Problem Set #4 – (Level 1 problems)

1. The pKa for ferric ion (Fe3+) is 2.   The pKa for cupric ion (Cu2+) is 9. 

       Which is a stronger Bronsted acid?  Give and explain two reasons for the ten   
       million fold difference in their acidities, using the concepts discussed in class.

Since ferric ion has a smaller pKa and pKa = -logKa, it is the stronger Bronsted acid. The two reasons for its greater acidity are:

• higher oxidation which leads to a higher charge density on Fe(III). This increases the polarity of Fe-O-H bonds, weakening the O-H.
• Fe(III) is a hard cation and will make stable interactions with hard anions, like -OH-, the product of deprotonated H2O.

2. My mother always told me to take iron vitamins with orange juice because of its high concentration of citric acid (structure below).   Analyze her suggestion as a chemist:  what aspects of the citric acid would be helpful to taking an iron nutritional supplement?

Three aspects of citric acid will favor its binding by Fe(III) or Fe(II).

• Each of its 3 carboxylate groups, -COOH, are acidic andcan be deprotonated to make anionic Lewis bases that will bond to any metal ion.
• Fe ions particuarly like anionic O-donor ligands.
• The multiple carboylate sites means citrate is a polydentate chelate ligand.

3. Will substituting all the water ligands in [Ni(H2O)6]2+. by ammonia ligands to form [Ni(NH3)6]2+.
     cause a change in spin state for nickel?  Explain you answer with appropriate CFT arguments.

The change in ligands will have not effect on the Ni(II) spin state. This is because Ni(II) is d8 and in a Oh crstal field htere is only one way to put 8 electrons into the five d orbitals: all the stabilized T2g orbitals (xz,yz and xy) are filled and there is one electron in each of the destabilized Eg orbitals (z2, x2-y2). The difference will show up in the CFSE where the crystal field splitting parameter is much larger for ammonia ligands.

4.   (a) Draw and label all the isomers possible for the complex Ru(phen)2(Br)2. (see picture below)
      (b) What is the formal oxidation state of ruthenium in the complex Ru(phen)2(Br)2? --- Ru(2+)
      (c) What are the valence electrons on ruthenium in the complex Ru(phen)2(Br)2 ? -- 4d6
      (d) What is the spin state of Ru(phen)2(Br)2? --Low spin; all 6 electrons in T2g orbitals, because 2 x phen ligands are strong field in addition to 2nd row Ru causes large crystal friled splitting.
      (e) What is the CFSE for Ru(phen)2(Br)2? CFSE = 6 x -0.4 delta + 3 P
      (f) Does the replacement of Br as a ligand by DMSO (dimethyl sulfoxide) create additional
           isomerization possibilities?  Explain. Yes, it creates more isomers for two reasons:

• DMSO can coordinate throguh the S or the O atom, and that creates linkages isomers, in addition to the cis/trans and enantiomers ( below).
• Also, ionization isomers are now possible for [Ru(phen)2(DMSO)2](Br)2. That is, in solution the formula [Ru(phen)2(DMSO)2](Br)2 could be either a dicationic [Ru(phen)2(DMSO)2]2+ with 2 Br- counter ions OR [Ru(phen)2(DMSO)(Br)]+ (Br-) . DMSO , etc.

LEVEL 2 PROBLEMS

1. The plot below shows the log of the formation constants logKf, vs. some first row metals for the formation of [M(L-L)3], from M(2+) and bidentate ligand L-L where L-L is one of the four following bidentate ligands: oxalate, glycine, ethylenediammine, aminothiolate.

Here are the questions:

(a) Write the reaction that corresponds to Kf.

M(2+) + 3 (L-L) <==> M(L-L)3

(b) Using the concepts we covered in Chapter 7, explain the trends observed in the magnitude of Kf for the four different bidentate ligands as they make complexes with the metal ions indicated in the plot.

• The early metals (Ba, Sr, Ca, Mg) are hard. They make the most stable complexes ( largest Kf) with the hard O-donor ligand oxalate, and less stable complexes with glycinate. They do not make stable complexes with the softest ligand aminothiolate.
• The metals get increasingly soft going from Fe, Co, Ni to Cu. They make the most stable complexes ( largest Kf) with the softest S-donor ligand aminothiolate.Their complexes with ethylenediamine (a intermediate to soft base) are less stable but more stable than for complexes of oxalate.

(c) Using the concepts introduced in Chapter 21, what is special about Mn(2+) that all the lines cross there?

Mn(II) is a d5 ion and as such has a particular stability with all 5 electrons unpaired in a high spin state and so it tends to form high spin complexes with most O- and N- donor ligands. With one electron in each d-orbital, CFSE =0 and so there is no difference in stability (according to CFT) in complexes of Mn(II) with different ligands.

(d) What is special about Zn(2+) that all the lines have a sharp decrease there?

Zn(II) is a d10 ion and has all d orbitals completely filled. CFSE =0 and so there is NO extra CF stability for Zn(II) complexes, as compared to Cu(II) or other M(II) complexes with the same ligands. Therefore its overall Kf values in the plot drop as compared to Cu(II).

2.  (a) What is the most likely geometry for PdBr2(NH3)2 ?  Defend your answer with appropriate CFT arguments.

Because Pd is a 2nd row metal d8 ion, it has a large CF splitting and is ALWAYS square planar, never tetrahedral.

(b) One isomer of PdBr2(NH3)2 is unstable with respect to a second isomer.  The isomerization process can be followed by IR spectroscopy.  The IR of the first isomer shows absorptions at 480 and 460 cm-1 assigned to Pd-N stretching modes. During the isomerization the band at 460 gradually disappears and the band at 480 shifts to 490 cm-1.  Explain these observations.

The square planar complex can occur as cis and trans isomers. The first complex is the cis isomer becuase it shows two Pd-N stretches ( symmetric 480 cm-1 and antisymmetric 460 cm-1 ). This isomer slowly rearranges to teh trans isomer which only has one IR active antisymmertic Pd-N stretch at 490 cm-1.