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An approximate scheme for constructing an atomic electron configuration by filling lower-energy atomic orbitals before higher-energy ones, subject to the Pauli exclusion principle. Orbital ordering can change on ionization and has exceptions in interacting chemical atoms. Selecting the free-ion ground term within a specified open-shell configuration additionally requires Hund's rules or a more detailed interaction calculation.

An effective local exchange coupling can favor parallel electron spins in different orbitals on the same atom. The displayed sign convention represents that preference. It underlies the high-spin tendency in Hund's rules and the alignment of itinerant and core spins used in double exchange. The coupling is an effective description of intra-atomic Coulomb and exchange energies, rather than a separate classical magnetic force.

For two equivalent $p$ electrons, antisymmetry permitsTheir $9+5+1=15$ states exhaust $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$. Hund's rules first maximize spin, then orbital angular momentum; the less-than-half-filled $p$ shell puts the smallest $J$ lowest. Hence the ground level is ${}^3{P}_0$. This is the allowed LS terms of a p-squared configuration.^[ref]

Closed $1s^2$ and $2s^2$ subshells have $L=S=0$. In the open $2p^4$ subshell, three electrons can have parallel spins in $m_{\ell }=-1,0,1$, and the fourth must have the opposite spin, giving maximum $S=1$. Treating the shell as two $p$ holes gives the same allowed terms as $p^2$; at this spin, antisymmetry permits $L=1$, rather than $L=2$. One can also see $M_L=1$ by placing the fourth electron in $m_{\ell }=1$.

Hund's rules therefore select the ${}^{3}P$ term. Since four electrons exceed half filling of the six-state $p$ subshell, its lowest atomic fine-structure level has the largest $J=L+S=2$:Its zero-field magnetic degeneracy is $2J+1=5$. The experimentally tabulated neutral-oxygen ground level has precisely this term and $J$.^[ref]

The open shell contains one $f$ electron, so $L=3$ and $S=1/2$. For a less-than-half-filled shell, Hund's rules select the smaller total angular momentum, $J=L-S=5/2$, giving the free-ion ground level ${}^2{F}_{5/2}$. The Landé g-factor isIn a magnetic field weak compared with the spin-orbit splitting, the electronic Zeeman shifts are $\delta E_{m_J}=g_J{\mu }_{B}B{m}_J$, for $m_J=-5/2,-3/2,\dots ,5/2$. Thus six equally spaced magnetic substates appear. At $B=1\mathrm{T}$, the adjacent-level spacing isThe complete lowest-to-highest multiplet spread is five spacings,Giving both quantities removes the ambiguity in the word splitting. This is the free-ion, weak-field result; no crystal-field doublet or hyperfine structure is specified.

The closed $1s^2$ and $2s^2$ subshells contribute zero total orbital angular momentum and spin, so only the two $2p$ electrons matter. The first of Hund's rules maximizes their total spin, giving $S=1$. The Pauli exclusion principle then requires an antisymmetric spatial wavefunction. Two parallel-spin electrons cannot occupy the same $m_l$ orbital: the greatest possible orbital projection is $1+0=1$, so the maximal allowed $L$ within this triplet is $L=1$.

The ground term is therefore ${}^{3}P$, in term symbol notation ${}^{2S+1}{L}_J$. Since the $p^2$ subshell contains fewer than half of its six possible electrons, the third of Hund's rules places the smallest $J$ lowest. Addition of angular momentum gives $J=0,1,2$, in that energy order within the ground term. ThusThese are the fine structure partners of the same electronic term, with magnetic multiplicities $1,3,5$, rather than different principal-shell excitations. The ordering agrees with the tabulated lowest neutral-carbon levels.^[ref]

The closed $1s^{2}2s^2$ shells have $L=S=0$, leaving the equivalent $2p^2$ electrons. Addition of angular momentum and the Pauli exclusion principle permit the atomic terms ${}^{3}P,{}^{1}D,{}^{1}S$. Their term symbols use ${}^{2S+1}{L}_J$. Hund's rules put the triplet state below the singlet states; for a less-than-half-filled $p$ shell the smallest $J$ in ${}^{3}P$ is lowest. ThusAll have even parity. These assignments agree with independent tabulated carbon energy levels.^[ref] The rounded source values should not be forced to satisfy an exact Landé interval rule; pure first-order LS coupling need not account for every measured fine structure correction.

