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Chem 263 Winter 2020
Problem Set #2
Due: February 12
1. Use size considerations to predict the crystal structures of PbF2, CoF2, and BeF2. Do your
predictions agree with the actual structures of these three compounds? Why or why not?
Use the radius ratio rules (rcation/ranion and compare to 1) to help predict the structures. The radii
we have are:
Radii found between the slides with the Shannon-Prewitt graphs or the ionic radii periodic table
or you can look them up on Wikipedia on the โIonic radiusโ page or found on โWeb Elementsโ
webpage
Atom Ion Size CN: 4 Ion Size CN: 6 Ion Size CN: 8
F- 1.17 ร 1.19 ร
Pb2+ 1.33 ร 1.43 ร
Be2+ 0.41 ร 0.59 ร
Co2+ 0.72 ร 0.79 ร low spin
0.885 ร high spin
1.04 ร
You should try the radius ratio rules for all possible CNโs. Iโll just use the 1.19 ร for Fluorine,
since the ionic radii between CN:4 and CN: 6 are almost the same
For PbF2, trying 1.19/1.33 = 0.895 meaning CN: 8, cubic (always make the radius ratio <1, so
flip the cation and anion radii if necessary)
Or trying 1.19/1.43 = 0.83, still cubic
For BeF2, 0.41/1.19 = 0.34 meaning CN: 4, tetrahedral or 0.496/1.19=0.417 which means CN: 6
octahedral (but barely, so the structure more likely has tetrahedral coordination)
For CoF2, 0.72/1.19 = 0.605 or 0.79/1.19=0.664 meaning CN: 6, octahedral or 0.885/1.19=0.744
or 1.04/1.19=0.874 meaning CN: 8, cubic.
Actual structures are:
PbF2 is fluorite which is cubic, so radius ratio rules agree.
BeF2 occurs as cristobalite as well as other SiO2-analogue polymorphs such as ฮฑ-quartz which is
trigonal and has tetrahedral coordination so again they agree.
CoF2 is rutile which has an octahedral coordination so they agree.
2. Consider the following thermochemical data for calcium and chlorine in their standard states:
S = +201 kJ/mol
D = +242 kJ/mol
IE1 = +590 kJ/mol
IE2 = +1146.4 kJ/mol
EA = -349 kJ/mol
First, use the Born-Landรฉ equation to estimate the lattice energies for CaCl and CaCl2.
Specify and justify any assumptions you needed to make to do this.
๐ = โ๐+๐โ๐2๐๐ด
๐๐โ (1 โ
1
๐)
is the Born-Lande equation where Z+ and Z- are the absolute value of the charges on the cation
and anion, respectively, e=1.6022x10-19 C, N=6.023x1023 atoms/mole, A is Madelungโs constant
(dependent on the crystal structure), n is the Born-Exponent
ACaCl = 1.748 (treat CaCl as a rock salt structure since Ca+ is very similar to K+)
ACaCl2 = 2.408 (rutile structure)
n = [Ar][Ar]=(9+9)/2=9
rca+ is not available, but you can estimate it by knowing that it should be smaller than the atomic
radius of Ca (which is 1.97 วบ) and should be slightly larger than the radius of K+ (1.52 วบ), so
Iโm going to use 1.60 วบ.
rCa2+=1.14 วบ and rCl-=1.67 วบ
re= 4ฯฮต0r0 = 1.112x10-10 C2/J.m * r0 and r0 for CaCl = 1.60 + 1.67= 3.27 วบ and r0 for CaCl2 is 1.14
+ 1.67 = 2.81 วบ
So finally,
๐๐ถ๐๐ถ๐ = โ(+1)(1)(1.6022๐ฅ10โ19)2(6.023๐ฅ1023)(1.748)
(1.112๐ฅ10โ10)(3.27๐ฅ10โ10)โ (1 โ
1
9)
= โ686489๐ฝ
๐๐๐รท 1000 = โ๐๐๐. ๐ ๐๐ฑ/๐๐๐
and
๐๐ถ๐๐ถ๐2 = โ(+2)(1)(1.6022๐ฅ10โ19)2(6.023๐ฅ1023)(2.408)
(1.112๐ฅ10โ10)(2.81๐ฅ10โ10)โ (1 โ
1
9) รท 1000
= โ๐๐๐๐. ๐ ๐๐ฑ/๐๐๐
Second, compare your Born-Landรฉ estimates against values from the Kapustinskii equation.
