Inorganic materials chemistry and functional materialsfolk.uio.no/ravi/cutn/scm/symmetry_and... ·...

Inorganic materials chemistry

and functional materials

Helmer Fjellvåg and Anja Olafsen Sjåstad

Lectures at CUTN spring 2016

Chemical bonding

CRYSTALLOGRAPHY - SYMMETRY

Symmetry

NATURE IS BEAUTIFUL

• ”The chief forms of beauty in nature are order, symmetry and definiteness.” Aristotle, 350 BC

• ”The scientist does not study nature because it is useful. He studies it because he takes pleasure in it, and he takes pleasure in it because it is beautiful.” Henry Poincaré, 1908

Symmetry

Symmetry

MOLECULAR SYMMETRY

• Molecules can be classified in terms of their symmetry

• What types of symmetry can we find in a molecule?

• Rotational axes, mirror planes, inversion centers

Symmetry

MOLECULAR SYMMETRY

Symmetry

ROTATION

• Described with symmetry element Cn, where the angle of rotation is 360°/n.

• If more than one rotational axis is present, we identify the principal axis as the one with the highest symmetry; the highest n.

• Ex: A square planar molecule with four equal substituents has a principal C4-axis.

Symmetry

ROTATION

Symmetry

MIRROR PLANE

• Mirror planes are denoted with a small sigma; σ.

• We can identify three types of mirror planes:

• Planes perpendicular to the principal axis: σh.

• Planes containing the principal axis: σv.

• Planes containing the principal axis, but bisecting the angle between two adjacent 2-fold axes: σd.

Symmetry

MIRROR PLANE

Symmetry

INVERSION CENTRE

• If all points in a molecule can be reflected through the centre and this produces an indistinguishable configuration, it contains a centre of inversion.

• Denoted i.

Symmetry

IMPROPER ROTATION

• If a rotation about an n-fold axis followed by a reflection through a plane perpendicular to this axis produces an indistinguishable configuration, we call this an improper rotation, denoted Sn.

Symmetry

IDENTITY

• For mathematical reasons, a last operator is needed, an operator that leaves the molecule unchanged.

• All molecules have the identity operation.

• We denote this operator: E.

Symmetry

SUCCESSIVE OPERATORS

• With all operators defined, we can permute a molecule with successive symmetry operations.

• Successive operations can be added together, and can correspond to other operations.

• Ex. Rotating 180° two times around a C2-axis will bring the molecule back to the original state. C2 x C2 = C2

2 = E.

Symmetry

GROUP THEORY

• A mathematical group is a set of elements that are related through given rules.

• The product of any two elements in the group must be an element of the group.

• There must be an identity element; an element that commute with all other elements and leave them unchanged.

• The associative law of multiplication must hold.

• All elements must have a reciprocal element which is also an element of the group. The product of an element and its reciprocal element must always be the identity element.

Symmetry

MATHEMATICAL GROUPS

• An example of a mathematical group is the set of all integers ℤ, with standard addition as the operator. This group is denoted (ℤ,+).

• Adding an integer to another integer will always give another integer.

• Adding 0 to any element will give that element, so 0 is the identity element.

• a+(b+c) = (a+b)+c for all integers a,b,c.

• The reciprocal element for all elements is it’s negative counterpart, as a + (-a) = 0.

Symmetry

MATHEMATICAL GROUPS

• As an example of something similar that is not a group is the set of all integers ℤ, with standard multiplication as the operator, (ℤ,·).

• Multiplying an integer with another integer will give another integer.

• Multiplying any element with 1 will give that element, so 1 is the identity element.

• a(bc) = (ab)c for all integers a,b,c.

• There does not exist any integers that can be multiplied with another integer to produce 1.

Symmetry

GROUP THEORY IN CHEMISTRY • Why do you need this?

• The set of symmetry operations that can be applied to a molecule form a mathematical group!

• An indefinite number of such point groups exist.

• Out of these only 32 groups are relevant for crystalline solids

• These 32 are termed: Crystallographic point groups

Symmetry

MOLECULES versus SOLIDS - I Inorganic molecules: rather small in size Interactions between molecules ar weak: - Dispersion forces (induced dipols; dipol – dipol;

hydrogen bonds) Molecular solids Crystalline solids with strong bonding in 1, 2 or 3 dimensions (chains, layers, network) - Strong ionic, covalent, or metallic bonding

Symmetry

MOLECULES versus SOLIDS - II Crystalline solids with strong bonding in 1, 2 or 3 dimensions (chains, layers, network) - Strong ionic, covalent, or metallic bonding Periodic structures; unit cells repeating in 3D - Metric of unit cell - Mathematical lattice for the structure - Symmetry operations compatible with a lattice - Crystal system, Bravais lattice, crystallographic point

group, space group Local symmetry for a given atom in the structure

The Unit Cell

smallest repeating unit that shows full symmetry of the crystal structure.

By stacking the «boxes» (unit cells) in three dimensions, a complete

description of the obtained structure is obtained.

3D stacking of

unit cells

crystallite

Unit cell

- Has a metric

- Contains atoms

- Shows various symmetries

Symmetry Symmetry

Unit Cell

Definition of relations between

angles (, , ) and axes (a, b, c) in

unit cell.

: angle between a and b

: angle between b and c

: angle between a and c

A unit cell in 3D is defined by three pairs of parallel planes

Symmetry Symmetry

Units

for unit cell dimensions and atomic/ionic radii

Ångström unit is frequently used:

The Ångström unit is in the range of the size of atoms and

molecules, and microscopic biological structures. It

is useful for discussing chemical bonding and arrangement

of atoms in crystals.

Å is not an “SI-unit”

1 Å = 0.1 nm = 100 pm = 10-10 m

Symmetry Symmetry

What unit cell is correct for the 2D lattice in (a)?

