Q4(c). Two lone pairs on Xe in XeF₄ from valence bond theory

Xe8 valence eHYBRIDsp³d² (octahedral set)SHAPEsquare planar, D₄ₕ

1. Ground state of Xe: [Kr]4d¹⁰ 5s²5p⁶. Every valence orbital is filled, so no bond can form.
2. Promotion. Two 5p electrons are promoted to two empty 5d orbitals (second excited state): 5s² 5pₓ² 5p_y¹5p_z¹ 5d¹5d¹. There are now four unpaired electrons, one for each Xe–F bond.
3. Hybridisation. One 5s, three 5p and two 5d orbitals mix to give six equivalent sp³d² hybrids pointing to the corners of an octahedron.
4. Filling the hybrids. Xe has 8 valence electrons. Four hybrids carry one electron each and overlap with the half-filled 2p orbitals of four F atoms: four Xe–F σ bonds. The remaining 4 electrons fill two hybrids: these are the two lone pairs. Count: 8-4 (used in bonds) =4 electrons =2 pairs.
5. Shape. The two lone pairs repel each other most strongly, so they take trans (axial) positions, 180° apart. The four F atoms lie in the equatorial plane: XeF₄ is square planar, F–Xe–F 90°, point group D₄ₕ, no dipole moment.
6. VSEPR agrees: steric number =4 bond pairs +2 lone pairs =6, AX₄E₂ ⇒ square planar.
7. Energy cost. The 5p→5d promotion is repaid by forming two extra bonds to the very electronegative F, which contracts the 5d orbitals. This is why xenon forms compounds only with the most electronegative elements, F and O.
8. Modern view: calculations show little d-orbital use; the bonding is also described by two three-centre four-electron F–Xe–F bonds formed from Xe 5p orbitals, which leaves the same two lone pairs (in 5s and in 5p_z, the 5p_z pair above and below the plane).

Application. XeF₄ is made directly: Xe + 2F₂ → XeF₄ at 600°C and 6 atm (Xe: F₂ = 1: 5). It is a strong fluorinating agent (XeF₄ + Pt → Xe + PtF₄), and with fluoride ion it gives the pentagonal planar XeF₅⁻, again with two trans lone pairs.