CHEM 548: Materials Chemistry
Lecture 7: Important Ternary and Quaternary Structures
Wed 09/16/2026
Spinel, garnet, perovskite, and Goldschmidt tolerance factors
Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
To finish our survey of the structures of extended solids, we're going to look at three more structures.
Two ternary structures , seen mostly for oxides, and one quaternary structure They house families of compounds with many important functional properties
Reading : Chapter 1, section 1.5, in Woodward, Karen, Evans, and Vogt
Let me lay out what you should be able to do by the end.
Learning Objectives
Spinel, garnet, perovskite recognize the structural features of the three lattices
Site preferences how tetrahedral versus octahedral cation preferences shape these structures
Functional properties magnetism, lasers, ferroelectricity, piezoelectricity
Parsimony and coordination apply Pauling’s rule of parsimony and coordination-number analysis to complex solids
Tolerance factor predict perovskite stability and distortions
Here are the five things you should be able to do by the end of the hour.
Recognize the structural details of the spinel, garnet, and perovskite latticesSay how cation site preferences , tetrahedral against octahedral, shape those structures
Relate the three structures to properties : magnetism, lasers, ferroelectricity, piezoelectricityApply Pauling's rule of parsimony and coordination number analysis to solids this complicated
Use the tolerance factor to predict perovskite stability and distortions
We start with spinel, because its site occupations make all of this concrete.
Spinel (MgAl2 O4 )
Magnetite Fe₃O₄ = Fe²⁺Fe³⁺₂O₄ AB₂O₄ with A and B both iron; the oldest technologically used magnetic material, in compasses over 2000 years ago
Site occupation Mg on tetrahedral sites, Al on octahedral sites
Compass (circa 4th century BC)
Tetrahedron Octahedron Example
1 3, 4 LiMn₂O₄ (a = 8.245 Å) 2 3, 3 ZnFe₂O₄ (a = 8.442 Å) 2 4, 2 Fe₂TiO₄ (a = 8.521 Å) 3 2, 3 Fe₃O₄ = Fe²⁺Fe³⁺₂O₄ (a = 8.394 Å) 4 2, 2 Ni₂SiO₄ (a = 8.045 Å) 5 1, 2 LiZnNbO₄ (a = 6.082, c = 8.403 Å) 6 1, 1 Na₂WO₄ (a = 9.108 Å)
Per Solid State Materials Chemistry , Table 1.7, §1.5.1.
The spinel structure, named after the mineral spinel, magnesium aluminum oxide, is a cubic oxide framework found in many important compounds.
Magnetite, Fe₃O₄, is AB₂O₄ with A and B both iron : one Fe²⁺ and two Fe³⁺, so Fe²⁺Fe³⁺₂O₄Earliest known magnetic material : Chinese compasses more than 2,000 years agoSpinels remain widely used for their magnetic properties
Here magnesium is tetrahedral , aluminum octahedral
Across the AB₂O₄ family : tetrahedral sites take +1, such as lithium, up to +6, such as tungsten , and the octahedral charge adjusts to keep the balance
That flexibility in site occupancy is central to the chemical and physical diversity of spinel compounds
Next we build the structural description one feature at a time.
Spinel Structure
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
Space group Fd3 m, a = 8.086 Å
Atom Site Symmetry x y z
Mg 8a 4 3m0 0 0 Al 16d m3 5/8 5/8 5/8 O 32e m 0.384 0.384 0.384
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
Mg O O O O Al Al
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
Mg O O O O Al Al
$\mathrm{Mg}^{\textcolor{#8d90e8}{[4]}}\mathrm{Al}_2^{\textcolor{#8d90e8}{[6]}}\mathrm{O}_4^{\textcolor{#8d90e8}{[1,3]}}$
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
$A^{\textcolor{#8d90e8}{[4]}}B_2^{\textcolor{#8d90e8}{[6]}}O_4^{\textcolor{#8d90e8}{[1,3]}}$
Normal spinel (e.g. MgAl₂O₄) A tetrahedral (Mg), B octahedral (Al)
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
$A^{\textcolor{#8d90e8}{[4]}}B_2^{\textcolor{#8d90e8}{[6]}}O_4^{\textcolor{#8d90e8}{[1,3]}}$
Normal spinel (e.g. MgAl₂O₄) A tetrahedral (Mg), B octahedral (Al)
Inverse spinel (e.g. TiFe₂O₄) B tetrahedral (Fe); B and A randomly octahedral (Fe, Ti)
The spinel unit cell contains 56 atoms, as we can see from its Wyckoff positions.
Cubic close-packed oxygens : aluminum fills half the octahedral holes, magnesium one eighth of the tetrahedralTetrahedral framework , aluminum removed: isolated magnesium-centered tetrahedra on the carbon positions of diamond , same space group Fd3̅mOctahedral framework , magnesium removed: AlO₆ octahedra share edges into chains in multiple directionsFilling every octahedral site would give rock salt , where each octahedron shares all its edges
In spinel not all edges are shared , still a network of edge-sharing octahedra
Site symmetry 4̅3m for magnesium, corresponding to an ideal tetrahedron
Let me draw the bonding out as a graph.
[advance: the spinel bond graph]
We could draw a bond graph for the spinel structure, which I've done here.
The graph obeys Pauling's rule of parsimony
We knew that from the Wyckoff sites : all the oxygens sit on a single site
So they are all crystallographically and chemically equivalent , and the same for magnesium and aluminum
Oxygen makes one bond to magnesium, three to aluminum
Those coordination numbers are what the crystal chemical formula records.
[advance: the crystal chemical formula]
So we could write the crystal chemical formula in this way.
Magnesium CN 4 , aluminum CN 6 Oxygen has four neighbors : one magnesium and three aluminum
Now let's look at how the two cations divide between those two sites.
[advance: the normal spinel row]
The mineral spinel, like many other spinels, is what we call a normal spinel structure.
Normal : one magnesium and two aluminumsThe element present in the smaller amount , magnesium, goes onto the tetrahedral site
The other cation, two per formula unit , occupies the octahedral sites
That is not the only way the two cations can be distributed.
[advance: the inverse spinel row]
Oftentimes we'll find what are called inverse spinels.
Inverse , like TiFe₂O₄: the minority cation, titanium, prefers the octahedral site The majority cation is then split between octahedral and tetrahedral
A disordered or random distribution of two different cations on the octahedral sites
That extra site complexity leads us into garnet, which has three different cation sites.
Garnet (e.g. Mg3 Al2 Si3 O12 )
Nd³⁺:YAG Y₃Al₅O₁₂, a popular laser crystal emitting 1064 nm light.
YIG Y₃Fe₅O₁₂, an important magnetic material.
Cr³⁺ for Al³⁺ precious gemstones similar to ruby.
Cube Octahedron Tetrahedron Example
2 6 2 Ca₃Te₂Zn₃O₁₂ (a = 10.930 Å) 3 3 3 Y₃Fe₂Fe₃O₁₂ (a = 12.376 Å) 2 3 4 Mg₃Al₂Si₃O₁₂ (a = 11.459 Å) 1 3 5 Na₃Sc₂V₃O₁₂ (a = 10.913 Å)
Per Solid State Materials Chemistry , Table 1.8, §1.5.2.
