= Solution
{c}
With $J=4$, the measured <effective paramagnetic moment> gives the <Landé g-factor>
$$
\widetilde g_{\rm exp}=\frac{m_{\rm eff}}{\sqrt{4(4+1)}}
\simeq\frac{3.5\text{–}3.6}{\sqrt{20}}\simeq0.78\text{–}0.81.
$$
For the free-<ion> <Hund's rules> values $S=1$, $L=5$, $J=4$, the stated formula yields
$$
\widetilde g_{\rm free}=\frac32+\frac{2-30}{2\times20}
=\frac32-\frac7{10}=\frac45.
$$
Thus
$$
\boxed{\widetilde g_{\rm exp}\simeq0.8\simeq\widetilde g_{\rm free},\qquad
m_{\rm eff}^{\rm free}=\frac45\sqrt{20}=3.578.}
$$
The source fit is consistent with the unquenched $4f$ orbital <magnetic moment> of the free <praseodymium> <ion>, within graphical accuracy. Small deviations need not indicate a change of <electron configuration>: <crystal field splitting>, interactions and the fit range can affect a measured <magnetic susceptibility>.
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