VSEPR: shapes of ClF₃, SF₄ and XeF₄, and where the lone pairs go

Electron pairs on the central atom

Pairs = ½(valence electrons of central atom + number of monovalent ligands).

• $\ce{ClF3}$: ½(7 + 3) = 5 → 3 bond pairs + 2 lone pairs; trigonal bipyramidal ($sp^3d$).
• $\ce{SF4}$: ½(6 + 4) = 5 → 4 bp + 1 lp; trigonal bipyramidal ($sp^3d$).
• $\ce{XeF4}$: ½(8 + 4) = 6 → 4 bp + 2 lp; octahedral ($sp^3d^2$).

Where the lone pairs go

Repulsion order: lp–lp > lp–bp > bp–bp, and only 90° interactions matter. In a trigonal bipyramid an equatorial site has two 90° neighbours, an axial site three. So lone pairs take equatorial sites:

• $\ce{ClF3}$: both lp equatorial → T-shaped.
• $\ce{SF4}$: lp equatorial → see-saw.

In an octahedron all sites are equivalent; two lone pairs go trans (180°) so that no lp–lp contact is at 90° → $\ce{XeF4}$ is square planar.

Bond angles

Lone pairs push the bonds together: in $\ce{ClF3}$ $\angle \text{F}_{ax}\text{–Cl–F}_{eq} \approx 87.5^\circ$ (bent T, not 90°); in $\ce{SF4}$ $\angle \text{F}_{ax}\text{–S–F}_{ax} \approx 173^\circ$ and $\angle \text{F}_{eq}\text{–S–F}_{eq} \approx 102^\circ$ (ideal 180° and 120°). In $\ce{XeF4}$ the trans lone pairs cancel, so F–Xe–F stays exactly 90°.

Conclusion. Shape follows the arrangement of all electron pairs, but is named from the atoms only; lone pairs choose the sites with the fewest 90° contacts.