Match the column -I & Column-II

Column-I

Column-II
(A)
$sp^{3} d$ hybridised central atom
(P) $BF_{3}$
(B) $sp^{3} d^{2}$ hybridised central atom
(Q) $H_{3} PO_{2}$
(C) $sp^{2}$ hybridised central atom
(R) $SO_{4}^{2 -}$
(D) $sp^{3}$ hybridised central atom
(S) $ClF_{3}$

Question

Match the column -I & Column-II

Column-I	Column-II
(A) $sp^{3} d$ hybridised central atom	(P) $BF_{3}$
(B) $sp^{3} d^{2}$ hybridised central atom	(Q) $H_{3} PO_{2}$
(C) $sp^{2}$ hybridised central atom	(R) $SO_{4}^{2 -}$
(D) $sp^{3}$ hybridised central atom	(S) $ClF_{3}$

TestHub · Accepted Answer

(A) sp³d → ClF₃
Chlorine has 3 bond pairs and 2 lone pairs = 5 electron pairs.
Thus, geometry is trigonal bipyramidal and hybridization is sp³d.

(B) sp³d² → XeO₆⁴⁻
Xenon forms 6 σ-bonds with oxygen atoms = 6 electron pairs.
Hence, octahedral geometry and sp³d² hybridization.

(C) sp² → BF₃
Boron forms 3 σ-bonds and has no lone pair = 3 electron pairs.
So, trigonal planar geometry and sp² hybridization.

(D) sp³ → H₃PO₂, SO₄²⁻

Chemistry - CHEMICAL BONDING Question with Solution | TestHub