Space group Ia3 d, a = 11.459 Å
Atom Site Symmetry x y z
Al 16a 3 0 0 0 Mg 24c 222 1/8 0 1/4 Si 24d 4 3/8 0 1/4 O 96h 1 0.034 0.050 0.654
Space group Ia3 d, a = 11.459 Å
Atom Site Symmetry x y z
Al 16a 3 0 0 0 Mg 24c 222 1/8 0 1/4 Si 24d 4 3/8 0 1/4 O 96h 1 0.034 0.050 0.654
$\mathrm{Mg}_3^{\textcolor{#8d90e8}{[8]}}\mathrm{Al}_2^{\textcolor{#8d90e8}{[6]}}\mathrm{Si}_3^{\textcolor{#8d90e8}{[4]}}\mathrm{O}_{12}^{\textcolor{#8d90e8}{[2,1,1]}}$
The garnet structure, X₃Y₂Si₃O₁₂, is even more complex than spinel.
Three site sizes : larger cations such as magnesium on distorted cubic sites, aluminum octahedral, the smallest, such as silicon, tetrahedralMany technological applications
YAG , Y₃Al₅O₁₂: aluminum on both tetrahedral and octahedral positions; with neodymium for some of the yttrium, the most widely used laser crystal , 1064 nm infrared, frequency-doubled to 532 nm greenYIG , iron for the aluminum: an important magnetic materialGemstones : chromium for aluminum gives red garnets often mistaken for rubies, the Black Prince's Ruby among themMany variants , properties set by the cations on the cubic, octahedral, and tetrahedral sites
Now let me put some numbers on those three sites.
[advance: the garnet Wyckoff sites]
If we were to look just a little bit more carefully at the structure, here are the details for magnesium aluminum silicate.
Ia3̅d , space group number 230, the very last one, and you might argue the most symmetric in the international tables160 atoms in the unit cell, but amazingly each atom on a single Wyckoff site All the silicons are crystallographically and chemically equivalent , and the same for aluminum, magnesium, and oxygen
So once again we obey Pauling's rule of parsimony
So let me go straight to the crystal chemical formula for garnet.
[advance: the garnet formula]
The bond graph is a bit complicated to draw because there's so many different atoms, but we can just write the crystal chemical formula, which is shown here.
Magnesium 8-coordinate , aluminum 6 , silicon 4 Oxygen once again has 4 bonds to the cations : 2 to magnesium, 1 to silicon, 1 to aluminum
Perovskite is structurally simpler than either of these, so it gives us a clean way to study what ionic size does.
Perovskites
Space group Pm3 m, a = 4.0 Å
Atom Site Symmetry x y z
Sr 1b m3 m 1/2 1/2 1/2 Ti 1a m3 m 0 0 0 O 3d 4/mmm 1/2 0 0
Space group Pm3 m, a = 4.0 Å
Atom Site Symmetry x y z
Sr 1b m3 m 1/2 1/2 1/2 Ti 1a m3 m 0 0 0 O 3d 4/mmm 1/2 0 0
$\mathrm{Sr}^{\textcolor{#8d90e8}{[12]}}\mathrm{Ti}^{\textcolor{#8d90e8}{[6]}}\mathrm{O}_3^{\textcolor{#8d90e8}{[4,2]}}$
Space group Pm3 m, a = 4.0 Å
Atom Site Symmetry x y z
Sr 1b m3 m 1/2 1/2 1/2 Ti 1a m3 m 0 0 0 O 3d 4/mmm 1/2 0 0
$\mathrm{Sr}^{\textcolor{#8d90e8}{[12]}}\mathrm{Ti}^{\textcolor{#8d90e8}{[6]}}\mathrm{O}_3^{\textcolor{#8d90e8}{[4,2]}}$
The perovskite structure, ABO₃, is simpler than spinel or garnet, with only five atoms per unit cell.
Small B-site cation, titanium here, at the cube corners ; oxygen at the edge midpoints; larger A-site cation, strontium here, at the body center
Remove the A cation: corner-connected TiO₆ octahedra , an infinite three-dimensional network, equivalent to ReO₃
Put the A cation back and it stuffs that framework , fitting the cavity formed by the corner-sharing octahedra
A is 12-coordinate in a cubic environment, the number of cube edges
Let me write those coordination numbers out as the crystal chemical formula.
[advance: the perovskite formula]
So the crystal chemical formula is strontium 12-coordinate, titanium 6-coordinate, and then the oxygen has four strontiums around it and two titanium.
Strontium 12 , titanium 6 Oxygen: four strontiums and two titanium
Each of those numbers is a coordination environment we can draw
Let me show you all three.
[advance: the three coordinations]
Here would be another view of the coordination environments of the three ions in the perovskite structure.
A site : a cubo-octahedron of twelve oxygensB site : an octahedron of sixAnion : between four strontiums and two titaniums
Those coordination environments make ionic size the next question.
Tolerance Factor
Tolerance factor t Goldschmidt’s simple, accurate geometric description of the perovskite structure; it predicts the likelihood of an octahedral tilting distortion.
Cell face: Ti at the corners, O on the edges
Cell face: Ti at the corners, O on the edges
a√2 Cell centre: Sr in the middle, O at the corners
Cell face: Ti at the corners, O on the edges
a√2 Cell centre: Sr in the middle, O at the corners
along the edge
$a = 2(r_{\mathrm{Ti}} + r_{\mathrm{O}})$
along the face diagonal
$a\sqrt{2} = 2(r_{\mathrm{Sr}} + r_{\mathrm{O}})$
Cell face: Ti at the corners, O on the edges
a√2 Cell centre: Sr in the middle, O at the corners
for any ABO₃: Sr is the A cation, Ti the B cation
$t = \dfrac{r_A + r_O}{\sqrt{2}\,(r_B + r_O)}$
$t > 1$ A too big
$t = 1$ perfect fit
$t < 1$ A too small
Per Solid State Materials Chemistry , Ch. 1 §1.5.
Goldschmidt developed a simple yet accurate geometric description of the perovskite structure, which allows us to predict the likelihood of an octahedral tilting distortion.
He called the parameter the tolerance factor, t
One parameter , the cubic cell edge, but two constraints : the optimal B to O distance, titanium to oxygen, and the optimal A to O distance, strontium to oxygenHow far both are met is what the tolerance factor expresses
Introduced by Goldschmidt about a century ago, who also synthesized some of the first perovskites
The mineral perovskite itself is calcium titanate, CaTiO₃
It comes out of the packing drawing here, so let me cut through the middle of the cell and set the geometry up.
[advance: the cut through the cell]
If we were to take a cut right out of the middle of the unit cell, through the titanium and the oxygen, the titanium sits at the corners and the oxygens sit on the edges, so along the cell edge the titanium and the oxygen touch.
Cut through Ti and O : Ti at the corners, O on the edges, touching along the edge
Now let me take the second cut, through the face of the cell.
