 Solution

For chromium the configuration is $[\mathrm{Ar}]3d^{5}4s^1$. The half-filled $3d$ subshell has $S_d=5/2$ and $L_d=0$; aligning the $4s$ spin gives $S=3$, while $L=0$. Therefore the ground-state term symbol is $\mathrm{Cr}:{}^7{S}_3$. For cobalt, $[\mathrm{Ar}]3d^{7}4s^2$ has a closed $4s^2$ pair. The $d^7$ subshell has the same possible $L,S$ values as three $d$ holes; maximizing spin and then orbital angular momentum gives $S=3/2$, $L=3$. Because it is more than half full, $J=3+3/2=9/2$, giving $\mathrm{Co}:{}^4{F}_{9/2}$. For copper, $[\mathrm{Ar}]3d^{10}4s^1$ has a closed $d$ subshell, leaving $L=0$, $S=1/2$ and $\mathrm{Cu}:{}^2{S}_{1/2}$. These are neutral-atom terms, rather than the terms of the common metal ions. The tabulated zero-energy levels independently confirm all three assignments.^{[ref][ref][ref]}