The rate of reduction of [Co(NH_3)_5(H_2O)]^3+ by [Cr(H_2O)_6]^2+ is seven order

Question

The rate of reduction of [Co(NH_3)_5(H_2O)]^3+ by [Cr(H_2O)_6]^2+ is seven orders of magnitude slower than reduction of its conjugate base, [Co(NH_3)_5(OH)]^2+. For the corresponding reductions with [Ru(NH_3)_6]^2+ instead of [Cr(H_2O)_6]^2+ as the oxidant, the two reactions differ by less than a factor of 10. What do these observation suggest about the mechanisms in reductions by [Cr(H_2O)_6]^2+ and [Ru(NH_3)_6]^2+ respectively? For the following 18 electron complexes, determine the unknown quantity in the chemical formula. What is the charge, z, of the complex? How many CO ligands, x, are present? What is the first-row (3d) transition metal, M? The reactions of Ni(CO)_4 in which phosphine or phosphites replace CO to give the family Ni(CO)_3L occur at the same rate for different phosphates or phosphites. Is the reaction an associative or dissociative mechanism? Explain

Explanation / Answer

1. The large difference in the rates of the reactions of [Co (NH3)5(H2O)]3+ with [Cr(H2O)6]2+ and

[Co (NH3)5(OH)]2+ with [Cr(H2O)6]2+ is due to a change in the reaction mechanism.

[Cr (H2O)6]2+ is a labile complex, in which metal-ligand bonds are easily broken. Although [Cr (H2O)6]2+ is labile, there is no ligand on the other complex in the reaction which can form an inner sphere reaction. Electron transfer must occur through the slower, outer sphere route.

[Co (NH3)5(OH)]2+ contains the OH- ligand which is able to form an additional bond to a second metal. The reaction of these two complexes proceeds via the formation of an inner sphere complex:

[Co (NH3)5(OH)]2+ with [Cr(H2O)6]2+ [Co(NH3)5-OH-Cr(H2O)5]4+

The close approach of the two metals ions facilitates e- transfer which proceeds via the orbitals of the ligand.

[Ru (NH3)6]2+ contains a low spin d6 Ru2+ ion which has t2g6 electron configuration. The lower pairing energy and larger oct values for 4d and 5d metals ensure that all their complexes are low spin. This leads to a large LFSE.As this LFSE is partially lost in exchanging ligands,so this complex is inert. Although

[Co (NH3)5(OH)]2+ possesses a ligand capable of forming a bridged complex, the other complex in the reaction is inert and no inner sphere complex forms. Electron transfer must occur through the slower outer sphere route for both [Co (NH3)5(OH)]2+ and [Co(NH3)5(H2O)]3+ and there is very little difference in the rate.

The small difference in the rate is due to the higher charge of Co (NH3)5(H2O)]3+ which leads to greater repulsion with the [Ru(NH3)6]2+ ion.

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