— Inorganic Chemistry Chemistry Question
Ammine complexes of transition metals
The synthesis of chromium(III) ammine complexes usually starts from a freshly prepared in situ solution of a chromium(II) salt. How can one prepare such a solution using metallic chrome? Specify the conditions.
Model Answer
Chrome is dissolved in a diluted sulfuric or hydrochloric acid:
Cr + 2 HCl → CrCl2 + H2
The experiment is conducted under inert atmosphere.
To the solution of a chromium(II) salt, the solution of ammonia and a solid ammonium chloride are added. Then a stream of air is passed through the solution. The red precipitate is formed that contains 28.75 % by mass of N. Determine the composition of the precipitate and give the reaction equation.
Model Answer
4 [Cr(NH3)6]Cl2 + 4 NH4Cl + O2 → 4 [Cr(NH3)5Cl]Cl2↓ + 4 NH3 + 2 H2O
The formula of the precipitate is CrCl3N5H15.
What oxidizer can be used instead of oxygen to obtain the same product? Justify the choice.
Model Answer
H2O2. The compound [Cr(NH3)5Cl]Cl2 is formed because the oxidation takes place via the η2-bridging peroxocomplex, followed by the hydrolysis when the leaving peroxo-group is replaced by the chloride-ion from the solution.
What product will be formed if the experiment described above is performed under inert atmosphere without oxygen? Give the equation.
Model Answer
2 [Cr(NH3)6]Cl2 + 2 NH4Cl → 2 [Cr(NH3)6]Cl3 + H2 + 2 NH3
Explain why the ammine complexes of chromium(III) cannot be prepared by the action of water ammonia on a solution of chromium(III) salt.
Model Answer
The chromium(III) complexes are inert, thus the substitution process occurs slowly. This is due to the d3 configuration.
Arrange the hexammine complexes of iron(II), chromium(III) and ruthenium(II) in a row of increasing stability towards the acidic water solutions. Explain your choice.
Model Answer
Fe(NH3)6 2+ < Ru(NH3)6 2+ < Cr(NH3)6 2+
The coordinated ammonia has no vacant electron pair and therefore cannot interact with a proton. The iron(II) complex is labile, that is, ammonia ligands can be easily substituted by water molecules, which have a free electron pair even when linked to a metal atom. The ruthenium(II) complex is inert, but due to high atomic radius of ruthenium has a possibility to form an intermediate complex with an enhanced coordination number. The chromium(III) complex is inert and has no possibility to bind a proton. Therefore it is the most stable complex in the acidic media.
In the case of [Ru(NH3)6] 2+ the hydrolysis rate increases upon the addition of an acid. Propose a mechanism and derive the rate law.
Model Answer
[Ru(NH3)6] 2+ + H2O + H+→ [Ru(H2O)(NH3)5] 2+ + NH4 +
[Ru(NH3)6] 2+ + H+ → [RuH(NH3)6] 3+
[RuH(NH3)6] 3+ + H2O + H+→ [RuH(NH3)5(H2O)]3+ + NH4 + (fast)
[RuH(NH3)5(H2O)]3+→ [Ru(NH3)5(H2O)]2+ + H+
r = k [H+] [RuH(NH3)6 2+]
See J.D. Atwood, Inorganic and organometallic reaction mechanisms, 2nd edition, Wiley-VCH, pp.85-86 and P.C. Ford et al, Inorg. Chem., 1968, 7, 1976.