Second Edition (
2001 McGraw-Hill)
Chapter 1
1.2 Ionic bonding and NaCl
The interaction energy between Na+ and Cl- ions in the NaCl crystal can be written as
E(r)= −4.03×10−28
r +
6.97×10−96
r8
where the energy is given in joules per ion pair, and the interionic separation r is in meters. Calculate the binding energy and the equilibrium ionic separation in the crystal; include the energy involved in electron transfer from Cl- to Na+. Also estimate the elastic modulus Y of NaCl given that
Y ≈ 1 6ro d2E dr2 r=ro
Solution
The PE curve for NaCl is given by E(r)= −4.03×10
−28
r +
6.97×10−96
r8
where E is the potential energy and r represents interionic separation.
We can differentiate this and set it to zero to find the minimum PE, and consequently the minimum (equilibrium) separation (ro).
dE(ro) dro = 0 ∴ −55.76 1 1096ro9 + 4.03 1 1028ro2 = 0 isolating ro: ro = 2.81 × 10-10 m or 2.81 Å
Let q = 1.602×10-19 J/eV. Then the apparent “ionic binding energy” (E
b) for the ions alone, in eV, is Eb(eV)= −E (ro) q = − −4.03×10−28 r + 6.97×10−96 r8 1 q o o ∴ Eb(eV)= − −4.03×10−28 2.81×10−10 m
(
)
+ 6.97×10 −96 2.81×10−10 m(
)
8 1 1.602×10−19 J/eV ∴ Eb(eV) = 7.83 eVNote however that this is the energy required to separate the Na+ and Cl- ions in the crystal and then to take the ions to infinity, that is to break up the crystal into its ions. The actual bond energy involves taking the NaCl crystal into its constituent neutral Na and Cl atoms. We have to transfer the
electron from Cl- to Na+. The energy for this transfer, according to Figure 1Q2-1, is -1.5 eV (negative represents energy release). Thus the bond energy is 7.83 - 1.5 = 6.33 eV as in Figure 1Q2-1.
Cl– Na+ ro = 0.28 nm 6 –6 0 –6.3 0.28 nm
Potential energy E(r), eV/(ion-pair)
Separation, r 1.5 eV r= Cl r= Na Na+ Cl– Figure 1Q2-1
Sketch of the potential energy per ion pair in solid NaCl. Zero energy corresponds to neutral Na and Cl atoms infinitely separated.
If r is defined as a variable representing interionic separation, then Young’s modulus is given
by: Y = 1 6r d dr dE(r) dr ∴ Y=1 6 −8.06 1 1028r3 + 501.84 1 1096r10 r = 8.364 ×10 −95 1 r11−1.3433×10 −28 1 r4
Substituting the value for equilibrium separation (ro) into this equation (2.81 × 10-10 m),
Y = 7.54 × 1010 Pa = 75 GPa
This value is somewhat larger than about 40 GPa in Table 1.2 (in the textbook), but not too far out.
Author’s Note: As explained in the textbook’s CD, in “Bonding in Molecules and Solids” , in general, the bulk modulus K can be found from the PE curve E = E(r) using
K = 1 9cro d2E dr2 r= ro o
where c is a constant of the order of unity that depends on the crystal structure. (For example, for the CsCl, c = 1.5 and NaCl, c = 2.) The elastic modulus Y and the bulk modulus K are related through the Poisson’s ratio υ,
K = Y/[3(1 − 2υ)]
For many ceramics υ ≈ 1/4 so that Y ≈ (3/2)K and hence
Y= 1 6cr d2E dr2 r= ro
A detailed analysis would make c = 2 for the NaCl structure, and hence would cut the calculate Y = 75 GPa to Y = 37.5 GPa close to the experimental value.
