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?


1 Answer 1


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
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.


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