[advance: the two cuts]
In that one the strontium and the oxygen touch along this face diagonal.
The cubic cell edge can be defined from the titanium and oxygen radius or from the strontium and oxygen radius
Let me write both of those definitions out.
[advance: the two cell edges]
In the upper cut, along the cell edge, we have one radius of titanium, two radii of oxygen, and one more radius of titanium, so the cell edge is twice the titanium radius plus the oxygen radius.
Edge : Ti, O, O, Ti in a row, so a = 2(r_Ti + r_O) Face diagonal : Sr touches the O at the corners, so a√2 = 2(r_Sr + r_O) Diagonal of a square is the edge times root two
Now we have two expressions for what the unit cell edge should be, and we can use those to calculate the tolerance factor.
[advance: the tolerance factor]
For a general perovskite ABO₃ we write A for the large cation and B for the small one, so strontium is A and titanium is B.
The ratio : the cell size that best fits the A cation over the size that best fits the B cation
Two values match: t = 1 , a perfect fit
t > 1 : A too large, stretches the framework of corner-sharing octahedrat < 1 : A too small, sits loosely in the cavity, leaving extra space inside
We can now put some real compounds through that ratio.
Size Mismatch and ABO3 Compounds
Cell edge each cation would prefer, and their ratio
Compound a (Å) A-cation a (Å) B-cation Tolerance factor t
BaMnO₃ 4.17 3.81 4.17/3.81 = 1.09 BaTiO₃ 4.17 3.93 1.06 SrTiO₃ 3.93 3.93 1.00 CaMnO₃ 3.72 3.81 0.976 CaTiO₃ 3.72 3.93 0.946 MgTiO₃ 3.33 3.93 0.847
Not perovskite
Distorted perovskite
Let's look at some examples of tolerance factors.
Strontium titanate : essentially a perfect fit, t = 1 Larger barium for the strontium takes it above one, to 1.06 ; smaller manganese(IV) for the titanium, further to 1.09
The other way: calcium for the strontium gives 0.946 ; magnesium for the calcium, about 0.85
At those extremes, magnesium titanate and barium manganate, no longer a true perovskite : the corner-connected octahedral network cannot hold
Intermediate values keep the perovskite framework but show distortions
Let me take the ones above one first, where the framework is stretched.
ATiO3 Structures (t
SrTiO₃
BaTiO₃
BaMnO₃
As the tolerance factor rises above 1.00 the connectivity of the octahedral corner sharing network starts to break down, first subtly and then more dramatically.
When the tolerance factor is slightly greater than one, as in barium titanate, the octahedra are stretched and the titanium atom shifts off-center.
That displacement creates a local dipole moment : barium titanate is the classic ferroelectric
Similar distortions matter for piezoelectric and dielectric applications, many of them relying on perovskites with tolerance factors just above one
Larger still, barium manganate at 1.09 : the A cation is too big for the corner-sharing network, so the structure becomes a hexagonal perovskite with different connectivity
Let me show you what those hexagonal and rhombohedral connectivities look like.
“Hexagonal/Rhombohedral Perovskites”
Hexagonal and rhombohedral perovskites form when cubic close-packed AO₃ layers are partially replaced by hexagonal close packing.
2H , full hexagonal packing: octahedra share faces into infinite chains that are not interconnected, leaving space for large cations such as bariumMixed cubic and hexagonal stacking: the connectivity gets more complex4H , the high-pressure form of barium ruthenate: four layers repeat, dimers of face-sharing octahedra linked by corners9R , barium ruthenate: nine layers repeat, trimers of face-sharing octahedra connected by corners to other trimers6H , barium titanate under high pressure: corner-sharing octahedra together with dimers of face-sharing onesThese hexagonal perovskites appear when t is greater than one
Now let me turn the factor the other way and make the A cation too small.
Crystal Structure - CaTiO3
1 > t > 0.9 the undersized A-site cation causes an octahedral tilting distortion
Ti environment still an octahedron
Ca coordination reduced to 8
When the tolerance factor drops below one, the A-site cation becomes too small and does not contact all of its neighbors.
The structure responds with an octahedral tilting distortion
Calcium titanate , at a tolerance factor between one and 0.9, is the case drawn hereBO₆ octahedra stay nearly ideal around the smaller cation, such as titaniumThe A-site environment is significantly altered : some oxygens closer, others farther away
A-site coordination number drops from twelve to about eight
If the factor falls further still, we get a different ordered structure altogether.
ATiO3 Structures (t
SrTiO₃
CaTiO₃
MgTiO₃
As the tolerance factor falls below 1.00 the octahedra begin to rotate so that some of the anions can maintain contact with the A-site cation. This is called an octahedral tilting distortion. The ilmenite structure emerges when the tolerance factor decreases beyond a certain point.
If the tolerance factor decreases far enough, even distorted perovskites become unstable.
The structure often transforms into ilmenite , as in magnesium titanate at t = 0.85
Ilmenite is closely related to corundum , aluminum oxide, which has pairs of face-sharing octahedra
In ilmenite those octahedra are ordered : titanium in one, magnesium in the other, alternating layers
The alkaline earth coordination number drops to six , more appropriate for a small ion such as magnesium
Some perovskites stay cubic, but a much larger fraction are distorted like calcium titanate or barium titanate
The family is vast, thousands of members , and comes back whenever we discuss functional properties
Let me pull the three structures together before we go to the practice.
Summary
Spinel, garnet, and perovskite compared on where the cations sit, and perovskite on how the framework responds when the A cation does not fit.
Structure Formula Sites and packing Applications
Spinel AB₂O₄ ccp O²⁻ lattice; normal versus inverse site distributions magnetism (Fe₃O₄) Garnet X₃Y₂Si₃O₁₂ cubic (8), octahedral (6), tetrahedral (4) sites lasers (Nd:YAG), magnets (YIG), gemstones Perovskite ABO₃ corner-sharing BO₆ octahedra, A in 12-coordination; fit measured by the tolerance factor t ferroelectric BaTiO₃; piezoelectrics and dielectrics
Let me summarize the three structures.
Spinel, AB₂O₄ : cubic close-packed oxide lattice, normal and inverse site distributions, important for magnetism , for example magnetiteGarnet, X₃Y₂Si₃O₁₂ : three cation sites, cubic 8, octahedral 6, tetrahedral 4; uses in lasers, magnetic materials, gemstones Perovskite, ABO₃ : corner-sharing octahedra, A in 12-coordination; the tolerance factor says whether A fitst = 1 : SrTiO₃ cubic; just above 1 : Ti off-centre, BaTiO₃ ferroelectric ; 1.09 : BaMnO₃ hexagonal, octahedra face to faceBelow 1 : octahedra tilt , CaTiO₃; 0.85 : MgTiO₃ is ilmenite
The rest of the hour is structure-drawing practice, and we start with fluorite.
