CHEM 481 Assignment 3 Answers
22. In the standard notation of closest-packing of spheres, the letters A, B and C refer to close-packed layers. Which of the
following sequences describe closest-packing in 3 dimensions, and which do not. State your reasoning explicitly:
(i) ABCABC....
Close-packed CCP.
(ii) ABACABA....
Close-packed, but irregular (mixture of CCP and HCP.
(iii) ABBABABBA....
Not close packed - a "B" layer cannot lie directly over a "B" layer.
23. Draw one layer of close-packed spheres. On this layer mark the positions of the centers of the B layer atoms using the
symbol ⊗ and, with the symbol ∅ mark the positions of the centers of the C layer atoms of an FCC lattice.
My graphic is shaded, i.e. white = ⊗; light grey = ∅
24. Determine the maximum size of (a) the tetrahedral (Td) 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 Td
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 which is half the
angle subtended from an apex of the tetrahedron to the center of
the hole to another apex. Then:
r
, so rh = r (1.225 − 1) ) = 0.225r , the fraction of
r + rh =
sin(54.25)
radius.
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
Geometry of Td hole
Geometry of Oh hole
be solved by pythagorus law, so that:
2
2
( r + rh ) = r + r = 2r = 1414
.
r so that rh = 0.414r .
25. Calcium metal crystallizes in a face-centered cubic unit cell. The density of the solid is 1.54 g/cm3 . What is the radius of a
calcium atom?
An FCC unit cell contains 4 atoms of calcium (just by counting atoms at the corners, faces and edges). The radius of an
FCC atom is ¼ the length of the face diagonal, and that is √2 of the edge length. We get the edge length by using the
density to get the volume of the unit cell, and then taking √3 of the volume to get the edge length. This same principle can
be used for any cubic unit cell, adjusting only for the internal geometry and atom count.
m
4atoms × 40.078 g / mol
= 557
. ⋅ 10 − 8 cm = 5.57Å. Thus r = √2/4×5.57 Å = 1.97 Å.
=3
a=3 V =3
. g / cm 3 × 6.022 ⋅ 10 23 atoms / mol
d
154
3
26. Vanadium metal has a density of 6.11 g/cm3 . Assuming the vanadium atomic radius is 1.32 Å, is the vanadium unit cell
simple cubic, body-centered cubic, or face-centered cubic?
This is partly the reverse calculation to #4. Again we can calculate the edge length a, except that we do not know how
many atoms to multiply the mass by. This requires us to modify the first equation:
50.9415 g / mol
m
a = 3 V = 3 no. atoms × 3 = 3 no. atoms × 3
= 3 no. atoms × 2.4 Å. There are only three
3
611
d
. g / cm × 6.022 ⋅ 10 23 atoms / mol
possibilities to consider: SC, 1 atom and r = a/2; BCC, 2 atoms and r = √3a/4; FCC, 4 atoms and r = √2a/4. If SC, a = 2.4, r =
1.2 Å, which is not right. If BCC, a = 1.26 × 2.4 = 3.02 r = 1.31 Å, which is a good fit to the measured radius. If FCC, a = 1.59
× 2.4 = 3.81, and r = 1.35 Å, which does not fit as well as the BCC. So its BCC!
27. 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.77 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. Since the anion is 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.
28. Calculate the maximum fraction of available volume occupied by hard spheres on (a) the simple cubic, (b) the BCC, and
(c) the FCC lattices.
The general approach is the same in each case. First study the unit cell diagrams of the three cell types:
"Sliced" view of Simple Cubic unit cell
"Sliced" view of BCC
"Sliced" view of FCC
We know that the volume of one unit cell is given by VBOX = a 3 for each cube. The volume of the spheres is given by 4πr3/3.
If we can relate the radius to a, we can ratio the two volumes and calculate the fraction of volume occupied. For SC, the
radius r = a/2, and there is one sphere per unit cell. Thus VSPHERE = 4πa3/(3 × 8) = 0.52 a 3. For BCC, the radius r = √3 × a/4,
and there are 2 spheres per unit cell.
3
4  3a 
Thus V


=
2
×
π
= 0.68a 3 . For FCC the radius r = √2 × a/4, and there are four atoms per unit cell. Thus
SPHERES
3  4 
3
4  2a 
 = 0.74a 3 . We see that the FCC = CCP structure is the densest of the three, and is in fact the most
VSPHERES = 4 × π 
3  4 
dense possible way to pack equal sized spheres into 3-D space. It is almost 50% more efficient than SC, so no wonder that
the latter is rarely seen for pure elements. On the other hand, if a suitably sized smaller sphere is added to fill the gaps
between the large spheres, higher densities are possible. Thus for example, the ionic CsCl structure is based on a SC
arrangement of anions (large) with somewhat smaller cation fitting into the big central gap.
29. 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.
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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 = M 2; 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 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 Ba 2+ 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 ZrO6 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.
30. The metal hydride LiH has a density of 0.77 g/cm3 . The edge of the unit cell is 4.086 Å. If it is assumed that the H– ions
define the lattice points, does the compound have a face-centered cubic or a simple cubic unit cell?
This problem, for an ionic solid, is solved much like #5, except that the LiH unit is counted as if it were the "atom" in the
stoichiometric calculation. With this information supplied, a slightly shorter approach is to simply calculate the density
from the crystallographic information, and compare our results with the known density. For a simple cubic, there is only one
7.95 g / mol
; whereas for FCC there are four times as many units
LiH per cell, so d = m =
.
