Magnetic Properties and Ground-State Configurations

Pour liquid oxygen between the poles of a strong magnet and it clings there. It hangs as a pale blue liquid suspended against gravity, until it boils away. Nitrogen poured over the same magnet falls straight through, indifferent to the field. Both are diatomic molecules of second-row elements. Both are held together by a handful of shared electrons. Yet one is magnetic and the other is not. Lewis structures cannot explain this. A textbook double bond O=O pairs every electron. It predicts a diamagnetic molecule, exactly the wrong answer. The resolution is one of the great early triumphs of Homonuclear Diatomics: MO Diagrams for Li2 to F2 — molecular orbital theory. It turns on a single degenerate pair of antibonding orbitals. Lewis structures have no way to represent this pair. This article is about reading magnetism directly off a molecular orbital diagram. It is also about the deeper bookkeeping — term symbols. This bookkeeping organizes the electronic states these configurations produce. The question "is this molecule paramagnetic?" is really the question "does the ground-state configuration leave unpaired spins?" That in turn is a question about how electrons fill degenerate orbitals. This follows Hund's rule, exactly as in atomic configurations. But now the orbitals are built from linear combinations of atomic $2s$ and $2p$ functions, rather than pure atomic orbitals. The reasoning is completely mechanical once the orbital ordering is fixed. That is why this topic sits naturally after the MO diagrams. It also follows the s-p Mixing in Second-Row Diatomics correction. That correction reorders the $\sigma_g(2p)$ and $\pi_u(2p)$ levels partway through the second row. The stakes go beyond a party trick with liquid oxygen. Whether a molecule has unpaired electrons determines its reactivity as a diradical. It also determines its response in an EPR spectrometer. It shapes its role as a ligand in transition-metal chemistry. For O $_2$ specifically, it governs the entire kinetics of combustion and respiration. Ground-state triplet oxygen reacts sluggishly with closed-shell singlet organic molecules, by spin-forbidden pathways. Excited singlet oxygen, generated photochemically, reacts explosively fast instead. Getting the ground state right is not a footnote. It is the gateway to understanding why O $_2$ behaves the way it does at all. By the end of this article you will be able to build the correct valence MO configuration. This applies to any homonuclear diatomic from Li $_2$ to Ne $_2$ , and several heteronuclear analogues. You will apply Hund's rule inside the frontier degenerate orbitals to get the ground-state spin. You will translate that configuration into a full molecular term symbol. You will use bond order and term symbol together to rationalize magnetic behavior. This includes bond length trends, and even electronic spectroscopy of species like O $_2$ , O $_2^+$ , and O $_2^-$ .

Key takeaways

A molecule is paramagnetic if its MO ground-state configuration has any unpaired electrons; otherwise it is diamagnetic.
Degenerate orbitals (the $\pi$ pairs) are filled one electron at a time with parallel spins first — Hund's rule for molecules, exactly as for atoms.
O $_2$ has two electrons in the degenerate $\pi_g^*(2p)$ pair; Hund's rule puts them in different orbitals with parallel spins, giving a paramagnetic $^3\Sigma_g^-$ ground state.
N $_2$ fills every bonding level and leaves the $\pi_g^*$ pair empty, giving a diamagnetic $^1\Sigma_g^+$ closed-shell ground state.
Molecular term symbols $^{2S+1}\Lambda_{g/u}^{(\pm)}$ encode total spin, orbital angular momentum about the bond axis, and symmetry — atomic term symbol logic applied to a linear molecule.
Bond order, magnetism, and term symbol all come from the same configuration, so removing or adding electrons (O $_2^+$ , O $_2^-$ , O $_2^{2-}$ ) predictably changes all three together.
Key idea: paramagnetism is not a property of bond order or electron count alone — it is a property of how electrons are forced to occupy degenerate orbitals, so you must build the full MO configuration, not just count bonding minus antibonding electrons.

Where this lives today

The paramagnetism of triplet ground-state O $_2$ governs the entire kinetics of combustion and of aerobic respiration: because ground-state O $_2$ is a triplet ( $S=1$ ) while almost all stable organic molecules are closed-shell singlets ( $S=0$ ), direct reaction between them to form new bonds is formally spin-forbidden, which is a major reason organic matter does not spontaneously combust in air at room temperature despite the thermodynamic favorability of oxidation. Photosensitizers used in photodynamic cancer therapy work by exciting molecular oxygen to its diamagnetic $^1\Delta_g$ singlet state (the same excited term derived above), which reacts with biomolecules far faster than ground-state triplet oxygen because the spin-forbiddenness is lifted, driving oxidative damage selectively in illuminated tumor tissue. Superoxide (O $_2^-$ ) and peroxide (O $_2^{2-}$ ) chemistry, rationalized entirely by the bond-order/term-symbol series worked out above, underlies both industrial bleaching and antiseptic chemistry (hydrogen peroxide, sodium peroxide) and the biochemistry of oxidative stress, where superoxide dismutase enzymes catalytically disproportionate the paramagnetic superoxide radical anion into ordinary O $_2$ and diamagnetic peroxide, a reaction whose necessity follows directly from superoxide's unpaired-electron reactivity. Electron paramagnetic resonance (EPR) spectroscopy is the direct experimental tool for everything discussed here: any species predicted to have unpaired electrons by this MO analysis — O $_2$ itself (though its zero-field splitting makes conventional EPR difficult), superoxide radicals, nitric oxide (NO, isoelectronic with O $_2^-$ in relevant respects), and countless transition-metal complexes analyzed with closely related ligand-field arguments — gives an EPR signal, while species predicted diamagnetic (peroxide, N $_2$ , CO) give none, providing a direct experimental verification route for every prediction this framework makes. Materials scientists and inorganic chemists use exactly this degenerate-shell logic to predict magnetism in dioxygenyl salts (O $_2^+$ AsF $_6^-$ and related "superacid" oxidizers), in alkali-metal superoxides and peroxides used as oxygen sources in life-support systems, and — extended to heteronuclear diatomics such as NO — in atmospheric and combustion chemistry, where NO's odd-electron, paramagnetic ground state (isoelectronic in frontier-orbital pattern with O $_2^-$ ) governs its reactivity as a radical signaling molecule in biology and as a pollutant intermediate in combustion.