The $4f^7$ shell carries a large spin (its ground atomic term under Hund's rules has $L=0,S=7/2$), and the $6s$ electron penetrates the inner shells. Its nonzero probability density at the atomic nucleus supplies a potentially large Fermi contact interaction with the nuclear magnetic dipole moment. The $6s$ shell here is singly occupied, so the obligatory paired-$6s^2$ cancellation is absent. Both stable europium isotopes have nonzero nuclear spin, as the spectrum below also establishes.

The term symbol ${}^8{D}_{11/2}$ specifies $S=7/2$, $L=2$, $J=11/2$. Its large $J$, combined with appreciable nuclear spin, permits a wide range of total angular momentum $F$ and hence a substantial total hyperfine structure span. For $I=5/2$, $F=3,\dots ,8$ and the span is $30|a|$. The penetrating $6s$ contribution and the available range of $F$ explain the large hyperfine structure; the term symbol alone does not determine its constant, which also depends on electronic recoupling, radial wavefunctions and configuration interaction.

Let $P_{12}$ exchange the complete labels of the two electrons. The spatial factors obey $P_{12}{\Phi }_{\pm }=\pm {\Phi }_{\pm }$. The singlet state spin factor is antisymmetric, while each triplet state spin factor is symmetric. Thus in all four casesas required by the Pauli exclusion principle. In particular, the $M_S=0$ triplet state remains a triplet despite having opposite measured spin projections in each product term.

For a spin-independent Hamiltonian and the same fixed spatial atomic orbitals, direct evaluation giveswhere $J_{ab}$ is the direct Coulomb integral and $K_{ab}>0$ is the exchange integral for the helium $1s,2s$ pair. ThereforeThe antisymmetric spatial wavefunction vanishes when the two positions coincide and reduces their Coulomb repulsion. The lower energy of the maximum-spin member of this fixed electron configuration is the first of Hund's rules.

Closed subshells have $L=S=0$, so only the open $p$ or $d$ subshell determines each atomic term. Apply Hund's rules in order: maximize $S$, maximize $L$ at that $S$, then choose the lowest $J$ below half filling or the highest $J$ above half filling.

For carbon, the two $2p$ electrons occupy different magnetic atomic orbitals with parallel spins, giving $S=1$. Their largest allowed $M_L$ at that spin is $1+0=1$, so $L=1$, a $P$ term. The $p^2$ subshell is less than half full, giving $J=|1-1|=0$. The ground level is ${}^3{P}_0$.

For vanadium, three parallel-spin $3d$ electrons give $S=3/2$. The largest $M_L$ compatible with the Pauli exclusion principle is $2+1+0=3$, hence $L=3$, an $F$ term. The $d^3$ subshell is less than half full, so $J=3-3/2=3/2$. The ground level is ${}^4{F}_{3/2}$.

For cobalt, the $d^7$ subshell is equivalent to three $d$ holes for the allowed $L,S$ terms, so its maximum-spin, maximum-$L$ term is again ${}^{4}F$. Explicitly, five spin-up electrons fill all $m_l$ values and two spin-down electrons can occupy $m_l=2,1$, giving $M_S=3/2$ and $M_L=3$. Since $d^7$ is more than half full, $J=3+3/2=9/2$. The ground level is ${}^4{F}_{9/2}$.

For $n$ unpaired electrons, the spin-only magnetic moment iswhere ${\mu }_{\mathrm{B}}$ is the Bohr magneton and the second expression uses $g\simeq 2$.

For the square, a large separation of the highest $B_{1g}$ level favours pairing in the lower four molecular orbitals: $S=0$, $n=0$, ${\mu }_{\mathrm{eff}}=0$. If the cost of occupying $B_{1g}$ were smaller than the relevant electron pairing energy, a high-spin complex with two unpaired electrons and $S=1$ could instead occur, giving $\sqrt{8}{\mu }_{\mathrm{B}}\simeq 2.83{\mu }_{\mathrm{B}}$.

