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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]

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 $\sqrt{24}{\mu }_B\simeq 4.90{\mu }_B$.