8.4^8.4. Electronegativity and Polarity^308^313^,,^18708^18937%
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8.4
Electronegativity and Polarity

So far, we have described chemical bonds as either ionic, when they occur between a metal and a nonmetal, or covalent, when they occur between nonmetals. In fact, ionic and covalent bonds are simply the extremes in a spectrum of bonding. Bonds that fall between these two extremes are polar, meaning that electrons are shared but are not shared equally. Such bonds are referred to as polar covalent bonds. The following shows a comparison of the different types of bonds, where M and X represent two different elements:

Student Annotation: In fact, the transfer of electrons in NaF is nearly complete. Even in an ionic bond, the electrons in question spend a small amount of time near the cation.

To illustrate the spectrum of bonding, let's consider three substances: H2, HF, and NaF. In the H2 molecule, where the two bonding atoms are identical, the electrons are shared equally. That is, the electrons in the covalent bond spend roughly the same amount of time in the vicinity of each H atom. In the HF molecule, on the other hand, where the two bonding atoms are different, the electrons are not shared equally. They spend more time in the vicinity of the F atom than in the vicinity of the H atom. (The δ symbol is used to denote partial charges on the atoms.) In NaF, the electrons are not shared at all but rather are transferred from sodium to fluorine.

One way to visualize the distribution of electrons in species such as H2, HF, and NaF is to use electrostatic potential models (Figure 8.5). These models show regions where electrons spend a lot of time in red, and regions where electrons spend very little time in blue. (Regions where electrons spend a moderate amount of time appear in green.)

Figure 8.5
Electron density maps show the distribution of charge in a covalent species (H2), a polar covalent species (HF), and an ionic species (NaF). The most electron-rich regions are red; the most electron-poor regions are blue.
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Electronegativity

Electronegativity is the ability of an atom in a compound to draw electrons to itself. It determines where electrons in a compound spend most of their time. Elements with high electronegativity have a greater tendency to attract electrons than do elements with low electronegativity. Electronegativity is related to electron affinity and ionization energy. An atom such as fluorine, which has a high electron affinity (tends to accept electrons) and a high ionization energy (does not lose electrons easily), has a high electronegativity. Sodium, on the other hand, has a low electron affinity, a low ionization energy, and therefore a low electronegativity.

Electronegativity is a relative concept, meaning that an element's electronegativity can be measured only in relation to the electronegativity of other elements. Linus Pauling2 devised a method for calculating the relative electronegativities of most elements. These values are shown in Figure 8.6. In general, electronegativity increases from left to right across a period in the periodic table, as the metallic character of the elements decreases. Within each group, electronegativity decreases with increasing atomic number and increasing metallic character. The transition metals do not follow these trends. The most electronegative elements (the halogens, oxygen, nitrogen, and sulfur) are found in the upper right-hand corner of the periodic table, and the least electronegative elements (the alkali and alkaline earth metals) are clustered near the lower left-hand corner. These trends are readily apparent in the graph in Figure 8.7.

Figure 8.6
Electronegativities of common elements.
Figure 8.7
Variation of electronegativity with atomic number.
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Electronegativity and electron affinity are related but distinct concepts. Both indicate the tendency of an atom to attract electrons. Electron affinity, however, refers to an isolated atom's ability to attract an additional electron in the gas phase, whereas electronegativity refers to the ability of an atom in a chemical bond (with another atom) to attract the shared electrons. Electron affinity, moreover, is an experimentally measurable quantity, whereas electronegativity is an estimated number that cannot be measured directly.

Atoms of elements with widely different electronegativities tend to form ionic compounds with each other because the atom of the less electronegative element gives up its electrons to the atom of the more electronegative element. Atoms of elements with comparable electronegativities tend to form polar covalent bonds, or simply polar bonds, with each other because the shift in electron density from one atom to the other is usually small. Only atoms of the same element, which have the same electronegativity, can be joined by a pure covalent bond.

