Match List I with List II :

List I (Complex/ion)	List II (Shape/geometry)
A. [Pt(Cl₂)(NH₃)₂]$[Pt(Cl_2)(NH_3)_2]$	(I) Octahedral
B. [Co(NH₃)₆]Cl₃$[Co(NH_3)_6]Cl_3$	(II) Trigonal bipyramidal
C. [NiCl₄]²⁻$[NiCl_4]^{2-}$	(III) Square planar
D. [Fe(CO)₅]$[Fe(CO)_5]$	(IV) Tetrahedral

Choose the correct answer from the options given below :

Core Logic

(A) [Pt(Cl₂)(NH₃)₂]$[Pt(Cl_2)(NH_3)_2]$: Pt(II) is a 5d⁸$5d^8$ system. 4d⁸$4d^8$ and 5d⁸$5d^8$ complexes are almost always square planar with dsp²$dsp^2$ hybridization regardless of ligand strength. (Matches III) (B) [Co(NH₃)₆]Cl₃$[Co(NH_3)_6]Cl_3$: Co(III) with Coordination Number 6. Has d²sp³$d^2sp^3$ hybridization, making it octahedral. (Matches I) (C) [NiCl₄]²⁻$[NiCl_4]^{2-}$: Ni(II) is a 3d⁸$3d^8$ system. With weak field ligand Cl^-$Cl^-$, it undergoes sp³$sp^3$ hybridization, forming a tetrahedral geometry. (Matches IV) (D) [Fe(CO)₅]$[Fe(CO)_5]$: Fe(0) is a 3d⁸ 4s²$3d^8 4s^2$ system. Under the influence of strong field ligand CO, electrons pair up to form 3d¹⁰$3d^{10}$, leaving 4s$4s$ and 4p$4p$ empty for dsp³$dsp^3$ hybridization. Shape is trigonal bipyramidal. (Matches II)

Step 1: Final Match

A arrow$\rightarrow$ III B arrow$\rightarrow$ I C arrow$\rightarrow$ IV D arrow$\rightarrow$ II

Pattern Recognition

Pt(II) and Pd(II) complexes with CN=4 are universally square planar. Ni²⁺$Ni^{2+}$ with weak ligands (Cl^-$Cl^-$) is tetrahedral, but with strong ligands (CN^-$CN^-$) is square planar. Fe with 5 CO ligands is the classic dsp³$dsp^3$ trigonal bipyramidal example.

Chapter Mix

Class 12 Chemistry: Coordination Compounds