combines in-phase and out-of-phase. This gives one bonding and one antibonding MO. Orbitals of σ symmetry along the internuclear axis combine separately from orbitals of π symmetry. This is because they belong to different irreducible representations. They cannot mix by symmetry alone. Electron filling then follows the aufbau principle and Hund's rule, exactly as in atoms. The resulting bond order,

BO=12(nbonding−nantibonding),BO = \frac{1}{2}\left(n_{\text{bonding}} - n_{\text{antibonding}}\right),

predicts trends in bond length and bond dissociation energy. This holds across the whole row with essentially no adjustable parameters. By the end you will be able to draw the correct MO diagram. This applies to any Period 2 homonuclear diatomic or its common ion. You will justify why the σ(2p)/π(2p) crossover happens at N₂/O₂. You will predict paramagnetism from the diagram alone. You will also connect bond-order changes on ionization (e.g. O₂ → O₂⁺) to measured changes in bond length. This sets up s-p Mixing and Orbital Ordering, which explains why the crossover happens in orbital-energy terms. It also leads to Heteronuclear Diatomic MOs and Magnetism from MO Diagrams. These generalize the method to unlike atoms and to quantitative magnetic moments.

Key takeaways
  • Combining nn atomic orbitals gives exactly nn molecular orbitals: half (roughly) bonding, half antibonding, split by symmetry into σ and π sets.
  • H₂ through F₂ are built from 1s (H, He) or 2s/2p (Li–F) valence orbitals; only orbitals of matching symmetry and comparable energy combine effectively.
  • For Li₂ through N₂ the order is σ(2s) < σ(2s) < π(2p) < σ(2p) < π(2p) < σ*(2p); for O₂, F₂, Ne₂ the σ(2p) drops below π(2p).
  • Bond order BO=12(nb−na)BO = \tfrac12(n_b - n_a) tracks measured bond length (shorter for higher BO) and bond energy (stronger for higher BO) across the whole row.
  • O₂'s ground state has two electrons in degenerate π*(2p) orbitals, unpaired by Hund's rule — this is the MO-theory explanation of O₂ paramagnetism.
  • Removing or adding electrons from real antibonding vs bonding orbitals (O₂ → O₂⁺ or O₂⁻) changes bond order and is directly measurable as a bond-length change.
  • Key idea: electrons fill molecular orbitals built from all the atoms' valence atomic orbitals, not two separate atomic electron clouds — bond order, magnetism, and bond length all follow from where the electron count lands in that shared orbital ladder.

Where this lives today

The O₂ paramagnetism result is not a museum piece — it is central to why O₂ activation chemistry in biology and catalysis is hard. Because ground-state O₂ is a triplet (two unpaired, parallel-spin electrons) while almost all organic molecules are closed-shell singlets, direct reaction of O₂ with organic substrates is spin-forbidden and kinetically slow at room temperature — this is precisely why life needs metalloenzymes (cytochrome P450s, hemoglobin's partner oxidases, and related metal centers) to first bind and activate O₂, converting it to a reactive singlet-compatible or metal-bound form before it reacts efficiently with organic carbon frameworks. The MO diagram derived above is the entire reason this spin barrier exists. The bond-order/bond-length correlation demonstrated for N₂/N₂⁺/N₂⁻ underlies interpretation of gas-phase photoelectron spectroscopy and mass spectrometry of diatomic radical cations and anions used in atmospheric and combustion chemistry — N₂⁺ is a genuine, spectroscopically characterized species in auroral emission and in plasma physics, and its bond length and vibrational frequency are used as textbook calibration points for exactly this MO bond-order logic. Superoxide (O₂⁻) and peroxide (O₂²⁻) chemistry, predicted correctly above from bond order and magnetism, are central to oxidative stress biochemistry, battery chemistry (metal–air batteries generate and consume superoxide/peroxide intermediates), and industrial bleaching and oxidation processes. More broadly, this same LCAO/aufbau/Hund machinery, generalized to unlike atoms, is the direct route into Heteronuclear Diatomic MOs (needed for CO, NO, and HF, all of industrial or biological importance — CO as a ligand in hemoglobin poisoning and in industrial carbonylation catalysis, NO as a biological signaling radical), and generalized to quantitative spin counting it becomes Magnetism from MO Diagrams, used routinely to predict and rationalize the magnetic behavior of molecular radicals and small inorganic clusters in materials chemistry.

