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:
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:
Show Hint
- A-II, B-III, C-IV, D-I
- A-IV, B-I, C-II, D-III
- A-I, B-II, C-III, D-IV
- A-III, B-I, C-IV, D-II
The Correct Option is A
Approach Solution - 1
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
Approach Solution -2
To solve this problem, we need to understand the hybridisation of each of the given chemical species and match them accordingly.
- 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.
- 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.
- 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.
- [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.
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