Physical Chemistry I. Crystal Structure

Physical Chemistry I Crystal Structure

Crystal Structure Introduction Crystal Lattice Bravis Lattices Crytal Planes, Miller indices Distances between planes Diffraction patters Bragg s law X-ray radiation Metal Crystals Ionic Crystals Quasi Crystals

Introduction States of Matter: Gas, liquid, solid Solids: Aggregated atoms, ions or molecules What is the difference between crystalline solids and amorphous solids Crystalline solids: Highly ordered array of atoms/ions Amorphous solids: Without order

Crystal Lattice Array of atoms, ions or molecules is called a lattice

Bravais Lattices Each unit cell is composed of lengths (a,b,c) and angles between them (α, β, γ) a b c β α γ

Types of Unit Cells Simple Cubic a = b = c α = β = γ = 90 Tetragonal a = b c α = β = γ = 90 Orthorhombic a b c α = β = γ = 90 Rhombohedral a = b = c α = β = γ 90 Monoclinic a b c α = γ = 90, β 90 Triclinic a b c α β γ 90 Hexagonal a = b c α = β = 90, γ = 120

Atom Positions in Cells In addition to atoms/ions (lattice points) being located at the corners they can also be located at the unit face or center A cell with lattice points only at the corner is called simple

Atom Positions in Cells In addition to atoms/ions (lattice points) being located at the corners they can also be located at the unit face or center A cell with lattice points also at the faces is called face-centered

Monoclinic C Triclinic Hexagonal Trigonal Rhombic 14 Bravais Lattices 7 crystal systems with 3 types of crystal (simple) P, (face-centered) F and (body centered) I generate 14 Bravais Lattices Cubic P Cubic I Cubic F Tetragonal P Tetragonal I Orthorhombic P Orthorhombic C Orthorhombic I Orthorhombic F Monoclinic P

1a 3a Crystal Planes Lattice Vectors There are multiple ways to connect lattice points to give crystal planes -1a 1b 1b a 2b 1b

Crystal Planes Lattice Vectors The connecting points are denoted as (1 a,1 b) or just (1, 1) +x +y (1,1) What crystal planes are these? (3,2) (-1,1)

Crystal Planes Lattice Vectors The connecting points are denoted as (1 a,1 b) or just (1, 1) +x What crystal plane is this? +y (,1)

3D Lattice Miller Indices Extending to the third axis (z) is done by adding a third index 1a-1b is (1,1, ) Typically in crystallography, miller indices are used instead of lattice vectors In this system infinity is more easily represented Miller indices are the reciprocal of lattice vectors They are typically multiplied such that the greatest common divisor is 1 Negative numbers such as (-1,1,0) are represented as (1,1,0)

Miller Indices Lattice Vector Miller Indices (1,1, ) 1 1, 1 1, 1 (1 1 0) (3,2, ) 1 3, 1 2, 1 (2 3 0) (-1,1, ) 1 1, 1 1, 1 (1 1 0) (,1, ) 1, 1 1, 1 (0 1 0)

Distance Between Planes For cubic lattices we can work out the distance between any two planes defined in miller indices using the following: 1 d 5 = h5 a 5 + k5 b 5 d = 1 h 5 a 5 + k5 b 5 d

Distance Between Planes To explain this we will start from the plane in miller indices (h, k, 0) = (2, 3, 0) Recall that h and k are inverse lattice vectors We can get the x, y distance between planes with x = a h, y = b k b k a h

Distance Between Planes Because of the cubic cell angles θ are the same The plane distance d can be found via trigonometry, b k θ d a h θ

Distance Between Planes For this upper triangle we use cos θ Since we have the hypotenuse and we want the adjacent b k d = b cos θ k θ d a h θ

Distance Between Planes For this lower triangle we use sin θ Since we have the hypotenuse and we want the opposite d = a sin θ h b k θ d a h θ

Distance Between Planes Rather than trying to work out θ (which you can do with cosine and sine rules), we use the following relationship: cos 5 θ + sin 5 θ = 1 b k θ d a h θ

Distance Between Planes We then do a little bit of algebra d = b dk cos θ cos θ = k b d = a h dh a sin θ sin θ = dh a 5 1 d 5 = + dk b h a 5 5 + k b = 1 5 b k θ d a h θ

Distance Between Planes In 3D, using (h,k,l), the solution is very similar for the cubic lattice: 1 d 5 = h5 a 5 + k5 b 5 + l5 c 5

Distance Between Planes Working through an example: For the plane (123) and (246) if an orthorhombic cell has the parameters a = 0.82nm b = 0.82nm and c = 0.75nm what is the lattice separation 1 d 5 = h5 a 5 + k5 b 5 + l5 c 5 1 d 5 = 15 0.82 5 + 25 0.82 5 + 35 0.75 5 = 22.0nm 1 d = 1 22 = 0.21nm

Distance Between Planes Working through an example: For the plane (123) and (246) if an orthorhombic cell has the parameters a = 0.82nm b = 0.82nm and c = 0.75nm what is the lattice separation 1 d 5 = h5 a 5 + k5 b 5 + l5 c 5 1 d 5 = 25 0.82 5 + 45 0.82 5 + 65 0.75 5 = 88nm 1 d = 1 88 = 0.10nm

Diffraction Patterns Waves such as electromagnetic waves interfere with each other in constructive or destructive manners

Diffraction Patterns When light passes through a slit larger than its wavelength, it will diffract. A diffraction pattern of light is made from the constructive and destructive interference To get a diffraction pattern of a crystal lattice we need to use a wave with a sufficiently small wave length.

