Download Chem 222 Intro to Inorganic Chemistry Summer 2011 Problem Set 3

Chem 222
Intro to Inorganic Chemistry
Problem Set 3-Answer Key
Summer 2011
Tuesday, June 21, 2011
1. (box 6.4 H and S) List the coordination number of the cation, the coordination
geometry, the expected r+/r- range and the overall stoichiometry for CsCl and
CaF2
Coordination
number of cation
Coordination
geometry
Expected
r+/r-
Stoichiometry
cation: anion
CsCl
8
cubic
> 0.73
1:1
CaF2
8 (anion 4)
cubic
>0.73
1:2
2. Calculate the the r+/r- for LiCl, NaBr, Na2O, What structure will each of these
adopt? Identify the elements on the appropriate lattice structure.
Salt
Cation (r+)
Anion (r-)
r+/r-
Structure
LiCl
76
181
0.420
NaCl
NaBr
102
196
0.520
NaCl
Na2O
102
140
0.728
Antifluorite
Fluorite structure (green=anions, red= Zinc Blende (ZnS with Zn and S on
cations) or antifluorite structure (green=
alternating sites) aka Diamond (with C
cations, red=anions)
at all sites)
Na2O green=Na and red=O
Chem 222
Intro to Inorganic Chemistry
Summer 2011
NaCl structure (anions are green, cations CsCl structure
are blue, note that this just reflects the
relative sizes, as the positions work out
to the same crystal structure, regardless)
LiCl, Cl=green, Li=blue
NaBr Br=green, Na=blue
3.
4. (H and S 6.15) Using data from the Appendices (of H and S) and the fact that
ΔfH° (298 K) = -859 kJmol-1, calculate a value for the lattice energy of BaCl2.
Outline any assumptions that you have made.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
-2050 kJ/mol
-assume ΔU (0 K) is approximately equal to ΔH(lattice) (298 K)
5. a) Give a balanced reaction for the burning of methanol in air.
2CH3OH + 3O2  2CO2 + 4H2O
b) Give two half reactions and the overall reaction for a direct methanol fuel cell
(methanol instead of H2 is used as a fuel and a proton exchange membrane is still
present)
anode (oxidation reaction)
4(CH3OH +H2O  CO2 + 6H+ + 6e-)
cathode (reduction reaction)
6(4H+ + O2 + 4e-  2H2O)
overall reaction
4CH3OH +4H2O + 24H+ + 6O2 + 24e-  4CO2 + 24H+ + 24e- + 12H2O
cancel the e- and H+ and H2O from both sides to get:
4CH3OH + 6O2  4CO2 + 8H2O
which simplifies to:
2CH3OH + 3O2  2CO2 + 4H2O (same as direct combustion)
6. Draw an MO diagram for the oxygen moiety in peroxide. Comment on the
magnetism.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
The extra two electrons (in red) go into the anti-bonding orbitals, weakening the oxygenoxygen bond. This explains the higher reactivity of peroxides. The resulting complex
is diamagnetic (very, very weak force) and will therefore be weakly repelled by a
magnetic field.
7. Sulphur miners often have lung problems after a few years of working. Write
balanced reactions starting from S8 to account for this damage. Suggest a
compound that could be put in a respirator to mitigate this reaction.
This is the same chemistry associated with acid rain, so:
S8 + 8O2  8SO2 (in air, aided by elevated temperatures)
2SO2 + O2  2SO3 (in air, aided by elevated temperatures and surfaces)
SO3 + H2O  H2SO4 (water in lungs plus SO2 gas)
H2SO4 H+ + HSO4- (dissociation in lung)
Chem 222
Intro to Inorganic Chemistry
Summer 2011
(Recall that for industrial synthesis of H2SO4, it is necessary to bubble the SO2 through
H2SO4 to avoid formation of a thick mist. However, due to the low concentrations of
H2SO4 in the air, this is not required in this case)
Any base in a respirator would work to react with SO2 vapour or droplets of H2SO4,
before they react with the lungs, but a weak base would be best, so that there is less
concern with inhaling caustic.
8. (H and S, section 6.11) Sulphur and selenium have very similar redox activities.
For the following (unbalanced) sulphur chemistry, give the analogous reactions
for selenium and determine the oxidation state of the selenium in each compound.
[SO4]2- H2SO3SH2S
[SeO4]2- H2SeO3SeH2Se
Assume O is O2- (most common) and H is H+ (as it is reacting with a non metal),
therefore the oxidation states are 6+, 4+, 0 and 2- for both S and Se.
