Match the LIST-I with LIST-II:

LIST-I (Molecules/ion)	LIST-II (Hybridisation of central atom)
(A) PF₅$\mathrm{PF_5}$	(I) dsp²$dsp^2$
(B) SF₆$\mathrm{SF_6}$	(II) sp³d$sp^3d$
(C) Ni(CO)₄$\mathrm{Ni(CO)_4}$	(III) sp³d²$sp^3d^2$
(D) [PtCl₄]²⁻$[\mathrm{PtCl_4}]^{2-}$	(IV) sp³$sp^3$

Choose the correct answer from the options given below:

Related Formula

Core Logic

A. PF₅$\mathrm{PF_5}$: 5σ$5\sigma$-bonds, 0$0$ lone pairs $\implies$ Steric number 5 sp³d$5 \implies sp^3d$ (II)

B. SF₆$\mathrm{SF_6}$: 6σ$6\sigma$-bonds, 0$0$ lone pairs $\implies$ Steric number 6 sp³d²$6 \implies sp^3d^2$ (III)

C. Ni(CO)₄$\mathrm{Ni(CO)_4}$: Ni(0) = 3d⁸ 4s² CO (strong field) 3d¹⁰$\mathrm{Ni}(0) = 3d^8 4s^2 \xrightarrow{\text{CO (strong field)}} 3d^{10}$, tetrahedral sp³$\implies sp^3$ (IV)

D. [PtCl₄]²⁻$[\mathrm{PtCl_4}]^{2-}$: Pt²⁺ = 5d⁸$\mathrm{Pt}^{2+} = 5d^8$, square planar for 5d transition metal dsp²$\implies dsp^2$ (I)

Step 1: Final Match

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

Pattern Recognition

5 bonds = sp³d$sp^3d$, 6 bonds = sp³d²$sp^3d^2$, Ni(CO)₄$\mathrm{Ni(CO)_4}$ = sp³$sp^3$, 5d metal complexes (Pt²⁺$Pt^{2+}$) = always square planar dsp²$dsp^2$.

Chapter Mix

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