FIRST MIDTERM EXAM
Chemistry 465
Professor Buhro
1 March 2011
__________________________________
Signature
__________________________________
Print Name Clearly
ID Number:_______________________
Information. This is a closed-book exam; no books, notes, other students, other student exams, or any
other resource materials may be consulted or examined during the exam period. Calculators are
permitted. Partial credit will be given for partially correct reasoning in support of incorrect or correct
final answers. Additional space for answers is provided at the end of this exam; please clearly label any
answers you place there. Please find “Potentially Useful Information” attached as the last pages of
this exam.
1.
____________________________ (15 pts)
2.
____________________________ (15 pts)
3.
____________________________ (15 pts)
4.
____________________________ (15 pts)
5.
____________________________ (15 pts)
6.
____________________________ (10 pts)
7.
____________________________ (15 pts)
______________________________________________________
Total
____________________________ (100 pts)
1
1. 15 total pts. An incomplete Mooser-Pearson plot for MX2 compounds is given below. A version of
this plot was given on the practice exam and key, although the answer given was partially incorrect.
There are four structure fields (regions) on this plot, as shown below. One field (region) contains
compounds having the fluorite (CaF2) structure, one contains compounds having the rutile structure,
and one contains compounds having either CdI2 or CdCl2 structures. The remaining field contains the
compound SiO2 having the -quartz or related structures in which Si atoms have a coordination
number of 4 and the O atoms have a coordination number of 2. Please assign each of these four fields
by adding the labels “CaF2,” “rutile,” “CdI2,” and “SiO2.” Then, in the space below, please write a
brief justification for the assignments you have made.
[2 pts. for each correct label]
CaF2
CdI2
n
rutile
SiO2

The fields for highest cation/anion coordination numbers and ionicity are at the upper
right, and those for lowest cation/anion coordination numbers and highest covalency are at
the lower left. Thus CaF2 (CNCa = 8; CNF = 4) is at the upper right [2 pts]. The bonding in
SiO2 is highly covalent, and the coordination numbers for this compound are the smallest
(CNSi = 4; CNO = 2), so it is at the lower left [2 pts]. Rutile and CdI2 have the same
intermediate coordination numbers (CNM = 6; CNX = 3), but compounds with the rutile
structure are more ionic (larger ), and compounds with the CdI2 structure are less ionic
(smaller ). Thus CdI2 is at the middle left and rutile is at the middle right [3 pts].
2
2. 15 total pts. The XRD pattern below was obtained from a powdered metallic specimen having a
conventional crystal structure for a metal. Note that the sine-squared-theta ratios are also given.
Please assign Miller indices to all the reflections in the pattern, identify the crystal structure of the
metal by name (type), and calculate the lattice parameter or parameters. Show your work, using this
and the following page.
110
[9 pts.]
211
200
321
310
220
411 or 330
222
Line No.
1
2
3
4
5
6
7
8
9
2
24.068
34.297
42.338
49.288
55.576
61.421
66.956
72.272
77.434
‡
sin2() mhkl/m100* mhkl/m100† mhkl/m100
3
0.043469
1
2
0.086936 1.999958 3.999917 5.999875
0.130408 3.000019 6.000039 9.000058
0.173871 3.999899 7.999797 11.99970
0.217344 4.999976 9.999951 14.99993
0.260815 6.000029 12.00006 18.00009
0.304281 6.999962 13.99992 20.99989
0.347751 7.999979 15.99996 23.99994
0.391218 8.999939 17.99988 26.99982
mhkl = (h2 + k2 +l2)
*Assuming first reflection is 100
†
Assuming first reflection is 110
‡
Assuming first reflection is 111
3
400
110
200
211
220
310
222
321
400
411/330
2. (cont.)
The crystal structure of this metal is bcc. [3 pts.]
Calculate the lattice parameter a:
d110 

2 sin(2 )
2

1.542 Å
 3.6980 Å

24.608
2 sin(
)
2
rearrange the d-spacing formula:
a  d110 (12  12  02 )1 2  2(3.6980 Å)  5.2280 Å
4
[3 pts.]
3. 15 total pts. The XRD pattern below was obtained from a powdered metallic specimen having a
conventional crystal structure for a metal. Note that the sine-squared-theta ratios are also given.
Please assign Miller indices to all the reflections in the pattern, identify the crystal structure of the
metal by name (type), and calculate the lattice parameter or parameters. Show your work, using this
and the following page.
101
[5 pts.]
002
102
100
Line No.
1
2
3
4
5
2
39.222
41.681
44.65
58.564
71.089
‡
sin2() mhkl/m100* mhkl/m100† mhkl/m100
0.112649
1
2
3
0.126571 1.123583 2.247166 3.370749
0.144293 1.280911 2.561822 3.842732
0.239227 2.123648 4.247296 6.370944
0.337950 3.000028 6.000055 9.000083
mhkl = (h2 + k2 +l2)
*Assuming first reflection is 100
†
Assuming first reflection is 110
‡
Assuming first reflection is 111
5
110
The structure is not cubic; it must
be hcp [1 pt.]. If so, line 3 must be
the 101 reflection. Assume that
line 1 is 100, line 2 is 002, and
calculate the lattice parameters a
and c. Then confirm that the 101
reflection appears at the correct
2 value. See the work on the next
page.
3. (cont.)
Use the first two reflections to solve for the lattice parameters a and c:
d 

