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 Pb 2+ 1.33 Å 1.43 Å Be 2+ 0.41 Å 0.59 Å Co 2+ 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
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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 “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