Chemistry 341 Section X - Mid-term Exam - 1½ hour

October 16th, 2001

Answer all questions. Please be sure to provide adequate explanations of you answers where it is appropriate, including diagrams. All questions carry equal weight.

    1. The hydrated chromium chloride that is available commercially has the overall composition CrCl3.6H2O. In solution it is violet, and has a molar conductivity similar to that of [Co(NH3)6)]Cl3. In contrast, CrCl3.5H2O is green and has a lower molar conductivity in solution. If a diluted solution of this green complex is allowed to stand for several hours it turns violet. Interpret these observations with structural diagrams.

      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 CrCl3.6H2O has a conductivity similar to that of [Co(NH3)6)]Cl3 i.e. it is probably also a 1:3 cation:anion electrolyte. So, the real formula is probably [Cr(H2O)6)]Cl3.

      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 Cr3+ 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(H2O)5)Cl]Cl2 or [Cr(H2O)4)Cl2]Cl

      In aqueous solution, the coordinated Cl- is gradually replaced by H2O, and the violet hexaquo complex is formed.

    2. The reaction of [ZrCl4(dppe)] with Mg(CH3)2 gives a compound [Zr(CH3)4(dppe)] where the NMR spectrum shows that the CH3 groups are all equivalent. Which geometry, octahedral, or trigonal prismatic, is consistent with this observation. (The ligand dppe is F2PCH2CH2PF2.)

      The two structures would be:

      The prismatic structure (on the left) has two planes of symmetry which render the 4 CH3 groups equivalent. The octahedral structure has a two fold axis only and there are two CH3's cis to both the P's, and two trans to one P and cis to the other, so they are inequivalent.

  1. Determine the d-electron configuration, the number of unpaired electrons, and the ligand field stabilization energy (LFSE) in terms of Do or Dt as appropriate for each of the following complexes. Please make it clear how you make any necessary decisions:

    1. [Fe(CN)6]4-

      This is surely a strong field d6 case because of the combination of Fe2+ and CN-.

      LFSE = 6 x 0.4Do = 2.4Do (The negative sign is "understood" since it is a stabilization energy.)


    2. [Fe(H2O)6]2+

      This is probably a weak field d6 case (Fe2+ and H2O)

      LFSE = (4 x 0.4 - 2 x 0.6)Do = 0.4Do

    3. [NiCl4]2-
      Chloride is quite low in the spectrochemical series, so this Ni2+ d8 complex will probably be tetrahedral.

      LFSE = (4 x 0.6 - 4 x 0.4)Dt = 0.8Dt


    4. [Ni(CN)4]2-
      Cyanide is high in the spectrochemical series so this is Ni2+ d8 complex will probably be square-planar.

      LFSE = (4 x 0.514 + 2 x 0.428 - 2 x 0.228)Do = 2.245Do

      To work this out you would have to remember the not very memorable levels: my mistake!


    5. [Ni(CO)4]


      This is a Ni0 d10 complex and the ligand field stabilization energy is zero in any geometry, so there is no reason for it not to be tetrahedral.


    1. The rate constants for the formation of of [CoX(NH3)5]2+ from [Co(NH3)6)3+ for X = Cl-, Br-, N3-, and SCN- differ by no more than a factor of two. What mechanism of substitution does this observation support?

      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!

    2. The equilibrium constants for the successive reactions of ethylene diamine with Co2+, Ni2+, and Cu2+ are as follows:

      [M(H2O)6]2+   +   en      [M(H20)4en]2+   +   2H2O      K1

      [M(H2O)4en]2+   +   en      [M(H20)2(en)2]2+   +   2H2O      K2

      [M(H2O)2(en)2]2+   +   en      [M(en)3]2+   +   2H2O      K3

      Ion log K1 Log K2 Log K3
      Co2+ 5.89 4.83 3.10
      Ni2+ 7.52 6.28 4.26
      Cu2+ 10.55 9.05 -1.0

      1. Why is there a general trend: K1 > K2 > K3

        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.

