1
Electron Crystallography X

EL2006 Antwerp
Structure Determination of Li
2
NaTa
7
O
19
and
K
7
Nb
15
W
13
O
80
from HREM images
Sven Hovmöller, Structural Chemistry, Stockholm University, S

106 91 Stockholm, Sweden
______________________________svenh@struc.s
u.se____________________________
The aim of this lab is to learn how to determine structures from HREM images, to understand
the symmetry and the meaning of amplitudes and phases. Two different compounds, Li
2
NaTa
7
O
19
and K
7
Nb
15
W
13
O
80
, will be used.
The la
b consists of two parts:
1.
Solve the structure from an HREM image of Li
2
NaTa
7
O
19
taken at Scherzer defocus by
crystallographic image processing (CRISP)
2.
Correct for the effects of CTF and solve the structure of K
7
Nb
15
W
13
O
80
.
_________________________________
____________________________________________
Part 1. Solving structures from HREM images by crystallographic
image processing (CRISP
)

Li
2
NaTa
7
O
19
Crystallographic image processing (CIP) is a technique for
solving unknown crystal structures
from HREM da
ta only
. We will use CRISP for this purpose.
1. Load an image and select an area to process
a.
Start
CRISP
.
Load the image
Tant6900.PCX
from C:
\
Program files
\
Calidris
\
CRISP2
\
Sample images
\
by clicking on
.
This is an HREM image of the
metal oxide L
i
2
NaTa
7
O
19
taken
at 200kV with a JEOL 200CX
microscope, along the short c

axis.
Open the
Information
box,
available under
Tools
(or press
F7). You can store some data
about the microscope and how
the image was taken and
digitised. Chose the
microscope
JEOL
200CX.
Change the Scale
to
0.645
Å/pix. Click on
.
As you see, the crystal is very thin near the edge, but
quickly gets thicker.
2
b.
You can cut out the region you want to use by clicking
on Tools
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2. Fourier transformation
You can calculate the Fourier transform of a
square area, from 128x128 to 2048x2048
pixels, using the icons
.
Select the 512 area and then click
You will
see the FFT window appe
ar.
The horizontal and vertical lines are caused
by the FT of the square box. If we calculate
the FFT from a circular area, this cross
disappears. Right

click inside the 512x512
area on the EM image and select Circle.
You can move the selected area around
, by
holding the left mouse

button down and
moving the mouse. In this way you can
compare the quality of the crystal in different
areas of the EM image very fast.
3.
Lattice Refinement
c.
Lattice refinement is used to detect periodic diffraction spots in
the Fourier transform and to
determine accurately the lattice parameters thus defined. Click the icon
to open up the
tool for
Lattice Refinement.
Click inside the FFT window and see how the radius of the
ring can be changed. Only data inside the ring wil
l be used for further processing. Move the
ring in the FFT window until only the strong diffraction spots are included (about radius
150).
3
d.
Check that
Auto
matic Lattice Detection is active (indicated by the cross in
). Click
. A blue and a red arrow
will appear in the FT. Check that CRISP found the
correct lattice (the lattice crosses the diffraction spots).
Manual lattice refinement
If you are not satisfied with the result of the
Auto
matic Lattice Detection (for example if
the reciprocal lattice do
es not follow the symmetry), you may index the lattice manually.
Uncheck the
Auto
box
.
Click number 1 under
Refl
in the
Detect
Lattice
dialogue box, click on a reflection
in the FFT. When a reflection is selected, a
cross will appear on this reflection in
the
Fourier transform and a
will appear
under Refl. Specify the indices of this
reflection. An index can be changed by
pointing the cursor at the index number and
right

clicking the mouse button, then
selecting a positive or negative index and
left

clic
king on it. Repeat for a second
reflection. Finally click
.
Click
Swap AB
to make sure that a* axis (red) is the longer reciprocal axis (this is according
to the unit cell setting which was published,
a
= 15.23 Å,
b =
23.57 Å and
c
= 3.84 Å).
c.
Note how
far the diffraction peaks extend (how many diffraction orders) in the two
directions. The image is taken along the short c

axis. Estimate the resolution of this image
from these data.
Question 6:
Number of diffraction spots along h:………=> Resolution alon
g h:……………
Number of diffraction spots along k:………=> Resolution along k:……………
How does your result compare with the 2.5 Å Scherzer resolution of this
microscope?
4. Symmetry Determination and Origin Refinement
Start the tool for orig
in refinement and symmetry determination by clicking the
icon. You
will see a list of the 17 possible 2

dimensional plane group symmetries starting with
p1
and
ending with
p6m
(4 plane groups
pm, pg, cm
and
pmg
have two different settings, with
m
(mirror
) or
g
(glide plane) perpendicular to the x and y axes, respectively). There is also a map
for
p1
. This represents the result of
Lattice averaging
.
Lattice averaging means eliminating the non

