Use Hund's rule to derive the electronic configuration of $Ce^{3+}$ ion and calculate its magnetic moment on the basis of 'spin-only' formula.

(D) The atomic number of $Ce$ is $58$. The electronic configuration of $Ce$ is $[Xe] 4f^{1} 5d^{1} 6s^{2}$.
To form $Ce^{3+}$,three electrons are removed (two from $6s$ and one from $5d$ or $4f$ depending on the energy levels; for $Ce^{3+}$,the configuration is $[Xe] 4f^{1}$).
Thus,the electronic configuration of $Ce^{3+}$ is $[Xe] 4f^{1}$.
There is $n = 1$ unpaired electron in the $4f$ orbital.
The 'spin-only' magnetic moment formula is $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n = 1$:
$\mu = \sqrt{1(1+2)} = \sqrt{3} = 1.732 \ BM$.