CoreChem:6.1 Ionic Bonding
From ChemEd Collaborative
6.1 IONIC BONDING
Since ionic bonding requires that the atoms involved have unequal attraction for their valence electrons, an ionic compound must involve atoms of two different elements. The simplest example is provided by the combination of elements number 1 (H) and number 3 (Li) in lithium hydride, LiH. On a microscopic level the formula LiH contains four electrons. In separate Li and H atoms these electrons are arranged as shown in Fig. 6.1a. The H atom has the electron configuration 1s1, and Li is 1s22s1. When the two atoms are brought close enough together, however, the striking rearrangement of the electron clouds shown in Fig. 6.1b takes place. Here the color coding shows clearly that the electron density which was associated with the 2s orbital in the individual Li atom has been transferred to a 1s orbital surrounding the H atom. As a result, two new microscopic species are formed. The extra electron transforms the H atom into a negative ion or anion, written H– and called the hydride ion. The two electrons left on the Li atom are not enough to balance the charge of +3 on the Li nucleus, and so removal of an electron produces a positive ion or cation, written Li+ and called the lithium ion. The electron-transfer process can be summarized in of Lewis diagrams as follows:
The opposite charges of Li+ and H–attract each other strongly, and the ions form an ion pair in which the two nuclei are separated by a distance of 160 pm (1.60 Å).
Energy and the Formation of Ions
Formation of an ion pair by transfer of an electron from an Li atom to an H atom results in an overall lowering of the total energy of the two nuclei and four electrons involved. How and why this occurs is best seen if we break ion-pair formation into three simpler steps and consider the energy change involved in each. The three steps are
1 Removal of the 2s electron from an Li atom to form an Li+ ion.
2 Addition of that same electron to an H atom to form an H– ion.
3 The coming together of the two ions to form an ion pair.
The energy required in step 1 to remove an electron completely from an isolated atom is called the ionization energy. The ionization energy of lithium is 520 kJ mol–1. In other words, 520 kJ of energy is needed to remove a mole of 2s electrons from a mole of isolated lithium atoms in order to form a mole of isolated lithium ions. Alternatively we can say that 520 kJ is needed to ionize a mole of lithium atoms.
While energy is needed to accomplish step 1, we find that energy is released in step 2 when a hydrogen atom accepts an electron and becomes a hydride ion. The reason for this can be seen in Fig. 6.1. The second electron acquired by the hydrogen atom can pair up with the electron already in the 1s orbital without contradicting the Pauli exclusion principle. As a result, the new electron can move in close enough to the hydrogen nucleus to be held fairly firmly, lowering its energy significantly. Although the paired electrons repel each other somewhat, this is not enough to offset the attraction of the nucleus for both. Since the energy of the electron is lowered, the law of conservation of energy requires that the same quantity of energy must be released when a hydrogen atom is transformed into a hydride ion. The energy released when an electron is acquired by an atom is called the electron affinity. The electron affinity of hydrogen is 73 kJ mol–1 indicating that 73 kJ of energy is released when 1 mol of isolated hydrogen atoms each accepts an electron and is converted into a hydride ion.
Since 520 kJ mol–1 is required to remove an electron from a lithium atom, while 73 kJ mol–1 is released when the electron is donated to a hydrogen atom, it follows that transfer of an electron from a lithium to a hydrogen atom requires (520 – 73) kJ mol–1 = 447 kJ mol–1. At room temperature processes which require such a large quantity of energy are extremely unlikely. Indeed the transfer of the electron would be impossible if it were not for step 3, the close approach of the two ions. When oppositely charged particles move closer to each other, their potential energy decreases and they release energy. The energy released when lithium ions and hydride ions come together to 160 pm under the influence of their mutual attraction is 690 kJ mol–1, more than enough to offset the 447 kJ mol–1 needed to transfer the electron. Thus there is a net release of (690 – 447) kJ mol–1 = 243 kJ mol–1 from the overall process. The transfer of the electron from lithium to hydrogen and the formation of an Li+H– ion pair results in an overall lowering of energy.
In Fig. 6.2 the energy change in each step and the overall change are illustrated diagrammatically. As in the case of atomic structure, where electrons occupy orbitals having the lowest allowable energy, a collection of atoms tends to rearrange its constituent electrons so as to minimize its total energy. Formation of a lithium hydride ion pair is energetically “downhill” and therefore favored.
