Chapter 2
How big is an ion?
There is no single answer to this question.
Ionic radii are generally derived
from X-ray crystallographic structure determinations and it is assumed that that the
ions in the structure are spherical and touching.
The radius of one commonly
occurring ion, such as O2- is taken as a standard; other consistent radii can then be
derived.
Even with this limitation, cation radii in particular are sensitive to the
immediate surroundings and to the nominal ionic charge so a cation does not have
“a” radius, but several, depending upon the crystal site, often tetrahedral or
octahedral, that the ion occupies.
While the majority of ions can be considered to be spherical, the lone pair ions
In+, Tl+, Sn
2+
, Pb2+, Sb3+ and Bi3+ are definitely not so and it is hard to assign a
unique radius to such ions.
Anions, such as CO32-, NO3-, are also not spherical,
although at high temperatures, rotation often makes them appear so.
What is covalent bonding?
A covalent bond between two atoms forms when they share a pair of
electrons, one being donated from each atom.
There are two main theories of covalent bond formation.
Valence bond
theory starts with the idea that a half-filled atomic orbital on one of the atoms, most
importantly an s or p orbital, overlaps with a similar half-filled orbital on the other. If
the electrons have opposite spins, charge density builds up between the nuclei and
these are pulled together to form a covalent bond.
Molecular orbital theory starts with the idea of two empty atomic orbitals, one
on each atom. As two atoms come together the atomic orbitals interact and spread
over the whole of the pair to form molecular orbitals.
The calculations show that
when two atomic orbitals interact, two molecular orbitals form, one with a higher
energy than the original pair, called an antibonding orbital and one with a lower
energy than the original pair, called a bonding orbital. The available electrons are
fed into the bonding and antibonding orbitals and a covalent bond forms between the
atoms when more electrons occupy bonding orbitals than antibonding orbitals.
What are energy bands?
Energy bands are ranges of allowed energies that electrons can take in a
crystal. This idea is usually associated with metals. The concept can be explained
from two viewpoints. Tight-binding theory starts from an atomic orbital base. When
two atoms come together two separate atomic orbitals, one on each atom, give rise
to two molecular orbitals, one of lower energy and one of higher energy.
When a
large number of atoms are present, so many of these orbital pairs are created that
the energy range is spanned by an effective continuum, so as to form an energy
band.
In the nearly-free-electron approach electrons move around as if they were a
gas in the solid.
The potential that the electrons experience is due to the period
arrangement of the positive atomic nuclei that forms the crystal. The electrons are
found to take any energy over a range that depends upon the nature of the periodic
array of atoms.
However electron waves with wavelengths similar to the spacing of
the nuclei cannot move through the crystal and the energies associated with these
wavelengths are said to be forbidden.
In this model, the allowed energy bands are
broken up by a series of forbidden energy ranges.
The allowed energy bands are
separated from each other by bands of forbidden energies.
Quick quiz
1. c; 2. b; 3. a; 4. b; 5. b; 6. c; 7. b; 8. a; 9. b; 10. a; 11. a; 12. c; 13. c;
14. b; 15. b; 16. c; 17. a; 18. b; 19. c; 20. b; 21. c; 22. c; 23. a; 24. b;
25. c; 26. a; 27. a; 28. a.
Calculations and questions
1. Cl-, [Ar]; Na+, [Ne]; Mg2+, [Ne]; S2-, [Ar]; N3-, [Ne]; Fe3+, [Ar] 3d5.
2. F-, [Ne]; Li+, [He]; O2-, [Ne]; P3-, [Ar]; Co3+, [Ar] 3d6.
3. O2-; H+; Na+; Ca2+; Zr4+; W 6+, W 4+.
4. Fe3+, Fe2+; Cl-; Al3+; S2-; La3+; Ta5+.
5. 3600 kJ mol-1.
6. NaCl, 769 kJ mol-1; KCl, 688 kJ mol-1.
7. NaBr, 725 kJ mol-1, KBr, 657 kJ mol-1.
8. 5.88 x 1028 m-3.
9. 1.83 x 1029 m-3.
10. 8.46 x 1028 m-3.
11. 8.61 x 1028 m-3.
12. 1.70 x 1029 m-3.
13. 1.81 x 10-33 J; kT = 4.14 x 10-21 J at 300 K.
14. 8.81 x 10-19 J.
15. 5.04 x 10-19 J.
16. 7.04 x 10-19 J.
17. 1.87 x 10-18 J.
