Which one of the following complex ions gives tetrahedral geometry ? a) [Ni(CN)4]^2 b) [NiCl4]^2 c) [PtCl4]^2 d) [CoF6]^3

The geometry of a complex ion depends on the hybridization and number of electron pairs around the central metal ion.

• $[\mathrm{Ni}(\mathrm{CN}){}_4]{}^{2-}$: Ni is in +2 oxidation state with $d^8$ configuration. CN⁻ is a strong field ligand, causing pairing of electrons. Hybridization is $ds{p}^2$ → square planar.
• $[\mathrm{NiCl}{}_4]{}^{2-}$: Cl⁻ is a weak field ligand. Ni²⁺ ($d^8$) has two unpaired electrons, leading to tetrahedral geometry (sp³ hybridization).
• $[\mathrm{PtCl}{}_4]{}^{2-}$: Pt²⁺ ($d^8$) with Cl⁻ ligands adopts square planar geometry due to strong ligand field splitting.
• $[\mathrm{CoF}{}_6]{}^{3-}$: Co³⁺ ($d^6$) with F⁻ ligands forms an octahedral complex (sp³d² hybridization).

Thus, the correct answer is (b) $[\mathrm{NiCl}{}_4]{}^{2-}$.