= Solution
The open shell contains one $f$ <electron>, so $L=3$ and $S=1/2$. For a less-than-half-filled shell, <Hund's rules> select the smaller total angular <momentum>, $J=L-S=5/2$, giving the free-ion ground level $ {}^2F_{5/2}$. The <Landé g-factor> is
$$
g_J=\frac32+\frac{(1/2)(3/2)-3(4)}{2(5/2)(7/2)}
=\boxed{\frac67}.
$$
In a <magnetic field> weak compared with the spin-orbit splitting, the electronic Zeeman shifts are $\delta E_{m_J}=g_J\mu_BBm_J$, for $m_J=-5/2,-3/2,\ldots,5/2$. Thus \b[six equally spaced magnetic substates] appear. At $B=1\,\mathrm T$, the adjacent-level spacing is
$$
\boxed{\Delta E_{\rm adjacent}=\frac67\mu_B B
=7.95\times10^{-24}\,\mathrm J
=4.96\times10^{-5}\,\mathrm{eV}.}
$$
The complete lowest-to-highest multiplet spread is five spacings,
$$
\boxed{\Delta E_{\rm full}=5\Delta E_{\rm adjacent}
=3.97\times10^{-23}\,\mathrm J
=2.48\times10^{-4}\,\mathrm{eV}.}
$$
Giving both quantities removes the ambiguity in the word splitting. This is the free-ion, weak-field result; no crystal-field doublet or <hyperfine structure> is specified.
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