JEE Main 2004 — Chemical Bonding And Molecular Structure Question with Solution
From: AIEEE 2004
Question
Which one of the following has the regular tetrahedral structure?
(Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)
(Atomic nos : B = 5, S = 16, Ni = 28, Xe = 54)
Choose an option
Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
Regular Tetrahedral structure is possible in sp3 hybridization where central atom has 4 bond pair and no lone pair.
(a) XeF4 is sp3d2 hybridised and structure is square planar.
(b) [Ni(CN)4]2 is coordinate compound and oxidation number of Ni is +2.
Electronic configuration of Ni+2 is [Ar]3d8
But because of CN ion which is a strong field ligand , it can perform pairing of electron.
And the structure of dsp2 hybridization is square planar.
(C) BF, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral.
(d)
SF4 is sp3d hybridised and structure is see-saw.
(a) XeF4 is sp3d2 hybridised and structure is square planar.
(b) [Ni(CN)4]2 is coordinate compound and oxidation number of Ni is +2.
Electronic configuration of Ni+2 is [Ar]3d8
But because of CN ion which is a strong field ligand , it can perform pairing of electron.
And the structure of dsp2 hybridization is square planar.
(C) BF, 4 bond pair present so angle is 109o 28' and sp3 hybridised. So structure is regular tetrahedral.
(d)
SF4 is sp3d hybridised and structure is see-saw.
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This is a previous-year question from JEE Main 2004, 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.