Part 2 - Defects, extended structures and ordered vacancy structures

Question 1

What types of point defects can there be in crystalline elemental solids?

Answer 1

Vacancies, interstitials or substitutional disorder.

Question 2

C, N, O and H are often found as impurities in transition metals. What kind of defects are they typically found as?

Answer 2

Interstitials.

Question 3

Name a couple of technological uses of interstitial impurities.

Answer 3

Interstitial carbon in iron increases mechanical strength.

Interstitial hydrogen in Pd can be used as hydrogen storage.

Question 4

What is an aliovalent atom?

Answer 4

A dopant atom of a different valence than the atom that originally occupied whichever site it is substituted to.

Question 5

What is an isovalent atom?

Answer 5

A dopant atom of the same valence as the atom that originally occupied the site it is substituted to.

Question 6

Name a technological use of substitutional doping.

Answer 6

Doping of elemental Si with Al or P leading to p- and n-type semiconductors.

Question 7

What is the difference between point defects in crystalline elemental solids and in crystalline ionic compounds?

Answer 7

In ionic compounds there is the added constraint of total electroneutrality in the crystal.

Question 8

What are intrinsic defects?

Answer 8

Defects that can occur in pure materials (nothing coming in from the outside)

Question 9

What is a Schottky defect?

Answer 9

A Schottky defect is a cation vacancy compensated with the appropriate amount of anion vacancies.

Question 10

Upon creation of a Schottky defect, what could happen with the structure?

Answer 10

The negative and positive charges will attract each other, which can lead to vancacy clustering.

The surrounding structure may also relax or distort to accommodate the defect.

Question 11

What is a Frenkel defect?

Answer 11

A Frenkel defect is a vacancy and a corresponding interstitial ion (of the same charge).

Question 12

What types of ions can Frenkel defects occur?

Answer 12

For both cations (cation-Frenkel) and anions (anion-Frenkel).

Question 13

What is a colour centre?

Answer 13

A colour centre can be described as electrons trapped on vacant anion sites (trapped by the positive charge of the vacancy).

Question 14

What kind of colour centres are there?

Answer 14

F-centre: isolated trapped electron
M-centre: pair of electrons trapped
R-centre: triplet of electrons trapped

Question 15

Why are colour centres called colour centres?

Answer 15

Because the potential holding the electrons trapped give rise to s- and p-like states. Upon excitation and relaxation of these electrons from one state to another they frequently give rise to characteristic colours.

Some examples of this are blue kinds of calcite and feldspar and green diamonds.

Question 16

How can we with a simple lattice model consider the formation of a vacancy?

Answer 16

We can consider a simple lattice with N_0 lattice sites. We then look at the formation of the vacancy by moving one atom from the bulk to the surface, so that we now have N_0 + 1 lattice points. We then consider what the Gibbs free energy of creating n such isolated, non-interacting vacancies.

Question 17

What is the main driving force behind vacancy creation?

Answer 17

Increase in configurational entropy.

Question 18

What is the configurational entropy term of the Gibbs free energy, and how does it look for n vacancies?

Answer 18

-kT lnΩ per n vacancies, where Ω is the number of ways to arrange n vacancies of N_0 + n sites.

Ω = (N_0 + n)! / (N_0!n!)

Question 19

What happens to the total ΔH during vacancy formation?

Answer 19

Formation of each vacancy costs energy, and the more vacancies formed, the greater the enthalpy increase is. This is counteracted by the increase in entropy (ΔG = ΔH - TΔS), and an equilibrium is found for the intrinsic number of vacancies.

The positive enthalpy change comes from the fact that the chemical bonds broken during formation of vacancies are only partly compensated by the fewer bonds created at the surface.

Question 20

How is the dependence of n for ΔH, ΔS_vib and ΔS_conf for low vacancy concentrations?

Answer 20

For ΔH and ΔS_vib, it is linearly dependent on n. ΔS_conf is not linearly dependent, but dependent through -kt ln Ω where Ω = (N_0 + n)! / (N_0!n!)

Question 21

What is the consequence of the non-linear dependence on n of ΔS_conf?

Answer 21

That the equilibrium number of vacancies will always be non-zero (alternatively, that ΔG always occurs at non-zero value of n).

Question 22

How can we calculate the equilibrium number of vacancies (n_eq) formed by thermodynamic considerations?

Answer 22

We differentiate the term for ΔG wrt n and solve for dΔG/dn = 0. We use the following formula:

d/dn [nΔH - nTΔS - kTln( (N_0+n)! / N_0! n!)]

To differentiate this, we use Stirling’s apporximation, that says ln(x!) ≈ xlnx - x.

Question 23

How is the fractional concentration of vacancies dependent on temperature, ΔH and ΔS_vib? What favours a large concentration of defects?

