JEE Main 2016 — Chemical Bonding And Molecular Structure Question with Solution
From: JEE Main 2016 (Online) 9th April Morning Slot
Question
The group of molecules having identical shape is :
Choose an option
Show full solutionCorrect option: B
Correct answer
BClF3 , XeOF2 , XeF
Step-by-step explanation
H = [VE + MA – c + a]
ClF3 , H = (7 + 3 – 0 + 0) = 5 (sp3d)
XeOF2 , H = (8 + 2 – 0 + 0) = 5 (sp3d)
XeF3+ , H = (8 + 3 – 1 + 0 ) = 5 (sp3d)
All molecules have sp3d hybridization and 3 bond pair + 2 lone pairs. Hence all have identical stucture (T-shape).
ClF3 , H = (7 + 3 – 0 + 0) = 5 (sp3d)
XeOF2 , H = (8 + 2 – 0 + 0) = 5 (sp3d)
XeF3+ , H = (8 + 3 – 1 + 0 ) = 5 (sp3d)
All molecules have sp3d hybridization and 3 bond pair + 2 lone pairs. Hence all have identical stucture (T-shape).
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This is a previous-year question from JEE Main 2016, 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.