Answer all questions. Please be sure to provide adequate explanations of you answers where it is appropriate, including diagrams. All questions carry equal weight.
The sort of formulae given in the question might be found on the label on the bottle, or in the chemical supplier catalogs. They do not contain any reliable structural information!You are told that in solution, violet CrCl_{3}.6H_{2}O has a conductivity similar to that of [Co(NH_{3})_{6})]Cl_{3} i.e. it is probably also a 1:3 cation:anion electrolyte. So, the real formula is probably [Cr(H_{2}O)_{6})]Cl_{3}.
The green compound starts out with less water in the solid state. If it too is octahedral there must be at least one Cl^{-} in the Cr^{3+} coordination sphere. The "lower conductivity" (it does not say how much lower) could correspond to either a 1:2 or 1:1 electrolyte, i.e. either [Cr(H_{2}O)_{5})Cl]Cl_{2} or [Cr(H_{2}O)_{4})Cl_{2}]Cl
In aqueous solution, the coordinated Cl^{-} is gradually replaced by H_{2}O, and the violet hexaquo complex is formed.
The two structures would be:
The prismatic structure (on the left) has two planes of symmetry which render the 4 CH_{3} groups equivalent. The octahedral structure has a two fold axis only and there are two CH_{3}'s cis to both the P's, and two trans to one P and cis to the other, so they are inequivalent.
This is surely a strong field d^{6} case because of the combination of Fe^{2+} and CN^{-}.
LFSE = 6 x 0.4D_{o} = 2.4D_{o} (The negative sign is "understood" since it is a stabilization energy.)
This is probably a weak field d^{6} case (Fe^{2+} and H_{2}O)
LFSE = (4 x 0.4 - 2 x 0.6)D_{o} = 0.4D_{o}
LFSE = (4 x 0.6 - 4 x 0.4)D_{t} = 0.8D_{t}
LFSE = (4 x 0.514 + 2 x 0.428 - 2 x 0.228)D_{o} = 2.245D_{o}
To work this out you would have to remember the not very memorable levels: my mistake!
This is a Ni^{0} d^{10} complex and the ligand field stabilization energy is zero in any geometry, so there is no reason for it not to be tetrahedral.
If the rate is only weakly related to the nature of the entering ligand, then the mechanism is dissociative.A later section of the course examines the mechanisms of substitution reactions in much greater detail. Expect questions of a more complex nature on the final exam!
[M(H_{2}O)_{6}]^{2+} + en [M(H_{2}0)_{4}en]^{2+} + 2H_{2}O K_{1}[M(H_{2}O)_{4}en]^{2+} + en [M(H_{2}0)_{2}(en)_{2}]^{2+} + 2H_{2}O K_{2}
[M(H_{2}O)_{2}(en)_{2}]^{2+} + en [M(en)_{3}]^{2+} + 2H_{2}O K_{3}
Ion log K_{1} Log K_{2} Log K_{3} Co^{2+} 5.89 4.83 3.10 Ni^{2+} 7.52 6.28 4.26 Cu^{2+} 10.55 9.05 -1.0
Statistically, molecules with fewer coordinated water molecules are less likely in aqueous solution: a molecule where all the water has been removed and replaced by ethylene diamine is the least likely species.
When an equilibrium constant seems particularly "out of line" like this, the most probable explanation is a change in the nature of the species involved in the equilibrium. In this case, there is a tendency for Cu(II) (d9) to exhibit a large Jahn-Teller distortion to the point of being virtually square-planar. Therefore the concentration of [Cu(H_{2}O)_{2}(en)_{2}]^{2+} is actually much lower than one would calculate using only the above sequence of equilibria, and the concentration of [Cu(en)_{3}]^{2+} would be correspondingly low. We have to consider the competing equilibrium:[Cu(H_{2}O)_{2}(en)_{2}]^{2+} [M(en)_{2}]^{2+} + 2H_{2}O K_{4}
All other factors being equal, the stability of the ethylene diamine complexes relative to the aquo complexes will be determined by the relative change in ligand field stabilization energies (LFSE). The absolute values of the LFSE i.e. D_{o}, will be greater for the smaller Ni^{2+} ion, so the differences should also be greater, hence the greater stability.
