cation vs anion

7/30/2019 cation vs anion

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Chem 484 Lecture Notes, Weeks 1-2

1. Lattice:

A lattice is aperiodic array of dots (or lattice points). It is a mathematic abstraction used to

describe the translational symmetry of a periodic (or extended) structure. The translationalsymmetry implies that the system has infinite repetition. In reality, we have to deal with crystalswith finite sizes.

Lattice parameters: constants (a, b, c) and angles (, , ).

We have three types of period systems:

One-dimensional (1-D), two-dimensional (2-D), and three-dimensional (3-D).

Crystal Structure = Lattice + Basis of the Lattice Point

2. Unit Cell:

The simplest portion of a lattice that can be repeated by translation to cover the entire 1-D, 2-D,or 3-D space.

The primitive cell (P) only contains one lattice point.

In general, we choose the unit cell such that it will reflect the symmetry of the lattice.

3. Two-Dimensional Lattices and Unit Cells:

We only have five different types of unit cells, four of them are primitive cells, and the other oneis centered rectangular.

a

a

90o

a

b

90o

a

b

90o

square lattice rectangular lattice rectangular latticesquare unit cell rectangular unit cell centered rectangular unit cell


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a a120

o a

b

hexagonal lattice, unit cell parallelogram lattice, unit cell

Example#1: a tile floor with the chess board pattern

Example#2: graphite sheet

http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Strucsol.html

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http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Coords/2Dhex.cmdf


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4. Three-Dimensional Lattices and Unit Cells:

Depending on the combination of different a, b, c; and , , , we only have 7 different crystalsystems: cubic, tetragonal, orthorhombic, hexagonal, trigonal, monoclinic, and triclinic.

We can only have 14 types of unique lattices, termed theBravais lattices. Seven of them areprimitive cells (P), while the body-centered (I) cell contains two lattice points; the face-centered(F) cell contains 4 lattice points; and the base-centered (C) cell contained 2 lattice points.

Crystal system Bravais latticescubic P, I, Ftetragonal P, Iorthorhombic P, I, F, Chexagonal Ptrigonal Pmonoclinic P, C

triclinic P

*Note that the tetragonal C unit cell can be reduced to a tetragonal P cell; the tetragonal F unitcell can be reduced to a tetragonal I cell.

The 14 Types of Bravais Lattices (or Unit Cells)

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5. The Crystal Structures of Metals:

Approximately two thirds of the elements in the periodic table are metals. Many of them havefound technologically important applications.

Because the interactions between metal atoms are highly isotropic, each atom prefers to havemaximum contact with the neighbors. The maximum number of an atom that can be in contactwith its neighbors is 12 (6 in the same plane, 3 above the plane, and 3 below the plane)

Metals are crystallized in four crystal structures: simple cubic (sc); body-centered cubic (bcc);face-centered cubic (fcc) or cubic-close-packing (ccp); and hexagonal-close-packing (hcp).

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6. Simple Cubic (sc):

The unit cell is a primitive one, with one lattice point (atom) per unit cell.

If we assume the radius of the atom is r and the lattice constant is a, we have:

a = 2 r (i.e., the adjacent atoms touch each other along the edge of the unit cell)

Vatom = 8 (1/8) (4r3)/3 = (4r3)/3

Vcell = (2 r)3 = 8 r3

The packing efficiency (or fraction of packing):

Vatom / Vcell = (4r3/3) / (8 r3) = /6 = 52.4%

The only example is polonium (Po). In this metal, 47% of the space is occupied by Po atoms, therest (47.6%) are void spaces.

(unit cell)

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7. Body-Centered Cubic (bcc):

The lattice is not a primitive one since there are 2 lattice points (atoms) per unit cell, one at thevertices (one eighth at each vertex) and other at the center of the cubic cell.


3 a = 4 r (i.e., the adjacent atoms touch each other along the body diagonal of the unit cell)

Vatom = 2 (4r3)/3 = 8r3/3

Vcell = (a)3 = (4r/3)3


Vatom / Vcell = (8r3

/3) / (4r/3)3

= 3/8 = 68.0%

A typical example is iron (Fe), which is a magnetic material.

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8. Cubic-Close-Packing (ccp) and Face-Centered Cubic (fcc):

The lattice is not a primitive one since there are 4 lattice points (atoms) per unit cell, one at thevertices (one eighth at each vertex) and three other on the 6 faces.


