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 “Ionic radius” page or found on “Web Elements”

webpage

Atom Ion Size CN: 4 Ion Size CN: 6 Ion Size CN: 8

F- 1.17 Å 1.19 Å

Pb2+ 1.33 Å 1.43 Å

Be2+ 0.41 Å 0.59 Å

Co2+ 0.72 Å 0.79 Å low spin

0.885 Å high spin

1.04 Å

You should try the radius ratio rules for all possible CN’s. I’ll just use the 1.19 Å 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 α-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-Landé equation to estimate the lattice energies for CaCl and CaCl2.

Specify and justify any assumptions you needed to make to do this.

𝑈 = −𝑍+𝑍−𝑒2𝑁𝐴

𝑟𝑒∗ (1 −

1

𝑛)

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 Madelung’s 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 Ǻ) and should be slightly larger than the radius of K+ (1.52 Ǻ), so

I’m going to use 1.60 Ǻ.

rCa2+=1.14 Ǻ and rCl-=1.67 Ǻ

re= 4πε0r0 = 1.112x10-10 C2/J.m * r0 and r0 for CaCl = 1.60 + 1.67= 3.27 Ǻ and r0 for CaCl2 is 1.14

+ 1.67 = 2.81 Ǻ

So finally,

𝑈𝐶𝑎𝐶𝑙 = −(+1)(1)(1.6022𝑥10−19)2(6.023𝑥1023)(1.748)

(1.112𝑥10−10)(3.27𝑥10−10)∗ (1 −

1

9)

= −686489𝐽

𝑚𝑜𝑙÷ 1000 = −𝟔𝟔𝟎. 𝟕 𝒌𝑱/𝒎𝒐𝒍

and

𝑈𝐶𝑎𝐶𝑙2 = −(+2)(1)(1.6022𝑥10−19)2(6.023𝑥1023)(2.408)

(1.112𝑥10−10)(2.81𝑥10−10)∗ (1 −

1

9) ÷ 1000

= −𝟐𝟏𝟏𝟖. 𝟐 𝒌𝑱/𝒎𝒐𝒍

Second, compare your Born-Landé estimates against values from the Kapustinskii equation.

𝑈 = −1200.5𝑉𝑍+𝑍−

𝑟𝑐+𝑟𝑎∗ (1 −

0.345

𝑟𝑐+𝑟𝑎) is the Kapustinskii equation where V is the # of ions in the

formula and 𝑟𝑐, 𝑟𝑎are the radii of the cation and anion respectively in Angstroms.

So we have,

𝑈𝐶𝑎𝐶𝑙 = −1200.5(2)(1)(1)

3.27∗ (1 −

0.345

3.27) = −𝟔𝟓𝟔. 𝟖 𝒌𝑱/𝒎𝒐𝒍

and we have

𝑈𝐶𝑎𝐶𝑙2 = −1200.5(3)(2)(1)

2.81∗ (1 −

0.345

2.81) = −𝟐𝟐𝟒𝟖. 𝟔 𝒌𝑱/𝒎𝒐𝒍

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.

∆𝐻𝑓 = 𝑆 + 𝐷 + 𝐼𝑃 + 𝐸𝐴 + 𝑈

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

∆𝐻𝑓𝑜𝑓 𝐶𝑎𝐶𝑙 = 201 + 121 + 590 − 349 − 656.8 = −𝟗𝟑. 𝟖 𝒌𝑱/𝒎𝒐𝒍

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

∆𝐻𝑓𝑜𝑓 𝐶𝑎𝐶𝑙2 = 201 + 242 + 590 + 1146.4 − 2 ∗ 349 − 2248.6 = −𝟕𝟔𝟕. 𝟐 𝒌𝑱/𝒎𝒐𝒍

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 we’d expect that CaCl is not stable and would disproportionate to CaCl2 and Ca.

