Concept:Complexes with a coordination number of 4 can be either square planar (\(dsp^2\) hybridization) or tetrahedral (\(sp^3\) hybridization), largely depending on the \(d\)-electron count and the strength of the ligand field.
Solution:- \([Fe(CO)_5]\): Coordination number 5 \(\implies\) Trigonal Bipyramidal.
- \([Au(Cl)_4]^-\): \(Au^{3+}\) is a \(5d^8\) system. Heavy \(4d\) and \(5d\) metals almost universally form low-spin, Square Planar complexes.
- \([Pt(NH_3)_4]^{2+}\): \(Pt^{2+}\) is a \(5d^8\) system \(\implies\) Square Planar.
- \([Cu(CN)_4]^{2-}\): The explanatory notes of the textbook specifically classify this complex as Tetrahedral. (Note: While the test prep Answer Key printed "B", the detailed explanatory note explicitly corrects this to "C" as Tetrahedral).
Why other options are incorrect:- \(5d^8\) metals (\(Pt, Au\)) strongly prefer square planar geometries due to immense crystal field splitting energy.
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