CHEMISTRY THE CENTRAL SCIENCE

8 BASIC CONCEPTS OF CHEMICAL BONDING

8.4 BOND POLARITY AND ELECT RONEGATIVITY

When two identical atoms bond, as in Cl2 or H2, the electron pairs must be shared equally. When two atoms from opposites sides of the periodic table bond, such as NaCl, there is relatively little sharing of electrons, which means that NaCl is best described as composed of Na+ and Cl- ions. The 3s electron of the Na atom is, in effect, transferred completely to chlorine. The bonds that are found in most substances fall somewhere between these extremes.

Bond polarity is a measure of how equally or unequally the electrons in any covalent bond are shared. A nonpolar covalent bond is one in which the electrons are shared equally, as in Cl2 and N2. In a polar covalent bond, one of the atoms exerts a greater attraction for the bonding electrons than the other. If the difference in relative ability to attract electrons is large enough, an ionic bond is formed.

Electronegativity

We use a quantity called electronegativity to estimate whether a given bond is nonpolar covalent, polar covalent, or ionic. Electronegativity is defined as the ability of an atom in a molecule to attract electrons to itself. The greater an atom's electronegativity, the greater its ability to attract electrons to itself. The electronegativity of an atom in a molecule is related to the atom's ionization energy and electron affinity, which are properties of isolated atoms. An atom with a very negative electron affinity and a high ionization energy both attracts electrons from other atoms and resists having its electrons attracted away; it is highly electronegative.

Electronegativity values can be based on a variety of properties, not just ionization energy and electron affinity. The American chemist Linus Pauling (1901-1994) developed the first and most widely used electronegativity scale, which is based on thermochemical data. As FIGURE 8.7 shows, there is generally an increase in electronegativity from left to right across a period—that is, from the most metallic to the most nonmetallic elements. With some exceptions (especially in the transition metals), electronegativity decreases with increasing atomic number in a group. This is what we expect because we know that ionization energies decrease with increasing atomic number in a group and electron affinities do not change very much.

You do not need to memorize electronegativity values. Instead, you should know the periodic trends so that you can predict which of two elements is more electronegative.

GIVE IT SOME THOUGHT

How does the electronegativity of an element differ from its electron affinity?

GO FIGURE

For the group 6A elements, what is the trend in electronegativity with increasing atomic number?

FIGURE 8.7 Electronegativity values based on Pauling's thermochemical data.

Electronegativity and Bond Polarity

We can use the difference in electronegativity between two atoms to gauge the polarity of the bond the atoms form. Consider these three fluorine-containing compounds:

In F2 the electrons are shared equally between the fluorine atoms and, thus, the covalent bond is nonpolar. A nonpolar covalent bond results when the electronegativities of the bonded atoms are equal.

In HF the fluorine atom has a greater electronegativity than the hydrogen atom, with the result that the electrons are shared unequally—the bond is polar. In general, a polar covalent bond results when the atoms differ in electronegativity. In HF the more electronegative fluorine atom attracts electron density away from the less electronegative hydrogen atom, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the fluorine atom. We can represent this charge distribution as

The δ+ and δ− (read “delta plus” and “delta minus”) symbolize the partial positive and negative charges, respectively.

In LiF the electronegativity difference is very large, meaning that the electron density is shifted far toward F. The resultant bond is therefore most accurately described as ionic.

The shift of electron density toward the more electronegative atom in a bond can be seen in the results of calculations of electron density distributions. For the three species in our example, the calculated electron density distributions are shown in FIGURE 8.8. You can see that in F2the distribution is symmetrical, in HF the electron density is clearly shifted toward fluorine, and in LiF the shift is even greater. These examples illustrate, therefore, that the greater the difference in electronegativity between two atoms, the more polar their bond.

FIGURE 8.8 Electron density distribution. This computer-generated rendering shows the calculated electron-density distribution on the surface of the F2, HF, and LiF molecules.