*1.3 van der Waals bonding
Below 24.5 K, Ne is a crystalline solid with an FCC structure. The interatomic interaction energy per atom can be written as
E(r)= −2ε 14.45 σ r 6 −12.13 σ r 12 (eV/atom)
where ε and σ are constants that depend on the polarizability, the mean dipole moment, and the extent of overlap of core electrons. For crystalline Ne, ε = 3.121 × 10-3 eV and σ = 0.274 nm.
a. Show that the equilibrium separation between the atoms in an inert gas crystal is given by ro = (1.090)σ. What is the equilibrium interatomic separation in the Ne crystal?
b. Find the bonding energy per atom in solid Ne.
c. Calculate the density of solid Ne (atomic mass = 20.18 g/mol).
Solution
a Let E = potential energy and x = distance variable between the atoms. The energy E is given by E(x)= −2ε 14.45 σ x 6 −12.13 σ x 12
The force F on each atom is given by
F(x)= −dE(x) dx = 2ε 145.56 σ σ x 11 x2 −86.7 σ σ x 5 x2 ∴ F(x)= 2ε 145.56σ 12 x13 − 86.7 σ6 x7
When the atoms are in equilibrium, this net force must be zero. Using ro to denote equilibrium separation, F(r )o = 0 2ε 145.56σ 12 r13 ∴ −86.7σ6 r7 o = 0 o 145.56 ∴ σ1213 = 86.7σ67 ro ro ∴ ro13 7 ro = 145.56 86.7 σ 12 6 σ ∴ ro = 1.090σ
For the Ne crystal, σ = 2.74 × 10-10 m and ε = 0.003121 eV. Therefore,
b Calculate energy per atom at equilibrium: E(ro)= −2ε 14.45 σ ro 6 −12.13 σ ro 12 ∴ E(ro)= −2 0.003121 eV
(
)
(
1.602×10−19 J/eV)
14.45 2.74×10-10 m 2.99×10-10 m 6 −12.13 2.74×10-10 m 2.99×10-10 m 12 ∴ E(ro) = -4.30 × 10-21 J or -0.0269 eVTherefore the bonding energy in solid Ne is 0.027 eV per atom.
c To calculate the density, remember that the unit cell is FCC, and density = (mass of atoms in the unit cell) / (volume of unit cell). There are 4 atoms per FCC unit cell, and the atomic mass of Ne is 20.18 g/mol. a a a 2R Figure 1Q3-1
Left: An FCC unit cell with close-packed spheres.
Right: Reduced-sphere representation of the FCC unit cell.
Examples: Ag, Al, Au, Ca, Cu, γ-Fe (>912 °C), Ni, Pd, Pt, Rh.
a
Since it is an FCC crystal structure, let a = lattice parameter (side of cubic cell) and R = radius of atom. The shortest interatomic separation is ro = 2R (atoms in contact means nucleus to nucleus
separation is 2R (see Figure 1Q3-1).
R = ro/2 and 2a2 = (4R)2 ∴ a= 2 2R = 2 2 ro 2 = 2 2.99 ×10
(
−10 m)
∴ a = 4.228 × 10-10 mTherefore, the volume (V) of the unit cell is:
V = a3 = (4.228 × 10-10 m)3 = 7.558 × 10-29 m3
The mass (m) of 1 Ne atom in grams is the atomic mass (Mat) divided by NA, because NA number of atoms have a mass of Mat.
m = Mat / NA
∴ m=
(
20.18 g/mol)
(
0.001 kg/g)
6.022×1023 mol-1 = 3.351 ×10 −26
There are 4 atoms per unit cell in the FCC cell. The density (ρ) can then be found by:
ρ = (4m) / V = [4 × (3.351 × 10-26 kg)] / (7.558 × 10-29 m3)
∴ ρ = 1774 kg/m3
In g/cm3 this density is:
ρ= 1774 kg/m3 100 cm/m
(
)
3 × 1000 g/kg(
)
= 1.77 g/cm3
The density of solid Ne is 1.77 g cm-3.
*1.14 BCC and FCC crystals
a. Consider iron below 912 °C, where its structure is BCC. Given the density of iron as 7.86 g cm-3 and its atomic mass as 55.85 g/mol, calculate the lattice parameter of the unit cell and the radius of the Fe atom.
b. At 912 °C, iron changes from the BCC (α-Fe) to the FCC (γ-Fe) structure. The radius of the Fe atom correspondingly changes from 0.1258 nm to 0.1291 nm. Calculate the density of γ-Fe and explain whether there is a volume expansion or contraction during this phase change.
