CHEMISTRY THE CENTRAL SCIENCE

8 BASIC CONCEPTS OF CHEMICAL BONDING

8.5 DRAWING LEWIS STRUCTURES

Lewis structures can help us understand the bonding in many compounds and are frequently used when discussing the properties of molecules. For this reason, drawing Lewis structures is an important skill that you should practice. To do so, you should use the following procedure.

1. Sum the valence electrons from all atoms. (Use the periodic table to help you determine the number of valence electrons in each atom.) For an anion, add one electron to the total for each negative charge. For a cation, subtract one electron from the total for each positive charge. Do not worry about keeping track of which electrons come from which atoms. Only the total number is important.

2. Write the symbols for the atoms, show which atoms are attached to which, and connect them with a single bond (a dash, representing two electrons). Chemical formulas are often written in the order in which the atoms are connected in the molecule or ion. The formula HCN, for example, tells you that the carbon atom is bonded to the H and to the N. In many polyatomic molecules and ions, the central atom is usually written first, as in  and SF4. Remember that the central atom is generally less electronegative than the atoms surrounding it. In other cases, you may need more information before you can draw the Lewis structure.

3. Complete the octets around all the atoms bonded to the central atom. Remember, however, that a hydrogen atom has only a single pair of electrons around it.

4. Place any leftover electrons on the central atom, even if doing so results in more than an octet of electrons around the atom.

5. If there are not enough electrons to give the central atom an octet, try multiple bonds. Use one or more of the unshared pairs of electrons on the atoms bonded to the central atom to form double or triple bonds.

SAMPLE EXERCISE 8.6 Drawing a Lewis Structure

Draw the Lewis structure for phosphorus trichloride, PCl3.

SOLUTION

Analyze and Plan We are asked to draw a Lewis structure from a molecular formula. Our plan is to follow the five-step procedure just described.

Solve

First, we sum the valence electrons. Phosphorus (group 5A) has five valence electrons, and each chlorine (group 7A) has seven. The total number of valence electrons is therefore

5 + (3 × 7) = 26

Second, we arrange the atoms to show which atom is connected to which, and we draw a single bond between them. There are various ways the atoms might be arranged. In binary compounds, however, the first element in the chemical formula is generally surrounded by the remaining atoms. Thus, we begin with a skeleton structure that shows a single bond between the P atom and each Cl atom:

(It is not crucial that the Cl atoms be left of, right of, and below the P atom—any structure that shows each of the three Cl atoms bonded to P will work.)

Third, we complete the octets on the atoms bonded to the central atom. Placing octets around each Cl atom accounts for 24 electrons (remember, each line in our structure represents two electrons):

Fourth, recalling that our total number of electrons is 26, we place the remaining two electrons on the central atom, completing the octet around it:

This structure gives each atom an octet, so we stop at this point. (In checking for octets, remember to count both electrons in a single bond twice, once for each atom in the bond.)

PRACTICE EXERCISE

(a) How many valence electrons should appear in the Lewis structure for CH2Cl2?

(b) Draw the Lewis structure.

Answers: 

SAMPLE EXERCISE 8.7 Lewis Structure with a Multiple Bond

Draw the Lewis structure for HCN.

SOLUTION

Hydrogen has one valence electron, carbon (group 4A) has four, and nitrogen (group 5A) has five. The total number of valence electrons is, therefore, 1 + 4 + 5 = 10. In principle, there are different ways in which we might choose to arrange the atoms. Because hydrogen can accommodate only one electron pair, it always has only one single bond associated with it. Therefore, C—H—N is an impossible arrangement. The remaining two possibilities are H—C—N and H—N—C. The first is the arrangement found experimentally. You might have guessed this because (a) the formula is written with the atoms in this order and (b) carbon is less electronegative than nitrogen. Thus, we begin with the skeleton structure

H—C—N

The two bonds account for four electrons. The H atom can have only two electrons associated with it, and so we will not add any more electrons to it. If we place the remaining six electrons around N to give it an octet, we do not achieve an octet on C:

We therefore try a double bond between C and N, using one of the unshared pairs we placed on N. Again we end up with fewer than eight electrons on C, and so we next try a triple bond. This structure gives an octet around both C and N:

The octet rule is satisfied for the C and N atoms, and the H atom has two electrons around it. This is a correct Lewis structure.