๐ = โ1200.5๐๐+๐โ
๐๐+๐๐โ (1 โ
0.345
๐๐+๐๐) is the Kapustinskii equation where V is the # of ions in the
formula and ๐๐, ๐๐are the radii of the cation and anion respectively in Angstroms.
So we have,
๐๐ถ๐๐ถ๐ = โ1200.5(2)(1)(1)
3.27โ (1 โ
0.345
3.27) = โ๐๐๐. ๐ ๐๐ฑ/๐๐๐
and we have
๐๐ถ๐๐ถ๐2 = โ1200.5(3)(2)(1)
2.81โ (1 โ
0.345
2.81) = โ๐๐๐๐. ๐ ๐๐ฑ/๐๐๐
The values agree very well. For CaCl they agree within less than 1% and for CaCl2 they agree
within about 6%.
Third, calculate the heats of formation for CaCl and CaCl2.
โ๐ป๐ = ๐ + ๐ท + ๐ผ๐ + ๐ธ๐ด + ๐
For this, I just used the calculated lattice energies found by the Kapustinskii equation. So for
CaCl, we only need the first IP for Ca and only 1/2D for the single Cl atom and we get
โ๐ป๐๐๐ ๐ถ๐๐ถ๐ = 201 + 121 + 590 โ 349 โ 656.8 = โ๐๐. ๐ ๐๐ฑ/๐๐๐
For CaCl2, we need both IP to get to Ca2+, the full dissociation energy since we have 2 Cl atoms
and we need to multiply the EA by 2 to account for both Cl atoms so we get
โ๐ป๐๐๐ ๐ถ๐๐ถ๐2 = 201 + 242 + 590 + 1146.4 โ 2 โ 349 โ 2248.6 = โ๐๐๐. ๐ ๐๐ฑ/๐๐๐
Why are they so different? Based on these numbers, do you expect CaCl to be stable or to
disproportionate to CaCl2 and Ca?
The heats of formation are different because the lattice energy is more negative for the CaCl2
since the sum of the radii is smaller and the charges on the atoms are larger. Based on these
numbers weโd expect that CaCl is not stable and would disproportionate to CaCl2 and Ca.
3. Gersten Smith 3.10 parts a and b:
๐1 = 2ฯ๐2 ร ๐3
๐1 โ (๐2 ร ๐3), ๐2 = 2ฯ
๐3 ร ๐๐
๐1 โ (๐2 ร ๐3), ๐๐ = 2ฯ
๐1 ร ๐๐
๐1 โ (๐2 ร ๐3)
So,
๐1 = 2ฯ(๐๐โ3
2(โ๐ ร ๐) +
๐๐2
(๐ ร ๐))
(๐โ3
2๐ข +
๐2
๐ฃ) โ (๐๐โ3
2(โ๐ ร ๐) +
๐๐2
(๐ ร ๐))
= 2ฯ(๐๐โ3
2(๐) +
๐๐2
(๐))
(๐โ3
2๐ข +
๐2
๐ฃ) โ (๐๐โ3
2(๐) +
๐๐2
(๐))
= 2ฯ(๐๐โ3
2(๐) +
๐๐2
(๐))
(๐2๐โ3
4+
๐2๐โ34
)
= 2ฯ(โ3(๐) + (๐))
๐โ3=
2ฯ
๐โ3๐ +
2ฯ
๐๐
Similarly, since the denominator stays the same we get
๐2 = 2ฯ(๐๐โ3
2(๐ ร ๐) +
๐๐2
(๐ ร ๐))
(๐2๐โ3
2)
= 2ฯ(โ3 (๐) + (โ๐))
(โ3๐)=
2ฯ
๐โ3(โ๐) +
2ฯ
๐๐
And
๐3 = 2ฯ(๐23
4(๐ ร โ๐) +
๐2โ34
(๐ ร ๐) +๐2โ3
4(๐ ร โ๐) +
๐2
4(๐ ร ๐))
(๐2๐โ3
2)
= 2ฯ(โ32 ๐ +
โ32 ๐)
๐โ3=
2ฯ
๐๐
Part b)
The reciprocal lattice of the hexagonal lattice is also a hexagonal lattice. The first Brillouin zone
is the Wigner-Seitz cell of the reciprocal lattice. The first Brillouin zone is also a hexagonal
prism. For a simple hexagonal Bravais lattice with lattice constants โaโ and โcโ, the reciprocal
lattice is a simple hexagonal lattice of lattice constants 2ฯ/c and 4๐
๐โ3 rotated by 30ยฐ relative to the
direct lattice.