Ask the tutor: For a heteronuclear diatomic like NO, how do you decide which atomic orbitals combine and in what proportions, given that the two atoms are no longer equivalent?

The physical picture — thorough prose

Start from the qualitative molecular orbital diagram for a homonuclear diatomic built from valence $2s$ and $2p$ atomic orbitals, as established when constructing MO diagrams across the second row. Combining two $2s$ orbitals gives $\sigma_g(2s)$ and $\sigma_u^*(2s)$ . Combining the two $2p_z$ orbitals (axis along the bond) gives $\sigma_g(2p)$ and $\sigma_u^*(2p)$ . Combining the two perpendicular pairs, $2p_x$ and $2p_y$ , gives a degenerate pair of bonding orbitals $\pi_u(2p)$ and a degenerate pair of antibonding orbitals $\pi_g^*(2p)$ . Degenerate means the two $\pi$ orbitals of a pair (one built from $p_x$ , one from $p_y$ ) have identical energy by the cylindrical symmetry of the molecule — there is no physical distinction between "x" and "y" once you have chosen the bond axis as $z$ . For B $_2$ through N $_2$ , $s$ – $p$ mixing pushes $\sigma_g(2p)$ above $\pi_u(2p)$ in energy; for O $_2$ , F $_2$ , and Ne $_2$ the mixing is too weak to invert the naive order, and $\sigma_g(2p)$ sits below $\pi_u(2p)$ . This ordering swap, covered in detail elsewhere, matters enormously for magnetism at exactly one molecule: B $_2$ . With the mixed ordering, B $_2$ 's two valence electrons beyond the $2s$ shell go into the degenerate $\pi_u(2p)$ pair, one each, parallel spins — paramagnetic, exactly as observed. With the naive (unmixed) ordering they would pair up in a single $\sigma_g(2p)$ orbital, wrongly predicting a diamagnetic ground state. B $_2$ is the classic evidence that $s$ – $p$ mixing is real. The magnetism rule itself is simple once the orbital ordering and filling are fixed: fill orbitals from lowest energy up, respecting the Pauli principle (two electrons per orbital, opposite spin), and whenever electrons are being placed into a set of degenerate orbitals, spread them out with parallel spin before pairing any — Hund's rule, applied to molecules exactly as to atoms. Any electron left without a partner of opposite spin in the same spatial orbital makes the molecule paramagnetic; if every orbital is either empty or doubly occupied, the molecule is diamagnetic. O $_2$ is the flagship case. Its sixteen electrons fill, in order: $\sigma_g(1s)^2\,\sigma_u^*(1s)^2\,\sigma_g(2s)^2\,\sigma_u^*(2s)^2\,\sigma_g(2p)^2\,\pi_u(2p)^4\,\pi_g^*(2p)^2$ . The last two electrons must occupy the doubly-degenerate $\pi_g^*(2p)$ level. Hund's rule places one electron in each of the two degenerate $\pi_g^*$ orbitals with parallel spins rather than forcing both into one orbital paired up. The result is a triplet ground state with two unpaired electrons — paramagnetic, matching experiment exactly, and matching the classic liquid-oxygen-in-a-magnet demonstration.

Figure

Filling the frontier π orbitals of O₂: Hund's rule forces two unpaired spins

Contrast N $_2$ , one shell lighter. Its fourteen electrons fill up through $\sigma_g(2s)^2\,\sigma_u^*(2s)^2\,\pi_u(2p)^4\,\sigma_g(2p)^2$ — note the mixed ordering places $\sigma_g(2p)$ above $\pi_u(2p)$ for N $_2$ — and the $\pi_g^*(2p)$ level is completely empty. Every occupied orbital is doubly filled, so N $_2$ is diamagnetic, with bond order $(8-2)/2=3$ , consistent with its famously short, strong triple bond and its chemical inertness.

