Out of the following which will be see-saw shaped?
A. \(XeO_2F_2\)
B. \(SF_4\)
C. \(BrF_3\)
D. \(XeO_4\)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A and B only
A see-saw shape arises when the central atom has four bond pairs and one lone pair (five electron pairs, sp3d): the lone pair takes an equatorial position of the trigonal bipyramid.
A. XeO2F2 — True. Xe has 8 valence electrons; two Xe=O and two Xe–F bonds use 6 of them (4 bonding directions) and leave 1 lone pair. 4 bp + 1 lp: see-saw.
B. SF4 — True. S has 6 valence electrons; four S–F bonds use 4 and leave 1 lone pair. 4 bp + 1 lp: see-saw.
C. BrF3 — False. Br has 7 valence electrons; three Br–F bonds leave 2 lone pairs. 3 bp + 2 lp: T-shaped.
D. XeO4 — False. All 8 electrons of Xe are used in four Xe=O bonds; with no lone pair the shape is tetrahedral.
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.
NCERT (Pre-rationalised edition, 2022-23):
Reactivity towards the Halogens – “… Amongst tetrafluorides, SF4 is a gas, SeF4 a liquid and TeF4 a solid. These fluorides have sp3d hybridisation and thus, have trigonal bipyramidal structures in which one of the equatorial positions is occupied by a lone pair of electrons. This geometry is also regarded as see-saw geometry.”