1) Smallest unit that shows full lattice symmetry

2) Choice of origin – optional (inversion center and/or atoms on cell corners, edges)

Symmetry Symmetry

7 crystal classes/systems and their 7 primitive unit cells

Symmetry Symmetry

RHOMBOHEDRAL R

a = b ≠ c

= = 90o

= 120o

a ≠ b ≠ c

= = 90o

≠ 90o

Seven crystal systems

Hexagonal Cubic

Tetragonal

Trigonal Orthorombic

Monoclinic

Triclinic

Hierarchy of crystal systems

Hierarchy implies:

Higher crystal system contains at least

one symmetry element that the lower

system does not contain.

Symmetry Symmetry

7 crystal systems and their 7 primitive Bravais lattices

What is a space lattice/lattice?

Mathematically repeating pattern of points.

REQUIREMENT:

All lattice points have identical surroundings

This is exactly the cases for our 7 lattices

These contain 1 mathematical point (8 x 1/8)

per unit cell

Are termed primitive Bravais lattices P

Consider now JUST mathematical points.

If one such point is placed at origin for the seven

different types of unit cells shown for seven

crystal systems, we then have unit cells for seven

Mathematical lattices; Bravais lattices

Symmetry Symmetry

a ≠ b ≠ c

= = 90o

≠ 90o

A point inside one unit cell 1 point per unit cell

A face point is shared between 2 cells 1/2 point per unit cell

An edge-point is shared between 4 cells 1/4 point per unit cell

A corner-pont is shared between 8 cells 1/8 point per unit cell

Counting of points of the unit cell

Symmetry Symmetry

Lattices with more than one point in the unit cell

P = Primitive

1 lattice point

I = Body centered

2 lattice points

F = Face centered

4 lattice points

C = Side centered

2 lattice points

Centering is on the

side normal to the

c-axis

Other options:

A, B centering

Question: are there more ways to arrange mathematical points in space and

obtain a repeating lattice with identical surroundings at each lattice point?

Symmetry Symmetry

A unit cell in 3D is defined by three pairs of parallel planes

7 crystal systems

4 lattice types P,F,I,C

14 Bravais Lattices

Ma

All lattice points have identical surroundings.

Symmetry Symmetry

a ≠ b ≠ c

= = 90o

≠ 90o

Exercise 7 (p. 14) “Draw a tetragonal primitive (P) lattice. Place an extra lattice point in ½, ½, 0; i.e. make a

C centring of the lattice. Show that a tetragonal C lattice does not exist as such,

i.e. , that it is equivalent to another tetragonal P lattice.”

Tetragonal - P

a = b c

= = = 90o

Symmetry Symmetry

A = (0.5,0.5,0.5)

B = (0.5,1.5,0.5) ( A = (0.5, 1.5-1, 0.5))

C = (0.5,2.5,0.5) ( 0.5, 2.5-2, 0.5)

Positioning of atoms

– atomic coordinates

Can add and subtract whole numbers at will.

-Values of coordinates (x, y, z) are always between 0 and 1 with

respect to the unit cell; i.e. we use fractional coordinates.

- The unit cell is repeating; the structure is periodic any integer

multipla of the unit vectors can be added/substracted to get identical

points in other unit cells

(0,0,0)

a

c

b

The unit cell vectors (axes) are basis for these translations

a b

c

(1,0,0)

Symmetry Symmetry

-The unit cell defines the mathematical room (Bravais lattice)

- The room then has to be filled with its basis/motif

(basis = non-equivalent atoms = atoms with different surroundings)

- «Crystal structure = lattice + basis/motif»

From mathematical

Bravais lattice to

crystal structure

2D unit cell

Mathematical lattice

Filling the unit cell

with “atoms” (motif)

Symmetry Symmetry

Example 7 (p. 14)

Strategy:

1) See that the spheres in corners are different from

the sphere in centre (i.e. disconnected)

2) Add several unit cells, to find the repeating unit

that applies for both Cs+ and Cl.

3) Recognize that the repeating unit for both Cs+

and Cl are identical; i.e. same Bravais lattice

CsCl structure:

Bravais lattice + basis

cubic-primitive + basis [Cs (0,0,0) and Cl (½, ½, ½)]

Identify lattice and basis for the CsCl type structure

Symmetry Symmetry

Differences between CsCl- and bcc structure

Cs+

Cl

CsCl type structure

= cubic primitive lattice (P) + basis

Basis = Cs (0,0,0) and Cl (½, ½, ½)

bcc (W) =

cubic body centred lattice (I) + basis

Basis = W (0,0,0)

Tungsten (W) is I-centered (bcc) since W atoms in

(0,0,0) and (½, ½, ½) have identical surroundings.

Symmetry Symmetry

Unit cell; Graphite

Symmetry Symmetry

Two-dimesnional sheets; graphene sheets; are stacked ABAB... along

the c-axis. Strong bonding within sheets; van der Waals forces between sheets

Graphene sheets

Graphene sheets

Graphene sheets

Hexagonal, P

ALL atoms listed

Symmetry operations

Point groups

Space groups

Assembly of symmetry operations

Lattice +Translation

Two systems for nomenclature (areas of use) – we learn both now:

1. Schönflies (spectroscopy)

2. Hermann-Mauguin (crystallography)

Symmetry Symmetry

Point groups

POINT GROUPS

• There are indefinite number of point groups

• There are 32 crystallographic point groups

• We will first use Schönflies notation to describe some molecules

• We will use Hermann Maugin notation to put point group symmetry into the frame of periodic structures

• You should be able to assign the correct point group to ANY molecular object

Point groups

Point groups

POINT GROUP DECISION TREE

• Makes point group assignment relatively straightforward

• Follow the tree along the correct branches, and you’ll find your point group

• Practice at some everyday examples

Point groups

AMERICAN FOOTBALL

Point groups

PYRAMID

Point groups

BOOMERANG

Point groups

POCl3

Point groups

S8

Point groups

BF3

Point groups

Allene

Point groups

THE CHARACTER TABLE

• Every point group has a corresponding character table

• This is a set of irreducible representations that span the group

• Let us see what this means

Point groups

THE WATER MOLECULE

• Symmetry operations?