Figure

N₂ versus O₂: same recipe, different frontier occupancy

The full symmetry label for a diatomic electronic state — the molecular term symbol — packages three pieces of information: total spin ( $S$ ), total orbital angular momentum projected onto the bond axis ( $\Lambda$ ), and inversion/reflection symmetry ( $g/u$ and $\pm$ ). It is built the same way atomic term symbols are built from atomic configurations, but the quantum number that plays the role of $L$ is now $\Lambda = |M_L|$ , the absolute value of the projection of orbital angular momentum along the internuclear axis, because in a linear molecule only that projection (not the full vector $\mathbf{L}$ ) is conserved.

The core ideas

For a linear molecule, the single-electron orbital angular momentum projection along the bond axis is labeled $\lambda$ ( $\lambda = 0,1,2,\dots$ for $\sigma,\pi,\delta,\dots$ orbitals), and the total is

\Lambda = \left| \sum_i m_{\lambda,i} \right|, \qquad \Lambda = 0,1,2,\dots \;\to\; \Sigma,\Pi,\Delta,\dots

where: $m_{\lambda,i}$ is the signed projection quantum number of electron $i$ (e.g. $\pi$ orbitals carry $m_\lambda=\pm1$ ), and the Greek-letter term label $\Sigma,\Pi,\Delta$ is the molecular analogue of atomic $S,P,D$ .

Reading: just as atomic terms sum individual $m_\ell$ values to get $M_L$ and take $L=|M_L|_{\max}$ , molecular terms sum the axial projections $m_\lambda$ of the open-shell electrons to get $\Lambda$ . Filled orbitals (any $\sigma^2$ or $\pi^4$ shell) contribute zero net $\Lambda$ and zero net $S$ , so only open shells matter for the term symbol.

S = \sum_i m_{s,i}\Big|_{\max}, \qquad \text{multiplicity} = 2S+1

where: $S$ is the maximum total spin obtainable from the open-shell electrons consistent with the Pauli principle, exactly as in atomic Hund's-rule ground states.

Reading: two electrons in a degenerate $\pi^*$ pair can couple to $S=1$ (parallel, different orbitals) or $S=0$ (paired or antiparallel in different orbitals); Hund's rule for molecules says the maximum-S term lies lowest in energy for the ground configuration, the same energetic logic (reduced electron–electron repulsion when spins are parallel and electrons occupy different orbitals) that drives atomic Hund's rule. The full term symbol is written

{}^{2S+1}\Lambda_{g/u}^{(\pm)}

where: the left superscript is the spin multiplicity, $\Lambda$ is replaced by its Greek letter ( $\Sigma,\Pi,\Delta,\dots$ ), the subscript $g$ or $u$ records the parity of the overall electronic wavefunction under inversion through the molecular center (only meaningful for homonuclear or otherwise centrosymmetric species), and the superscript $\pm$ (attached only to $\Sigma$ states) records the sign under reflection through a plane containing the bond axis.

Reading: $g/u$ of the total wavefunction is obtained by multiplying the parities of all singly-occupied orbitals (filled shells are always net $g$ ); the $\pm$ label distinguishes two ways two electrons in a degenerate $\pi$ shell can combine — symmetric combinations give $\Sigma^-$ for a triplet from $\pi^2$ , antisymmetric spatial combinations give $\Sigma^+$ or $\Delta$ singlets, by an argument fully parallel to the atomic $p^2$ term-symbol derivation ( $^3P,\,^1D,\,^1S$ ). Bond order, the simpler and more familiar quantity, is

\text{Bond order} = \frac{n_{\text{bonding}} - n_{\text{antibonding}}}{2}

where: $n_{\text{bonding}}$ and $n_{\text{antibonding}}$ count electrons in bonding and antibonding molecular orbitals respectively (core $1s$ orbitals are conventionally included but contribute zero net bond order since they are equally bonding and antibonding in pairs).

Reading: bond order tracks bond strength and length trends but says nothing about spin — two different configurations can share a bond order while differing completely in magnetism, which is exactly why O $_2^-$ (bond order 1.5) and B $_2$ (bond order 1) can both be paramagnetic while having very different bond orders, and why bond order alone is an incomplete diagnostic.

Derivation AGround-state term symbol of O₂ from the π_g\*² configuration

Goal. Show explicitly why the open-shell configuration $(\pi_g^*)^2$ produces a triplet $^3\Sigma_g^-$ ground state for O $_2$ , rather than a singlet, and enumerate the full set of terms this configuration can generate — the molecular analogue of deriving $^3P,\,^1D,\,^1S$ from an atomic $p^2$ configuration.

Step 1 — identify the open shell and the single-particle quantum numbers. The two π_g^\* orbitals are built from $2p_x$ and $2p_y$ on each oxygen; in the molecular basis they carry axial angular momentum projections $m_\lambda = +1$ and $m_\lambda = -1$ (a $\pi$ orbital, like an atomic $p$ orbital rotated about the axis, is doubly degenerate with $m_\lambda=\pm1$ ). Two electrons must be distributed over these two degenerate spatial orbitals, each of which can hold up to two electrons (opposite spin), giving four spin-orbitals total: $(m_\lambda=+1,\uparrow), (m_\lambda=+1,\downarrow), (m_\lambda=-1,\uparrow), (m_\lambda=-1,\downarrow)$ .