GIVE IT SOME THOUGHT

Based on differences in electronegativity, how would you characterize the bonding in sulfur dioxide, SO2? Do you expect the bonds between S and O to be nonpolar, polar covalent, or ionic?

SAMPLE EXERCISE 8.4 Bond Polarity

In each case, which bond is more polar: (a) B—Cl or C—Cl, (b) P—F or P—Cl? Indicate in each case which atom has the partial negative charge.

SOLUTION

Analyze We are asked to determine relative bond polarities, given nothing but the atoms involved in the bonds.

Plan Because we are not asked for quantitative answers, we can use the periodic table and our knowledge of electronegativity trends to answer the question.

Solve

(a) The chlorine atom is common to both bonds. Therefore, the analysis reduces to a comparison of the electronegativities of B and C. Because boron is to the left of carbon in the periodic table, we predict that boron has the lower electronegativity. Chlorine, being on the right side of the table, has a higher electronegativity. The more polar bond will be the one between the atoms having the lowest electronegativity (boron) and the highest electronegativity (chlorine). Consequently, the B—Cl bond is more polar; the chlorine atom carries the partial negative charge because it has a higher electronegativity.

(b) In this example phosphorus is common to both bonds, and the analysis reduces to a comparison of the electronegativities of F and Cl. Because fluorine is above chlorine in the periodic table, it should be more electronegative and will form the more polar bond with P. The higher electronegativity of fluorine means that it will carry the partial negative charge.

Check

(a) Using Figure 8.7: The difference in the electronegativities of chlorine and boron is 3.0 − 2.0 = 1.0; the difference between chlorine and carbon is 3.0 − 2.5 = 0.5. Hence, the B—Cl bond is more polar, as we had predicted.

(b) Using Figure 8.7: The difference in the electronegativities of chlorine and phosphorus is 3.0 − 2.1 = 0.9; the difference between fluorine and phosphorus is 4.0 − 2.1 = 1.9. Hence, the P—F bond is more polar, as we had predicted.

PRACTICE EXERCISE

Which of the following bonds is most polar: S—Cl, S—Br, Se—Cl, or Se—Br?

Answer: Se—Cl

Dipole Moments

The difference in electronegativity between H and F leads to a polar covalent bond in the HF molecule. As a consequence, there is a concentration of negative charge on the more electronegative F atom, leaving the less electronegative H atom at the positive end of the molecule. A molecule such as HF, in which the centers of positive and negative charge do not coincide, is a polar molecule. Thus, we describe both bonds and entire molecules as being polar and nonpolar.

We can indicate the polarity of the HF molecule in two ways:

In the notation on the right, the arrow denotes the shift in electron density toward the fluorine atom. The crossed end of the arrow can be thought of as a plus sign designating the positive end of the molecule.

Polarity helps determine many properties we observe at the macroscopic level in the laboratory and in everyday life. Polar molecules align themselves with respect to one another, with the negative end of one molecule and the positive end of another attracting each other. Polar molecules are likewise attracted to ions. The negative end of a polar molecule is attracted to a positive ion, and the positive end is attracted to a negative ion. These interactions account for many properties of liquids, solids, and solutions, as you will see in Chapters 11,12, and 13.

How can we quantify the polarity of a molecule? Whenever two electrical charges of equal magnitude but opposite sign are separated by a distance, a dipole is established. The quantitative measure of the magnitude of a dipole is called its dipole moment, denoted μ. If two equal and opposite charges Q+ and Q− are separated by a distance r, as in FIGURE 8.9, the magnitude of the dipole moment the product of Q and r:

GO FIGURE

If the charged particles are moved closer together, does μ increase, decrease, or stay the same?

FIGURE 8.9 Dipole and dipole moment. When charges of equal magnitude and opposite sign Q+ and Q- are separated by a distance r, a dipole is produced.

This expression tells us that dipole moment increases as the magnitude of Q increases and as r increases. For a nonpolar molecule, such as F2, the dipole moment is zero because there is no charge separation.

