CM3020 Solid State Chemistry – Crystal Chemistry
Tutorial Sheet (weeks 1-4)
1. What symmetries, if any, do the following molecules possess:
methyl chloride
dichloromethane
methane
[Hint: to find all the symmetries in methane, it may be useful to imagine a tetrahedron as a cube with
alternate corners missing]
2. Copper metal is face centred cubic with a=3.615Å. Assuming the atoms touch along the face diagonal,
calculate the metallic radius of copper.
3. NaCl is face centred cubic with a = 5.64Å. Assuming the radius of the chloride ion is 1.80Å, calculate
the radius for Na+. Are the chloride ions in contact in NaCl?
4. Calculate the d-spacing of the (1 2 3) planes in a cubic crystal with a=b=c=8Å. Calculate the d-spacing
of the same planes in an orthorhombic crystal with a=7Å, b=8Å and c=9Å.
5. In 1989, there was much controversy over the possibility of “cold fusion” involving Pd/D. Pd is a face
centred cubic metal, 3.89Å. The deuterium atoms could enter either tetrahedral or octahedral sites.
Calculate the Pd-D bond distances for each. If “cold fusion” occurs, it may be because D atoms become
too close to each other.
Calculate the shortest distance between:
a) Two tetrahedral sites
b) Two octahedral sites
c) An octahedral site and a tetrahedral site
Comment on your results! The bond length in the deuterium molecule is 0.71Å. More recently, it was
discovered that the addition of aluminium improved the yield. If it is indeed the D-D distance which is
critical, speculate on why this might be the case!
6. Li2O has a cubic structure with unit cell edge 4.61Å.
The atomic coordinates are:
Li
¼ ¼ ¼
¾ ¾ ¾
¼ ¼ ¾
O
¼ ¾ ¾
¼ ¾ ¼
¾ ¼ ¾
¾ ¼ ¼
¾ ¾ ¼
0 0 0
½ ½ 0
0 ½ ½
½0 ½
Draw a projection of the structure onto the xy plane. Indicate the coordination environments of Li and
O. Calculate the Li-O bond length. What is the name for this structure type?
7. CsCl is primitive cubic with unit cell a = 4.115Å. Using the d-spacing equation, calculate the spacing of
(a) the 110 planes and (b) the 111 planes.
8. X-rays of wavelength =1.5Å are reflected from the (2 2 2) planes of a cubic crystal with unit cell a =
5Å. Calculate the Bragg angle, , for n=1.
Self-test questions to help with the maths used in this course are available at:
http://www.abdn.ac.uk/~ltu003/files/physics.htm
Web pages for this section of the course (will be added as material is covered):
http://www.abdn.ac.uk/~che241/y3-2002/index.htm
CHEMISTRY CM3020 CRYSTAL CHEMISTRY
1) Methyl chloride:
ANSWERS
one 3-fold rotation axis along the C-Cl bond
three mirror planes
NO 2-fold rotation axes
H
Cl
3
C
C
H
H
H
H
H
m
3-fold axis
mirror plane (3 equivalents)
2) Dichloromethane:
NO 3-fold axes
one 2-fold axis
two mirror planes
2-axis out of page
H
Cl
C
Cl
m
H
m
3) Methane:
4 3-fold axes
3 2-fold axes
six mirror planes
3-axes down body diagonals
2-fold axes along face centres
Mirrors bisect face diagonals
(3 faces therefore 6 diagonals)
None of the molecules has a centre of symmetry
H
H
C
H
H
2) In fcc Cu, the metal atoms are in contact along the face diagonals of the unit cell; this is the
key point.
Atomic radius r
Therefore face diagonal = 4r
Cell edge = a = 3.615Å
Pythagoras:
a2 + a2 = (4r)2
 r = 1.28Å
3) In NaCl, anions and cations are in contact along the cell edges:
a = 2(rNa + rCl)
5.64 = 2(rNa + 1.80)
Na+
rNa = 1.02Å
Cl+
Cell face diagonal = 2  a = 7.98Å
But if Cl- ions are in contact along the face diagonal, this length should equal 4rCl- i.e. 7.2Å.
Conclude that the Cl- ions are not touching.
4)
1 h 2  k 2  l2 1  4  9
64


