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An electron configuration specifies the occupied one-electron subshells and their populations, such as $1s^{2}2s^{2}2p^2$, without fixing how all orbital and spin angular momenta are coupled.

An atomic term groups states of a configuration with the same $L$ and $S$. An atomic fine-structure level additionally specifies $J$; spin-orbit coupling can split one term into several such levels. In zero external field, a level retains $2J+1$ magnetic substates.

The term symbol iswith letters $S,P,D,F,G,\dots$ denoting $L=0,1,2,3,4,\dots$. The superscript is the spin multiplicity. Without the $J$ subscript, ${}^{2S+1}L$ identifies an atomic term, rather than an individual fine-structure level.

In LS coupling, add all orbital and spin angular momenta separately:The quantum numbers specifyThus $M_L=-L,\dots ,L$ and $M_S=-S,\dots ,S$. The addition of angular momentum rule givesIn a product basis with individual projection labels,The magnitudes $L$ and $S$ are not generally the arithmetic sums of the individual magnitudes; they result from angular-momentum coupling. The sum rules for projections remain valid for every product component of a coupled angular momentum eigenstate.

Differentiate the Lennard-Jones potential. Its minimum satisfiesFor dilithium, the two wells consequently haveUnder an additional same-orbital, Hund-type exchange comparison, assign $(3.4,0.06)$ to the triplet and $(4.9,0.05)$ to the singlet: the lower-energy minimum is associated with the exchange-stabilized triplet state, as in part (a).

This assignment needs that extra orbital-parentage assumption. A scalar Lennard-Jones potential does not itself encode spin, and a common atomic dissociation limit does not specify a unique molecular electronic configuration or spatial symmetry. The two curves even cross at about $5.006$ bohr: neither lies below the other at every separation. Thus the minima and well depths are determined unambiguously; the spin labels are a conditional qualitative inference, rather than a unique consequence of the four supplied parameters. Neither “a triplet must always have a longer bond” nor a universal ordering of unrelated excited states is justified.

The electronic transition starts from $A_{1g}$, so the electric-dipole test reduces to whether the final electronic irreducible representation occurs among $A_{2u}+E_{1u}$. The excited $E_{1u}$ state matches the $(x,y)$ dipole components and is therefore electric-dipole allowed.

The $B_{2u}$ state matches neither $z$ nor $(x,y)$ and is electric-dipole forbidden at the equilibrium geometry. Both excited states are singlets, so the difference is spatial symmetry rather than the spin selection rule. Inversion parity alone does not decide the result: both final states are ungerade, but only the matching vector representation is allowed.

Take the energy of an isolated, field-free gas atom as zero, and let $E_b>0$ be the field-independent binding energy at one adsorption site. The two adsorbed spin energies areHere ${\mu }_{\mathrm{B}}$ is the Bohr magneton. The lower state is stabilized by increasing $|B|$; the assignment of the two signs to names for the spin projections is a convention.

Use the ideal Langmuir adsorption model: identical independent sites, at most one atom per site, no lateral interactions, and no additional thermally active adsorbed degrees of freedom beyond the two spin states. Treat the atoms as indistinguishable and localized after adsorption; only site occupancy and spin choices distinguish configurations.

There are $\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ ways to select $N$ occupied sites. For any such selection, independent spin sums contribute $z^N$. ThusWith $q_{\mathrm{ads}}=Mz$ as defined in part (b), this isEquivalently, starting from labelled particles gives $M!/(M-N)!$ distinct assignments to different sites; division by $N!$ removes particle relabellings. The $M^{-N}$ removes the site multiplicity already included in each full one-particle function.

If $q_{\mathrm{ads}}$ denotes the one-site internal factor instead, $f(M,N)=\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ and $Q_N=\left(\genfrac{}{}{0ex}{}{M}{N}\right)q_{\mathrm{ads}}^N$. These are identical physical results, not different counting models. In particular, the complete-function convention gives $f(M,1)=1$ and hence $Q_1=q_{\mathrm{ads}}$.

For the adsorbed lattice gas, the Helmholtz free energy isUse Stirling's approximation with $M,N,M-N$ large, then differentiate at fixed $M,T,B$. Writing surface coverage as $\theta =N/M$ gives the adsorption chemical potentialThe occupation ratio comes from site-mixing entropy; it is essential near saturation.

Let $N_g$ be the number of free atoms and $n_g=N_g/V$. In the field-free gas, both spin states have zero energy, so the one-atom partition function is $q_g=2V/{\Lambda }^3$, where $\Lambda$ is the thermal de Broglie wavelength. For a dilute ideal gas,ThereforeThe last equality uses $p=n_g{k}_{\mathrm{B}}T$. The bulk spin degeneracy is two even though the bulk magnetic field vanishes: degeneracy is not the absence of spin states.

At adsorption equilibrium, ${\mu }_{\mathrm{ads}}={\mu }_g$. The expressions from part (d) giveDefineSolving the occupation ratio yieldsThe coefficient $b$ has units of inverse pressure, so $bp$ is dimensionless. The two spin states contribute a factor two on both sides of the adsorption equilibrium and cancel; retaining the surface factor but dropping the bulk degeneracy would incorrectly double $b$.

