Lewis acids and bases
Shortly after Bronsted and Lowry proposed their definition of acids and bases, the American chemist Gilbert Newton Lewis, building upon his new understanding of the nature of the chemical bond, proposed the definition that an acid is an electron pair acceptor while a base is an electron pair donor. This definition is more general than the Bronsted-Lowry definition and builds upon earlier work (1916) by Lewis that the electron pair is the major instrument of bond formation. A simple reaction that illustrates this new definition is the reaction of a proton with a hydride ion.
The proton, due to its positive charge and the absence of electrons around the proton, is capable of accepting a pair of electron. The hydride ion has two electrons (a lone pair, or non bonded pair) and can therefore donate these two electrons to the proton. In the hydrogen molecule, the product of the reaction, these two electrons are shared between the two hydrogen atoms. The product of this sharing of a pair of electrons is called an adduct (later we will see that other terms such as "complex" are also used for this product).
Let’s consider another simple example: the reaction between the proton and a water molecule. The water molecule has two pairs of electrons to share and the hydronium ion--the product of the reaction--contains a new covalent bond.
Notice that only one of the lone pairs on the water molecule has been used to form the covalent bond to the proton. It is also important to recognize that the hydronium ion has a positive charge because of the formation of the bond to the positively charged proton. Because sharing of electrons are involved in all reactions that involve covalent molecules or ions, it should come as no surprise to find that this new definition is one of the most important in chemistry. In fact, we prefer to call Lewis acid-base reactions electron-sharing reactions.
In our examples above the Lewis acid was the same—the proton—but the base was the hydride ion in the first example and water in the second example. The nature of Lewis bases is, for the most part, not troublesome, because they are the same bases that we are accustomed to in the Bronsted-Lowry definition--those species with one or more lone pairs of electrons. Lewis acids, on the other hand, are a bit different. According to Lewis an acid must be capable of accepting a pair of electrons. The most obvious candidates for Lewis acids, therefore, are those species like the proton that are cations: they have a positive charge and an empty orbital that can hold a pair of electrons.
Let’s take a closer look at a few cations. The sodium ion has a total of 11 protons in its nucleus and 10 electrons outside the nucleus; the positive charge on the atom is due to the fact that the charge on the nucleus (+11) is not completely neutralized by the charge of the ten electrons (-10). The charge itself, however, does not make the sodium ion a Lewis acid. In order to be a Lewis acid there must be an orbital or a rearrangement of electrons that can accommodate a pair of electrons. If you are unfamiliar with the concept of orbitals, please go to our “Metaphor for orbitals.”
For simple monatomic (one atom) cations, probably the simplest way to verify that an empty orbital is present is to write the electron configuration of the neutral atom. For the sodium atom the electron configuration is:
When we ionize the atom to form the Na+ cation, we remove the electron with the highest amount of energy, in this case the 3s electron. Because we removed an electron from the 3s orbital we know that this orbital is empty and is available to hold two electrons (of opposite spin, of course). Hence, Na+ is a potential Lewis acid, but does it in fact accept a pair of electrons? Forty or more years ago, chemists believed that the sodium cation was the only stable ion formed by sodium metal. It is now known that the sodide ion— Na- -- exists, though, as you might expect, it is very reactive. The sodide ion can be considered to be the product of an electron-transfer reaction (another name for an oxidation-reduction reaction) to Na+. In other words, although the reaction
Na+ + 2 e- à :Na-
shows Na+ behaving as a Lewis acid, it does not appear to involve the sharing of electrons. Thus, the equation above is better thought of as the reduction part of an oxidation-reduction reaction. The sodium ion does, however, take part in Lewis reactions if the Lewis base contains several electron rich atoms in a ring. For example, the sodium ion will react to form an adduct with a crown ether such as
In the adduct the sodium ion sits in the center of the crown ether, surrounded by electron density from the lone pair electrons on the six oxygen atoms.
Generally, the cations formed by the alkali and alkaline earth metals are not good Lewis acids unless the cation has a high charge density. Charge density is defined as the charge on the cation divided by the volume of the cation. Table 1 contains the volumes of some cations and allows us to calculate the charge density.
