Worked Example 4.1 Molecular Compounds: Octet Rule and Covalent Bonds

1. Worked Example 4.1 Molecular Compounds: Octet Rule and Covalent Bonds Look at Figure 4.3 and tell whether the following molecules are likely to exist. (a) (b) (c) (d) Analysis Count the number of covalent bonds formed by each element and see if the numbers correspond to those shown in Figure 4.3. Solution (a) No. Carbon needs 4 covalent bonds but has only 3 in . (b) Yes. Both iodine and chlorine have 1 covalent bond in ICl. (c) No. Fluorine only needs 1 covalent bond to achieve an octet. It cannot form more than 1 covalent bond because it is in the second period and does not have valence d orbitals to use for bonding. (d) Yes. Sulfur, which is in group 6A like oxygen, often forms 2 covalent bonds.

2. Worked Example 4.2 Molecular Compounds: Electron-Dot Symbols Using electron-dot symbols, show the reaction between a hydrogen atom and a fluorine atom. Analysis The electron-dot symbols show the valence electrons for the hydrogen and fluorine atoms. A covalent bond is formed by the sharing of unpaired valence electrons between the 2 atoms. Solution Draw the electron-dot symbols for the H and F atoms, showing the covalent bond as a shared electron pair.

3. Worked Example 4.3 Molecular Compounds: Predicting Number of Bonds What are likely formulas for the following molecules? (a) (b) (c) Analysis The numbers of covalent bonds formed by each element should be those shown in Figure 4.3. Solution (a) Silicon typically forms 4 bonds: (b) Hydrogen forms only 1 bond: HBr (c) Phosphorus typically forms 3 bonds:

4. Worked Example 4.4 Molecular Compounds: Multiple Bonds The compound 1-butene contains a multiple bond. In the following representation, however, only the connections between atoms are shown; the multiple bond is not specifically indicated. Identify the position of the multiple bond. Analysis Look for 2 adjacent atoms that appear to have fewer than the typical number of covalent bonds, and connect those atoms by a double or triple bond. Refer to Figure 4.3 to see how many bonds will typically be formed by hydrogen and carbon atoms. Solution

5. Worked Example 4.5 Multiple Bonds: Electron-Dot and Line Structures Draw the oxygen molecule (a) using the electron-dot symbols, and (b) using lines rather than dots to indicate covalent bonds. Analysis Each oxygen atom has 6 valence electrons and will tend to form 2 covalent bonds to reach an octet. Thus, each oxygen will need to share 4 electrons to form a double bond. Solution or

6. Worked Example 4.6 Multiple Bonds: Electron Dots and Valence Electrons Draw a Lewis structure for the toxic gas hydrogen cyanide, HCN. The atoms are connected in the order shown in the preceding sentence. Analysis Follow the procedure outlined in the text. Solution STEP 1: Find the total number of valence electrons: H = 1, C = 4, N = 5 Total number of valence electrons = 10 STEP 2: Draw a line between each pair of connected atoms to represent bonding electron pairs: H—C—N 2 bonds = 4 electrons; 6 electrons remaining STEP 3: Add lone pairs so that each atom (except H) has a complete octet: STEP 4: All valence electrons have been used, and so step 4 is not needed. H and N have filled valence shells, but C does not.

7. Worked Example 4.6 Multiple Bonds: Electron Dots and Valence Electrons Continued STEP 5: If the central atom (C in this case) does not yet have an octet, use lone pairs from a neighboring atom (N) to form multiple bonds. This results in a triple bond between the C and N atoms, as shown in the electron dot and ball-and-stick representations below: We can check the structure by noting that all 10 valence electrons have been used (in 4 covalent bonds and one lone pair) and that each atom has the expected number of bonds (1 bond for H, 3 for N, and 4 for C).

8. Worked Example 4.7 Lewis Structures: Location of Multiple Bonds Draw a Lewis structure for vinyl chloride, a substance used in making polyvinyl chloride, or PVC, plastic. Analysis Since H and Cl form only 1 bond each, the carbon atoms must be bonded to each other, with the remaining atoms bonded to the carbons. With only 4 atoms available to bond with them, the carbon atoms cannot have 4 covalent bonds each unless they are joined by a double bond. Solution STEP 1: The total number of valence electrons is 18; 4 from each of the 2 C atoms, 1 from each of the 3 H atoms, and 7 from the Cl atom. STEP 2: Place the 2 C atoms in the center, and divide the 4 other atoms between them: The 5 bonds account for 10 valence electrons, with 8 remaining. STEP 3: Place 6 of the remaining valence electrons around the Cl atom so that it has a complete octet, and place the remaining 2 valence electrons on one of the C atoms (either C, it does not matter):

9. Worked Example 4.7 Lewis Structures: Location of Multiple Bonds Continued When all the valence electrons are distributed, the C atoms still do not have a complete octet; they each need 4 bonds but have only 3. STEP 4: The lone pair of electrons on the C atom can be used to form a double bond between the C atoms giving each a total of 4 bonds (8 electrons). Placement of the double bond yields the Lewis structure and ball-and-stick model for vinyl chloride shown below: All 18 valence electrons are accounted for in 6 covalent bonds and three lone pairs, and each atom has the expected number of bonds.

