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Mar 24, 2023 at 13:27 comment added Proscionexium I was saying in the light of Valence Bond Theory. But now I realise that is not the center of the discussion.
Mar 24, 2023 at 13:08 comment added orthocresol @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.
Mar 24, 2023 at 3:29 comment added Proscionexium 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.
Sep 8, 2020 at 14:25 comment added Nilay Ghosh Update: a low spin cobalt(II) complex with a distorted tetrahedral geometry. You can see the complex here. Another example is $\ce{Cr[N(SiMe3)2]3[NO]}$
May 10, 2019 at 6:28 comment added Greg CFT contains zero reference to hybridization. It is ridiculous to force into it.
May 9, 2019 at 11:22 comment added orthocresol 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.
May 9, 2019 at 11:10 comment added evamPUNdit It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..
May 9, 2019 at 11:09 vote accept evamPUNdit
May 9, 2019 at 11:05 history answered orthocresol CC BY-SA 4.0