Plan View Diagram
To construct the first Brillouin zone, take perpendicular bisectors and enclose
4. A & M 6.1
A (FCC) B (BCC) C (diamond)
ฮ sin2 ฮ ratio ฮ sin2 ฮ ratio ฮ sin2 ฮ ratio
21.1 0.129 3 14.4 0.0618 2 21.4 0.133 3
24.6 0.173 4 20.5 0.1226 4 36.6 0.353 8
36 0.343 8 25.4 0.184 6 44.5 0.4913 11
43.6 0.476 11 29.8 0.247 8 57.5 0.7113 16
5. Using a computer, draw to scale the first five rings of the X-ray ring patterns produced by
diffraction from powders of SC, FCC, BCC, and diamond crystal structures of lattice constant a
= 5.0 ร .
6. The first four XRD reflections of an unknown cubic alkali halide occur at 2ฮธ = 23.38, 27.07,
38.65, and 45.66ยฐ (Cu Kฮฑ radiation). Assign d-spacings and Miller indices to these reflections
and calculate the lattice constant. The density of the alkali halide is 2.80 g/cm3 at room
temperature. Identify the substance.
To find the d-spacing, just use Braggโs Law and solve for d.
๐ =๐ โ ๐
2 โ ๐ ๐๐ฮ
where n=1 by conventional XRD, ฮป=1.54 วบ which is the length of Cu Kฮฑ radiation
2ฮ
(degrees)
ฮ
(degrees)
d-
spacing
(วบ)
d2 Ratios
of d2
Whole
integer
ratios
Miller indices Lattice
constant
values (วบ)
23.38 11.69 3.80 14.44 1 3 (111) 6.58
27.07 13.535 3.29 10.82 1.33 4 (200) 6.58
38.65 19.325 2.33 5.43 2.66 8 (220) 6.59
45.66 22.83 1.98 3.92 3.68 11 (311) 6.57
By referring to the slides on systematic absences, we can see by looking at the whole integer
ratios of the d-spacings that the cubic structure must be FCC, so now we can assign Miller
indices based on the structure factors and knowing that for FCC h,k, and l must all be even or all
odd.
Since we know the structure is cubic, we can calculate the lattice constant (a) by the cubic d-
spacing formula
๐ =๐
โโ2 + ๐2 + ๐2
The lattice constant values should all be relatively the same as long as the Miller indices were
properly assigned to the correct d-spacing values.
Now we can find out the substance by
๐๐๐๐ ๐๐ก๐ฆ = ๐๐๐ก๐๐๐
๐๐๐๐๐= (
๐๐ด
๐๐ด) โ (
1
๐๐๐๐๐)
Where n is the number of atoms per unit cell which we know to be 4 for an FCC lattice, A is the
atomic mass which we will solve for, NA is Avogadroโs number and Vcell is equal to 6.583 since
the unit cell is cubic (need to change angstroms to cm).
2.80๐/๐๐3 = (4 ๐๐ก๐๐๐ โ ๐ด (
๐๐๐๐
)
6.023๐ฅ1023๐๐ก๐๐๐ /๐๐๐) โ (
1
(6.58 โ 10โ8 ๐๐)3)
A = 120.11 g/mol, so by some trial and error and by looking at the alkalis and halides we can see
that the mass of RbCl is 120.921 g/mol which is nearly the same as what we calculated.
