Question

Using the valence bond approach, deduce the shape and magnetic character of [Cr(CO)₆].

Answer 1

1. Using the valence bond approach, the complex [Cr(CO)₆] can be understood in terms of hybridisation and electron configuration.
2. In this complex, chromium is in the zero oxidation state (Cr⁰), with an electronic configuration of [Ar] 3d⁵ 4s¹.
3. The ligand carbon monoxide (CO) is a strong field ligand, which causes pairing of the 3d electrons.
4. The central chromium ion uses six hybrid orbitals to bond with the six CO ligands.
5. As a result, all five 3d electrons pair up in the lower energy orbitals, making room for hybridisation.
6. Chromium then uses two 3d, one 4s, and three 4p orbitals to form d²sp³ hybrid orbitals, which accommodate electron pairs from six CO ligands.
7. The geometry of the complex is octahedral because the six ligands (CO molecules) are arranged symmetrically around the central atom in an octahedral fashion.
8. In an octahedral field, the five degenerate ddd-orbitals of the metal ion split into two sets of energy levels:
   A lower-energy set of t_2g​ orbitals (three orbitals).
   A higher-energy set of e_g​ orbitals (two orbitals).
9. The CO ligand is a strong field ligand, which means it causes pairing of electrons in the lower-energy t_2g​ orbitals.
10. Since chromium has 6 electrons, the electrons will be arranged in the t_2g​ orbitals, and due to the strong field nature of CO, all the electrons will be paired in the t_2g​ orbitals. Therefore, there are no unpaired electrons.

Thus, according to valence bond theory, [Cr(CO)₆] is octahedral in shape and the complex is diamagnetic because there are no unpaired electrons.