JEE Main202331 Jan 2023Morning ShiftChemistryChemical Bonding and Molecular StructureActual
Match List I with List II List I List II A. XeF 4 I.See-saw B. SF 4 II. Square planar C. NH 4 + III. Bent T - shaped D. BrF 3 IV. Tetrahedral Choose the correct answer from the options given below :
Options
- AA - IV ,   B - III ,   C - II ,   D - I
- BA - II ,   B - I ,   C - III ,   D - IV
- CA - IV ,   B - I ,   C - II ,   D - III
- DA - II ,   B - I ,   C - IV ,   D - III
Correct answer
D. A - II ,   B - I ,   C - IV ,   D - III
Step-by-step solution
(A) XeF 4 : Xe has 4 bond pairs along with 2 lone pairs in the XeF 4 . Thus, the hybridisation of Xe is sp 3 d 2 . Now since it contains two lone pairs it will show square planar geometry. (B) SF 4 : The lone pair is an equatorial position, and there are two lone-pair—bond pair repulsions. Hence, it is more stable. So, the shape is described as a distorted tetrahedron, a folded square or a see-saw. (C) NH 4 + : In this case steric number = lone pair + sigma bond = 0 + 4 = 4. So, it is sp 3 hybridisation. Ther