Coordination Compounds Part 4

Concept:Unpaired electrons in coordination complexes depend on the oxidation state, d-electron count, and geometry (tetrahedral or square planar) influenced by ligand field strength.Explanation:For $[\mathrm{Ni}(\mathrm{CO})_4]$: Ni is in $0$ oxidation state ($3d^8 4s^2$).CO is a strong field ligand, causing pairing of electrons.Hybridisation: $sp^3$ (tetrahedral).Unpaired electrons = $0$.For $[\mathrm{NiCl}_4]^{2-}$: Ni is $+2$ ($3d^8$).Cl⁻ is a weak field ligand, no pairing.Hybridisation: $sp^3$ (tetrahedral).Unpaired electrons = $2$.For $[\mathrm{PtCl}_2(\mathrm{NH}_3)_2]$: Pt is $+2$ ($5d^8$).Strong field ligands (Cl⁻ and NH₃) cause pairing; square planar geometry ($dsp^2$).Unpaired electrons = $0$.For $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$: Ni is $+2$ ($3d^8$).CN⁻ is a strong field ligand, forces pairing; geometry is square planar ($dsp^2$).Unpaired electrons = $0$.For $[\mathrm{Pt}(\mathrm{CN})_4]^{2-}$: Pt is $+2$ ($5d^8$).Strong field ligand CN⁻ leads to pairing; square planar ($dsp^2$).Unpaired electrons = $0$.Total unpaired electrons = $0 + 2 + 0 + 0 + 0 = 2$.Answer:$2$ (Option A)