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The zinc ion in its +2 state can show no more pairing, since all the electrons are already paired - so in any complex it forms it will have the same magnetic moment as it does right now. $\ce{Ti}$ in its +3 state will show both outer and inner orbital complexes without any electrons pairing.

So while calling these complexes high spin or low spin complexes is not useful, since there is no possibility of pairing involved in either - but all the same, won't they be called high spin complexes, since no lowering of magnetic moment occurs?

My teacher said, however, that since $\ce{Zn}$ in +2 state forms only outer orbital complexes, its complexes will be called high spin - and $\ce{Ti}$ in its +2 state similarly when it forms inner orbital complexes (which it does without pairing), will be called low spin because the complex is inner orbital. But shouldn't such a complex of $\ce{Ti}$ be a high spin complex? (I'm saying this in reference to VBT).

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Are you familiar with ligand field stabilisation energy? This approach answers your question. Interaction of the central metal ion with the surrounding ligands causes splitting of the $d$ orbitals into an essentially upper and lower energy level, the form of the splitting is rather complex and depends directly upon the symmetry point group that the ligand complex belongs to.

Either way, high spin complexes only form if the reduction in inter electron repulsion energy is greater than the energy splitting of the orbitals, with the definition of high spin being a none aufbau filling of the split orbitals. It does not make sense to describe the electronic structure of a $\ce{Zn^2+}$ ion in a complex as high or low spin, there is only one configuration available as there are 10 $d$ electrons present, and the 5 $d$ orbitals split to form 5 new orbitals of varying degeneracies (dependent upon the point group of the complex and the symmetry of the coordination environment).

$\ce{Ti^2+}$ has 2 valence d electrons available to be involved in bonding with the ligand environment, however there is no coordination environment that I am aware of (The splitting would have to be extremely small and other thermodynamic factors such as the ionic contribution to the Gibbs energy of complex formation would mean that the complex would never be observed anyway - such a low splitting would only be the result of very poor overlap which would be caused by large separation between the central ion and the ligands, resulting in a lower ionic term) that causes splitting where the magnitude of the splitting is less than the repulsion term invoked when both electrons occupy the same molecular orbital, be it degenerate or not.

The $\ce{Ti^3+}$ ion has only 1 $d$ electron, it will definitely reside in the lowest available molecular orbital. However, the first excited states, ignoring hyperfine splitting, of both titanium ions will be high spin - the magnitude of the splitting usually corresponds to wavelengths of light in the visible region of the spectrum, hence many transition metal complexes are coloured.

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  • $\begingroup$ I didn't understand why you said Ti forms high spin complexes in +3 state. Are you talking about the splitting according to the Crystal Field Theory? I think the d electron will go to the lowest orbital - though this really isn't non aufbau filling, since no pairing occurs - I think it should be high spin. High spin means complexes in which no pairing occurs and magnetic moment of the metal ion doesn't reduce on complex formation - am I right? This is what I learnt. $\endgroup$ – Charles Apr 26 '15 at 7:36

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