 d2 
2
2
64
14
d
a
1 h 2 k 2 l2




d 2 a 2 b2 c2

d = 2.14Å
1
4 9


 d 2  (0.194) 1  5.154  d=2.27Å
49 64 81
5) I have used a projection (below) showing the positions of the fcc Pd atoms and the possible
positions of ALL interstitial sites (tetrahedral and octahedral). This helps to visualise the
positions relative to one another - but if this doesn't help you, find a 3d model in the lab, or
make your own sketches.
For octahedral D, the Pd-D distance is a/2 (i.e. half the unit cell edge)
= 1.945Å
For tetrahedral D, the Pd-D distance is (a3)/4 (i.e. ¼ of the body diagonal of the cube) =
1.684 Å
a) tetrahedral sites are separated by a/2 = 1.945Å
b) octahedral sites are separated by ½ a2 = 2.751Å (half a face diagonal)
c) octahedral and tetrahedral sites are separated by (a3)/4 = 1.684Å (i.e. ¼ of the body
diagonal)
Pd ½
oct 0,1
Pd 0,1
oct ½
tet ¼, ¾
tet ¼, ¾
Pd ½
oct 0,1
Pd ½
oct 0,1
Pd 0,1
oct ½
tet ¼, ¾
tet ¼, ¾
Pd 0,1
oct ½
Pd 0,1
oct ½
Pd ½
oct 0,1
Pd 0,1
oct ½
Even if all tetrahedral and octahedral sites were occupied by D atoms, the closest separation is
1.684Å, much greater than in D2. Some other explanation for cold fusion would be required,
therefore.
The increase in yield on introduction of Al could be explained by the reduction in the lattice
size and therefore the D-D distance. Pd radius = 1.00Å c.f. Al = 0.675Å.
But we just concluded that the separation of D-D could not be responsible, so this result may
contradict that!
6) It is useful to remember the following routine:
 decide which axes to draw. If cubic, then it doesn’t matter. If tetragonal, then it makes
sense to draw ac or bc. Lower symmetry becomes more a matter of choice – it is often
easier to choose the two longest axes.
 Draw the unit cell, mark the origin
 Now it is just like marking points on a graph – take your two coordinates, plot the point,
then mark on the third coordinate as the “height”.
 Any atoms on cell corners or edges should be marked in equivalent positions – i.e. corners
should be at all corners, edges should be at the opposite edge.
Some extra, simple examples for projections are provided as a supplemental aid (next page).
In this exercise, you should be able to draw a projection on graph paper, labelling atom heights, and
from that deduce the coordination. This part can be difficult in terms of visualising the
structure in 3d.
Li – tetrahedral
O - 8 coordinate
Li-O = a3/4 = 1.99Å
Antifluorite structure
O½
O 0,1
O½
Li ¼, ¾
O½
O 0,1
O 0,1
Li ¼, ¾
O½
O 0,1
Li ¼, ¾
O 0,1
Li ¼,O¾½
Li ¼, ¾
O 0,1
Extra projection examples:
MgO (Periclase)
Unit cell: Cubic, face-centred, a = 4.22 Å
Atoms positions
Mg:
O:
(0, 0, 0)
(½, ½, ½)
(½, ½, 0)
(0, 0, ½)
(½. 0, ½)
(0, ½, 0)
(0, ½, ½)
(½, 0, 0)
SnO2 (Tin (IV) dioxide)
Unit cell: Tetragonal, primitive, a = 4.74, c=3.19Å
Atom positions:
Sn: (0, 0, 0) (½, ½, ½)
O:
(0.307, 0.307, 0)
(0.693, 0.693, 0)
(0.193, 0.807, ½)
ZrO2 (Tetragonal zirconia)
Unit cell: Tetragonal, primitive, a = 3.64, c=5.27Å
Atom positions:
Zr:
O:
(0, 0, 0)
(½, ½, ½)
(0, ½, 0.188) (0, ½, 0.688) (½, 0, 0.312) (½, 0, 0.812)
(0.807, 0.192, ½)
7) The d-spacing equation for orthogonal systems is written:
1 h2 k 2 l 2

 
d 2 a2 b2 c2
This can be simplified for cubic where a=b=c to:
1 h2  k 2  l 2

d2
a2
So for the plane 110 this becomes:
1
12  12  0
2


2
2
16.933
d
4.115
16.933
 d2 
 8.467
2
 d  8.467  2.910 A
8)
1 4  4  4 12


25
d2
52
d2 
25
12
d = 1.443Å
=2dsin
  
  sin 1  
 2d 
 1.5 
  sin 1 

 2  1.443 
 = 31.3º
For the plane 111 this becomes:
1
12  12  12
3


2
2
16.933
d
4.115
16.933
 d2 
 5.644
3
 d  5.644  2.376 A