= Solution
The closed argon core contributes $L=S=0$. In the half-filled $3d^5$ shell, <Hund's rules> put one parallel-spin electron in each $m_\ell=-2,-1,0,1,2$ orbital. This gives $S_d=5/2$ and the orbital term $L_d=0$. Adding the parallel $4s$ electron, with $\ell=0$ and $s=1/2$, gives
$$
\boxed{S=3,\qquad L=0,\qquad J=3,\qquad {}^{2S+1}L_J={}^7S_3.}
$$
The <spin multiplicity> is seven. The magnetic projection labels run $M_S,M_J=-3,-2,\ldots,3$, while $M_L=0$. The <term symbol> agrees with the observed neutral-<chromium> ground level.<Energy levels of neutral chromium in the NIST Handbook>[Cr I, ground level a7S3]{ref}
Back to article page