9. Draw the bonding and anti-bonding molecular orbitals arising from py-py
interactions and pz-pz interactions.
Note that px-px bonding and antibonding orbitals would be the same shape and
perpendicular to the py orbitals.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
10. (H&S 2.9) (a) Use MO theory to determine the bond order in each of [He2]+ and
[He2]2+.
He has electron configuration 1s2, and has only the 1s orbital in its valence shell. The
MO-diagram needed to examine the bonding in the two cations is:
u* (1s)
1s
E
n
e
rg
y
1s
He
+
g (1s)
He
The [He2] cation has three electrons instead of the four electrons for neutral [He2], and
the [He2]2+ cation has only two electrons. This gives the following MO diagrams:
Chem 222
Intro to Inorganic Chemistry
Summer 2011
u* (1s)
1s
E
n
e
rg
y
1s
g (1s)
He
He
[He2]+
u* (1s)
1s
E
n
e
rg
y
1s
He
g (1s)
He
[He2]2+
Based on these diagrams, [He2]+ would have a bond order of (2–1)/2 = 0.5, and [He2]2+
would have a bond order of (2–0)/2 = 1.
(b) Does the MO picture of the bonding in these ions suggest that they
are viable species?
Both are “viable” although the monocationic species would have a fairly weak sigma
bonding interaction and a longer He-He distance.
(c) Is either of these ions paramagnetic?
The monocation [He2]+ has one unpaired electron, so it is paramagnetic.
11. (H&S 2.25c) Account for the observation that, in the salt formed from the
reaction of Br2 and SbF5, the Br-Br bond distance in the Br2+ ion is 215 pm, i.e.
shorter than in Br2.
We didn’t talk about MO pictures for n=4 elements, and it’s not really within the scope of
this course to generate such a diagram. However, based on the MO pics we’ve looked
at for the first row diatomics (n=1,2, F2 in particular, see H&S Fig 2.9, p.39), we can
make an educated guess about what is going on here and thereby develop a reasonable
approximation. If oxidation of Br2 to [Br2]+ (i.e. removal of one electron from this
diatomic molecule) causes the Br-distance to get shorter, that suggests that the bond
gets stronger - i.e. the bond order increases on removal of an electron. The HOMO for
Br2 must have antibonding character, which is reduced when one electron is removed.
12. (H&S 15.20) Assuming that it has similar molecular orbital energies to those of
NO, use an MO approach to show the bonding in CO.
The following MO diagram results. The strength of the bonding is the same (filled versus
unfilled) but the correct relative MO ordering will affect its behavior as a ligand. Also
note that for exams, you would be given the filling order for any heteroatomic
molecular orbitals.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
13. (RC&O 15.5) Contrast the behavior of nitrogen and carbon by comparing the
properties of (a) methane and ammonia, and (b) ethene (ethylene) and hydrazine.
(a) Methane and ammonia are both colourless gases, in which the central atom is sp3
hybridized. Because nitrogen has one more valence electron than carbon, this gives it
one filled sp3 orbital – a stereochemically active lone pair. Thus methane, CH4, is nonpolar and insoluble in water, while ammonia, NH3, is polar and dissolves in water to
form NH4OH, which then dissociates to give a basic solution. Methane has no odour,
while ammonia has a distinctive, pungent odour.
(b) The same stability versus reactivity is seen in ethylene versus hydrazine. Both ethene
and hydrazine contain an element-element bond: H2EEH2. However, carbon can
readily re-hybridize from sp3 to sp2 in this structure to offset its lower than optimum
valency of three. This places the remaining valence electron on each carbon in a porbital perpendicular to the C-C bond axis. The side-on, π-overlap of these orbitals
gives a π-bond, which results in the double bond character of ethane: it has a shortened
Chem 222
Intro to Inorganic Chemistry
Summer 2011
C-C bond distance, a stronger C-C bond, and is non-polar and reasonably inert (unless
you ignite it).
¹ -bond
H
H
C
C
H
H
sp2
H
N
H
H
H
H
sp3
H
N
H
H
view down the
N-N bond axis
Because each nitrogen in hydrazine has an extra electron relative to carbon, such rehybridization is much more difficult. This would place two electron pairs in adjacent
p-orbitals, a high energy structure because of the electron repulsion between these two
lone pairs. Even when the nitrogens in hydrazine remain sp3 hybridized, the single
bond twists to rotate the lone pairs away from each other. From H&S p.447:
Hydrazine is a clear, colourless liquid (mp 275K, bp 386K) that dissolves in water to
give weakly basic solutions (the hydrazinium ion, N2H5+ forms). It is corrosive, and its
vapour forms explosive mixtures with air (no spark needed).
14. How are quartz glass and common soda-lime glass different?
Both of these materials are silica-based, comprised of the network covalent solid SiO2.
Quartz glass is pure silica that has been heated above its glass transition temperature
(Tg) to give a molten liquid, and then cooled to give a random spatial arrangement of
SiO4 tetrahedral units. Common soda-lime glass is prepared in the same way except
that about 13% of sodium oxide (Na2O) and 11% of calcium oxide (CaO), with traces
of K2O and other oxide impurities, are added to the melt before cooling. This changes
the physical properties of the resulting glass:
Quartz glass is very strong and has low thermal expansivity, which means it doesn’t
expand or contract too much when you heat or cool it. Soda-lime glass is not quite as
strong, although it is strong enough to hold carbonated beverages under pressure. It
has higher thermal expansivity, which is why you can’t make tea in an empty pop
bottle. The main benefit of soda-lime glass, from a manufacturing point of view, is that
it has a much lower Tg than does quartz glass (573°C vs 1140°C), so it is easier and
cheaper to “melt” for shaping into the required forms.
15. (H and S, 6.6) What do you understand by the band theory of metals?
The band theory of metals allows us to apply quantum mechanics and the concept of the
linear combination of atomic orbitals (LCAO) to make molecular orbitals and extend this
concept to a much larger number of atoms (as in a moles worth of atoms). The band
theory of metals explains their electronic conductivity, as the electrons can very easily be
excited into an unfilled orbital (conduction band) and is therefore delocalised. This sea
of delocalised electrons also explains why metals are malleable and ductile, as the
electrons can immediately adjust to any deformation in the crystal structure. The
shininess of metals results from absorbing and re-emitting photons, as the electrons move
between the (very accessible) energy levels.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
16. Chemical analysis of a germanium crystal reveals indium at a level of 0.0003
atomic percent.
a) Assuming the concentration of thermally excited charge carriers from the Ge
matrix is negligible, calculate the density of free charge carriers (carriers/cm3) in
this Ge crystal.
b) Draw a schematic energy band diagram for this material and label all critical
features.
17. a) Aluminium Phosphide (AlP) is a semiconductor with a band gap Eg of 3.0 eV.
Sketch the absorption spectrum of this material (absorption vs. wavelength)
Chem 222
Intro to Inorganic Chemistry
Summer 2011
b) Aluminium antimonide (AlSb) is also a semiconductor. Do you expect the band
gap of this material to be greater or less than the bangap of AlP? Explain.
18. You wish to make n-type germanium.
a) Name a suitable dopant.
You will need to dope with an electron (n-type =negative) donor, which means it must be
from group 16, giving us P, As and Sb as reasonable choices.
b) Name the majority charge carrier.
The majority charge carrier is the electron.
c) Draw a schematic energy band diagram of the doped material. Label the
valence band, conduction band and any energy levels associated with the
presence of the dopant.
Chem 222
Intro to Inorganic Chemistry
Summer 2011
19. (H&S 14.16a) Comment on the observation that the pyroxenes CaMgSi 2O6 and
CaFeSi2O6 are isomorphous.
These two silicate minerals have identical empirical formulas, except that the Mg2+ ions
in one are replaced by Fe2+ ions in the other. If they are isomorphous it means they
have the same structure: for silicates this means that the arrangement of the SiO4
tetrahedral units is identical (these ones have the chain structure shown for [SiO3]n2n- ),
as is the placement of the metal cations in interstitial holes throughout the array.
Presumably the only difference is that the specific holes containing Mg2+ in one
mineral contain Fe2+ for the other. This is reasonable, based on the ionic radii reported
for Mg2+ (72 pm for CN=6) and Fe2+ (78 pm for high spin and CN=6) (see Appendix 6
in H&S). [Note: “high spin” means that the Fe2+ ion, in an Oh environment for which
its nearest neighbours are oxygens, has a relatively low value of oct. This will be
discussed when transition metals are considered ]
Definitions/Concepts: glass, silica, silicates, silicone, molecular orbitals, -bond, πbond, nodal plane, bond order, , pnictogens, chalcogens, catenation,, fuel cell
hygroscopic, deliquescent, desiccant, dehydration, dehydrogenation, diamagnetism,
paramagnetism, ferromagnetism, allotrope, photosynthesis, amphoteric, basic oxide,
acidic oxide, orbital hybridization, n-type semiconductor, p-type semiconductor,
compound semiconductor, metal, metalloid, non metal, band theory.