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3810 Problem set 6 answers in 2003

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3810 Problem set 6 answers in 2003 Powered By Docstoc
					Chemistry 3810                              Answers to Problem Set #6

Topic: Chemistry of the Group 13 Elements. You are responsible for Ch. 10, sections 10.1-10.6 of Shriver-Atkins

1.   Preferably without consulting reference material, list the elements in Groups 13 and 14 and indicate (a) the metals and
     nonmetals, (b) those that can crystallize in the diamond structure, (c) the most oxophilic elements in these groups.




2.   Draw the B12 unit that is a common motif of boron structures; take a viewpoint along a C2 axis.



                                                                              C2
                                                                 C2



     The C2 axes are perpendicular to the principal C5 axis of the icosahedron. The first picture shows the location of the axis, and
     the second view, twisted about the vertical by 90°, shows the two-fold rotation relationship by peering perpendicular to the C2,
     whose location is indicated by the blue dot.

3.   Give balanced chemical equations and conditions for the recovery of boron from its ores.
     Boron is recovered from the mineral borax, Na2B4O5(OH)4⋅8H2O, by formation of B2O3 through acidification and dehydration,
     followed by treatment with magnesium, as shown in the exothermic reaction:
                                            B2O3     + 3 Mg       → 2 B + 3 MgO

4.   Arrange the following in order of increasing Lewis acidity: BF3, BCl3, SiF4, AlCl3–. (b) In the light of this order, write balanced
     chemical reactions (or NR) for
     (i) SiF4N(CH3)3 + BF3 →                (ii) BF3N(CH3)3 + BCl3 →              (iii) BH3CO + BBr3 →
5.   Using BCl3 as a starting material and other reagents of your choice, devise a synthesis for the Lewis acid chelating agent, F2B–
     C2H4–BF2.




6.   State the coordination numbers of boron, carbon, and silicon in their oxoanions and devise plausible explanations for the
     differences based on Lewis electron dot structures.




7.   Which boron hydride would you expect to be more thermally stable, B6H10 or B6H12? Give a generalization by which the
     thermal stability of a borane can be judged.




8.   (a) Give a balanced chemical equation (including the state of each reactant and product) for the air oxidation of
     pentaborane(9). (b) Describe the probable disadvantages, other than cost, for the use of pentaborane as a fuel for an internal
     combustion engine.




9.   (a) From its formula, classify B10H14 as closo, nido, or arachno.
     (b) Use Wade's rules to determine the number of framework electron pairs for decaborane(14).
     (c) Verify by detailed accounting of valence electrons that the number of cluster valence electrons of B10H14 is the same as that
     determined in (b).
10. Starting with B10H14 and other reagents of your choice, give the equations for the synthesis of [Fe(nido-B9C2H11)2 ] 2–, and
    sketch the structure of this species.




11. (a) What are the similarities and differences in the structures of layered BN and graphite? (b) Contrast their reactivity with Na
    and Br2 (c) Suggest a rationalization for the differences in structure and reactivity.




12. Devise a synthesis for the borazines (a) Ph3N3B3Cl3 and (b) Me3N3B3H3, starting with BCl3 and other reagents of your choice.
    Draw the structures of the products.
13. Give the structures and names of B4H10, B5H9, and 1,2-B10C2H12.




14. Arrange the following boron hydrides in order of increasing Brønsted acidity, and draw a structure for the probable structure
    of the deprotonated form of one of them: B2H6, B10H14, B5H9.




15. The lightest p-block elements often display different physical and chemical properties from the heavier members. Discuss the
    similarities and differences by comparison of:
    (a) The structures and electrical properties of boron and aluminum.
    (b) The structures of the halides of boron and aluminum.
    (a) The various allotropes of elemental boron consist of icosahedral B12 cages connected together in various ways. In contrast,
        aluminum is a cubic close-packed metal at all temperatures. In harmony with the more localized bonding in boron, it is a
        semiconductor whereas aluminum is a good metallic conductor. Thus the metalloid boron has distinctly covalent bonds in
        its elemental form, while the metal aluminum has a metallic lattice.
    (b) The boron halides are trigonal planar molecules that are gases or volatile liquids at ambient conditions. In contrast,
        aluminum halides are solids with extended lattices for X = F and Cl and with dimeric Al2X6 molecules for X = Br and I.
        The differences between halides of boron and aluminum can be traced to their relative sizes. Since boron is so small it (i)
        resists formation of the halide bridges present in the AlX3 structures and (ii) can form bonds with halogens that have partial
        double bond character, in marked contrast with Al–X bonds, which have little or no double bond character.
                                            Br        Br                    Br
                                                                                       Br
                                                  B              Br
                                                                        Al      Al
                                                  Br             Br                    Br
                                                                            Br
                                             monomeric BBr3            dimeric "AlBr3"


