Can someone explain hybridization of elements?

[AFB]

Like Sulfur Hexafluoride (SF6)? I know it involves the d orbitals , but why doesn't it do that when it's normally bonding? Why does it sometimes form Sulfur Tetrafluoride and sometimes Sulfur Hexafluoride?

 

[AFB]

I know we have chemists here.

 

[CycloWizard]

Let me know if this makes any sense. If not, I'll try to explain it another way.

To determine whether or not hybridization will occur, you first need to examine whether the bonding elements have any unpaired orbitals. If it has no unpaired electrons, then it must hybridize to allow bonding. For example, if you want to form BeF2, Be has no unpaired electrons (both 1s and 2s orbitals contain two electrons each). So, to bond, it must bump an electron up to the next orbital (2p) from the 2s to make space. This creates two unpaired electrons - each may accept the lone electron from one fluorine atom. The electrons are accepted as one electron into the 2p orbital and one into the 2s orbital. Thus, the resulting hybrid orbitals are called 'sp hybrid orbitals'. The Be now has 1s and sp orbitals, where the sp orbital is a hybridization of the 2s and 2p orbitals.

If you need more, let me know. Don't want to go into too much detail. I'm an engineer, not a chemist. 😛

 

[CycloWizard]

More info:

To extend this to higher orbitals (sp3, d, and so on), determine the number of shared atoms in the resultant molecule. For example, SF4 results in sp3d hybridization (five sp3d orbitals) because four electrons are shared between the four fluorine and the one sulfur.

The easiest way to do it I guess is to write down how many electrons are in each orbital. You know that each shell of each orbital can have two electrons. In the p orbital, each electron will be single until all three suborbitals have at least one electron. So, for example, SF4 has 2 e- in the 3s, then one in each suborbital of the 3p orbital. So, to accept four electrons, it is clear that one electron must be promoted to the d orbital. This promoted electron must come from a pair, so it is promoted from 3s. It can now accept all four electrons from the four fluorine atoms, and results in sp3d hybridization.

For SF6, same thing only you have to promote one additional electron, so you end up with sp3d2 hybridization.

 

[AFB]

Originally posted by: CycloWizard
More info:

To extend this to higher orbitals (sp3, d, and so on), determine the number of shared atoms in the resultant molecule. For example, SF4 results in sp3d hybridization (five sp3d orbitals) because four electrons are shared between the four fluorine and the one sulfur.

The easiest way to do it I guess is to write down how many electrons are in each orbital. You know that each shell of each orbital can have two electrons. In the p orbital, each electron will be single until all three suborbitals have at least one electron. So, for example, SF4 has 2 e- in the 3s, then one in each suborbital of the 3p orbital. So, to accept four electrons, it is clear that one electron must be promoted to the d orbital. This promoted electron must come from a pair, so it is promoted from 3s. It can now accept all four electrons from the four fluorine atoms, and results in sp3d hybridization.

For SF6, same thing only you have to promote one additional electron, so you end up with sp3d2 hybridization.

Thanks 😀