GIVE IT SOME THOUGHT

Chlorine monofluoride, ClF, and iodine monofluoride, IF, are interhalogen compounds—compounds that contain bonds between different halogen elements. Which of these molecules has the larger dipole moment?

Dipole moments are usually reported in debyes (D), a unit that equals 3.34 × 10−30 coulomb-meters (C-m). For molecules, we usually measure charge in units of the electronic charge e, 1.60 × 10−19 C, and distance in angstroms. This means we need to convert units whenever we want to report a dipole moment in debyes. Suppose that two charges 1 + and 1 − (in units of e) are separated by 1.00 Å. The dipole moment produced is

Measurement of the dipole moments can provide us with valuable information about the charge distributions in molecules, as illustrated in Sample Exercise 8.5.

SAMPLE EXERCISE 8.5 Dipole Moments of Diatomic Molecules

The bond length in the HCl molecule is 1.27 Å. (a) Calculate the dipole moment, in debyes, that results if the charges on the H and Cl atoms were 1+ and 1-, respectively. (b) The experimentally measured dipole moment of HCl(g) is 1.08 D. What magnitude of charge, in units of e, on the H and Cl atoms leads to this dipole moment?

SOLUTION

Analyze and Plan We are asked in (a) to calculate the dipole moment of HCl that would result if there were a full charge transferred from H to Cl. We can use Equation 8.11 to obtain this result. In (b), we are given the actual dipole moment for the molecule and will use that value to calculate the actual partial charges on the H and Cl atoms.

Solve:

(a) The charge on each atom is the electronic charge, e = 1.60 × 10-19 C. The separation is 1.27 Å. The dipole moment is therefore

(b) We know the value of μ, 1.08 D, and the value of r, 1.27 Å. We want to calculate the value of Q:

We can readily convert this charge to units of e:

Change in 

Thus, the experimental dipole moment indicates that the charge separation in the HCl molecule is

Because the experimental dipole moment is less than that calculated in part (a), the charges on the atoms are much less than a full electronic charge. We could have anticipated this because the H—Cl bond is polar covalent rather than ionic.

PRACTICE EXERCISE

The dipole moment of chlorine monofluoride, ClF(g), is 0.88 D. The bond length of the molecule is 1.63 Å. (a) Which atom is expected to have the partial negative charge? (b) What is the charge on that atom in units of e?

Answers: (a) F, (b) 0.11-

TABLE 8.3 • Bond Lengths, Electronegativity Differences, and Dipole Moments of the Hydrogen Halides

TABLE 8.3 presents the bond lengths and dipole moments of the hydrogen halides. Notice that as we proceed from HF to HI, the electronegativity difference decreases and the bond length increases. The first effect decreases the amount of charge separated and causes the dipole moment to decrease from HF to HI, even though the bond length is increasing. Calculations identical to those used in Sample Exercise 8.5 show that the charges on the atoms decrease from 0.41+ and 0.41- in HF to 0.057+ and 0.057- in HI. We can visualize the varying degree of electronic charge shift in these substances from computer-generated renderings based on calculations of electron distribution, as shown in FIGURE 8.10. For these molecules, the change in the electronegativity difference has a greater effect on the dipole moment than does the change in bond length.

GIVE IT SOME THOUGHT

The bond between carbon and hydrogen is one of the most important types of bonds in chemistry. The length of a H—C bond is approximately 1.1 Å. Based on this distance and differences in electronegativity, do you expect the dipole moment of an individual H—C bond to be larger or smaller than that of the H—I bond?

Before leaving this section, let's return to the LiF molecule in Figure 8.8. Under standard conditions, LiF exists as an ionic solid with an arrangement of atoms analogous to the sodium chloride structure shown in Figure 8.3. However, it is possible to generate LiF molecules by vaporizing the ionic solid at high temperature. The molecules have a dipole moment of 6.28 D and a bond distance of 1.53 Å. From these values we can calculate the charge on lithium and fluorine to be 0.857+ and 0.857-, respectively. This bond is extremely polar, and the presence of such large charges strongly favors the formation of an extended ionic lattice in which each lithium ion is surrounded by fluoride ions and vice versa.

