Hybridisation

Question

The hybridisation of the central atom in XeF_4 and the predicted molecular geometry of this molecule are respectively:

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Xenon in XeF_4 has the electronic configuration with 8 valence electrons. In XeF_4, xenon forms four bonds with four fluorine atoms and retains two lone pairs, giving a total of six electron pairs around the central atom. Six electron pairs require sp^3d^2 hybridisation, which arranges them in an octahedral geometry. However, the two lone pairs adopt axial positions (to maximise their separation from each other), and the four bonding pairs occupy the equatorial plane, resulting in a square planar molecular geometry. Option 'sp^3, tetrahedral' is wrong because that accounts for only four electron pairs, ignoring the two lone pairs. Option 'sp^3d, trigonal bipyramidal' is incorrect as it corresponds to five electron pairs (e.g., PCl_5). Option 'sp^3d^2, octahedral' correctly identifies hybridisation but misidentifies the molecular shape; octahedral shape requires six bonding pairs with no lone pairs. This is a standard NCERT Class 11 noble gas compound question testing the difference between electron geometry and molecular geometry. Verification: Xe has 8 valence electrons; 4 are used for bonding, leaving 4 electrons (2 lone pairs), confirming sp^3d^2 hybridisation and square planar shape.

Hybridisation

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