There is no sharp distinction between nonpolar covalent and polar covalent or between polar covalent and ionic, but the following guidelines can help distinguish among them:

Sometimes chemists describe bonds using the term percent ionic character [ Page 312]. A purely ionic bond would have 100 percent ionic character (although no such bond is known). A purely covalent, nonpolar bond has 0 percent ionic character.

Sample Problem 8.4 shows how to use electronegativities to identify a chemical bond as nonpolar, polar, or ionic.

Sample Problem 8.4

Classify the following bonds as nonpolar, polar, or ionic: (a) the bond in ClF, (b) the bond in CsBr, and (c) the carbon-carbon double bond in C2H4.

Strategy

Using the information in Figure 8.6, determine which bonds have identical, similar, and widely different electronegativities.

Setup

Electronegativity values from Figure 8.6 are: Cl (3.0), F (4.0), Cs (0.7), Br (2.8), C (2.5).

Solution

(a) The difference between the electronegativities of F and Cl is 4.0 − 3.0 = 1.0, making the bond in ClF polar.

(b) In CsBr, the difference is 2.8 − 0.7 = 2.1, making the bond ionic.

Think About It

By convention, the difference in electronegativity is always calculated by subtracting the smaller number from the larger one, so the result is always positive.

(c) In C2H4, the two atoms are identical. (Not only are they the same element, but each C atom is bonded to two H atoms.) The carbon-carbon double bond in C2H4 is nonpolar.

Practice Problem A

Classify the following bonds as nonpolar, polar, or ionic: (a) the bonds in H2S, (b) the H—O bonds in H2O2, and (c) the O—O bond in H2O2.

Practice Problem B

Electrostatic potential maps are shown for HCl and LiH. Determine which diagram is which. (The H atom is shown on the left in both.)

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Dipole Moment, Partial Charges, and Percent Ionic Character

The shift of electron density in a polar bond is symbolized by placing a crossed arrow (a dipole arrow) above the Lewis structure to indicate the direction of the shift. For example,

The consequent charge separation can be represented as

Student Annotation: The distance, r, between partial charges in a polar diatomic molecule is the bond length expressed in meters. Bond lengths are usually given in angstroms (Å) or picometers (pm), so it is generally necessary to convert to meters.

A quantitative measure of the polarity of a bond is its dipole moment (μ), which is calculated as the product of the charge (Q) and the distance (r) between the charges:

Equation 8.1

For a diatomic molecule containing a polar bond to be electrically neutral, the partial positive and partial negative charges must have the same magnitude. Therefore, the Q term in Equation 8.1 refers to the magnitude of the partial charges and the calculated value of μ is always positive. Dipole moments are usually expressed in debye units (D), named for Peter Debye.3 In terms of more familiar SI units,

Student Annotation: We usually express the charge on an electron as −1. This refers to units of electronic charge. However, remember that the charge on an electron can also be expressed in coulombs [ Section 2.2]. The conversion factor between the two is necessary to calculate dipole moments: 1e = 1.6022 × 10−19 C.

where C is coulombs and m is meters. Table 8.5 lists several polar diatomic molecules, their bond lengths, and their experimentally measured dipole moments.

TABLE 8.5
Bond Lengths and Dipole Moments of the Hydrogen Halides
Molecule Bond Length (Å) Dipole Moment (D)
HF 0.92 1.82
HCL 1.28 1.08
HBr 1.41 0.82
HI 1.61 0.44

Sample Problem 8.5 shows how to use bond lengths and dipole moments to determine the magnitude of the partial charges in a polar diatomic molecule.

Student Annotation: Hydrofluoric acid has several important industrial applications, including the etching of glass and the manufacture of electronic components.
Sample Problem 8.5

Burns caused by hydrofluoric acid [HF(aq)] are unlike any other acid burns and present unique medical complications. HF solutions typically penetrate the skin and damage internal tissues, including bone, often with minimal surface damage. Less concentrated solutions actually can cause greater injury than more concentrated ones by penetrating more deeply before causing injury, thus delaying the onset of symptoms and preventing timely treatment. Determine the magnitude of the partial positive and partial negative charges in the HF molecule.