• C2 rotation axis

• Two vertical mirror planes

• Identity operations

• Has C2v symmetry

Point groups

THE WATER MOLECULE

• The hydrogen atoms are labeled for clarity and a coordinate system is added.

Point groups

THE WATER MOLECULE

• As the result of symmetry operations, all atoms are shifted to new coordinates

• In such a 3D-system, mapping a point to a new position can be described fully by a 3x3 matrix

Point groups

THE WATER MOLECULE

• The character of a matrix is the sum of the numbers on the diagonal from upper left to lower right

Point groups

THE WATER MOLECULE

• The set of character forms a representation, in this case a reducible representation

• This can be further reduced, first by block diagonalization

Point groups

THE WATER MOLECULE

• By doing this, we diagonalize the x, y and z coordinates and they become independent of each other:

Point groups

THE WATER MOLECULE

• We are closing in on the character table, which is the complete set of irreducible operations for the group

• Mathematics sets some important constraints on how to form such a complete set

• 1. The number of symmetry operations in the group is the order of the group.

Point groups

THE WATER MOLECULE

• 2. Symmetry operations are arranged in class, all operations in one class have identical transformation matrix characters

• 3. The number of irreducible representations equals the number of classes

• 4. The sum of the squares of the characters under the identity operation equals the order of the group

Point groups

THE WATER MOLECULE • 5. For any irreducible representation, the sum of

the squares of the characters multiplied with the number of operations in the class equals the order of the group

• 6. Irreducible operations are orthogonal to each other

• 7. All groups include a totally symmetric representation, with characters of 1 for all operations

Point groups

THE WATER MOLECULE

• By using the three existing representations, we can now make the required fourth representation and write up the complete character table for the point group C2v

Point groups

CHARACTER TABLES

• All point groups have their own character table

• Why do we need this?

• Symmetry is an important property of a system that can be used to deduce physical properties and response to certain characterization techniques. The character table summarizes this symmetry in a simple way.

Point groups

CHARACTER TABLES – and spectroscopy

• IR: Change in dipole moment during vibration means IR-activity. How is this described in the character table?

• Raman: Change in polarizability tensor during vibration means Raman-acivity. How is this described in the character table?

Symmetry operations

4-fold rotation axis

= 360o/n = 360o/4 = 90o

1. Rotation axis, n-fold (rotation by 360o/n)

Crystals (periodic lattices ) only display rotational symmetries of 2, 3, 4 and 6

Molecules can in addition have n = 5, 7, ………

Point groups; rotation

2. Identity, i (rotation axis n = 1; 360o/1)

(x, y, z) (x, y, z) (i.e. the object is transferred into itself)

(Required operation according to group theory; and

for us the identity operation has no other practical use.)

Point groups; identity

3. Mirror plane, m

1

1

‘ 1 1

‘

A mirror plane transfers a right handed object to a left handed object

horizontal

vertical

Use the symbol (,) to

indicate the mirror image

+

+ +

Point groups; mirror planes

4. Inversion, 1

(x, y, z) (x, y, z)

1 +

1 -

‘

Inversion symmetry transfers a right handed object to a left handed object

The object with inversion symmetry is said to be centrosymmetric

Point groups; inversion

5. Roto-inversion axis, n

Rotation-inversion axis

Rotation 360o/n + inversion

Methane

CH4

4

Is a combined symmetry operation: rotation 360o/n followed directly by an inversion

Point groups; rotation-inversion

Point symmetry elements according to Hermann-Mauguin notation

Point groups - notation

Exercise 10 (p. 16)

Which symmetry operations can you identify for the following

objects:

H2O NH3 CO2 CH4 dxy orbital

Hint:

Visit “Symmetry@Otterbein; http://symmetry.otterbein.edu/tutorial/

EXAMPLE - EXERCISE

32 crystallographic point groups

(classified according to their crystal system)

(Hermann-Mauguin notation)

Note:

Usually the point

group symbol starts with the

highest or characteristic

symmetry (n)

(see p. 18:

not fully consistent)

Point groups - symbols

Exercise 13 (p. 19)

Which symmetry elements can you

identify in the following point group

symbols: –1, 3, 6/m, –4/m ?

–1 Inversion (centrosymmetry) Triclinic

3 1 three-fold rotation axis Trigonal

6/m 1 six-fold rotation axis +

mirror plane normal to 6-fold rot.axis

Hexagonal

–4/m 1 fourfold roto-inversion axis +

mirror plane normal to –4 rot-inv. axis

Tetragonal


Rules for the first, second and third symbol (point group)

Symbol nb. 1 2 3

j k l Fictive point group:

Symbol 1: Symmetry direction 1

(Characteristic symmetry

element of the crystal system)



1 2 3

(for mirror planes: normal vector)


•Surround the object/crystal by a sphere

•Project the object down on a xy-equatorial plane

•The projected point is determined by the intersection of a connection line from the

point of interest to the pole of the opposite side.

•The two halves of the spheres are distinguished by assigning + and – , and/or by

the use filled and open symbols

Pole

Pole

xy-plane

+

Construction of stereographic projections - I

• A thin circle limits the stereograms in the paper plane, representing the intersection of

the xy-plane with the sphere

• Symmetry elements are drawn using relevant graphical symbols (see Table)

• The highest rotation axis is always chosen to be perpendicular to the projection plane

• Vertical mirror planes - drawn as thick lines

• Horizontal mirror plane drawn as a thick line around the circle periphery (in xy-plane)

Construction of stereographic projections - II

Point groups – stereographic projections

Exercise: Draw the point groups mmm and 4mm in stereographic projections.

b

a

Exercise: Are there more symmetry elements in the projection drawn for mmm?

