Square planar geometry is typically seen with strong field ligands like cyanide, while tetrahedral geometry is common with weak field ligands like chloride.
Step 1: Understanding the geometries.
- \([Ni(CN)_4]^{2-}\) typically forms a square planar geometry due to the strong field ligand \( \text{CN}^- \), which causes the d-orbitals to undergo pairing and leads to a square planar arrangement.
- \([NiCl_4]^{2-}\) generally forms a tetrahedral geometry because chloride ions are weaker field ligands, which do not cause significant pairing of d-electrons. Step 2: Conclusion.
Thus, \([Ni(CN)_4]^{2-}\) has square planar geometry and \([NiCl_4]^{2-}\) has tetrahedral geometry.
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Approach Solution -2
Step 1: [Ni(CN)₄]²⁻ has cyanide (CN⁻), a strong-field ligand, which pairs up nickel's d-electrons and gives a flat, square planar shape.
Step 2: [NiCl₄]²⁻ has chloride, a weak-field ligand, which doesn't force pairing, so it keeps a tetrahedral shape.
Step 3: So [Ni(CN)₄]²⁻ is square planar, and [NiCl₄]²⁻ is tetrahedral.
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Approach Solution -3
Comparing the two ligands: Cyanide is a strong-field ligand high on the spectrochemical series, so it forces the nickel ion's d-electrons to pair up rather than spread across more orbitals, and with all electrons paired, a four-coordinate complex favors a square planar geometry. Chloride sits much lower on the spectrochemical series and doesn't force pairing, so the nickel ion keeps its natural arrangement, and a four-coordinate complex like this adopts a tetrahedral shape instead. So \( [Ni(CN)_4]^{2-} \) is square planar and \( [NiCl_4]^{2-} \) is tetrahedral.
