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As I was going through Concise Inorganic Chemistry by J. D. Lee, I realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex? Usually low-spin complexes are in $\mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $\mathrm{sp^3}$.

orthocresol
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evamPUNdit
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1 Answers1

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Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.

The reason why low-spin $T_\mathrm d$ complexes are rare is because the splitting parameter, $\Delta_t$, is significantly smaller than the corresponding octahedral parameter $\Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)

$$\Delta_t = \frac{4}{9}\Delta_o$$

For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.

Since the splitting $\Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $\mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $\mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.

orthocresol
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    It's just that we just went through the basics of CFT. I can't think without using hybridisation right now.. – evamPUNdit May 09 '19 at 11:10
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    I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too. – orthocresol May 09 '19 at 11:22
  • CFT contains zero reference to hybridization. It is ridiculous to force into it. – Greg May 10 '19 at 06:28
  • Update: a low spin cobalt(II) complex with a distorted tetrahedral geometry. You can see the complex here. Another example is $\ce{Cr[N(SiMe3)2]3[NO]}$ – Nilay Ghosh Sep 08 '20 at 14:25
  • There are very rare exceptions to this rule. Only very strong ligands can form low spin complexes keeping the shape still tetrahedral. For example, Ni(CO)4 is diamagnetic in which 2 electrons of 4s are demoted to pair up with the 2 unpaired electrons of 3d. – Proscionexium Mar 24 '23 at 03:29
  • @Priyansh2003 Not sure what you're getting at here. Ni(0) is $d^{10}$, so the terms high- or low-spin are meaningless. Those terms aren't to do with distribution of electrons in 4s/3d, it's about the distribution of electrons in $e_g$/$t_{2g}$ or $e$/$t_2$ orbitals, and for a $d^{10}$ complex both the $e$ and $t_2$ orbitals are fully filled, so there's no distribution to talk of. – orthocresol Mar 24 '23 at 13:08
  • I was saying in the light of Valence Bond Theory. But now I realise that is not the center of the discussion. – Proscionexium Mar 24 '23 at 13:27