Atomic Structure, arrangement of atoms

Atoms bond to lower their energy. Bonds are primary (strong, 100-1000 kJ/mol) or secondary (weak, below about 40 kJ/mol).

Primary bonds

• Ionic bond: electron transfer between a metal and a non-metal gives opposite ions held by electrostatic attraction (NaCl, MgO). Bond is non-directional, hard, brittle, high melting, insulating in solid state.
• Covalent bond: atoms share electrons in a fixed direction (diamond, Si, SiC). Very hard, high melting, poor ductility.
• Metallic bond: valence electrons form a free "electron cloud" shared by positive ion cores (Cu, Fe). Gives ductility, good electrical and thermal conductivity, lustre.

Secondary bonds

• Van der Waals forces between temporary or permanent dipoles, and hydrogen bonds (H bonded to O, N, F). Responsible for bonding between polymer chains and in ice; they explain the low melting point of polymers.

Force and energy curves

Attractive force dominates at large distance and repulsive force at small distance. At equilibrium spacing $r_0$ the net force is zero and the potential energy is minimum.

 E
 |\                 
 | \             ___------  r
 0--\-----------/----------->
 |   \         /
 |    \_______/ <- minimum E at r0
 |            |
 |         -Eb (bond energy)

$$E = -\frac{A}{r^m} + \frac{B}{r^n}, \qquad F = \frac{dE}{dr} = 0 \ \text{at}\ r = r_0$$

Bond (binding) energy $E_b$ is the energy needed to separate two atoms from the equilibrium spacing $r_0$ to infinity, i.e., the depth of the energy well. Materials with deep, narrow wells have high melting point, high modulus and low thermal expansion.

A crystalline solid has atoms arranged in a regular, repeating three-dimensional pattern with long-range order; an amorphous solid has only short-range order.

Space lattice

A space lattice is an infinite three-dimensional array of points in which every point has identical surroundings. Each point is a lattice point; atoms are placed at or around these points.

Unit cell

The unit cell is the smallest group of atoms (smallest parallelepiped) that, when repeated in three dimensions, builds the whole crystal. It is described by the edge lengths $a, b, c$ and the angles $\alpha, \beta, \gamma$ between them. There are 7 crystal systems and 14 Bravais lattices; most metals crystallise as BCC, FCC or HCP.

Atomic packing factor (APF) is the fraction of the unit-cell volume occupied by atoms:

$$\text{APF} = \frac{n \times \frac{4}{3}\pi R^3}{a^3}$$

BCC (e.g., $\alpha$-Fe, Cr, W)

Atoms at the 8 corners and one at the body centre. Atoms touch along the body diagonal.

• Atoms per cell: $8 \times \tfrac{1}{8} + 1 = 2$
• Body diagonal $= \sqrt{3}\,a = 4R$, so $R = \dfrac{\sqrt{3}\,a}{4}$
• Coordination number = 8

$$\text{APF} = \frac{2 \times \frac{4}{3}\pi\left(\frac{\sqrt{3}a}{4}\right)^3}{a^3} = \frac{\sqrt{3}\,\pi}{8} = 0.68$$

FCC (e.g., $\gamma$-Fe, Cu, Al, Ni)

Atoms at the 8 corners and the centre of each of the 6 faces. Atoms touch along the face diagonal.

• Atoms per cell: $8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4$
• Face diagonal $= \sqrt{2}\,a = 4R$, so $R = \dfrac{\sqrt{2}\,a}{4}$
• Coordination number = 12

$$\text{APF} = \frac{4 \times \frac{4}{3}\pi\left(\frac{\sqrt{2}a}{4}\right)^3}{a^3} = \frac{\pi}{3\sqrt{2}} = 0.74$$

HCP (e.g., Zn, Mg, Ti, Co)

Two hexagonal layers (top and bottom) with a central atom each, plus a triangular layer of 3 atoms midway between them. Atoms per cell = $12 \times \tfrac{1}{6} + 2 \times \tfrac{1}{2} + 3 = 6$. Ideal axial ratio $c/a = 1.633$, coordination number 12, APF = 0.74.

FCC and HCP are both close-packed, differing in stacking sequence (ABCABC for FCC, ABABAB for HCP).