CaF2 (fluorite)
Space group Fm3 m (#225) Lattice parameter a = 5.46 Å Atom Wyckoff site x y z Ca 4a 0 0 0 F 8c 1/4 1/4 1/4
Space group Fm3 m (#225) Lattice parameter a = 5.46 Å Atom Wyckoff site x y z Ca 4a 0 0 0 F 8c 1/4 1/4 1/4
Ca (4a) 0,0,0 ½,½,0 ½,0,½ 0,½,½ F (8c) ¼,¼,¼ ¾,¾,¼ ¾,¼,¾ ¼,¾,¾ ¼,¼,¾ ¾,¾,¾ ¾,¼,¼ ¼,¾,¼
Here is the mineral fluorite, alongside the data we will enter in VESTA.
Fluorite, CaF₂ : space group Fm3̅m, number 225, a = 5.46 ÅCalcium on 4a at zero, zero, zeroFluorine on 8c at one quarter, one quarter, one quarter
Let's use these two rows and the symmetry to place every atom in the unit cell.
[advance: the fluorite unit cell]
Here is the unit cell with every atom that lies fully or partially inside it.
Calcium at the origin , and the generators place others at (½,½,0), (½,0,½), and (0,½,½)Remaining calcium atoms from translational symmetry : integer multiples of the unit cell vectors added to those four positions
We can always add or subtract integers to bring atoms back inside, or partly inside, the cell
Now let's build fluorite in VESTA using these data.
Start a new structure
File → New Structure… (Ctrl+N)
The New Data dialog opens. Use its tabs in this order:
Phase : name the structureUnit cell : set the symmetry and lattice parametersStructure parameters : enter the atoms
Go to Unit cell first.
We are going to build fluorite in VESTA using its space group and two rows of atom data.
File → New Structure , or Ctrl+N : VESTA opens the New Data dialogTabs in data-entry order: Phase for the name, Unit cell for symmetry and lattice parameters, Structure parameters for atoms
Unit cell first , to set symmetry before entering atom positions
First, we'll choose the crystal system and space group.
Set the crystal system and space group
System Cubic Space group Fm3 m (#225)
Choose Cubic to limit the list to cubic space groups.
Scroll to 225 and click it. Check that the Setting panel shows setting 1, F m -3 m .
Fluorite is cubic, so choose Cubic in the System list on the left.
Middle column updates immediately; only cubic space groups selectable
Use our knowledge of symmetry to choose the group
Scroll to 225, Fm3̅m , and click it
Check the Setting panel for setting 1, Fm3̅m
Next, we'll set the size of the unit cell.
Set the lattice parameter
Set a = 5.46 in the a field at the bottom of the Unit cell tab and press Return. Because the cell is cubic, VESTA sets b and c automatically and greys out those fields. The angles remain 90°.
Fluorite's lattice parameter is 5.46 angstroms.
a = 5.46 at the bottom of the Unit cell tab, then ReturnCubic cell: b and c automatically set to 5.46 and greyed out; angles remain 90°
Lattice parameter required to define the cell and interpret the drawing; return here to check it if the structure looks wrong
Now we can enter the atoms.
Add the calcium site
Symbol Label x y z Occ. Ca 4a 0 0 0 1
On the Structure parameters tab, press New and enter the values above. You can choose any Label, but the Wyckoff symbol identifies the site in each row.
Click the Structure parameters tab and press New to add a row.
Symbol Ca , Label 4a , x = y = z = 0 Any label allowed ; the Wyckoff symbol identifies the site in each rowCheck the row : Ca, 4a, three zeros, occupancy 1
Fluorine is the second and last site we need to enter.
Add the fluorine site
Symbol Label x y z Occ. F 8c 1/4 1/4 1/4 1
Press New again. The new row contains the calcium values, so clear each field before typing . Enter 1/4 for each coordinate; VESTA stores it as 0.250000. Then press OK .
Press New again to add fluorine.
Clear each field first : VESTA copies the calcium values and adds typed text to the existing valueSymbol F , Label 8c , x = y = z = 1/4 1/4 converts to 0.250000 on leaving the field; fractions useful later for coordinates of one third
Press OK to draw the structure from these two rows.
Apply and look along the axes
1 Use a, b, c to look along each axis. For fluorite, the views are identical because cubic symmetry makes the axes equivalent.
2 Choose Standard orientation to tilt the cell and see every atom at once.
1 2
Large spheres are Ca; small spheres are F.
VESTA now displays the fluorite cell.
Top-left buttons for views along a, b and c ; identical for fluorite because cubic symmetry makes the axes equivalent
Next button: standard orientation , cell tilted so every atom is visible
Large spheres Ca, small spheres F ; each F at the centre of a small cube of four Ca
Now let's count the atoms in the drawing.
Read the structure output
Check the structure data in the output panel. The last two columns list the generated Wyckoff sites ; the last line gives the number of atoms drawn .
Count the Ca and F atoms in one unit cell.
4 Ca and 8 F.
VESTA includes atoms on the cell boundary, so it draws 14 Ca:4 .8 .
Let's check the structure data in the panel below the drawing.
Space group Fm3̅m, 225 ; cell edges all 5.46 Å
Last two columns: 4a, m3̅m and 8c, 4̅3m
Last line: 22 atoms
Count how many calcium and fluorine atoms belong to one unit cell.
[advance: the count]
One unit cell contains four calcium atoms and eight fluorine atoms.
VESTA includes every atom touching the cell boundary , giving 14 Ca plus 8 F
8 corners × ⅛ + 6 faces × ½ = 4 Ca All 8 F entirely inside , each counting as one
Two input rows; remaining positions generated by VESTA using symmetry
Next, we'll use the same crystallographic description to draw the sodalite network.
Sodalite – Na8 Al6 Si6 O24 Cl2
Cage packing the cages share hexagonal faces (6-rings) in a body-centered cubic arrangement
Openings 6 tetrahedra around the ring, about 2.6 Å; guest molecules cannot get in or out (not microporous)
Space group P4̅3n, a = 8.875 Å
Na1 8e 0.1777(2) 0.1777(2) 0.1777(2) Al1 6d 1/4 0 1/2 Si1 6c 1/4 1/2 0 O1 24i 0.1387(3) 0.1484(3) 0.4375(2) Cl1 2a 0 0 0
The sodalite network takes its name from the mineral sodalite, a sodium aluminum silicate with chloride in it.
Cages shaped like truncated octahedra , resembling blue soccer balls
Hexagonal faces (six-rings) and square/diamond faces (four-rings), matching the sodalite net topology from earlierThe cages share those hexagonal faces in a body-centered cubic arrangement
In sodalite itself, chloride sits at the cage centers ; the sodium and oxygen are not drawn here
The building unit of many zeolites , but here the cages are fused non-porous , ring openings about 2.6 Å, so molecules cannot get in or out
Cubic , space group P4̅3n, number 218 , a = 8.875 Å; one Wyckoff position each: Na1, Al1, Si1, O1, Cl1
Let's review how the space-group symbol tells us about the symmetry.
