From the spin-only formula we can predict that for tetrahedral cobalt (II) complexes $$m_{eff} = 3.87 \mu_B $$
This ignores orbital angular momentum effects, which result in higher magnetic moments for the tetrahedral halide complexes (Hund's 3rd rule results in spin-orbit coupling "together"). Why do the Co(II) cyanide and Co (III) alkyl complexes have lower magnetic moments than the spin only formula predict ? I have quoted values of 2.15 $\mu_B$ and ~3 $\mu_B$ for these cases.