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Put $\mathbf{J}={\mathbf{S}}_1+{\mathbf{S}}_2$. Angular momentum addition gives $j=0,1,2,3$, and the spin-dot-product eigenvalue isThus the spin-three-halves pair spectrum isFor identical fermions, exchange parity of two coupled identical spins makes $j=0,2$ spin-antisymmetric and $j=1,3$ spin-symmetric. Corresponding spatial exchange symmetries make the full state antisymmetric; without a specified spatial state, none of these total-spin sectors is excluded.

Adopt $\mathcal{H}=-2J({\mathbf{S}}_1+{\mathbf{S}}_2)\cdot \mathbf{s}-2J^{\prime }{\mathbf{S}}_1\cdot {\mathbf{S}}_2$, with $S_1=S_2=1$ and ligand spin $s=1/2$. Let ${\mathbf{S}}_{12}={\mathbf{S}}_1+{\mathbf{S}}_2$ and $\mathbf{S}={\mathbf{S}}_{12}+\mathbf{s}$. The angular momentum addition identities giveThe five allowed multiplets areTheir degeneracies $2,2,4,4,6$ sum to 18, matching $3\times 3\times 2$. For dominant ferromagnetic $J$, the ground multiplet has $S_{12}=2$, $S=5/2$. A weak antiferromagnetic $J^{\prime }$ does not overcome the two strong metal–radical bonds.

For a single-photon transition, angular momentum addition requires $|J_i-J_f|\le L\le J_i+J_f$, $L\ge 1$, together with the electric/magnetic parity rules above. Thus a known final state and a measured multipole restrict both the spin difference and parity of the emitting level. For example, an E1 line reverses parity and allows $\Delta J=0,1$ except $0\to 0$; a mixed M1+E2 line preserves parity and must satisfy both angular triangles. An E2 line preserves parity and allows spin differences up to two.

Line energies identify connected levels, while calibrated intensities and coincidences test cascade flow and branching. If a proposed assignment permits a fast low-order transition to a lower state, an experimentally stringent absence of that line argues against the assignment. Nevertheless, an allowed transition can have a small or vanishing nuclear matrix element. Pure E2 radiation alone does not prove that M1 is forbidden by angular momentum. Detection sensitivity, efficiency and overlap must be considered, and in real nuclei internal conversion can remove photons; this question explicitly excludes that competing channel. Lifetimes or reduced-strength estimates make a missing-line argument substantially stronger than a bare selection-rule statement.

The spin Hilbert space is a tensor product of two two-dimensional spaces and therefore has dimension four. Angular momentum addition decomposes it asThese are three orthogonal triplet states and one singlet state. For identical electrons the complete wavefunction must also satisfy fermionic exchange symmetry: a symmetric triplet spin factor accompanies an antisymmetric spatial factor, whereas the antisymmetric singlet spin factor accompanies a symmetric spatial factor. If both electrons occupy the same spatial orbital, only the singlet spin factor is allowed. The four-dimensional spin space does not assert four allowed spin states for that particular spatial occupation.

For $s_1=s_2=3/2$, angular momentum addition gives one copy of $j=0,1,2,3$. The identity $2{\mathbf{S}}_1\cdot {\mathbf{S}}_2={\mathbf{J}}^2-{\mathbf{S}}_1^2-{\mathbf{S}}_2^2$ gives the four eigenvalues in the title, with state degeneracy $1,3,5,7$. These exhaust the $4\times 4$ spin space. If the particles are identical fermions, an appropriate spatial exchange parity accommodates each sector; only a specified spatial restriction can remove sectors.