Alfred Werner used the technique of 'isomer counting' to deduce the structure of metal complexes wit — Physical Chemistry — Kinetics Chemistry Question
Transition metal complexes
Alfred Werner used the technique of 'isomer counting' to deduce the structure of metal complexes with coordination number six. Three of the shapes he considered are shown below.
In each structure, the empty circle shows the location of the central metal atom and the filled circles show the locations of the ligands. Octahedral complexes are far more common than trigonal prismatic. Werner isolated five compounds C – G containing Co(III), Cl, and NH3 only, each of which contained one octahedral complex. (There is actually a sixth compound but Werner could not isolate it.) compound, H, which contained no carbon atoms. The compound, H, is composed of only cobalt, ammonia, chloride and an oxygen species which could be either H2O, or HO− or O2−. The compound contains octahedrally coordinated cobalt ions. All of the chloride is easily removed from the compound by titration with aqueous silver nitrate. A 0.2872 g sample of H (containing no water of crystallization) required 22.8 cm3 of a silver nitrate solution (c = 0.100 mol dm–3) to exchange all of the chloride. ... of potassium iodide was added, and then the mixture was acidified with aqueous HCl. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate (c = 0.200 mol dm–3) and required 21.0 cm3 for complete reaction.
Fill in the table below to indicate how many geometrical isomers may be formed for each structure X, Y, and Z as the monodentate ligands A are substituted by mono-dentate ligands B or by symmetrical bidentate ligands, denoted C–C. Bidentate ligand C–C can only link between two atoms on adjacent positions, i.e. those positions connected by a line in the structures X, Y, and Z.
In each case write the number of geometrical isomers in the space provided. If one of the isomers exists as a pair of enantiomers, include an asterisk, *, in the box. If two exist as two pairs of enantiomers, include two asterisks and so on. For example, if you think there are five geometrical isomers of a particular structure, three of which exist as pairs of enantiomers, write 5 *
Number of predicted geometrical isomers
Hexagonal planar X Trigonal Prismatic Y Octahedral Z
MA6 1 1 1
MA5B
MA4B2
MA3B3
Model Answer
Number of predicted geometrical isomers
Hexagonal planar X Trigonal Prismatic Y Octahedral Z
MA6 1 1 1
MA5B 1 1 1
MA4B2 3 3* 2
MA3B3 3 3* 2
MA4(C-C) 1 2 1
MA2(C-C)2 2 4* 2*
M(C-C)3 1 2 1*
For each of the splitting patterns shown below label which d orbitals are which.
The two complexes [Mn(H2O)6]2+ and [Mn(CN)6] 2– are both octahedral. One has a magnetic moment of 5.9 BM, the other has a magnetic moment of 3.8 BM but you must decide which is which.
On the diagram below, draw the electronic arrangements for each of the complexes.
Model Answer
[Mn(CN)6] 2- [Mn(H2O)6] 2+
The magnetic moments of complexes A and B shown below have been measured and found to be 1.9 and 2.7 BM but you must decide which is which.
Draw the orbital splitting diagrams for the two complexes, including the arrangements of the electrons.
Model Answer
A B
Werner isolated five compounds C – G containing Co(III), Cl, and NH3 only, each of which contained one octahedral complex. Draw the structures of these compounds.
Model Answer
E and F could be either way round
Calculate the percentage, by mass, of chloride in H.
Model Answer
n(Ag+) = 0.100 mol dm–3 × 0.0228 dm3 = 2.28 ⋅ 10–3 mol
n(Cl–) = 2.28 ⋅ 10–3 mol
m(Cl) = 8.083 ⋅ 10–2 g
% Cl = (8.083 ⋅ 10–2 g / 0.2872 g) × 100 = 28.1 %
Calculate the percentage, by mass, of ammonia in H.
Model Answer
n(KOH) = 0.0124 mol
n(HCl) neutralised by ammonia = 0.025 mol – 0.0124 mol = 0.0126 mol
m(NH3) = 17.034 g mol–1 × 0.0126 mol = 0.2146 g
% NH3 = (0.2146 g / 0.7934 g) × 100 = 27.1 %
Give the equation for the reaction of cobalt(III) oxide with potassium iodide in aqueous acid.
Model Answer
Co2O3 + 2 KI + 6 HCl → 2 CoCl2 + I2 + 3 H2O + 2 KCl
Calculate the percentage, by mass, of cobalt in H.
Model Answer
2 Na2S2O3 + I2 → 2 NaI + Na2S4O6
n(Na2S2O3) = 0.200 mol dm–3 × 0.021 dm3 = 4.20 ⋅ 10–3 mol
n(I2) = 2.10 ⋅ 10–3 mol
n(Co2+) = 4.20 ⋅ 10–3 mol
m(Co) = 4.20 ⋅ 10–3 mol × 58.93 g mol–1 = 0.2475 g
% Co by mass = (0.2475 g / ...) × 100
Calculate the identity of the oxygen species contained in H. Show your working.
Give the empirical formula of H.
Model Answer
Empirical formula of H: Co2 N6 H21 O3 Cl3
Suggest a structure for the chiral compound H.
Model Answer
Structure must fit the empirical formula worked out above, contain only octahedral cobalt, and be chiral. Some marks deducted if chloride is directly coordinated to cobalt, or if any single ammonia molecule is coordinated to more than one cobalt atom.