# Cu2+ octahedral complexes

$\ce{Cu}^{2+}$ has nine d-electrons, regardless of the ligand field strength it will have one free electron (so it is paramagnetic. Are $\ce{Cu}^{2+}$ octahedral complexes high spin or low spin?

• I dont really see what the question whether $\ce{Cu^{2+}}$ complexes are high spin or low spin has to do with the fact that $\ce{Cu^{2+}}$ will have at least one unpaired d electron. Could you explain why you think this is important? – Philipp Dec 6 '14 at 12:20

• Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, $\Delta_O$
In the complexes where $\Delta_O$ is higher than the pairing energy, the result will be a high spin configuration (it's easier for the electrons to become promoted to a higher level than pairing them with another electron), and if it is smaller, the result will be the opposite. But this only applies for d4, d5, d7 and d8.
Now think about d1 d2 and d3: It's nonsense to put the electrons on the upper level, when they can be all by their own in lower energy orbitals, so there are no high spin/low spin possibilities, but only one possible case, the three electrons one on each $T_2g$ orbital.