Timeline for Why are low spin tetrahedral complexes so rare?
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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 |