We are now ready to move to the monoclinic system. There are 13 monoclinic space groups. - PowerPoint PPT Presentation

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We are now ready to move to the monoclinic system. There are 13 monoclinic space groups.

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We are now ready to move to the monoclinic system. There are 13 monoclinic space groups. Monoclinic crystals belong to crystal classes 2, m or 2/m. – PowerPoint PPT presentation

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Title: We are now ready to move to the monoclinic system. There are 13 monoclinic space groups.


1
We are now ready to move to the monoclinic
system. There are 13 monoclinic space groups.
Monoclinic crystals belong to crystal classes 2,
m or 2/m. In this section well consider space
groups P2, C2, and P21. Well start with space
group P2, No. 3. In this case we have a
primitive monoclinic unit, with a twofold axis.
2
Primitive Monoclinic (Click to rotate) 1
C-Centered Monoclinic (Click to rotate)1
In the monoclinic system, a ? b ? c ? ? 90º,
and ? ? 90º. The b axis is perpendicular to a
and c, and we thus call it the unique axis. It is
only possible to place a 2 along the b axis, and
only possible to place a mirror plane (m)
perpendicular to the b axis. 1http//phycomp.tech
nion.ac.il/sshaharr/intro.html
3
Bonjour, Professeur! If I think about the
monoclinic system, I simply do not see why I
cannot place a twofold axis along a or c, or a
mirror plane perpendicular to a or c !!
A. Étudiant
Ah, my young friend, you must do more than
thatyou must make a sketch, and then you will
see that incorrect placement of these symmetry
elements will, in effect, be incompatible with
the translational symmetry of the unit cell!
Auguste Bravais
4
First, lets look at placement of a single
twofold axis along the b direction, at the
intersection of four unit cells in the ac plane





Clearly, the twofold axis reproduces an infinite
set of unit cells in the ac plane. (Keep in mind
that the b axis is perpendicular to the page, and
collinear with the twofold axis).
Finally, it should also be evident that there
will be additional twofold axes generated by
translation, and they are also compatible with
the infinite array of unit cells.
5
Now, lets try putting the twofold axis along a
it turns out that whatever we learn here will be
equally applicable to our locating the axis along
c.
If this array is rotated 180º about a, the result
is incompatible with the translation repetition
of the monoclinic lattice.
6
So, lets construct P2, space group number 3.
Well put in a 2 parallel to b, thus coming out
of the page, and again use that 2 and the
translation operators to derive the complete
mathematical group. We did this before with P1
and P1-bar.
What are the properties of a group? Youve
thought about this before when you learned about
point groups. First, there is an operation that
is a trivial one of making no change (the
identity). Second, closure applies, i.e., the
product of any two members of the set is already
a member of the set. Third, each operation has
an inverse which is a member of the set. Fourth,
the associative law holds A(BC) (AB)C.
7
Z2
8

Z2
Is this group enantiomorphous or
non-enantiomorphous?
Enantiomorphous
9

Z2
Is this group centrosymmetric or
non-centrosymmetric?
Non-centrosymmetric
10
Z2
Note that there are two molecules per unit cell,
and that other 2-fold rotation axes were
generated at a/2 and b/2 as we added atoms or
groups.
It turns out that there are four independent
rotation axes where are they?
(0, y, 0) (0, y, ½) (½, y, 0) (½, y, ½) Note
that (0, y, 0) represents a line rather than a
point!
11
These four independent twofold rotation axes
correspond to the four separate special positions
for P2, each with a multiplicity (Msp) of 1