PRACTICE EXERCISE

Draw the Lewis structure for (a) NO+ ion, (b) C2H4.

Answers: 

SAMPLE EXERCISE 8.8 Lewis Structure for a Polyatomic Ion

Draw the Lewis structure for the  ion.

SOLUTION

Bromine (group 7A) has seven valence electrons, and oxygen (group 6A) has six. We must add one more electron to our sum to account for the 1- charge of the ion. The total number of valence electrons is, therefore, 7 + (3 × 6) + 1 = 26. For oxyanions—, and so forth—the oxygen atoms surround the central nonmetal atom. After following this format and then putting in the single bonds and distributing the unshared electron pairs, we have

Notice that the Lewis structure for an ion is written in brackets and the charge is shown outside the brackets at the upper right.

PRACTICE EXERCISE

Draw the Lewis structure for .

Answers: 

Formal Charge and Alternative Lewis Structures

When we draw a Lewis structure, we are describing how the electrons are distributed in a molecule or polyatomic ion. In some instances we can draw more than one Lewis structure and have all of them obey the octet rule. All these structures can be thought of as contributing to the actualarrangement of the electrons in the molecule, but not all of them will contribute to the same extent. How do we decide which of several Lewis structures is the most important? One approach is to do some “bookkeeping” of the valence electrons to determine the formal charge of each atom in each Lewis structure. The formal charge of any atom in a molecule is the charge the atom would have if all the atoms in the molecule had the same electronegativity (that is, if each bonding electron pair in the molecule were shared equally between its two atoms).

To calculate the formal charge on any atom in a Lewis structure, we assign electrons to the atom as follows:

1. All unshared (nonbonding) electrons are assigned to the atom on which they are found.

2. For any bond—single, double, or triple—half of the bonding electrons are assigned to each atom in the bond.

The formal charge of each atom is calculated by subtracting the number of electrons assigned to the atom from the number of valence electrons in the neutral atom.

Let's practice by calculating the formal charges in the cyanide ion, CN, which has the Lewis structure

For the C atom, there are two nonbonding electrons and three electrons from the six in the triple bond  for a total of five. The number of valence electrons on a neutral C atom is four. Thus, the formal charge on C is 4 – 5 = –1. For N, there are two nonbonding electrons and three electrons from the triple bond. Because the number of valence electrons on a neutral N atom is five, its formal charge is 5 – 5 = 0:

Notice that the sum of the formal charges equals the overall charge on the ion, 1 —. The formal charges on a neutral molecule must add to zero, whereas those on an ion add to give the charge on the ion.

If we can draw several Lewis structures for a molecule, the concept of formal charge can help us decide which is the most important, which we shall call the dominant Lewis structure. One Lewis structure for CO2, for instance, has two double bonds, as we saw on page 298. However, we can also satisfy the octet rule by drawing a Lewis structure having one single bond and one triple bond. Calculating formal charges in these structures, we have

Note that in both cases the formal charges add up to zero, as they must because CO2 is a neutral molecule. So, which is the more correct structure? As a general rule, when more than one Lewis structure is possible, we will use the following guidelines to choose the dominant one:

1. The dominant Lewis structure is generally the one in which the atoms bear formal charges closest to zero.

2. A Lewis structure in which any negative charges reside on the more electronegative atoms is generally more dominant than one that has negative charges on the less electronegative atoms.

Thus, the first Lewis structure of CO2 is the dominant one because the atoms carry no formal charges and so satisfy the first guideline. The other Lewis structure shown (and the similar one that has a triple bond to the left O and a single bond to the right O) do contribute to the actual structure but to a much smaller extent.

Although the concept of formal charge helps us to arrange alternative Lewis structures in order of importance, it is important that you remember that formal charges do not represent real charges on atoms. These charges are just a bookkeeping convention. The actual charge distributions in molecules and ions are determined not by formal charges but by a number of other factors, including electronegativity differences between atoms.

GIVE IT SOME THOUGHT

Suppose a Lewis structure for a neutral fluorine-containing molecule results in a formal charge on the fluorine atom of +1. What conclusion would you draw?

