It seems rather arbitrary to me that we have decided that nitrogen sp3 hybridizes in ammonia to stay consistent with VSEPR, yet that fluorine uses its 2p orbital to bond with another fluorine in F2 (I.e. it does not hybridize). I understand that sp3 hybridization is associated with the tetrahedral shape, but what is to say that F2 does not undergo hybridization? Both of the fluorine atoms technically have a steric number of four. F2 will inherently have a linear geometry since it is a diatomic molecule. I do not understand why the lone pair on nitrogen in ammonia is equal in energy to the bonding electrons, yet one of the three lone pairs on each of the fluorines exists at a lower energy (2S orbital). How do we know that the orbitals do not hybridize in this case? And why not?

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    $\begingroup$ Hybridization occurs in our heads. $\endgroup$ Mar 27 at 7:48
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    $\begingroup$ And our heads suppose hybridization occurs when there are appropriate bond angles, which do not exist with only two atoms. $\endgroup$ Mar 27 at 10:03

Whether you say if $\ce{F2}$ hybridises or not you reach the same answer - the molecule is linear.

As indicated in the comments, hybridisation isn't an accurate reflection of how bonding actually works, however it is a good enough approximation to determine bond angles for simple molecules like ammonia.

Note: to add some detail on your comment of the lone pair energies in these molecules. the lone pairs in $\ce{F2}$ are higher in energy than the bonding electrons in the molecule. This is a failure of VSEPR theory. The lone pair also being higher in energy than bonding electrons in NH3 is also true.


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