Reading a short symbol
Crystal system Shape Its point groups What the positions mean, in order
Triclinic all skew 1, 1̄ no direction to name Monoclinic one lean 2, m, 2/m one direction: b Orthorhombic a brick 222, mm2, mmm a , then b , then c Tetragonal square column 4, 4̄, 4/m, 422, 4mm, 4̄2m, 4/mmm c ; then a and b ; then the in-plane diagonalsTrigonal & hexagonal 120° base 3, 3̄, 32, 3m, 3̄m; 6, 6̄, 6/m, 622, 6mm, 6̄m2, 6/mmm c ; then the cell edges; then the in-plane diagonalCubic a die 23, m3̄, 432, 4̄3m, m3̄m the cell edges; then the body diagonals; then the face diagonals
Per Solid State Materials Chemistry , §1.1.3, Table 1.1 and §1.1.7, Table 1.3.
1. Crystal system write every glide as m and every screw as its rotation, then match what is left against a row above: mma becomes mmm, orthorhombic. A 3 in the first position is trigonal, in the second cubic
2. Bravais lattice the leading letter is the centering, and system plus letter name the lattice; a trigonal group under P sits on the primitive hexagonal lattice, under R on the rhombohedral
3. Point group the replaced symbol of step 1, without its lattice letter, is the point group
4. Symmorphic? if you replaced anything in step 1, the symbol carries a glide or a screw and the group is non-symmorphic
Let's review how to read a short space-group symbol.
The table is the one from the middle of the meeting , extended: every system, triclinic included , with a shape to remember it by : the names already tell you the shape
Monoclinic is one lean , orthorhombic a brick , tetragonal a square column , cubic the die ; triclinic is all skew , no direction to name, point groups 1 and 1̄
Next to the shapes, the point groups that belong to each system and what each position points along
Under it, the recipe arrives one step at a time , and four steps take any short symbol apart
[advance: step one, the crystal system]
Step one turns the symbol into a point group you can recognize: write every glide as a plain m and every screw as its plain rotation , then find the row whose point groups contain what is leftThe a in Pmma is a glide and counts as a plane, so what is left is mmm , and mmm sits in the orthorhombic row
Two systems need a tiebreak, and it is where the 3 sits : first position, along c, is trigonal ; second position, along the body diagonals, is cubic
[advance: step two, the Bravais lattice]
Step two : the leading letter is the centering , P, C, I, F or R, and the crystal system says which of the fourteen lattices the pair namesOne exception, worth memorizing, and it catches everybody : a trigonal symbol starting with P is described on the primitive hexagonal lattice, and only one starting with R uses the rhombohedral
[advance: step three, the point group]
Step three costs nothing , because step one already did the work: take the replaced symbol , drop the lattice letter , and that is the point groupPmma gave mmm, so the point group is mmm
[advance: step four, symmorphic or not]
Step four : if step one had to replace anything , the symbol carries a glide or a screw and the group is one of the 157 non-symmorphic onesIf nothing needed replacing , it is one of the 73
Let's return to VESTA and draw the sodalite cage as tetrahedra.
Show one sodalite cage
Start with the sodalite structure entered above.
Open Objects → Boundary… . Set all three ranges to 0–1 , then click OK .
Use one cell for this view. A 3 × 3 × 3 block contains many overlapping cages.
Let's draw the sodalite cage as connected tetrahedra.
Objects → Boundary… : x, y and z each 0–1 , then OKOne cell for the cage view; return from 3 × 3 × 3 if necessaryOxygen atoms outside the cell included through the bond settings
Now we'll specify the aluminum–oxygen and silicon–oxygen bonds.
Define the two tetrahedral units
Open Edit → Bonds… . For this view, keep only these two entries; use New to add any missing row.
A1 A2 Min. Å Max. Å Al O 0 2.0 Si O 0 2.0
For both rows, tick Show polyhedra and select Search A2 bonded to A1 .
Under Boundary mode, choose Search additional atoms if A1 is included in the boundary . Click OK .
Each tetrahedron has aluminum or silicon at its centre and oxygen at its corners.
Keep only Al–O and Si–O ; New for missing rows, Delete for other bond rows
Both rows: 0–2.0 Å ; Search by label unchecked
A1 = Al or Si , A2 = O; Search A2 bonded to A1; Show polyhedra checkedBoundary mode: Search additional atoms if A1 is included in the boundary
Four oxygen neighbours per centre; complete tetrahedra at the boundary
We can now switch on the polyhedral display.
Switch to the polyhedral display
In the left panel, open Style and select Polyhedral .
Check that both AlO₄ and SiO₄ tetrahedra appear.
If one set is missing, return to Edit → Bonds… and check that row’s 2.0 Å limit and Show polyhedra box.
The bond settings define the tetrahedra, and the Style tab controls how we display them.
Style → Polyhedral Check for both AlO₄ and SiO₄
Missing tetrahedra: check the row’s 2.0 Å cutoff and Show polyhedra
Na and Cl spheres still obscure parts of the cage
Let's hide those spheres so we can see the framework.
Hide the ions inside the cages
Open the Objects tab in the left panel and select Atoms .
In the V column, clear the checkboxes for Na and Cl . Keep Al, Si and O visible.
To hide the cell edges, open Properties… → General and choose Unit cell → Do not show .
This changes the display, not the structure.
We can hide sodium and chloride without removing them from the structure.
Objects tab → Atoms ; V column controls visibilityClear Na and Cl ; retain Al, Si and O
Properties → General → Unit cell → Do not show removes cell edgesAtoms remain in the data; boundary changes reset visibility
Next, we'll give the two kinds of tetrahedra different colours.
Distinguish aluminum and silicon
Open Properties… → Polyhedra . Under Planes , choose each Central atom and enter its Color (RGBA) .
Centre R G B A Al 160 160 160 255 Si 170 0 170 255
AlO₄ is grey ; SiO₄ is purple . An alpha value of 255 makes the faces opaque.
Keep Show polyhedral edges checked, then click OK .
Different colours make the alternating aluminum and silicon sites easier to follow.
Properties → Polyhedra → Planes , select Central atomAl RGBA: 160, 160, 160, 255 ; Si: 170, 0, 170, 255
Grey AlO₄, purple SiO₄ ; alpha 255 for opaque facesEdit Planes colour ; retain Show polyhedral edges
Now we'll turn the cage to see its openings.
Look through a hexagonal opening
Click Orientation… . Select Project along [u v w] and set the projection vector to 1, 1, 1 .
Leave the upward vector at 0, 1, 0 , then click OK .
To separate the front and back tetrahedra, set Step (°) = 15 and click the first curved rotation arrow once.
Looking along the body diagonal gives us a view through a hexagonal cage opening.
Orientation → Project along [u v w] Projection [111] ; upward vector [010]
Step (°) = 15 ; first curved rotation arrow onceSmall tilt separates the front and back tetrahedra
We can now follow the tetrahedra around the cage.
The sodalite cage in VESTA
Grey AlO₄ and purple SiO₄ tetrahedra share red oxygen corners.
Na and Cl are hidden in this view. Save the display with File → Save As… → VESTA format .
This is the sodalite cage drawn from the structure data we entered.
