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$\ce{Cu}^{2+}$ 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?

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  • $\begingroup$ 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? $\endgroup$ – Philipp Dec 6 '14 at 12:20
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There are not two possibilities for low spin and high spin complex for a d9 metal ion. The only cases where you can have both is for d4, d5, d6 and d7.

Remember that both configurations arise from the possibility of organizing the d electrons in a more stable way, this is, with the lowest possible energy. You need extra energy for doing two things:

  • Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, $\Delta_O$
  • Puting another electron on a semioccupied orbital (pairing energy, due to repulsions)

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.

Now d8 d9 and d10. There is once again, only one possibility to put the electrons. For your case, d9, you have nine electrons, so 6 of then will be filling the lower energy orbitals for sure, and the other 3, on the upper orbitals. You may think, well, there is another possibility, promote one of the lower energy electrons to the hole on the higher energy orbital. But this is again nonsense: you are looking for the most energy favorable situation, so, as long as you only have to add to the system the pairing extra energy, why would you also add the promoting energy instead of filling a semioccupied lower energy orbital? This would result in a excited state of the complex, but this is not what you're looking for when you talk about high/low spin. The same applies to d8 and d10, there is only one possible situation, and no high/low spin possibilities.

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