Chapter 5: Bonding in solids
5.1 Metallic bonding: introducing bands
The conductivity of a metallic conductor decreases with increasing T, while the conductivity of a semiconductor increases with increasing T. Bands can explain these observations and other properties of metals.
5.1.1 Bands in one dimension
The electronic structure of a solid is characterised by a series of bands (energy levels so close in energy that they form a continuum), and band gaps (energies which do not correspond to any molecular orbitals (MO)).
Molecular Orbital theory, discussed in Chapter 3, quite readily explains the occurrence of bands. Figure 5.1, shows that this can be explained by using the simplest analogy: a one-dimensional solid, or a line of N atoms. Allowing overlap of only s orbitals yields N MOs and N energy levels:
Figure 5.1, p. 181.
Energies of atomic orbitals that constructively and destructively interfere with increase number of atoms in chain.
Figure 5.2, p. 181.
Density of states is greatest at the highest and lowest energies.
Note the development of the band as N increases in Figure 5.1 from 1 to 2 (e.g., diatomic molecule) to 3 (e.g., triatomic molecule) etc. Bands in solids can, to a first approximation, be identified with individual overlap of s, p or d orbitals. The widths, band gaps, and even band overlaps determine the electronic properties of a solid.
Figure 5.2 shows that that the number of energy levels in a small range of energies (density of states) increases significantly at the highest and lowest energies of the bands as the numbers of atoms in the chain increases.
Figure 5.3, p. 182
Limitations on bandwidth due to repulsions.
The most puzzling feature though is the width of the band (i.e., the energy difference between most antibonding and most bonding orbitals). This is never infinite, even for an effectively infinite (say 1 mole) number of atoms. The bandwidth is governed by the strength of the interaction between orbitals, which in turn depends on their shape, size and internuclear separation. This is shown in Figure 5.3.
5.1.2 Conduction of electricity
Figure 5.4, p. 183.
Flow of electrons in partially filled bands.
Figure 5.4 demonstrates how partially filled bands can conduct electricity whereas filled bands cannot.
5.1.3 Bands in three dimensions
Figure 5.5, p. 184 .
Band overlap leading to stronger bonds.
Bands in three dimensions can explain complex properties in solids. For example, bonding is strengthened and elements with filled bands become conducting when antibonding Crystal Orbitals (COs) become bonding COs when two bands overlap. This is shown in Figure 5.5.
5.1.4 Bands formed from MOs
Bands also form by the overlap of molecular orbitals. That is, in molecular solids, such H2O. As a consequence the formation of solids, such as low-temperature H2, and the conductivity of high-pressure solid H2, can be explained. This is shown in Figure 5.6 for molecular hydrogen.
Figure 5.6, p. 185.
Bands formed by overlapping MO’s. That is, σ and σ** MOs for H2 molecules rather than atoms.
Figure 5.7, p. 186.
Band gap, absence of molecular orbitals between the conduction and valence bands.
5.1.5 Band gaps and semiconductors
Bands are separated by band gaps as shown in Figure 5.7. An insulator can be regarded simply as a semiconductor with a large band gap, and where the occupied (valence) band is full and the unoccupied (conduction) band is empty. If the band gap is sufficiently large and the temperature low enough, no electrons can be promoted to higher levels - the conduction band - and hence the solid does not conduct electricity. As the band gap decreases, increasing numbers of electrons become "mobile", and conductivity increases with increasing T - semiconductivity.
Figure 5.8, p. 186.p>
Schematic of band gaps for Group 14 elements..
Semiconductors are mainly characterised by the size of the band gap. Typical band gaps at 25oC are 580, 107, 64.2 and 7.7 kJ mol-1 for diamond, silicon, germanium and a-tin, respectively.
Figure 5.8, p. 186.p>
Schematic of band gaps for Group 14 elements.
Notice that the band gap for diamond - an insulator - is very large (580 kJ mol-1), while for materials normally classified as semiconductors, the band gap is much smaller - 107 kJ mol-1 for Si and only 64.2 kJ mol-1 for Ge. This is shown in Figure 5.8.
5.1.6 Graphite
Diamond and graphite are the two most common crystalline forms of elemental carbon. Diamond is effectively an electrical insulator and the hardest known substance, while graphite is impure, a good conductor and is ‘slippery’. The structures and bonding in the two allotropes are the origins of these widely differing physical properties. The structure of graphite is shown in Figure 5.9
Figure 5.9, p. 188.