Grey AlO₄ alternating with purple SiO₄ Shared corners are oxygen atoms , shown in red
Na and Cl hidden ; framework continues into neighbouring cagesFile → Save As → VESTA format retains the display
We finish by counting layers instead of cages, in the sphere packing problem from the book.
Problem 1.18: stacking in terbium
With a structure-drawing program, determine the stacking sequence in Tb. Suggest the Ramsdell symbol and the Jagodzinski–Wyckoff notation.
P6₃/mmc (194), hexagonal
Atom Site x y z Tb 2b 0 0 ¼ Tb 4f ⅓ ⅔ 0.083
Build the structure, look down c, then turn the cell to see the layers along c. Label their lateral positions A, B and C.
SSMC, Problem 1.18; data reproduced in the Lecture 7 VESTA tutorial.
Now use the same structure-building method to work out how terbium packs.
Build Tb : group 194; two independent sitesDown c : close-packed layers; side view: stacking orderPause for a sketch : A/B/C repeat and both notationsConnection: 2H, 4H, 9R and 6H perovskites Start by building the hexagonal terbium cell.
Build the hexagonal Tb cell
Choose File → New Structure… , then Unit cell .
Select Hexagonal and space group 194, P6₃/mmc .
Parameter Value a = b 3.068 Å c 14.87 Å α = β 90° γ 120°
Enter a and c. The symmetry fixes b and the angles.
Start with the cell given in Problem 1.18.
New Structure → Unit cell Hexagonal; 194, P6₃/mmc a = b = 3.068 Å ; c = 14.87 Å Angles 90°, 90°, 120° Now enter the two independent terbium sites.
Enter two independent Tb sites
Open Structure parameters . Click New for each row.
Label Symbol x y z Tb1 Tb 0 0 0.25 Tb2 Tb 1/3 2/3 0.083
Leave both occupancies at 1 , then click OK .
Check the Output pane: Tb1 is 2b , Tb2 is 4f .
2 + 4 = 6 Tb atoms per cell.
Two input rows generate six terbium atoms per cell.
Tb1: (0, 0, ¼), occupancy 1Tb2: (⅓, ⅔, 0.083), occupancy 1Output: 2b + 4f → six atoms Cell-edge copies can inflate the displayed count Trim those boundary copies before reading the stack.
Keep one representative of each position
Open Objects → Boundary… .
Range Minimum Maximum x 0 0.99 y 0 0.99 z 0 1
Leave Cutoff planes empty and click OK .
Select Style → Ball-and-stick .
The display now contains six atoms . Each represents one layer in the stack.
We want one representative atom from each layer.
Boundary: x/y 0–0.99; z 0–1 No cutoff planes Six representatives; cell-edge copies removed Ball-and-stick ; no bonds neededLook down c to distinguish the three lateral positions.
Look down c to label A, B and C
Click the c button on the top toolbar to view along [001] .
Use the mouse wheel or + to enlarge the model.
Position x y A 0 0 B 1/3 2/3 C 2/3 1/3
The three spots are columns of overlapping Tb atoms . This view identifies positions, but hides their order along c.
A, B and C name lateral positions within the basal plane.
Toolbar c → view [001] A (0,0); B (⅓,⅔); C (⅔,⅓) All spots are Tb Two heights overlap at each spot; order hidden Turn to a side view to recover the order hidden by this projection.
Turn the six layers upright
Click Orientation… and choose Project along [u v w] .
Vector Values Projection 1, 0, 0 Upward 0, 0, 1
Click OK . Use the top-toolbar fit-to-screen button if needed.
View along a, [100]. c, [001], points up.
The (001) layers are now seen edge-on.
We now look perpendicular to the stacking direction.
Orientation → Project along [u v w] View [100] ; upward [001] c vertical ; basal layers edge-onFit model to window; read bottom to top Read the six heights from the bottom of the cell to the top.
Read the stacking from bottom to top
View [100]; c points up.
z Position 0.083 B 0.250 A 0.417 B 0.583 C 0.750 A 0.917 C
Follow the column positions upward:
B → A → B → C → A → C
Each atom represents a whole close-packed layer. The next cell repeats the sequence.
Reading upward gives B, A, B, C, A, C.
Bottom to top: B A B C A C A: z = ¼, ¾ from 2bB/C: four heights from 4fSix layers per repeat; two independent sites Extend the display slightly so the neighbours at both ends are visible.
Include neighbours across the cell boundary
Return to Boundary… . Keep x and y unchanged; extend z from −0.10 to 1.10 .
Click OK and fit the model to the window.
Layer to classify Below Above First B, z = 0.083 C, z = −0.083 A, z = 0.250 Last C, z = 0.917 A, z = 0.750 B, z = 1.083
The two extra atoms belong to adjacent cells . They do not increase the six-layer repeat.
The first and last layers also have neighbours on both sides.
Boundary: z −0.10 to 1.10 First B: C below, A above Last C: A below, B above Extra end atoms are periodic neighbours ; repeat stays six Use the same neighbour test on every layer.
Assign h or c to each layer
Compare the layers immediately below and above the layer being classified .
Same position → h. For B–A –B, the middle A is h.Different positions → c. For A–B –C, the middle B is c.
z Below Layer Above Label 0.083 C B A c 0.250 B A B h 0.417 A B C c 0.583 B C A c 0.750 C A C h 0.917 A C B c
Bottom to top: c h c c h c .h c c h c c = (hcc)₂ .
The h or c label belongs to the middle layer of each three-layer group.
Middle layer gets the labelB–A –B and C–A –C: h Unlike neighbours: c c h c c h c ; start at A → (hcc)₂ Same h/c test used for perovskite polytypes Now combine the six-layer repeat with the hexagonal lattice to write the Ramsdell symbol.
Problem 1.18: check the two symbols
Ramsdell symbol 4H
Jagodzinski–Wyckoff notation (hc)₂
ICSD #102655 P6₃/mmc (194) hexagonal, a = 3.770 Å, c = 12.159 Å; La on 2a (0, 0, 0) and 2c (1/3, 2/3, 1/4)
Tb
C c
A h
C c
B c
A h
B c
6H · (hcc)₂
Per Solid State Materials Chemistry , Figure 1.22 and Problem 1.18.
The lanthanum example has four layers per repeat; terbium has six.
α-La : four-layer repeat, 4H, (hc)₂A/B/C : lateral positions; h/c : neighbouring layersh : same neighbours; c : different neighboursUse periodic neighbours at the cell boundary Let's label each terbium layer using its neighbours.
[advance: the terbium layer labels]
Reading up the terbium cell gives B A B C A C.
Bottom to top: B A B C A C Both A layers: h ; all B/C layers: c c h c c h c ; start at A to write h c c h c c We can now write the Ramsdell and Jagodzinski–Wyckoff symbols together.
[advance: the two terbium symbols]
Terbium is 6H, with Jagodzinski–Wyckoff notation (hcc)₂.
6H : six layers, hexagonal lattice(hcc)₂ : two copies of the local hcc sequenceCoordinates → stacking → polytype labels Next, we'll use the spinel formula to count occupied holes.
Problem 1.31: hole filling in spinel
Analyze hole filling in the spinel structure of cubic close-packed oxygens.