Solutions
1 Write out the electron configuration of the ions: Cl-, Na+, Mg2+, S-2, N-3, Fe3+.
Cl-: 1s2 2s2 2p6 3s2 3p6 or [Ar]
Na+: 1s2 2s2 2p6 or [Ne]
Mg2+: 1s2 2s2 2p6 or [Ne]
S2-: 1s2 2s2 2p6 3s2 3p6 or [Ar]
N3-: 1s2 2s2 2p6 or [Ne]
Fe3+: 1s2 2s2 2p6 3s2 3p6 3d5 or [Ar] 3d5
2 Write out the electron configuration of the ions: F-, Li+, O2-, P3-, Co2+.
F-: 1s2 2s2 2p6 or [Ne]
Li+: 1s2 or [He]
O2-: 1s2 2s2 2p6 or [Ne]
P3-: 1s2 2s2 2p6 3s2 3p6 or [Ar]
Co3+: 1s2 2s2 2p6 3s2 3p6 3d6 or [Ar] 3d6
3
Write the symbols of the ions formed by oxygen, hydrogen, sodium, calcium,
zirconium, tungsten.
O2-, H+, Na+, Ca2+, Zr4+, W 6+ and W 4+
4
Write the symbols of the ions formed by iron, chlorine, aluminium, sulphur,
lanthanum, tantalum.
Fe3+ and Fe2+, Cl-, Al3+, S2-, La3+, Ta5+
5
Calculate the lattice energy of the ionic oxide CaO, which has the halite (NaCl)
structure.  = 1.75, n = 9, r0 = 0.24 nm.
UL = [NA  Z2 e2 / 4  0 r0] (1 – 1 / n)
= [6.0224 x 1023 x 1.75 x 22 x(1.60218 x 10-19)2] / 4  x 8.85419 x 10-12 x 0.24
x 10-9] (1 – 1/9)
= 3.60 x 106 J mol-1 = 3600 kJ mol-1
6
Calculate the lattice energy of the ionic halides NaCl and KCl, which have the
halite (NaCl) structure.  = 1.75, n = 9, r0 (NaCl) = 0.281 nm, r0 (KCl) = 0.314 nm.
UL = [NA  Z2 e2 / 4  0 r0] (1 – 1 / n)
For NaCl:
=
[6.0224 x 1023 x 1.75 x 12 x (1.60218 x 10-19)2] / 4  x 8.85419 x 10-12 x
0.281 x 10-9] (1 – 1/9)
769 kJ mol-1
For KCl:
=
[6.0224 x 1023 x 1.75 x 12 x (1.60218 x 10-19)2] / 4  x 8.85419 x 10-12 x
0.314 x 10-9] (1 – 1/9)
688 kJ mol-1
NB. UL (KCl) =
UL (NaCl) x 0.281 / 0.314
7
Calculate the lattice energy of the ionic halides NaBr and KBr, which have the
halite (NaCl) structure.  = 1.75, n = 9, r0 (NaBr) = 0.298 nm, r0 (KBr) = 0.329 nm.
As in Q 6:
UL (NaBr) =
UL (NaCl) x 0.281 / 0.298
= 769 x 0.281 / 0.298 = 725 kJ mol-1
UL (KBr) =
UL (NaCl) x 0.281 / 0.329
= 769 x 0.281 / 0.329 = 657 kJ mol-1
8
Determine the number of free electrons in gold, assuming that each atom
contributes one electron to the “electron gas”.
The molar mass of gold is 0.19697
kg mol-1, and the density is 19281 kg m-3.
1 m3 of gold = 19261 kg = 19261 / 0.19697 mol
1 mol contains NA (6.02214 x 1023) atoms and each atom contributes 1 electron,
hence:
Number of free electrons = (19261 / 0.19697) x 6.02214 x 1023 m-3
= 5.88 x 1028 m-3
From this example we can write:
number of free electrons per cubic metre =
density (kg m-3) x NA x number of electrons donated per atom / molar mass (kg mol-1)
9
Determine the number of free electrons in nickel, assuming that each atom
contributes two electrons to the “electron gas”. The molar mass of nickel is 0.05869
kg mol-1, and the density is 8907 kg m-3.
Using the general formula above:
Number of free electrons = (8907 x 6.02214 x 1023 x 2 / 0.05869) m-3
= 1.83 x 1029 m-3
10
Determine the number of free electrons in copper, assuming that each atom
contributes one electron to the “electron gas”. The molar mass of copper is 0.06355
kg mol-1, and the density is 8933 kg m-3.
Using the general formula above:
Number of free electrons = (8933 x 6.02214 x 1023 x 1 / 0.06355) m-3
= 8.46 x 1028 m-3
11 Determine the number of free electrons in magnesium, assuming that each atom
contributes two electrons to the “electron gas”.
The molar mass of magnesium is
0.02431 kg mol-1, and the density is 1738 kg m-3.
Using the general formula above:
Number of free electrons = (1738 x 6.02214 x 1023 x 2 / 0.02431) m-3
= 8.61 x 1028 m-3
12
Determine the number of free electrons in iron, assuming that each atom
contributes two electrons to the “electron gas”.
The molar mass of iron is 0.05585
kg mol-1, and the density is 7873 kg m-3.