Ask the tutor: How exactly does the shrinking 2s–2p energy gap across Period 2 cause σ(2p) to drop below π(2p) at O₂, in terms of orbital mixing coefficients?

The physical picture

Start from the rule that governs every diatomic in this row: atomic orbitals combine only with partners of the same symmetry along the bond axis, and mixing is strongest when the two atomic orbitals are close in energy. Along the internuclear (z) axis, an s orbital and a pzp_z orbital both have cylindrical symmetry with no nodal plane containing the axis — both are labeled σ. The pxp_x and pyp_y orbitals on each atom have one nodal plane containing the axis and are labeled π; they combine only with π-symmetry partners on the other atom, never directly with σ orbitals, because a σ- and π-symmetry orbital product integrates to zero by symmetry (their overlap integral vanishes exactly, independent of the energies involved). For H₂ there is only one valence orbital per atom (1s), so the whole molecule reduces to the textbook two-orbital problem: in-phase combination gives σ₁ₛ (bonding, lower in energy, node-free between nuclei), out-of-phase combination gives σ₁ₛ (antibonding, higher in energy, with a nodal plane bisecting the bond). Both electrons of H₂ occupy σ₁ₛ, giving bond order 1 — a single covalent bond, matching the Lewis picture exactly. For Period 2, each atom contributes four valence atomic orbitals: 2s, 2px2p_x, 2py2p_y, 2pz2p_z (taking z as the internuclear axis). Four orbitals per atom, eight atoms' worth total, means eight molecular orbitals must appear. The 2s orbitals combine to give σ(2s) and σ(2s). The two 2pz2p_z orbitals, pointing directly along the bond, combine head-on to give σ(2p) and σ(2p). The 2px2p_x orbitals combine side-on to give one bonding π(2p) and one antibonding π(2p); the 2py2p_y orbitals do the same, giving a second, energy-degenerate π(2p)/π(2p) pair. So the π(2p) bonding level is doubly degenerate, and so is π(2p) — each holds up to four electrons total (two orbitals × two electrons), which is exactly why Hund's rule becomes decisive for O₂. The one subtlety that changes the whole story partway across the row is s–p mixing: the σ(2s), σ(2s) orbitals are not perfectly "pure" 2s combinations, and the σ(2p) orbital is not pure 2pz2p_z. Because σ-symmetry MOs built from 2s and from 2pz2p_z have the same* symmetry label (σg\sigma_g or σu\sigma_u depending on convention), they are allowed to mix with each other, and they do so more strongly when the 2s–2p energy gap on the atom is small. Early in the period (Li, Be, B, C, N) the 2s–2p gap is small, mixing is large, and it pushes σ(2p) up above π(2p). Late in the period (O, F, Ne) the 2s–2p gap has grown large enough that mixing is weak, and σ(2p) drops back below π(2p), which is where a naive "no-mixing" MO diagram would have placed it all along. This crossover is the single most commonly mis-drawn feature of Period 2 MO diagrams, and it is developed in full in s-p Mixing and Orbital Ordering; here it is simply built in as an empirical, well-verified fact for each element.

