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)
In the valence bond theory the metal uses a set of equivalent hybrid orbitals, and the number and type of orbitals mixed decide the shape. The hybrid orbitals point to the corners of a regular figure, and the ligands sit at those corners.
(A) sp3 — This matches (IV) Tetrahedral. One s and three p orbitals give four orbitals at 109.5°, as in \( \ce{[NiCl4]^2-} \).
(B) dsp2 — This matches (III) Square planar. One inner d (\( d_{x^2-y^2} \)), one s and two p orbitals give four orbitals in one plane at 90°, as in \( \ce{[Ni(CN)4]^2-} \).
(C) sp3d — This matches (I) Trigonal bipyramidal. Five orbitals: three in a plane at 120° and two along the axis above and below.
(D) sp3d2 — This matches (II) Octahedral. Six orbitals pointing to the corners of an octahedron, as in \( \ce{[CoF6]^3-} \).
Shortcut: (A)-(IV) appears in Options (3) and (4); (B)-(III) then leaves only Option (4).
NCERT (2026-27 edition):
Valence Bond Theory – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | 16-05 S1 | 19-07 S1 | – | – |