7. Derive the simplified structure factor equation for (a) cubic SrTiO3 and (b) CaF2. For what
combinations of hkl will systematic absences occur in these two crystals?
The total structure factor is
๐๐ = โ ๐๐(๐ฒ)๐๐๐ฒโ๐ ๐
๐
By using the B-cell of the perovskite structure, we can describe SrTiO3 as simple cubic with a 5-
atom basis:
Ti (0, 0, 0), Sr ( ยฝ, ยฝ, ยฝ) and O ( ยฝ, 0, 0), (0, ยฝ, 0) and (0, 0, ยฝ) and using ๐ฒ =2๐
๐(โ๐ + ๐๐ +
๐๐)
๐๐ = ๐๐๐๐๐(
2๐๐
(โ๐+๐๐+๐๐))โ(๐(0โx+0โy+0โz)) + ๐๐๐๐๐(2๐๐
(โ๐+๐๐+๐๐))โ(๐(12
โx+12
โy+12
โz))
+ ๐๐[๐๐(
2๐๐
(โ๐+๐๐+๐๐))โ(๐(12
โx+0โy+0โz))
+ ๐๐(
2๐๐
(โ๐+๐๐+๐๐))โ(๐(0โx+12
โy+0โz))+ ๐
๐(2๐๐
(โ๐+๐๐+๐๐))โ(๐(0โx+0โy+12
โz))
Which simplifies down to
๐๐ = ๐๐๐ + ๐๐๐๐๐๐(โ+๐+๐) + ๐๐[๐๐๐(โ) + ๐๐๐(๐) + ๐๐๐(๐)], where ๐๐๐ = โ1
Without knowing the form factors, no systematic absences will occur for SrTiO3, but these
situations will occur for various values of h, k, and l:
All h, k, and l are odd: ๐๐๐ โ ๐๐๐ โ 3๐๐
All h, k, and l are even: ๐๐๐ + ๐๐๐ + 3๐๐
When 1 is odd/2 are even: ๐๐๐ โ ๐๐๐ + ๐๐
When 2 are odd/1 is even: ๐๐๐ + ๐๐๐ โ ๐๐
For CaF2, you can also see this worked out in Ch. 5 of the Old West book (pg. 160-161). We
have an FCC lattice of Ca and an 8-atom basis of F, so we should expect to have 12 terms in the
structure factor equation.
The basis positions are:
Ca (0, 0, 0), ( ยฝ, ยฝ, 0), ( ยฝ, 0, ยฝ), (0, ยฝ, ยฝ) and F ( ยผ, ยผ, ยผ), ( ยผ, ยผ, ยพ), ( ยผ, ยพ, ยผ), ( ยพ, ยผ, ยผ),
( ยพ, ยพ, ยผ), ( ยพ, ยผ, ยพ), ( ยผ, ยพ, ยพ), ( ยพ, ยพ, ยพ)
๐๐ = ๐๐ถ๐[1 + ๐๐๐(โ+๐) + ๐๐๐(๐+๐) + ๐๐๐(โ+๐)] + ๐๐น[๐๐๐2
(โ+๐+๐)+ ๐๐
๐2
(โ+๐+3๐)+ ๐๐
๐2
(โ+3๐+๐)
+ ๐๐๐2
(3โ+๐+๐)+ ๐๐
๐2
(3โ+3๐+๐)+ ๐๐
๐2
(3โ+๐+3๐)+ ๐๐
๐2
(โ+3๐+3๐)
+ ๐๐๐2
(3โ+3๐+3๐)]
We already know that for FCC, h, k, and l must all be even or all be odd to observe a reflection.
Turns out, itโs still true for the fluorite structure.
The structure factor will be:
All odd h,k, and l: 4๐๐ถ๐
For h, k, and l all even, with sum 2x an odd number: 4๐๐ถ๐-8๐๐น
For h, k, and l all even, with sum 2x an even number: 4๐๐ถ๐+8๐๐น
Otherwise = 0