Hexagonal structure of graphite.
Electrical properties are a consequence of the different band structures, due to the different three-dimensional structures. For graphite, conductivity and reactivity is related to the structure of the delocalized π bonds. Conductivity perpendicular to the planes is low and increases with increasing temperature, and so is a semiconductor in that direction. Electrical conductivity is much higher parallel to the planes, but decreases as the temperature is raised (graphite is a semi-metal in that direction). This anisotropy of the conductivity is consistent with bands, where the electrons are in half-full π bands.
5.2 Ionic solids
Ionic bonding is inherently simpler conceptually than covalent bonding, because it can be adequately described by a purely electrostatic approach - ions are located in a lattice in such a way as to maximise attractions and minimise repulsions.
The lattice enthalpy, ΔHlattice is defined as the molar enthalpy change for the formation of an ionic solid from gaseous ions:
M+(g) + X-(g) ) → MX(s)
The experimentally measured value is generally the standard molar lattice enthalpy ΔHolattice, which is approximately equal to the standard internal energy change, Elattice at 0 K. Lattice enthalpies are invariably exothermic. It is not possible to design an experiment to measure lattice enthalpies directly and so 'experimental' values are calculated from Born-Haber cycles. The next section describes this method, which is not discussed in the textbook. (Sections 5.2.1 – 5.2.4 describe methods to estimate lattice enthalpies using electrostatic models.)
Lattice enthalpy: the Born-Haber cycle
Lattice enthalpies can be experimentally determined using Hess’ Law, provided all other thermodynamic quantities in a Born-Haber cycle are known. Hess’ law states that the enthalpy change for the process reactants → products is independent of the pathway taken between the initial and final states. Thus the overall reaction can be broken into steps, and summed to give the overall reaction. The overall energy change and hence the lattice enthalpy, will also be the sum of the individual reaction energy terms.
Consider the example of LiF. To calculate the enthalpy of formation (ΔfHo) of LiF, that is the enthalpy change for the process
Li(s) + ½ F2(g) → LiF(s) | ΔfHo |
the following 5 reaction steps can be considered:
Step 1: sublimation of solid lithium
Li(s) → Li(g) | ΔsubHo(Li) = +161 kJ mol-1 |
Step 2: ionisation of lithium atoms to form Li+ ions in the gas phase
Li(g) → Li+(g) + e- | ΔionHo(Li) = +520 kJ mol-1 |
Step 3: dissociation of fluorine molecules
½ F2(g) → F(g) | ½ΔdisH°(F2) = 77 kJ mol-1 |
Step 4: formation of F– ions from F atoms in the gas phase
F(g) + e- → F-(g) | ΔegHo(F) = -328 kJ mol-1 |
(ΔH for this reaction is minus the electron affinity of fluorine)
Step 5: formation of solid LiF from the gaseous Li+and F– ions
Li+(g) + F-(g) → LiF(g) | ΔHolattice = -1047 kJ mol-1 |
Since the sum of these five processes yields the desired overall reaction:
| Step | Process | Enthalpy change (kJ mol-1) |
|---|---|---|
| 1 | Li(s) → Li(g) | 161 |
| 2 | Li(g) → Li+(g) + e- | 520 |
| 3 | ½ F2(g) → F(g) | 77 |
| 4 | F(g) + e- → F-(g) | -328 |
| 5 | Li+(g) + F-(g) → LiF(g) | -1047 |
| overall: | Li(s) + ½ F2(g) ® LiF(s) | -617 |
This is perhaps more easily shown graphically:
In this example, the enthalpy of formation was determined from other known quantities; clearly, cycles such as this may be used to determine the value of any one quantity, provided all others are known.
Values of lattice enthalpies determined in this manner can be used to determine whether the substance may be best regarded as ionic. To do this requires a comparison against the predictions made by the ionic model, and that is the subject of the next section.
The following describes methods to estimate lattice energies using electrostatic models.