periodic (random) noise by averaging over
all the unit cells tha
t we selected (in our case here around 20 by 5 = 100 unit cells). This often
has a great effect for protein crystals but practically no effect for inorganic crystals. How is it in
this case? In general
HREM images of inorganic compounds usually contain ver
y little
random noise.
This is in contrast to images of proteins and other biological specimens, which
are very noisy.
The main problems in HREM on inorganic specimens are systematic errors in
the images
caused by electron optical distortions, multiple sca
ttering and crystal tilt
.
With
Crystallographic Image Processing, these systematic distortions can be corrected to a smaller or
larger extent, as you will see in the following pages.
4
You can create the
p1
image by clicking
.
Can you see the structure imag
e consisting
of 5

fold stars of black TaO
6

octahedra,
? Probably not at this stage.
Determine the symmetry
by clicking
. CRISP will now test each of the possible 21 2D
symmetries, and give
Figures of merit
for each one, in the form of R

values
on
amplitudes
(
RA%
) and Phase residuals
on phases (
Res).
Note that the resulting map is changing as
different symmetries are imposed.
5
This part may be the most specialised crystallographic part of CIP. Note t
he crystallographic
structure factor
phase info
rmation is present in EM images
.
In fact the phases are not only
present, they are usually of much higher quality than the amplitudes.
Question 6:
Compare the phase residuals for the 3 orthorhombic symmetries:
pmm (or p222)
……….
pmg (or p222
1
)……….
pgg
(or
p22
1
2
1
)……….
What is the symmetry of this projection?………………
The best (= lowest)
phase residual is
obtained for
pgg
. This is
also the correct
(projected) symmetry of
this crystal. Select
pgg
and
.
Notice
the significant
improvement of the
pgg
map, compa
red with the
p1
map; now all metal

oxide octahedra are
resolved, and you can
easily see the 5

fold
stars of black TaO
6
octahedra.
6
It is possible to change the scale of the density
map by right

clicking on the density map and
selecting
Tune
.
A w
indow called
Tune Density
Map
will be opened. You can scale the map by
changing the number in
Cells.
The unit cells can
also be translated and rotated. Click
to
see what happens. You can toggle the unit
Cell
edge
on/off. By double

clicking in the map it
self
you can toggle its frame on/off. Click
and move the map on top of the original image to
make an inset of the processed structure.
5. Inspect the amplitudes and phases numerically
a. Click
and a window called
HK Edit
containing
h, k, Amp, Amp
S, Pha, PhaS Err
will appear. It contains amplitudes and phases both before (Amp, Pha) and after imposing
the
s
ymmetry (Amp
S
, Pha
S
) for all reflections. Symmetry

related reflections are grouped
together.
Question 7:
What are the amplitude and phase relat
ions for
pgg
?
Amp(h k) and Amp(h

k):
Pha (h k) and Pha (h

k):
Question 8:
How well does the original data fit the symmetry, i.e. how much did CRISP have to
change the input phase values Pha to make them conform to the rules of
pgg
,
namely that all phases m
ust be 0 or 180 degrees and such that symmetry

related
pairs of reflections (
hk
) and (
h
–
k
) have allowed phase relations?
b. b. You can investigate the effect of
changing the phase of a single strong
reflection, for example (1 2) by
clicking first on that
reflection, then
Inverse
. Look at the map in the
Origin Refinement window

it is
changing as you proceed!
Click
again to get back to the original
phase
.
Try reversing some other
strong reflections. You will notice
that all the ten strongest reflections
h
ave to have correct phases if the map
shall look good.
It has been said that
a phase is worth at least twice as
much as an amplitude.
c. Select in turn each of the following 7 plane group symmetries:
p1, p2, pm, pg, pmm, pmg
and pgg
, and go to
to st
udy their phase and amplitude relationships.
Question 9:
a)
In which of these seven plane groups are (
h,k
) and (
h,

k
) symmetry

related?…………………
7
b)
In the orthorhombic plane groups (
pmm
,
pmg
and
pgg
) the amplitudes of pairs of reflections
(
h,k
) and (
h,

k
) sh
ould be identical. What is the main reason they are not identical in
experimental HREM images?………………..
c)
What about the phases of symmetry related reflections? Should they also be identical?
Yes, for symmetry/ies…………………. No for symmetry/ies……………………….
d)
W
hat is the rule for phase relations in the plane group
pgg?
Phase of (h k) = Phase of (h

k) +
………………
.
6. Determine the Atomic Co

ordinates
a. Activate the
pgg
Density Map
. Select the tool
.
A window called
Atomic Meter
will be
opened and a cross w
ill appear on the density map. Put the cross at the center of an atom
(dark density), and its fractional atomic co

ordinates are given in the table. CRISP may also
refine
the atomic position if you check the box
.
b. Select
to add a new atom and m
ove the cross to another peak. Find its co

ordinates.
Try to find all the unique peaks inside the unit cell, marked in the map in the same way as
above.
The
coordinates obtained by x

ray powder diffraction can be found in the paper:
Synthesis and
Crysta
l Structure of Li
2
NaTa
7
O
19
by Jekabs Grins et al.,
J. Mater. Chem.
4
(1994) 445