16. Boron-11 NMR is an excellent spectroscopic tool for inferring the structures of boron compounds. Under conditions in which
    the 11B–11B coupling is absent, it is possible to determine the number of attached H atoms by the multiplicity of a resonance: BH
    gives a doublet, BH2 a triplet, and BH3 a quartet. Also, B atoms on the closed side of nido and arachno clusters are generally
    more shielded than those on the open face. Assuming no B-B or B-H-B coupling, predict the general pattern of the 11B-NMR
    spectra of (a) BH3CO, (b) [B12H12] 2–, and (c) B4H10. Then sketch predicted 1H NMR spectra for molecules (a), (b) and (c).
    Terminal hydrogens on boranes do not show resolved coupling to other hydrogen nuclei. Also bridging hydrogen atoms
    typically appear as singlets with no resolved coupling to any nucleus, because of chemical exchange.
17. Identify the incorrect statements in the following description of Group 13 chemistry and provide corrections along with
    explanations of principles or chemical generalizations that apply.
     (a) All the elements in Group 13 are nonmetals.
          False. Boron is a non-metal. But the rest from aluminum to thallium are all metals.
     (b) The increase in chemical hardness on going down the group is illustrated by greater oxophilicity and fluorophilicity for
          the heavier elements.
          False. This is the reverse of the actual trend.
    (c) The Lewis acidity increases for BX3 from X = F to Br and this may be explained by stronger Br–B π bonding.
          False. The Lewis acidity does increase for BX3 from X = F to Br and this may be explained by weaker Br–B π bonding
          compared to B–F.
     (d) Arachno-boron hydrides have a 2(n + 3) skeletal electron count and are more stable than nido-boron hydrides.
          False. Arachno-boron hydrides do have a 2(n + 3) skeletal electron count and are less stable than nido-boron hydrides.
     (e) In a series of nido-boron hydrides acidity increases with increasing size.
          True. This is explained by an increase in electron delocalization, stabilizing the anions that develop upon donation of H+.
     (f) Layered boron nitride is similar in structure to graphite and, because it has a small separation between HOMO and
         LUMO, is a good electrical conductor.
         Boron nitride is a poor conductor, and in fact has a very large HOMO-LUMO gap.

18. Acetylcholine, [(CH3)3N(CH2)2OC(O)CH3] +, is an important neurotransmitter, and the physiological properties of the neutral
    boron analog (CH3)2(BH3)N(CH2)2OC(O)CH3 are of interest. (B.F Spielvogel, F.U. Ahmed, and A.T. McPhail, Inorg. Chem.
    25, 4395 (1986).) Devise a method of synthesis for this analog starting with [(CH3)2HN(CH2)2OC(O)CH3] + and other reagents
    of your choice.
                          H
                     H3C     +         O        CH3                          H3C ⋅ ⋅         O      CH3
                           N                                 NaOH                   N
                             CH3             O                                      CH3              O


                                                                                             BH3.OEt2


                                                                                  H3B
                                                                                        +        O        CH3
                                                                            H3C      N
                                                                                     CH3             O
     The method shown here depends on simple acid base chemistry, to remove the proton and hence reform the lone pair on
     nitrogen. This makes the amine a strong Lewis base. This donor group is now set up to form a complex with the acceptor
     borane. Borane can be supplied in a number of useful forms, the simplest being in solution using commercially available
     adducts with Lewis bases. Hence the choice of reagent and solvent for the reaction must be a species which is a considerably
     weaker Lewis base than the amine we wish to form the adduct with.
19. Bonding in a nido cluster. The nido cluster pentaborane(9) is formally derived from
    B6H62– by the removal of B–H2+ (i.e. with only two electrons), which formally produces
    the B5H54– cluster (not a known species, and indeed is highly unlikely to exist with such
    a huge charge). This species then formally picks up 4 H+, to produce the nido cluster as
    a neutral molecule. Its structure is shown at right.
    Study the bonding of this cluster as follows: consider the radial and tangential orbitals
    developed in the lecture notes (e.g. on page 100), and sketch what the result is of
    removing the bottom B–H unit. Consider only the filled orbitals for the sake of brevity.
    Using these reactive fragments, add in the four hydrogen 1s orbitals to them to create
    bonding molecular orbitals for the bridging H atoms. Consult the SAO table for the correct symmetries of four H 1s orbitals in
    the C4v point group. Then build the structure of B5H9 in HyperChem and minimize it at the AM1 level (neutral molecule).
    Identify the MO’s in the cluster that most closely resemble the orbitals that you have sketched out using the above approach.
    How many bonding MO’s do you expect? Once you have identified this number of MO’s among the filled levels in the AM1
    calculation, answer the following question: Does the AM1 method provide a bonding description consistent with the newly
    added, bridging, hydrogen atoms being the most reactive, indeed even acidic, hydrogen atoms in the cluster?

