molecular crystals
DESCRIPTION
Molecular Crystals. Molecular Crystals: Consist of repeating arrays of molecules and/or ions. C 17 H 24 NO 2 + Cl - . 3 H 2 O. Although Z = 2, the unit cell contains portions of a number of molecules. Cl -. Cl -. Cl -. H 2 O. Hydrogen bonds. Cl OH 2. Cl -. H 2 O. - PowerPoint PPT PresentationTRANSCRIPT
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-
Cl-
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
Unit cell volume = V
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
Note: text page 807 may not be correct.
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1 - cos2
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1 - cos2 sin x = 1 - cos2 x
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1 - cos2 sin x = 1 - cos2 x
V = abc sin
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
cos 90o = 0
V = abc
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
a = b
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
a = b
V = a2c
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = a3
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
V = a2c 1- cos2 - cos2 - cos2 + 2 cos cos cos
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:
Cell volume and cell contents:
A unit cell will usually contain an
integral number of formula units.
Cell volume and cell contents:
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
Space group General Positions Z
P1 x, y, z 1
Space group General Positions Z
P1 x, y, z 1
P1 x, y, z 2 -x, -y, -z
Unit Cell
Unit cell volume = V
V = abc 1- cos2 - cos2 - cos2 + 2 cos cos cos
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.
These sets of planes can act as a
diffraction grating for waves of the
proper wavelength.
d
These sets of planes can act as a
diffraction grating for waves of the
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?
1 Å
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
hot filament – e- source
metal target
hot filament – e- source
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
hot filament – e- source
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’
If (B B’)-(E E’) =an integral # ofwavelengths, 100%reinforcement.
2dsin = n
B’
E E’
If (B B’)-(E E’) =an integral # ofwavelengths, 100%reinforcement.
2dsin = nBragg’s Law
B’
E E’
If (B B’)-(E E’) =an integral # ofwavelengths, 100%reinforcement.
2dsin = nBragg’s Law
= wavelengthn = integer (order of diffraction)
2dsin = n
Bragg’s Law= wavelengthn = integer (order of diffraction)
2dsin = n
Bragg’s Law= wavelengthn = integer (order of diffraction)
If d becomes larger, must
decrease.
2dsin = n
Bragg’s Law= wavelengthn = integer (order of diffraction)
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
is set by the X-ray target
can be measured by determining the angle between the direct anddiffracted beam.
2dsin = n
d*
n = 0 1 2 3 4
is set by the X-ray target
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.
Note that intensities of thediffraction spots vary.
Diffraction intensities tend to decease
as increases.
Note that intensities of thediffraction spots vary.
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.
Note that intensities of thediffraction spots vary.
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.