10. Energetics of ionic bonding and introduction to Lewis structures
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In this part of the course, we will set aside quantum mechanics briefly and discuss the so-called classical theory of chemical bonding. In our crude classical picture, electrons are viewed as point particles that move in stable classical orbits determined by Newton's second law around the atoms in a molecule. Because the nuclei are several thousand times more massive than electrons, we regard the nuclei as fixed on the time scale of significant ``classical'' electron motion.
Despite our crude classical picture, however, a conceptually useful definition of chemical bonding can be stated that relies on the relatively high mobility of the electrons compared to the nuclei:
Energetic considerations of chemical bonding
Based on our crude, conceptual definition of chemical bonding, it is clear that an important issue concerns the ability of atoms to give up electrons to or accept electrons from other atoms.
In order to quantify this notion, there are two energies that are relevant. These are:
Let us examine these in greater detail:
Ionization energy and the shell model of the atom
Consider the reaction describing the loss of an electron by a neutral atom:
\[ X \rightarrow X^+ + e^-\]
Elements toward the left of the periodic table have low electron affinities, so they tend to act as electron donors, while atoms to the right of the periodic table have high electronegativities, and they tend to act as electron acceptors.
The Lewis dot model
When atoms are brought into some arrangement, the problem of determining the chemical bonding pattern amounts to figuring out where the electrons are and what they look like. The solution of this problem is known as electronic structure theory and is a quantum mechanical problem. All of chemical bonding can be explained by solving the Schrödinger wave equation, which we will discuss in the next chapter.
A simple model, developed by Lewis in 1916, exists that can be used to gain a qualitative understanding of chemical bonding and electronic structure. This is the Lewis dot model.
The Lewis model involves recognizing that not all electrons in an atom play a role in chemical bonding. Electrons that are closest to the nucleus for a set called the core electrons. These are generally chemically inert and not important for chemical bonding.
The outermost electrons of an atom, called the valence electrons are chemically, important, however, and are treated within the Lewis model.
It is worth noting that for quantitative accuracy, the effects of core electrons does need to be included in some approximate way. This can amount to a particularly difficult problem, especially in heavier elements, where the charge on the nucleus is large, and hence the forces on core electrons cause them to move at speeds near light speed. In that case, a fully relativistic treatment of core electrons is necessary, so if you thought your study of relativity theory in physics was useless, think again.
For elements in groups I-VIII in the periodic table,
\[\# \ valence \ electrons \ = \ group \ \# \ in \ periodic\ table\]These atoms have multiple valencies, hence the roman numeral, which is used to distinguish between the different ions. The roman numeral is taken to be equal to the charge on the ion. The reason for the large number of valence electrons has to do with the unfilled 
\(d\) orbitals of these atoms (to be discussed in chapters 15 and 16) next semester.
There are several polyatomic cations and anions that are important and will be encountered later. They are:
Energetics of ionic bonding
Consider the KF molecule. K has a low ionization energy and can act as an electron donor according to the reaction:
Chemical bonds are formed either by one atom's transferring electrons to another atom or by two atoms' sharing electrons between them, or something in between these two situations.
- The electronegativity measures the ability of an atom to draw electrons to itself in a chemical bond. While it is reported as an atomic property, it is concerned with bonding within a molecule. One definition of electronegativity that was proposed by Mulliken in 1934 is particularly simple, being a simple arithmetical average:
\[\Chi_{mulliken} \propto \frac{1}{2}(IE_1 +EA)\]
The larger the electronegativity, the greater the tendency for an atom to draw electrons to itself in a chemical bond.
In general,
- if \(IE_1\) and \(EA\) are both large, giving up an electron is unlikely, but gaining an electron is likely, and the atom tends to act as an electron acceptor, or is ``electronegative.''
- if \(IE_1\) is small, and \(EA\) is small or negative, giving up an electron is likely, but gaining an electron is unlikely, and the atom tends to act as an electron donor, or is ``electropositive.'' Helium is an exception to this rule, having only 2 valence electrons.