     Here are the B5H54– fragment orbitals:                              And here the SAO’s for Hydrogen in C4v.




           #5                 #6              #7




           #2              #3             #4




                              #1
     Now ask yourself, which fragment orbital is able to match with the SAO’s. This is also a hint as to the probable symmetry labels
     of the fragments in the reduced symmetry C4v point group. Just looking at it, and avoiding all purely orthogonal relationships,
     and placing the new hydrogen atoms in the correct location below the plane of the cluster, and between the boron atoms, it
     seems to me that #6 is ideally set up to match to B1, and #5 and #7 fit to the E set. Also, #3 and #4 clearly cannot overlap with
     any of the hydrogen SAO’s (orthogonal interactions), However, both #1 and #2 are capable of forming bonds to the A1 SAO,
     and this will clearly be a recipe for mixing of these wavefunctions. We might expect the net bonding interactions therefore to
     look something like:



                                                       +
                         #5               #7                         E                                   E




                                                   +
                                   #6                      B1                                       B1




                                               +                     +
                                   #2                      #1                A1
                                                                                                    A1
                                                                4–
     We thus expect there to be four filled orbitals in the B5H5 molecular model (not a realistic species because of the high charge)
     resembling fragments (mix of #1 and #2), #6, #5 and #7. Then on adding the four H+ (i.e. protons) in the indicated locations, we
     expect to find bonding orbitals resembling the right hand set of four MO’s. There are of course many empty, out-of-phase
     combinations as well, but we will ignore these in this problem. We now turn to AM1 calculations to check our ideas.
Structure of B5H54–                    Highest filled orbitals, energy and topology:




These fragments are from bottom to top labeled as 4a1, 1b1 and 3e. The positive energies reflect the excess of electrons over
nuclear charge. There is indeed a remarkable agreement between these four MO’s and the four we constructed on the previous
page. Now let’s see what happens when we construct the orbitals of B5H9 by addition of four H and setting the charge to
neutral. Note that the structure I use has been minimized, and so the radial B–H orbitals are tipped up a bit, etc.

Structure of B5H9                      Highest filled orbitals, energy and topology:




The orbitals we discover as the highest filled orbitals in the neutral molecule are labeled from the bottom up as 3a1, 1b1, 2e, 4a1
and 3e. Of these, 4a1 is extremely similar to the orbital of the same label in the fragment. Compared to it, both 2e and 1b1 have
dropped lower in energy. Thus the best bridging-hydrogen bonding MO is certainly 1b1. The character of 2e is quite different
from that of the e fragment orbital, which more strongly resembles the HOMO of the molecule, 3e, and this is primarily a B-B
bonding orbital along the unbridged edges of the cluster. 2e is a fairly even mixture of terminal and bridging B-H bonding. This
difference suggests once again the limitation of using a “partial MO” description by arbitrarily picking only the inward-pointing
radial orbitals and ignoring those that involve bonding to the terminal H’s. 3a1 ends up having no contribution from the
bridging H-atoms, so that indeed of our original fragments, it turns out that #2 was more important to bridge bonding than #1.
The reason must be that #1 is considerably lower in energy and thus does not interact as strongly with the weakly-held bridging
H atoms.

As to reactivity, the AM1 calculations indeed show that some of the highest lying (but not really the HOMO itself) are B-H-B
bridge bonding MO’s (such as 4a1). This lends support to the notion that the most acidic H atoms are those in the bridging
positions. The HOMO itself is very similar to one of the t2g set of B-B edge bonding MO’s as seen in the closo B6H62– structure
(notes, p. 100).
20. Develop a bonding model for the closo-B5H52– cluster, using only the radial and tangential cluster-forming orbitals. Use a
    HyperChem AM1 calculation to guide you, but remember that it is not possible to reliably derive the orbitals shapes without
    reference to the symmetry adapted sets of orbitals. These orbitals in a trigonal bipyramid geometry are just a combination of a
    linear diatomic, now arranged vertically, and a trigonal planar set, i.e. SAO’s that we are extremely familiar with in this
    course. However, the linear orbitals need to be renamed in the D3h point group, and that has been done below.