Step 2 — enumerate all Pauli-allowed two-electron microstates. Choosing 2 of these 4 spin-orbitals gives $\binom{4}{2}=6$ microstates. Their $(M_L, M_S)$ values, with $M_L = m_{\lambda,1}+m_{\lambda,2}$ and $M_S = m_{s,1}+m_{s,2}$ , are:

(+1_\uparrow,-1_\uparrow):\,(M_L,M_S)=(0,+1) \qquad (+1_\downarrow,-1_\downarrow):\,(0,-1)

(+1_\uparrow,-1_\downarrow):\,(0,0) \qquad (+1_\downarrow,-1_\uparrow):\,(0,0)

(+1_\uparrow,+1_\downarrow):\,(+2,0) \qquad (-1_\uparrow,-1_\downarrow):\,(-2,0)

This table is the exact molecular counterpart of the atomic $p^2$ microstate table, with $m_\lambda$ replacing $m_\ell$ .

Step 3 — group microstates into terms. The $M_L=\pm2$ states with $M_S=0$ (two states) form a $\Lambda=2$ , $S=0$ term: a $^1\Delta$ term (5-fold orbital degeneracy in atoms becomes 2-fold here since only $M_L=\pm2$ exist, i.e. a single $\Delta$ level with two components). One combination among the $M_L=0,\,M_S=0$ microstates, together with $M_L=0,M_S=\pm1$ , forms a 3-fold spin-degenerate $S=1$ term with $\Lambda=0$ : a $^3\Sigma$ term. The remaining single $M_L=0,M_S=0$ combination is a $^1\Sigma$ term. Bookkeeping check: total microstates accounted for is $2(^1\Delta) + 3(^3\Sigma) + 1(^1\Sigma) = 6$ , matching Step 2.

Step 4 — assign parity labels using orbital symmetry. Each $\pi_g^*$ orbital is individually $g$ (gerade), inherited from the antibonding combination of $p$ orbitals under the inversion center of a homonuclear diatomic. Two electrons both in $g$ orbitals give an overall spatial parity of $g\times g = g$ for every term arising from this configuration: $^3\Sigma_g,\,^1\Delta_g,\,^1\Sigma_g$ all carry $g$ subscripts.

Step 5 — assign the $\pm$ label to the $\Sigma$ terms. For two electrons in a degenerate $\pi$ shell, the spin-triplet spatial wavefunction must be antisymmetric under exchange of the two electrons' spatial labels (since the full wavefunction is antisymmetric and the triplet spin part is already symmetric). An antisymmetric combination of the $\Sigma$ -type spatial functions built from $\pi\pi$ products transforms as $\Sigma^-$ under the reflection operation, while the symmetric combination transforms as $\Sigma^+$ . Group-theoretic reduction of $\pi_g \times \pi_g$ gives $\Sigma_g^+ + \Sigma_g^- + \Delta_g$ ; matching multiplicities to spatial symmetry (antisymmetric spatial ↔ triplet, symmetric spatial ↔ singlet) assigns the triplet to $\Sigma^-$ and the two singlets to $\Sigma^+$ and $\Delta$ .

Step 6 — apply Hund's rule to rank the terms and select the ground state. Just as atomic $p^2$ gives $^3P$ below $^1D$ below $^1S$ , the molecular $\pi^2$ configuration gives

{}^3\Sigma_g^- \;<\; {}^1\Delta_g \;<\; {}^1\Sigma_g^+

in order of increasing energy — maximum spin multiplicity lowest, consistent with minimized electron–electron repulsion when the two electrons occupy different $\pi_g^*$ lobes with parallel spins. The ground state of O $_2$ is therefore $^3\Sigma_g^-$ , exactly matching the paramagnetic triplet found by the elementary aufbau argument, while $^1\Delta_g$ (radiationless-decay-populated "singlet oxygen," about 94 kJ mol $^{-1}$ above the ground state) and $^1\Sigma_g^+$ (about 157 kJ mol $^{-1}$ above) are the two lowest excited states, both diamagnetic.

Go deeper · Derivation B · 2 minBond order and magnetism across the isoelectronic O₂/O₂⁺/O₂⁻/O₂²⁻ series

Derivation BBond order and magnetism across the isoelectronic O₂/O₂⁺/O₂⁻/O₂²⁻ series

Goal. Track how removing or adding electrons to the O $_2$ frontier orbital changes bond order, spin, and term symbol together, since these are not independent quantities but three faces of the same configuration.

Step 1 — write the reference configuration. Neutral O $_2$ (16 electrons) has the valence configuration derived above, with the frontier shell $(\sigma_g 2p)^2(\pi_u 2p)^4(\pi_g^* 2p)^2$ . Define the bond order using only the valence $2p$ -derived orbitals (the $1s$ and $2s$ shells contribute a net bonding/antibonding cancellation and can be dropped from the running comparison since they are unchanged across this series):

\text{BO} = \frac{2 + 4 - n_{\pi^*}}{2} = 3 - \frac{n_{\pi^*}}{2}

where $n_{\pi^*}$ is the number of electrons in the $\pi_g^*(2p)$ pair (0 to 4 possible) — this is just the general bond-order formula specialized to the O $_2$ frontier, since $\sigma_g(2p)^2\pi_u(2p)^4$ contributes a fixed $6/2=3$ before any $\pi^*$ electrons are added.

