CH511 Problem Set 2

1. A pellet formed from a pure solid powder is attached to 2 electrodes and the electrical conductivity measured at different temperatures. The data obtained are listed below:

        What do you now know about the band structure of this solid? If it is a pure element, which can it be?

ln σ =  ln A - E_g / 2kT

 

so an Arrhenius plot of ln σ vs. 1/T should be linear and have slope = -E_g / 2kT. Linear regression gives a slope = -4.1 E+3 K^-1, and solving,  E_g = 0.67 eV. This is close to the reported value for Ge (s).

2.     HF autoionizes to form H₂F⁺(solv) and HF₂^-(solv). Draw a molecular orbital diagram for the linear bifluoride anion, giving the expected electronic configuration and the bond order. Compare this model with the Lewis structure for the anion.

                

You  can draw the Lewis structure as a resonance between the fluoride anion and HF

F-H    F^-    <-->    F^-   H-F

(this doesn't show the lone pairs, the F  ion has 4, the F bound to H has 3)

This suggests the bond order is 1/2, which agrees with the MO diagram.

3.    Shriver/Atkins Problem 4.3 (p. 139)

CO₂ (g) + H₂O (liq)  =  H₂CO₃ (aq)                                                (1)

M₂SiO₄ (aq) + 2H₂CO₃ (aq)  =  Si(OH)₄ (s)  +  M₂CO₃ (aq)         (2)

Adding M₂SiO₄ to a solution will reduce the carbonic acid concentration via equilibrium (2). This allows the first equilibrium to move to the right, thereby dissolving more CO₂ (g).

4.    Shriver/Atkins Problem 4.5 (p. 139)

As we go left to right, the metal ion size decreases (for a given charge) due to increased effective nuclear charge. Shorter, stronger M-O bonds are expected going left to right across a period, this is observed by an increased frequency in the M-O bond stretching mode. (higher frequency indicates a stronger bond). The increased charge density at the metal cation also results in a more acidic complex. Ni²⁺ (aq) has the highest M-O (stretch) frequency and is the most acidic complex in the series.

5.

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Discuss the three Pourbaix diagrams above for Ti, Cu, and Au in terms of (1) periodic trends, and (2) the naturally occurring forms of these elements.

Ti is an early TM, readily oxidized to the group number. We see the only thermodynamically stable oxidation state in aqueous solution is 4+. Cu is a late TM, much high effective nuclear charge, much harder to oxidize. We see oxidation numbers 0 (Cu meta), +1 (except at low pH), and 2+ are stable in aqueous solutions. For Au, the heaviest congener in the Cu triad, oxidation is much less favored, and only Au metal is thermodynamically stable in aqueous solution. The chemistry reflects these properties. Ti is found only as TiO₂ or a titanate naturally, and the reduction to form Ti(m) is energy intensive. Cu occasionally is found in metallic form naturally, but is more common as a cuprate ore. Au is found in metallic form. However, in the presence of an aqueous halide solution (for example seawater), Au⁺(aq) can be stable.

6.    Shriver/Atkins Problem 5.9 (p. 168)

Hg²⁺ and Hg₂²⁺ are both good oxidizing agents. Conversely, Hg (m) is a very mild reducing agent. Hg₂²⁺ will not undergo disproportionation in acidic aqueous solutions.

7.    Shriver/Atkins Problem 5.12 (p. 168)  In addition, discuss the effect of pH on the stability of permanganate solutions.

the balanced half-reaction is:

            3e^-  +  MnO₄^- (aq)  +  4 H⁺ (aq)   →   MnO₂ (s)  + 2H₂O (l)   

            and E⁰  =  1.69 V at pH = 0

Note that you cannot balance the half-reaction with OH^- and also use the Nernst equation with E⁰ at pH = 0, that will give the wrong answer.

            the Nernst equation predicts that:

            E  =  E⁰  -  (0.059 V / 3) log ( 1 / [H⁺]⁴)  =  1.69 V - (-4)(0.020 V)(pH)

            at pH = 9, E = 1.69 - (4)(0.020 V)(9) = + 0.98 V

The potential is less favorable at the pH increases (MnO₄^- is as weaker oxidant). This makes sense since H⁺ is a reactant in the half-reaction.

          The oxidation potential of water at pH = 0 is -1.23 V, i.e. this is the potential for the half-reaction:

          2 H₂O (l)    →   O₂ (g)  +  4 H⁺ (aq)  +  4 e^-

Since permanganate reduction at this pH has E for= +1.69, the hydrolysis of permanganate, i.e. the redox reaction where permanganate is reduced and water is oxidized, will be favorable, with E = +1.69 - 1.23 = +0.46. The reaction occurs at a slow rate because is involves multiple electron transfer, and also involves breaking/making strong bonds, with a potential is less than +0.6 V. At pH = 9, the potential for both half-reactions changes. As we saw above, permanganate reduction now has E = +0.98. For water oxidation, E = -1.23 V + (0.059 V)(pH) = -1.23 V + 0.53 V = -0.70 V. The potential for the redox reaction of water and permanganate therefore is E = 0.98 V - 0.70 V = +0.28 V at pH = 9. This reaction is still thermodynamically  favorable at pH = 9, but the potential is lower and permanganate is more kinetically stable. To confirm this, look at the Pourbaix diagram for Mn in the notes. As pH increases, the blue line indicating the MnO₄^- / MnO₂ equilibrium is always above, but becomes closer to, the red line for the O₂ / H₂O equilibrium.

8.    Describe the expected ¹⁹F NMR spectra for PF₂Cl₃ and PF₃Cl₂ at temperatures above and below that required for rearrangement due to a Berry-type mechanism.  Include peak splittings due to F-F and P-F coupling where appropriate.

PF₂Cl₃  :  both F will be axial in the most stable structure.  Either at high or low temp., the F's are equivalent, so only one NMR signal for F should be expected. The peak position would shift, depending on the contribution from equitorial site occupancy (none at low T, 3/5 at high T). The peak will be split by ³¹P-F coupling to give a doublet.

PF₃Cl₂    :   at lower T, where no rearrangement can occur, we will see 2 peaks, in a 2:1 intensity ratio, corresponding to axial and equitorial F's. F-F coupling will cause the larger peak to be a doublet, the smaller peak to be a triplet. Each of these will be split by P-F coupling, yielding a doublet of doublets, and a doublet of triplets, respectively. At higher T, where these can exchange, all F are equivalent and we would see only a doublet (due to P-F coupling).