February, 17th 1999, 18:05 - 20:30    Instructor: P.H. Bird

Instructions: THIS IS A CLOSED BOOK EXAM and NO MATERIALS ARE TO BE SHARED. Answer ONLY 5 questions. They carry equal marks.

  1. Respecting the normal valencies of the elements concerned, and showing all the non-bonding electrons, deduce the main structural features of:

    1. K3[B3O6]   (2½ marks)


      The group 1 and 2 metals Na to Cs and Ca to Ba
      do not form covalent bonds in their compounds.

      Do not attach these metals to the counter ions
      by lines indicating covalent bonding!

    2. Ca[B2O4] (in which the boron is involved in "infinite" chains)   (2½ marks)

    3. Na3[P3O9]   (2½ marks)

    4. K5[P3O10]   (2½ marks)

    1. Write a pair of equations to illustrate the amphoteric nature of ZnO.   (5 marks)

      An "amphoteric" compound will react as an acid or a base (with either a base or an acid, respectively, but generally not just water):

      With acid:   ZnO   +   2H3O+      Zn2+   +   3H2O

      With base:   ZnO   +   2OH-      ZnO22-   +   H2O

    2. Draw canonical structures indicating how nitrate ion might act as a monodentate ligand, as a chelating ligand, and as a bridging ligand.   (5 marks)

  2. Draw their structures, and classify the following acids as "very stong", "strong", "weak" or "very weak":   (5 marks for the structures, and 5 for the strengths with an explanation)

    The strength of the oxyacids correlates roughly with the number of =O's they have. If there are none, the acid will be very weak (e.g. HOCl). With three =O groups, the acid becomes very strong. This happens because the negative charge generated by the departure of the proton can be spread over the available (electronegative) oxygen atoms in a series of equivalent canonical structures.

    In the case of the fluorosulphonic acid (e), the fluoride also boosts the acid strength since it is a strongly electron withdrawing element.

  3. Using an appropriate acid/base definition (say which, and describe it in each case) indicate which is the base and which the acid among the reactants in the following:   (2 marks each part)

    1. B(OR)3   +   NaH      Na[BH(OR)3]

      This equation can be re-written:

    2. B(OR)3   +   Na+ :H-      Na+ [BH(OR)3]-

      By the Lewis definition the B(OR)3 is the electon-pair acceptor i.e. the acid, and the hydride ion is the electron-pair donor i.e. the base.

      Note: Just because you see an "H" does not mean you are dealing with a Brønsted/Lowry acid! Remember, compounds with Na- ions are virtually unknown, but saline hydrides of the group 1 and 2 metals are well characterized.

    3. 2HF   +   PF5      H2F+   +   PF6-

      Hydrogen fluoride is a useful solvent. Its autoionization is:

      3HF      H2F+   +   HF2-

      In the "general solvent system" definition of acid and bases, and acid is anything which increases the concentration of the characteristic cation of the solvent. By this definition, then, PF5 is acting as an acid. The base will be a molecule of HF.

    4. 1/2Al2Cl6   +   PF3      Cl3Al:PF3

      Cl3Al:PF3 is a "classic" Lewis adduct. In the Lewis definition, AlCl3, derived from Al2Cl6, would be the electron acceptor or acid, and the PF3 is the electron donor or base.

    5. SiO2   +   Na2O      Na2SiO3

      This reaction is best classified by the Lux/Flood definition, in which an acid is an oxide ion acceptor, while a base is an oxide ion donor. SiO2 will be the acid, and Na2O the base.

    6. NOF   +   ClF3      NO+   +   ClF4-

      ClF3 is a liquid capable of some autoionization:

      2ClF3      ClF2+   +   ClF4-

      Since NOF results in an increase in the ClF4- concentration it would be classified in the general solvent system definition as a base.

      In the Lewis definition, NOF is acting as a source of F-, an base because it has 4 lone-pairs of electrons. The ClF3 acts as an electron acceptor, i.e. the acid.

