molecular crystals

Molecular Crystals

Molecular Crystals:

Consist of repeating arrays

of molecules and/or ions.

C17H24NO2+ Cl- . 3 H2O

Although Z = 2, the unit cell containsportions of a number of molecules.

Cl-

H2O

Cl-

H2O

Hydrogen bondsCl OH2

Hydrogen bond

Model with atoms having VDW radii.

C17H24NO2+ Cl- . 3 H2O

Although this material is ionic, the + and - chargesare not close enough tocontribute to the formationof the crystal.

Molecular crystals tend to be

held together by forces weaker than

chemical bonds.

van der Waal’s forces are always

a factor.

Hydrogen bonding is often present.

A layer in an ionic solid with ionsof similar radii.

Metallic crystal – single layer of like

sized atoms forms hexagonal array.

Second layer can start at a point designated

b or c.

At this point, the third layer can repeat the

first and start at a or it can start at c.

Third layer repeats

first layer.

Unit Cell

Unit cell volume = V

Unit Cell


V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos

Note: text page 807 may not be correct.


cos 90o = 0


cos 90o = 0

V = abc 1 - cos2


cos 90o = 0

V = abc 1 - cos2 sin x = 1 - cos2 x


cos 90o = 0

V = abc 1 - cos2 sin x = 1 - cos2 x

V = abc sin


cos 90o = 0


cos 90o = 0

V = abc


a = b


a = b

V = a2c


V = a3

V = a2c 1- cos2 - cos2 - cos2 + 2 cos cos cos

V = a2c 1- cos2

V = a2c 1- cos2

V = a2c sin

V = a2c 1- cos2

V = a2c sin = a2c sin 120o

Cell volume and cell contents:


A unit cell will usually contain an

integral number of formula units.


A unit cell will usually contain an

integral number of formula units.

The number of formula units in the

cell is often related to the symmetry

of the cell.

The number of formula units in the

unit cell is designated by Z.

Space group General Positions Z


P1 x, y, z 1


P1 x, y, z 1

P1 x, y, z 2 -x, -y, -z

Unit Cell



Note: text page 807 may not be correct.

Triclinic

P1

Z = 2

If Z = 2 then the total mass in the

unit cell is the formula weight x 2.

If Z = 2 then the total mass in the

unit cell is the formula weight x 2.

If the volume is V then the density

of the crystal is formula wt. X 2

V x No

Triclinic cell:

a = 6.8613 Å = 74.746o

b = 9.1535 Å = 81.573o

c = 16.8637 Å = 73.339o

V = 974.45 Å3

C17H24NO2+ Cl- . 3 H2O FW = 363.87 g/mol

V = 974.45 Å3

C17H24NO2+ Cl- . 3 H2O FW = 363.87 g/mol

Z = 2

Density =363.87 g (2)

974.45 Å3 x 6.02 x 1023

V = 974.45 Å3

C17H24NO2+ Cl- . 3 H2O FW = 363.87 g/mol

Z = 2

Density =363.87 g (2)

974.45 Å3 x 6.02 x 1023

1Å = 1 x 10-8 cm

V = 974.45 Å3

C17H24NO2+ Cl- . 3 H2O FW = 363.87 g/mol

Z = 2

1Å = 1 x 10-8 cm

Density =363.87 g (2)

974.45 x 10-24 x 6.02 x 1023

Density =363.87 g (2)

974.45 Å3 x 6.02 x 1023

V = 974.45 Å3

C17H24NO2+ Cl- . 3 H2O FW = 363.87 g/mol

Z = 2

Density =727.74 g

5866.19 x 10-1

Density =363.87 g (2)

974.45 x 10-24 x 6.02 x 1023

V = 974.45 Å3

C17H24NO2+ Cl- . 3 H2O FW = 363.87 g/mol

Z = 2

Density =727.74 g

5866.19 x 10-1cm3

Density =363.87 g (2)

974.45 x 10-24 x 6.02 x 1023

= 1.241 g/cm3

Infinitely repeating lattices

Three possible unit cells; one lattice.

Crystal lattices include a large number

of repeating sets of planes.

These sets of planes can act as a

diffraction grating for waves of the

proper wavelength.



proper wavelength.

d



proper wavelength.

d = < 1 to 250 Å

When radiation on the order of

1 ångstöm wavelength interacts

with a crystal lattice having

interplanar spacings on the order

of ångstöms, diffraction occurs.

Where do we find 1 ångstöm

Wavelength radiation?

What is the source of 1 Å

radiation?

1 Å

Emission spectrum for hydrogen in visible range

Electron transitionsfor H atom.

Electron transitionsfor H atom.

Transitions in visibleregion.

It is possible to cause certain metals

to emit X-rays by temporarily removing

a core electron.

X-ray emission

e-

+-HV

e-

+- HV

If the potential difference is large enough,core electrons will be ejected from the metal.

source ofelectrons

Metal target

hot filament – e- source


metal target


metal target

+ -

Accelerating potential

1 x 10 mm

1 x 10 mm

KV = 50

mA = 40

1 x 10 mm

KV = 50

mA = 40

= 2000 watts


metal target

+ -

Accelerating potential

X-ray scattering is due to the interaction ofX-rays and the electron density around atoms.

d = < 1 to 250 Å

d = < 1 to 250 Å

B’

E E’

d = < 1 to 250 Å

B’

E E’

If (B B’)-(E E’) =an integral # ofwavelengths, 100%reinforcement.

B’

E E’


2dsin = n

B’

E E’


2dsin = nBragg’s Law

B’

E E’


2dsin = nBragg’s Law

= wavelengthn = integer (order of diffraction)

2dsin = n

Bragg’s Law= wavelengthn = integer (order of diffraction)

2dsin = n


If d becomes larger, must

decrease.

2dsin = n


If d becomes larger, must decrease.

There is a reciprocal relationship betweenThe crystal lattice and the diffraction pattern.

2dsin = n

d*

n = 0 1 2 3 4

2dsin = n

d*

n = 0 1 2 3 4

is set by the X-ray target

2dsin = n

d*

n = 0 1 2 3 4


can be measured by determining the angle between the direct anddiffracted beam.

2dsin = n

d*

n = 0 1 2 3 4


can be measured by determining the angle between the direct anddiffracted beam.

Unit cell can be determinedfrom this data.

Note that intensities of thediffraction spots vary.


Diffraction intensities tend to decease

as increases.


The derivation of Bragg’s Law is correctbut the conditions are more complicated.Each diffraction spot is the sum of the wavesfrom all atoms in the unit cell.


The derivation of Bragg’s Law is correctbut the conditions are more complicated.Each diffraction spot is the sum of the wavesfrom all atoms in the unit cell. This includesa significant amount of destructive interference.

dd

Each diffraction maximum includes

information on the electron density

in the repeat distance.

Conversion of X-ray intensities to

electron densities is a very complicated

process.

Conversion of X-ray intensities to

electron densities is a very complicated

process.

A major step is determining the atomic

coordinates for a model.

Once the coordinates for a model are

determined, it is possible to calculate

what the intensity data for that model

would look like.

Once the coordinates for a model are

determined, it is possible to calculate

what the intensity data for that model

would look like. The observed intensities

and the model intensities are compared.

A least-squares refinement of model

intensities against observed intensities

allows the model structure to become

the actual crystal structure.

molecular crystals

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cos cos cos cos

cos2 cos2

cos cos cos v

cos cos cos note

cos2 v

cos2 sin x

abc sin v

a2c sin v