Using the general formula above:
Number of free electrons = (7873 x 6.02214 x 1023 x 2 / 0.05585) m-3
= 1.70 x 1029 m-3
13
Calculate the lowest energy level of a free electron in a cube of metal 1 cm on
edge and compare it with thermal energy at room temperature, given by kBT, where
kB is Boltzmann’s constant and T is the absolute temperature.
The energy of the free electron is given by:
E = [h2 / 8 me a2] (nx2 + ny2 + nz2) in a cube of side a.
Taking the rest mass of the electron as me = 9.109 x 10-31 kg, the lowest energy level
is given by nx = ny = nz = 1:
E = (6.626 x 10-34)2 (1 + 1 + 1) / [8 x 9.109 x 10-31 x (10-2)2]
= 1.807 x 10-33 J
Thermal energy at room temperature, say 300 K is:
E = 1.381 x 10-23 x 300 = 4.14 x 10-21 J
Thermal energy is far greater than the energy of electrons in the ground state.
14 Estimate the Fermi energy of silver. The molar mass of silver is 0.1079 kg mol 1
, the density is 10500 kg m-3, and each silver atom contributes one electron to the
“electron gas”.
The Fermi energy is given by:
EF = (h2 / 8 me)(3 N / ( V)2/3
to calculate N / V, use the general formula given above (Q8):
N / V = (10500 x 6.02214 x 1023 x 1 / 0.1079) m-3
= 5.86 x 1028 m-3
EF = [(6.626 x 10-34)2 / ( 8 x 9.109 x 10-31)] x [ (3 x 5.86 x 1028 / )]2/3
= 8.81 x 10-19 J
15 Estimate the Fermi energy of sodium. The molar mass of sodium is 0.02299 kg
mol-1, the density is 966 kg m-3, and each sodium atom contributes one electron to
the “electron gas”.
Using the method in the previous question:
N / V = (966 x 6.02214 x 1023 x 1 / 0.02299) m-3
= 2.53 x 1028 m-3
EF = [(6.626 x 10-34)2 / ( 8 x 9.109 x 10-31)] x [ (3 x 2.53 x 1028 / )]2/3
= 5.04 x 10-19 J
16
Estimate the Fermi energy of calcium.
The molar mass of calcium is 0.04408
kg mol-1, the density is 1530 kg m-3, and each calcium atom contributes two
electrons to the “electron gas”.
Using the method in the previous question:
N / V = (1530 x 6.02214 x 1023 x 2 / 0.04408) m-3
= 4.18 x 1028 m-3
EF = [(6.626 x 10-34)2 / ( 8 x 9.109 x 10-31)] x [ (3 x 4.18 x 1028 / )]2/3
= 7.04 x 10-19 J
17
Estimate the Fermi energy of aluminium.
The molar mass of aluminium is
0.02698 kg mol-1, the density is 2698 kg m-3, and each aluminium atom contributes
three electrons to the “electron gas”.
Using the method in the previous question:
N / V = (2698 x 6.02214 x 1023 x 3 / 0.02698) m-3
= 1.807 x 1029 m-3
EF = [(6.626 x 10-34)2 / ( 8 x 9.109 x 10-31)] x [ (3 x 1.807 x 1029 / )]2/3
= 1.87 x 10-19 J
18 Draw the wave functions and the probability of locating an electron at a position
x for an electron trapped in a one-dimensional potential well, with internal potential 0
and external potential infinity.
The solution to the Schrödinger equation for an electron in a 1-dimensional potential
well is:
 = [(2 / a)] sin (n  x / a)
n = 1, 2, 3 …
The functions will have maxima and minima of [(2 / a)] when sin (n  x / a) = 1:
n = 1: 1 maximum at x / a = ½
n = 2: 1 maximum at x / a = ¼, 1 minimum at x / a = ¾
n = 3: 2 maxima at x / a = 1/6, 5/6, 1 minimum at x / a = 3/6 = ½
the probability of findig the electron in a small volume dx is given by 2 dx. Plots of
2 are always positive with peaks given by:
n = 1: 1 maximum at x / a = ½
n = 2: 2 maxima at x / a = ¼, ¾
n = 3: 3 maxima at x / a = 1/6, 3/6, 5/6
the graphs, drawn with a = 100 are:
n=1
n=2
n=3
19
Make accurate plots of the Fermi function, which expresses the probability of
finding an electron at an energy E, for temperatures of 0, 300, 1000 and 5000 K.
The Fermi function is given by:
P(E) = 1 / {exp [(E – EF)/ kBT] – 1}
Representative points are given by:
E = 0, P(E) = 1; E = EF, P(E) = ½; E >> EF, P(E) = 0;
The plots given below were made using the representative values EF = 10 eV, kB =
8.617 x 10-5 eV K-1.
E / eV
E / eV
E / eV