1

In this section we’ll consider space groups Pm, Cm, Pc and Cc. All four belong to crystal class m.

Glide planes (more details to follow) involve a reflection to the plane followed by a specified translation. In the monoclinic system with b as the unique axis, the possible glide planes are "a", "c" and "n". The "a" and "c" glide planes have translational components of ½ along the a and c directions, respectively, while an "n" glide plane has a translation along the ac diagonal, viz., (½, 0, ½).

2

Now we will build Pm, space group number 6. We’ll put in an m to b, thus in the plane of the page, and again use that m and the translation operators to derive the complete mathematical group. We’ll use as the symbol for an m in the page.

And if this array is reflected through the mirror shown, the result is incompatible with the translation repetition of the monoclinic lattice.

Again, we can examine the effect of “mis-placing” the mirror; if we place it in the ab or bc planes:

a

bFor a mirror plane

page, we'll use a solid red or black line

3

Since the mirror plane is in the plane of the screen (viz., the ac plane), we need a device or trick to remind ourselves that the reflected molecule:

(a) is directly behind the molecule at (x,y,z)

(b) is of opposite chirality to the molecule at (x,y,z)

We’ll do this by taking our original:

+,-

+

After applying the mirror operation, the new molecule appears behind the old one, and we indicate this using a vertical bar (splitting the original group), a comma, and a “-” to show chirality and position:

4

Is this space group enantiomorphous or non-enantiomorphous?

Non-enantiomorphous

),,( zyxZ=2;

6.NoPm

+,-

+,-+,-

),,( zyx

+,-HCE

HeightChiralityElements

a

c

5

),,( zyx),,( zyxZ=2;

6.NoPm

+,-

+,-+,-

+,-

Is this space group centrosymmetric or non-centrosymmetric?

Non-centrosymmetric

a

c

6

6.NoPm

The two independent mirror planes at y = 0 and ½ are the two special positions for Pm with Msp = 1 and coordinates

Msp Element Coordinates

+,-

+,-+,-

+,-

1 m (x, 0, z) (x, ½, z)Mirror planes gen'd at intervals of b/2

Note that (x, 0, z);… represents a plane rather than a point!

a

c

7

Hopefully you have discerned that a mirror plane is generated at y = ½. This is hard to see at this stage: we need a lesson to make ourselves aware of it!

Let’s do a quick derivation of Pm again, but this time we’ll use an ab projection:

+ ),,( zyx+ ,

),1,( zyx

+ ,),,( zyx

b

a

And, now we see the mirror at y = ½. Often it is very useful to redraw a space group in a more user-friendly orientation!

8

How could we have found the mirror plane at y = ½ only from the Pm diagram in the ac plane?

One unit cell above the molecules in the top left corner of the ac diagram lies another pair at (x, 1+y, z) and (x, 1-y, z). The location of the mirror planes is easy to deduce by averaging the coordinates of the corresponding pairs of molecules. Thus, (x, y, z) and (x, -y, z) are related by a mirror plane at [ y + (-y) ] / 2 = 0, and (x, y, z) and (x, 1-y, z) are related by a mirror plane at [ y+1-y ] / 2 = ½ !!!

+ ),,( zyx+ ,

),1,( zyx

+ ,),,( zyx

b

a

+,-

+,-

+,-

+,-

a

c

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Next we’ll build space group Pc, number 7. The c-glide plane is placed in the ac plane, b, just as we did with the mirror plane in the case of space group Pm.

A c-glide plane, like the 21 screw axis, is a two-step operation. In the first step, the atom or molecule is reflected b, and then is translated in the c-direction by c/2. (This operation, also like the 21, is commutative…the order in which we do the two steps makes no difference to the final result).

Unlike screw axes, which never change the chirality of the molecule under consideration, glide planes always produce a molecule of opposite chirality.

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Let's look at the glide operation in a perspective view. The overall result is shown above. The unfilled atom is first reflected through a mirror in the ac plane, followed by a translation of c/2. Click ahead to see the stepwise process.

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First, the atom is reflected through the ac plane.

12

Then, the molecule is translated by c/2.