Table 1. The volumes of some cations, nm3 (From Marcus, et al. 2002)
Li+ 0.0014 Ca2+ 0.0042
Na+ 0.0044 Cu2+ 0.0016
Be2+ 0.00027 Al3+ 0.00062
Mg2+ 0.0016 Fe3+ 0.0012
Now let’s look at the charge densities of the Na+, Li+, Ca 2+, Be2+, and the Cu2+ ions. Using Na+ as an example, we first calculate the charge density of the ions as:
charge density = +1/0.0044 nm3
= 2.3 x 102 nm-3
The charge densities of several cations, calculated in this way, are: Li+, 7.1 x 102 nm-3; Be2+, 7.4 x 103 nm-3; Mg2+, 1.2 x 103 nm-3; Ca2+, 4.8 x 102 nm-3; Al3+, 4.8 x 103 nm-3; Cu2+, 6.2 x 102 nm-3. (We could also calculate this using the actual charge on the sodium ion as 1.6 x 10-19 coulombs.)
The charge density is a measure of the amount of charge per unit volume, similar to the density of a solid. [Gold has a density of about 19 grams per cm3, while aluminum metal has a density of about 3 g per cm3. Thus, if you had the same volume of each metal, gold would weigh about 6 times more than the aluminum.] If you compare Mg2+ and Li+, which have similar volumes, you see that the charge density of Mg2+ is about twice that of Li+. This is due to the fact that Mg2+ has twice the charge of Li+. A comparison of Be2+ with Ca2+, both of which have a +2 charge, shows that beryllium has a charge density 15 times higher than that of Ca2+ due to its much smaller size. Both Mg2+ and Be2+ are relatively good Lewis acids and react with water molecules to form electron-sharing adducts. In the reaction of Be2+ shown below, Be2+ accepts electron density from four water molecules to form a tetrahedral aquo complex. The wide bold lines between the Be and the water molecules represent the electrons that are shared. Notice that the adduct has a +2 charge, as it must in order to balance the equation.
A good question at this point is “Why does Be2+ accept four pairs of electrons (one from each of four water molecules) rather than just one? Actually, the Be2+ ion also forms adducts with just one water molecule [(Be-OH2)+2], with two water molecules [(H2O-Be-H2O)+2], and with three water molecules [ (H2O)3Be+2)]. The maximum number of coordinated (just a fancy word for “bonded”) water molecules is four, which is a result of the number of energetically available orbitals on Be2+ for occupation by the lone pairs from the water molecules. The electron configuration of Be2+ is 1s2, which leaves all four of the orbitals in the n=2 quantum level open for electrons. These four orbitals are the 2s and the three 2p orbitals. Importantly, these orbitals are at a low enough energy to allow them to be easily used in the bonding to the lone pair of electrons from the water molecules. The orbital overlap diagram below shows four hybrid sp3 orbitals overlapping with an sp3 hybrid orbital on each of the water molecules. Each set of two overlapped sp3 orbitals is occupied by two electrons that are shared by the oxygen of water and the beryllium.
Another good question is does Mg2+, the second member of the alkaline earth metals, also function as a Lewis acid? The charge density of Mg2+ is a good bit smaller than that of Be2+ due to the larger size of the Mg atom. This difference in charge density suggests that electrons are not as strongly attracted to the Mg nucleus and therefore Mg2+ is a weaker Lewis acid. It does form an aquo adduct with water that is usually written as Mg(OH2)6+2. The magnesium has a total of six orbitals that can be used to form those Mg-O bonds. This “expansion” of the octet relative to beryllium is reasonable because Mg is a third row element that has access to a 3s, three 3p and five 3d orbitals. For six-coordination the 3s and 3p and two 3d orbitals are used for the bonding. Of course, the bonding between Mg and O is weaker than that between Be and O because of the lower charge density of Mg2+.