Step 2 — cation O $_2^+$ (15 electrons, remove one electron from $\pi_g^*$ ). Now $n_{\pi^*}=1$ :

\text{BO}(\mathrm{O_2^+}) = 3 - \frac{1}{2} = 2.5

A single electron in a doubly-degenerate level is automatically unpaired — O $_2^+$ is paramagnetic (one unpaired electron, a doublet). Its term symbol is $^2\Pi_g$ (the single hole in $\pi_g^*$ carries $\Lambda=1$ and the shell is a hole in a $g$ orbital, which remains $g$ ). Experimentally, removing an antibonding electron shortens the bond (110.7 pm in O $_2^+$ vs. 120.7 pm in neutral O $_2$ ) and raises the stretching frequency, consistent with the bond-order increase from 2 to 2.5.

Step 3 — anion O $_2^-$ (17 electrons, add one electron to $\pi_g^*$ ). Now $n_{\pi^*}=3$ , i.e., one $\pi_g^*$ orbital is doubly occupied and the other singly occupied:

\text{BO}(\mathrm{O_2^-}) = 3 - \frac{3}{2} = 1.5

One unpaired electron remains — superoxide is paramagnetic (doublet, $^2\Pi_g$ again, now a single electron rather than a single hole, but same $\Lambda$ bookkeeping). The bond order drops from 2.0 to 1.5, and the experimental bond length lengthens accordingly (about 128 pm), consistent with weakening the bond by adding an antibonding electron.

Step 4 — peroxide dianion O $_2^{2-}$ (18 electrons, fill the entire $\pi_g^*$ shell). Now $n_{\pi^*}=4$ , both $\pi_g^*$ orbitals doubly occupied:

\text{BO}(\mathrm{O_2^{2-}}) = 3 - \frac{4}{2} = 1

Every orbital is now filled or empty, so peroxide is diamagnetic, a closed-shell $^1\Sigma_g^+$ ground state, with a single O–O bond just as its Lewis structure (O–O with two lone pairs on each oxygen, formal single bond) correctly predicts. This is the case where the naive Lewis picture and MO theory agree, because peroxide has no partially-filled degenerate shell left to produce Hund's-rule surprises.

Step 5 — collect the series and note the pattern. Tabulating:

Species	$n_{\pi^*}$	Bond order	Unpaired e⁻	Magnetism	Term symbol
O $_2^+$	1	2.5	1	paramagnetic	$^2\Pi_g$
O $_2$	2	2.0	2	paramagnetic	$^3\Sigma_g^-$
O $_2^-$	3	1.5	1	paramagnetic	$^2\Pi_g$
O $_2^{2-}$	4	1.0	0	diamagnetic	$^1\Sigma_g^+$

Bond order decreases monotonically and smoothly by half-integer steps as electrons fill the degenerate antibonding pair one at a time, while magnetism switches off only at the very last step, when the shell finally closes — a direct, checkable illustration that bond order is a smooth scalar while magnetism is an on/off consequence of shell closure.

Worked examples

Example 1 — Foundational: build the full ground-state configuration and term symbol for C $_2$ . Carbon has 6 electrons, so C $_2$ has 12 total electrons, 8 of them valence (beyond the two $1s^2$ cores). Using the "mixed" ordering appropriate for B $_2$ through N $_2$ (since $s$ – $p$ mixing is strong enough to invert $\sigma_g(2p)$ above $\pi_u(2p)$ for elements this light), the valence filling order is $\sigma_g(2s), \sigma_u^*(2s), \pi_u(2p), \sigma_g(2p)$ . Filling 8 valence electrons:

\sigma_g(2s)^2\,\sigma_u^*(2s)^2\,\pi_u(2p)^4

All eight valence electrons are exhausted after filling $\sigma_g(2s)^2\sigma_u^*(2s)^2$ (4 electrons) and then completely filling both $\pi_u(2p)$ orbitals (4 more electrons, 2 each) — there are no electrons left for $\sigma_g(2p)$ , which remains empty. Every orbital occupied is completely filled: C $_2$ is diamagnetic, a genuinely surprising result at first glance because a naive Lewis-style expectation (isoelectronic reasoning from N $_2$ , or drawing C=C) might suggest unpaired electrons or a simple double bond, but the actual ground state has zero unpaired spins and a $^1\Sigma_g^+$ term symbol. Bond order: bonding electrons are $\sigma_g(2s)^2 + \pi_u(2p)^4 = 6$ , antibonding is $\sigma_u^*(2s)^2 = 2$ , so

\text{BO(C}_2) = \frac{6-2}{2} = 2

giving a formal double bond, though the four $\pi_u$ electrons and zero $\sigma_g(2p)$ electrons mean this "double bond" is built from two $\pi$ bonds and no $\sigma$ bond in the valence MO picture — an unusual electronic structure that has made C $_2$ 's true bonding character (some argue for partial quadruple-bond character via near-degenerate configurations) a live topic of discussion, but the diamagnetism itself is unambiguous and experimentally confirmed.