Figure
Atomic orbital overlap: σ from head-on p, π from side-on p
σ(2p): head-on overlap along bond axis bonding σ(2p): in-phase, no node between nuclei antibonding σ*(2p): node between nuclei π(2p): side-on overlap, off-axis lobes bonding π(2p): lobes reinforce above/below axis antibonding π*(2p): sign-mismatched lobes cancel overlap
Figure
Two MO orderings across Period 2: with and without s–p mixing
Li₂ – N₂ (strong s–p mixing) O₂, F₂, Ne₂ (weak s–p mixing) σ(2s) σ*(2s) π(2pₓ,2p_y) [degenerate] σ(2p) π*(2pₓ,2p_y) σ*(2p) σ(2s) σ*(2s) σ(2p) π(2pₓ,2p_y) π*(2pₓ,2p_y) σ*(2p) σ(2p) above π(2p): small 2s–2p gap ⇒ strong mixing σ(2p) below π(2p): large 2s–2p gap ⇒ weak mixing

The core ideas

The bond order relationship already stated is the single quantitative handle on everything else:

BO=12(nbonding−nantibonding)BO = \frac{1}{2}\left(n_{\text{bonding}} - n_{\text{antibonding}}\right)

where:

  • nbondingn_{\text{bonding}} = number of electrons in bonding MOs (σ and π combined),
  • nantibondingn_{\text{antibonding}} = number of electrons in antibonding MOs,
  • non-bonding electrons (rare in this row, but relevant once core orbitals are included loosely) do not enter the count.

Reading: each bonding electron pair contributes a full bond; each antibonding electron pair cancels one. A bond order of zero, as for He₂ or (hypothetically) Ne₂, means no net stabilization — the molecule is not expected to exist as a stable species relative to separated atoms. Bond order in turn correlates with two measurable quantities. Bond length shortens as bond order rises across an isoelectronic-type series because more bonding character pulls the nuclei closer, and — treating the bond loosely as a stretched spring for comparison purposes — bond dissociation energy rises with bond order because more net bonding electron density must be removed to separate the atoms:

re  decreases,De  increases,as BO increases (within one row, same atoms).r_e \; \text{decreases}, \qquad D_e \; \text{increases}, \qquad \text{as } BO \text{ increases (within one row, same atoms).}

where:

  • rer_e = equilibrium bond length,
  • DeD_e = bond dissociation energy (or bond enthalpy, in the thermochemical convention),
  • the comparison is only valid for the same pair of atoms — comparing bond order across different elements (e.g., N₂ vs O₂) still tracks length/energy well empirically in Period 2, but the correlation is not a rigorous law across arbitrary atom pairs.

A second core relationship is simply electron bookkeeping — the orbital count conservation that makes the whole method predictive rather than descriptive:

NMO=NAO,A+NAO,BN_{\text{MO}} = N_{\text{AO,A}} + N_{\text{AO,B}}

where:

  • NMON_{\text{MO}} = total number of molecular orbitals produced,
  • NAO,A,NAO,BN_{\text{AO,A}}, N_{\text{AO,B}} = number of valence atomic orbitals contributed by atom A and atom B respectively.

Reading: for Period 2 homonuclear diatomics, 4+4=84+4=8 molecular orbitals always appear (σ2s, σ2s, σ2p, π2p ×2, π2p ×2, σ*2p), regardless of which element or how many electrons are actually available to fill them. This is what makes the diagram a fixed scaffold that you simply pour electrons into, element by element, exactly as with atomic orbital diagrams and the aufbau principle.

Derivation ABuilding the bond order for N₂ from first principles, and why it is 3

Goal: show explicitly how the electron count of N₂ maps onto the fixed eight-orbital scaffold, and derive BO=3BO=3 without simply quoting the Lewis triple bond.

Step 1 — count valence electrons. Each N atom has the configuration 1s22s22p31s^2 2s^2 2p^3; the 1s1s electrons are core and are conventionally excluded from the valence MO diagram (they form a nonbonding σ₁ₛ/σ*₁ₛ pair so close in energy and so localized that they cancel exactly and are dropped). Valence electrons per N atom: 5. Total valence electrons for N₂: 2×5=102 \times 5 = 10.

Step 2 — fix the orbital ladder. Nitrogen sits early enough in Period 2 that the 2s–2p atomic energy gap is small (as it is for Li, Be, B, C, N), so s–p mixing is strong and the diagram uses the "mixed" ordering:

σ(2s)<σ∗(2s)<π(2px)=π(2py)<σ(2p)<π∗(2px)=π∗(2py)<σ∗(2p).\sigma(2s) < \sigma^*(2s) < \pi(2p_x)=\pi(2p_y) < \sigma(2p) < \pi^*(2p_x)=\pi^*(2p_y) < \sigma^*(2p).

