Question

The complexes [Co(NH3)6]3+ and [Mo(CO)6] are isoelectronic and diamagnetic. The first complex is orange and the second complex is white. What can you deduce about the value of Δ in both of these complexes? Select one: a. Δ is larger in [Mo(CO)6] than in [Co(NH3)6]3+ b. Δ is smaller in [Mo(CO)6] than in [Co(NH3)6]3+ c. Δ is the same in the two complexes d. Δ is zero in [Mo(CO)6]

Answer 1

Electronic configuration of Mo(0) is [Kr] 4d⁶ for bonding purposes.
Electronic configuration of Co(0) is Ar 3d⁷ 4s², so Co(III) is [Ar] 3d⁶ and is always low spin except for [CoF₆]^3- and [CoF₃(H₂O)₃. In the octahedral ligand environment (octahedral crystal field if you wish), the 3d atomic orbitals are split into a three atomic orbitals low-energy set, the t₂g AOs, and a higher energy set of two 3d AOs, the eg set.The filling is as following: t₂g(↑↓)(↑↓)(↑↓)→Δoct→(0)(0)eg A t₂g e⁻ can absorb a photon often in the visible region and be promoted to the eg AO: t₂g(↑↓)(↑↓)(↑↓) (0)(0)eg →Δ/hν → t₂g(↑↓)(↑↓)(↑) (↓)(0)eg the energy of the photon absorbs depends on the Δoct and you see the complementary color to that which is absorbed.
[Co(NH₃)₆]³⁺ is yellow-red, so it absorbs in the green-blue. Mo(CO)6 (Cr(CO)6 is a nice stable that is white and should be the example used) has no absorptions in the visible; its lowest energy e⁻ transition is in the high-energy UV (mixed up with other transitions). This is consistent with CO being, contrary to what you might expect, a strong field ligand. So, answer is: 3. Δ is larger in [Mo(CO)₆] than in [Co(NH₃)₆]³⁺. Why CO (0.112 D dipole moment) binds with Mo(0) (i.e., zero charge) is all explained by Molecular Orbital theory (LFT).