Hybridisation

Question

In the molecule BrF₃, the central bromine atom undergoes sp³d hybridisation. The molecular geometry of BrF₃ is:

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1. **Count Valence Electrons**: Br has 7 valence electrons and each F contributes 7, totalling $7 + 3 	imes 7 = 28$ electrons.
2. **Assign Bonds and Lone Pairs on Br**: Three Br–F bonds use 6 electrons. Each F gets 3 lone pairs (18 electrons total). Remaining on Br: $28 - 6 - 18 = 4$ electrons = **2 lone pairs**.
3. **Determine Electron-Pair Geometry**: 3 bond pairs + 2 lone pairs = 5 electron domains → trigonal bipyramidal electron-pair arrangement (sp³d hybridisation).
4. **Position of Lone Pairs**: Lone pairs preferentially occupy equatorial positions in a trigonal bipyramid to minimise 90° lp–bp repulsions. Both lone pairs sit in the equatorial plane.
5. **Resulting Shape**: The three F atoms occupy the two axial positions and one equatorial position, giving a **T-shaped** molecular geometry with bond angles slightly less than 90° and ~180° between axial atoms.
6. **Why Not Others?** Trigonal planar requires 0 lone pairs (e.g., BF₃). Trigonal pyramidal requires 1 lone pair with tetrahedral e-pair geometry (e.g., NH₃). Bent requires 2 lone pairs on tetrahedral arrangement (e.g., H₂O).

Hybridisation

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