Yes, there are three 2-fold rotation axes

All three are normal to one of the mirror

planes

The full point group symbol is therefore

2/m 2/m 2/m

EXAMPLE - EXERCISE

General and special positions

- A general position is not situated on any symmetry element.

- A special position is situated on one or more symmetry elements.

If a point is situated on a symmetry element, the latter operations

will transfer the point just onto itself.

General and special positions

(x, y, z) generates 4 equivalent points (x, y, 0) generates 2 equivalent points

(0, 0, z) generates 2 equivalent points (0, 0, 0) generates 1 equivalent point

For convenience we select symmetry direction 1 to be II to c

General and special positions for 2/m

General position Special position

Special position Special position

Glide planes and screw axes Going from point symmetry to symmetry for solids, two new

symmetry elements appear

Screw axis (nm) = rotation (n) + translation (m/n)

Glide plane (a, b, c, n, d) = mirroring + ½ translation along an axis

(or more axes)

Symmetry planes Symbol Translation

Axial glide a a/2

b b/2

c c/2

Face diagonal glide n (a+b)/2, (b+c)/2, (a+c)/2

(a+b+c)/2 for cubic and tetragonal only

Diamond glide d (a±b)/4, (b±c)/4, (a±c)/4

(Body diagonal) (a±b±c)/4 for cubic and tetragonal only

Symmetry – operations with translation

0

Screw axis 21

a

Two-fold rotation (360o/2 = 180o)

+

Translation (1/2) parallel to rotation axis

Symmetry – screw axis

Comparison

with 2-fold

rotation

a

0

Glide plane

Example:

a-glide

Mirroring + ½ glide along a

Symmetry – glide planes

Comparison

with just

mirroring

230 space-groups

Space group symbol: Xefg

Lattice:

P (R)

F, I

A, B, C

Characteristic symmetry elements

(dependent on crystal system)

Symmetry operations without translation:

Inversion -1,1

Rotation n

Mirror m

Rotation-inversion n

Symmetry operations with translation:

Screw-axis nm, 21,63, etc.

Glideplane a,b,c,n,d

Symmorphic space groups (73 groups)

Non-symmorphic space groups (157 groups)

Used to describe periodic objects – solids

Combination of 32 point groups with the

14 Bravais lattices, glide plans and screw axes

Space groups

To identify the corresponding point group of a space group one must:

1) Remove symbol for the Bravais lattice (but note whether it is primitive or centered)

2) Exchange symbols for symmetry elements with translation with corresponding

symbols for symmetry elements without translation

nm n for a screw axis (go to rotation axis)

a,b,c,d or n m for glide plane (go to a mirror plane)

Examples:

P63/mmc 63/mmc 6/mmm

Pnma nma mmm

Information from space group on point group

Why? E.g. be able to find number of general positions for a structure

Space groups

How to read International tables of crystallography

See hand-outs, or page 31 in Compendium

Space groups

Exercise 1 (p. 5)

a

b

Cu(1)

O(2)

O(1)

Draw a 2-dimensional orthogonal unit cell with cell edges

a = 3.80 Å and b = 3.90 Å.

Place the following atoms in the unit cell:

Cu(1) (0,0); O(1) (0,0.5); O(2) (0.5,0)

Calculate Cu-Cu distances and Cu-O distances.

How do the calculated Cu-O distances fit your expectations of ionic/covalent bonding?

Cu-Cu:

a2 + b2 = (Cu-Cu)2

Cu-Cu = (3.802 + 3.902) = 4.72 Å

From Shannon:

Cu(II) CN = 4 0.71 Å

Cu(II) CN = 6 0.87 Å

O(II) (CN = 6) 1.26 Å

Cu-O1 = 1/2b = 1.95 Å

Cu-O2 = 1/2a = 1.90 Å

Cu-O = 1.97 Å

Cu-O = 2.13 Å

EXAMPLE - EXERCISE

EXAMPLE - EXERCISE

How to draw a structure: CuO

a = 465 pm, b = 341 pm, c = 511 pm, = 99.5°

Space group C2/c

Cu in 4c

O in 4e y = 0.416

Crystal system: a b c, = = 90°, 90° ( hence monoclinic)

Bravais-lattice: C monoclinic, side centered (in ab plane) – two lattice points

Corresponding crystallographic point group: 2/m (4 points)

Maximum general points for the space group is therefore 2 x 4 = 8

Number of formula units (CuO) per unit cell = 4 (4c and 4e positions are occupied)

Given the information:

Question: What can you conclude directly from the given information?

EXAMPLE - EXERCISE

EXAMPLE - EXERCISE

From

International

Tables of

Crystallography

O

Cu

EXAMPLE - EXERCISE

C2/c

O

Cu

a = 465 pm, b = 341 pm, c = 511 pm, = 99.5°

Space group C2/c

Cu in 4c

O in 4e y = 0.416

1. (0.00, 0.416, 0.25) 3. (0.00, 0.584, 0.75)

2. (0.50, 0.916, 0.25) 4. (0.50, 0.084, 0.75)

1. (0.25, 0.25, 0.00) 3. (0.75, 0.25, 0.50)

2. (0.75, 0.75, 0.00) 4. (0.25, 0.75, 0.50)

EXAMPLE - EXERCISE

O

Cu

1. (0.00, 0.416, 0.25) 3. (0.00, 0.584, 0.75)

2. (0.50, 0.916, 0.25) 4. (0.50, 0.084, 0.75)

