JEE Main 2019 — Chemical Bonding And Molecular Structure Question with Solution
From: JEE Main 2019 (Online) 10th January Morning Slot
Question
The type of hybridisation and number of lone pair (s) of electrons of Xe in XeOF4, respectively, are:
Choose an option
Show full solutionCorrect option: D
Correct answer
Dsp3d2 and 1
Step-by-step explanation
H = (V + M - c + a)
H = (8 + 4) = 6
From structure, it is clear that it has five bond pairs and one lone pair.
H = (8 + 4) = 6
From structure, it is clear that it has five bond pairs and one lone pair.
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This is a previous-year question from JEE Main 2019, 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.