= 0194
−8
V (4.086 ⋅ 10 cm) 3 × 6.022 ⋅ 10 23 atom / mol
in the same volume, so the density works out to be four times as great, d = 0.194 × 4 = 0.77 g/cm3.
31. Confirm that in (a) rutile and (b) perovskite the stoichiometry is consistent with the structure. The unit cells are shown
below:
Perovskite
Rutile
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.
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32. 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
ZnS has 4 Zn in centre, 8 × 1/8th + 6 ×1/2 = 4 S.
33. 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 lattice energy of CsCl. The enthalpy of formation of CsC1 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 dissociation
energy of Cl2 is +243 kJ/mol of Cl2 molecules and the first electron attachment enthalpy of Cl is –349 kJ/mol of Cl atoms.
243
∆H diss
V = − ∆H 0f + ∆H subl + ∆H IE +
+ ∆H EA = +443 + 78 + 375 +
− 349 = +668.5 kJ / mol
2
2
b) Use the following data to calculate the lattice energy of CaO.
The enthalpy of formation of CaO is –636 kJ/ mol. The enthalpy of
V
O2-(g)
CaO (c)
Ca2+(g) +
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 O2, is +494 kJ/mol of O2 molecules, the 2 n d I E 2 n d E A
first electron attachment enthalpy of O is –141 kJ/mol of O atoms, and
Ca+(g)
O-(g)
the second electron attachment enthalpy of O is +845 kJ/mol of O–
ions.
r e v e r s e o f
1 s t I E
1st EA
o f f o r m a t i
∆H diss
0
V = − ∆H f + ∆H subl + ∆H1stIE + ∆H 2ndIE +
+ ∆H 1stEA + ∆H 2ndEA
Ca (g)
O (g)
2
494
s u b l .
½ BE
= +636 + 192 + 590 + 1145 +
− 141+ 845 = 3514 kJ / mol
2
Ca (s) +
½ O2 (g)
c) Use the following data to calculate the enthalpy of formation of
Rb2O. 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 of O2 molecules, the first
electron attachment enthalpy of O is –141 kJ/mol of O atoms, and the second electron attachment enthalpy of O is +845
kJ/mol of O- ions. The lattice energy, V, of Rb2O is +2250 kJ/mol.
∆Hdiss
V = − ∆H 0f + 2 ∆H subl + 2 ∆H1stIE +
+ ∆H1stEA + ∆H2 ndEA = +2250
2
494
+ 2250 = −∆H f0 + 2 × 82 + 2 × 403 +
− 141+ 845; ∆H f0 = −329 kJ / mol
2
d) Use the following data to calculate the enthalpy of formation of SrCl2. 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 of Cl2 molecules, and the first electron attachment enthalpy of Cl is –349
kJ/mol of Cl atoms. The lattice energy, V, of SrCl2 is +2150 kJ/mol.
V = − ∆H 0f + ∆Hsubl + ∆H1stIE + ∆H2 ndIE + ∆Hdiss + 2 ∆H1stEA = +2150
+ 2150 = −∆H f0 + 164 + 549 + 1064 + 243 − 2 × 349; ∆H 0f = −828 kJ / mol
34. Calculate the lattice energ y 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.
∆H°sublimation(Tl) = +182 kJ mol–1
∆H°f(TlCl) = -204 kJ mol –1
∆H°Ist I.E.(Tl) = +589 kJ mol–
1
∆H°bond dissociation(Cl 2) = +243 kJ mol –1
∆H°electron attachment(Cl) = -349 kJ mol –1
a) V = − ∆H 0 + ∆H + ∆H + ∆Hdiss + ∆H = +204 + 182 + 589 + 243 − 349 = +747.5 kJ / mol
f
subl
IE
EA
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
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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
1389 Z A Z B  1 
1389 × 1 × 1 
1 
VAB = A ×
1 −  = 176267
.
×
1 −
 = 652 kJ / mol
 n
rAB
(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.
35. Calculate the enthalpy of formation of the hypothetical compound KF2 assuming a fluorite structure. Use the Born-Meyer
equation to obtain the lattice energy and estimate the radius o f K2+ by extrapolation of the data in the table of ionic
radii. Both SAL and K&T list 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, using the method outlined in 12(d) above. We find that the radius of
fluoride is 1.17 for CN4, while that of K2+ is probably similar to Ca 2+ with CN8 in the fluorite structure, thus 1.26 Å. Thus:
1389 Z AZ B  1 
1389 × 2 × 1  1 
VAB = A ×
1 −  = 2.51939 ×
 1 −  = 2520 kJ / mol So far this looks great. But we go on.
 n
rAB
(117
. + 126
. )  8
V = − ∆H 0f + ∆H subl + ∆H1stIE + ∆H2 ndIE + ∆Hdiss + 2 ∆H1stEA = +2520
+ 2520 = −∆H 0f + 89 + 419 + 3051 + 158 − 2 × 328; ∆H 0f = +541 kJ / mol
.
The formation of KF2 is endothermic by over 500 kJ/mol. Comparison to SrCl2 in 12(d) shows that the main contributor to
this unfavorable heat of formation is the large size of the 2nd 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.
36. 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.
37. 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 NaBF4 is predicted to be more soluble in water.
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