Given the general formula [tetrahedron]₁[octahedron]₂O₄, calculate the total fractions of tetrahedral and octahedral holes filled. Use a structure-drawing program to orient the structure and identify the type of filled holes between layers of densely packed oxygen atoms. Determine the sequence and the fraction of each hole filling. Use the MgAl₂O₄ structure from the start of this lecture. Look along [111] to identify the oxygen layers, then turn the model to see the spaces between them.
SSMC, Problem 1.31; Lecture 7 VESTA tutorial.
Now count the occupied holes in the oxygen packing of spinel.
MgAl₂O₄ : one tetrahedral and two octahedral cations per O₄For N oxygens : N octahedral and 2N tetrahedral holes Pause : calculate totals, then inspect the interlayer sequenceDistinguish whole-structure fractions from plane fractions First, let's count the available and occupied holes per four oxygens.
Spinel Structure
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
Space group Fd3 m, a = 8.086 Å
Atom Site Symmetry x y z
Mg 8a 4 3m0 0 0 Al 16d m3 5/8 5/8 5/8 O 32e m 0.384 0.384 0.384
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
Mg O O O O Al Al
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
Mg O O O O Al Al
$\mathrm{Mg}^{\textcolor{#8d90e8}{[4]}}\mathrm{Al}_2^{\textcolor{#8d90e8}{[6]}}\mathrm{O}_4^{\textcolor{#8d90e8}{[1,3]}}$
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
$A^{\textcolor{#8d90e8}{[4]}}B_2^{\textcolor{#8d90e8}{[6]}}O_4^{\textcolor{#8d90e8}{[1,3]}}$
Normal spinel (e.g. MgAl₂O₄) A tetrahedral (Mg), B octahedral (Al)
MgO₄ tetrahedra not connected
ccp array of O; Al in 1/2 octahedral holes; Mg in 1/8 tetrahedral holes
AlO₆ octahedra share edges
$A^{\textcolor{#8d90e8}{[4]}}B_2^{\textcolor{#8d90e8}{[6]}}O_4^{\textcolor{#8d90e8}{[1,3]}}$
Normal spinel (e.g. MgAl₂O₄) A tetrahedral (Mg), B octahedral (Al)
Inverse spinel (e.g. TiFe₂O₄) B tetrahedral (Fe); B and A randomly octahedral (Fe, Ti)
The spinel unit cell contains 56 atoms, as we can see from its Wyckoff positions.
Cubic close-packed oxygens : aluminum fills half the octahedral holes, magnesium one eighth of the tetrahedralTetrahedral framework , aluminum removed: isolated magnesium-centered tetrahedra on the carbon positions of diamond , same space group Fd3̅mOctahedral framework , magnesium removed: AlO₆ octahedra share edges into chains in multiple directionsFilling every octahedral site would give rock salt , where each octahedron shares all its edges
In spinel not all edges are shared , still a network of edge-sharing octahedra
Site symmetry 4̅3m for magnesium, corresponding to an ideal tetrahedron
Let me draw the bonding out as a graph.
[advance: the spinel bond graph]
We could draw a bond graph for the spinel structure, which I've done here.
The graph obeys Pauling's rule of parsimony
We knew that from the Wyckoff sites : all the oxygens sit on a single site
So they are all crystallographically and chemically equivalent , and the same for magnesium and aluminum
Oxygen makes one bond to magnesium, three to aluminum
Those coordination numbers are what the crystal chemical formula records.
[advance: the crystal chemical formula]
So we could write the crystal chemical formula in this way.
Magnesium CN 4 , aluminum CN 6 Oxygen has four neighbors : one magnesium and three aluminum
Now let's look at how the two cations divide between those two sites.
[advance: the normal spinel row]
The mineral spinel, like many other spinels, is what we call a normal spinel structure.
Normal : one magnesium and two aluminumsThe element present in the smaller amount , magnesium, goes onto the tetrahedral site
The other cation, two per formula unit , occupies the octahedral sites
That is not the only way the two cations can be distributed.
[advance: the inverse spinel row]
Oftentimes we'll find what are called inverse spinels.
Inverse , like TiFe₂O₄: the minority cation, titanium, prefers the octahedral site The majority cation is then split between octahedral and tetrahedral
A disordered or random distribution of two different cations on the octahedral sites
That extra site complexity leads us into garnet, which has three different cation sites.
Build spinel with origin choice 1
Choose File → New Structure… → Unit cell .
Select Cubic , then 227, Fd3̅m .
Choose Origin choice 1 before entering the atoms.
Enter a = 8.086 Å . The other edges follow; all angles are 90°.
The coordinates on the next slide use this origin.
Let's build the spinel structure before counting its occupied holes.
File → New Structure → Unit cell Cubic; 227, Fd3̅m; origin choice 1 a = 8.086 Å ; three right anglesCoordinates must match the origin choice Now enter the magnesium, aluminum and oxygen sites.
Enter the three independent sites
Open Structure parameters . Click New for each row; keep occupancy at 1 .
Atom Site x = y = z Mg 8a 0 Al 16d 5/8 O 32e 0.384
Click OK , then the standard-orientation button.
One cell contains 8 Mg, 16 Al and 32 O , or eight MgAl₂O₄ formula units.
Three independent sites generate the spinel unit cell.
Mg 8a : 0, 0, 0; Al 16d : 5/8, 5/8, 5/8O 32e : 0.384, 0.384, 0.384; all occupancies 1Cell: 8 Mg + 16 Al + 32 O = 8 MgAl₂O₄ Boundary copies are shared Next, define the bonds that identify the coordination polyhedra.
Add Mg–O and Al–O bonds
Open Edit → Bonds… . Keep only these two rows.
A1 A2 Min. Å Max. Å Mg O 1.5 2.2 Al O 1.5 2.2
For each row: Search A2 bonded to A1 , Show polyhedra , and Search additional atoms if A1 is included in the boundary .
Click OK , then Style → Polyhedral .
Magnesium and aluminum occupy different kinds of oxygen hole.
Only Mg–O and Al–O , both 1.5–2.2 Å A1 is the cation; A2 is oxygen Show polyhedra ; complete boundary coordination shellsStyle → Polyhedral Check the number of oxygen corners around each kind of cation.
Check the coordination polyhedra
Orange MgO₄ tetrahedra have four oxygen corners.
Blue AlO₆ octahedra have six oxygen corners.
Red spheres are oxygen. These identify the filled tetrahedral and octahedral holes .
For 32 O: 64 tetrahedral holes and 32 octahedral holes .⅛ ; Al fills 16/32 = ½ .
The orange polyhedra have four oxygen corners, and the blue ones have six.
Orange MgO₄ : tetrahedral holesBlue AlO₆ : octahedral holesRed O : the close-packed arrayCheck coordination before counting occupations Let's compare the occupied sites with all the holes provided by thirty-two oxygens.
[advance: the overall hole counts]
Thirty-two oxygens provide sixty-four tetrahedral holes and thirty-two octahedral holes.
