2D Plane Groups Escher

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2-D Plane Groups --- Escher
Homework Exercise 3.40 Huheey
What symmetry elements can you find?
Answer
4 different mirror planes
Mirror plane periodic translation, t,
generates mirror plane at t/2
?All four mirrors are generated by two
perpendicular mirrors (mm) two periodic
translations, a and b
Plane group pmm
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Depth by Escher 3-D arrayof fish
Each fish is found at the intersection of three
lines of fish, all of which cross each other at
right angles.
This gives unit cells, each of which contains one
molecule (fish). If the eyes of the fish are
ignored, each fish, and ? each unit cell, has C2v
(mm2) symmetry. The space group would then be
Pmm2
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Homework Exercise Huheey 3.41
Among the thirteen possible monoclinic space
groups are P21, P21/m, and P21/c. Compare these
space groups by listing the symmetry elements for
each.
These are all primitive space groups with a two
fold screw axis (21). In P21/m there is a mirror
plane perpendicular to 21. In P21/c there is a
glide plane perpendicular to that axis with the
glide translation parallel to the crystal c axis.
As for the point group twofold axis, a 21 with a
mirror plane or a glide plane perpendicular to
it creates a center of inversion.
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(a) Label the unit cell axes and designate the
unit cell edge lengths and angles between the
edges in the diagrams for each of the unit cells.

crystal 1
C2/c
C centering
b
a
c
? 116.8
90
b
a
90
a 30.22Å
b 21.10Å
c 27.15Å
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crystal 2
P21/n
b
a
c
? 112.7
90
b
a
90
b 19.15Å
c 20.47Å
a 17.53Å
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Aba2 (A centered)
crystal 3
all angles 90
c
b
a
c
b
a 21.24Å
b 29.48Å
c 19.42Å
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(b) Given the calculated density (rcalc) and the
number of formula weights/unit cell (Z),
determine the formula weights of the molecules.
Does this agree with the molecular formulas given
for the molecules that make up these two crystals
if the solvate molecule is included?
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RECALL
Z number of formula units (molecules) per unit
cell
Generally, Z is the number of equivalent
positions for the space group, but may be some
simple fraction (special positions) or simple
multiple of that number
Z can be derived if the density of crystal is
determined
M atomic mass of molecule in amu V volume of
unit cell in Å3
where M is the molecular or formula weight
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crystal 1
M molecular weight
?calc 1.658 g/cm3 1.660 ? 4 ? M / 15424
M 3851 C124H192F32Mn12O56 , which agrees
with table
No, this does not agree with the molecular
formula given in the abstract, for which M
3763.
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crystal 2
M molecular weight
?calc 1.376 g/cm3 1.660 ? 2 ? M / 6336
M 2625 agrees with formulas given in the
abstract and table
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crystal 3
M molecular weight
?calc 1.399 g/cm3 1.660 ? 2 ? M / 12157
M 2561 does not agree with abstract formula
Conclusions 1) The crystal contains solvent
molecules that were not detected in the X-ray
diffraction experiment 2)The chemical analyses
need to be redone by the authors.
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c) Explain why Z 2 for crystal 2 (space group
P21/n).
The molecule must have a center of inversion that
coincides with a crystallographic center of
inversion. The centers of inversion of the two
molecules must be located at Wyckoff a, b, c, or
d sites in the crystal. This is simply a choice
of origin, so we will chose Wyckoff a, (0,0,0)
and (1/2,1/2,1/2).