SAMPLE EXERCISE 8.9 Lewis Structures and Formal Charges

Three possible Lewis structures for the thiocyanate ion, NCS-, are

(a) Determine the formal charges in each structure. (b) Based on the formal charges, which Lewis structure is the dominant one?

SOLUTION

(a) Neutral N, C, and S atoms have five, four, and six valence electrons, respectively. We can determine the formal charges in the three structures by using the rules we just discussed:

As they must, the formal charges in all three structures sum to 1-, the overall charge of the ion. (b) The dominant Lewis structure generally produces formal charges of the smallest magnitude (guideline 1). That rules out the left structure as the dominant one. Further, as discussed in Section 8.4, N is more electronegative than C or S. Therefore, we expect any negative formal charge to reside on the N atom (guideline 2). For these two reasons, the middle Lewis structure is the dominant one for NCS-.

PRACTICE EXERCISE

The cyanate ion, NCO-, has three possible Lewis structures. (a) Draw these three structures and assign formal charges in each. (b) Which Lewis structure is dominant?

(b) Structure (iii), which places a negative charge on oxygen, the most electronegative element in the ion, is the dominant Lewis structure.

A CLOSER LOOK
OXIDATION NUMBERS, FORMAL CHARGES, AND ACTUAL PARTIAL CHARGES

In Chapter 4 we introduced the rules for assigning oxidation numbers to atoms. The concept of electronegativity is the basis of these numbers. An atom's oxidation number is the charge the atom would have if its bonds were completely ionic. That is, in determining oxidation number, all shared electrons are counted with the more electronegative atom. For example, consider the Lewis structure of HCl in FIGURE 8.11 (a). To assign oxidation numbers, both electrons in the covalent bond between the atoms are assigned to the more electronegative Cl atom. This procedure gives Cl eight valence electrons, one more than in the neutral atom. Thus, its oxidation number is − 1. Hydrogen has no valence electrons when they are counted this way, giving it an oxidation number of +1.

To assign formal charges in this molecule, we ignore electronegativity and assign the electrons in bonds equally to the two bonded atoms. In the case of HCl, we divide the bonding pair of electrons equally between H and Cl, as in Figure 8.11(b). In this case Cl has seven assigned electrons, the same as that of the neutral Cl atom, and H has one assigned electron. Thus, the formal charges of both Cl and H in this compound are 0.

Neither oxidation number nor formal charge gives an accurate depiction of the actual charges on atoms because oxidation numbers overstate the role of electronegativity and formal charges ignore it. It seems reasonable that electrons in covalent bonds should be apportioned according to the relative electronegativities of the bonded atoms. From Figure 8.7 we see that Cl has an electronegativity of 3.0, while that of H is 2.1. The more electronegative Cl atom might therefore be expected to have roughly 3.0/(3.0 + 2.1) = 0.59 of the electrical charge in the bonding pair, whereas the H atom has 2.1/(3.0 + 2.1) = 0.41 of the charge. Because the bond consists of two electrons, the Cl atom's share is 0.59 × 2e = 1.18e, or 0.18e more than the neutral Cl atom. This gives rise to a partial negative charge of 0.18- on Cl and a partial positive charge of 0.18+ on H. (Notice again that we place the plus and minus signs before the magnitude in writing oxidation numbers and formal charges but after the magnitude in writing actual charges.)

The dipole moment of HCl gives an experimental measure of the partial charge on each atom. In Sample Exercise 8.5 we saw that the dipole moment of HCl indicates a partial charge of 0.178+ on H and 0.178- on Cl, in remarkably good agreement with our simple approximation based on electronegativities. Although our approximation method provides “ballpark” numbers for the magnitude of charge on atoms, the relationship between electronegativities and charge separation is generally more complicated. As we have already seen, computer programs employing quantum mechanical principles have been developed to obtain more accurate estimates of the partial charges on atoms, even in complex molecules. A computer-graphical representation of the charge distribution in HCl is shown in Figure 8.11(c).

RELATED EXERCISES 8.8, 8.49, 8.50, 8.51, 8.52, 8.86, 8.87, 8.90, and 8.91

FIGURE 8.11 Oxidation number, formal charge, and electron density distribution for the HCl molecule.