SPACE GROUP (cont).Lecture 3(cont) 092809

If space group is P21: monoclinic – non-centrosymmetric. 2 equivalent positions in the cell, related by 21 screw axis:

With Z = 2, then pt. group of an individual molecule is 1

With Z = 4, then pt. group of 2 molecules is 1

With Z = 8, then pt. group of 4 molecules is 1 – most probably twinned!!!

With Z = 1, wrong space group!!

If space group is P21/c: monoclinic – centrosymmetric. 4 equivalent positions/cell with a center of symmetry at ½, ½, ½ and also at all the other centers.

With Z = 4, then pt. group of an individual molecule is 1

With Z = 2, then pt. group of an individual molecule is -1, with ½ of a molecule

related to the other ½ through a center of symmetry (molecule itself

MUST have a center of symmetry = 1- ).

With Z =1, wrong space group!!

Example of structure inference from cell dimensions, space group info. and Z:

Anthracene: a ≠ b ≠ c, α = γ = 90o, β > 90o; therefore, monoclinic. Space group = P21/c. From the volume and the measured density (can guess the density to be somewhere between 0.9 and 1.3 gcm-3), we can calculate Z = 2. Therefore, the molecule must have a center of symmetry, -1 !!!!!

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Interference: physical basis of diffraction phenomenon is interference of electromagnetic

waves.

Interference patterns:

a)completely constructive:

b)completely destructive:

c)out-of-phase (phase difference):

If the phase difference is n, then get constructive interference.

Diffraction Experiment:

P1

S1

P2

Point

oP0 SCREEN

Source

P2’

S2

P1’

2-Slit diffraction experiment: Distances oS1 = oS2. Choose P0 such that S1P0 = S2P0. When this is the case, point P0 is very bright and exhibits constructive interference.

Also get bright spots at P1 and P1’ because S1P1 and S2P1 differ by an integral number of n

But, at points P2 and P2', S1P2 and S2P2 differ by /2; this is complete destructive interference and therefore the point will be black.

Another analogy: drop a rock in water. Drop a 2nd rock in at exactly the proper time, and get constructive interference; at the “wrong” time, get destructive interference.

Sir William Bragg and his son Laurence Bragg began work in England analyzing NaCl, KCl and ZnS in 1912, using X-rays. They initially took stationary photos like Max von Laue. But, they recognized that a single crystal of the material was acting like a 3-D grating and was diffracting the X-rays. They considered the single crystal to be made up of planes of atoms (consisting of the electrons which truly diffract the X-rays). So, they moved the crystal around into different orientations while exposing the film, and saw the diffraction patterns.

Compare 2nd order diffraction of the (010) with the 1st order diffraction of the (020) reflection:

When n = 2: 22d010 sin010When n = 1: 12d020 sin020

dividing this by 2yields: 1d010 sin010

But, 2d020 = d010

Therefored010 sin020

This is the same diffraction maximum!Therefore, the 2nd order diffraction of (010) is the same as the 1st order diffraction of (020).

If use Miller indices, n2dhkl sin hkl

If use general indices, 2dhkl sin hkl



Can we determine the distance, a, between the lines on a diffraction grating? YES.

o1st order

oxtan  = x/r

o0th order  = tan-1 (x/r)

o r

sin na

or a = nsin  nsin[tan-1 (x/r since

Introduce the reciprocal lattice (r.l.): Consider the normals to all the planes to radiate from a lattice point which is the origin of the unit cell. Let this normal be 1/dhkl in length and let it terminate in a point.

d100

d200

d300 000 100 200 300 400 500

d400

Planes in real space => points in reciprocal space

b*

↑

040 o o o o o o o

440

030 o o o o o o o

230

020 o o o o o o o

520

010 o o o o o o o

610

o o o o o o o

000 100 200 300 400 500 600 ------→ a*

This 2-D lattice is all when c* (l) = 0. Of course, there is a similar 2-D lattice just above this one with l = 1, then another with l = 2, etc. Also, there are other lattices with l = -1,

l = -2, l = -3, etc., below the plane Therefore, it is easy to see that there are planes (in real space) and data (in reciprocal space) with indices of 123, 3-42, -5-2-1, 612, 666, etc.

Consider an orthorhombic cell a ≠ b ≠ c,  =  =  = 90o. Because all of the angles are 90o, the (100), (010) and (001) planes are all perpendicular to the a, b and c axes, respectively. Therefore, their normals are along the axes: a & a* are collinear, b and b* and c & c* are also collinear.

And, a* = 1/a; b* = 1/b; c* = 1/c

In the monoclinic system, the b axis is unique since it has two 90o angles around it. Therefore, b* = 1/b. However, a* ≠ 1/a and c* ≠ 1/c, because of the  angle.

a* = 1/a sin and c* = 1/c sin  Also, 

See the sheets on FORMULAE for the most general case = triclinic, to calculate ALL the reciprocal cell dimensions and angles.

d*hkl = 1/d hkl = [h2a*2+ k2b*2+ l2c*2+2hka*b*cos*+2hla*c*cos*+2klb*c*cos*]1/2

Therefore, knowing the reciprocal cell dimensions for any system, one may then easily calculate the d*hkl for any reciprocal lattice point and therefore one may calculate the angle  for diffraction.

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