We now turn to the Cu +2 ion. Copper and calcium are in the same row of the periodic chart, and their charge densities are similar, 4.8 x 102 nm-3 for Ca+2 and 6.2 x 102 nm-3 for Cu+2 (the charge density for copper is a bit higher because the ion is smaller). We might therefore expect their Lewis acidities to be similar. But, it turns out that Cu+2 is a much stronger Lewis acid. It forms adduct with a huge variety of Lewis bases such as the halides (F-, Cl-, Br-, I-), NH3, PH3, H2O, CN-, OH-, and so on. The calcium ion does not even form stable adducts with water or ammonia (NH3). The difference in Lewis acidities is partly due to a difference in charge densities, but mainly to the presence of low energy d-orbitals for the copper ion.
In order to understand why the copper ion and the other transition metal ions are outstanding Lewis acids, we need to look more carefully at the energy of an electron in an orbital. We will try to convince you that the d-orbitals of copper are lower in energy than those of calcium, even though both elements are in the same row of the periodic chart. To begin with, the copper ion is smaller than the calcium ion: the ionic radii are 1.00 Å for Ca+2, and 0.73 Å for Cu2+. The smaller size of the copper ion is explained by the nine additional protons (relative to calcium) in its nucleus, which exert a greater force on all of the copper electrons. Moreover, the outermost electrons in Cu2+ reside in 3d-orbitals, while the electrons in Ca+2 are in the s- and p- orbitals of the n=3 quantum levels. Thus, the electrons of both ions reside in the n=3 quantum level, but with different value of angular momentum quantum number (s, p, and d-orbitals have different values for their angular momentum).
Most importantly, the outermost electrons of copper are in d-orbitals, whereas the outermost electrons of calcium are in p orbitals. This difference is important because s- and p- electrons in their travels (this is using a clearly particle view of the electron) around the nucleus can approach the nucleus more closely than is possible for d-electrons. The closer approach of the s- and p-electrons means that these electrons essentially cancel a portion of the nuclear charge so that the remaining electrons experience the attraction of (effectively) fewer protons. The lesser attraction due to (effectively) fewer protons means that energy of the remaining electrons is higher (see Figure 1). The lower energy of the outermost orbitals in the copper ion means that any electrons that enter that orbital will be held more tightly to the nucleus. Because transition metal cations generally have these less-well shielded d-electrons, they have lower energy orbitals that can be used to hold electrons that are shared with a Lewis base. The greater attraction of these shared electrons to the more exposed nucleus strengthens the bond and makes transition metals generally good Lewis acids.
Figure 1. Shielding of outer electrons in Ca+2 and Cu2+.
Some of the most colorful Lewis adducts are those formed with transition metal cations. For example, the Cu2+ adduct formed with four ammonia molecules Cu(NH3)42+ is a dark blue color in aqueous solution, the Ni+2 complex with 6 ammonia molecules is green, whereas the Ag+ complex with two molecules of ammonia is colorless.
Cu2+ + 4 NH3 --> Cu(NH3)42+
Ni+2 + 6 NH3 --> Ni(NH3)62+
Ag+ + 2 NH3 --> Ag(NH3)2+
Now, let's take a look at another type of Lewis acid, the molecular compounds that can function as Lewis acid, in many cases, but not all, by expanding their octet. Boron trifluoride, BF3, accepts a pair of electrons very readily, and this pair of electrons makes its attack on the central atom--the boron. The boron atom has a low electronegativity of 2.0, while the attached fluorines have the highest possible electronegativity of 4.0 on the Pauling electronegativity scale. Remembering that Pauling defined electronegativity as the power of an atom to attract electrons within a molecule, we can easily imagine that the fluorine atoms have much more of an attraction (than the boron atom) for the electrons that are shared between the boron and each fluorine atom. This results in an unequal sharing of electrons that produces the computer-generated figure shown below:
Figure 2. Picture of electron density in BF3. Blue area indicates deficiency of electron density, red areas indicate excess electron density. Taken from Casetechnology.com Gaussian image.
The situation in BF3 is exacerbated by the fact that there are three B-F bonds, each of which is polar. This means that the boron has considerably less electron density than it would if the electron density were equally shared.
