## An Introduction to Structures and Types of Solids. Solids

An Introduction to Structures and Types of Solids Solids • • Amorphous Solids: – Considerable disorder in structure Crystalline Solids: – Highly regu...
Author: Oscar Ramsey
An Introduction to Structures and Types of Solids Solids • •

Amorphous Solids: – Considerable disorder in structure Crystalline Solids: – Highly regular arrangements of their components (atoms, ions, molecules) ⇒ Ordered Structures – Lattice: 3-dim system of points designating positions of components – Unit cell: smallest repeating unit of the lattice

Generating a Lattice from a Pattern

In (a) select any point in pattern, in (b) move out through the pattern, marking additional points and in (c) discard the original pattern and keep the grid work of points

Crystal Lattice and the Unit Cell

A primitive unit cell (red) contains 4 x ¼ = 1 lattice point. The centered unit cell (blue) contains two lattice points

Checkerboard: 2-dim analogy

There are 7 crystal systems and 14 types of unit cells in nature

Three Cubic Unit Cells and the Corresponding Lattices Note that only parts of spheres on the corners and faces of the unit cells reside inside the unit cell, as shown by the cutoff versions.

To define structure we need: a) type of crystal lattice – we concern only with cubic lattices here b) number of atoms of each kind in the unit cell – composition of unit cell c) coordination number for each atom – number of nearest neighbors Cell Type Simple cubic (sc) Body-centered cubic (bcc) Face-centered cubic (fcc)

Atoms per unit cell 1 2 4

Simple cubic: (8 corners of a cube)(1/8 of each corner atom within a unit cell)=1 net atom per unit cell. Body-centered cubic: it has one additional atom within the unit cell at the cube’s center. Face-centered cubic: on the faces of the cube, (6 faces of a cube)(1/2 of an atom within a unit cell) = 3 net face-centered atoms within a unit cell. Plus one more atom contributed by the corner atoms.

Face-Centered Cubic

Probing the Structure of Condensed Matter A well-defined beam of x-rays beam strike a thin film of a crystalline solid. Some of the x-rays are absorbed, some pass through unchanged and others are scattered at various 2θ angles. The 2θ scattering angle is the angle between the direction of the diffracted beam and the direction of the straight beam.

X rays scattered from two different atoms may reinforce (constructive interference) or cancel (destructive interference) one another

Incident

Reflected

Bragg Equation

xy + yz = nλ Reflection of X rays of wavelength λ from a pair of atoms in two different layers of a crystal. The lower wave travels an extra distance equal to the sum of xy and yz. If this distance is an integral number of wavelengths (n = 1, 2, 3, . . .), the waves will reinforce each other when they exit the crystal.

xy + yz = 2d sin θ

nλ = 2d sin θ n = integer (1,2,3,… λ= wavelength of the X rays d = distance between the atoms θ = angle of incidence and reflection

Example In a crystal of Ag, planes of Ag atoms are separated by a distance of 1.4446 Å. Compute the angles θ for all possible Bragg reflections for these planes if the wavelength of the incident X-radiation is 0.7093 Å.

Types of Crystalline Solids • • •

Ionic Solids – ions at the points of the lattice that describes the structure of the solid. Molecular Solids – discrete covalently bonded molecules at each of its lattice points. Atomic Solids – atoms at the lattice points that describe the structure of the solid.

Structure and Bonding in Metals Metals • High thermal conductivity • High electrical conductivity • Malleability (can be shaped into something else without breaking) • Ductility (can be hammered thin or stretched into wire without breaking) A metallic crystal can be pictured as containing spherical metal atoms packed together and bonded to each other equally in all directions. We can model such a structure by packing uniform hard spheres in a manner that most efficiently fills in the space. (closest packing)

The Closest Packing Arrangement of Uniform Spheres • •

abab packing – the 2nd layer is like the 1st but it is displaced so that each sphere in the 2nd layer occupies a dimple in the 1st layer. The spheres in the 3rd layer occupy dimples in the 2nd layer so that the spheres in the 3rd layer lie directly over those in the 1st layer.