13

Finally, note that the reflected molecule is of opposite chirality.

We can now attempt our derivation of space group Pc.

14

+

),,( zyx

-,

)21z,y,x( Z=2;

+

+ +

-, )21z,y,x1(

7.NoPc

Is this space group enantiomorphous or non-enantiomorphous?

Non-enantiomorphous

HCE

c-glide in-plane symbol

a

c

15

+

),,( zyx

-,

)21z,y,x( Z=2;

+

+ +

-, )21z,y,x1(

a

c

7.NoPc

Is this space group centrosymmetric or non-centrosymmetric?

Non-centrosymmetric

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It should be clear, once again, that space group symmetry elements which combine translations with rotation or reflection will not be special positions in the unit cell. A point corresponding to a location on a glide plane will not be mapped onto itself by any further operation of the group (cf. slide 16 in the first Chapter!).

Thus, in Pc, if we place an atom or molecule on the glide plane at (x, 0, z), another is generated at (x, 0, ½+z). No symmetry restrictions are placed on the molecule, and it does not have a reduced multiplicity. The characteristics of the point (x, 0, z) are no different from those of the point (x, y, z).

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Recall again that we can obtain the crystal class by removing the translations from the symbol; thus a c glide plane becomes an m.

To get the crystal class for any space group, we remove the translational symmetry, and ignore the lattice type. Thus, space groups Pm and Cm belong to crystal class m, as do space groups Pc and Cc.

As we move forward, recall that the C-centering operation involves a (½, ½, 0)+ translation applied to all primitive lattice points.

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To construct a diagram for Cm, we’ll start by regenerating our previous Pm diagram, and then add the molecules generated by the C centering operation. At the end of the process, what extra symmetry will be generated?…look for it just after the last molecule is added.

Sue N. Smart

Sir, I’m not certain what it will be, but I think it will be some sort of glide plane…..because C-centering and a mirror plane must generate another mirror, with translational symmetry!

Erwin Félix Lewy-Bertaut

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a

c

Z=4;

8.NoCm

Is this space group enantiomorphous or non-enantiomorphous?Non-enantiomorphous

)0,21,21(),21,21( zyx

½+

),21,21( zyx

½-,

½-,

½+

+,-

+,-+,-

),,( zyx )0,0,0(

+

),,( zyx

,-

HCE

¼

An a-glide!

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This is very nice! There is indeed an a glide plane generated. Hmmm…if I think about this intuitively, I could say that the C centering interacts with the mirror plane at (x, 0, z) to generate another mirror at (½+x, ½, z)…but an a glide plane at y = ¼ would also do exactly that!

Sue N. Smart

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a

c

Z=4;

¼ 8.NoCm

)0,21,21(),21,21( zyx ),21,21( zyx

),,( zyx )0,0,0(),,( zyx

½+½-,

½-,

½+

+,-

+,-+,-

+,-

Is this space group centrosymmetric or non-centrosymmetric?

Non-centrosymmetric

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a

c

Z=4;

¼ 8.NoCm

)0,21,21(),21,21( zyx ),21,21( zyx

),,( zyx )0,0,0(),,( zyx

½+½-,

½-,

½+

+,-

+,-+,-

+,-

2 m (x, 0, z)

Msp Element Coordinate (x, 0, z) and (½+x, ½, z)

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Aha, this will generate an n glide plane at y = ¼. I’ll have to watch for that!

Hmm…an n glide plane b moves a point from (x y,z) to (½+x, -y, ½+z), but if the glide plane

is at y = ¼…..

Now I’m ready to do Cc. I’ll start with Pc and a c glide plane at y = 0. From before, I know that this generates a second equivalent position at (x, -y, ½+z). The C centering generates another point at (½+x, ½-y, ½+z).

A. Genius

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-,

)21,,( zyx Z=4;

-,

+

++ ½ +

9.NoCc

Non-centrosymmetricNon-enantiomorphous

Special Positions: none

a

c

+

),,( zyx )0,0,0(

-½,

)21,21,21( zyx

½ +

),21,21( zyx )0,21,21(

HCE

¼

An n-glide!

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End of Section 3, Pointgroup m