X-ray Diffraction X-rays have a wavelength of the order of 100 pm, at this wavelength we can observe spacing between lattice points. As a result when x-rays pass through a crystal lattice a diffraction pattern is observed http://photojournal.jpl.nasa.gov/jpeg/pia16217.jpg

Bragg s Law We can use the diffraction pattern to discern the lattice spacing d d

Bragg s Law The extra distance the light travels is given by 2l which is called the path-length θ l θ l d

Bragg s Law When the diffraction pattern is constructive then the extra distance the light travels is a multiple of the wavelength nλ 2l = nλ From the trigonometry it can be seen that: 2l = 2d sin θ Therefore nλ = 2l = 2d sin θ We then include n in d and get Bragg s law λ = 2d sin θ θ l θ l d

Bragg s Law λ = 2d sin θ As an example, a reflection on the (111) plane of a cubic cell was observed at a glancing angle of 11.2 when the wavelength was 154pm, what is the length of the unit cell a d = λ 2 sin θ d = 154 2 sin 11.2 = 396.4

Bragg s Law λ = 2d sin θ To get the unit cell distance a we combine d with our previous equation 1 d 5 = h5 + k 5 + l 5 a 5 1 d 5 = 15 + 1 5 + 1 5 a 5 = 3 a 5 a = 3d a = 396.4 3 a = 687pm

Metal Crystal The atoms in metals tend to pack as close as possible in a close packed structure These structures typically have a coordination number of 12 (or close to 12) Two lattice structures form a close packed structure, cubic close close pack (ccp) and hexagonal close pack (hcp) Cubic close packing (AKA, face centered cubic FCC) Hexagonal close packing (AKA, body centered hexagonal)

Metal Crystal Due to their structure CPP has 8 slip plans whist HPC has 1 slip plane This means: CPP metals are malleable HPC metals are brittle Cubic close packing (AKA, face centered cubic FCC) Hexagonal close packing (AKA, body centered hexagonal)

Ionic Crystals The different sizes of the atoms in ionic crystals and the need for charge neutrality means that close packing structures are not possible Ionic crystals tend to have coordination numbers of 8 (with the cesium chloride structure) and 6 (with the rock salt structure) Cesium-Chloride structure Rock-salt structure

Ionic Radii The ratio of radii of the cations and anions can be estimated from the structure of ionic solids r T The distance labelled here as l V spans the radii of r U, r T, r T, r U a r U l V r T a r T a

Ionic Radii The ratio of radii of the cations and anions can be estimated from the structure of ionic solids r T We can find this l V with some trigonometry a r U l V d V a d V = a 5 + a 5 = 2a l V = a 5 + 2a 5 r T a r T a l V = a 5 + 2a 5 = 3a 5 = 3a

Ionic Radii The ratio of radii of the cations and anions can be estimated from the structure of ionic solids r T The distance l 5 is a which spans the radii r U, r U a r U r T a l 5 r T a

Ionic Radii The ratio of radii of the cations and anions can be estimated from the structure of ionic solids a r T r U a l 5 l V rt r T a l V = 2r T + 2r U = 3a l 5 = 2r T = a 2r U + 2r T = 2 3r T r T + r U = 3 r T r T 1 + r U r T = 3 r U r T = 3 1 = 0.732

Ionic Radii For rock-salt we can perform the following. The distance labelled here as l V spans the radii r U, r U and a spans r U, r T r T l V = 2a 5 = 2a = 2r T a = 2 2 r T = 2r T l V r U r T a a a = r U + r T 2r T = r U + r T 2 = r T + r U r T r T r U = 2 1 = 0.414 r T

Ionic Radii Ionic crystals with radii ratios of: W X W Y > 0.732 have CsCl structure 0.732 > W X W Y > 0.414 have rock-salt structure For even more extreme ratios i.e. W X W Y < 0.414 then the crystal adopts the Zinc blende structure with coordination of 4 Zinc blende structure

Quasicrystals Quasicrystals have order but are not periodic This means to translation can match the original shape exactly Aluminum allows such as Al-Li-Cu, Al- Mn-Si, Al-Ni-Co etc have these structures These are two types of quasicrystal, polygonal and icosahedral quasicrystals

Questions 1. From these lattice vectors what are the miller indices 1. 1a 3b 1c 2. 2a 3b c 3. -1a -1b 2c 2. For a cubic lattice where a = 150nm and the following planes what are the lattice separations d. 1. ( 1 1 1 ) 2. ( 2 3 1) 3. ( 0 1 1)

Questions 3. A x-ray is being aimed at a lattice with wavelength 150nm, which of the following path lengths will have completely constructive interference 1. 300nm 2. 100nm 3. 150nm 4. Given the ionic radii, Li + = 90pm, Na + = 116pm, K + = 152pm, F - = 119pm, Cl - = 167pm and Br - = 182pm. Based on the ratio of ionic radii what structure will these ionic lattices take 1. LiF, 2. NaCl, 3. KBr, 4. LiCl, 5. LiBr, 6. NaBr

Questions 5. An x-ray, of wavelength 150pm aimed at a cubic lattice with the following incidence angles and planes, what are the lattice separations d and the lattice parameters a 1. 10.9 1 1 0 2. 11.3 1 1 1 3. 15 2 3 0 4. 10.9 ( 1 1 1 )

Homework Question 1. For a lattice with the following parameters, a = 700pm, b = 649pm, c = 800pm, α = 95, β = 85, γ = 120 what is the lattice separation distance d for the plane 2 3 0. Show your workings. Hint, the formulae shown earlier only works for angles of 90. Think of this as a 2D lattice and use cosine and sine rules