So, you ask, how does this make BF3 a Lewis acid? A good way to think about this goes back to the Lewis model for bonding, which relies on the idea that many atoms and ions in the main group elements should have eight electrons in their valence shell. Following the octet rule, then, for BF3 we would find the electron-dot formulas
The choice of a particular F for the double bond is arbitrary, which leads to the three resonance forms indicated by the double-head arrows. Each of these formulas clearly gives the boron a total of 8 shared electrons. Because a neutral boron atom has only 3 electrons in its valence shell these electron dot formulas indicate that the boron is electron rich (the boron has a formal charge of -1). These electron dot formulas therefore give us no way to explain the Lewis acidity of BF3.
What we are looking for is some formula that appears to be electron-deficient; that is, a formula that, for example, would have a boron atom that appears to be “eager” to acquire a pair of electrons. This formula is shown below:
In this formula all of the fluorine have an octet of electrons, but the boron does not. The boron has only six electrons in its valence shell, but according to the Lewis bonding model, it needs to share eight, not six electrons. We usually refer to this electron dot formula as an electron-deficient formula. Now let’s use this formula to show how BF3 shares an electron pair with a fluoride ion.
The equation above uses an arrow to show how the fluoride ion donates an electron pair to the electron-deficient boron to form the tetrafluoroborate adduct. Notice that in the adduct the boron has an octet of electrons.
Although we now recognize that we can think of BF3 as an electron-deficient molecule, we must always think about what orbital the Lewis acid uses to share the electron pair. BF3 is a trigonal planar molecule (the boron lies at the center of a triangle and the fluorine atoms occupy the corners of the triangle). As shown below, the plane of the molecule is generally designated as the xy plane.
The s and px and py orbitals on the boron are used to form hybrid (sp2) orbitals that form bonds to the fluorine atoms in this plane, leaving the pz orbital without any bonding function. Moreover, in the electron dot formulas that we have written above, the boron has only six electrons in its valence shell. The pz orbital on boron is therefore empty and can be used as a temporary repository of the electrons from the base. When the electron density begins to form a covalent bond the geometry at the boron changes to tetrahedral. This change in geometry also influences the type of orbital that the boron uses to bond with the fluorine. The valence bond model, which describes the bonding in terms of the overlap of orbitals, describes the bonding as involving sp3 hybrid orbitals on the boron.
The Lewis acid SnCl4 contains a central atom that resides fairly low in the periodic chart and has empty d orbitals in its valence shell. The tin atom can therefore use these d-orbitals to expand its valence shell beyond an octet. SnCl4 is the homolog of CCl4 because both tin and carbon are in the same group on the periodic chart. Although CCl4 is not a Lewis acid, SnCl4 can react with a variety of Lewis bases. For example, SnCl4 will react with the chloride ion, Cl- to form the adduct SnCl5-
or with two chloride ions to form the adduct SnCl62-.
Now, lets look at a more complicated Lewis acid--CO2. This acid is extremely important in a number of ways: a) it dissolves in water to form carbonic acid, H2CO3, which forms a buffer system with the hydrogen carbonate ion, HCO3-, to maintain our blood at a constant pH, b) in the ocean the carbonic acid reacts with the carbonate of coral and other carbonate deposits to form a buffer system that controls the pH of the oceans and maintains marine life. c) as one of the greenhouse gases in the atmosphere, it serves to reflects heat back to the earth, thereby maintaining a relatively constant (though slightly increasing) temperature on the earth's surface, and d) as a part of the carbon cycle it is utilized by plants and algae to make carbohydrates and it is produced by respiration and combustion.
Let's first examine the CO2 molecule for an electron-deficient center. The greater electronegativity of the oxygen atoms relative to the central carbon produces a distribution of electron density in the molecule that has more electron density at the oxygen atoms than at the carbon atom. This effect can be shown by the electron density map (red indicates higher density of electrons) below and also represented by the diagra
Figure 3. A depiction of the electron density in the CO2 molecule. Taken from https://www.quora.com
where the Greek delta + indicates a deficiency of electrons and a delta - indicates that the atom has more electron density than it would if the electrons were shared equally between the atoms. It appears that the carbon is somewhat electron-deficient. Thus, it is reasonable to believe that the carbon center of CO2 can be attracted to an electron pair of a Lewis base. But, where is the orbital that will hold the electron pair provided by the base? For this we need the molecular orbital model of bonding.
