Hybridization Question

[CJhooper123]

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Hi all. Quick question from my TPR FL.

What is the hybridization of XeF4?

Answer is sp3d2, but why? I am pretty comfortable with sp, sp2, and sp3.... when d orbitals get involved it gets a little hairy.

 

[aldol16]

Draw out the Lewis structure of XeF4. You'll find that it has four fluoride ligands bound to it, along with two lone pairs. That gives a total of six"things" around the Xe center, thereby requiring six hybridized orbitals. Start with s, p, and then d. So the s hybridizes, along with the 3 p orbitals, giving you four hybrid orbitals. You need two more, so you take 2 d orbitals as well to yield a total of 6 hybrid orbitals.

 

[CJhooper123]

Draw out the Lewis structure of XeF4. You'll find that it has four fluoride ligands bound to it, along with two lone pairs. That gives a total of six"things" around the Xe center, thereby requiring six hybridized orbitals. Start with s, p, and then d. So the s hybridizes, along with the 3 p orbitals, giving you four hybrid orbitals. You need two more, so you take 2 d orbitals as well to yield a total of 6 hybrid orbitals.

Thank you. I guess the noble gas configuration of Xenon tripped me up. Didn't consider it's expanded role in the octet rule.


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[aldol16]

Yes, always remember your procedures for drawing Lewis structures. You know that Xenon has 8 valence electrons and fluorine has 7, so there are a total of 36 electrons in this system. The only scenario where you can distribute all 36 electrons is having two lone pairs on the Xe center.