     We start by taking just the framework orbitals from among the full set of SAO’s. This effectively equals the formation of sp
     hybrid orbitals from the s and the radial p orbitals, and then using only the inward pointing hybrids. The outward pointing
     hybrids are used to form the low-energy B–H bonds. This is done on the left hand side of the diagram below, using the SAO’s
     provided with the question (see end of this answer guide for that original diagram).

     Then we combine the in-plane and out-of-plane SAO’s according to their symmetry labels, remembering that each time two
     SAO’s are combined, two new ones must be formed. This is accomplished by combining them in-phase and out-of-phase. The
     results of these operations are depicted on the right hand side of the diagram. The order of the labels are chosen by considering
     the number of nodes introduced upon formation of the combined SAO’s. The orbitals have also been grouped according to
     whether they are radial, mixed or tangential, remembering that in closo-B6H62– the most stable filled cluster MO’s were pure
     radial, then came the mixed, and finally the pure tangential were highest and formed the HOMO. Based on the valence electron
     calculation, we expect the closo-B5H52– cluster to have [22 – 10] = 12 cluster electrons, or six filled cluster MO’s (and of course
     five primarily B–H bonding MO’s at lower energy.
  Cluster-building orbitals for B5H52- (radial and tangential)
   radial                                                   pure radial         •                     •


          A1'                             •
                             D                       •             1A1'         •         2A1'        •
                               3h
                                             •             •
                                       A1'       A2"                                 •                 •
  E' set
                                                               mixed radial and tangential
                                             •             •


   tangential                                                             1A2"               2A2"

           A2"                                                                   •       •                            •        •

                                             •         •
  E" set                                                        1E' set          •                    3E' set         •

                                             •         •
                                    E' set                                                                            •        •
                                                                                 •       •

                                             •         •
                                                                pure tangential
           A2'                                         •
                                             •
                                                                2E' set                                         A2'
  E' set                            E" set   •         •



                                             •         •


                                                               1E" set                              2E" set



     Can we guess which these filled orbitals are? We must consider both the shape of
     the orbitals (degree of overlap) and the number of nodes. Remember that nodes
     between nuclei are the really destructive ones; nodes at the nucleus of a boron atom
     from a p-wavefunction are not inimical to bonding (though obviously not as good as
     pure s orbitals, but we have done away with those by using hybrid orbitals in the first
     place! My choice based on these considerations is as follows:
                                  1A1’ < 1A2” < 1E’ < 1E”
     I would not want to choose beyond that, but if correct, those will accommodate the
     12 electrons. Now we perform a HyperChem AM1 calculation on the dianion, and
     find the expected 11 filled orbitals, of which the upper six are displayed at right, and
     indeed they are from the bottom 3a1’, 2a1”, 2e’, 1e”. When the MO’s are plotted,
     they are remarkably similar to the predicted SAO’s (after making allowance for the
     arbitrary interpretation of the degenerate e sets by the software.) The major
     difference is that all except the pure tangential 1e”, all these MO’s have also got B–H
     bonding character. This discrepancy in viewpoint was developed in the notes.
21. In lecture we derived a bonding scheme for borazine, which also applies to other 13/15 element combinations. Write the Lewis
    structure of borazine, and compare it to the MO description. Use HyperChem to model the structure using AM1. Calculate the
    atomic charges, and use charge, bond order and orbital topology to develop as complete an understanding of the bonding in
    such 13/15 “heteroaromatic” six-membered rings.