Step 3 — fill electrons via aufbau/Hund. Pour in the 10 valence electrons from the bottom:

σ(2s)2 σ∗(2s)2 π(2px)2π(2py)2 σ(2p)2.\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^2\pi(2p_y)^2\,\sigma(2p)^2.

That accounts for 2+2+2+2+2=102+2+2+2+2 = 10 electrons exactly, with nothing left over to populate π(2p) or σ(2p).

Step 4 — compute bond order. Bonding electrons: σ(2s) has 2, π(2p) pair has 4, σ(2p) has 2, total nbonding=8n_{\text{bonding}} = 8. Antibonding electrons: σ*(2s) has 2, and nothing else is filled, so nantibonding=2n_{\text{antibonding}} = 2. Then

BO=12(8−2)=12(6)=3.BO = \frac{1}{2}(8 - 2) = \frac{1}{2}(6) = 3.

This reproduces the Lewis triple bond, but now with an explicit statement of why — the σ(2s)/σ(2s) pair cancels to zero net bonding (2 bonding − 2 antibonding = 0), and the three remaining bonding pairs (one π(2p_x), one π(2p_y), one σ(2p)) are unopposed by any antibonding occupation, giving three whole bonds. This is also why N₂ has the shortest, strongest bond in the entire row (bond length 109.8 pm, bond energy 945 kJ mol⁻¹) — it is the last element before antibonding π orbitals start to be occupied.

Limiting-case check: if instead the "unmixed" ordering (σ(2p) below π(2p)) were used for N₂, the same 10 electrons would fill σ(2s)²σ(2s)²σ(2p)²π(2p)⁴, giving the identical bond order of 3 — the two orderings agree on bond order for N₂ (and for all species with fully filled or fully empty π(2p)/σ(2p) shells). The orderings only diverge in predicted level degeneracies and photoelectron spectra* for species like N₂ itself and B₂, C₂, which is exactly the subtlety explored in s-p Mixing and Orbital Ordering — bond order survives either way, but which orbital is highest occupied (HOMO) does not.

Go deeper · Derivation B · 2 minWhy O₂ must be paramagnetic: Hund's rule applied to degenerate π* orbitals
Derivation BWhy O₂ must be paramagnetic: Hund's rule applied to degenerate π* orbitals

Goal: derive the ground-state electron configuration of O₂ and show that two unpaired electrons are unavoidable — not a special feature of oxygen, but a direct consequence of degeneracy plus electron count.

Step 1 — count valence electrons. Each O atom is 1s22s22p41s^2 2s^2 2p^4; valence electrons per atom = 6 (excluding core 1s). Total for O₂: 2×6=122\times 6 = 12.

Step 2 — fix the orbital ladder for O₂. Oxygen is late enough in the period that the 2s–2p gap has widened and s–p mixing is weak, so the ordering reverts to the "unmixed" form:

σ(2s)<σ∗(2s)<σ(2p)<π(2px)=π(2py)<π∗(2px)=π∗(2py)<σ∗(2p).\sigma(2s) < \sigma^*(2s) < \sigma(2p) < \pi(2p_x)=\pi(2p_y) < \pi^*(2p_x)=\pi^*(2p_y) < \sigma^*(2p).

Step 3 — fill 12 electrons from the bottom.

σ(2s)2 σ∗(2s)2 σ(2p)2 π(2px)2π(2py)2 π∗(2px)1π∗(2py)1.\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p)^2\,\pi(2p_x)^2\pi(2p_y)^2\,\pi^*(2p_x)^1\pi^*(2p_y)^1.

Running total: 2+2+2+2+2+1+1=122+2+2+2+2+1+1 = 12, exactly matching the electron count.