2 a = 4 r(i.e., the adjacent atoms touch each other along the face diagonal of the unit cell)

Vatom = 4 (4r3)/3 = 16r3/3

Vcell = (a)3 = (4r/2)3


Vatom / Vcell = (16r3/3) / (4r/2)3 = 2/6 = 74.0%

This is the highest packing efficiency for a cubic lattice. This is also known as cubic-close-packing (or ccp) or cubic-closest-packed (or ccp) structure.

A typical example is silver (Ag), which is an important material in photography. Cubic-close-packing is also favored by inert gases when they are cooled down to crystallize (e.g., neon below25 K and argon below 84 K). This is because the van der Waals forces between these atoms arealso highly isotropic.

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9. Hexagonal-Close-Packing (hcp):

The lattice is not a primitive one since there are two lattice points (atoms) per unit cell, one at thevertices (one eighth at each vertex) and the other within the cell.

If we assume the radius of the atom is r and the lattice constant is a in the lateral directionand c in the vertical direction, we have:

a = 2 r (i.e., the adjacent atoms touch each other along the horizontal side the unit cell)

c = (4 2 r)/3

Vatom = 2 (4r3)/3 = 8r3/3

Vcell = (a) (a3/2) c = 82 r3


Vatom / Vcell = (8r3/3) / (82 r3) = /(32) = 74.0%

This is also called hexagonal-close-packing (hcp).

A typical example is cobalt (Co), which is also a magnetic material.


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http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Coords/Met-hcp.cmdf


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10. The Crystal Structures of Ionic Crystals:

We only focus on binary compounds formed between a metal and a nonmetal.

Note that the anions are negatively charged, and need cations between them to stable the lattice.

Because anions are much larger in size as compared to cations, the anions will form a simplecubic (sc), cubic-close-packed (ccp), or hexagonal-close-packed (hcp) lattice, while the cationswill be situated in the interstitial sites (i.e., the holes formed between anions).

The specific structure (or lattice) formed and the location of cation (or type of hole) are mainlydetermined by the radius ratio () between the cation and the anion: =r+/r-

type of hole coordination number (CN)

0.22 0.41 tetrahedral 4 (ccp or hcp)

0.41 0.73 octahedral 6 (ccp or hcp)

0.73 1.00 cubic 8 (sc)

*Coordination number (CN) is defined as the total number of the nearest neighbors in theopposite charge.

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11. Derivation of the Size Range for Cubic Hole:

Note that the cation and anion are in physical contact along the body diagonal direction. So wehave:

(2 r-)

2

+ (

2 2 r-)

2

= (2 r+ + 2 r-)

2

therefore

(3 -1) r- = r+

= r+/r- = 3 -1 = 0.73

This result indicates that the cation has to be at least (0.73r-) in radius in order to make physicalcontact with the 8 anions around it, and thus to stabilize the structure.

The color figure is a cross-sectional view of the cubic lattice along the face diagonal direction.For the rectangular unit cell, the length of short axis is (2r-), which is the same as the edge lengthof the cubic cell; the length of long axis is (2 2 r-), which is the face diagonal of the cubic cell.

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12. Derivation of the Size Range for Octahedral Hole:

Note that the cation and anion are in physical contact along the diagonal direction in the colorfigure (below). So we have:

(2 r-)

2

+ (2 r-)

2

= (2 r+ + 2 r-)

2

therefore

(2 -1) r- = r+

= r+/r- = 2 -1 = 0.41

This result indicates that the cation has to be at least (0.41r-) in radius in order to make physicalcontact with the 6 anions around it, and thus to stabilize the structure.

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13. Derivation of the Size Range for Tetrahedral Hole:

Put the vertices of a tetrahedra at the four corners of a cube, and the face diagonal of such a cubewill be (2r-). The edge length of such a cubic cell will be: a = (2r-)sin45

o = 2 r-.

The body diagonal of such a cubic cell will be

3a =

6 r-.

Note that the cation ion and anion are in physical contact along the body diagonal of this cube, sowe have:

(r+ + r-) = (6 r-)/2

and = r+/r- = 6/2 -1 = 0.22

14. The Symmetries of a Lattice:

All lattices have translational symmetry: that is, all the lattice points can be generated when theunit cell is shifted in space by ua, vb, wc, or a combination of them (here u, v and w are arbitraryintegers;a,b, andc are the translational vectors of the unit cell, with their absolute values beingthe lattice constants).