5.2.1 The ionic model for lattice enthalpies
The total Coulombic contribution to the lattice energy of an idealised ionic solid can be calculated by summing over all potential energy terms arising from the interaction between point charges. The total potential energy per mole of formula units is
| ECoulomb = -NAA | z2e2 |
| 4πε0r |
where NA is Avogadro's number, A is the Madelung constant, z are the charges on the ions, e.g. +1 for Na+ and -2 for O2-, ε0 is the permittivity of a vacuum (ie a constant = 8.85 7 × 10-12 F m-1), e is the electronic charge, and r the separation between nearest neighbours. Rather than insert values of fundamental constants into this expression each time it is used, we can absorb them into a single constant, and arrive at a more useable form in kJ mol-1:
| ECoulomb = -1389.4 | A z2 |
| r |
The Madelung constant A arises from the summation of all electrostatic interactions for each ion over the lattice, and so the sum of an infinite series whose terms depend only on the geometry of the lattice. Table 5.2 on page 191 lists values of A for common structures.
Lattice energies are thus proportional to the product of ionic charges and inversely proportional to inter-ionic distances (which can be obtained from x-ray crystallography).
| lattice enthalpy ∝ | z2 |
| r |
Real ions are of course not point charges - they have a considerable electron distribution. Overlap between ions in close proximity leads to repulsion between their electron clouds, which can be described by the following equation
| E(repulsions) = | NAC |
| rn |
where C = repulsion coefficient and n = Born exponent. Values of the Born exponent are dependent on the electronic configurations of the ions in the lattice, while the repulsion coefficient is equal to:
| C = A | z2e2ren-1 |
| 4πε0n |
where re = equilibrium separation of the ions.
The lattice energy then results by combining the attractive Coulombic and repulsive Born interactions (shown in Figure 5.10):
| Eequil. = -NAA | z2e2 | (1 - | 1 | ) = -1389.4 | A z2 | ( 1 - | 1 | ) |
| 4πε0re | n | re | n |
Example 5.1, p. 192 is an example using this equation.
Figure 5.10, p. 189.
Attractive, repulsive and total energies as a function of distance between ions.
For three-dimensional ionic crystals such as those shown in Figure 5.12 a minor modification on the lattice energy is required; the charges z are generalized to z+ and z– and the absolute value of these quantities are used.
Figure 5.12, p. 193
Unit cells of four ionic crystals.
The lattice enthalpy in three dimensions is therefore approximated by the equation:
| ΔHolattice = -NAA | z+z-e2 | (1 - | 1 | ) = -1389.4 | A z+z- | (1 - | 1 | ) |
| 4πε0re | n | re | n |
5.2.2 Ionic radii
It is not possible to assign a unique radius to every ion that is independent of the other ions in the structure (contrary to what you may have learned in first year). Ionic radii are derived from known structures, by sometimes quite arbitrary partitioning of the distances between nuclei. Different internally consistent sets have been derived, and so do not mix values for ions from different sets.
Make sure you understand the following general trends:
- Ionic radii usually increase on going down a group.
- Ionic radii the same charge decrease across a period.
- For ions in environments with different coordination numbers, its radius increases as the coordination number increases. This is because the anions will need to retreat from the metal ion to avoid anion-anion repulsions in higher coordination solids.
- For cations with different charge numbers (oxidation state), and a given coordination number, its radius decreases with increasing charge number. This is because increasing nuclear charge and fewer electrons will result in a smaller ion
- Cations are usually smaller than anions with similar atomic numbers, because a positive charge indicates a reduced number of electrons, and hence a more dominant nuclear attraction.
5.2.3 The Kapustinskii equation
From the Coulombic lattice energy calculations, it is possible to predict with a fair degree of accuracy the existence or not of any ionic compound. However, a more rough and ready estimation can often be used, and it turns out to be particularly useful for ionic solids incorporating polyatomic anions and cations.
The Kapustinskii equation is based on the hypothesis that there exists a hypothetical rock salt structure, which is energetically equivalent to the actual structure of any ionic solid. In the equation
| Elattice/kJ mol-1 ≈ - | nion|z+||z-| | 107 × 105 |
| (r+ + r-)/pm |
where
Because of the inherent covalency within polyatomic ions, a prediction of a standard ion "size" from one ionic compound to the next is difficult. Nevertheless, on a more general basis, these values are accurate enough for lattice energy calculations and structure predictions.
