Question

For heterobimetallic complex M(dppf)Cl2, what are the possible geometries of metal center M? Draw the d-orbital splitting diagrams for those geometries and fill them with electrons. What difference do you observe that can be used to distinguish between the two geometries without using X-ray crystallography?

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Answer 1

​​​​​​Considering the d-electron count for M(dppf)Cl2 it can be seen that both Ni and Pd have the same oxidation state (2+) and d-electron count (d8) as they both belong to the same group viz. group 10 in the periodic table. Upon filling the two d-orbital splitting diagrams with 8 electrons, the square planar geometry results in a diamagnetic complex, whereas the MO diagram for td geometry results in a paramagnetic species. Which geometry is energetically favored depends on various electronic and steric factors. The square planar geometry contains fewer electrons in anti-bonding orbitals, which indicates that the square planar geometry is more electronically favored. Considering the electron pairing energy it is observed that the pairing energy in square planar molecules is higher than that in tetrahedral molecules, which have more incompletely filled orbitals. Finally, considering the destabilisation energy, Larger metal atoms have greater spatial overlap with ligands, resulting in higher energy σ* d-orbitals.

Again, Tetrahedral geometry is more sterically favoured (with angles of 109.5 °) compared to square planar geometry (90 °) from extent of steric interaction point of view.

these two geometries can be distinguished using NMR. If the molecule is square planar, a 1H NMR of a diamagnetic species will be observed. paramagnetic signals in the 1H NMR will result for tetrahedral geometry. Using the Evans method the solution magnetic moment of the paramagnetic species can be determined for the confirmation of geometry.