      2. Why is the value of K3 for Cu2+ so different?

        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(H2O)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(H2O)2(en)2]2+      [M(en)2]2+   +   2H2O      K4

      3. Why are the corresponding values for Ni2+ greater than those for Co2+?

        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. Do, will be greater for the smaller Ni2+ ion, so the differences should also be greater, hence the greater stability.

    1. Deduce the Russell-Saunders terms for the configuration p2.

      The table below shows the 15 microstates possible for p2 in no particular order.

      ml
       ML  1 0 -1
      2     MS = 0
      0    
      -2    
      1   MS = 1
      0  
      -1  
      1   MS = -1
      0  
      -1  
      1   MS = 0
      0  
      -1  
      1   MS = 0
      0  
      -1  

      The next table contains a count of the number of microstates with the same ML and MS values that can be found in the table above.

      MS
      1 0 -1
      ML 2 0 1 0
      1 1 2 1
      0 1 3 0
      -1 1 2 1
      -2 0 1 0

      The maximum value of ML for which a microstate exists is 2 (in red), which must be part of a D term (i.e. with L = 2 so ML can also be 1,0,-1 and -2), and the only MS value for which this term is found is 0 so it is part of a 1D term. (S = 0.) Now remove the 5 microstates corresponding to this from the above table to get the next.

      MS
      1 0 -1
      ML 2 0 0 0
      1 1 1 1
      0 1 2 1
      -1 1 1 1
      -2 0 0 0

      The maximum value of ML for which a microstate exists is 1 (in red) which must be part of a P term (i.e. with L = 1 so ML can also be 0 and -1), and the largest MS value for which this term is found is 1, so it is part of a 3P term. (i.e. S = 1, so MS can also be 0 and -1.) Now remove the 9 microstates corresponding to this from the above table to get the next.

      MS
      1 0 -1
      ML 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 ML = 0 and MS = 0 which is a 1S term. (i.e. L = 0 and S = 0)

    2. Using the information supplied, estimate or calculate Do and the Racah parameter, B, for [Ni(H2O)6]2+, which has d-d absorptions at 8 500, 15 400 and 26 000 cm-1.

      The complex contains Ni(II), a d8 ion in octahedral coordination. As the question implies, there are two methods for solving this problem; graphically, and more precisely by calculation:

    3. Estimated from Tanabe-Sugano diagram

      The diagram for d8 shows that the ground state is 3A1g with spin-allowed transitions (in order of increasing frequency or wavenumber) to 3T2g, 3T1g (from the 3F term) and 3T1g (from the 3P 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 Do.

      To find B from the Tanabe-Sugano diagram, note that the experimental ratio n2/n1 = 1.72, and find the point along the Do/B axis that yields the same ratio. By trial and error, I found that at Do/B » 7.1, the required ratio is obtained. Therefore, B = 8500/7.1 = 1197 cm-1. (The free ion value for Ni2+ is 1041 cm-1.


    4. Calculation Method This method is applicable if all three d - d transitions are identifable, which is the case here.


      This diagram is constructed using the information from Huheey Table 9.26 (distributed with the exam), and the knowledge that the 3P term is 15B above the 3F (which is also evident from the Tanabe-Sugano diagram).

      The diagram for d8 shows that the ground state is 3A1g with spin-allowed transitions (in order of increasing frequency or wavenumber) to 3T2g, 3T1g (from the 3F term) and 3T1g (from the 3P term).


      The following equations can be obtained from the spacings marked on the diagram:

      Do   =   8500 cm-1         (1)

      1.8Do   -   x   =   15400         (2)

      15B   +   x   +   1.2Do   =   26000         (3)

      Adding (2) and (3) gives:

      15B   +   3.0Do   =   41400

      and substituting from (1) gives:

      B = 1060 cm-1