At fixed temperature,Thus magnetic-field-dependent Langmuir adsorption is even in $B$ and increases with $|B|$. At fixed nonzero gas pressure, stronger field increases surface coverage; in an isotherm plot, the half-coverage pressure $p_{1/2}=1/b$ moves to lower pressure. The saturation coverage remains one because the number of sites is unchanged.

For weak fields, with ${\theta }_0=b(0)p/[1+b(0)p]$,There is no linear small-field change. At large $|x|$, $\cosh x\sim e^{|x|}/2$, and at fixed $p>0$ the coverage approaches one. The field lowers the free energy of the surface spin ensemble relative to the unsplit bulk; simply averaging the two surface energies to $-E_b$ would miss this equilibrium effect.

For monatomic species, there are no molecular rotational or vibrational factors. Assume excited electronic states are negligible, so the internal partition sums are the ground-state degeneracies. A free electron has two spin states, givingwith the analogous expression for the ion. Since losing one electron changes the atomic mass negligibly, $(m_{{\mathrm{M}}^{+}}/m_{\mathrm{M}}{)}^{3/2}\simeq 1$. HenceThis is the ground-state form of the Saha ionization equation. Its assumptions include a dilute, nondegenerate ideal mixture, negligible interparticle interactions, thermal and chemical equilibrium, no additional ionization stages, and consistent treatment of internal degeneracies. Nuclear-spin factors may be omitted when they cancel between the atom and ion; they must not be counted on only one side.

In atomic units, $\hslash =m_e=e=4\pi {\epsilon }_0=1$, and the hydrogen atom hasThe allowed quantum numbers areHere the fixed electron spin quantum number is $s=1/2$. The energy eigenvalues are in hartrees; restoring units gives ${\hslash }^{2}l(l+1)$ and $\hslash m$ for the orbital operators and the corresponding factors for spin. The spin-independent hydrogen Hamiltonian does not split the two spin projections.

Keep one normalized spin function fixed and write $A=1/\sqrt{24}$ and $f_m(\theta ,\varphi )=\sin \theta e^{im\varphi }$, with $m=\pm 1$. Overall constants and the spin factor pass unchanged through the spatial operators.

First, in atomic units, ${\hat{l}}_z=-i{\partial }_{\varphi }$, soFor the squared orbital angular momentum, ${\partial }_{\varphi }^2{f}_m=-f_m$ andTherefore the two terms of the angular operator giveusing $1-{\cos }^2\theta ={\sin }^2\theta$. Thus ${\hat{l}}^2\Psi =2\Psi$.

Finally, the spatial hydrogen Hamiltonian isFor $R=Ar{e}^{-r/2}$,Substitute the already established angular eigenvalue $l(l+1)=2$:Hence $E=-1/8$ hartree, consistently with $n=2$. The angular calculation must precede the radial substitution if one wants to avoid differentiating the entire product repeatedly.

Denote the two complex spatial atomic orbitals by ${\psi }_{+}$ and ${\psi }_{-}$, retaining the same spin state. With the phases specified by the supplied spherical harmonics, chooseThe two original atomic orbitals are orthonormal, so the squared coefficients add to one and these combinations are normalized and mutually orthogonal. Explicitly,which are real spatial wavefunctions. An overall sign on either orbital has no physical consequence.

These are the familiar $2p_x$ and $2p_y$ p orbitals. Each has two lobes of opposite wavefunction sign, with an atomic-orbital nodal plane through the atomic nucleus: $x=0$ for $2p_x$ and $y=0$ for $2p_y$. The sign is a phase convention, not a positive or negative electric charge; the probability density is nonnegative on both lobes.

There are $L$ primitive cells and therefore $L$ lattice points. The specified reciprocal-space density $L{A}_c/(2\pi {)}^2$ counts distinct allowed wavevectors; each supports two electron spin states. Filling the disk at zero temperature givesUsing $A_c=2a^2$ and $k_F^2={\pi }^2/(2a^2)$,The circle fills a fraction $\pi /4$ of the first-zone area, while one spin-degenerate band has two states per primitive cell; these give the same result. A single-spin count would instead be $\pi /4$, omitting the physical twofold spin degeneracy of this model. The circle has only just reached the sides, so the zone corners remain unfilled: first contact does not mean filling the entire first band.

The Pauli principle requires the complete wavefunction of identical fermions to be antisymmetric under exchanging the full coordinates, including spin:It is the total spatial-and-spin state that changes sign; the spatial factor alone need not be antisymmetric.

In a metallocene, several orbitals with substantial metal $d$ character are nonbonding molecular orbitals or only weakly bonding or antibonding with respect to the strong Cp–metal framework. Adding or removing a small number of electrons from them can therefore change oxidation state or spin without destroying the principal ring–metal bonds. This allows stable compounds on either side of the 18-electron count: for example, the usual neutral counts for chromocene, ferrocene, cobaltocene and nickelocene are respectively 16, 18, 19 and 20. The 18-electron rule remains a useful reference, but occupation of a weakly antibonding orbital is much less disruptive than occupation of a strongly antibonding orbital. Redox stability also depends on geometry, substituents and reaction conditions, so this is a bonding explanation of the observed flexibility rather than a guarantee of equal stability for all counts.

