The geometry of the electron pairs around sulphur in SF4 is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Trigonal bipyramidal
Sulphur has 6 valence electrons. In \( \ce{SF4} \) four of them form four S–F bonds and the remaining two form one lone pair, so there are 5 electron pairs around S:
4 bond pairs + 1 lone pair = 5 pairs ⇒ sp3d hybridisation ⇒ trigonal bipyramidal arrangement
The lone pair takes an equatorial position of the trigonal bipyramid, so the four F atoms alone are said to have a see-saw shape.
Why the other options are incorrect
Option (1) Square planar — Incorrect. Needs 6 electron pairs (4 bond pairs + 2 lone pairs, sp3d2), as in \( \ce{XeF4} \).
Option (2) Tetrahedral — Incorrect. Needs only 4 electron pairs (sp3), as in \( \ce{SiF4} \); it ignores the lone pair on S.
Option (4) Octahedral — Incorrect. Needs 6 bond pairs, as in \( \ce{SF6} \).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.