3. Gersten Smith 3.10 parts a and b:

𝐛1 = 2π𝐚2 × 𝐚3

𝐚1 ⋅ (𝐚2 × 𝐚3), 𝐛2 = 2π

𝐚3 × 𝐚𝟏

𝐚1 ⋅ (𝐚2 × 𝐚3), 𝐛𝟑 = 2π

𝐚1 × 𝐚𝟐

𝐚1 ⋅ (𝐚2 × 𝐚3)

So,

𝒈1 = 2π(𝑎𝑐√3

2(−𝒊 × 𝒌) +

𝑎𝑐2

(𝒋 × 𝒌))

(𝑎√3

2𝐢 +

𝑎2

𝐣) ⋅ (𝑎𝑐√3

2(−𝒊 × 𝒌) +

𝑎𝑐2

(𝒋 × 𝒌))

= 2π(𝑎𝑐√3

2(𝒋) +

𝑎𝑐2

(𝑖))

(𝑎√3

2𝐢 +

𝑎2

𝐣) ⋅ (𝑎𝑐√3

2(𝒋) +

𝑎𝑐2

(𝒊))

= 2π(𝑎𝑐√3

2(𝑗) +

𝑎𝑐2

(𝒊))

(𝑎2𝑐√3

4+

𝑎2𝑐√34

)

= 2π(√3(𝒋) + (𝑖))

𝑎√3=

2π

𝑎√3𝒊 +

2π

𝑎𝒋

Similarly, since the denominator stays the same we get

𝒈2 = 2π(𝑎𝑐√3

2(𝒌 × 𝒊) +

𝑎𝑐2

(𝒌 × 𝒋))

(𝑎2𝑐√3

2)

= 2π(√3 (𝒋) + (−𝒊))

(√3𝑎)=

2π

𝑎√3(−𝒊) +

2π

𝑎𝒋

And

𝒈3 = 2π(𝑎23

4(𝒊 × −𝒊) +

𝑎2√34

(𝒊 × 𝒋) +𝑎2√3

4(𝒋 × −𝒊) +

𝑎2

4(𝒋 × 𝒋))

(𝑎2𝑐√3

2)

= 2π(√32 𝒌 +

√32 𝒌)

𝑐√3=

2π

𝑐𝒌

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 “a” and “c”, the reciprocal

lattice is a simple hexagonal lattice of lattice constants 2π/c and 4𝜋

𝑎√3 rotated by 30° 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)

Θ sin2 Θ ratio Θ sin2 Θ ratio Θ sin2 Θ 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 Å.

6. The first four XRD reflections of an unknown cubic alkali halide occur at 2θ = 23.38, 27.07,

38.65, and 45.66° (Cu Kα 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 Bragg’s Law and solve for d.

𝑑 =𝑛 ∗ 𝜆

2 ∗ 𝑠𝑖𝑛Θ

where n=1 by conventional XRD, λ=1.54 Ǻ which is the length of Cu Kα radiation

2Θ

(degrees)

Θ

(degrees)

d-

spacing

(Ǻ)

d2 Ratios

of d2

Whole

integer

ratios

Miller indices Lattice

constant

values (Ǻ)

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

𝑑 =𝑎

√ℎ2 + 𝑘2 + 𝑙2

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

𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑚𝑎𝑡𝑜𝑚𝑠

𝑉𝑐𝑒𝑙𝑙= (

𝑛𝐴

𝑁𝐴) ∗ (

1

𝑉𝑐𝑒𝑙𝑙)

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 Avogadro’s number and Vcell is equal to 6.583 since

the unit cell is cubic (need to change angstroms to cm).

