# Is cobalt (III) hexaaqua paramagnetic or diamagnetic?

The complex ion $$\ce{[Co(H2O)6]^3+}$$ has $$\ce{Co}$$ in the $$+3$$ oxidation state, meaning it has an electron configuration of $$[Ar] 4s^0 3d^6$$. Therefore it has 4 unpaired electrons and would be paramagnetic.

However, in the octahedral complex ion, the d orbitals split into two levels, with three lower-energy orbitals and two higher-energy ones. The six d electrons would therefore be in the lower set, and all paired. Following this logic, the $$\ce{Co}$$ atom would be diamagnetic.

Which is correct?

Quoting Housecroft and Sharpe "Inorganic Chemistry" (second edition): "The blue, low-spin $$\ce{[Co(H2O)6]^3+}$$ ion can be prepared in situ by ...".
Greenwood and Earnshaw says about $$\ce {Co}^{3+}$$ complexes "these are virtually all low-spin and octahedral" and "Even $$\ce{[Co(H2O)6]^3+}$$ is low spin".
Thus $$\ce{[Co(H2O)6]^3+}$$ is $$d^6$$, octahedral and low spin. It is thus diamagnetic.