      A few students, evidently anxious to make sure all the options were covered, invoked the Bronsted/Lowry definition on reactions where there was no hydrogen at all. Think!

  4.    (5 marks each part)
    1. Science fiction writers occasionally invoke life-forms based on silicon. What differences between the chemistry of carbon and silicon make this scenario very unlikely?

      The molecules involved in biological systems have several characteristics relevant to this question. They are large and complex, featuring long and often unsaturated chains. They are also relatively stable to decomposition, notably by water.

      Carbon readily forms double and triple bonds with itself and the other non-metallic atoms of the first short period which allows for an effectively unlimited number of different molecules. Because its normal valence is limited to four i.e. it cannot form more than four 2-centre, 2-electron bonds, its compounds can be relatively stable to attack.

      Silicon has limited ability to form chains of Si atoms singly bonded to each other because the Si-Si bond is relatively weak, and it does not form multiple bonds to itself or other atoms under normal circumstances because of poor p-bonding overlap. On the other hand, it has a great affinity for oxygen, as displayed by the great variety of robust silicates, and silicones known.

      In addition many silicon compounds are susceptible to chemical attack through 5 and 6-coordinated intermediates (see part (b)) which limits the chemical stability of many of its compounds.

    2. SiCl4 undergoes rapid hydrolysis with water while CCl4 is inert. Write an overall equation for the reaction of water with SiCl4 and some possible mechanistic steps to explain the difference.

      The overall reaction of SiCl4 would produce silicic acid Si(OH)4 via a series of steps involving 5 or 6-coordinated silicon. These higher coordination numbers are available if silicon uses its empty d-orbitals to form sp3d or sp3d2 hybrid orbitals. They are not available to carbon which has no additional low-lying orbitals to use.

      (The reaction would not really yield pure silicic acid because Si-O-Si links rapidly form by loss of water:

      (HO)3Si-O-H   +   H-O-Si(OH)3      (HO)3Si-O-Si(OH)3   +   H2O   etc

      The result is a gel containing very ill-defined compounds.)

  5.    (5 marks each part)
    1. What is the "inert pair effect? Use group 14 to illustrate your answer.

      For the p-block elements, the electron configuration is [core]ns2npx, where x = 1 to 6. The inert pair effect refers to a tendency, which increases down a group, for the 2 electrons in the ns orbitals not to be involved in bonding. This arises because, the promotion energy needed to form hybrids is not compensated for by the energy recovered by the extra bond formation, and also because the gap between the ns and np subshells gradually increases due to the effect of the increasing number of electrons in the core. Recall that in a one-electron system e.g. hydrogen, the subshells of the same n, but different l, have the same energy, but for multi electron atoms, differential penetration into the core by s, p and d electrons leads to the observed energy differences.

      Group 14 includes C, Si, Ge, Sn and Pb. Carbon is tetravalent in virtually all its stable compounds, but as we go down the group, the divalent state, characterised by primarily ionic M2+ compounds, becomes more and more important. At lead, the tetravalent state is rare. The following reactions illustrate this:

      GeCl2   +   Cl2      GeCl4      Very rapid at 25 oC.

      SnCl2   +   Cl2      SnCl4      Slow at 25 oC.

      PbCl2   +   Cl2      PbCl4      Requires heat and/or pressure.

    2. What are the normal oxidation states of the group 17? Give real examples of a selection of neutral molecules and anions for each, showing their structures.

      The ground state configuration of the group 17 elements (F, Cl, Br, I) is [core]ns2np5nd0. In this configuration oxidation states of I or -I are possible. Examples are HF, ClF and Br-.

      The possible excited states, for all except fluorine, are:

      ns2np4nd1 leading to the III state, for example ClF3 and IF4-

      ns2np3nd2 leading to the V state, for example BrF5 and IF6-

      ns1np3nd3 leading to the VII state, for example IF7

      The range of compounds is limited by steric effects.