447. The
coordinates obtained from HREM differ on average by 0.2 Å compared to those obtained from
x

ray powder diffraction.
Question 10:
What do you expect that the peaks
(black) correspond to chemically? ……………
8
Part 2. Correcting for the effects of CTF and solving the structure
of K
7
Nb
15
W
13
O
80
Open the image KNB.PCX,
select a 256x256 area on the
thin part of the crystal and
process it
to get a density map.
This is an HREM
image taken
at 200kV. The resolution of the
image is 2.5 Å.
Question 11:
Which is the
correct symmetry for this
crystal? …………
=
=
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Information
box from
Tools

Information
.
Change
Def. spread
to 100, and
Update
.
N
ow process the KNBCTF image and get the
density map.
You should get density maps similar to these below if you have applied the correct symmetry:
KNB
KNBCTF
9
This crystal is a niobium oxide. Niobium oxides very often contain
atomic arrangements wi
th local 5

fold symmetry. Each black blob
represents a NbO
6
octahedron. The resolution is not sufficient to resolve
the oxygen atoms.
Question 12:
Which of the two images gives us directly the structure?…
=
=
=
=
=
Question 13:
Look at the FFT of the two i
mages. Which image was taken near Scherzer
resolution and which one not?………………………………….
Correct for the CTF
Activate the FFT of
KNB

CTF. Right

click inside the FFT
and select
Tune.
Make the FFT
brighter so that you
can see the Thon
rings.
Click on
. A
Filter window opens
up. Check the box
to open up
the CTF correction
dialogue. Check the
box
.
10
A red ellipse appears
on the FFT. Adjust the
shape and size of this
ellipse to coincide
with the black Thon
ring.
Then adjust the
second red elli
pse to
fit with the second
Thon ring, (which is
only visible at some
places).
Click
in the Filter window.
CRISP now corrects for the CTF. This is
done by multiplying all points between the
first and second red ring by
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–
1. This is equivalent to
adding a phase shift of 180
o
to all those points. As you can see in the CTF function, the points
between the first and second cross

over have reversed co
ntrast: black becomes white and white
becomes black. After correcting for the oscillating behaviour of the CTF, the contrast is correct
everywhere.
The only remaining problem is that some reflections come very close to the CTF cross

overs, and thus are mu
ltiplied by almost zero. If there are not too many diffraction spots very
close to the CTF cross

overs, we may still be able to reconstruct the correct structure. An
alternative is to take another HRTEM image, with a different defocus, and then add in data
from
that second image.
Activate the Inverse FFT, click
Edit

Copy
data
and then
Edit

Paste
. An inverse FFT that
we can work on has been created.
Notice how different this CTF

corrected
image is, compared to the original image. In the
original image KNB

CTF it was hard to see if
the atoms should be black or white, but in the
corrected image, it clearly looks as if the black
dots correspond to atoms.
Process this image as usual, until you get a
density map. Are you surprised that the FFT
looks just the sa
me as before, even though it has
been corrected?
Question 14: W
hy is it similar?………….
=
=
11
The density maps in
p1
and the correct symmetry will hopefully look something like this:
p1
pmg
If you didn’t get such a nice result, check the symmetry and/or
go back to the step where you
adjust the ellipses for the CTF

correction. You will notice that some diffraction points are very
close to the red ellipses. It is very critical to get the CTF correction very close to correct, or else
some of the reflections
close to the cross

over points will get wrong phases.
from KNBCTF.PCX
from KNB.PCX
If you compare the maps (a) and (b), you will notice that the map from the image taken at non

Scherzer defocus and suffering astigmatism (a) is in fact even a lit
tle better than that from the
image taken at the Scherzer defocus (b)! This means that although the image KNBCTF had so
severe distortions that it was impossible to see anything in the original image, the information
was not at all scrambled in some uncont
rolled random way. Rather there was only one problem;
the focus conditions. But we could compensate for those and restore a beautiful image of the
structure!
Notice that in the CTF

corrected image there seems to be some weak atoms in the
S

shape
d tunnels between the pentagons (see arrow). It is known from X

ray crystallography that
there are potassium atoms just there!
a
b
12
Determination of atomic co

ordinates
As a final step you may determine the
atomic co

ordinates of the atoms.
Open the
Atomic meter by clicking on the
icon. Click on atoms and then on
Add
.
You get fractional atomic co

ordinates, with an
accuracy of 0.1 to 0.2 Ångström. Obviously the
electron microscopes are not the limiting factors
in structure determination by E
M, but the
radiation sensitive samples are limiting.
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