5.2.4 Validity of the ionic model
Measured and calculated lattice enthalpies are compared in the table below. Calculated values use the more exact equation on page 194.
| Compound | ΔHolattice | calc/expt (%) | |
|---|---|---|---|
| Calculated | Experimental | ||
| LiFa | 1033 | 1037 | 99.6 |
| LiCla | 845 | 852 | 99.2 |
| LiBra | 798 | 815 | 97.9 |
| LiIa | 740 | 761 | 97.2 |
| CsFa | 748 | 750 | 99.7 |
| CsClb | 652 | 676 | 96.4 |
| CsBrb | 632 | 654 | 96.6 |
| CsIbb | 601 | 620 | 96.9 |
| AgFa | 920 | 969 | 94.9 |
| AgCla | 833 | 912 | 91.3 |
| AgBra | 816 | 900 | 90.7 |
a Rock-salt structure; b Cesium-chloride structure
From this table it is readily seen that the best agreement is obtained for systems with large electronegativity differences (LiF, LiCl, CsF), while much worse agreement results for systems like AgCl, AgBr where the differences are smaller.
Good agreement between Born-Haber and calculated values for the lattice enthalpy - as observed in the previous example for NaCl - suggests mostly ionic character; poor agreement suggests polarisation of ions and significant covalent character in the bonding. When there is poor agreement between Born-Haber and calculated lattice enthalpies this does not mean that the 'experimental' Born-Haber value is the correct value as, when significant covalent character is present in a compound, the interpretation of lattice energy becomes more complex and a much less useful concept.
Because lattice energy is an indication of the stability of ionic compounds, its value can help us rationalise the formulas of these compounds. Consider magnesium chloride as an example. The ionisation energy of an element increases rapidly as successive electrons are removed from its atom. For example, the first ionisation energy of magnesium is 738 kJ mol-1, and the second ionisation energy is 1450 kJ mol-1, almost twice the first. We might ask why, from the standpoint of energy, magnesium does not prefer to form unipositive ions in its compounds. Why doesn't magnesium chloride have the formula MgCl (containing the Mg+ ion) rather than MgCl2 (containing the Mg2+ ion)? Admittedly, the Mg2+ ion has the noble gas configuration [Ne], which represents stability because of its completely filled shells. But the stability gained through the filled shells does not, in fact, outweigh the energy input needed to remove an electron from the Mg+ ion. The solution to this puzzle lies in the extra stability gained when magnesium chloride forms in the solid state. The lattice energy of MgCl2 is 2527 kJ mol–1. This is more than enough to compensate for the energy needed to remove the first two electrons from a Mg atom (738 kJ mol-1 + 1450 kJ mol-1 = 2188 kJ mol-1).
What about sodium chloride? Why is the formula for sodium chloride NaCl and not NaCl2 (containing the Na2+ ion)? Although Na2+ does not have the noble gas electron configuration, we might expect the compound to be NaCl2 because Na2+ has a higher charge and therefore the hypothetical NaCl2 should have a greater lattice energy. Again the answer lies in the balance between energy input (that is, ionisation energies) and the stability gained from the formation of the solid. The sum of the first two ionisation energies of sodium is
496 kJ mol-1 + 4560 kJ mol-1 = 5056 kJ mol-1
Since NaCl2 does not exist, we do not know its lattice energy value. However, if we assume a value roughly the same as that for MgCl2 (2527 kJ mol-1), we see that it is far too small an energy yield to compensate for the energy required to produce the Na2+ ion.
What has been said about the cations applies also to the anions. For example, the electron affinity of oxygen is considerable, meaning the following process releases energy (and is therefore favourable):
| O(g) + e- → O–(g) | ΔegHo(O) = -141.0 kJ mol-1 |
As we would expect, adding another electron to the O- ion
| O-(g) + e– → O2–(g) | ΔegHo(O-) = +844.2 kJ mol-1 |
would be unfavourable because of the increase in electrostatic repulsion. Indeed, the electron affinity of O– is positive. Yet compounds containing the oxide ion (O2-) do exist and are very stable, whereas compounds containing the O– ion are not known. Again, the high lattice energy realised by the presence of O2- ions in compounds such as Na2O or MgO far outweighs the energy needed to produce the O2- ion.
The text discusses the deviation from the Kapustinskii equation as shown in Figure 13.
Figure 5.13, p. 196.
Correlation of experimental lattice enthalpies with the inverse of the sum of the ionic radii.