For ideal tetrahedral $d^8$, the occupation is $e^4{t}_2^4$. Hund's rules distribute four electrons over the three degenerate $t_2$ molecular orbitals as one pair and two single occupations. Thus $S=1$ and ${\mu }_{\mathrm{eff}}\simeq 2.83{\mu }_{\mathrm{B}}$. Merely increasing the $e$–$t_2$ splitting does not create a conventional paired tetrahedral $d^8$ alternative: four electrons must still occupy $t_2$.

The closed argon core contributes $L=S=0$. In the half-filled $3d^5$ shell, Hund's rules put one parallel-spin electron in each $m_{\ell }=-2,-1,0,1,2$ orbital. This gives $S_d=5/2$ and the orbital term $L_d=0$. Adding the parallel $4s$ electron, with $\ell =0$ and $s=1/2$, givesThe spin multiplicity is seven. The magnetic projection labels run $M_S,M_J=-3,-2,\dots ,3$, while $M_L=0$. The term symbol agrees with the observed neutral-chromium ground level.^[ref]

The $2s$ atomic orbital is spherically symmetric and has one radial node. A real $2p$ atomic orbital has two lobes of opposite quantum phase, one atomic-orbital nodal plane through the nucleus and no radial node. The three spatial orientations $2p_x,2p_y,2p_z$ remain degenerate in a spherically symmetric atom when external fields and fine structure are neglected.

In a many-electron atom, $2s$ penetrates the screened core more effectively than $2p$, experiences a stronger attraction and is lower in energy: ${\epsilon }_{2s}<{\epsilon }_{2p}$. The hydrogen-like ion degeneracy therefore does not survive screening. Both valence orbital types are generally more tightly bound and spatially contracted in nitrogen than in carbon, because of the larger effective nuclear charge. These are central-field orbital statements; the detailed atomic multiplet structure also involves electron repulsion and Hund's rules.

Take the bond axis as $z$. The heavy-atom $2s$ and $2p_z$ atomic orbitals can mix with hydrogen $1s$ to make three sigma-symmetry molecular orbitals. In the qualitative valence ordering, these are a low predominantly heavy-atom $2s$ level, a bonding $\sigma$ level and a higher ${\sigma }^{\ast }$ level. The perpendicular $2p_x,2p_y$ orbitals form a degenerate approximately nonbonding $\pi$ pair because hydrogen has no valence orbitals of the matching sideways symmetry. The filled heavy-atom $1s$ core is omitted from the diagram.

Counting valence electrons givesThus both exhibit paramagnetism: CH has one unpaired electron and NH has two in its ground state. For NH, Hund's rules favour separate parallel-spin occupation of the two degenerate $\pi$ orbitals. The corresponding observed ground terms are $X{}^2\Pi$ and $X{}^3{\Sigma }^{-}$; the total bond order is approximately one in this simple sigma-bond/nonbonding-pi description.^[ref]

The nitrogen-derived orbital energies lie lower than the corresponding carbon-derived ones on an approximate common energy scale, while the isolated hydrogen $1s$ reference is unchanged. The bonding orbitals consequently have different polarizations and mixing coefficients; drawing identical energies for C and N would miss the screening trend. The figure specifies ordering and occupancy, not measured gaps or a numerical prediction from (a).

The terms are ${}^3{\Sigma }_g^{-}$, ${}^1{\Delta }_g$ and ${}^1{\Sigma }_g^{+}$. Closed molecular orbital shells contribute no net spin angular momentum or axial orbital projection, so only the two equivalent ${\pi }_g$ electrons need be coupled. Their individual projections are $\lambda =\pm 1$.

Putting both electrons in the same projection gives $|\Lambda |=2$ and forces opposite spins by the Pauli exclusion principle, yielding ${}^1{\Delta }_g$. With one electron in each projection, the spin-adapted symmetric and antisymmetric spatial combinations give a singlet $\Sigma$ and a triplet $\Sigma$. Their reflection parities are $+$ and $-$, respectively. Both open-shell electrons are gerade, so their product is gerade. The dimensions $3+2+1=6$ exhaust the $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ microstates. By Hund's rules, the triplet is the lowest term, and ${}^1{\Delta }_g$ is the lower singlet.

