Solving crystal structures

Question 1

What does a diffractometer measure?

Answer 1

The intensity of the reciprocal lattice points diffracted by a single crystal
Each of these corresponds to a set of Miller planes in direct space
The corrected intensity (after subtraction of background radiation) is equal to the square of the experimental/observed structure factor for the associated Miller index

Question 2

Phase problem

Answer 2

Experimental data provides the magnitude but not the sign of the structure factor values
F(hkl) contains information about the positions (xj, yj, zj) of all atoms from all symmetry related objects within the unit cell, according to the space group
Therefore once the positions of all atoms in the unit cell are known, the structure factors can be calculated for every Miller index

Question 3

Fourier series for electron density

Answer 3

Represents the electron density at any location in the 3D unit cell
Equation

Question 4

How do we find out where the asymmetric objects in the unit cell are located relative to each other?

Answer 4

From the space group equivalent positions, which are determined by the pattern(s) of absences in the data set

Question 5

What information do we need to calculate the structure factors?

Answer 5

Where the asymmetric objects in the unit cell are located relative to each other

Question 6

What are calculated structure factors (and their phases) used for?

Answer 6

To construct an electron density map for the asymmetric objet located at x, y, z
This is in the knowledge that the remaining electron density in the unit cell is located at the other equivalent positions for the space group
(which is why it’s so important to get the space group correct!)

Question 7

What must be overcome before the crystal structure can be solved?

Answer 7

The phase problem

Question 8

Strategies for overcoming the phase problem

Answer 8

Depend on the sample composition

1. Direct methods - for a purely organic molecule
2. Patterson methods - for a structure containing heavy atoms (e.g. organometallic). Direct methods can also be used

Question 9

Patterson methods

Answer 9

For heavy-atom structures only
Uses a Patterson map, which is also a Fourier series but based on (Fobs)^2 rather than F(hkl) (write out equation)
This Patterson series is generated only ONCE (i.e. as the initially step) in the process of solving a crystal structure

Question 10

Peaks in Patterson series

Answer 10

Have no physical meaning
Each peak corresponds to an interatomic vector between 2 atoms, with one atom placed at the origin
i.e. therefore a Patterson map is just a plot of all the interatomic vectors in the molecule being studied
Each peak is, however, proportional to the squares of the atomic numbers of the atoms giving rise to them
This means Patterson peaks involving a heavy atom can be easily identified

Question 11

Using a heavy atom to calculate structure factors

Answer 11

Once a heavy atom (e.g. Ni) has been identified in the Patterson map, the SHELX program can work out the coordinates of this atom, as well as the location of other Ni atoms in the unit cell, based on equivalent position information
The atomic coordinates for all heavy atoms in the unit cell can then be input into the structure factor equation and used to calculated the structure factors (and their phases) for each Miller index

i.e. at this point, the Fcalc(hkl) values are ONLY based on the heavy atom positions in the unit cell

Question 12

Why can we be confident that the phases of the majority of the Fcalc(hkl) values will be correct after this first iteration using the heavy atom?

Answer 12

Because heavy atoms are so efficient at scattering X-rays

Question 13

Least-squares iterations

Answer 13

The Fcalc(hkl) values calculated from the heavy atom information can be input into the Fourier series for electron density to produce an electron density map
A parallel calculation is also carried out using the phases of the Fcalc(hkl) values but the magnitudes of the Fobs(hkl) values
Can then ‘subtract’ the electron density map calculated from the Fcalc(hkl) values from that produced from the Fobs(hkl) values and the difference plotted
Other regions of electron density will be evident from this subtraction, which can be assigned to other atoms in the molecule
Using these new atom positions, and the ‘old’ heavy atom positions, all Fcalc(hkl) values can be recalculated
This iterative process continues until all electron density has been accounted for and all atoms located
Each cycle of this process involves a ‘least-squares fit’ of the proposed atomic positions to the data that refines these positions

Question 14

Magnitude of all Fcalc(hkl) values after all atomic positions have been identified

Answer 14

Will be very close to the corresponding experimental Fobs(hkl) values
Therefore the refinement factor (“R factor”) will be very low (2-6%)

Question 15

R factor equation

Answer 15

See flashcard

Question 16

Direct methods strategy

Answer 16

For when there is no heavy atom present in a molecule
Involves initially guessing the phases of some of the structure factors - those that have the strongest intensity
(As with the Patterson method, this ‘guessing’ strategy is only used once as the initial step when trying to solve a light atom structure)
The best guess at the phases of the strongest reflections is then input into the Fourier series for electron density to generate an electron density map, that will usually contain a large recognisable fragment of the object (which is usually a molecule)
The positions of the correctly assigned atoms in this fragment can then be inserted into the structure factor equation to calculated the signs of Fcalc(hkl)
Least-squares strategy from this point on is identical to that for a heavy atom structure

Question 17

Guessing in direct methods

Answer 17

Guessing is based on examining linked mathematical inequalities
e.g. if strong reflections are assigned positive phases, this will have mathematical implications on how strong some other reflections are
The validity of these implications can be tested by examining the magnitudes of the experimental structure factors Fobs(hkl)

Question 18

Atomic displacement parameters

Answer 18

ADPs
A measure of the probability of the atomic electron cloud sitting within the ellipsoid envelope
Bigger ellipsoids mean there is a lot of vibration (thermal motion) in the solid

Question 19

Experimental protocols

Answer 19

Experiments typically conducted using a low temperature device, where the sample is bathed in a beam of cold N2 (150K is a common temp)

Question 20

Reasons for low temperature data collection

Answer 20

Helps with:
Unstable compounds
Compounds with low melting points (maintains their crystallinity)
Stabilising thermal vibrations where there are long aliphatic chains
Crystal stability in cases where solvent evaporating from the lattice would cause the crystalline structure itself to breakdown

Question 21

Information gained from an X-ray crystallography experiment

Answer 21

Identification of all atom positions (xj, yj, zj) in the asymmetric object (which is usually a molecule)
Identification of all atom types (fj) in the asymmetric unit
3D spatial representation of the molecule (e.g. space-filling)
Insight into how the molecules pack together in the unit cell (from the space group equivalent positions)
Bond distances and angles