Solution
Given:Density of iron at room temperature: ρ = 7.86 × 103 kg/m3 Atomic mass of iron: Mat = 55.85 g/mole
a For the BCC structure, the density is given by:
ρ= 2Mat × 10 −3 kg/g
(
)
NA a3Thus the lattice parameter a is:
a= 1 500 g/kg
(
)
NMatρ 1 3 A ∴ a= 1 500 g/kg(
)
(
6.022×1023(
mol55.85 g/mol−1)
(
7.86)
×103 kg/m3)
1 3 = 2.87 × 10-10 mThe radius of the Fe atom, R, and the lattice parameter, a, are related.
a = 4R R= 3 3 ∴ = 1.24 × 10-10 m 4 a= 3 4 2.87×10 −10 m
(
)
b Fe has a BCC structure just below 912 °C (α-Fe). An Fe atom in the α-Fe state has a radius of
ρBCC= 2Mat × 10 −3 kg/g
(
)
NA 4RBCC 3 3 = 2(
55.85 g/mol)
10 −3 kg/g(
)
6.022×1023 mol−1(
)
4 0.1258(
×10−9 m)
3 3= 7564 kg/m3 (less than at room temperature)
Fe has a FCC structure just above 912 °C (γ-Fe). An Fe atom in the γ-Fe state has a radius of
RFCC = 0.1291 × 10-9 m (Remember that for a FCC structure, a 2= 4RFCC). The density of γ-Fe is therefore: ρFCC= 4Mat× 10 −3 kg/g
(
)
NA 4RFCC 2 3 = 4(
55.85 g/mol)
10 −3 kg/g(
)
6.022×1023 mol−1(
)
4 0.1291(
×10−9 m)
2 3 = ρFCC = 7620 kg/m 3As the density increases, the volume must contract (the Fe retains the same mass).
1.16 Diamond and zinc blende
Si has the diamond and GaAs has the zinc blende crystal structure. Given the lattice parameters of Si and GaAs, a = 0.543 nm and a = 0.565 nm, respectively, and the atomic masses of Si, Ga, and As as 28.08 g/mol, 69.73 g/mol, and 74.92 g/mol, respectively, calculate the density of Si and GaAs. What is the atomic concentration (atoms per unit volume) in each crystal?
Solution
Referring to the diamond crystal structure in Figure 1Q16-1, we can identify the following types of atoms
8 corner atoms labeled C,
6 face center atoms (labeled FC) and 4 inside atoms labeled 1,2,3,4.
The effective number of atoms within the unit cell is: (8 Corners) × (1/
8 C-atom) + (6 Faces) × (1/2 FC-atom) + 4 atoms within the cell (1,2,3,4) = 8
a C C C a a 1 2 4 3 C C C FC FC FC FC FC FC
Figure 1Q16-1 The diamond crystal structure.
The lattice parameter (length of a cube side) of the unit cell is a. Thus the atomic concentration in the Si crystal (nSi) is nSi = 8 a3 = 8 (0.543×10−9m)3= 5.0 × 10 28 atoms per m-3
If Mat is the atomic mass in the Periodic Table then the mass of the atom (mat) in kg is
mat = (10-3 kg/g)Mat/NA (1)
where NA is Avogadro’s number. For Si, Mat = MSi = 28.09 g/mol, so then the density of Si is
ρ = (number of atoms per unit volume) × (mass per atom) = nSi mat
or ρ= 8 a3 (10 −3 kg/g)M Si NA i.e. ρ= 8 0.543×10−9 m
(
)
3 (10 −3 kg/g) 28.09 g mol(
-1)
6.022×1023 mol−1(
)
= 2.33 × 103 kg m-3 or 2.33 g cm -3In the case of GaAs, it is apparent that there are 4 Ga and 4 As atoms in the unit cell. The concentration of Ga (or As) atoms per unit volume (nGa) is
nGa = 4
a3 =
4
(0.565×10−9m)3 = 2.22 × 10
28 m-3
Total atomic concentration (counting both Ga and As atoms) is twice nGa.