2.80𝑔/𝑐𝑚3 = (4 𝑎𝑡𝑜𝑚𝑠 ∗ 𝐴 (

𝑔𝑚𝑜𝑙

)

6.023𝑥1023𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙) ∗ (

1

(6.58 ∗ 10−8 𝑐𝑚)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

𝜑𝑘 = ∑ 𝑓𝑗(𝑲)𝑒𝑖𝑲∙𝒅𝒋

𝑗

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 ( ½, ½, ½) and O ( ½, 0, 0), (0, ½, 0) and (0, 0, ½) and using 𝑲 =2𝜋

𝑎(ℎ𝒙 + 𝑘𝒚 +

𝑙𝒛)

𝜑𝑘 = 𝑓𝑇𝑖𝑒𝑖(

2𝜋𝑎

(ℎ𝒙+𝑘𝒚+𝑙𝒛))∙(𝑎(0∗x+0∗y+0∗z)) + 𝑓𝑆𝑟𝑒𝑖(2𝜋𝑎

(ℎ𝒙+𝑘𝒚+𝑙𝒛))∙(𝑎(12

∗x+12

∗y+12

∗z))

+ 𝑓𝑂[𝑒𝑖(

2𝜋𝑎

(ℎ𝒙+𝑘𝒚+𝑙𝒛))∙(𝑎(12

∗x+0∗y+0∗z))

+ 𝑒𝑖(

2𝜋𝑎

(ℎ𝒙+𝑘𝒚+𝑙𝒛))∙(𝑎(0∗x+12

∗y+0∗z))+ 𝑒

𝑖(2𝜋𝑎

(ℎ𝒙+𝑘𝒚+𝑙𝒛))∙(𝑎(0∗x+0∗y+12

∗z))

Which simplifies down to

𝜑𝑘 = 𝑓𝑇𝑖 + 𝑓𝑆𝑟𝑒𝑖𝜋(ℎ+𝑘+𝑙) + 𝑓𝑂[𝑒𝑖𝜋(ℎ) + 𝑒𝑖𝜋(𝑘) + 𝑒𝑖𝜋(𝑘)], where 𝑒𝑖𝜋 = −1

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: 𝑓𝑇𝑖 − 𝑓𝑆𝑟 − 3𝑓𝑂

All h, k, and l are even: 𝑓𝑇𝑖 + 𝑓𝑆𝑟 + 3𝑓𝑂

When 1 is odd/2 are even: 𝑓𝑇𝑖 − 𝑓𝑆𝑟 + 𝑓𝑂

When 2 are odd/1 is even: 𝑓𝑇𝑖 + 𝑓𝑆𝑟 − 𝑓𝑂

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), ( ½, ½, 0), ( ½, 0, ½), (0, ½, ½) and F ( ¼, ¼, ¼), ( ¼, ¼, ¾), ( ¼, ¾, ¼), ( ¾, ¼, ¼),

( ¾, ¾, ¼), ( ¾, ¼, ¾), ( ¼, ¾, ¾), ( ¾, ¾, ¾)

𝜑𝑘 = 𝑓𝐶𝑎[1 + 𝑒𝑖𝜋(ℎ+𝑘) + 𝑒𝑖𝜋(𝑘+𝑙) + 𝑒𝑖𝜋(ℎ+𝑙)] + 𝑓𝐹[𝑒𝑖𝜋2

(ℎ+𝑘+𝑙)+ 𝑒𝑖

𝜋2

(ℎ+𝑘+3𝑙)+ 𝑒𝑖

𝜋2

(ℎ+3𝑘+𝑙)

+ 𝑒𝑖𝜋2

(3ℎ+𝑘+𝒍)+ 𝑒𝑖

𝜋2

(3ℎ+3𝑘+𝑙)+ 𝑒𝑖

𝜋2

(3ℎ+𝑘+3𝑙)+ 𝑒𝑖

𝜋2

(ℎ+3𝑘+3𝑙)

+ 𝑒𝑖𝜋2

(3ℎ+3𝑘+3𝑙)]

We already know that for FCC, h, k, and l must all be even or all be odd to observe a reflection.

Turns out, it’s still true for the fluorite structure.

The structure factor will be:

All odd h,k, and l: 4𝑓𝐶𝑎

For h, k, and l all even, with sum 2x an odd number: 4𝑓𝐶𝑎-8𝑓𝐹

For h, k, and l all even, with sum 2x an even number: 4𝑓𝐶𝑎+8𝑓𝐹

Otherwise = 0