Magnetic moments of tetrahedral Cobalt (II) and (III) complexes

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.

The question isn't really clear exactly what complexes you are referring to. However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin while the 3.87 value is for high spin (3/2). I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar?