Chemistry 3840
Answers to Assignment #2
Topic: Ionic Solids
1.
Determine the maximum size of (a) the tetrahedral (T d) hole and (b) the octahedral (Oh) hole in a close-packed structure.
Express your result as a decimal fraction of the radii of the close-packed spheres, rs. (Hints: it is valid to think of the T d
hole as a small central atom of a methane-like molecule; for the
Oh hole, consider the four circles obtained by cutting a plane
through the hole and the surrounding square array of spheres, i.e.
treat is as FCC rather than ABC CP layers.)
Consider the pictures at right which set up the trigonometric
parameters needed to solve the problem. First look at the Td
hole. For this system we can define a right triangle dissecting the
edge of the tetrahedron with the hypotenuse being (r + rh). The
angle in this triangle opposite to the side, which is defined as the
distance from an apex of the tetrahedron to the center of the hole
Geometry of Td hole
Geometry of Oh hole
(r + rh), is half of the tetrahedral angle (109.5°). Then:
r , so r  r1.225  1)  0.225r .
sin(54.25C) 
h
r  rh
For the Oh hole the geometry is quite a bit simpler. Shown in the picture is half the face of an FCC unit cell. This right
triangle can be solved by pythagoras law, so that:
r  rh  
r 2  r 2  2r  1.414r
so that rh  0.414r .
2.
Depending on temperature, RbCl can exist in either the rock-salt or cesium-chloride structure. (a) What is the coordination
number of the anion and cation in each of these structures? (b) In which of these structures will Rb have the larger
apparent radius? Compare your answer with the radii data collected by Shannon (Table on p.167 of the notes).
The rock salt (NaCl) structure has CN = 6, while the CsCl has CN = 8. We can answer this by considering the ideal
radius ratio's of the two structures, i.e., r +/r– = 0.414 for rock salt and 0.732 for CsCl. If the anion radius stays the same,
this means that the cation radius will "seem" larger in the CsCl structure type. The data tables have Rb: 1.66(6), 1.75(8),
which fits.
3.
The ReO3 structure is cubic with Re at each corner of the unit cell and one O atom on each unit cell edge midway between
the Re atoms. Sketch this unit cell and determine (a) the coordination number of the cation and anion and (b) the identity
of the structure type that would be generated if a cation were inserted in the center of the ReO3 structure.
(a) The original structure is shown at right. To properly answer the question, you need to carefully consider the
placement of the neighbouring unit cells. From this it is clear that the CN of the small grey spheres representing the Re
atoms will be 6 (octahedral), while the coordination number of each oxygen is only two (linear).
The stoichiometry of this unit cell is 8  1/8 = 1 Re ion; 12  ¼ = 3 oxide ions, hence ReO3.
ReO3
With Ba ion at center
Conventional Perovskite setting
(b) Adding a second cation at the center changes the formula to M 2O3 (8 corner plus one central cation = M2; still the
same 12 edge anions for X3.) Such structures are known, but are never observed when the two metal ions are of the same
type. However, when a different, and indeed larger ion, is placed in the center of the unit cell, we generate a unit cell of
Perovskite. For example, BaZrO3 has this structure. The more conventional unit cell choice for a Perovskite is shown at
far right. Imagine starting from our modified structure with the barium ion at the middle, and adding the required 6
1-4
oxygen ions to it. Compare the unit at the corner (the “growth”) and notice that it is the same as the central Zr ion in the
standard Perovskite. The central Ba2+ ion becomes then the new corner of a unit cell that has the eight barium ions at
each of the eight corners. Many ionic structures can be described in two alternative forms like this.
Another way to view such a composite structure is the view that the ZrO 6 units are octahedral packed around the large
barium ions. Each octahedron consists of a central Zr ion with six oxygen ions associated, thus bearing an overall 2–
charge. There is then one Ba+2 ion per unit cell to compensate the [ZrO6] 2– unit, which being at the corner is also one per
unit cell.
4.
Confirm that in (a) rutile (TiO2) and (b) perovskite (CaTiO3) the stoichiometry is consistent with the structure. The unit
cells are shown below:
Rutile
Perovskite
Rutile: 8 Ti  1/8th + 1 full Ti = 2 Ti; 4 O  ½ + 2 full O = 4 O. Thus 2  TiO2.
Perovskite: 1  Ti = Ti at centre; 8 Ca  1/8th = 1 Ca; 6 O  ½ = 3 O. Thus CaTiO3.
5.
How many cesium and chloride ions are there in a single unit cell of CsCl? How many zinc and sulfide ions are there in a
single unit cell of zinc blende?
CsCl has 1 Cs at centre; 8  1/8th = 1 Cl
6.
ZnS has 4 Zn in centre, 8  1/8th + 6 1/2 = 4 S.
Born-Haber Cycles require a lot of practice to master, although it is a straightforward application of Hess’ Law of Heat
Summation. Most of the data needed to set-up and solve Born-Haber cycles is contained within the problems. Additional
data can be taken from standard data tables, either in SAL or a General Chemistry book. Make sure you can do these
calculations!
Be sure that you can construct coherent B-H cycles, as in the example provided for (b). The complete energy calculation
formula is provided in each case, but you should derive each one from its respective cycle:
a) Use the following data to calculate the enthalpy of lattice formation of CsCl. The enthalpy of formation of CsCl is –
443 kJ/ mol. The enthalpy of sublimation of Cs is +78 kJ/mol, and the first ionization energy of Cs is +375 kJ/mol. The
bond dissociation enthalpy of Cl2 is +243 kJ/mol and the first
enthalpy of electron attraction of Cl is –349 kJ/mol of Cl atoms.
 H
 lf H   f H 0   sublH  I1  BD   EA H
2
243kJ / mol
 443kJ / mol  78kJ / mol  375kJ / mol 
 349kJ / mol
2
 668.5kJ / mol
b) Use the following data to calculate the lattice energy of CaO.
The enthalpy of lattice formation of CaO is –636 kJ/ mol. The
enthalpy of sublimation of Ca is +192 kJ/mol, the first ionization
energy of Ca is +590 kJ/mol, and the second ionization energy of
Ca is +1145 kJ/mol. The enthalpy of dissociation of O 2, is +494
kJ/mol, the first enthalpy of electronic attraction of O is –141
kJ/mol of O atoms, and the enthalpy of electronic attraction affinity
2-4
of
O–
the
ion
is
845
kJ/mol.
 BD H
 lf H   f H   sublH  I1  I 2 
  EA H (1)   EA H (2)
2
494kJ / mol
 636kJ / mol  192kJ / mol  590kJ / mol  1145kJ / mol 
 141kJ / mol  845kJ / mol  3514kJ / mol
2
c) Use the following data to calculate the enthalpy of formation of Rb 2O. The enthalpy of sublimation of Rb is +82
kJ/mol, and the first ionization energy of Rb is +403 kJ/mol. The enthalpy of dissociation of O2 is +494 kJ/mol, the
enthalpy of electronic attraction of O is –141 kJ/mol, and the enthalpy of electronic attraction of the O- ion is 845 kJ/mol.
The enthalpy of lattice formation, ΔlfH, of Rb2O is –2250 kJ/mol.
0
 BD H
  EA H (1)   EA H (2)  2250
2
494kJ / mol
 2250  H 0f  2  82kJ / mol  2  403kJ / mol 
 141kJ / mol  845kJ / mol; H 0f  329kJ / mol
2
d) Use the following data to calculate the enthalpy of formation of SrCl 2. The enthalpy of sublimation of Sr is +164
kJ/mol, the first ionization energy of Sr is +549 kJ/mol, and the second ionization energy of Sr is +1064 kJ/mol. The
enthalpy of dissociation of Cl2 is +243 kJ/mol, and the enthalpy of electronic attraction of Cl is +349 kJ/mol. The enthalpy
of
lattice
formation,
ΔlfH,
of
SrCl2
is
–2150
kJ/mol.
 lf H   f H 0   sublH  I1  I 2   BD H  2 EA H  2150kJ / mol
 lf H   f H 0  2 sublH  2 I1 
 2150kJ / mol  H 0f  164kJ / mol  549kJ / mol  1064kJ / mol  243kJ / mol  2  349kJ / mol; H 0f  828kJ / mol
7.
Calculate the lattice energy of TlCl by (a) a thermochemical cycle and (b) using the Born-Mayer equation. (c) Discuss the
relationship between these two numbers. Make use of all the concepts developed in the course which are relevant to this
discussion.
sublH(Tl) = +182 kJ mol–1
fH(TlCl) = -204 kJ mol–1
I1(Tl) = +589 kJ mol–1
BDH(Cl2) = +243 kJ mol–1
a)
lf H   f H 0   sibl H  I1 
ΔEAH(Cl) = -349 kJ mol–1
 BD H
243kJ / mol
  EA H  204kJ / mol  182kJ / mol  589kJ / mol 
 349kJ / mol  747.5kJ / mol
2
2
b) To use the Born-Meyer equation, we need to know the inter-ionic distance, but also what value of the Madelung constant
to use. In the absence of any better information, we must use the radius ratio rule to predict which structure to use. We
need to use the radii data from the table of ionic radii. When we don’t know the correct structure type, a reasonable
procedure
is
to
first
try
the
CN6
radii,
such
that:

r
164
.
This predicts the CsCl structure. Ideally we should now check the answer using CN8 radii, but there

 0.98
r  167
.
is no data available for chloride with this radius, so we will assume it is indeed CsCl. The closest past noble gas
configuration for thallium is Xe. For Cl it is Ar. We take the average of the n values of these two configurations, or ½
(9+12) = 10.5
1389mkJmol1Z A Z B  1 
1389mkJmol1  1 1 
1 
1    1.76267 
1 
  652 kJ / mol
rAB
n
(
1
.
73

1
.
67
)
10
.5 



c) We see that the experimental lattice energy is 14% higher than the calculated value. This is a large discrepancy, and
suggests additional covalent bonding beyond the ionic lattice forces. This is not terribly surprising, since Tl in Group 13
is a very soft acid, and chloride is a borderline base. So we expect a lot of covalency in their compounds.
lf H  A 
8.
Calculate the enthalpy of formation of the hypothetical compound KF 2 assuming a fluorite structure. Use the Born-Meyer
equation to obtain the lattice energy and estimate the radius of K2+ by extrapolation of the data in the table of ionic radii.
Your textbook lists the other required thermochemical data. Is the lattice energy favorable for this compound? Why then
can it not exist?
It is hypothetical, so we must use the Born-Meyer equation to estimate its lattice energy. This estimate is then used in a BH cycle in order to estimate the enthalpy of formation. We find that the radius of fluoride is 1.17 for CN4, while that of K 2+
is probably similar to Ca2+ with CN8 in the fluorite structure, thus 1.26 Å.
Thus:
3-4
1389mkJmol1Z A Z B  1 
1389mkJmol1  2  1  1 
1    2.51939 
1    2520 kJ / mol
rAB
(1.17  1.26)
 n
 8
But
we
go
lf H   f H 0   sublH  I1  I 2   BD H  2 EA H  2520kJ / mol
lf H  A 
So far this looks great.
 2520kJ / mol    f H 0  89kJ / mol  419kJ / mol  3051kJ / mol  158kJ / mol  2  328kJ / mol; H 0f  541kJ / mol
on.
.
The formation of KF2 is endothermic by over 500 kJ/mol. Comparison to SrCl 2 in 6(d) shows that the main contributor to
this unfavorable heat of formation is the large size of the 2 nd ionization energy. For potassium, this second ionization
involves the opening of the core orbitals, i.e. those of the argon configuration. This costs a lot of energy, more than the
lattice energy of a typical MX2 salt can supply. Thus KF2 cannot exist.
9.
Which one of each of the following pairs of isostructural compounds is likely to undergo thermal decomposition at a lower
temperature? Give your reasoning.
(a) MgCO3 and CaCO3 (decompose to the metal oxide and carbon dioxide). The answer hinges on the relative size of the
carbonate (reactant) and oxide (product) lattice energies. For Mg, the lattice energy with the big carbonate is going to be
less favorable than for calcium, while for the oxide, the smaller Mg will be more stable than the larger calcium. Both
effects contribute to making the magnesium carbonate the material which decomposes more easily on heating.
(b) CsI3 and NMe4+I3– (decomposition products are MI + I2; tetramethyl ammonium is a much larger ion than Cs+. Again,
the larger triiodide ion is more compatible with the large tetramethylammonium cation, while the smaller iodide ion in the
product will give a bigger lattice energy with the smaller cesium ion. We predict that cesium triiodide will be the less
thermally stable.
10. Which member of each pair is likely to be the more soluble in water:
(a) SrSO4 or MgSO4 We predict a more stable lattice energy for strontium sulfate than for magnesium sulfate, since the
larger Sr2+is better matched to the large SO42–. Thus the strontium salt will be less soluble in water, and the magnesium
more soluble.
(b) NaF or NaBF4? We predict a more stable lattice with the compatible-sized ions sodium and fluoride than between
sodium and the large tetrafluoroborate anion. Thus the NaBF 4 is predicted to be more soluble in water.
4-4