nTotal = 2nGa = 4.44 × 1028 m-3
There are 2.22 × 1028 Ga-As pairs per m3. We can calculate the mass of the Ga and As atoms from their relative atomic masses in the Periodic Table using Equation (1) with Mat = MGa = 69.72 g/mol for Ga and Mat = MAs = 74.92 g/mol for As. Thus,
ρ= 4 a3 (10 −3 kg/g)(M Ga + MAs) N A or ρ= 4 (0.565×10−9 m)3 (10 −3 kg/g)(69.72 g/mol+ 74.92 g/mol) 6.022×1023 mol−1 i.e. ρ = 5.33 × 103 kg m-3 or 5.33 g cm-3
1.18 Crystallographic directions and planes
Consider the cubic crystal system.a. Show that the line [hkl] is perpendicular to the (hkl) plane. b. Show that the spacing between adjacent (hkl) planes is given by
d = a
Solution
This problem assumes that students are familiar with three dimensional geometry and vector products.
Figure 1Q18-1(a) shows a typical [hkl] line, labeled as ON, and a (hkl) plane in a cubic crystal.
ux, uy and uz are the unit vectors along the x, y, z coordinates. This is a cubic lattice so we have Cartesian coordinates and ux⋅ux = 1 and ux⋅uy = 0 etc.
N O α az1 ay1 ax1 β γ ah ak al A B C
u
xu
yu
z(a)
D O α az1 ay1 ax1 β γ A B C(b)
Figure 1Q18-1 Crystallographic directions and planes
a Given a = lattice parameter, then from the definition of Miller indices (h = 1/x1, k = 1/y1 and l = 1/z1) , the plane has intercepts: xo = ax1 =a/h; yo = ay1 = a/k; zo = az1 = a/l.
The vector ON = ahux + akuy + aluz
If ON is perpendicular to the (hkl) plane then the product of this vector with any vector in the (hkl) plane will be zero. We only have to choose 2 non-parallel vectors (such as AB and BC) in the plane and show that the dot product of these with ON is zero.
AB = OB − OA = (a/k)uy − (a/h)ux
ON•AB = (ahux + akuy + aluz) • ( (a/k)uy − (a/h)ux) = a2 − a2 = 0 Recall that uxux = uyuy =1 and uxuy = uxuz = uyuz= 0
Similarly, ON•BC = (ahux + akuy + aluz) • ( (a/l)uz − (a/k)uy) = 0 Therefore ON or [hkl] is normal to the (hkl) plane.
b Suppose that OD is the normal from the plane to the origin as shown in Figure 1Q18-1(b). Shifting a plane by multiples of lattice parameters does not change the miller indices. We can therefore assume the adjacent plane passes through O. The separation between the adjacent planes is then simply the distance OD in Figure 1Q18-1(b).
Let α, β and γ be the angles of OD with the x, y and z axes. Consider the direction cosines of the line OD: cosα = d/(ax1) = dh/a; cosβ = d/(ay1) = dk/a; cosγ = d/(az1) = dl/a
But in 3 dimensions, (cosα)2 + (cosβ)2 + (cosγ)2 = 1 Thus, (d2h2/a2) + (d2k2/a2) + (d2l2/a2) = 1 Rearranging, d2 = a2 / [h2 + k2 + l2]
or, d= a/ [h2 + k2 + l2]1/2
1.19 Si and SiO
2a. Given the Si lattice parameter a = 0.543 nm, calculate the number of Si atoms per unit volume, in
nm-3.
b. Calculate the number of atoms per m2 and per nm2 on the (100), (110) and (111) planes in the Si crystal as shown on Figure 1Q19-1. Which plane has the most number of atoms per unit area?
c. The density of SiO2 is 2.27 g cm-3. Given that its structure is amorphous, calculate the number of molecules per unit volume, in nm-3. Compare your result with (a) and comment on what happens when the surface of an Si crystal oxidizes. The atomic masses of Si and O are 28.09 g/mol and 16 g/mol, respectively.
a a
(100) plane (110) plane (111) plane
a
Figure 1Q19-1 Diamond cubic crystal structure and planes. Determine what portion of a black-colored atom belongs to the plane that is hatched.
