= Solution
<Hund's rules> maximizes total <spin> within the partially filled shell: parallel <spins> require an antisymmetric spatial wavefunction, reducing Coulomb overlap through exchange. Among terms with that <spin>, the second rule selects the largest total <orbital angular momentum>. The third rule, from <spin-orbit coupling> within the selected term, gives $J=|L-S|$ below half filling and $J=L+S$ above half filling, with the half-filled case treated directly.
For seven $4f$ electrons, put one parallel-spin electron in each $m_l=-3,-2,-1,0,1,2,3$ orbital. Their orbital projections sum to zero and form the half-filled $L=0$ shell; all seven <spins> add. Thus
$$
\boxed{S=\frac72,\qquad L=0,\qquad J=\frac72,\qquad{}^{2S+1}L_J={}^8S_{7/2}.}
$$
There is no ambiguity from applying the below- or above-half-filled formula because $L=0$. The absence of a first-order orbital moment is important for the weak <magnetic anisotropy> discussed in <3/c>. These are isolated-ion term rules; a crystal field or strong covalency can alter other ions' orbital behavior, but the half-filled Eu$^{2+}$ shell is unusually robust.
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