Choose x along O–O. The valence orbital order appropriate to O${}_2$ is approximatelyIts twelve valence electrons occupyThe two degenerate pi-antibonding orbitals are singly occupied with parallel spins according to Hund's rules. Thus $S=1$, multiplicity $2S+1=3$, with three possible $M_S$ values. A triplet is a many-electron spin classification, not three electrons in one orbital. The resulting ground term is ${}^3{\Sigma }_g^{-}$. Exchange stabilization favours this triplet over the singlets formed within the same frontier-orbital manifold.

Within a specified electron configuration in the LS coupling regime, Hund's rules select the lowest atomic term as follows. First maximize the total electron spin $S$ subject to the Pauli exclusion principle. Among the terms having that $S$, choose the largest total orbital angular momentum $L$. Finally, spin-orbit coupling places the smallest allowed $J=|L-S|$ lowest for a subshell less than half full, and the largest $J=L+S$ lowest for a subshell more than half full. At half filling the maximum-spin term has $L=0$, hence $J=S$. These rules rank terms within configurations; the exceptional neutral-atom configurations of chromium and copper must also be used.

A $3d^9$ shell has one hole in the otherwise filled $d$ shell. Hund's rules give $S=1/2$, $L=2$, and, for a more-than-half-filled shell, $J=L+S=5/2$: the free-ion term is ${}^2{D}_{5/2}$. Its Landé factor isDistinguish the Curie effective moment from a saturated projection:The question then explicitly asks us to use orbital quenching in the crystal. Ligand fields split and mix orbital states, suppressing their freely orientable angular momentum. The spin-only approximation has $g\simeq 2$, $S=1/2$, so ${\mu }_{\mathrm{eff}}=\sqrt{3}{\mu }_B$ but the two field-axis projections are $\pm {\mu }_B$. The latter, not $\sqrt{3}{\mu }_B$, belong in the tanh argument and saturation magnetization.

The sulfate and water environment supplies mainly closed-shell, weakly magnetic surroundings, stabilizes localized Cu2+ sites and separates their moments so that exchange can be weak on the temperature scale considered. Paramagnetism here is an approximation of weakly interacting local spins, not a claim that dilution guarantees no magnetic ordering at any temperature. Elemental copper has a filled $3d$ band and itinerant conduction electrons rather than one localized Cu2+ spin per atom. Pauli spin response and orbital/core contributions then replace a local-spin Curie law; ordinary pure copper has a weak net diamagnetic response.^[ref]

There is one Cu2+ per formula unit, hence $n_s=N_A/V_m=5.5148\times 10^{27}$ m${}^{-3}$. The spin-only weak-field SI volume susceptibility isThese neglect exchange, anisotropic $g$ factors and background diamagnetism. At high field and low temperature, saturation requires the nonlinear response in a; a field-independent Curie value then describes only the zero-field slope.

An $f$ shell has orbital quantum number $l=3$ and seven orbital states with $m_l=-3,-2,\dots ,3$. The first of Hund's rules maximizes total spin subject to the Pauli exclusion principle. Seven electrons occupy the seven orbitals singly with parallel spins, givingTheir orbital projections sum to zero, so the half-filled shell also has total orbital angular momentum $L=0$, total $J=7/2$, and Landé g-factor $g_J=2$. These further values identify the magnetic moment used below.

For a fixed open-shell electron configuration in the LS coupling regime, Hund's rules select the free-ion ground term in the following order:

The first two rules come from the Coulomb interaction between electrons within the partially filled subshell, including its exchange interaction. Parallel spins require an antisymmetric spatial wavefunction, reducing close approach and often lowering the repulsive Coulomb interaction energy. Among the allowed high-spin terms, the angular dependence of the Coulomb interaction generally favours the largest $L$; a simple magnetic alignment argument alone does not derive this second ordering rule.

The weaker spin-orbit coupling then splits a selected $L,S$ term. In LS coupling, its term-level energy is proportional toThe effective sign appropriate to less-than-half versus more-than-half filling gives the third rule. These rules assume the electrostatic term separation dominates spin-orbit coupling; strong crystal field splitting or other coupling regimes require further analysis.