 Solution

The $4f^7$ shell carries a large spin (its ground atomic term under Hund's rules has $L=0,S=7/2$), and the $6s$ electron penetrates the inner shells. Its nonzero probability density at the atomic nucleus supplies a potentially large Fermi contact interaction with the nuclear magnetic dipole moment. The $6s$ shell here is singly occupied, so the obligatory paired-$6s^2$ cancellation is absent. Both stable europium isotopes have nonzero nuclear spin, as the spectrum below also establishes.

The term symbol ${}^8{D}_{11/2}$ specifies $S=7/2$, $L=2$, $J=11/2$. Its large $J$, combined with appreciable nuclear spin, permits a wide range of total angular momentum $F$ and hence a substantial total hyperfine structure span. For $I=5/2$, $F=3,\dots ,8$ and the span is $30|a|$. The penetrating $6s$ contribution and the available range of $F$ explain the large hyperfine structure; the term symbol alone does not determine its constant, which also depends on electronic recoupling, radial wavefunctions and configuration interaction.