Final answer:
The electron configuration of the square planar complex [PdCl4]2− is dsp², corresponding to option D, due to the d⁸ electron configuration of palladium in its +2 oxidation state and the large crystal field splitting causing a low-spin and dsp² hybridization.
Step-by-step explanation:
The electron configuration of the square planar complex [PdCl4]2− can be determined by looking at the central metal ion, which is palladium (Pd) in this case, and considering the ligand field strength. Palladium (Pd) has a d10 electron configuration. However, in the complex [PdCl4]2−, palladium is in the +2 oxidation state, which corresponds to a d8 electron configuration.
Since Cl− is a weak-field ligand, we might expect the complex to be high spin. However, for square planar complexes, the crystal field splitting is large, which results in a low-spin complex. This means electrons will pair up in the lower-energy orbitals first before occupying the higher-energy ones. In a square planar complex, one of the d orbitals (dx2-y2) is at much higher energy compared to the others, and it becomes energetically favourable for electrons to avoid this orbital, resulting in dsp2 hybridization.
Thus, the correct electron configuration for the [PdCl4]2− complex is dsp2, which corresponds to option D.