The molecular orbital model assumes that there are orbitals that extend over a molecule; in other words, these orbitals are analogous to atomic orbitals, except that they belong to the whole molecule. This model is a bit too complex for the present essay, but it turns out that CO2 has an orbital called an antibonding molecular orbital that lies directly above the highest occupied orbital shown in the diagram below (Figure 4). In this diagram the energies of the molecular orbitals are shown in the middle of the figure and are labeled MO1, MO2, …. The orbitals that contain the electrons are either bonding orbitals (ones that have electron density between the carbon and oxygen atoms and therefore provide the electronic “cement” that holds the atoms together) like MO1, MO4, MO5 and MO7, or non-bonding orbitals (the electron density in these orbitals resides mainly on the oxygen atoms) like MO11 and MO12. The antibonding orbitals, as their name suggests, are high in energy and, when these orbitals are populated, actually detract from the electronic cement that holds the atoms together. MO2 is the lowest unoccupied molecular orbital (the LUMO) and it is the orbital that CO2 uses to hold the pair of electrons donated by a Lewis base.
Figure 4. Molecular orbital diagram for CO2. From https://commons.wikimedia.org/wiki/File%3AMO_Diagram_CO2.jpg
Now, let’s take a look at the flow of electrons that is involved in adduct formation of CO2 with the hydroxide ion. We start with an equation for the reaction
Carbon dioxide accepts an electron pair from the hydroxide ion in order to form the adduct--the bicarbonate ion. Next, we write (Figure 5) a slightly different equation that shows how the electrons can be imagined to move as adduct formation takes place.
Figure 5. Depiction of changes in electron density as hydroxide ion attacks CO2.
On the left you can see that the head of the arrow that starts at the lone pair of electrons on the hydroxide ion is directed to the electron-deficient carbon. This pair of electrons makes its way into the antibonding LUMO of the CO2. Then the electrons begin to rearrange. This rearrangement of electrons is perhaps initiated by the decrease in stability of the CO2 due to the presence of antibonding electron density. Frequently, this rearrangement is rationalized not by using MO theory but by recognizing that the octet of electrons around the carbon would be violated (see the middle depiction above). The arrow that starts at the double bond of the C=O bond shows that a pair of electrons in the double bond moves to the oxygen as the bond to the hydroxide is formed. In other words, the arrows show the flow of electron density as the reaction proceeds. On the right in figure 5 is the adduct formed in the Lewis acid-base reaction.
The reactions that we have discussed thus far, such as the reaction of BF3 with the fluoride ion, all have the form A + B --> C and are called addition reactions. The reaction of Cu+2 with ammonia is also an addition reaction, even though four ammonia molecules eventually bond to the copper ion. This reaction has the form A + 4 B --> C. Any Lewis reaction that produces only one product is an addition reaction. The word addition results from the fact that the base adds to the acid and nothing is displaced from either the acid or base. Another type of Lewis acid-base reaction results in the displacement of an acid or base from an adduct. Displacement reactions have the form
A + B --> C + D and can generally be thought of as the reaction of a Lewis acid or base with an adduct to produce another adduct plus a displaced Lewis acid or base. For example, silver ion reacts with water molecules to form the adduct Ag(OH2)2+:
The adduct Ag(OH2)2+ formed in this addition reaction can now undergo a displacement reaction by adding a Lewis base like ammonia (NH3):
Here the base NH3 displaces the water molecules in the adduct Ag(OH2)2+ to form a new adduct Ag(NH3)2+ . Because one base displaces another base, this is a base displacement reaction.
There are also acid displacement reactions in which a Lewis acid reacts with an adduct to form another adduct with the displacement of a Lewis acid. The reaction below is used in the introductory chemistry laboratory in the qualitative analysis of cations:
Ag(NH3)2+ + 2 H+ --> 2 NH4+ + Ag+
In this reaction the Lewis acid, H+, reacts with the adduct (Ag(NH3)2+) to form the adduct NH4+ with displacement of the Lewis acid Ag+.
Marcus, Y., Jenkins, H. D. B., Glasser, L. Ion volumes: a comparison, J. Chem. Soc., Dalton Trans., 2002, 3795 – 3798.