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Cu+2 ion has total of 9 electrons in its 3d orbital. When it undergoes d-d splitting an electronic transition can take place between t2 and e as in {Cu(CN)4}-2 complex because of the presence of one unpaired electron and it will contribute to its colour of the complex. Ti+3 has also only one unpaired electron in its d orbital so it can also undergo transition when d-d splitting occurs.

My question is that why would Ti+3 undergo d-d splitting in the first place when it has only one electron in only one of its degenerate d orbital. I mean d-d splitting occurs because of the repulsion between electrons of the approaching ligand and electrons in the d orbital and here only one of the d orbitals has one electron other degenerate d orbitals are empty so why would d-d splitting occur in the first place?? I will be very thankful if someone answers this.

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  • $\begingroup$ Is the Ti(III) in the gas phase? If not, what are the solvent/ligand/counterion? $\endgroup$
    – Andrew
    Feb 19 at 12:57

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With only one $d$ orbital occupied by an electron in an ocathedral Ti(III) complex the remaining $d$ orbitals are virtual states only, so to speak of splitting between them is not really proper. But we can say that because of the crystal field or ligand field effect there is some distinguishing of $d$ orbitals into which the electron may be placed. We can tell the $t_{2g}$ states from the $e_g$ states by their different energy levels; when the electron is in a $t_{2g}$ state the complex is more stable than would be predicted from a model based on spherically symmetric charge or electron distributions.

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