The Closest Packing Arrangement of Uniform Spheres •

abca packing – the spheres in the 3rd layer occupy dimples in the 2nd layer so that no spheres in the 3rd layer lie above any in the 1st layer. The 4th layer is like the 1st.

Hexagonal Closest Packing

Cubic Closest Packing

fcc

The Indicated Sphere Has 12 Nearest Neighbors

Each sphere in both ccp and hcp has 12 equivalent nearest neighbors.

Silver crystallizes in a cubic closest packed structure. The radius of Ag atom is 1.44 Å (144 pm). Calculate the density of solid silver.

Body-centered Cubic Packing

Spheres touch along the body diagonal

unit cell with the center sphere deleted

One face of the bodycentered cubic unit cell. By the Pythagorean theorem, f2 = e2 + e2 =2 e2

The relationship of the body diagonal (b) to the face diagonal (f) and edge (e)

Bonding Models for Metals Consistent with main properties of metals (high thermal and electric conductivity, malleability and ductility) ⇒ strong and non-directional

Electron Sea Model

A regular array of cations in a “sea” of mobile valence electrons

21

Bonding Models for Metals •

Band Model or MO Model

Electrons are assumed to travel around the metal crystal in molecular orbitals formed from the valence atomic orbitals of the metal atoms.

The molecular orbital energy levels produced when various numbers of atomic orbitals interact. Note that for two atomic orbitals two rather widely spaced energy levels result. As more atomic orbitals become available to form MOs, the resulting energy levels become more closely spaced, finally producing a band of very closely spaced orbitals.

Mg crystal (hcp) 1s22s22p63s2

Bonding Models for Metals ←Conduction Band ← Valence Band

Thermal energy can excite electron from the filled MOs called (valence band) to the empty MOs (conduction band)

A representation of the energy levels (bands) in a magnesium crystal. The electrons in the 1s, 2s, and 2p orbitals are close to the nuclei and thus are localized on each magnesium atom as shown. However, the 3s and 3p valence orbitals overlap and mix to form MOs. Electrons in these energy levels can travel throughout the crystal.

Metal Alloys (contains a mixture of elements and has metallic properties) •

Substitutional Alloy – some of the host metal atoms are replaced by other metal atoms of similar size. ~1/3 of Cu replaced by Zn

White gold (75% Au, 25% Ag) •

Interstitial Alloy – some of the holes in the closest packed metal structure are occupied by small atoms. Carbon inclusions provide directional bonding to the undirectional iron, giving strength Mild steel: < 0.2% C (ductile and malleable) Medium steel: 0.2 -0.6% C (structural steel beams) High-C steel: 0.6 – 1.5% C (springs, cutlery, tools)

Network Atomic Solids • Contain strong directional covalent bonds Carbon allotropes (different forms): graphite, diamond, fullerenes (molelular solid)

all electrons are tied up in sp3 covalent bonds

Diamond

Different Metal

“band gap” diamond = 6eV no thermal excitation possible at 300 K. insulator 1 eV = 1.6 x10-19 J

Graphite (layers of C atoms) sp2 hybridized

Leftover unhybridized 2p forms π MOs

benzene

Directional conductivity

Molecular Solids Fullerenes, C60 , 60 carbon atoms, arranged as 12 pentagons and 20 hexagons

• Intermolecular forces: dipole-dipole, London dispersion and H-bonds. • Weak intermolecular forces give rise to low melting points.

CO2 I2, P4, S8

Ionic Solids • Ionic solids are stable, high melting point substances held together by the strong electrostatic forces that exist between oppositely charged ions. Three Types of Holes in Closest Packed Structures Trigonal holes are formed by three spheres in the same layer.

Tetrahedral holes are formed when a sphere sits in the dimple of three spheres in an adjacent layer.