The table below shows the 15 microstates possible for p^{2} in no particular order.
m_{l} M_{L} 1 0 -1 2 M_{S} = 0 0 -2 1 M_{S} = 1 0 -1 1 M_{S} = -1 0 -1 1 M_{S} = 0 0 -1 1 M_{S} = 0 0 -1 The next table contains a count of the number of microstates with the same M_{L} and M_{S} values that can be found in the table above.
M_{S} 1 0 -1 M_{L} 2 0 1 0 1 1 2 1 0 1 3 0 -1 1 2 1 -2 0 1 0 The maximum value of M_{L} for which a microstate exists is 2 (in red), which must be part of a D term (i.e. with L = 2 so M_{L} can also be 1,0,-1 and -2), and the only M_{S} value for which this term is found is 0 so it is part of a ^{1}D term. (S = 0.) Now remove the 5 microstates corresponding to this from the above table to get the next.
M_{S} 1 0 -1 M_{L} 2 0 0 0 1 1 1 1 0 1 2 1 -1 1 1 1 -2 0 0 0 The maximum value of M_{L} for which a microstate exists is 1 (in red) which must be part of a P term (i.e. with L = 1 so M_{L} can also be 0 and -1), and the largest M_{S} value for which this term is found is 1, so it is part of a ^{3}P term. (i.e. S = 1, so M_{S} can also be 0 and -1.) Now remove the 9 microstates corresponding to this from the above table to get the next.
M_{S} 1 0 -1 M_{L} 2 0 0 0 1 0 0 0 0 0 1 0 -1 0 0 0 -2 0 0 0 All that is left is one microstate with M_{L} = 0 and M_{S} = 0 which is a ^{1}S term. (i.e. L = 0 and S = 0)
The complex contains Ni(II), a d^{8} ion in octahedral coordination. As the question implies, there are two methods for solving this problem; graphically, and more precisely by calculation:
The diagram for d^{8} shows that the ground state is ^{3}A_{1g} with spin-allowed transitions (in order of increasing frequency or wavenumber) to ^{3}T_{2g}, ^{3}T_{1g} (from the ^{3}F term) and ^{3}T_{1g} (from the ^{3}P term).
Reference to Huheey Table 9.26 (distributed with the exam) indicates that the first of these transitions at 8 500 cm^{-1} corresponds to D_{o}.
To find B from the Tanabe-Sugano diagram, note that the experimental ratio n_{2}/n_{1} = 1.72, and find the point along the D_{o}/B axis that yields the same ratio. By trial and error, I found that at D_{o}/B » 7.1, the required ratio is obtained. Therefore, B = 8500/7.1 = 1197 cm^{-1}. (The free ion value for Ni^{2+} is 1041 cm^{-1}.
This diagram is constructed using the information from Huheey Table 9.26 (distributed with the exam), and the knowledge that the ^{3}P term is 15B above the ^{3}F (which is also evident from the Tanabe-Sugano diagram).
The diagram for d^{8} shows that the ground state is ^{3}A_{1g} with spin-allowed transitions (in order of increasing frequency or wavenumber) to ^{3}T_{2g}, ^{3}T_{1g} (from the ^{3}F term) and ^{3}T_{1g} (from the ^{3}P term).
The following equations can be obtained from the spacings marked on the diagram:
D_{o} = 8500 cm^{-1} (1)Adding (2) and (3) gives:1.8D_{o} - x = 15400 (2)
15B + x + 1.2D_{o} = 26000 (3)
15B + 3.0D_{o} = 41400and substituting from (1) gives:
B = 1060 cm^{-1}