Note that the group 18 elements, known as the noble gases, have 8 valence electrons (except He, which has only 2) and are highly non-reactive. We can surmise that 8 valence electrons is a kind of magic number giving special stability to the electronic structure of the atom. We call this an octet, which plays an important role in the Lewis model.
As an example of identifying the number of valence electrons, consider the second period: - Figure 2:
For this period, we see that
\[\# \ core \ electrons \ = \ atomic \ number \ - \ \# valence \ electrons\]
In the Lewis dot model, valence electrons are represented by dots, while core electrons are not shown explicitly. As an example of this representation, consider the first two periods:
- Figure 3:
Notice how the dots are drawn as single electrons first before pairs of electrons are drawn. Again, the exception to this is Helium, whose two electrons form a closed valence shell and thus must be drawn as a pair. The element Ne also has a closed valence shell, but with eight electrons drawn as four pairs.Ionic bondingAccording to our crude, conceptual definition, chemical bonds can form either by electron transfer between atoms or by sharing of electrons. The former case constitutes a kind of bond known as an ionic bond. The reason for this is that if one atom transfer electrons to another, the donor is left with a net positive charge, while the acceptor is left with a net negative charge. According to Coulomb's law, these charged species should attract each other, and this attraction is what stabilizes the bond.
When atoms lose or gain electrons, they become what are called ions. Loss of electrons leaves an atom with a net positive charge, and the atom is called a cation. Gain of electrons leaves an atom with a net negative charge, and the atom is called an anion.
Some examples are:
\[Na\cdot \rightarrow Na^+ +e^-\]
\[Na\cdot \ sodium \ atom\]
\[Na^+ \ sodium \ cation\]
\[\cdot Ca \cdot \ calcium \ atom\]
\[Ca^{2+} \ calcium \ cation\]
\[\stackrel{..}{\stackrel{:F\cdot}{\stackrel{..}{}}} \ Fluorine \ atom\]
\[\stackrel{..}{\stackrel{:F:}{\stackrel{..}{}}}^- \ Fluorine \ anion\]
\[\stackrel{..}{\stackrel{\cdot S\cdot}{\stackrel{..}{}}}^{2-} \ Sulfur \ atom\]
\[\stackrel{..}{\stackrel{:S:}{\stackrel{..}{}}}^{2-} \ Sulfur \ anion\]
Special stability results when an atom, by losing or gaining electrons has as many valence electrons as a noble gas atom. In this case, we say that an atom or ion has a completed octet. \(Na^+\) has as many electrons as \(Ne\), which explains why it loses only 1 electron, while \(Ca^{2+}\) also has as many electrons as \(Ne\), but since it is a Group II, element, it must lose 2 electrons to achieve an octet configuration.
The basic assumption of the Lewis model is that atoms will tend to form chemical bonds in order to complete their valence shells and achieve completed octets (or doublets in the case of H and He). This is known as the octet rule.
Consider the loss of an electron by sodium:
\[Na \rightarrow Na^+ + e^-\]
By losing the electron, \(Na\) achieves octet stability, since \(Na^+\) is isoelectronic with \(Ne\). Similarly, chlorine favors gaining an electron:
\[e^- +\stackrel{..}{\stackrel{:Cl\cdot}{\stackrel{..}{}}}\rightarrow \stackrel{..}{\stackrel{:Cl:}{\stackrel{..}{}}}^-\]
Thus, \(Na\) can donate an electron, which \(Cl\) can accept and an ionic bond can form between them. This is expressed chemically as
\[Na\cdot +\stackrel{..}{\stackrel{:Cl\cdot}{\stackrel{..}{}}}\rightarrow Na^+ +\stackrel{..}{\stackrel{:Cl:}{\stackrel{..}{}}}^- \rightarrow Na^+ \stackrel{..}{\stackrel{:Cl:}{\stackrel{..}{}}}^-\]
which is a molecule of sodium chloride. Atoms that can lose two electrons, such as Calcium, can form molecules with two chemical bonds:
\[\cdot Ca\cdot +2\stackrel{..}{\stackrel{:Br\cdot}{\stackrel{..}{}}}\rightarrow Ca^{2+} +2\stackrel{..}{\stackrel{:Br:}{\stackrel{..}{}}}^- \rightarrow Ca^{2+}\left ( \stackrel{..}{\stackrel{:Br:}{\stackrel{..}{}}}^- \right )_2\]
In transition metal atoms and in many heavier elements, the octet rule does not apply.