Depending on the crystal system and lattice type, a lattice may also have other symmetries suchas rotational axis, mirror plane, and inversion center. An n-fold rotational axis is defined suchthat the lattice points will be superimposed when the lattice (or unit cell) is rotated by 360o/ndegrees against the axis.

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15. Fractional Coordinates of Ions:

It is a simple way to specify the locations of ions in a unit cell. In a typical procedure, the ion isprojected onto the three axes (a,b,c) of the unit cell, and the intercepts (x, y, z) are measured interms of the lattice constants a, b, and c. The fractional numbers x, y, and z are placed in curved

brackets and referred to as the fractional coordinates of the ion. As a general rule, we only needto specify the fractional coordinate of one of the many ions that are related by the symmetry ofthe crystal lattice.

For CsCl crystal, we only need to provide Cs+(, , ) and Cl-(0, 0, 0). It is not necessary toprovide fractional coordinates for other chloride ions such as Cl-(1, 0, 0), Cl-(0, 1, 0), Cl-(0, 0, 1),Cl-(1, 1, 0), Cl-(1, 0, 1), Cl-(0, 1, 1), and Cl-(1, 1, 1), because these ions can be generated throughthe translational operation. Note: Do not forget to put a comma between the fractional numbers,otherwise, it will be confused with the Miller indices used to specify a crystal plane.

16. Miller Indices:

It is introduced to specify the plane or direction in a crystal lattice. For a crystal plane, the Millerindices can be obtained through the following three steps:

a) Find the intersections of the plane on the three axes (a,b, andc) of the unit cellb) Represent the intersections as fractions of the lattice constant, e.g., a/h; b/k; c/lc) Take reciprocals of the fractions, and put them in a pair of curved brackets (hkl)

Note that there is no comma between the numbers. The notion of (hkl) represent a set of parallelplanes with their interplane spacing being dhkl. Do not multiply or divide the numbers h, k, and lby/with a common factor as (100) and (200) represent different crystal planes. One should usethe translational symmetry of a lattice to avoid planes that pass through the origin of the unit cell.For planes that are equivalent to each other due to the symmetry of the lattice, they can be simplydescribed by placing the numbers in a pair of curly brackets: {hkl}. For example, {100}represents all the six faces of a cubic lattice, not just one of them.

For different directions in a crystal, we can use a similar system to index them. In this case, wedraw a line passing through the origin of the unit cell, pick any point (other than the origin) fromthe line and find the fractional coordinates of this point x, y, and z. By multiplication or divisionwith a common factor, find the set of smallest possible integers h, k, and l, and put them in a pairof squared brackets: [hkl]. Note that [110] and [330] represent the direction and we commonlyuse [110]. For directions related to the symmetry of the lattice, we put the numbers in anglebrackets: .

17. The d-Spacing (dhkl)

It gives the separation between (hkl) planes. For an orthorhombic lattice with lattice constants ofa, b, and c, the formula can be written as: (1/dhkl

2) = h2/a2 + k2/b2 + l2/c2This equation can also be extended to other orthogonal lattices such as cubic and tetragonallattices.

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16. Examples:

CsCl Crystal: r(Cs+) = 1.69 r(Cl-) = 1.81 = 0.93

So the anions of this crystal are packed in a single cubic (sc) structure, with Cs+ cations situated

in the cubic holes. The cubic hole has to expand a little bit in order to host the Cs+

cation.

NaCl Crystal: r(Na+) = 1.16 r(Cl-) = 1.81 = 0.64

So the anions of this crystal form a cubic-close-pacing structure, with Na+ cations situated in theoctahedral hole. The octahedral hole has to expand in order to host the Na+ cation.

ZnS Crystal: r(Zn2+) = 0.88 r(S2-) = 2.19 = 0.40

The cations are situated in the tetrahedral holes, while the S2- anions are packed into a ccp or hcplattice, which is called zinc blende or wurtzite structure, respectively. The tetrahedral hole has to

expand in order to host the Zn2+

cation.

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CsCl Crystal Structure

Note that this is a single cubic (sc) lattice; with each lattice point represents the unit CsCl.

NaCl Crystal Structure

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http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Coords/NaCl.cmdf


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ZnS Zinc Blende (Sphalerite) Structure

ZnS Wrtzite Structure


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http://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Coords/ZnBlende.cmdfhttp://www.chem.ox.ac.uk/icl/heyes/structure_of_solids/Coords/ZnWurz.cmdf

cation vs anion

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