Question

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. PF5	I. dsp2
B. SF6	II. sp3d
C. Ni(CO)4	III. sp3d2
D. [PtCl4]2-	IV. sp3

Choose the correct answer from the options given below:

• A. A-II, B-III, C-IV, D-I
• B. A-IV, B-I, C-II, D-III
• C. A-I, B-II, C-III, D-IV
• D. A-III, B-I, C-IV, D-II

Hint:

Determine the hybridization of the central atom in each molecule/ion by counting the number of sigma bonds and lone pairs.

Answer 1

The Correct Option is A

PF₅:
5σ + 0 lone pair ⇒ sp³d hybridisation

SF₆:
6σ + 0 lone pair ⇒ sp³d² hybridisation

Ni(CO)₄:
Ni oxidation state = 0
In presence of ligand field:
Ni(0): [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ _ _ _ _
Orbitals: 3d, 4s, 4p
⇒ sp³ hybridisation

[PtCl₄]^2-:
Pt oxidation state = +2
In presence of ligand field:
Pt²⁺: [Kr] ↑↓ ↑↓ ↑↓ ↑↓ _ _ _
Orbitals: 5d, 6s, 6p
⇒ dsp² hybridisation

Answer 2

To solve this problem, we need to understand the hybridisation of each of the given chemical species and match them accordingly.

1. PF₅:
   • The structure of phosphorus pentafluoride (PF₅) is trigonal bipyramidal.
   • The central atom Phosphorus (P) is bonded to five fluorine atoms.
   • The hybridisation of the central atom is \(sp^3d\), corresponding to option II.
2. SF₆:
   • The structure of sulfur hexafluoride (SF₆) is octahedral.
   • The central atom Sulfur (S) is bonded to six fluorine atoms.
   • The hybridisation of the central atom is \(sp^3d^2\), corresponding to option III.
3. Ni(CO)₄:
   • Nitric tetracarbonyl [Ni(CO)₄] has a tetrahedral geometry.
   • The central atom Nickel (Ni) uses four CO ligands to form this complex.
   • The hybridisation of the central atom is \(sp^3\), corresponding to option IV.
4. [PtCl₄]^2-:
   • The complex ion [PtCl₄]^2- has a square planar structure.
   • The central atom Platinum (Pt) bonds with four chlorine atoms in a square planar manner.
   • The hybridisation of the central atom is \(dsp^2\), corresponding to option I.

Therefore, the correct match is:

• A. PF₅ - II. \(sp^3d\)
• B. SF₆ - III. \(sp^3d^2\)
• C. Ni(CO)₄ - IV. \(sp^3\)
• D. [PtCl₄]^2- - I. \(dsp^2\)

Thus, the correct answer is: A-II, B-III, C-IV, D-I.