Step 4 — apply Hund's rule to the last two electrons. The π(2p_x) and π(2p_y) orbitals are degenerate by symmetry (both come from side-on 2p overlap, related to each other by a 90° rotation about the bond axis, so nothing distinguishes their energies). Exactly as in filling the degenerate 2p atomic orbitals of carbon, placing two electrons into two degenerate orbitals minimizes electron–electron repulsion by keeping spins parallel and occupying different orbitals — one electron in π(2p_x), one in π(2p_y), both spin-up. This is forced by the same physics as Hund's first rule in atoms; there is no alternative low-energy configuration once the electron count lands exactly on "two electrons, two degenerate empty orbitals."

Step 5 — read off bond order and magnetism. Bonding electrons: σ(2s) 2, σ(2p) 2, π(2p)×2 gives 4, subtotal nbonding=8n_{\text{bonding}} = 8. Antibonding electrons: σ(2s) 2, π(2p) 2, subtotal nantibonding=4n_{\text{antibonding}} = 4. Then

BO=12(8−4)=12(4)=2,BO = \frac{1}{2}(8-4) = \frac{1}{2}(4) = 2,

consistent with the Lewis O=O double bond — but now the diagram also predicts, with no extra input, that O₂ has two unpaired electrons and is paramagnetic. This is the calculation that Lewis structures cannot do: a double bond drawn as two shared pairs implies no unpaired electrons, yet liquid O₂ is experimentally paramagnetic and is pulled into a magnetic field. MO theory gets this right because it fills electrons into molecular, not atomic, orbitals, and two of those molecular orbitals happen to be degenerate at exactly the electron count oxygen supplies.

Limiting-case check: compare to the isoelectronic superoxide ion O₂⁻ (13 valence electrons) and peroxide O₂²⁻ (14 valence electrons). Adding one electron to O₂ must go into a π(2p) orbital (the next available slot), giving configuration …π(2p_x)²π(2p_y)¹ (or the symmetric alternative), one unpaired electron, BO=12(8−5)=1.5BO = \tfrac12(8-5) = 1.5; adding a second electron pairs up the π set completely, …π(2p_x)²π(2p_y)², zero unpaired electrons (diamagnetic), BO=12(8−6)=1BO = \tfrac12(8-6)=1. Both predictions match experiment: O₂⁻ is paramagnetic with a bond order of 1.5 and a bond length (~135 pm) intermediate between O₂ (121 pm, BO 2) and O₂²⁻ (149 pm, BO 1), exactly tracking the bond-order/bond-length trend derived in the core relationships above.

Worked examples

Example 1 (foundational): Li₂, the simplest Period-2 case. Each Li atom is 1s22s11s^2 2s^1; only the single valence 2s electron per atom matters (1s is core, dropped as a canceling nonbonding pair as before). Total valence electrons for Li₂: 2×1=22\times 1 = 2. These two electrons fill the lowest available molecular orbital, σ(2s), completely: configuration σ(2s)². No electrons reach σ*(2s) or anything higher. Bonding electrons nb=2n_b = 2; antibonding electrons na=0n_a = 0. Then

BO=12(2−0)=1.BO = \frac{1}{2}(2-0) = 1.

Li₂ is predicted to have a single bond, and indeed it is a known (weakly bound, gas-phase) diatomic molecule with bond length 267 pm and bond energy about 105 kJ mol⁻¹ — much longer and weaker than N₂'s triple bond, exactly as expected for a bond order of only 1 built from large, diffuse 2s orbitals with poor overlap. As a limiting check: Be₂ (four valence electrons, filling σ(2s)²σ*(2s)²) gives BO=12(2−2)=0BO = \tfrac12(2-2) = 0; consistent with this, Be₂ is only very weakly bound (a van der Waals-scale complex in its ground state) rather than a normal covalent diatomic, confirming that the "cancel to zero" prediction is not just a bookkeeping curiosity but matches the real absence of a robust Be–Be bond.