1. (0.25, 0.25, 0.00) 3. (0.75, 0.25, 0.50)

2. (0.75, 0.75, 0.00) 4. (0.25, 0.75, 0.50)

a = 465 pm

= 99.9 o

a = 465 pm, b = 341 pm, c = 511 pm, = 99.5°

is the angle between a and c axis

b-axis normal to ac plane

Draw -angle in the paper plane

Project along short axis

EXAMPLE - EXERCISE

O

Cu

1. (0.00, 0.416, 0.25) 3. (0.00, 0.584, 0.75)

2. (0.50, 0.916, 0.25) 4. (0.50, 0.084, 0.75)

1. (0.25, 0.25, 0.00) 3. (0.75, 0.25, 0.50)

2. (0.75, 0.75, 0.00) 4. (0.25, 0.75, 0.50)

a = 465 pm

= 99.5 o

a = 465 pm, b = 341 pm, c = 511 pm,

= 99.5°

0.416 0.416

0.584 0.584 0.084

0.916

0.25

0.25

0.75

075

0.25 0.75

EXAMPLE - EXERCISE

What is the coordination number (CN) of Cu and O

in this crystal structure?

Do we know some other compounds that could be similar to CuO?

NiO: NaCl-type structure; Ni in the center of octahedra

MnO

CoO

Could CuO take a similar structure?

Space group and crystal system do indicate some issues. Why?

Cu(II): 3d9 Jahn-Teller ion expect deformation of octahedron

EXAMPLE - EXERCISE

O Cu a = 465 pm

= 99.5 o a = 465 pm, b = 341 pm, c = 511 pm, = 99.5°

0.416 0.416

0.584 0.584

0.084

0.916

0.25

0.25

0.75

075

0.25 0.75

What is the CN of Cu in this crystal structure?

-0.084

1.084 -0.416

-0.584

What is expected Cu-O bond?

Ionic radius Cu(II)?

Ionic radius (O-II)?

What is around Cu within a

radius of approx. 200 pm?

EXAMPLE - EXERCISE

a

c -0.084

0.916 -0.584

0.416

-0.416

0.584 1.084

0.084

EXAMPLE - EXERCISE

a

c -0.084

0.916 -0.584

0.416

-0.416

0.584

1.084

0.084

EXAMPLE - EXERCISE

a

c -0.084

0.916

-0.584

0.416

-0.416

0.584

1.084

0.084 -0.084 0.416

0.084 0.584

Coordinates along b-axis given (normal to paper)

Change color of atoms with shortest

Cu – O bonds)

0.25

EXAMPLE - EXERCISE

-0.084

0.416

0.584 0.084

a

c

0.584 1.084

0.416 0.916 -0.084 -0.584

-0.416 0.084

0.25

EXAMPLE - EXERCISE

Cu(II) is 6-coordinated (4+2) in a deformed octahedron

Cu(II) 3d9 – Jahn-Teller ion

a

c

Remember: Work 3D when

determining coordination number

EXAMPLE - EXERCISE

O Cu

a = 465 pm

= 99.9 o

0.416 0.416

0.584 0.584

0.084

0.916

0.25

0.25

0.75

075

0.25 0.75

What is the CN of O in this crystal structure?

m

x

XCN

MCN

XM xm

=)(

)(

Ionic compound MmXx

Refresh

EXAMPLE - EXERCISE

Crystallographic planes and directions

• Crystallographic planes and directions are given relative

to the coordinate system used to define the unit cell

• We use Miller indexes to define parallel planes (h k l)

• We use [u v w] to give directions in a crystal


Crystal planes - Miller indices ( h k l )

Recipe:

1. Identify the plane that is adjacent

to an equivalent plane that

passes through origin

2. Find intersection of this plane

with unit cell edges (a, b, c)

3. The reciprocals of these values

are the Miller indices of the planes

(1/ 0)

c

a

b

h

(001)

a b c

1

h k l

1/ 1/ 1/1

0 0 1


Crystal planes - Miller indices ( h k l )


Crystal planes – Miller indices ( h k l ) – parallell planes


[112]

(1,1,2)

(½, ½,1) (1,0,1)

(2,0,2)

[101] and [202]

[½ 0 ½] = [101] = [202] = n[101]

Directions 1) Directions are given as [uvw]

2) Directions in crystals are

defined from origo:

[100] II to a-axis

[010] II to b-axis

[001] II to c-axis

3) Parallel directions have same index

a

b

c


Crystal planes and crystal directions – more on notation

A given plane (h k l)

A set of equivalent planes {h k l} Cubic system (1 0 0), (0 1 0), (0 0 1)

{1 0 0}

A given direction [uvw]

A set of equivalent directions <uvw>

The equivalent planes and directions are a

result of the symmetry of the system

e.g. fcc <111>

[111] [111] [111] [111]

[111] [111] [111] [111]

[111]

c

a

b

[-111]


Collection of 32 point groups

Supplementary information


Triclinic

Monoclinic

http://metafysica.nl/derivation_32.html


Orthorombic

Tetragonal


Trigonal http://metafysica.nl/derivation_32.html


Hexagonal http://metafysica.nl/derivation_32.html


Cubic


What is a solid solution?

Water Alcohol Mixing on the molecular scale

Cu Ni Cu-Ni alloy

Mixing on the atomic scale

in a crystalline solid

Solid solutions

What types of solid solutions exist?

Cu-Ni alloy

Mixing on the atomic scale

in a crystalline solid

Solid solutions

Substitutional solid solution

Interstitial solid solution

Aliovalent substitutions – defect generating solid solution

Metals: substitutional solid solution

Hume-Rothery substitutional solubility rules

1. The crystal structure of each element of the involved pair is the same

2. The atomic sizes of the involved atoms do not differ more than 15%

3. The elements do not differ greatly in electronegativity

4. Elements should have same valence

Solid solutions

Solid solutions and defects Intrinsic defects associated with stoichiometric and pure crystals

Extrinsic defects associated with substitutants or impurities (0.1 – 1 %)

What about substitutants > 1% ???

