Inorganic Chemistry
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Inorganic Chemistry


DisciplinaQuímica Inorgânica I3.901 materiais31.960 seguidores
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halides that are in reasonably close agreement with the \u2018experimental\u2019 
values from the Born-Haber cycle. Some examples are shown in Table 1. A strict comparison 
requires some corrections. Born-Haber values are generally enthalpies, not total energies, and are 
estimated from data normally measured at 298 K not absolute zero; further corrections can be made, 
for example, including van der Waals\u2019 forces between ions. 
Table 1. Comparison of lattice energies (all kJ mol\u22121) determined by different methods 
 
Compound Born-Haber cycle Born-Landé Extended Kapustinskii 
NaCl 772 757 770 765 
CsCl 652 623 636 617 
CaF2 2597 2594 2610 2779 
AgCl 902 \u2013 833 732 
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When these extended calculations are compared with experiment many compounds agree well (see 
Table 1). Significant deviations do occur, however; for example, in compounds of metals in later 
groups where bonding is certainly less ionic (e.g. AgCl). 
One of the disadvantages of the fully theoretical approach is that it is necessary to know the 
crystal structure and the interionic distances to estimate the lattice energy. The Kapustinskii 
equations overcome this limitation by making some assumptions. The Madelung constant A and the 
repulsive parameter n are put equal to average values, and it is also assumed that the interionic 
distance can be estimated as the sum of anion and cation radii r+ and r\u2212 (see Topic D4). The simpler 
of the Kapustinskii equations for a binary solid is 
 
(1) 
\u3bd is the number of ions in the formula unit (e.g. two for NaCl, three for MgF2 and five for Al2O3) 
and C is a constant equal to 1.079×105 when UL is in kJ mol
\u22121 and the radii are in pm. The 
Kapustinskii equation is useful for rough calculations or where the crystal structure is unknown. It 
emphasizes two essential features of lattice energies, which are true even when the bonding is not 
fully ionic: 
Calculations can be extended to complex ions such as carbonate and sulfate by the use of 
thermochemical radii, chosen to give the best match between experimental lattice energies and 
those estimated by the Kapustinskii equation. 
Applications 
Even though ionic model calculations do not always give accurate predictions of lattice energies (and 
especially when the approximate Kapustinskii equation is used) the trends predicted are usually 
reliable and can be used to rationalize many observations in inorganic chemistry. 
(i) Group oxidation states 
The occurrence of ions such as Na+, Mg2+ and Al3+ depends on the balance between the energies 
required to form them in the gas phase and the lattice energies that stabilize them in solids. Consider 
magnesium. The gas-phase ionization energy (IE) required to form Mg2+ is considerably greater than 
for Mg+. However, the lattice energy stabilizing the ionic structure MgF2 is much larger than that of 
MgF, and amply compensates for the extra IE. It is possible to estimate the lattice energy of MgF, 
and (depending on what assumptions are used about the ionic radius of Mg+) its formation from the 
elements may be exothermic. However, the enthalpy of formation of MgF2 is predicted to be much 
more negative, and the reason why MgF(s) is unknown is that it spontaneously disproportionates: 
 
Ionization beyond the closed-shell configuration Mg2+ involves the removal of a much more tightly 
bound 2p electron (see Topics A4 and A5). The third IE is therefore very large and can never be 
compensated by the extra lattice energy of a Mg3+ compound. 
\u2022 lattice energies increase strongly with increasing charge on the ions; 
\u2022 lattice energies are always larger for smaller ions. 
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(ii) Stabilization of high and low oxidation states 
When an element has variable oxidation states, it is often found that the highest value is obtained 
with oxide and/or fluoride (see, e.g. Topic H4). The ionic model again suggests that a balance 
between IE and lattice energy is important. Small and/or highly charged ions provide the highest 
lattice energies according to Equation 1, and the increase in lattice energy with higher oxidation state 
is more likely to compensate for the high IE. 
By contrast, a large ion with low charge such as I\u2212 is more likely to stabilize a low oxidation state, 
as the smaller lattice energy may no longer compensate for high IE input. Thus CuF is not known but 
the other halides CuX are. Presumably the lattice energy increase from CuF to CuF2 is sufficient to 
force a disproportionation like that of MgF but this is not so with larger halide ions. By contrast, 
CuX2 is stable with X=F, Cl and Br, but not I. 
(iii) Stabilization of large onions or cations 
It is a useful rule that large cations stabilize large anions. Oxoanion salts such as carbonates are 
harder to decompose thermally when combined with large cations (see Topic B6). It is also found 
that solids where both ions are large are generally less soluble in water than ones with a large ion and 
a small one. These trends are sometimes erroneously ascribed to \u2018lattice packing\u2019 effects, with the 
implication that two large ions together have a larger lattice energy than a large and a small ion. 
Theoretical (and experimental) estimates of lattice energies contradict this view, and a satisfactory 
explanation depends on a balance of energies (see also Topic E4). 
Consider the decomposition of a group 2 metal carbonate MCO3: 
 
Figure 2 shows a thermochemical cycle, which predicts that the enthalpy change in this reaction is 
 
(2) 
where X is enthalpy input required for the gas-phase decomposition of and HL are the lattice 
enthalpies. X is positive, but according to Equation 1 the lattice energy of MO will always be larger 
than that of MCO3 because the oxide ion is smaller. The difference of lattice energies in Equation 2 
therefore gives a negative contribution to the overall \u394H. If we have a larger M2+ ion, both lattice 
energies become smaller, but the important thing is that their difference becomes smaller. Thus 
smaller M2+ gives a less endothermic decomposition reaction, which is therefore possible at a lower 
temperature. 
 
Fig. 2. Thermochemical cycle for the decomposition of MCO3. 
 
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Section D\u2014Structure and bonding in solids 
D7 
ELECTRICAL AND OPTICAL PROPERTIES OF 
SOLIDS 
The band model 
The band model of solids is an extension of the molecular orbital (MO) method (see Topics C4\u2013
C7). The overlap of atomic orbitals in an extended solid gives rise to continuous bands of electronic 
energy levels associated with different degrees of bonding. In a simple monatomic solid the bottom 
of the band is made up of orbitals bonding between all neighboring atoms; orbitals at the top of the 
band are antibonding, and levels in the middle have an intermediate bonding character. Different 
atomic orbitals can, in principle, give rise to different bands, although they may overlap in energy. 
The fundamental distinction between metallic and nonmetallic solids arises from the way in 
which orbitals are filled (see Fig. 1). Metallic behavior results from a band partially occupied by 
electrons, so that there is no energy gap between the top filled level (known as the Fermi level) and 
the lowest empty one. On the other hand, a nonmetallic solid has a bandgap between a completely 
filled band (the valence band VB) and a completely empty one (the conduction band CB). In a 
filled band the motion of any electron is matched by another one moving in the opposite direction, so 
that there is no