Solution
a Si has the diamond crystal structure with 8 atoms in the unit cell, and we are given the lattice parameter a = 0.543 × 10-9 m and atomic mass M
at = 28.09 × 10-3 kg/mol. The concentration of atoms per unit volume (n) in nm-3 is therefore:
n= 8 a3 1 109 nm/m
(
)
3 = 8 0.543×10−9 m(
)
3 1 109 nm/m(
)
3= 50.0 atoms/nm3If desired, the density ρ can be found as follows:
ρ= 8Mat NA a3 = 828.09×10 −3 kg/mol 6.022×1023 mol−1 0.543×10 m−9
(
)
3 = 2331 kg m-3 or 2.33 g cm-3b The (100) plane has 4 shared atoms at the corners and 1 unshared atom at the center. The corner atom is shared by 4 (100) type planes. Number of atoms per square nm of (100) plane area (n) is shown in Fig. 1Q19-2:
(100) a a A B C D E a Ge a a A B C D E
Figure 1Q19-2 The (100) plane of the diamond crystal structure. The number of atoms per nm2, n100, is therefore:
n100 = 4 1 4 +1 a2 1 109 nm/m
(
)
2 = 4 1 4 + 1 0.543×10−9 m(
)
2 1 109 nm/m(
)
2 ∴ n100 = 6.78 atoms/nm2 or 6.78 × 1018 atoms/m2The (110) plane is shown below in Fig. 1Q19-3. There are 4 atoms at the corners and shared with neighboring planes (hence each contributing a quarter), 2 atoms on upper and lower sides shared with upper and lower planes (hence each atom contributing 1/2) and 2 atoms wholly within the plane.
(110) a a¦2 A B C D (110) A B C D
Figure 1Q19-3 The (110) plane of the diamond crystal structure. The number of atoms per nm2, n110, is therefore:
n110 = 4 1 4 + 2 12 + 2 a a 2
( )
[
]
(
109 nm/m1)
2 ∴ n110 = 4 1 4 + 2 12 + 2 0.543×10−9 m(
)
(
(
0.543×10−9 m)
2)
[
]
(
109 nm/m1)
2 ∴ n110 = 9.59 atoms/nm2 or 9.59 × 1018 atoms/m2This is the most crowded plane with the most number of atoms per unit area. The (111) plane is shown below in Fig. 1Q19-4:
A C B D a¦2 a¦2 a¦2 2 a¦22 60° 60° 30° 30° a¦3 ¦2 A B C D
Figure 1Q19-4 The (111) plane of the diamond crystal structure The number of atoms per nm2, n111, is therefore:
n111 = 3 60 360 + 3 12 2 1 2 a 22 a 3 2 1 109 nm/m
(
)
2 ∴ n111 = 3 60 360 + 3 12 2 1 2 0.543 ×10(
−9 m)
22 0.543 ×10−9 m(
)
3 2 1 109 nm/m(
)
2 ∴ n111 = 7.83 atoms/nm2 or 7.83 × 1018 atoms/m2 c Given:Molar mass of SiO2: Mat = 28.09 × 10-3 kg/mol + 2 × 16 × 10-3 kg/mol = 60.09 × 10-3 kg/mol Density of SiO2: ρ = 2.27 × 103 kg m-3
Let n be the number of SiO2 molecules per unit volume, then:
ρ= nMat NA ∴ n= NAρ Mat = 6.022×1023 mol−1
(
)
(
2.27×103 kg m−3)
60.09×10−3 kg/mol(
)
= 2.27 × 1028 molecules per m3Or, converting to molecules per nm3:
n=2.27×1028 molecules/m3 109 nm/m
(
)
3 =22.7 molecules per nm3Oxide has less dense packing so it has a more open structure. For every 1 micron of oxide formed on the crystal surface, only about 0.5 micron of Si crystal is consumed.
