Match the type of hybridisation given in List-I with their geometry given in List-II
|
List-I Type of hybridisation |
List-II Geometry |
|
(A) $sp^3$ |
(I) Trigonal bipyramidal |
|
(B) $dsp^2$ |
(II) Octahedral |
|
(C) $sp^3d$ |
(III) Square planar |
|
(D) $sp^3d^2$ |
(IV) Tetrahedral |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
|
List-I Type of hybridisation |
List-II Geometry |
|
(A) $sp^3$ |
(IV) Tetrahedral |
|
(B) $dsp^2$ |
(III) Square planar |
|
(C) $sp^3d$ |
(I) Trigonal bipyramidal |
|
(D) $sp^3d^2$ |
(II) Octahedral |
The correct answer is Option (4) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II).
Reasoning:
Geometry depends on the number of hybrid orbitals (steric number) formed around the central atom.
(A) sp³
sp³ means 1s + 3p = 4 hybrid orbitals → arrange to minimize repulsion → tetrahedral geometry → (IV)
(B) dsp²
dsp² means 1d + 1s + 2p = 4 hybrid orbitals arranged in one plane → square planar geometry → (III)
(C) sp³d
sp³d means 1s + 3p + 1d = 5 hybrid orbitals → trigonal bipyramidal geometry → (I)
(D) sp³d²
sp³d² means 1s + 3p + 2d = 6 hybrid orbitals → octahedral geometry → (II)
Quick memory trick:
-
4 orbitals → tetrahedral
-
4 (with d involvement, planar) → square planar
-
5 orbitals → trigonal bipyramidal
-
6 orbitals → octahedral
.