10. Worked Example 4.8 Lewis Structures: Octet Rule and Multiple Bonds Draw a Lewis structure for sulfur dioxide, . The connections are . Analysis Follow the procedure outlined in the text. Solution STEP 1: The total number of valence electrons is 18, 6 from each atom: STEP 2: Two covalent bonds use 4 valence electrons. STEP 3: Adding three lone pairs to each oxygen to give each an octet uses 12 additional valence electrons. STEP 4: The remaining 2 valence electrons are placed on sulfur, but sulfur still does not have an octet. STEP 5: Moving one lone pair from a neighboring oxygen to form a double bond with the central sulfur gives sulfur an octet (it does not matter on which side the S O bond is written): NOTE: The Lewis structure for includes a single bond to one O and a double bond to the other O. It doesn’t matter which O has the double bond—both structures are equally acceptable. In reality, however, the bonds in this molecule are actually closer to 1.5, an average between the two possible structures we could draw. This is an example of resonance structures, or different Lewis structures that could be used to represent the same molecule.

11. Worked Example 4.9 Lewis Structures: Molecular Shape What shape would you expect for the hydronium ion, ? Analysis Draw the Lewis structure for the molecular ion, and count the number of charge clouds around the central oxygen atom; imagine the clouds orienting as far away from one another as possible. Solution The Lewis structure for the hydronium ion shows that the oxygen atom has four charge clouds (3 single bonds and one lone pair). The hydronium ion is therefore pyramidal with bond angles of approximately 109.5°.

12. Worked Example 4.10 Lewis Structures: Charge Cloud Geometry Predict the geometry around each of the carbon atoms in an acetaldehyde molecule, . Analysis Draw the Lewis structure and identify the number of charge clouds around each of the central carbon atoms. Solution The Lewis structure of acetaldehyde shows that the carbon has four charge clouds (4 single bonds) and the CHO carbon atom has three charge clouds (2 single bonds, 1 double bond). Table 4.2 indicates that the carbon is tetrahedral, but the CHO carbon is trigonal planar.

13. Worked Example 4.11 Electronegativity: Ionic, Nonpolar, and Polar Covalent Bonds Predict whether each of the bonds between the following atoms would be ionic, polar covalent, or nonpolar covalent. If polar covalent, which atom would carry the partial positive and negative charges? (a) C and Br (b) Li and CL (c) N and H (d) Si and I Analysis Compare the electronegativity values for the atoms and classify the nature of the bonding based on the electronegativity difference. Solution (a) The electronegativity for C is 2.5, and for Br is 2.8; the difference is 0.3, indicating nonpolar covalent bonding would occur between these atoms. (b) The electronegativity for Li is 1.0, and for Cl is 3.0; the difference is 2.0, indicating that ionic bonding would occur between these atoms. (c) The electronegativity for N is 3.0, and for H is 2.5; the difference is 0.5. Bonding would be polar covalent, with and . (d) The electronegativity for Si is 1.8, and for I is 2.5; the difference is 0.7. Bonding would be polar covalent, with , and .

14. Worked Example 4.12 Electronegativity: Polar Bonds and Polar Molecules Look at the structures of (a) hydrogen cyanide (HCN) and (b) vinyl chloride ( ), described in Worked Examples 4.6 and 4.7, decide whether or not the molecules are polar, and show the direction of net polarity in each. Analysis Draw a Lewis structure for each molecule to find its shape, and identify any polar bonds using the electronegativity values in Figure 4.6 . Then, decide on net polarity by adding the individual contributions. Solution (a) The carbon atom in hydrogen cyanide has two charge clouds, making HCN a linear molecule. The bond is relatively nonpolar, but the bonding electrons are pulled toward the electronegative nitrogen atom. In addition, a lone pair protrudes from nitrogen. Thus, the molecule has a net polarity: (b) Vinyl chloride, like ethylene, is a planar molecule. The and bonds are nonpolar, but the bonding electrons are displaced toward the electronegative chlorine. Thus, the molecule has a net polarity:

15. Worked Example 4.13 Naming Molecular Compounds Name the following compounds: (a) (b) (c) Solution (a) Dinitrogen trioxide (b) Germanium tetrachloride (c) Phosphorus pentachloride

16. Worked Example 4.14 Writing Formulas for Molecular Compounds Write molecular formulas for the following compounds: (a) Nitrogen triiodide (b) Silicon tetrachloride (c) Carbon disulfide Solution (a) (b)(c)