= Solution
{c}
The given <electron configuration> has an inert closed $[\mathrm{Xe}]$ core and two $4f$ <electrons>. The <Aufbau principle> identifies the occupied subshells; the free-<ion> term within the open $4f^2$ configuration is then selected by <Hund's rules>. A closed core contributes $L=S=0$.
Each $f$ <electron> has orbital <angular momentum quantum number> $l=3$. Place the two <spins> parallel, so $S=1$. The <Pauli exclusion principle> then requires distinct orbital projections. The largest allowed total projection is $M_L=3+2=5$, obtained from $m_l=3$ and $m_l=2$, giving the maximum allowed $L=5$. A larger $M_L=6$ would place two parallel-<spin> <electrons> in the same <atomic orbital> and is forbidden.
The $f$ subshell holds fourteen <electrons>; $f^2$ is less than half filled. The third of <Hund's rules> therefore selects $J=L-S=4$. Hence
$$
\boxed{L=5,\qquad S=1,\qquad J=4,\qquad \text{ground term }{}^3H_4.}
$$
The <term symbol> uses $2S+1=3$ and the spectroscopic letter $H$ for $L=5$. The <praseodymium> <ion> retains an orbital contribution, so a spin-only estimate would be inappropriate.
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