Match the LIST-I with LIST-II. Choose the correct answer from the options given below:

Core Logic

Let us evaluate each central atom configuration systematically:

• A. PF₅$\text{PF}_5$: Phosphorus has 5 valence electrons, forming 5σ$5\sigma$ bonds with zero lone pairs. Steric number = 5 sp³d$= 5 \implies sp^3d$ hybridisation.
• B. SF₆$\text{SF}_6$: Sulfur has 6 valence electrons, forming 6σ$6\sigma$ bonds with zero lone pairs. Steric number = 6 sp³d²$= 6 \implies sp^3d^2$ hybridisation.
• C. Ni(CO)₄$\text{Ni}(\text{CO})_4$: Nickel is in a 0 oxidation state (3d⁸ 4s²$3d^8 4s^2$). Carbon monoxide is a strong field ligand, forcing rearrangement into a filled 3d¹⁰$3d^{10}$ state. The vacant 4s$4s$ and three 4p$4p$ orbitals hybridise to give an sp³$sp^3$ configuration.

  

• D. [PtCl₄]²⁻$[\text{PtCl}_4]^{2-}$: Platinum is in the +2$+2$ oxidation state (5d⁸$5d^8$). Since it belongs to the 5d$5d$ transition series, all ligands behave as strong field elements, leading to interior spin-pairing and an inner orbital square-planar dsp²$dsp^2$ hybridisation state.

  

Pattern Recognition

Shortcut: Match main-group species first: PF₅ arrow sp³d (II)$\text{PF}_5 \rightarrow sp^3d\text{ (II)}$, SF₆ arrow sp³d² (III)$\text{SF}_6 \rightarrow sp^3d^2\text{ (III)}$. This immediately isolates Option (A) without needing to evaluate coordination fields.

Evaluation Rubric / Model Answer

Option (A)

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds