Question

Synthesis and Characterization of Mo₂(O₂CCH₃)₄ and K₄Mo₂Cl₈: Compounds that Contain Metal-Metal Quadruple Bonds

1. Include complete balanced reactions of both preparations, from starting materials to products.

2. Trimethylamine-N-oxide is used as an initiator for the Mo₂(O₂CCH₃)₄ preparation. Suggest a reaction by which Trimethyl-amine-N-oxide activates molybdenum hexacarbonyl.

3. Mo₂(O₂CCH₃)₄ has a bond order of four and is diamagnetic. If this compound was oxidized or reduced by one electron, what would be the resulting bond orders?

Answer 1

The reaction for the formation of Mo_2 (O_2 CCH_3)_4 is, The Mo_2 (O_2 CCH_3)_4 reacts with HCl and KCl gives K_4 Mo_2 Cl_8. The reaction is as follows: Mo_2 (O_2 CCH_3)_4 + 4 HCl + 4 KCl rightarrow K_4 Mo_2 Cl_8 + 4 CH_3 CO_2 H The Tri methylamine-N-oxide is generally used as decarbonylating agent in organometallic chemistry. The general reaction is, M(CO)_n + (CH_3)_3 NO + Ligand rightarrow M(CO)_n - 1 L + (CH_3)_3 CO_2 Therefore, the Trimethyl amine-N-oxide activates Mo(CO)_6 by removing one CO group from Mo(CO)_6 to produce coordinatively unsaturated intermediate which subsequently reacts with excess CH_3COOH gives final product. The activation reaction is as follows: The relative energy diagram of the d-orbital interaction is, Bond order = no of bonding - no of anti bonding electrons/2 = 8 - 0/2 = 4 From the above diagram, the HOMO is delta (bonding) orbital, Therefore, addition of an electron (reduction) will decrease the bond order