Solid solution


Aliovalent (heterovalent) substitution

Interstitial solid solution

Solid solutions

Requirements: for substitutional solid solution to form (for

ionic/polar covalent compounds)

1) The ions must be of same charge

2) The ions must be similar in size.

(15-20 % difference acceptable)

3) The crystal structures of the end members must be isostructural for

complete solid solubility

4) Partial solid solubility is possible for non-isostructural end members

Mg2SiO4 (Mg in octahedra) - Zn2SiO4 (Zn in tetrahedra)

5) The involved atoms must have preference for the same type of sites

Cr3+ only in octahedral sites, Al3+ both octahedral and tetrahedral sites

LiCrO2 - LiCr1-xAlxO2 - LiAlO2

Solid solutions

Al2O3 corundum Cr2O3 corundum (Al2-xCrx)O3 corundum

Statistic distribution of

Al(III) and Cr(III) atoms Al3+

covalent radius 1.18 Å Cr3+

covalent radius 1.18 Å


ionic compounds

Solid solutions

Hume-Rothery substitutional solubility rules

1. Crystal structure of each element in the pair is the same

2. Atomic sizes of the atoms do not differ more than 15%

3. The elements do not differ greatly in electronegativity

4. Elements should have same valence

Disordered Au0.5Cu0.5

bcc

Low temperature

Ordered Au0.5Cu0.5

primitive cubic

High temperature

Metals – substitutional solid solution

Effect of temperature

Solid solutions

Option -1

Layer -A

Layer-B

Layer-A

Hexagonal close-packed (hcp) (heksagonal tetteste kulepakking)

ABABAB…….

Layer-B

Where to put the third layer?

Degree of filling = 74 %

Each sphere have 12

nearest neighbors

CN = 12

Closest packing of spheres in 3D – hcp

Sphere packings - holes

Layer-A

Layer-B

Layer-C

Layer-A

Cubic close-packed (ccp/fcc); (kubisk tetteste kulepakking)

ABCABCABC…..

Option 2

Closest packing in 3D – ccp



Each sphere have 12

nearest neighbors

CN = 12


Voids/interstitial holes – interstital solid solutions

Example: ccp(fcc), hcp, bcc metals/alloys

Octahedral hole

A

B

rM/rX 0.41

Tetrahedral hole

Space for small

Interstitial atoms

rM/rX 0.22

Possible to fill

with in smaller

atom:

Solid solutions Solid solutions

For hcp and ccp:

125

Close packing and holes in hcp and ccp 74% of the volume is filled by the spheres 26% voids

Octahedral hole (1 per packing sphere)

Tetrahedral holes (2 per packing sphere)

T+ and T

hcp: Trigonal bipyramidal hole (5 spheres generating hole)

(can not be filled at the same time as the tetrahedral holes)


Metals – interstitial solid solutions

Small atoms can enter into the smaller holes in the metal structure

and create phases as:

Interstitial carbides – steel industry

Stoichiometric carbides Fe3C – cementite

Interstitial and stoichiometric hydrides

Hydrogen storage

For interstitial solid solutions, the Hume-Rothery rules are:

1. Solute atoms must be smaller than the interstitial sites in the solvent lattice.

2. The solute and solvent should have similar electronegativities.

Solid solutions

Aliovalent substitution – in ionic compounds AOx

Substitution by higher valence cations (B) into AOx

Cation vacancies Interstitial anions

Substitution by lower valence cations (C)

Anion vacancies Interstitial cations

Side note: Similar patterns can be made for anion substitution; but not included here

as anion substitution occurs to less extent in solid solutions

Defect formation; charge compensation

Too much positive charge

Too much negative charge

Solid solutions

Density

The unit cell: Contains Z number of formula units of the compound in question

Note: the cations and anions can take more than one Wückoff position each!

When checking/calculating/drawing: be aware of Wückoff site multiplicity

The unit cell volume and the mass of the atoms in the unit cell can be

calculated exact.

NOTE: you anticipate that the atomic arrangement is perfect

i.e. that there are no defects (interstitials; vacancies)

Aunitcell

rayXNvolumeUnitcell

cell/unitsofnumberweightFormula

V

m

=

=

The calculated density is sometimes called «X-ray density»

Comparison of

experimental (pycnometric)

calculated; X-ray density

x-ray > exp.

Density

Why???

Experimental density is burdened with errors:

- Poor wetting of the particles/material

- Inner porosity in the material not accessible by the liquid

Calculated density is not reflecting the fully true picture

- Major amounts of defects have not been accounted for

Applications of symmetry

REPETITION

• Let’s quickly take a look at the NH3-molecule

• What are the symmetry elements?

• What is the point group?

• How do we create the character table?


REPETITION

Block diagonalization, this time x and y are coupled:


REPETITION

Note the three classes of elements. There are two C3’s, both C3 and C32,

but the transformation matrix is the same so they belong to the same class.

There are three vertical mirror planes, but they are all invariant, so they

also belong to the same class.


REPETITION

• How do we find the last representation?

• Orthogonality and the fact that the character must be symmetric under E.


APPLICATIONS

• Let’s take it a step further and look at vibrational spectroscopy of the N2O4-molecule.

• Self test: What is the point group?


N2O4

• You might have learned that the number of vibrational modes of a molecule is 3N – 6 where N is the number of atoms.

• Is there a direct logic to this equation?