Answer 23

It depends solely on these through the relation:

x_v = n_eq / (N_0 + n_eq) = exp(ΔS_vib / k) exp(-ΔH/kT)

A not too large ΔH (weak bonding) and large T would favour vacancy formations.

Question 24

How does the thermodynamic expression for the formation of isolated interstitials compare to the expression for the formation of vacancies?

Answer 24

It is similar.

Question 25

How does the thermodynamic expression for the formation of Schottky defects compare to the expression for the formation of vacancies?

Answer 25

Schottky defects would need to take into considerations the number of ways to rearrange the cation vacancies and the anion vacancies (the configurational entropy term would be -2kTlnΩ).

We thus get the same expression, but the fraction of Schottky defects would go as:

x_S = exp(ΔS_vib / 2k) exp(-ΔH_S/2kT)

(extra factor of 1/2 in each exponential)

Question 26

What is an extrinsic defect?

Answer 26

An extrinsic defect is a defect involving new chemical species.

Question 27

What kind of charge compensation mechanism could we have from an donor dopant?

Answer 27

A donor dopant would have a higher valence, and we would need additional negative charge to compensate. This could be acheived either through formation of cation vacancies (1), addition of anions (2) or by redox compensation (3)

1) Ca2+ in NaCl -> Na1_2x Cax []x Cl
2) Y3+ in CaF2 -> Ca1-x Yx F2+x
3) La3+ in CaMnO3 -> Ca1-x Lax Mn4+1-x Mn3+x O3

Question 28

What is a donor dopant?

Answer 28

A donor dopant is a dopant with a higher valence. It refers to the elemental state, in that it can donate an extra electron.

Question 29

What is an acceptor dopant?

Answer 29

An acceptor dopant is a dopant with a lower valence. It refers to the elemental state, in that it can accept an extra electron.

Question 30

What kind of charge compensation mechanism could we have from an acceptor dopant?

Answer 30

An acceptor dopant would have a lower valence, and we would need additional positive charge to compensate. This could be achieved either through formation of anion vacancies (1), addition of cations (2) or redox compensation (3).

1) Ca2+ in ZrO2 -> Zr1-x Cax O2-x []x
2) Al3+ in SiO2 -> Lix Si1-x Alx O2
3) Ca2+ in LaMnO3 -> La1-x Cax Mn3+1-x Mn4+x O3

Question 31

How can we understand the charge balance in substitution of Sr2+ in La2CuO4 to La2-xSrxCuO4, without any additional ions or vacancies?

Answer 31

If we substitute Sr2+ in on La3+, we get La2-xSrxCuO4.

We can understand the charge balance from the change in average oxidation state of Cu (originally 2+) to a higher average oxidation state.

Question 32

What can happen with NiO upon oxidation?

Answer 32

Either increase oxygen content or create Ni-vacancies.

Question 33

What can happen with NiO upon reduction?

Answer 33

Either decrease oxygen content or excess Ni.

Question 34

What is the relative occurence of solid solutions based on anion substitution vs. cation substitution?

Answer 34

Solid solutions based on anion substitutions are rarer. This is mainly because the range of anions of similar size/charge is smaller than the range of metals.

Question 35

What is a solid solution?

Answer 35

A solid solution is when substitutions occur randomly throughout a crystal structure.

Question 36

What is a material that follows Vegard’s law?

Answer 36

A material where the unit cell parameters vary smoothly upon further substitution.

Question 37

For solid solutions following Vegard’s law, how can you determine the composition if you know the cell parameters of the compound in question and the cell parameters of the end members?

Answer 37

The cell parameter of the solid solution following Vegard’s law is given by a weighted sum of cell parameters of the end-members:

a(x) = xa1 + (1-x)a2

Solve this for x and we get:

x = (a(x)-a2)/(a1-a2)

Question 38

Why is it that Vegard’s law is not always followed precisely?

Answer 38

There can be both negative and positive deviations from compounds exhibiting Vegard’s law dependence. The departure is because atoms/ions only approximately can be treated as hard spheres and specific bonding interactions cause small deviations in volume.

Question 39

What is the solid-solubility limit?

Answer 39

The limit beyond which it is impossible to perform further substitution.

Question 40

What are two basic types of line defects?

Answer 40

Edge dislocation and screw dislocation.

Question 41

What is an edge dislocation?

Answer 41

An edge dislocation is a line defect where there is an extra plane of atoms that has been inserted into the crystal structure and terminated at some point. The termination edge is called a dislocation line.

This is analogous to an interstitial atom, but it extends over a two-dimensional plane.

Question 42

How can edge dislocations explain why metals can be easily deformed without cracking or failure?

Answer 42

If one applies a force perpendicular to the top half of the crystal (where the extra plane lies) then the extra plane of atoms can move to alleviate the stress. This can form the dislocation line to move across the crystal one lattice spacing at the time (across what is known as a slip plane). Each step requires relatively few bonds to be broken.

