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Inelastic neutron scattering resolves $E_i-E_f\ne 0$. Through conservation of energy, a peak at positive $\hslash \omega$ identifies an allowed excitation energy; through momentum conservation, its location in $\mathbf{Q}$ constrains the spatial variation of that excitation. A phonon is a quantum of lattice vibration; a magnon is a quantum of a spin wave. Their different couplings make energy transfer alone insufficient to identify which was created.

Bertram Neville Brockhouse's aluminium experiment rotated a large single crystal, scanning analyzer angle for neutron groups. His vanadium comparison exposed the elastic resolution. Later magnetite measurements used a magnetic field intensity test to identify magnons.^[ref]

Successive orientations sample different wave vectors; analyzer settings distinguish outgoing energies. This supplies points on a dispersion relation, rather than only an angular diffraction pattern. Finite resolution can merge neighboring excitations; background, multiple scattering and processes involving several phonons can complicate interpretation. A lattice model is a testable explanation of selected features, not a guarantee that every spectral peak has the same origin.

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 ${}^4{F}_{9/2}$.

The terminal nitrosyls are linear in both proposed electron-precise structures; the formal M–M bond orders are V–V double and Fe–Fe single. Use three-electron terminal linear NO donors, three-electron ${\mu }_2$-NO bridges, and five-electron neutral Cp donors in covalent counting.

For the vanadium dimer the two metal fragments together containA metal–metal bond of order $b$ adds $b$ to each local count, so $32+2b=36$ gives $b=2$ and 18 electrons per V. For an ionic cross-check, Cp is $-1$, each linear terminal NO is conventionally NO${}^{+}$, and each bridge is represented as NO${}^{-}$. ThusEach V then has $4+6_{\mathrm{Cp}}+2_{\mathrm{terminal}\mathrm{NO}}+4_{\mathrm{two}\mathrm{bridges}}+2_{\mathrm{V}=\mathrm{V}}=18$ electrons.

The iron species is monoanionic, and ${\mathrm{L}}^{-}$ is a pyrazolate bridge binding through its two nitrogens. This is a four-electron donor overall in ionic counting, or three overall in neutral counting. The carbonyl bridge supplies two electrons overall. The covalent fragment total is consequentlyTherefore $34+2b=36$ gives $b=1$. In the formal NO${}^{+}$ description,Each Fe receives $10+4_{\mathrm{two}\mathrm{NO}}+2_{\mathrm{pyrazolate}\mathrm{N}}+1_{\mu \mathrm{CO}}+1_{\mathrm{Fe}-\mathrm{Fe}}=18$ electrons. The negative formal oxidation state follows from the chosen nitrosyl assignment and the overall charge; omitting that charge would change the count.

Bent terminal NO ligands would supply fewer neutral-count electrons and demand much higher formal M–M bond orders to attain 18-electron counts. The linear assignments give the ordinary double and single bond solutions. Since NO is a non-innocent ligand, these oxidation states are formal bookkeeping limits, not measured metal charges; actual nitrosyl bending and M–M bond strengths require structural and spectroscopic evidence.^[ref]

Vanadium is the chemical element with atomic number 23 and symbol $\mathrm{V}$. It is a transition metal. It supports several oxidation states, producing differently colored ions in solution. Vanadium-containing alloys strengthen steel, while vanadium oxides participate in oxidation catalysis; the same element therefore plays distinct structural and chemical roles.^[ref]

Vanadium(II) oxide is the monoxide conventionally written $\mathrm{VO}$. The name assigns formal oxidation state +2 to vanadium in the ideal one-to-one composition.

Real vanadium monoxide can be nonstoichiometric, with substantial lattice vacancies and composition-dependent ordering. Diffraction studies show that the distribution of occupied and vacant sites requires more information than a single ideal formula.^[ref]