1. Name the following complexes.  Do they follow the 18 electron rule?

 

h6-benzenetricarbonyl chromium(0); yes

 

tetrachloro(h2-ethylene) platinate(IV); no

 

  1. Calculate the ligand field stabilization energy (LFSE) for all cases from d1-d10, high and low spin.  Which of these have orbital angular momentum?

 

I’m only going to show the octahedral case.  You should also be able to do the tetrahedral case.

 

LFSE       d1 = -0.4 ,  o.a.m   

                d2 = -0.8,   o.a.m

                d3 = -1.2 ,  no

high spin d4 = -0.6,   no

high spin d5 =   0,      no

high spin d6 = -0.4,   o.a.m.

high spin d7 = -0.8,   o.a.m.

               d8 = -1.2,    no

               d9 = -0.6,    no

               d10=   0,      no

low spin  d4 = -1.6,   o.a.m.

low spin  d5 = -2.0,   o.a.m

low spin d6 = -2.4,    no

low spin d7 = -1.8,    no

 

 

  1. Calculate the LFSE for the following complexes.  Will these complexes have orbital angular momentum?

 

[Mn(OH2)6]2+ 

 

high spin d5; -0.4(3) + 0.6(2) = 0; no orbital angular momentum

 

[Ru(NH3)6]3+

 

low spin d5; -0.4(5) + 0.6(0) = -2 Do; yes

 

[Mn(CN)6]3-

 

low spin d4; -0.4(4) + 0.6(0) = -1.6 Do; yes

 

 

 

  1. Explain the differences in the magnetic moments for these two iron complexes.

 

[Fe(C2O4)3]3-                 meff = 5.85-5.95 BM

 

[Fe(CN)6]3-                  meff = 2.3-2.4 BM

 

Use the spin only equation for magnetic moment for each case.

 

ueff = [n(n+2)]1/2

 

In the first case, iron is high spin and in the second case iron is low spin (look at the difference in the magnetic moment).  In the first case, high spin d5 results in no orbital angular momentum and the theoretical value of the magnetic moment calculated from the spin only equation (5.9 BM) matches well with the experimental value found.  In the second case, low spin d5 leads to one unpaired electron, but the theoretical value of the mag. moment (1.73 BM) is considerably lower than the experimental value.  Thus, we must invoke orbital angular momentum which is possible in the low spin d5 case.

 

  1. Explain the trends in figure 13-16 from the notes (both the trend with the different metals and with the different ligands).

 

The trend in log K1 vs. number of d electrons can be explained using Zeff, LFSE, and the Jahn-Teller effect.  The trend of increasing K1 with smaller ligands can be explained based on the ‘bite’ size of the ligands (the 3d metals prefer ligands that can form 4 or 5 membered rings over those which form 6 membered rings).

 

  1. Explain the trend in formation constants.  Calculate KTOT.

 

                                             K1

[Cu(H2O)4]2+ + NH3 --- [Cu(H2O)3(NH3)]2+ + H2O

                                   

                                    K2

                                    K3

 

                                           K4

[Cu(H2O)(NH3)3]2+ + NH3 ---[Cu(NH3)4]2+ + H2O

K1 = 1.66´104

K2 = 3.16´103

K3 = 8.31´102

K4 = 1.51´102

 

The formation constants decrease with each successive addition of NH3.  With each addition, the copper becomes less acidic, thus it becomes more difficult to add another ammine.

 

Ktot = K1K2K3K4 = 6.58 ´ 1012

 

  1. Explain the trend in free energy of formation

 

Metal                     Ligand              DG° (kcal/mole)

Hg2+                          F-                    -1.4

                              Cl-                   -9.19

                              Br-                   -12.8

I-                                           -17.5

Hg2+ can be classified as ‘soft’, and the ligands are arranged in increasing ‘softness’.  The softer the ligand, the more favourable the addition of the ligand to the ‘soft’ metal.

 

 

  1. Comment on the trends in K. 

 

K1

[M(H2O)6]2+ + en ® [M(H2O) 4(en)]2+ + 2H2O

 

                              K2

[M(H2O) 4(en)]2+ + en ® [M(H2O) 2(en)2]2+ + 2H2O

 

                              K3

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

 

M =                 Co2+                 Ni2+                  Cu2+                

 

K1                    5.89                 7.52                 10.55

 

K2                    4.83                 6.28                 9.05

 

K3                    3.10                 4.26                 -1.0

 

For each of the metals, the formation constant decreases with each additional ligand added (see question 6).  For K1, and K2, the constants increase from Co2+ to Ni2+ to Cu2+.  This can be explained by a number of factors: Zeff, LFSE, and Jahn-Teller effect (on a test, explain each factor in detail).  Finally, for K3, there is a drastic drop for Cu2+ due to the Jahn-Teller effect.