JEE Main 2026 — Chemical Bonding and Molecular Structure Question with Solution
JEE Main 2026 (22 January Shift 1)
Question
Two p-block elements X and Y form fluorides of the type . The fluoride compound is a Lewis acid and is a Lewis base. The hybridizations of the central atoms of and respectively are
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Correct answer
A and
Step-by-step explanation
XF₃ is a Lewis acid, indicating the central atom X lacks a complete octet and can accept electron pairs. This is characteristic of Boron (B).
Boron in BF₃:
Valence electrons: 3
Steric number: 3 (three B-F bonds, no lone pairs)
Geometry: Trigonal planar
Hybridization: sp²
YF₃ is a Lewis base, indicating the central atom Y has lone pair(s) available to donate. This is characteristic of Nitrogen (N).
Nitrogen in NF₃:
Valence electrons: 5
Steric number: 4 (three N-F bonds + one lone pair)
Geometry: Trigonal pyramidal
Hybridization: sp³
Therefore, the hybridizations are sp² for XF₃ (Boron) and sp³ for YF₃ (Nitrogen).
Boron in BF₃:
Valence electrons: 3
Steric number: 3 (three B-F bonds, no lone pairs)
Geometry: Trigonal planar
Hybridization: sp²
YF₃ is a Lewis base, indicating the central atom Y has lone pair(s) available to donate. This is characteristic of Nitrogen (N).
Nitrogen in NF₃:
Valence electrons: 5
Steric number: 4 (three N-F bonds + one lone pair)
Geometry: Trigonal pyramidal
Hybridization: sp³
Therefore, the hybridizations are sp² for XF₃ (Boron) and sp³ for YF₃ (Nitrogen).
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This is a previous-year question from JEE Main 2026, covering the Chemical Bonding and Molecular Structure chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.