Examples:- \(Cu^+\) ion has 10 valence electrons - copper(I) ion.
- \(Cu^{2+}\) ion has 9 valence electrons - copper(II) ion.
- \(Fe^{3+}\) ion has five valence electrons - iron(III) ion.
- \(Fe^{2+}\) ion has 6 valence electrons - iron(II) ion.
- \(NH_{4}^{+}\) - ammonium ion.
- \(H_3 O^+\) - hydronium ion (important in acidic solutions).
- \(CO_{3}^{-}\) - carbonate ion.
- \(NO_{2}^{-}\) - nitrite ion.
- \(NO_{3}^{-}\) - nitrate ion.
- \(ClO_{4}^{-}\) - perchlorate ion (forms interesting compounds with the ammonium and hydronium ions).
- \(OH^-\) - hydroxyl ion (important in basic solutions).
- \(F\) has a high electron affinity and can accept an electron according to the reaction:
- \[F+e^-\rightarrow F^- \ ; \ \Delta E_2 =-EA =-328 \ kJ/mol\]
so that \(EA=328 \ kJ/mol\). The total energy to form \(K^+\) and \(F^-\) far away from each other is just the sum of \(\Delta E_1\) and \(\Delta E_2\): - \[\Delta E_{tot}=\Delta E_1 +\Delta E_2 =IE_1 -EA=91 \ kJ/mol\]
which implies that there is an energy cost associated with the formation of these two ions. How, then, can the ionic bond form if there is an energy cost?
The answer lies in Coulomb's law, which states that if \(K^+\) and \(F^-\) are brought near each other, there is an attraction of opposite charges. Since the charge on \(K^+\) is \(e\) and the charge on \(F^-\) is -
\(-e\), if the bond length of KF is \(R_e\), the attractive energy between them due to Coulomb's law is \[\Delta E_{Coul}=-\frac{e^2}{4\pi \varepsilon_0 R_e}\]
which is negative and tends to lower the total energy and stabilize the bond.
Let us estimate the energy needed to form the KF bond from neutral atoms, given the above information. The total energy needed to form the ionic bond should be
- \[\Delta E_f \approx -\frac{e^2}{4\pi \varepsilon_0 R_e}-EA+IE_1\]
where the first term is the energy gained due to the Coulomb attraction, and the second and third terms constitute the energy needed to transfer the electron from K to F to form ions. Substituting in the numbers:
\[\begin{align*}IE_1 &= 419 \ kJ/mol\\ EA &= 328 \ kJ/mol\\ R_e &= 2.17\stackrel{\circ}{A} =2.17*10^{-10} \ m\\e&=1.602*10^{-19} \ C\end{align*}\]
gives
\[\Delta E_f\approx -\frac{(1.602*10^{-19}C)^2}{4\pi (8.854*10^{-12}C^2 J^{-1}m^{-1})(2.17*10^{-10}m)}*6.022*10^{23}mol^{-1}+9.1*10^4 \ J/mol\]
gives the result \(-5.49*10^5 \ J/mol\) or \(-549 \ kJ/mol\). The fact that the energy is negative tells us that the bond is stabilized by the Coulomb attraction. The energy needed to dissociate the bond into neutral species \(K\) and \(F\) is just
\[\Delta E_d =-\Delta E_f\]
which is \(549 \ kJ/mol\). This is the energy needed to overcome the Coulomb attraction and transfer an electron from \(F^-\) to \(K^+\) to form neutral species.
This estimated dissociation energy is in qualitative agreement with the experimental value of 498 kJ/mol. However, it can never be quantitative agreement because the energy really needs to be computed using the rules of quantum mechanics, with a proper accounting of the electronic wave functions in the bond.