Using an hkl generating program (e.g.
),
or otherwise, determine whether glide planes or screw axes or are present, and
hence determine the reflection conditions for this material.
[Hint: you can test for the presence of glide planes and screw axes separately.]
Give a logical explanation for your choice.
[10 Marks]
For glide planes to be present then one of the hkl indices has to be zero and one
of the other indices, or a summation of the other two indices, has to be even.
Reflections that have all even integer Miller indices should not be considered. The presence
of the 402 reflection provides no symmetry information. Using these criteria, only the 510 reflection identified
in question 2 is a candidate for evidence for a glide plane. For the 510 reflection, both h and k are odd
indicating that if there is a glide plane perpendicular to c then it cannot be
either a or b but it could be n (or d) - because h+k is even. However, for
C centring the reflection condition is hkl:
h + k = 2n and so the 510 reflection only appears to be a glide
plane.[This also applies to the 110 and 310 reflections identified in question 5]. This analysis indicates that glide planes are
not present in the material.
For screw axes to be present then two of the hkl indices need to be zero. The 002 reflection suggests the presence of screw axes.
However, the monoclinic unit cells only have twofold axes parallel to b and so the 002 reflection is probable from a sub-set
of a more general reflection - i.e. it satisfies the lattice centring reflection condition.
Thus, the reflection condition for the material is;
hkl:
h + k = 2n due to lattice centring
Hence suggest the most likely space group for this material. There
is something surprising about the answer given that the material is
the salt of a naturally-occurring organic material. Comment on this.
[5 Marks]
There are three possibilities for the space group; i) C2 - a C-centred lattice with a twofold rotation axis
parallel to b [Space Group 5] ii) Cm - a C-centred lattice with a mirror plane [Space Group 8]and iii) C2/m - a C-centred
lattice with a twofold rotation axis parallel to b with this axis perpendicular to a mirror plane [Space Group 12].
Consideration of the table in question 2 shows the presence of reflections -5 1 2 and 5 1 2. This suggests the presence of
a mirror plane perpendicular to the x-axis. [Similarly, my table shown in question 5 shows the presence of reflections
2 0 2 & -2 0 2, 4 0 2 & -4 0 2 and 1 1 2 & -1 1 2 - again, all suggesting the presence of a mirror plane
perpendicular to the x-axis.]. There appears to be a good evidence for a mirror plane, thus, I rule out option i) and suggest
that the space-group is either option ii) Cm or option iii) C2/m.
The mirror plane in iii) is perpendicular to the y-axis whereas the data indicates a mirror plane perpendicular to the x-axis.
Thus, I rule out option iii) and I suggest the most likely space group for the material is Cm.
Space group Cm has a symmetry element (mirror plane) that CAN change the handedness of an object. Thus it is not an
enantiomorphic space group. This is surprising since the 'naturally-occurring organic material' is probably enantiomorphic
in that it can probably occur in left-handed and right-handed forms. Thus, it appears that the salt can be both left and right handed
even though it parent 'organic material' was either all left handed or all right handed.
Finally, using a unit cell refinement program (e.g.
),
complete the following two tables:
| 2q |
h |
k |
l |
|
2q |
h |
k |
l |
|
2q |
h |
k |
l |
| 7.50 |
2 | 0 | 0 | |
14.82 |
1 | 1 | 0 | |
16.90 |
-2 | 0 | 4 |
| 8.07 |
0 | 0 | 2 | |
15.05 |
4 | 0 | 0 | |
17.15 |
1 | 1 | 2 |
| 10.25 |
-2 | 0 | 2 | |
16.09 |
-4 | 0 | 2 | |
18.07 |
4 | 0 | 2 |
| 11.77 |
2 | 0 | 2 | |
16.64 |
-1 | 1 | 2 | |
18.27 |
3 | 1 | 0 |
| |
a (Å) |
b (Å) |
c (Å) |
a (°) |
b (°) |
g (°) |
| Value |
23.72856 |
6.17111 |
22.06438 |
90.0000 |
97.9013 |
90.0000 |
| ESD |
0.00822 |
0.00189 |
0.00907 |
0.0000 |
0.0157 |
0.0000 |
Briefly comment on the refined value of the 2q zero
offset.
Lower estimated standard deviations are usually obtained by including
higher angle data. Give two reasons why this is difficult for this
particular data set.
[Hint: you may want to cut and paste the HTML source from these tables
if you are using a plain text editor for your writing your answer.]
[10 Marks]
Zero Error
2qzero = -0.0135 ± 0.0047
The zero error indicates that the positions of the reflections have 2-theta angles that are slightly lower (0.0135°)
than the theoretically expected values. The standard deviation of 0.0047 indicates the uncertainty limits for
the 2-theta error. As expected, the error is low and confirms the assertion given in the question that the data is from
a well aligned instrument.
It is difficult to include high angle data for this data set because:
i) there are a lot of peaks at high angle leading to a lot of peak overlaps. It is therefore
difficult to get a prescise position for these reflections.
ii) even for relatively well defined peaks, a close inspection of data from an hkl generating program shows that there can be several
crystallographic reflections contributing to the peak. It is thus difficult to know which hkl value to assign to the reflection (a requirement of
unit cell refinement programs).