In the structure of SF4, the lone pair of electrons on S is in:
Chemical Bonding · Class 11 · JEE Main Previous Year Question
In the structure of , the lone pair of electrons on S is in:
- a✓
Equatorial position and there are two lone pair–bond pair repulsions at 90°
- b
Equatorial position and there are three lone pair–bond pair repulsions at 90°
- c
Axial position and there are three lone pair–bond pair repulsions at 90°
- d
Axial position and there are two lone pair–bond pair repulsions at 90°
Equatorial position and there are two lone pair–bond pair repulsions at 90°
🧠 In (steric number 5 = 4 bonds + 1 LP), the electron geometry is trigonal bipyramidal. Lone pairs always occupy the equatorial position because this minimises repulsion.
Why equatorial?
- Equatorial LP has 2 bond pairs at 90° (the 2 axial F atoms) and 2 bond pairs at ~120° (equatorial F atoms). Total 90° repulsions = 2.
- If LP were axial: it would have 3 bond pairs at 90° (the 3 equatorial F atoms). Total 90° repulsions = 3.
Fewer 90° repulsions → lower energy → LP goes equatorial.
Count at 90° for equatorial LP: the two axial S–F bonds are at 90° to the equatorial lone pair → 2 lone pair–bond pair repulsions at 90°.
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