The Ionic Crystal Lattice
Up to this point we have been considering what would happen if a single Li atom and a single H atom were combined. When a large number of atoms of each kind combine, the result is somewhat different.
Electrons are again transferred, and ions are formed, but the ions no longer pair off in twos. Instead, under the influence of their mutual attractions and repulsions, they collect together in much larger aggregates, eventually forming a three-dimensional array like that shown in Fig. 6.3. On the macroscopic level a crystal of solid lithium hydride is formed.
The formation of such an ionic crystal lattice results in a lower potential energy than is possible if the ions only group into pairs. It is easy to see from Fig. 6.3 why this should be so. In an ion pair each Li+ ion is close to only one H– ion, whereas in the crystal lattice it is close to no less than six ions of opposite charge. Conversely each H– ion is surrounded by six Li+ ions. In the crystal lattice therefore, more opposite charges are brought closer together than is possible for separate ion pairs and the potential energy is lower by an additional 227 kJ mol–1. The arrangement of the ions in a crystal of LiH corresponds to the lowest possible energy. If there were an alternative geometrical arrangement bringing even more ions of opposite charge even closer together than that shown in Fig. 6.3, the Li+ ions and H– ions would certainly adopt it.
Ions and Noble-Gas Electron Configurations
One further aspect of the formation of LiH needs to be explained. If the transfer of one electron from Li to H is energetically favorable, why is the same not true for the transfer of a second electron to produce Li2+H2–? Certainly the double charges on Li2+ and H2– would attract more strongly than the single charges on Li+ and H–, and the doubly charged ions would be held more tightly in the crystal lattice. The answer to this question can be found by looking back at Fig. 6.1. Removal of a second electron from Li would require much more energy than the removal of the first because this second electron would be a 1s electron rather than a 2s electron. Not only is this second electron much closer to the nucleus, but it also is very poorly shielded from the nucleus. It is not surprising, therefore, that the second ionization energy of Li (the energy required to remove this second electron) is 7297 kJ mol–1 almost 14 times as large as the first ionization energy! Such a colossal energy requirement is enough to insure that only the outermost electron (the valence electron) of Li will be removed and that the inner 1s kernel with its helium-type electron configuration will remain intact. A similar argument applies to the acceptance of a second electron by the H atom to form the H2– ion. If such an ion were to be formed, the extra electron would have to occupy the 2s orbital. Its electron cloud would extend far from the nucleus (even farther than for the 2s electron in Li, because the nuclear charge in H2– would only be +1, as opposed to +3 in Li), and it would be quite high in energy. So much energy would be needed to force second electron to move around the H nucleus in this way, that only one electron is H2– transferred. The ion formed has the formula H– and a helium-type 1s2 electronic structure. The simple example of lithium hydride is typical of all ionic compounds which can be formed by combination of two elements. Invariably we find that one of the two elements has a relatively low ionization energy and is capable of easily losing one or more electrons. The other element has a relatively high electron affinity and is able to accept one or more electrons into its structure. The ions formed by this transfer of electrons almost always have an electronic structure which is the same as that of noble gas, and all electrons are paired in each ion. The resulting compound is always a solid in which the ions are arranged in a three-dimensional array or crystal lattice similar to, though not always identical with, that shown in Fig. 6.3. In such a solid the nearest neighbors of each anion are always cations and vice versa, and the solid is held together by the coulombic forces of attraction between the ions of opposite sign. An everyday example of such an ionic compound is ordinary table salt, sodium chloride, whose formula is NaCl. As we shall see in the next section, sodium is an element with a low ionization energy, and chlorine is an element with a high electron affinity. On the microscopic level crystals of sodium chloride consist of an array of sodium ions, Na+, and chloride ions, Cl–, packed together in a lattice like that shown for lithium hydride in Fig. 6.3. The chloride ions are chlorine atoms which have gained an electron and thus have the electronic structure 1s22s22p63s23p6, the same as that of the noble-gas argon. The sodium ions are sodium atoms which have lost an electron, giving them the structure 1s22s22p6, the same as that of the noble-gas neon. All electrons in both kinds of ions are paired.