Example 2 — Applied: rationalize the trend in O–O bond dissociation energy and paramagnetism across O $_2$ , O $_2^+$ , superoxide (KO $_2$ ), and peroxide (Na $_2$ O $_2$ ) in inorganic chemistry. Superoxide salts like KO $_2$ (used in some self-contained breathing apparatuses, reacting with exhaled CO $_2$ and moisture to release O $_2$ ) contain the O $_2^-$ ion, which from the derivation above has bond order 1.5 and one unpaired electron. Its EPR spectrum shows a clean signal, directly confirming the unpaired spin predicted by the $(\pi_g^*)^3$ configuration; this is routinely used to identify superoxide radicals in biological oxidative-stress studies, since superoxide dismutase enzymes exist specifically to detoxify this same paramagnetic species. Peroxide salts like Na $_2$ O $_2$ contain O $_2^{2-}$ with bond order 1.0 and zero unpaired electrons (fully paired, EPR silent), matching the fact that peroxide is a normal closed-shell diamagnetic anion, isoelectronic in valence structure with F $_2$ (both have $(\pi_g^*)^4$ , bond order 1). The measured O–O distance in peroxide salts, near 149 pm, is longer than in superoxide (128 pm), which is longer than in neutral O $_2$ (121 pm), which is longer than in O $_2^+$ salts such as O $_2^+$ PtF $_6^-$ (dioxygenyl hexafluoroplatinate, 112 pm) — a monotonic bond-length lengthening exactly tracking the monotonic bond-order decrease $2.5 \to 2.0 \to 1.5 \to 1.0$ computed in the derivation, while the on/off magnetism (paramagnetic, paramagnetic, paramagnetic, diamagnetic) tracks shell closure rather than bond order directly. This is precisely why a chemist reasoning purely from "shorter bond = stronger bond, therefore more paired electrons" would be misled: O $_2^+$ has fewer unpaired electrons than O $_2$ despite both being paramagnetic, and the bond gets shorter and the molecule stays paramagnetic simultaneously, because bond order and magnetism are governed by different aspects of the same configuration (total antibonding population vs. degenerate-shell occupancy).

Example 3 — Exam/competition level: predict the ground-state term symbol and magnetic moment (in Bohr magnetons, spin-only) of the hypothetical ion N $_2^{2-}$ , and check the limiting case of the isoelectronic neutral species. N $_2$ has 14 electrons; N $_2^{2-}$ has 16 electrons, the same total electron count as neutral O $_2$ — this is deliberately an isoelectronic species to test whether the student applies configuration-based reasoning rather than memorized facts about specific molecules. Because N $_2^{2-}$ is built from two nitrogen nuclei (each with nuclear charge 7, versus 8 for oxygen), the atomic orbital energies going into the LCAO are different from those of O $_2$ , and in particular $s$ - $p$ mixing considerations for a highly reduced, electron-rich nitrogen-based diatomic should be re-examined rather than assumed identical to neutral N $_2$ . However, for the purposes of ordering valence orbitals in a dianion this electron-rich, the additional electrons occupy antibonding levels well above any $\sigma$ / $\pi$ crossover ambiguity, so the ordering question that matters (mixed vs. unmixed) only affects the lower orbitals, which are completely filled regardless of their relative order. Filling 16 electrons total (10 valence beyond four $1s$ core electrons is wrong bookkeeping — recount): N has $1s^22s^22p^3$ , so each N contributes 5 valence electrons, 10 valence electrons total in neutral N $_2$ ; N $_2^{2-}$ has 12 valence electrons. Filling:

\sigma_g(2s)^2\,\sigma_u^*(2s)^2\,\pi_u(2p)^4\,\sigma_g(2p)^2\,\pi_g^*(2p)^2

This is 2+2+4+2+2 = 12 valence electrons, correctly matching N $_2^{2-}$ 's electron count, and it places the last two electrons into the degenerate $\pi_g^*(2p)$ pair — structurally identical in shell-occupancy pattern to neutral O $_2$ 's ground-state configuration. By the same Hund's-rule argument used in the O $_2$ derivation, these two electrons occupy different $\pi_g^*$ orbitals with parallel spins, giving

\text{Term symbol: } {}^3\Sigma_g^-, \qquad S=1, \qquad \text{2 unpaired electrons}