• -3 because of translation

• -3 because of rotation


N2O4

• 12 vibrational modes should be possible for this molecule

• Point group: D2h

• In fact, we can now use the character table to determine the symmetry of all 18 motions of this molecule and assign them to translation, rotation or vibration


N2O4

• Let’s first label all the atoms:

• Now we need to see what happens to the axes under the different symmetry operations


N2O4

• We use characters and say that:

• If an atom moves, the character is 0.

• If an atom is stationary and the axis direction is unchanged, the character is 1.

• If an atom is stationary and the axis directions is reversed, the character is -1.


N2O4

• Now sum all characters in each class to make a reducible representation.


N2O4

• Now comes the tricky part. We have to reduce this by using the character table and the following rule:


N2O4

• Well, the order is 8. So we should have:


N2O4

• Now we use the character table to remove translations and rotations:


N2O4

• By looking at the character table we can easily imagine how some of these vibrations might look. Take the the three Ag’s for example, they have to be symmetric:


N2O4

• Can we deduce if the vibrations are IR active?

• The dipole moment must change, and only representations with x, y or z symmetry do this.


N2O4

• Can we deduce if the vibrations are IR active?

• The dipole moment must change, and only representations with x, y or z symmetry does this.


N2O4

• Can we deduce if the vibrations are Raman active?

• The polarizability of the molecule must change, and only representations that transform like products of x, y and z do this.

A two-fold rotation axis parallel with the z-axis is described as 2[001]

Crystallography – space groups


2/m is identical to the inversion operation (-1):

A b-axial glide plane with (y,z) reflection (mirror) is described as:


-

Diagonal glide plane, perpendicular to the c-axis:


Space group Pmm2: symmetry operations: m[100]; m[010]; 2[001]

All these have ORIGIN as a common point (see figure right)

P orthorhombic Bravais lattice; crystallographic space group: mm2


How many points are generated (to be drawn in left figure)?

Answer: = (number of equivalent points of crystallographic point group) x

(identical points of the Bravail lattice)

mm2 gives 4 identical points (cf. stereographic projection)

Bravais lattice is primitive

Hence: nb of points = 4 x 1 = 4


Triclinic crystal system space group P-1;

primitive Bravais lattice

crystallographic point group -1


Additional symmetry exists; in addition to the inversion center in origin:


Monoclinic crystal system: a ≠ b ≠ c

AND one angle deviating from 90o, Which one?

STANDARD setting: C-centering (if it does exist; and not use A or B centering)

Consequence: b-axis is the so-called unique axis; i.e. perpendiacular

to a- and c-axes; and ≠ 90o.


According to standard notation, ab-projection is drawn for the space group

Then practical to draw the deviating angle in the plane of projection.

Then c-axis is unique axis; and any centering (if existing) is chosen as B

Symmetry (from the space group symbol)

B Bravais lattice; 2 equi.points

Crystallographic space group: 2

Expects: 2 x 2 = 4 pointsto the drawn


Stepwise approach:

Sketching: points sketching symmetry operations

1

2

Add 2[001] O+ O+

O+ O+

3

Add B lattice O+ O+

O+ O+

a

b

O+½ O+½


3

O+ O+

O+ O+

O+½ O+½

4

O+ O+

O+ O+

O+½ O+½

Operate on the new point with the 2-fold rotation axis:

O+½ O+½

The unit cell contains now all 4 expected points

Now continue with analysis of symmetry elements being present


4

O+ O+

O+ O+

O+½ O+½ O+½ O+½

We have now the following points:

(1) x,y,z; (2) –x,-y,z; (3) x+½; y,z; (4) ½-x,-y,z

How are the points (1) and (4) connected?

We can add coordinates for more atoms in the sketch; e.g. (5) at –x,1-y,z and ask:

How is (1) and (5) connected by symmetry?

And so on....

In this way we identify all symmetry for this space gorup, and

become able to draw the complete figure of symmetry operations


Example:

How are the points z,y,z, and –x+1/2, -y, z+1/2 related?

We then write this in terms of the Seitz operator, and split the matrices:

We realize quickly that there is a two-fold rotation axis:

From the first comumn matrix we realize that this is actually a 21[001] axis.

What about the second column matrix? This tells you where this symmetry

operation is located in the unit cell!!

Lacoated at half the coordinate values: i.e. at 0.5 x (1/2 0 0) = (1/4, 0 0)


Two settings of the non-centrosymmetric

space group Bb or Cc

Respectively, with

c-unique and b-unique axis


We will now consider space group Pnma in detail and compare our

findings/procedure with what tabulated in the International Tables for

Crystallography; which is the following figures:


Pnma What can we read out of the space group symbol?

Primitive Bravais lattice; 1 lattice point (0,0,0)

Corresponding crystallographic point group: mmm

A point group in the orthorhombic crystal system

8 general points for mmm

Full symbol 2/m 2/m 2/m

Altogether 8 point symmetry operations

I, -1, m[100], m[010], m[001], 2[100], 2[010], 2[001],

Expected 8 x 1 = 8 points in the unti cell for Pnma

Pnma:

Essential operators are: n[100] and m[010] and a[001]

This is our starting point now.....


