Updated on August 6, 2022
Thank you. I really am lost with this.
SF4 – you might think that this is sp3d, but its not. There is little to no d-orbital involvement. Yes, I know that the majority of the freshman chem textbooks out there talk about sp3d and sp3d2 for trigonal bipyramidal and octahedral electron pair geometries, but the most recent work indicates that that isn’t the case. In the most recent edition of Brown, LeMay and Bursten they explain that sp3d and sp3d2 hybridizations do not exist.
So what do we have? In SF4, which according to VSEPR theory has trigonal pyramidal electron pair geometry and “see-saw” molecular geometry, we find sp3 hybridization for the electrons in the equatorial position, and 3-center-4-electron bonding for the F-S-F bonds in the axial position.
In the case of PF6^- there is no need for any hybridization to explain the geometry. Simply three sets of 3-center-4-electron bonds using p-orbitals will work nicely.
Check out this paper on three-center-four-electron (3C4E) bonding and molecular geometry and hybridization. http://www.springerlink.com/content/x9585476848636…
Hybridization is a way to explain e- pair geometry
1. HCN = sp, linear
2. SO32- = sp3, tetrahedral
3. SF4 = sp3 and 3C4E bonding
4. PF6- = no hybridization, 3C4E bonding
5. NH3Cl+ = sp3, tetrahedral
6. N3- = sp, linear
For those of you who are doing this question on mastering chemistry
the answer is: not applicalSource(s): Mastering Chemistry