Question 43

What is a slip plane?

Answer 43

The plane along which the dislocation line moves.

Question 44

Why can impurities significantly influence the mechanical properties of metals?

Answer 44

Through pinning, where impurities can trap dislocations at specific locations in a structure and prevent their motion.

Question 45

Why are ceramic materials usually harder, more brittle and more prone to cleavage than metals?

Answer 45

In metals, the nature of the metallic bond allows easy slip-plane formation. In an ionic solid the slippage of one atomic unit per step is unfavourable because of repulsive interactions between ions of like charge.

Question 46

What is a screw dislocation?

Answer 46

One can imagine that we slice a crystal to its centre and slide adjacent layers of atoms by one atomic plane. The dislocation line runs up the center of the spiral that is then created.

A screw dislocation can also move through a crystal under stress in a similar manner to edge dislocations.

Question 47

What are the four important categories of planar defects in materials?

Answer 47

Stacking faults, twin formation, antiphase boundaries and crystallographic shear structures.

Question 48

What are polytypes?

Answer 48

Polytypes are a type of stacking fault where ordered variants differing by their stacking sequences are formed. They are a combination of cubic and hexagonal closest-packing.

Question 49

What is turbostratic disorder?

Answer 49

Turbostratic disorder is a type of stacking faults. This can occur when the forces holding adjacent layers together are so weak that one layer can rotate with respect to the next. This is commonly found in graphite, layered clays and molecular intercalates.

Question 50

What is polymorphism?

Answer 50

Polymorphism is the phenomenon of the existence of more than one structural form of a given material.

Question 51

What is a twinned crystal?

Answer 51

A twinned crystal is a planer defect and is defined as an intergrowth of two or more individual crystals of the same species in which the different portions are related by a symmetry operation that doesn’t belong to the point group of the crystal.

Each individual is called a twin boundary (or twin domain) and are related by a twin law. The interface between is called a twin boundary or domain boundary.

Question 52

How can twinning be detected?

Answer 52

• Sometimes by the eye (if the twin domains form an angle between two domains that is macroscopically visible).
• In translucent crystals it can be seen using cross polarisers as different domains will show extinction at different angles.
• Single-crystal x-ray diffraction as each domain will give its own pattern that may or may not overlap (depending on the twin law).

Question 53

What are the three different types of twins?

Answer 53

Growth twins, deformation twins and transformation twins.

Question 54

What are growth twins?

Answer 54

Growth twins can arise during formation of a crystal if it grows from multiple nuclei.

Question 55

What are deformation twins?

Answer 55

Deformation twins can occur when a shear stress is applied to a crystal and causes each plane of atoms to move by a fraction of a unit cell relative to the layer below it.

Question 56

Explain how twinning can occur when the shape of the unit cell has a higher point symmetry than the atomic contents.

Answer 56

The analogy is parallelpiped bricks. They are easy to stack, even though you put some bricks upside down. This doesn’t affect the overall stacking, but the atomic contents might change direction.

This can happen when the energy penalty for placing the unit cell in an laternative orientation is low.

This commonly occurs when a struture undergoes a phase transition from high symmetry to a related low symmetry form.

Question 57

What is a transformation twin?

Answer 57

If minor structural changes occur during a phase transition from a high symmetry to a related low symmetry form, the “lost” symmetry element can act as a twin law giving rise to a transformation twin.

One example is a phase transition going from cubic to tetragonal. In this process the orientation of the elongated c-axis could be randomly oriented, cause twin domains.

Question 58

How are the twin domains related ina transformation twin?

Answer 58

By a rotational symmetry element present in the high-symmetry phase but absent in the low symmetry phase. They are thus associated with translationengleiche transitions (as rotational symmetry are dilluted in these).

Question 59

How can you predict the number of twin variants of a transofmration twin?

Answer 59

By looking at the dillution of rotational symmetry per lattice point.

Question 60

What are antiphase boundaries?

Answer 60

Antiphase boundaries are planar defects that are associated with site ordering in materials. When this site ordering happens from several sites during the ordering process, when two domains meet they may or may not be in phase. An antiphase boundary is then when the two domains are out-of-phase. This is energetically unfavourable, but the “cost of repair” is too great for that to happen.

These form when a subset of translational symmetry is gone in a transition, and will then be expected from klassengleiche transitions where translational symmetry is dilluted.

Question 61

What are crystallographic shear structures?

Answer 61

Crystallographic shear structures are structures with planar defects that can accommodate non-stoichiometry. They form a family of compounds with a discrete formula, even though they are very adaptive.

One can imagine a partial reduction of an ideal WO3-structure. We can imagine that we emove oxygen along a plane, after which the W on one of the sides of this O-plane shifts to form new octahedra that are now also edge-sharing.

Question 62

What are Magnéli phases?

