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

Lakshmi TripathiDropper Sample

6 Oct 2026 Β· 1 min read

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 XeF4XeF_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 XeF2XeF_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 XeF2XeF_2, XeF4XeF_4, XeF6XeF_6, XeO3XeO_3 and XeOF4XeOF_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.

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  • [+]Ishaan Bose11thSampleΒ· 13d ago

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

    • [+]Nisha Verma12thSampleΒ· 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 sp3d2sp^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 Bose11thSampleΒ· 13d ago

        ok fine that actually makes sense, I hate that it makes sense 😀

      • [+]Lakshmi TripathiDropperSampleΒ· 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 DubeyDropperSampleΒ· 13d ago

    The list with counting:

    MoleculeBond pairsLone pairsHybridisationShape
    XeF2XeF_223sp3dsp^3dlinear
    XeF4XeF_442sp3d2sp^3d^2square planar
    XeF6XeF_661sp3d3sp^3d^3distorted octahedral
    XeO3XeO_331sp3sp^3pyramidal
    XeOF4XeOF_451sp3d2sp^3d^2square pyramidal

    For XeF2XeF_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 YadavDropperSampleΒ· 13d ago

    Formation conditions from NCERT, as I remember them:

    • XeF2XeF_2: Xe in excess + F2F_2, 673 K, 1 bar
    • XeF4XeF_4: Xe : F2F_2 = 1 : 5, 873 K, 7 bar
    • XeF6XeF_6: Xe : F2F_2 = 1 : 20, 573 K, 60–70 bar

    Notice that more fluorine and higher pressure gives higher fluoride. Hydrolysis of XeF6XeF_6 gives XeO3XeO_3: XeF6+3H2O→XeO3+6HFXeF_6 + 3H_2O \rightarrow XeO_3 + 6HF. Partial hydrolysis gives XeOF4XeOF_4.

  • [+]Ishita GoyalDropperSampleΒ· 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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