Example 2 (applied): C₂ and the surprising bond order 2 vs the "expected" triple bond. Carbon is 1s22s22p21s^2 2s^2 2p^2; valence electrons per atom = 4, total for C₂ = 8. Carbon lies in the strong-mixing region of the period (small 2s–2p gap), so the ordering is the same "mixed" ladder used for N₂: σ(2s) < σ(2s) < π(2p) < σ(2p) < π(2p) < σ*(2p). Filling 8 electrons from the bottom:

σ(2s)2 σ∗(2s)2 π(2px)2π(2py)2.\sigma(2s)^2\,\sigma^*(2s)^2\,\pi(2p_x)^2\pi(2p_y)^2.

That is 2+2+2+2=82+2+2+2 = 8, exactly using up the supply — note that with the mixed ordering, π(2p) fills before σ(2p), so σ(2p) remains completely empty. Bonding electrons: σ(2s) contributes 2, π(2p) pair contributes 4, subtotal nb=6n_b = 6. Antibonding electrons: σ*(2s) contributes 2, subtotal na=2n_a = 2. Then

BO=12(6−2)=2.BO = \frac{1}{2}(6-2) = 2.

This is a genuine, experimentally confirmed prediction of MO theory that a simple bonding-pairs argument gets wrong if you assume σ(2p) fills before π(2p): C₂ is a double bond with both bonding pairs being π bonds and no σ(2p) bonding contribution, and — because both π(2p_x) and π(2p_y) are exactly filled (2 electrons each) — the molecule is diamagnetic, with no unpaired electrons, despite naive expectations (based on the O₂ pattern) that partially filled degenerate orbitals might be involved. As a limiting check, compare the isoelectronic CN⁻ or the cation C₂⁺: removing one electron from C₂ takes it out of π(2p) (the HOMO under mixed ordering), giving π(2p_x)²π(2p_y)¹, one unpaired electron, and BO=12(6−1−2)=1.5BO = \tfrac12(6-1-2) = 1.5 — a weaker, longer bond than neutral C₂, consistent with photoelectron spectroscopy assignments for this ordering.

Example 3 (exam/competition level): Rank N₂, N₂⁺, and N₂⁻ by bond order and bond length, and identify which ionization removes a bonding vs an antibonding electron. Starting point: neutral N₂ has 10 valence electrons and configuration σ(2s)²σ(2s)²π(2p)⁴σ(2p)² (mixed ordering, as derived above), BO=3BO=3, bond length 109.8 pm. For N₂⁺ (9 valence electrons), the electron removed on ionization comes from the highest-filled orbital, which under the mixed ordering is σ(2p) — a bonding orbital. New configuration: σ(2s)²σ(2s)²π(2p)⁴σ(2p)¹. Bonding electrons: 2+4+1=72 + 4 + 1 = 7; antibonding electrons: 22. Then

BO(N2+)=12(7−2)=2.5.BO(\text{N}_2^+) = \frac{1}{2}(7-2) = 2.5.

Removing a bonding electron weakens the bond, so N₂⁺ should be longer and weaker-bonded than N₂ — and indeed N₂⁺ has a bond length of about 111.6 pm, longer than N₂'s 109.8 pm, exactly tracking the drop from BO 3 to BO 2.5. For N₂⁻ (11 valence electrons), the added electron must go into the lowest unoccupied orbital, which is π(2p) — an antibonding orbital. New configuration: σ(2s)²σ(2s)²π(2p)⁴σ(2p)²π*(2p)¹. Bonding electrons: 2+4+2=82+4+2 = 8; antibonding electrons: 2+1=32 + 1 = 3. Then

BO(N2−)=12(8−3)=2.5.BO(\text{N}_2^-) = \frac{1}{2}(8-3) = 2.5.

