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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Charkol on October 01, 2010, 10:24:37 PM

Title: What is the orbital hybridization of SF[sub]6[/sub]?
Post by: Charkol on October 01, 2010, 10:24:37 PM

What is the orbital hybridization of SF₆? 

I understand SF₆ is octahedral, but I am not sure how it is able to form 6 bonds.  My natural assumption is that it is hybridized by spppdd, but I am told by other sources that there is absolutely no d orbitals on this energy level.

This goes against what I thought about Be, it has 3 p orbitals, but they don't get filled unless it is hybridized I believe, e.g. BeCl₂.

Please help me demystify this dilemma.

Title: Re: What is the orbital hybridization of SF[sub]6[/sub]?
Post by: ooosh on October 02, 2010, 01:01:41 AM

Yes,you are right.In SF6 the S atom is sp3d2 hybridized.

Title: Re: What is the orbital hybridization of SF[sub]6[/sub]?
Post by: Charkol on October 05, 2010, 09:47:12 AM

Thanks, it seems that an extraordinary amount of energy would be required for spppdd hybridation.

So, could the electron be written as this?
1s²
2s²
2p³
3s²
2d⁰
3p⁴

I understand this is not 'standard' notation.  But it might be more 'accurate'?

Title: Re: What is the orbital hybridization of SF[sub]6[/sub]?
Post by: ooosh on October 05, 2010, 11:30:24 AM

The inner shell's eletrons don't involve in the bond formation ,only the outermost valence electrons.And you got a mistake, there's no 2d orbital.  The right electron arrangement of S atom is 1s²2s²2p⁶3s²3p⁴,and S atom have 6 valence eletrons in 3s3p.There's no eletron in 3d before hybridization.

Title: Re: What is the orbital hybridization of SF[sub]6[/sub]?
Post by: Charkol on October 05, 2010, 02:25:57 PM

Okay, so there really is no 2d subshell.

But in electron configurations of elements beyond potassium, there is a 4s subshell which gets filled before the 3d subshell gets filled.

Could it be possible to write the electon configuration for sulfur as this, then:

1s2
2s2
2p6
3s2
3p4
3d0
...even though the 3d0 is redundant (i.e. there are no electrons in it), would it still be accurate to write this electron configuration.  To me this acknowledges the existance of the 3d subshell, which I think I was taking for granted.