Answer 62

Magnéli phases is another term for crystallographic shear structures.

Crystallographic shear structures are structures with planar defects that can accommodate non-stoichiometry. They form a family of compounds with a discrete formula, even though they are very adaptive.

One can imagine a partial reduction of an ideal WO3-structure. We can imagine that we emove oxygen along a plane, after which the W on one of the sides of this O-plane shifts to form new octahedra that are now also edge-sharing.

Question 63

How does one describe crystallographic shear structures in WO3?

Answer 63

We denote the fault plane with (hkl)-notation. This relates to the unit cell of the ideal WO3, and we would use a- and c-vectors going from (101) (diagonal of the original unit cell) to (001) (along c). Depending on the fault plane we get characteristic chemical formulae for the ordered phases (with regular spacing between fault planes).

E.g.: (102) shear plane gives the formula WnO3n-1, (103) gives WnO3n-2.

These can exist up to a threshold where the spacing becomes too close, and any further non-stoichiometry would instead be accommodated by another type of shear plane.

Question 64

What happens if shear planes occur in two perpendicular directions?

Answer 64

We can get a large family of block or column structures.

Question 65

What other common material can exhibit crystallographic shear structures (other than WO3)?

Answer 65

TiO2 rutile. However, these shear planes are harder to visualise as there are no convenient projection.

Question 66

How does the evolution of oxidation of hcp-Ti? (ie. increasing oxygen content from Ti to TiO2)

Answer 66

Ti metal is hcp. Initially half of the octahedral hole planes are filled up to Ti2O, where an anti-CdI2-type structure occurs. Vegard’s law is followed up to Ti3O.

At TiO, it most commonly forms a NaCl-structure with a significant number of defects. On both sides of this 1:1 stoichiometry there are a large number of vacancies. At 1:1 around 15% of both types of sites are vacant. These vacancies are ordered at low temperature.

At TiO1.2, there is also an ordering at low temperature where 4/5 of cation sites are occupied and all O are occupied.

At Ti2O3 it assumes the corundum structure, Ti3O5 it has several polymorphs and Ti4O7. After this we get the crystallographic shear-plane series TinO2n-1.

Question 67

How could we detect the number of vacancies in a 1:1 TiO compound?

Answer 67

Not from elemental analysis or refinement of XRD-data. Would have to do density measurements and compare with expected for a defect-free material.

Question 68

How can vacancies cluster in wüstite (FeO)?

Answer 68

FeO has a NaCl-type structure and will always have Fe2+-vacancies. These are charge compensated by Fe3+ in tetrahedral holes. These interstitials will attract nearby Fe2+-vacancies (with a formal charge of -2) and create clusters of Fe3+ surrounded by tetrahedra.

How the vacancies cluster on a larger scale is disputed and many cluster types are suggested.

Question 69

What are incommensurate structures?

Answer 69

Incommensurate structures are structures where there seems to be fully occupancy of sites, but the overall composition is still non-stoichiometry, owing to the fact that two parts of the structure have non-matching lattice parameters so that they never meet up (i.e. they fit in one direction, but not the other, and there is no simple integer ratio such at the parameters match).

One can use an analogy of tiling a bathroom floor with tiles that are of different lengths. The gaps between two tiles may never align (unless there is some definite ratio between the lengths).

Question 70

What are two different ways of describing incommensurate structures?

Answer 70

One way is to approximate it if there is an almost match of different unit cells. Then we can use a supercell to approximate this. This is, according to Pavel, very inelegant.

Another way is to use superspace groups and modulation-wave vectors. This is because structures that are not commensurate in 3D will become so in higher-dimensional space.

Question 71

What are infinitely adaptive structures?

Answer 71

Infinitely adaptive structures are structures where any small compositional change leads to a structure which is unique, even if closely related to those of neighbouring compositions. Each composition is a single phase and the compositions are so dense that there are effectively no two-phase regions, just one solid-solution range.

Question 72

How can we understand the infinitely adaptive Y(O,F)-structures from Y(O,F)2.13 -> Y(O,F)2.2?

Answer 72

These are synthesised by mixture of YOF and YF3.

We can understand these by considering the fluorite MX2-structure in terms of grids of X with M arranged in checkerboard pattern above and below. This forms edge-sharing XM4/4-tetrahedra.

We then have alternating layers of these MX-slabs, giving MX2 overall.

In our Y(O,F)-structure, we swap out the square grids of X with triangular grids of X, which gives a denser packing of X-atoms and give a hyperstoichiometry. The size of the triangular nets can then be adjusted to accommodate different stoichiometries.

Question 73

For what kind of materials can we get the infinitely adaptive Y(O,F)-kind of stacking?

Answer 73

Since the adaptiveness of the structure is dependent on a varying density in one layer, this leads to different cation coordination.

We thus need cations that can have variable coordination numbers.