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A high-spin complex retains more unpaired electrons when the energy required for pairing exceeds the relevant orbital splitting. An octahedral high-spin $d^6$ center has $t_{2g}^4{e}_g^2$ and $S=2$.

A heteronuclear three-valence-electron diatomic. A simple bonding/antibonding two-orbital description predicts bond order one-half and one unpaired electron. More electronegative beryllium carries the negative end of the bond dipole.

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 $t_2$.

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.

BeLi has three valence electrons: two occupy a bonding sigma orbital and the third its higher partner. The elementary bond order is therefore $1/2$, with one unpaired electron: expect a relatively weak, paramagnetic, doublet diatomic molecule. Because Be attracts valence charge more strongly than Li, the bond is polar toward Be. This is a qualitative prediction, not a numerical dissociation energy.

This is the elementary picture of the lithium beryllium diatomic molecule.

The 1922 interpretation concerned space quantization of atomic magnetic moments in the old quantum theory. The modern electron spin proposal came in 1925; the original observers did not identify intrinsic electron spin as the cause of their splitting. An orbital interpretation then seemed plausible, but the later spin explanation gives a different account of the same two-component evidence.^[ref]

Modern atomic data assign neutral silver the ground state electron configuration $[\mathrm{Kr}]4d^{10}5s$ and electronic term ${}^2{S}_{1/2}$. The filled electron shells contribute zero net electronic angular momentum; the outer unpaired electron has orbital angular momentum $L=0$, spin $S=1/2$, and electronic total angular momentum $J=1/2$. The magnetic quantum number has two values $m_J=\pm 1/2$.^[ref]

In the effective electronic description,The minus sign matters: for positive $G$, the component with $m_J=+1/2$ has negative electronic magnetic moment and deflects toward negative $z$ in our coordinates. Calling the upper spot “spin up” without specifying the magnetic field and channel calibration can therefore reverse the assignment. The total angular momentum operator here sums electronic orbital angular momentum and electronic spin; $J$ is not a claim that the atomic nucleus has no spin. Hyperfine structure and nuclear magnetic moments can require additional internal levels, so a two-channel electronic approximation must be justified for the field and resolution being used.

This is a neutral atomic beam, not a beam of free electrons. The deflected mass in the flight equations is the entire silver chemical atom; its dominant electronic magnetic moment supplies the state-dependent force. Replacing that atom by a free electron would also introduce the charged particle's Lorentz force and a different preparation problem. The ideal two-state model is useful because it isolates the connection between discrete internal components and spatially resolved outputs; it does not retroactively turn the 1922 deposit into a direct contemporary identification of intrinsic electron spin.

A coordination complex with four carbon-bound cyanide ligands and a formal nickel(II), $d^8$, centre. Square planar molecular geometry permits a low-spin complex with no unpaired electrons.