Msp Coordinates
1 (0, y, 0) (0, y, ½) (½, y, 0) (½, y, ½)
12
Each of the four twofold axes corresponds to a
special position in P2. Recall that special
positions always correspond to a point group
symmetry element, i. e. a rotation axis,
reflection, inversion or rotary inversion axis.
A special position always has reduced
multiplicity compared to the general position.
13
When we discussed special positions in P1bar, we
noted that the multiplicity (or number) of
molecules placed in the general position (Z)
would be equal to two, while the multiplicity
(the number) of molecules placed on each special
position would be equal to 1. You have just seen
that, in P2, the multiplicity, Z, of the general
position is equal to two, and the special
position again has a multiplicity of 1. Special
positions always have a reduced multiplicity
which is, e. g., Z/2, Z/4, etc., compared to that
for the general position. Recall that each point
of a special position is mapped onto itself by an
operation of the group.
Fritz Laves
14
Must a centrosymmetric space group be
non-enantiomorphous?
Just so. We must have pairs of molecules,
related by an inversion operation.
15
Must a non-centrosymmetric space group be
enantiomorphous?
Nein. For example, if we have a mirror plane,
there will be pairs of enantiomers but no center
of symmetry.
16
Number 4 is space group P21. The 21 is the
symbol for a screw axis in this case it must
again be along b (why?). The 21 is a commutative
two-step operation (from the symbol MN, a
rotation of (360/M)º followed by a translation of
N/M in fractional coordinates parallel to the
axis), and in the diagram on the following page
well use the symbol . Just for practice,
lets look at how a 21 operation would appear if
we have an ab or ac projection.
Symbol 21 page is a half-arrow
b 1
b 0
0.5
This operation would take a point from (x,y,z) to
(-x, ½y,-z), a rotation of 360o/2 and advancing
½ of the unit translation.
17
Let's also look at the 21 screw axis in
projection, with the crystallographic b axis
vertical and in the plane of the screen. The
operation is a rotation by 180, followed by a
translation of ½ along the b direction. As you
will see, the operation cycles on the second
repetition, giving the original point translated
one full unit cell along b, i.e., to the point
(x, 1 y, z).
18
Rotation is right-handed
19
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20
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21
Now, lets construct the International Tables
Diagram for space group P21.
22
Z2
Is this group enantiomorphous or
non-enantiomorphous?
Enantiomorphous
23
Will B. Learning
Helen D. Megaw
24
Z2
Is this group centrosymmetric or
non-centrosymmetric?
Non-centrosymmetric
25
Recall that we can obtain the crystal class by
removing the translations from the symbol thus a
21 becomes a 2. To get the crystal class for
any space group, we remove the translational
symmetry, and ignore the lattice type. Thus,
space groups P2 and C2 belong to crystal class 2,
as does space group P21.
The next space group is C2, number 5. The
rotational symmetry is just as you saw in the
previous rendering of P2. But here the
difference is in the lattice type. The C
refers to C-centering, i.e., centering on the
crystal ab face recall that the C-centering
operation involves a (½, ½, 0) translation
applied to all primitive lattice points.
26
Like the P or 2 that youve seen before, the
C-centering operation is also a symmetry
operation. As we noted for space group P1, in
the primitive lattice, we can think of the P
operation as a set of translation operators
(1,0,0), (0,1,0) and (0,0,1). Compared to a
primitive lattice, in lattice type C, the
translation operator (½,½,0) is additionally
applied to all of the general equivalent
positions. To construct a diagram for C2, well
start by regenerating our previous P2 diagram,
and then add the molecules generated by the C
centering operation.
27
Z4

Is this group enantiomorphous or
non-enantiomorphous?
Enantiomorphous
28
Z4

Is this group centrosymmetric or
non-centrosymmetric?
Non-centrosymmetric
29
The four independent 2-fold rotation axes are the
four special positions for C2, each with Msp 2
and separate coordinates
Msp Element Coordinates
2 2 (0, y, 0) (0, y, ½)
30
For the special positions in C2, Msp 2 tells us
that there is another, symmetry-equivalent
position for each of the two listed (½,½ y,0)
with (0,y,0) and (½,½ y,½) with (0,y,½).
For space group C2, we have another examplea
very significant oneof the process of
constructing a closed set of operations by visual
group multiplication weve generated a number of
21 axes displaced by ¼ along the a axis, relative
to the 2s in the unit cell, e.g., (¼,y,0),
(¼,y,½) etc. This suggests an interesting
question will we be attempting a derivation of
C21, following the analogy of P2 and C2 ? Read
on
31
I have looked very carefully in the International
Tables, but I cannot find any mention of space
group C21. I believe that there is an error. I
will write to the Editor!
A. Student
There is no error. We could have derived C21 by
starting from the P21 diagram, with the addition
of C-centering. The completed diagram would show
2s at (¼,y,0) and so on. It is the same space
group as C2, with the origin shifted by (¼,0,0)
not a new group). Do you see?
Dorothy Hodgkin
32
A. Student
33
Bonjour, Professeur! I have determined the
crystal structure of a new chiral compound, and
find it to be the D-form. It crystallizes in
space group P21, with Z 2. But, life is very
strange when I dissolve my compound and measure
the optical rotation, it is zero!
A. Étudiant
Ah, young man, if your compound is stable to
racemization, you must return to the microscope
and look very carefully at your crystals. You
should find two different crystals both D- and
L-forms will be present. This is called
spontaneous resolution the forms can be
separated by hand each set will have opposite
rotations if you have the patience to collect
enough for an experiment!
Louis Pasteur
34
End of Section 2, Pointgroup 2
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