The spin-only magnetic moment is

\mu_{\text{spin-only}} = \sqrt{n(n+2)}\ \mu_B = \sqrt{2(2+2)}\ \mu_B = \sqrt{8}\ \mu_B \approx 2.83\ \mu_B

where $n=2$ is the number of unpaired electrons and $\mu_B$ is the Bohr magneton — exactly the same spin-only moment computed for neutral O $_2$ itself, since both share an identical open-shell pattern despite being different molecules with different nuclear charges and different overall electron counts. This is the key check: the magnetic moment depends on the open-shell occupancy pattern, not on the absolute identity of the atoms, which is precisely the isoelectronic-analogy limiting case the question is designed to expose. (For comparison, the experimental spin-only estimate for O $_2$ itself using the same formula gives the identical $2.83\ \mu_B$ , and the measured magnetic moment of O $_2$ , about $2.6$ – $2.8\ \mu_B$ depending on method, confirms the spin-only approximation works well for this molecule since orbital angular momentum quenching is substantial along the $\Sigma$ axis.) Bond order for N $_2^{2-}$ : bonding electrons $2+4+2=8$ , antibonding $2+2=4$ (including the two new $\pi_g^*$ electrons), so

\text{BO}(\mathrm{N_2^{2-}}) = \frac{8-4}{2} = 2

— two full bond orders weaker than neutral N $_2$ 's triple bond, exactly the expected penalty for populating one full degenerate antibonding pair, and numerically identical to neutral O $_2$ 's bond order of 2, again because the two species share the identical valence shell-occupancy pattern once electron count is matched.

Limiting cases, scope & common traps

Situation	What happens	Why
Filled-shell configuration ( $\sigma^2$ or $\pi^4$ everywhere)	Diamagnetic, $^1\Sigma_g^+$ or $^1\Sigma^+$ ground state	No degenerate shell is partially occupied; Pauli principle forces all spins paired
Exactly one electron in a degenerate pair (radical cation/anion, e.g. O $_2^+$ , O $_2^-$ , NO)	Paramagnetic, doublet, $\Lambda\neq0$ term (e.g. $^2\Pi$ )	A lone electron cannot pair with itself
Exactly two electrons in a degenerate pair, shell less than half full or exactly half full (e.g. O $_2$ , B $_2$ )	Paramagnetic triplet ground state (Hund's rule)	Parallel spins in different degenerate orbitals minimize repulsion, exactly as in atomic $p^2$
Degenerate pair three-quarters filled (e.g. O $_2^-$ )	Paramagnetic doublet, one hole left unpaired	Symmetric to the one-electron case, now counting the hole
As bond order changes but shell stays fully closed (e.g. comparing N $_2$ to hypothetical fully-closed-shell dications)	Magnetism stays off regardless of bond order value	Magnetism tracks shell closure, not the bond-order scalar
Heteronuclear analogues (CO, NO, CN $^-$ )	Same aufbau/Hund logic applies, but $g/u$ labels disappear (no inversion center)	Term symbols become $^{2S+1}\Lambda^{(\pm)}$ without parity subscripts — the topic of heteronuclear diatomic MO diagrams

The chemist behind it

The molecular orbital explanation of O $_2$ paramagnetism is usually credited jointly to Robert Mulliken and Friedrich Hund, who independently developed the molecular orbital approach to chemical bonding through the 1920s, building directly on the group-theoretic and spectroscopic groundwork of atomic term symbols that Hund himself had helped formalize for atoms.

Origins · 1 minRobert S. Mulliken and how this was discovered

Robert S. Mulliken

1896–1986 · American physical chemist and spectroscopist, University of Chicago

Mulliken spent the 1920s and 1930s systematically translating diatomic band spectra into an orbital language, insisting that molecular electrons occupy delocalized orbitals spanning the whole molecule rather than localized bonds between specific atom pairs. He coined much of the vocabulary still used today — bonding and antibonding orbitals, the $\sigma$ / $\pi$ notation, and the correlation diagrams linking separated-atom orbitals to united-atom limits. His painstaking analysis of ultraviolet and visible absorption spectra of diatomic molecules, including O $_2$ 's Schumann–Runge bands, provided the spectroscopic term-symbol evidence that pinned down the ground and excited electronic states discussed here, well before any large-scale numerical calculation could have confirmed the ordering independently. He received the Nobel Prize in Chemistry in 1966 "for his fundamental work concerning chemical bonds and the electronic structure of molecules by the molecular orbital method."

History

- 1927: Friedrich Hund and Robert Mulliken begin developing molecular orbital theory as a rival framework to the valence-bond picture then dominant.

- 1928–1932: Mulliken systematically assigns term symbols to diatomic electronic states from observed band spectra, including O

_2

's ground and low-lying excited states.

- 1931: Erich Hückel extends molecular orbital ideas to π-electron systems in conjugated and aromatic organic molecules.

- 1939: Linus Pauling's The Nature of the Chemical Bond popularizes valence-bond theory even as spectroscopic evidence for MO theory's correct prediction of O

_2

paramagnetism continues to accumulate.

- 1966: Mulliken receives the Nobel Prize in Chemistry for the molecular orbital method, cementing MO theory's status as the standard framework for diatomic electronic structure.

Why doesn't a Lewis structure predict O2's paramagnetism?