32 O → 64 tetrahedral + 32 octahedral holes Mg: 8/64 = 1/8 ; Al: 16/32 = 1/2 Part a complete; next locate the occupied planes Turn the model so the oxygen layers run horizontally.
Turn the oxygen layers sideways
Select Style → Ball-and-stick , then click Orientation… .
Choose Project along [u v w] .
Vector Values Projection 1, −1, 0 Upward 1, 1, 1
Click OK . The stacking direction [111] now points up.
In Properties → General , hide the unit-cell outline with Do not show .
↑ [111] Up: layer stacking
→ [1 1 −2] Right: along a layer
View along [1 −1 0] (111) oxygen layers seen edge-on
The close-packed oxygen layers are perpendicular to a cubic body diagonal.
Ball-and-stick for cation planesView along [1 −1 0] ; screen up [111] ; screen right [1 1 −2] Oxygen layers horizontal; stacking vertical Optional: hide cell edges in Properties → General Use two cutoff planes to retain the layers we want to compare.
Cut out two interlayer regions
Open Objects → Boundary… . Set x, y and z to −1 through 2 .
Under Cutoff planes , click New twice and enter:
h, k, l Distance: × d 1, 1, 1 2.17 −1, −1, −1 −1.10
Use the × d field , not the Å field. Leave Apply symmetry operations unchecked.
Click OK . Use the mouse wheel to enlarge the retained strip.
Two parallel cutoff planes let us examine adjacent gaps between oxygen layers.
Boundary x/y/z: −1 to 2 Upper plane (111), 2.17 × d Lower plane (−1 −1 −1), −1.10 × d Times-d field ; symmetry operations uncheckedRetains 1.10 ≤ x+y+z ≤ 2.17 for these coordinates Now read the cation planes between the red oxygen layers.
Identify the alternating cation planes
Orange = Mg · Blue = Al · Red = O
The lower gap contains one Al plane .
The next gap contains Mg–Al–Mg , a tetrahedral–octahedral–tetrahedral sandwich.
Count each plane separately. Atoms at different depths overlap in this side view.
↑ [111] Up: layer stacking
→ [1 1 −2] Right: along a layer
View along [1 −1 0] (111) oxygen layers seen edge-on
The two gaps have different cation arrangements.
Same orientation: view [1 −1 0] , up [111] , right [1 1 −2] (111) oxygen layers seen edge-on; stacking runs up the screenLower gap: Al only , octahedral coordination Upper gap: Mg–Al–Mg , tetrahedral–octahedral–tetrahedral Follow the red oxygen layers as boundaries Overlapping atoms: use isolated face-on planes to count Before isolating a single cation plane, stop VESTA from adding atoms outside the slice.
Prepare to count one plane
Open Edit → Bonds… .
For both Mg–O and Al–O , change Boundary mode to Do not search atoms beyond the boundary .
Keep Ball-and-stick selected.
In Orientation… , set projection to 1, 1, 1 and upward to 0, 1, 0 .
We now look straight at an isolated (111) plane.
A narrow counting slice must contain only atoms inside its boundaries.
Both bond rows: Do not search atoms beyond the boundary Ball-and-stick ; isolated cations have no oxygen shellProjection [111] ; upward [010] Move the two cutoff planes to select one plane First isolate the aluminum plane in the gap that contains no magnesium.
Count the Al-only plane
In Boundary , select each cutoff row and change its × d value:
Plane × d (1, 1, 1) 1.40 (−1, −1, −1) −1.35
The marked repeat contains four octahedral sites : three have Al; one is empty.
Al-only plane: ¾ filled.
Solid rings mark occupied sites. The dashed ring marks an empty hole; it is an annotation, not an atom.
This aluminum plane fills three of every four octahedral holes.
Cutoffs: +1.40 and −1.35 in the times-d fields Al plane at x+y+z = 1.375 Four candidate sites: 3 occupied + 1 empty 3/4 filling ; count an interior repeatNow move to the aluminum plane inside the magnesium–aluminum–magnesium sandwich.
Count Al in the sandwich
Keep the face-on view and change only the two cutoff distances.
Plane × d (1, 1, 1) 1.90 (−1, −1, −1) −1.85
Now the four-site repeat contains one Al atom and three empty octahedral holes .
Sandwich Al plane: ¼ filled.
Together, the two Al planes average (¾ + ¼)/2 = ½ .
The aluminum plane inside the sandwich fills only one of every four octahedral holes.
Cutoffs: +1.90 and −1.85 Al plane at x+y+z = 1.875 Four sites: 1 occupied + 3 empty Overall octahedral fraction: (3/4 + 1/4)/2 = 1/2 The sandwich also contains two magnesium planes, one on each side of this aluminum plane.
Count both Mg planes
Lower Mg plane: +1.77 / −1.73
Upper Mg plane: +2.02 / −1.98
Each pair gives the (111) / (−1 −1 −1) cutoff distances in the × d fields.
Each Mg plane fills ¼ of its tetrahedral sites. The alternating Al-only gap has no Mg .
Both magnesium planes fill one quarter of their tetrahedral sites.
Lower Mg: +1.77 / −1.73 ; plane at x+y+z = 1.75 Upper Mg: +2.02 / −1.98 ; plane at x+y+z = 2 Each plane: 1 occupied + 3 empty Alternating Al-only gap: no Mg Overall tetrahedral fraction: (1/4 + 0)/2 = 1/8 We can now check the complete layer sequence against the spinel formula.
Problem 1.31: check the complete solution
Between close-packed oxygen layers, repeat:
Al-only gap ¾-filled octahedral plane
Mg–Al–Mg gap ¼ tetrahedral · ¼ octahedral · ¼ tetrahedral
Hole type From the planes From MgAl₂O₄ Octahedral (¾ + ¼)/2 = ½ 2 cations / 4 holes Tetrahedral (¼ + 0)/2 = ⅛ 1 cation / 8 holes
The empty holes are part of ideal spinel . In the next lecture, we will describe defects relative to its expected site occupations.
SSMC, Problem 1.31; Lecture 7 VESTA tutorial. Views reproduced in VESTA 3.5.8.
The plane counts and the formula give the same overall occupations.
Alternate Al-only and Mg–Al–Mg gaps Octahedral: 1/2 ; tetrahedral: 1/8 Plane counts agree with MgAl₂O₄ Symmetry → sites → coordination → occupations Next lecture: defects relative to ideal site occupations Use the practice questions to check these counts before we start defects.
Practice questions
Work the Lecture 7 practice questions
before the next class. They cover the spinel and garnet structures, cation site preferences,
the perovskite framework, and the Goldschmidt tolerance factor.
Open the Lecture 7 practice questions
Every question carries a worked
explanation, so you can check your reasoning as you go.
Your homework is the Lecture 7 practice questions, on the course website at chem548.wileylab.org, or through the Practice link in the nav.
They cover what we did today: the spinel and garnet structures , the cation site preferences , the perovskite framework , the Goldschmidt tolerance factor
Work them before the next class Every question carries a worked explanation , so you can check your reasoning as you go
Next class we take these perfect structures and start putting defects into them.
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