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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 $t_2$.

Charge balance fixes each metal as ${\mathrm{M}}^{\mathrm{II}}$. In an octahedral crystal field the lower $t_{2g}$ set is $(d_{xy},d_{xz},d_{yz})$ and the upper $e_g$ set is $(d_{z^2},d_{x^2-y^2})$. The measured spin-only magnetic moments select the following occupations:The result follows from $S=n_{\mathrm{unpaired}}/2$ and $2\sqrt{S(S+1)}=\sqrt{n_{\mathrm{unpaired}}(n_{\mathrm{unpaired}}+2)}$. Manganese(II) and iron(II) are high-spin complexes: their octahedral splitting is too small to compensate the additional pairing cost of their low-spin alternatives. Copper(II) is $d^9$, with one unpaired electron regardless of which member of the upper pair is singly occupied. spin-orbit coupling and exchange interactions can produce deviations from the spin-only estimate in real measurements; no such correction is required to explain the supplied values.

Both complexes have all their $d$ electrons paired, so $S=0$. Their spin-only magnetic moment is thereforeThey exhibit diamagnetism. The formula describes the spin contribution; it does not say that a closed-shell sample has no diamagnetic response to an applied field.

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]

Write the two allowed moment projections as $\pm \mu$, where $\mu =g{\mu }_B/2$ for spin one-half. Their Zeeman energies in the prescribed flux density are $\mp \mu B$. The single-particle partition function and the mean moment along the field areFor noninteracting moments, magnetization is moment per volume:For electron-like spin-only moments $g\simeq 2$, these become $M_s=N{\mu }_B/V$ and $\eta ={\mu }_B/k_B$.

Differentiate with respect to the specified $B$ to obtainThis $B$ derivative is not a dimensionless SI volume susceptibility. In the weak-feedback approximation $B\simeq {\mu }_{0}H$, the usual dimensionless differential susceptibility is ${\chi }_{\mathrm{diff}}\simeq {\mu }_0{\chi }_B$. The small-field limit gives Curie's law ${\chi }_0={\mu }_{0}N{\mu }^2/(V{k}_{B}T)$. At finite bias a secant response $M/H\simeq {\mu }_{0}M/B={\chi }_0\tanh (x)/x$ differs from the differential slope ${\chi }_0{\mathrm{sech}}^{2}x$. For an internal $B={\mu }_0(H+M)$ retained self-consistently, the exact local conversion is ${\chi }_H={\mu }_0{\chi }_B/(1-{\mu }_0{\chi }_B)$; demagnetizing geometry can further affect an applied-field measurement. Stating the field convention avoids a hidden factor of ${\mu }_0$ or confusion between a slope and a ratio.

For $B=5$ T, the same spin-only two-state probabilities apply, but the low-temperature argument is no longer small. At 300 K, $x=0.0111952$; at 2 K, $x=1.67928$. HenceThe 2 K sample is about $93.3\%$ of its saturation magnetization, with about $96.6\%$ of the moments in the aligned state. The two percentages describe different observables. A linear extrapolation would predict $M/M_s\simeq x=1.679$, exceeding saturation and therefore contradicting the bounded two-state population.

For completeness, at this 2 K, 5 T bias the dimensionless secant susceptibility is approximately ${\chi }_0\tanh x/x=0.01199$, while the differential susceptibility is ${\chi }_0{\mathrm{sech}}^{2}x=0.002805$ under the same prescribed-field approximation. Neither equals the zero-field Curie slope $0.02159$. Thus saturation reduces the incremental response even while the magnetization itself is large.

For $B=0.05$ T, use the two spin-only moment states. Their probabilities and normalized magnetization areAt 300 K, $x=1.11952\times 10^{-4}$; at 2 K, $x=0.0167928$. ThereforeThe phrase “fraction aligned” is ambiguous if it is used to mean net alignment. Half the moments are already in the parallel state at zero field, but the net magnetization is then zero because the antiparallel half cancels them. The table reports both the literal parallel-state population and the imbalance that produces magnetization. At this weak field even the 2 K sample is far from saturation, so the Curie approximation is accurate within this independent-spin model.

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.

The spin-only magnetic moment neglects orbital contributions and approximates $g$ by two: ${\mu }_{\mathrm{eff}}=g\sqrt{S(S+1)}{\mu }_B=\sqrt{n(n+2)}{\mu }_B$ for $n$ parallel unpaired electrons. Four unpaired electrons give $\sqrt{24}{\mu }_B\simeq 4.90{\mu }_B$.