Lewis structures assign electrons to two-center bonds and localized lone pairs, always in pairs, with no mechanism for describing degenerate molecular orbitals that extend over the whole molecule. O $_2$ 's paramagnetism arises specifically because its last two electrons must occupy a doubly-degenerate antibonding orbital pair ( $\pi_g^*$ ), and Hund's rule then spreads them into different orbitals with parallel spins rather than forcing a pair. Lewis theory, built from atomic valence and octet-filling rules, has no concept of orbital degeneracy or of a partially-filled shell surviving into the ground state; it always predicts fully paired electrons for a simple double bond. Molecular orbital theory captures this because it builds delocalized orbitals directly from the full symmetry of the diatomic core, correctly reproducing the degenerate $\pi$ pair that Lewis theory cannot represent at all.

Is bond order sufficient to predict magnetism on its own?

No — bond order is a single scalar summarizing bonding-minus-antibonding electron count, while magnetism depends on the detailed occupancy pattern within degenerate orbital sets. Two species can share identical bond order while differing in magnetism, or differ in bond order while sharing the same magnetic behavior (unpaired electron count). The safest procedure is always to build the explicit MO configuration, identify any orbitals that are part of a degenerate set (π or δ), and apply Hund's rule specifically to that set; bond order can then be read off afterward as a convenient bonus, but it should never be used as a magnetism shortcut on its own, since it discards the shell-structure information that magnetism actually depends on.

What is singlet oxygen, and how does it relate to the term symbols derived here?

Singlet oxygen refers to the excited $^1\Delta_g$ state of O $_2$ , lying about 94 kJ mol $^{-1}$ above the $^3\Sigma_g^-$ ground state, in which the same two π_g^\ electrons occupy the same* orbital lobe pattern but pair their spins (S=0) instead of remaining unpaired. It is diamagnetic, unlike ground-state O $_2$ , and reacts far more readily with ordinary closed-shell organic singlet molecules because the reaction is no longer spin-forbidden. Singlet oxygen is generated photochemically (using a sensitizer dye and light) or via certain chemical routes, and is central to photodynamic therapy, some autoxidation and photobleaching processes, and atmospheric photochemistry. Its existence and energy were established directly from the same term-symbol analysis of the π² configuration used to derive the ground state.

Does the same reasoning apply to heteronuclear diatomics like NO or CO?

Yes, with one structural change: heteronuclear diatomics lack an inversion center, so molecular orbitals and term symbols carry no g/u label, and orbitals mix more freely between the two different atoms' orbitals of matching symmetry (foreshadowing the treatment needed for HF and CO specifically). NO, for example, has 15 electrons and one unpaired electron in a π\* orbital analogous to O $_2^+$ 's frontier shell, making it a stable paramagnetic radical (a rare example of a thermodynamically stable odd-electron neutral molecule), with a $^2\Pi$ ground term. CO, by contrast, is isoelectronic with N $_2$ (14 electrons) and has a completely closed-shell, diamagnetic $^1\Sigma^+$ ground state, though the orbital energies and atomic contributions differ substantially from N $_2$ 's because carbon and oxygen have very different electronegativities.

Why is B2 the classic test case for s-p mixing rather than O2 or N2?

B $_2$ 's magnetism flips depending on whether σg(2p) or π_u(2p) is lower in energy, making it maximally sensitive to the s–p mixing correction. With the mixed ordering (σg(2p) above π_u(2p), correct for B $_2$ through N $_2$ ), B $_2$ 's two frontier electrons singly occupy the degenerate π_u(2p) pair with parallel spins, correctly predicting the experimentally observed paramagnetism. With the naive unmixed ordering, those same two electrons would instead pair up in a single σg(2p) orbital, wrongly predicting a diamagnetic ground state. O $_2$ and N $_2$ , by contrast, have their frontier electrons in configurations where the ordering ambiguity does not change the final answer, so they cannot distinguish the two ordering schemes as decisively as B $_2$ does.

How do you determine the g or u parity label for an open-shell configuration?

Multiply the individual parities of every singly-occupied orbital in the open shell (filled shells always contribute net g and can be ignored). A single electron in a g orbital gives an overall g state; a single electron in a u orbital gives an overall u state. For two electrons both in g orbitals (as in O $_2$ 's π_g\*² configuration), g × g = g, so all three resulting terms (³Σg⁻, ¹Δg, ¹Σg⁺) carry the g label. This rule is the direct molecular analogue of combining individual orbital parities to get the parity of a many-electron atomic or molecular term, and it only applies to species with a genuine inversion center, i.e., homonuclear diatomics or other centrosymmetric systems.

Why does the spin-only magnetic moment formula work well for O2 despite O2 having nonzero orbital angular momentum in principle?

The spin-only formula μ = √[n(n+2)] μB assumes orbital angular momentum is fully "quenched," contributing nothing to the observed moment, which is an excellent approximation whenever orbital angular momentum is not free to reorient in response to an external field. In O $_2$ 's ³Σg⁻ ground state, Λ=0 (a Σ state has zero net axial orbital angular momentum by definition), so there is no low-lying orbital contribution to quench in the first place, and the spin-only estimate of 2.83 μB compares very well with experiment. This is different from many transition-metal complexes with nonzero orbital angular momentum in the ground term, where spin-only estimates can be noticeably off and an orbital contribution correction becomes necessary.