Strategy

Rearrange Equation 8.1 to solve for Q. Convert the resulting charge in coulombs to charge in units of electronic charge.

Setup

According to Table 8.5, μ = 1.82 D and r = 0.92 Å for HF. The dipole moment must be converted from debye to C · m and the distance between the ions must be converted to meters.

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Solution

In coulombs:

In units of electronic charge:

Therefore, the partial charges in HF are +0.41 and −0.41 on H and F, respectively.

Think About It

Calculated partial charges should always be less than 1. If a “partial” charge were 1 or greater, it would indicate that at least one electron had been transferred from one atom to the other. Remember that polar bonds involve unequal sharing of electrons, not a complete transfer of electrons.

Practice Problem A

Using data from Table 8.5, determine the magnitude of the partial charges in HBr.

Practice Problem B

Given that the partial charges on C and O in carbon monoxide are +0.020 and −0.020, respectively, calculate the dipole moment of CO. (The distance between the partial charges, r, is 113 pm.)

Although the designations “covalent,” “polar covalent,” and “ionic” can be useful, sometimes chemists wish to describe and compare chemical bonds with more precision. For this purpose, we can use Equation 8.1 to calculate the dipole moment we would expect if the charges on the atoms were discrete instead of partial; that is, if an electron had actually been transferred from one atom to the other. Comparing this calculated dipole moment with the measured value gives us a quantitative way to describe the nature of a bond using the term percent ionic character, which is defined as the ratio of observed μ to calculated μ, multiplied by 100.

Equation 8.2

Figure 8.8 illustrates the relationship between percent ionic character and the electronegativity difference in a heteronuclear diatomic molecule.

 
Figure 8.8
Relationship between percent ionic character and electronegativity difference.

Sample Problem 8.6 shows how to calculate percent ionic character using Equation 8.2.

Sample Problem 8.6

Using data from Table 8.5, calculate the percent ionic character of the bond in HI.

Strategy

Use Equation 8.1 to calculate the dipole moment in HI assuming that the charges on H and I are +1 and −1, respectively; and use Equation 8.2 to calculate percent ionic character. The magnitude of the charges must be expressed as coulombs (1 e = 1.6022 × 10−19 C); the bond length (r) must be expressed as meters (1 Å = 1 × 10−10 m); and the calculated dipole moment should be expressed as debyes (1 D = 3.336 × 10−30 C · m).

Setup

From Table 8.5, the bond length in HI is 1.61Å (1.61 × 10−10 m) and the measured dipole moment of HI is 0.44 D.

Solution

The dipole moment we would expect if the magnitude of charges were 1.6022 × 10−19 C is

Converting to debyes gives

Think About It

A purely covalent bond (in a homonuclear diatomic molecule such as H2) would have 0 percent ionic character. In theory, a purely ionic bond would be expected to have 100 percent ionic character, although no such bond is known.

The percent ionic character of the H—I bond is

Practice Problem A

Using data from Table 8.5, calculate the percent ionic character of the bond in HF.

Practice Problem B

Using information from Figure 7.12, and given that the NaI bond has 59.7 percent ionic character, determine the measured dipole moment of NaI.

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Checkpoint 8.4 Electronegativity and Polarity
  • 8.4.1 In which of the following molecules are the bonds most polar?

    1. H2Se

    2. H2O

    3. CO2

    4. BCl3

    5. PCl5

  • 8.4.2 Using data from Table 8.5, calculate the magnitude of the partial charges in HI.

    1. 0.39

    2. 1.8

    3. 0.057

    4. 0.60

    5. 0.15

  • 8.4.3 Arrange molecules A through E in order of increasing percent ionic character.

    1. A < B < C < D < E

    2. A = B < C < D < E

    3. A = B < C < D = E

    4. A < B < C < D = E

    5. A < B = C = D < E

  • 8.4.4 Using data from Table 8.5, calculate the percent ionic character of HCl.

    1. 85.0 percent

    2. 6.10 percent

    3. 17.7 percent

    4. 2.03 percent

    5. 20.3 percent