In mathematical for, these tre essential symmetry operators are:

They have one point in common: the origin of the unit cell


Points Symmetry

m[010]

n[100] a[001]

O+ Ø+

Ø+1/2

Ø-

Coordinates for the transformed points are:

(1)x,y,z

(2) -x, y+½, z+½ n[100]

(3)x, -y, z m[010]

(4)y+½, y, -z a[001]

We have now located four out of the eight target positions


The next steps are now to adopt combinations of the three symmetry operators

(5) By operating m[010] on point (2) n m

(6) By operating a[010] on point (2) n a

(7) By operating a[010] on point (5) n m a

(8) By operating a[010] on point (3) m a

Points Symmetry

m[010]

n[100] a[001]

O+ Ø+

Ø+1/2

Ø-

(5): m[010] (-x,y+½,z+½) = (-x, ½-y, ½+z)

(6): a[001](-x,y+½,z+½) = (½-x, ½+y, ½-z)

(1)

(4)

(3) (2) O+1/2

O-1/2

(5)

(6)


The next steps are now to adopt combinations of the three symmetry operators

(5) By operating m[010] on point (2) n m

(6) By operating a[010] on point (2) n a

(7) By operating a[010] on point (5) n m a

(8) By operating a[010] on point (3) m a

(7): a[010] (-x,y-½,z+½) = (½-x, ½-y, ½-z)

(8): a[001](x,y,z) = (½+x, -y, -z)

Points Symmetry

m[010]

n[100] a[001]

O+ Ø+

Ø+1/2

Ø-

(1)

(4)

(3) (2) O+1/2

O-1/2

(5)

(6) Ø-1/2

O-

(7) (8)


Points Symmetry m[010]

n[100] a[001]

O+ Ø+

Ø+1/2

Ø-

(1)

(4)

(3) (2) O+1/2

O-1/2

(5)

(6) Ø-1/2

O-

(7) (8)

The next step is now to identify all other symmetry operations and draw them

into the upper right figure. We do this by again asking what type of symmetry

connects various points (beyound what considered already)

How is (1) and (5) connected?





(1) x,y,z

(2) -x, y+½, z+½

(3) x, -y, z

(4) y+½, y, -z

(5) (-x, ½-y, ½+z)

(6) (½-x, ½+y, ½-z

(7) (½-x, ½-y, ½-z)

(8) (½+x, -y, -z)





1 – 5: we see that there is a 2[001] transforming x, y into –x, -y: There is a

translational component of ½ along c; hence a 21[001] axis. Which is located at

(0,1/4,0)

1 – 6: we see that there is a 2[010] transforming x, z into –x, -z: There is a

translational component of ½ along b; hence a 21[010] axis. Which is located at

(1/4,0,1/4)

1 – 7: we see that there is inversion center transforming x,y,z into –x,-y,-z:

Which is located at (1/4,1/4,1/4)

1 – 8: we see that there is a 2[100] transforming y, z into –y, -z: There is a

translational component of ½ along a; hence a 21[100] axis. Which is located at

(0,0,0)


21[001] axis at (0,1/4,0)

21[010] axis at (1/4,0,1/4)

Inversion at (1/4,1/4,1/4)

21[100] axis at (0,0,0)

We will now draw these into the figure

Points Symmetry m[010]

n[100] a[001]

O+ Ø+

Ø+1/2

Ø-

(1)

(4)

(3) (2) O+1/2

O-1/2

(5)

(6) Ø-1/2

O-

(7) (8)

o

1/4 o 1/4

The final step will be to complete the figure (let symmetry act on what is drawn)


Note: we found inversion symmetry at (1/4,1/4,1/4)

A centrosymmetric spacegroup should have its origin at a point of inversion.

Hence, we now need to do a coordinate shift for having inversion at origin.

Thereafter, our figure will (when fully completed) be identical to that of

the International Tables of Crystallography.

We will discuss these representations in light of other information in the

International Tables of Crystallography.

Crystallography – phase transitions

Example on phase transition; NiAs – to MnP-type

NiAs-type structure

Hexagonal

As: in hcp

Ni: octahedral holes

a = b ≠ c

= = 90o

= 120o

Unit cells drawn as projection on the ab-plane

Adding atoms at a later stage

*

Compendium West

*

*

* *

*


Option -1

Layer -A

Layer-B

Layer-A

Hexagonal close-packed (hcp) (heksagonal tetteste kulepakking)

ABABAB…….

Layer-B



Each sphere have 12

nearest neighbors

CN = 12

Closest packing of spheres in 3D – hcp




NiAs-type structure

Hexagonal

As: in hcp

Ni: oct.holes

a = b ≠ c

= = 90o

= 120o

Unit cells drawn as projection on the ab-plane

Adding atoms at a later stage

Space group

P63/mmc

Primitive Bravais lattice

One point (0,0,0)

6/mmm

a

b


The orthorhombic MnP-type structure is structurally very related to the

hexagonal NiAs-type structure

A continuous, displacive, 2.order phase transition may occur for a given

compound, as function of pressure or temperature (or substitution)

We will explore this in detail, starting with defining relationship between

the two unit cells of these structures



NiAs-type structure

Hexagonal

As: in hcp

Ni: oct.holes

a = b ≠ c

= = 90o

= 120o

Space group

P63/mmc


One point (0,0,0)

6/mmm

ah

bh

MnP

Space group Pnma


One point (0,0,0)

mmm

bo co

Vectors



ah

bh

bo co

Vectors

ao = ch bo = ah co = 2bh + ah

In case of an orthorhombic structure, the requirement on a = b (for the hexagonal

structure) is no longer active. If this length requirement is still in place, we have so

far just descirbed the hexagonal structure in an orthorhombic setting. We call this unit

cell for an orthohexagonal unit cell. Then co/bo = √3 =1.732..

A consequence of the hexagonal orthorhombic phase transition is then the possibility

of co/bo ≠1.732.. (but |co/bo = √3| is not a good order parameter for the transition)

NOTE:

The orthorhombic

cell is A-centered

as long as

co/bo = √3 =1.732..



ah

bh

bo co

Vectors

Vector relations; describing orthorhombic unit cell in terms of the hexagonal one:

ao = ch bo = ah co = 2bh + ah

In a matrix representation we write (note the way of writing!):



ah

bh

bo co

Vectors

And then the opposite relationship:

Vector relations; describing hexagonal unit cell in terms of the orthorhombic one:

ah = bo bh = -1/2bo + 1/2co ch = ao

In a matrix representation we write (note the way of writing!):


See additional information for completeness

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