hybridization?

Question

Identify the hybridization of the central atom for both





XeF4 and PO4-3

Answer 1

Xe is 5s2-5p6 when it does not bond (2 + 6 = 8)
since it needs to bond 4 times it will end up having 2 pairs of non-bonding electrons. 4+ 8 = 12.
(↑↓)(↑↓)(↑↓)(↑↓)()()()()()
5s2///------5p6------/// 5d
You promote
(↑↓)-(↑↓)(↑↓)(↑)-(↑)(↑)()()()
5s2///------5p6------/// 5d
Therefore, you need 2 of the 5 d orbitals. You must hybridize to get all identical equivalent orbitals
(↑↓)(↑↓)(↑↓)(↑)(↑)(↑)
-----------sp3d2-------------
XeF4 has a plane square geometry (both non-bonding pairs of electrons on the axial positions (z axis)), Fs in the equatorial axis (N & S in y axis and W & E in the x axis ) and Xe in the middle of everything.
For PO4- (this is a special case)
You have 32 electrons in total. So, you just need 3 simple bonds and 1 double bond (resonance), having P 10 electrons instead of 8, and the hybridization is:
Basal P (3s2-3p3):
(↑↓)-(↑)(↑)(↑)
3s2///---3p3---
You just need to promote 1 electron
(↑)-(↑)(↑)(↑)-(↑)()()()()
5s///----3p---/// 5d
and hybridize
(↑)(↑)(↑)(↑)(↑)
------sp3d--------
Nevertheless, this is a compound with a tethraedral geometry, and they are only sp3. Recent research has found that the more suiting structure is the one that has 4 simple bonds and a P with an octet instead of 10 electrons, despite the fact that it would give higher formal charges for its atoms (-1,-1,-1,-1 (for O) and +1 for P against -1,-1,-1 and 0 for O and 0 for P). Therefore, the hybridization is sp3. Besides, a hybridization of sp3d would mean that P has 1 non-bonding pair of electrons, and that is false, so:
First, you promote from 3s to 3p
(↑)-(↑↓)(↑)(↑)
3s///---3p4---
And then you promote from 3p to 3d
(↑)-(↑)(↑)(↑)-(↑)()()()()
3s1///---3p3-//----3d1---
Finally, you hybridize, leaving the 3d electron you do not need apart (this electron is "given" to O in 1 coordinated bond. Remember, in this ion, P works with 5+ valence)
(↑)(↑)(↑)(↑)-(↑)()()()()
-----sp3------//-3d1---

Answer 2

The Xe is in d2sp3
The P is in sp3