Formula: theoretical density

$$\rho = \frac{n A}{V_c N_A}$$

where $n$ = atoms per unit cell, $A$ = atomic mass, $V_c = a^3$, $N_A$ = Avogadro's number.

(a) Copper (FCC, $n = 4$)

Atomic radius, from $4R = \sqrt{2}\,a$:

$$R = \frac{\sqrt{2}\times 0.3615}{4} = 0.1278\ \text{nm}$$

Cell volume: $a = 0.3615 \times 10^{-7}$ cm, so $V_c = (0.3615\times10^{-7})^3 = 4.724\times10^{-23}\ \text{cm}^3$.

$$\rho = \frac{4 \times 63.54}{(4.724\times10^{-23})(6.022\times10^{23})} = \frac{254.16}{28.45} = 8.93\ \text{g/cm}^3$$

(b) Tungsten (BCC, $n = 2$)

$a = 0.3165\times10^{-7}$ cm, so $V_c = 3.171\times10^{-23}\ \text{cm}^3$.

$$\rho = \frac{2 \times 183.85}{(3.171\times10^{-23})(6.022\times10^{23})} = \frac{367.7}{19.10} = 19.26\ \text{g/cm}^3$$

The copper value agrees with the handbook value of 8.94 g/cm$^3$ and tungsten with 19.25 g/cm$^3$.

Answer: (a) $R = 0.128$ nm, $\rho_{Cu} = 8.93$ g/cm³ (8930 kg/m³); (b) $\rho_{W} = 19.26$ g/cm³ (19 260 kg/m³).

Miller indices $(hkl)$ are a set of three smallest integers that describe the orientation of a crystal plane.

Procedure for planes

1. Find the intercepts of the plane on the $x, y, z$ axes in terms of lattice constants $a, b, c$ (use $\infty$ if parallel).
2. Take the reciprocals of the intercepts.
3. Clear fractions by multiplying by the smallest common factor.
4. Write the result in round brackets without commas, $(hkl)$; a negative index has a bar over it.

(a) Intercepts 2, 3, 6

Plane is $(321)$.

(b) Intercepts 1/2, 1, $\infty$

Plane is $(210)$.

Directions

Procedure: subtract the coordinates of the tail from the head, clear fractions to the smallest integers, and write in square brackets $[uvw]$.

• Origin to $(\tfrac12, 1, 0)$: components $(\tfrac12, 1, 0)$; multiply by 2, giving $[120]$.
• $(1,0,0)$ to $(0,1,1)$: components $(0-1,\ 1-0,\ 1-0) = (-1, 1, 1)$, written $[\bar{1}11]$.

Answer: (a) (321); (b) (210); directions [120] and $[\bar{1}11]$.

Planar density = atoms centred on the plane per unit area of plane; linear density = atoms centred on the line per unit length.

(a) Planar density

(100) plane: a square of side $a$. Four corner atoms each contribute $\tfrac14$, so 1 atom.

$$\text{PD}_{(100)} = \frac{1}{a^2} = \frac{1}{(0.2866)^2} = 12.17\ \text{atoms/nm}^2$$

(110) plane: a rectangle $a \times \sqrt{2}\,a$. Four corner atoms ($4\times\tfrac14 = 1$) plus the body-centre atom, which lies on this plane, give 2 atoms.

$$\text{PD}_{(110)} = \frac{2}{\sqrt{2}\,a^2} = \frac{2}{1.4142 \times 0.08214} = 17.22\ \text{atoms/nm}^2$$

(b) Linear density

For [111] the body diagonal passes through two corner atoms and the centre atom, so 2 atoms are counted in the repeat length $\sqrt{3}\,a$.

Close-packed plane and direction

The densest plane is (110) (17.22 atoms/nm$^2$) and the densest direction is [111] (4.03 atoms/nm). Slip needs the least shear stress on close-packed planes and along close-packed directions, because the atomic spacing is smallest along the direction (small Burgers vector) and planes are most widely spaced. Hence in BCC metals slip occurs on {110}<111> systems (and also on {112}, {123}).

Answer: PD(100) = 12.17 atoms/nm², PD(110) = 17.22 atoms/nm²; LD[100] = 3.49, LD[110] = 2.47, LD[111] = 4.03 atoms/nm; close-packed plane (110), direction [111].