Shapes of XeF2, XeF4, XeF6 and XeO3 by VSEPR, and why only xenon forms compounds easily

I don't post much, but this one has been bothering me for days. The noble gas chapter. We were told noble gases are "inert", and then straight away we're told xenon forms fluorides and oxides. Why xenon and not helium or neon? I read the ionisation enthalpies: He 2372, Ne 2080, Ar 1520, Kr 1351, Xe 1170 kJ/mol, so it's decreasing, but is that the only reason?

The bigger problem is shapes. I tried VSEPR for $XeF_4$ : Xe has 8 valence electrons, 4 are used for 4 bonds, which leaves 4 electrons = 2 lone pairs. So 6 electron pairs. In my head that's a "tetrahedral", but my notes say square planar. How? And I tried $XeF_2$ the same way and got linear, which matched, but I don't really know why.

Can someone give me the shape and hybridisation of $XeF_2$ , $XeF_4$ , $XeF_6$ , $XeO_3$ and $XeOF_4$ , with the reasoning, not just the list? The conditions for making the fluorides are in NCERT too (temperature, pressure, ratio), but I can't hold them in my head.

7 answers & replies

[+]Ishaan Bose11thSample· 13d ago

XeF4: Xe has 4 bonds, so 4 bond pairs, like methane, tetrahedral, sp3. Done. Chemistry people overcomplicate everything.

[+][–]Nisha Verma12thSample· 13d ago
Ishaan, 4 bond pairs isn't the full story, the lone pairs count for the geometry. Lakshmi's counting was right: 4 bond pairs + 2 lone pairs = 6 electron pairs. Six pairs arrange in an octahedron, so hybridisation is $sp^3d^2$ . The two lone pairs sit on opposite corners (trans), because that minimises their repulsion. Remove those two positions and what you see is a flat square: square planar. The molecular shape (atoms only) is different from the electron-pair geometry, that's the trick.
- [+][–]Ishaan Bose11thSample· 13d ago
  ok fine that actually makes sense, I hate that it makes sense 😤
- [+][–]Lakshmi TripathiDropperSample· 13d agoAuthor
  thank you nisha, the "lone pairs trans to each other" is the part I was missing. And rahul bhaiya, saving that table.

[+]Rahul DubeyDropperSample· 13d ago

The list with counting:

Molecule	Bond pairs	Lone pairs	Hybridisation	Shape
$XeF_2$	2	3	$sp^3d$	linear
$XeF_4$	4	2	$sp^3d^2$	square planar
$XeF_6$	6	1	$sp^3d^3$	distorted octahedral
$XeO_3$	3	1	$sp^3$	pyramidal
$XeOF_4$	5	1	$sp^3d^2$	square pyramidal

For $XeF_2$ : 5 pairs make a trigonal bipyramid. Lone pairs go to the equatorial positions (less repulsion), leaving the two F atoms at the axial positions, so the molecule is linear.

[+]Aman YadavDropperSample· 13d ago

Formation conditions from NCERT, as I remember them:

$XeF_2$ : Xe in excess + $F_2$ , 673 K, 1 bar
$XeF_4$ : Xe : $F_2$ = 1 : 5, 873 K, 7 bar
$XeF_6$ : Xe : $F_2$ = 1 : 20, 573 K, 60–70 bar

Notice that more fluorine and higher pressure gives higher fluoride. Hydrolysis of $XeF_6$ gives $XeO_3$ : $XeF_6 + 3H_2O \rightarrow XeO_3 + 6HF$ . Partial hydrolysis gives $XeOF_4$ .

[+]Ishita GoyalDropperSample· 13d ago

On "why xenon": ionisation enthalpy is lowest in the group for Xe (as you said), so the electron is easiest to pull away, and the bigger atom can accommodate more electron pairs around it. Helium and neon have very high IE and no accessible d orbitals/size for expansion, plus they're too small. Xenon also bonds only to the most electronegative elements, F and O, since only they can pull electron density away strongly enough. That is why you will never see xenon chloride in NCERT.

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