Octahedral holes are formed between two sets of three spheres in adjoining layers of the For spheres of a given diameter, the holes closest packed structures. increase in size in the order: trigonal < tetrahedral < octahedral

What is the Radius of the Octahedral Holes

The diagonal of the square, d = R + 2r + R where r is the radius of the octahedral hole and R is the radius of the packed spheres.

What is the Radius of the Tetrahedral Holes

The center of the tetrahedral hole (shown in red) is at the center of the body diagonal b The four spheres around a tetrahedral hole are (shown in purple). shown inscribed in a cube. The spheres are shown much smaller than actual size. They One packed sphere actually touch along the face diagonal f. and its relationship to the tetrahedral hole. Note that (body diagonal)/2 = R + r.

Guidelines for Filling Tetrahedral and Octahedral Holes

A simple cubic array with X- ions, with an M+ ion in the center (in the cubic hole).

The body diagonal b equals R + 2r + R, since Xand M+ touch along this body diagonal.

Structures of Actual Ionic Solids −

The locations (gray X) of the octahedral holes in the facecentered cubic unit cell

Representation of the unit cell for solid NaCl. The Cl- ions (green spheres) have a ccp arrangement, with Na+ ions (gray spheres) filling all the octahedral holes. This representation shows the idealized closest packed structure of NaCl. In the actual structure, the Cl- ions do not quite touch.

Empirical formula can be related to the structure by counting the number of cations and anions contained in one unit cell. Na+ contained in a unit cell: 1 Na+ in the center of the unit cell) + (1/4 of Na+ in each edge x 12 edges) = net 4 Na+ in a NaCl unit cell. We repeat the same for Cl and the ratio of Na+/ Cl - =1

Example A unit cell of perovskite is shown below. What is its formula? Method: Identify the ions present in the unit cell and their locations within the unit cell. Decide on the net number of ions of each kind of the cell. Ca2+ are in the corners of the cube, a Ti 4+ in the center o the cell and O , a Ti 4+ in the center o the cell and O in the face centers. Ca2+ ions: (8 Ca2+ ions at cube corners)(1/8 of each ion inside unit cell) = 1 net Ca2+ ion No. of Ti 4+: one ion is in the cube center No. of O2- ions: (12 O2- in cube edges)(1/4 of each ion inside cell) = 3 net O2- ions in cue edges)(1/4 of each ion inside cell) = 3 net O2-

CaTiO3

Structures of Actual Ionic Solids

The location (red X) of a tetrahedral hole in the facecentered cubic unit cell

One of the tetrahedral holes.

The unit cell for ZnS, where the S2- ions (yellow) are closest packed, with the Zn2+ ions (purple) filling alternate tetrahedral holes

The unit cell for CaF2, where the Ca2+ ions (purple) form a face-centered cubic arrangement, with the F- ions (yellow) in all of the tetrahedral holes.

Lattice Defects

Schottky defects (there are vacant sites)

Frenkel defects (an atom or an ion of either sign is present at an inappropriate site.

Sometimes involve impurities ⇒ nonstoichiometric compounds

Example Assume the two-dimensional structure of an ionic compound MxAy is shown below. What is the empirical formula of this compound?

Assuming the anions A are the larger circles, there are four anions completely in this repeating square. The corner cations (smaller circles) are shared by four different repeating squares. Therefore, there is one cation in the middle of the square plus 1/4 (4) = 1 net cation from the corners. Each repeating square has two cations and four anions. The empirical formula is MA2.

Example The unit cell for nickel arsenide is shown below. What is the formula of this compound?

The unit cell contains 2 ions of Ni and 2 ions of As, which gives a formula of NiAs. 8 corners × (1/8)/corner + 4 edges ×1/4/edge = 2 Ni ions 2 As ions inside the unit cell

Example Show that the net composition of each unit corresponds to the correct formula of the compound