So both ions land at the same bond order of 2.5 despite opposite processes (one removes a bonding electron, the other adds an antibonding one) — a good self-check, since removing a bonding electron and adding an antibonding electron both reduce net bond order by exactly 12\tfrac12 relative to the neutral. The predicted ranking of bond length is

r(N2)<r(N2+)≈r(N2−),r(\text{N}_2) < r(\text{N}_2^+) \approx r(\text{N}_2^-) ,

with N₂ shortest (highest BO = 3) and both ions longer (BO = 2.5) — the diagram distinguishes mechanism (bonding electron lost vs antibonding electron gained) while correctly predicting that the magnitude of the bond-order change, and hence the length change, is comparable either way.

Limiting cases, scope & common traps

SituationWhat the MO diagram predictsCheck against reality
Core 1s orbitals in Li₂–F₂Treated as nonbonding, canceling σ₁ₛ²σ*₁ₛ² pair, dropped from the valence diagramStandard simplification; core orbitals are too contracted and too far apart in energy to mix appreciably
He₂BO=12(2−2)=0BO = \tfrac12(2-2) = 0No stable He₂ molecule under ordinary conditions — matches BO = 0 prediction
Li₂ through N₂Mixed ordering: σ(2p) above π(2p)Confirmed by photoelectron spectroscopy ordering of ionization energies for these molecules
O₂, F₂, (Ne₂)Unmixed ordering: σ(2p) below π(2p)Confirmed by photoelectron spectroscopy; also required to get O₂ paramagnetism right
B₂Under mixed ordering: π(2p)¹π(2p)¹, two unpaired electrons, paramagneticExperimentally confirmed — B₂ is paramagnetic, a result the unmixed ordering would get wrong
Comparing bond order across different element pairsRough correlation with length/energy, not a strict lawValid as a trend across Period 2 homonuclear series; use caution comparing, e.g., a diatomic bond order to a very different bonding situation
Ne₂BO=12(8−8)=0BO = \tfrac12(8-8) = 0No stable covalent Ne₂; only a van der Waals complex exists

The chemist behind it

The MO picture of diatomics was not the work of one person alone; it emerged through the 1920s–30s from the convergence of quantum mechanics with spectroscopic data on diatomic molecules, and its two most influential early architects were Robert Mulliken and Friedrich Hund. Robert S. Mulliken spent much of his career turning band spectra of diatomic molecules into a systematic orbital-based language — the very σ/π, bonding/antibonding notation used throughout this article is largely his. He insisted on building molecular states from delocalized orbitals spanning the whole molecule, rather than from pre-formed atomic bonding pairs, which is precisely the leap that gets O₂'s magnetism right where a localized-pair Lewis picture cannot. Mulliken also pushed the LCAO approximation from a qualitative sketch toward a genuinely predictive, spectroscopically calibrated tool, cross-checking orbital orderings against measured ionization energies from early photoelectron and absorption spectroscopy — the same kind of check used above to justify the σ(2p)/π(2p) crossover.

Origins · 1 minRobert Sanderson Mulliken and how this was discovered
RS
Robert Sanderson Mulliken
1896–1986 · Physical chemistry, molecular spectroscopy, University of Chicago

Mulliken received the 1966 Nobel Prize in Chemistry "for his fundamental work concerning chemical bonds and the electronic structure of molecules by the molecular orbital method." Working largely from analysis of diatomic band spectra, he built the systematic MO framework — including the σ, π, g/u notation and the LCAO approach — that turned molecular orbital theory from a formalism into a working predictive tool for real molecules, diatomics chief among them.

History
- 1927 — Walter Heitler and Fritz London give the first quantum-mechanical treatment of the H₂ bond, launching quantum theories of the chemical bond.
- 1928–1932 — Friedrich Hund and Robert Mulliken independently develop the molecular orbital approach, treating electrons as delocalized over the whole molecule rather than localized between atom pairs.
- 1932 — Mulliken introduces systematic σ/π and bonding/antibonding orbital notation for diatomic molecules, calibrated against band spectroscopy.
- 1930s–1950s — Photoelectron and UV absorption spectra progressively confirm orbital energy orderings for Period 2 diatomics, including the s–p mixing crossover between N₂ and O₂.
- 1966 — Mulliken awarded the Nobel Prize in Chemistry for molecular orbital theory of chemical bonds and electronic structure.
Why do we drop the 1s core orbitals from Period 2 MO diagrams?

