= Solution
For <chromium> the configuration is $[\mathrm{Ar}]3d^54s^1$. 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 \b[$\mathrm{Cr}:{}^7S_3$]. For <cobalt>, $[\mathrm{Ar}]3d^74s^2$ 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 \b[$\mathrm{Co}:{}^4F_{9/2}$]. For <copper>, $[\mathrm{Ar}]3d^{10}4s^1$ has a closed $d$ subshell, leaving $L=0$, $S=1/2$ and \b[$\mathrm{Cu}:{}^2S_{1/2}$]. These are neutral-atom terms, rather than the terms of the common metal ions. The tabulated zero-energy levels independently confirm all three assignments.<Energy levels of neutral chromium in the NIST Handbook>[Cr I table, ground row 3d5(6S)4s 7S3]{ref}<Handbook of Basic Atomic Spectroscopic Data by Jean E. Sansonetti and William C. Martin/Energy Levels of Neutral Cobalt (Co I) by National Institute of Standards and Technology>[Co I table, ground row 3d7 4s2 4F9/2]{ref}<Handbook of Basic Atomic Spectroscopic Data by Jean E. Sansonetti and William C. Martin/Energy Levels of Neutral Copper (Cu I) by National Institute of Standards and Technology>[Cu I table, ground row 3d10(1S)4s 2S1/2]{ref}
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