Discussion
Which of the following identity is not correct as regards the hybridization of atomic orbitals to form MO :
Answer: C
The identity that is not correct is sp\(^{3}\)-square planar - PtCl\(_{4}^{2-}\).
Square planar complexes such as \(PtCl_{4}^{2-}\) use \(dsp^{2}\) hybrid orbitals: one d, one s and two p orbitals pointing to the corners of a square. \(sp^{3}\) hybridisation gives a tetrahedral arrangement, not a square planar one.
The others are correct: \(BCl_{3}\) is trigonal planar with \(sp^{2}\) boron, \(CO_{2}\) is linear with sp carbon, and \(CH_{4}\) is tetrahedral with \(sp^{3}\) carbon.
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