The 1s orbitals on Li through F are tightly contracted around each nucleus and lie enormously lower in energy than the valence 2s/2p orbitals, so their overlap with the other atom's orbitals is negligible and their energy separation from valence orbitals is huge. They form an essentially nonbonding σ₁ₛ/σ*₁ₛ pair that is always filled with all four core electrons (two bonding, two antibonding), contributing exactly 12(2−2)=0\tfrac12(2-2)=0 to bond order. Since they never affect bond order, bond length trends, or magnetism, they are conventionally omitted and only the valence 2s/2p-derived orbitals are drawn — exactly as core electrons are omitted from valence Lewis structures.

Is the σ(2p)/π(2p) crossover a sharp switch at one specific molecule, or a gradual trend?

It reflects a genuinely continuous trend in the atomic 2s–2p energy gap, which grows steadily heavier across Li→F as effective nuclear charge on the 2p electrons increases faster than on the more penetrating 2s electrons. In practice, though, the consequence for bond order and magnetism only becomes visible once you fill electrons all the way up to the π(2p)/σ(2p) region, which happens exactly at N₂ and O₂. So while the underlying orbital energies shift gradually, the empirically observed "switch" in diagrams is concentrated at that one point in the row because that is where the ordering starts to matter for which electrons get placed where.

Does the mixed vs. unmixed ordering ever change the predicted bond order, not just orbital identity?

For fully closed-shell configurations (all of σ(2p) and π(2p) either completely filled or completely empty), no — bond order comes out identical either way, because swapping the order of two completely-filled or completely-empty levels doesn't change how many total bonding vs antibonding electrons are counted. The two orderings diverge only in cases with partial filling straddling the crossover region, such as B₂ and C₂, where they predict different HOMOs, different degeneracies, and (for B₂) different magnetism — so the distinction matters for spectroscopy and for spin state, even when it happens not to matter for the simple bond-order number.

How does this MO approach relate to the simpler bonding/antibonding picture used for diatomic hydrogen?

It is the exact same framework scaled up. H₂'s MO diagram (σ₁ₛ bonding, σ*₁ₛ antibonding, two electrons filling only the bonding orbital) is the minimal case of the same LCAO, symmetry-matching, and aufbau-filling rules used here for eight-orbital Period 2 diagrams. The only genuinely new features introduced going from H₂ to Period 2 are (1) the σ/π symmetry split, which appears once p orbitals are available, and (2) s–p mixing, which has no counterpart in H₂ because there is only one valence orbital type (1s) to combine.

Why does bond order predict bond dissociation energy trends but not exact numerical energies?

Bond order is a purely orbital-occupation-based bookkeeping number; it says nothing about orbital overlap magnitude, electron correlation, or the detailed shape of the potential energy curve, all of which set the actual numerical value of DeD_e. What it reliably captures is the ordering: more net bonding character (higher BO) between the same pair of atoms means a deeper potential well and shorter equilibrium separation, which is why comparisons like N₂ vs N₂⁺ vs N₂⁻ (all built from the same two nitrogen atoms) track so well. Comparing bond order between different atom pairs (say, N₂ vs O₂) still trends correctly here empirically, but that is a feature of Period 2 specifically, not a universal law.

What happens to this whole framework once you go past F₂, to Ne₂?

Ne is 1s22s22p61s^2 2s^2 2p^6, giving 8 valence electrons per atom and 16 total for Ne₂ — enough to completely fill all eight molecular orbitals in the diagram, bonding and antibonding alike. That gives BO=12(8−8)=0BO = \tfrac12(8-8) = 0 exactly, predicting no net covalent bond, which matches the experimental absence of a stable, chemically bonded Ne₂ molecule (only an extremely weak van der Waals-bound species is observed at very low temperature). This is the same zero-bond-order outcome as He₂, and it marks the natural closing point of the Period 2 homonuclear diatomic series.

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