2 sin 
 4 
d100  

 3a 2 
combine :
1 2
3


a
2 sin 100
2
a

3 sin 100
3
a
2


1.542 Å
3 sin 39.222°

similarly :
 2 
d002   
 c 2 
1 2

2

 2.653 Å [2 pts.]
c


c

2 sin 002 
c
2(1.542 Å)


2 sin 002 2 sin 41.681°


 4.334 Å [2 pts.]
2
Now check the position of the 101 reflection:
1 2


4
2


 2.030 Å
d101  
 3(2.653 Å)2 (4.334 Å)2 
  
1  1.542 Å 

2101  2 sin 1 
  2 sin 
  44.64 , which matches line 3 [2 pts.]
d
2
2(2.030
Å)


 101 
Now guess that line 4 or 5 is the 110 reflection:
1 2


4
d110  
(12  12  11) 
 1.327 Å
 3(2.653 Å)2

 1.542 Å 

2110  2 sin 1 
  71.07 , which matches line 5 [2 pts.]
 2(1.327 Å) 
Line 4 must be the 111, 102, or 201 reflection (smallest available indices). Given the values
of a and c, the 102 will be at lowest 2 :
1 2


4
22

d102  

 1.576 Å
 3(2.653 Å)2 (4.334 Å)2 
 1.542 Å 

2102  2 sin 1 
  58.58 , which matches line 4 [1 pt.]
 2(1.576 Å) 
6
4. 15 total pts. The XRD pattern below was obtained from a powdered specimen having the CsCl
structure. Note that the sine-squared-theta ratios are also given. Please assign Miller indices to all the
reflections in the pattern, and calculate the lattice parameter or parameters. Show your work, using
this and the following page.
110
[11 pts.]
211
310
200
100
111
220
210
311
222
300 or 221
Line No.
1
2
3
4
5
6
7
8
9
10
11
2
20.707
29.499
36.274
42.133
47.391
52.238
61.106
65.255
69.269
73.179
77.010
‡
sin2() mhkl/m100* mhkl/m100† mhkl/m100
3
0.032300
1
2
0.064818 2.006771 4.013542 6.020312
0.096902 3.000088 6.000176 9.000263
0.129205 4.000213 8.000426 12.00064
0.161504 5.000195 10.00039 15.00059
0.193809 6.000343 12.00069 18.00103
0.258405 8.000248 16.0005 24.00074
0.29071 9.000419 18.00084 27.00126
0.32301 10.00043 20.00085 30.00128
0.355309 11.00041 22.00082 33.00123
0.38761 12.00045 24.0009 36.00135
mhkl = (h2 + k2 +l2)
*Assuming first reflection is 100
†
Assuming first reflection is 110
‡
Assuming first reflection is 111
7
100
110
111
200
210
211
220
300/221
310
311
222
4. (cont.)
Calculate the lattice parameter a:
d100 

2 sin(2 )
2

1.542 Å
 4.2900 Å

2 sin(20.707 )
2
rearrange the d-spacing formula:
a  d100 (12  02  02 )1 2  d100  4.2900 Å
[4 pts.]
8
5. 15 total pts. (a) 5 pts. The unit cell for the rock-salt (NaCl) structure is shown below. The smaller
cations are black and the larger anions are gray. Note that an anion is positioned at the origin of the
unit cell. Please resketch the cell after moving a cation to the origin, using the frame on the lower
right.
-1 pt. each missing ion, up to a max. of -2
(b) 5 pts. The unit cell for the zinc blende (ZnS) structure is shown below. The smaller cations are black
and the larger anions are gray. Note that an anion is positioned at the origin of the unit cell. Please
resketch the cell after moving a cation to the origin, using the frame on the lower right.
-1 pt. each missing ion, up to a max. of -2
Anions may occupy either T + or T –
sites (alternating octants must be
occupied).
9
5. (c) 5 pts. Although many crystal structures have “anti” relatives, such as fluorite and antifluorite,
other crystal structures do not. Please explain why there are no anti-rock-salt, anti-zinc-blende, or
anti-CsCl structures.
The anti structures in these cases do not exist because the structures generated by
exchanging the cation and anion positions are identical to the original structures.
6. 10 total points. (a) 5 pts. The cubic-close-packed crystal structure can be described by a cubic unit
cell or a hexagonal unit cell. Please explain briefly why the cubic unit cell is conventionally used.
The cubic cell is conventional because it has higher symmetry [5 pts.], or because it
reveals the full symmetry of the crystal structure [5 pts.].
(b) 5 pts. Please briefly explain the purpose of refining a fitted pattern to experimental XRD data. What
primary goals are achieved by refinement?
The two primary goals are to confirm the crystal structure [2.5 pts.] and to obtain precise
lattice parameters [2.5 pts.].
10
7. 15 total pts. Unit cells from a cubic lattice are depicted below. Please assign Miller indices to the
crystallographic directions depicted by each of the arrows placed within the unit cells. Write your
answers in the brackets provided.
[3 pts. for each correct set of indices]
c
b
a
[ 121 ]
[ 001 ]
[ 111 ]
[ 121 ]
[ 121 ]
11
12