     The bonding in borazine is developed in detail in the notes (see page 91). The π-MO’s of the molecule are also depicted there.
     However, the clearly present π-bonds in these diagrams suggest double bonds between B and N, and this will result in negative
     formal charge on the boron atoms. In the following diagram the Lewis diagram referred to here is shown, and also the actual
     calculated nuclear charges form an AM1 calculation.
                                            H                                    H
                                              _
                                   H        B + H                        H       B+0.15 H -0.01
                                        N+      N                            N-0.35 N
                                      _           _
                                        B + B                                B       B
                                   H        N        H                   H       N      H +0.22

                                            H                                   H
`    The Lewis structure clearly predicts the transfer of a unit of charge onto boron. However, the MO approach shows charge
     distributions in the molecule opposite in direction of charge transfer. Indeed this is the case, and while the π-system of the
     molecule transfers charge from nitrogen to boron, there must be a compensating mechanism operating among the σ-orbitals.
     Indeed, the σ contribution cannot be ignored if one is to fully understand the properties of this molecule. Based simply on
     electronegativity, we would predict that the nitrogen atom should be the more negative, and boron the more positive. The
     calculations do show net positive charge on the boron, and negative on nitrogen. However, the differences are not large, and
     this is thus explained by compensating σ and π charge drifts in this molecule. Thus the B–N bonds are unexpectedly non-polar,
     and H3BNH3 has very similar physical properties to that of benzene, its formal analogue among carbon compounds.
                H                                 H
                  _
      H         B + H                     H       B+0.15 H -0.01              Charge drift from single and double bond is opposite
           N+       N                         N-0.35 N                                 δ+ δ-                δ- δ+
         _                                                                             N B           vs.    N• B
                                                                                                              •
           B    + B_                          B        B
      H         N       H                 H       N        H +0.22

              H                                    H

22. Benzene is formally a trimer of ethyne. By analogy, borazine is a trimer of the linear H–B–N–H molecule with a formal B–N
    triple bond. Use AM1 to develop a full MO diagram for a linear neutral H–B–N–H. Then create an interaction diagram
    between three such molecules and the borazine diagram developed in 21. Discuss the changes in structure that accompany
    trimerization.
    The Lewis structures for linear H–B–N–H are the following resonance pair:
                                                                                                ••                  +
    An AM1 calculation on this system gives the energy level diagram below, and          H B N H           H B N H
    the MO’s depicted next to it. The orbitals are the familiar set for C∝v: 1σ, 2σ,
    3σ, 1π, 4σ, 2π, 5σ, 6σ. The lower five orbitals are all good bonding orbitals, and thus the BO = 5, consistent with a B≡N triple
    bond.
We now construct an interaction diagram between the π-orbitals of borazine and the appropriate orbitals in the monomer.
Which are these orbitals? Well, the borazine ring is flat, so that of the original π-orbitals of the monomer constituents, only one
per doubly degenerate set will form the π-system of the ring. In terms of concerted reaction theory, this can be diagramed as a
cycloadition reaction in which one π-bond from each triple bond becomes a σ-bond in the ring compound.


                                              +
                                                   H              H                                          H
                                             N                        B                                          _
                                        B                                 N
                                                                              +                 H            B + H
                                    H                                                                   N+     N
                                                                                  H                 _
                                                                                                        B    + B_
                                                                                                H            N    H
                                                        +                                                    H
                                             H      N       B    H

Thus, although in the interaction diagram there are 3 × 4 = 12 orbitals on the left, there are only 6 on the right. The remainder
become part of the σ-bond system of the ring. It is also very instructive to recognize that the net bonding energy of the π-
orbitals actually decreases on going from the borazine monomer to the trimer (six of the original 1π orbitals equals –72 eV,
while the sum of the two 1a” and 1e” orbitals is only –69 eV) by this calculation. There can be no doubt then that the driving
force for trimerization must be found in the σ-framework of the molecule. Nevertheless, the π-system is present, and among
other things, keeps the ring planar. We conclude that a puckered borazine would be even less favourable despite the obvious
fact that the planar ring is eclipsed and has higher substituent steric repulsion.

          H B N H                                borazine π orbitals
    +5     monomer 3 ×
                                                                  2a"
                       2π                                                                2a”
    eV                      •
                   •
                                                                  2e"


     0
                        y
                                                                                  2 e”                       and

           x                                D3h
               z
     -5




   -10                                             ↑↓            ↑↓               1e”                        and
               ↑↓↑↓ ↑↓                                            1e"
                       1π                                   ↑↓
                                                                  1a"
                   •            •
   -15


                                                                                          1a”
                         z
               D3h
         A1'                       y
                     x
                                       •             •




E' set                       A1'       •
                                           A2"       •



                                       •             •




         A1'                           •
                                                 •


                                       •             •
                             A1'           A2"
E' set
                                       •             •




         A2"
                                       •         •

E" set
                                       •         •
                             E' set
                                       •         •



         A2'                                     •
                                       •



E' set                       E" set    •         •



                                       •         •

				
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