A Co(III) centre has $d^6$, and charge balance requires one closed-shell catecholate ligand and one $S=1/2$ semiquinonate ligand. For low-spin Co(III), $t_{2g}^6(e_g^{\ast }{)}^0$, the metal has $S=0$, giving an independent-spin magnetic moment ofThus $A^1$ contains low-spin Co(III), $S_{\mathrm{Co}}=0$, with its observed moment residing on the free radical ligand. High-spin Co(III), $t_{2g}^4(e_g^{\ast }{)}^2$, would have $S=2$ and give $\sqrt{27}=5.20$ under the stated independent-spin approximation, so it is inconsistent with this measurement.

For Co(II), $d^7$, charge balance instead requires two $S=1/2$ semiquinonate ligands. High-spin Co(II) has $t_{2g}^5(e_g^{\ast }{)}^2$ and $S=3/2$, soTherefore $A^2$ contains high-spin Co(II), $S_{\mathrm{Co}}=3/2$. Low-spin Co(II), $t_{2g}^6(e_g^{\ast }{)}^1$, would give $\sqrt{9}=3.00$ when the two ligand spins are included. These four checks distinguish both oxidation state and metal spin.

The larger octahedral splitting at Co(III) exceeds the effective pairing energy and favours a low-spin complex; the smaller octahedral splitting at Co(II) favours a high-spin complex with occupation of both $e_g^{\ast }$ orbitals before further pairing. Interconversion also transfers an electron between cobalt and a dioxolene ligand, so this is valence tautomerism, rather than a metal-only spin change. Squared moments add here because the supplied model treats the centres as independent; adding the three spins first would be a different, exchange-coupled model.

The dimer has two unpaired electrons and a triplet ground state, $S=1$. Its approximately threefold-symmetric Fe${}_2$(CO)${}_3$ core supplies a pair of degenerate frontier molecular orbitals. The last two electrons occupy separate members with parallel spins, favored by exchange, rather than pairing in one. A useful qualitative picture is a sigma bond plus two perpendicular, singly occupied pi half-bonds: together these support a formal Fe–Fe double bond while retaining paramagnetism. The spin-only magnetic moment for $S=1$ is $\sqrt{8}{\mu }_B\simeq 2.83{\mu }_B$; real moments can include orbital and spin–orbit effects.

The formal 18-electron count does not require every molecular orbital to be doubly occupied. It is therefore compatible with an open-shell dimer. The primary structural/magnetic study establishes this unusual triplet state; an isolated-atom high-spin $d^7$ diagram would not describe the coupled metal–ligand system.^[ref]

The metal is Cr(II), $d^4$. A chloride ligand produces a relatively weak ligand field, so the octahedral center is a high-spin complex with $t_{2g}^3{e}_g^1$, $S=2$. The singly occupied $e_g$ pair is orbitally degenerate: $d_{z^2}$ and $d_{x^2-y^2}$ cannot retain identical occupancies once the distortion selects one. This is a strong setting for the Jahn-Teller effect.

On tetragonal elongation, the two axial Cr–Cl distances increase. Repulsion is reduced most for $d_{z^2}$, so it falls relative to $d_{x^2-y^2}$; $d_{xz},d_{yz}$ also fall relative to $d_{xy}$. The single $e_g$ electron occupies the lower $d_{z^2}$-derived level. The electronic stabilization offsets an elastic cost until a distorted equilibrium geometry is reached. The high-spin $d^4$ occupancy drives an elongated, rather than regular, octahedron. The three lower-set electrons remain one per orbital in this simple diagram; it is primarily the unequal $e_g$ occupancy that produces the strong distortion.

iron remains Fe(II), $d^6$, in both states; the change is a spin crossover, rather than a change in oxidation state. At the higher temperature the high-spin complex has $t_{2g}^4{e}_g^2$: four unpaired electrons, $S=2$, andThis matches the observed moment. At low temperature a low-spin complex with $t_{2g}^6{e}_g^0$ has $S=0$ and shows diamagnetism. Cooling switches high-spin $d^6$ to low-spin $d^6$. Emptying the metal–ligand antibonding $e_g$ orbitals strengthens and shortens the Fe–N bonds, accounting for the contraction by roughly $0.2\text{A˚}$.

The octahedral splitting and electron pairing energy must be sufficiently close for both states to be accessible. The low-spin state gains electronic energy, while the high-spin state has greater spin degeneracy and usually greater vibrational entropy because of its softer bonds. The resulting Gibbs free energy balance favors high spin at higher temperature. The two endpoint observations do not by themselves specify hysteresis or transition cooperativity; those require a temperature series. Coupled structural and magnetic measurements distinguish this explanation from a mere bond-length anomaly.