GO FIGURE

How do you interpret the fact that there is no red in the HBr and HI representations?

FIGURE 8.10 Charge separation in the hydrogen halides. In HF, the strongly electronegative F pulls much of the electron density away from H. In HI, the I, being much less electronegative than F, does not attract the shared electrons as strongly and, consequently, there is far less polarization of the bond.

Differentiating Ionic and Covalent Bonding

To understand the interactions responsible for chemical bonding, it is advantageous to treat ionic and covalent bonding separately. That is the approach taken in this chapter, as well as in most other undergraduate-level chemistry texts. In reality, however, there is a continuum between the extremes of ionic and covalent bonding. This lack of a well-defined separation between the two types of bonding may seem unsettling or confusing at first.

The simple models of ionic and covalent bonding presented in this chapter go a long way toward understanding and predicting the structures and properties of chemical compounds. When covalent bonding is dominant, more often than not we expect compounds to exist as molecules,*having all the properties we associate with molecular substances, such as relatively low melting and boiling points and nonelectrolyte behavior when dissolved in water. When ionic bonding is dominant, we expect the compounds to be brittle, high-melting solids with extended lattice structures and exhibiting strong electrolyte behavior when dissolved in water.

There are, of course, exceptions to these general characterizations, some of which we examine later in the book. Nonetheless, the ability to quickly categorize the predominant bonding interactions in a substance as covalent or ionic imparts considerable insight into the properties of that substance. The question then becomes the best way to recognize which type of bonding dominates.

The simplest approach is to assume that the interaction between a metal and a non-metal is ionic and that between two nonmetals is covalent. While this classification scheme is reasonably predictive, there are far too many exceptions to use it blindly. For example, tin is a metal and chlorine is a nonmetal, but SnCl4 is a molecular substance that exists as a colorless liquid at room temperature. It freezes at —33 °C and boils at 114 °C. Clearly this substance does not have the characteristics of an ionic substance. A more sophisticated approach is to use the difference in electronegativity as the main criterion for determining whether ionic or covalent bonding will be dominant. This approach correctly predicts the bonding in SnCl4 to be polar covalent based on an electronegativity difference of 1.2 and at the same time correctly predicts the bonding in NaCl to be predominantly ionic based on an electronegativity difference of 2.1.

Evaluating bonding based on electronegativity difference is a useful system, but it has one shortcoming. The electronegativity values given in Figure 8.7 do not take into account changes in bonding that accompany changes in the oxidation state of the metal. For example, Figure 8.7gives the electronegativity difference between manganese and oxygen as 3.5 − 1.5 = 2.0, which falls in the range where the bonding is normally considered ionic (the electronegativity difference for NaCl is 3.0 − 0.9 = 2.1). Therefore, it is not surprising to learn that manganese(II) oxide, MnO, is a green solid that melts at 1842 °C and has the same crystal structure as NaCl.

However, the bonding between manganese and oxygen is not always ionic. Man-ganese(VII) oxide, Mn2O7, is a green liquid that freezes at 5.9 °C, which indicates that covalent rather than ionic bonding dominates. The change in the oxidation state of manganese is responsible for the change in bonding. In general, as the oxidation state of a metal increases, so does the degree of covalent bonding. When the oxidation state of the metal is highly positive (roughly speaking, +4 or larger), we should expect significant covalency in the bonds it forms with nonmetals. Thus, with metals in high oxidation states we find molecular substances, such as Mn2O7, or polyatomic ions, such as MnO4− and CrO4, rather than ionic compounds.

GIVE IT SOME THOUGHT

You have a yellow solid that melts at 41 °C and boils at 131 °C and a green solid that melts at 2320 °C. If you are told that one of them is Cr2O3 and the other is OsO4, which one do you expect to be the yellow solid?