 Solution

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 $J=L-S=4$. HenceThe 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.