Class 12 NEET Chemistry Coordination Compounds

Concept:Geometry of coordination complexes depends on the coordination number (C.N.) and the hybridization of the central metal ion, which is influenced by oxidation state and ligand field strength.Chemical Equation / Formula:A: $[\mathrm{Pt}(\mathrm{Cl}_2)(\mathrm{NH}_3)_2]$B: $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$C: $[\mathrm{NiCl}_4]^{2-}$D: $[\mathrm{Fe}(\mathrm{CO})_5]$Explanation:For A: $\mathrm{Pt}^{2+}$ has $5d^8$ configuration; $\mathrm{Cl}^-$ and $\mathrm{NH}_3$ are strong field ligands (SFL). C.N. = 4, hybridization is $dsp^2$, giving square planar geometry. So A matches III.For B: $\mathrm{Co}^{3+}$ (from $[\mathrm{Co}(\mathrm{NH}_3)_6]\mathrm{Cl}_3$) has C.N. = 6; $\mathrm{NH}_3$ is SFL, hybridization is $d^2sp^3$, giving octahedral shape. So B matches I.For C: $\mathrm{Ni}^{2+}$ has $3d^8$; $\mathrm{Cl}^-$ is a weak field ligand (WFL), no pairing occurs, sp³ hybridization occurs, giving tetrahedral geometry. So C matches IV.For D: $\mathrm{Fe}$ in $[\mathrm{Fe}(\mathrm{CO})_5]$ has zero oxidation state; C.N. = 5; CO is strong field, hybridization is $dsp^3$, giving trigonal bipyramidal shape. So D matches II.Shortcut:For C.N. = 4: If metal is $d^8$ with strong field ligands → square planar; with weak field → tetrahedral. For C.N. = 5 → usually trigonal bipyramidal. For C.N. = 6 → octahedral.Answer:A-III, B-I, C-IV, D-II. Option (3).