1. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Since phosphorus is in Group 5A in the periodic table, it has 5 valence electrons. Represent these as five dots surrounding the symbol for phosphorus. EXAMPLE 10.1 Writing Lewis Structures for Elements Write a Lewis structure for Mg. SKILLBUILDER 10.1 Writing Lewis Structures for Elements FOR MORE PRACTICE Example 10.12; Problems 27, 28. Write a Lewis structure for phosphorus. Solution:

2. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Draw the Lewis structures of magnesium and oxygen by drawing two dots around the symbol for magnesium and six dots around the symbol for oxygen. EXAMPLE 10.2 Writing Ionic Lewis Structures Write a Lewis structure for the compound NaBr. SKILLBUILDER 10.2 Writing Ionic Lewis Structures FOR MORE PRACTICE Example 10.13; Problems 39, 40. Write a Lewis structure for the compound MgO. In MgO, magnesium loses its 2 valence electrons, forming a 2+ charge, and oxygen gains 2 electrons, forming a 2– charge and acquiring an octet. Solution:

3. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Draw the Lewis structures of calcium and chlorine by drawing two dots around the symbol for calcium and seven dots around the symbol for chlorine. EXAMPLE 10.3 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound Use Lewis theory to predict the formula for the compound that forms between magnesium and nitrogen. SKILLBUILDER 10.3 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound FOR MORE PRACTICE Example 10.14; Problems 41, 42, 43, 44. Use Lewis theory to predict the formula for the compound that forms between calcium and chlorine. Calcium must lose its 2 valence electrons (to effectively get an octet in its previous principal shell), while chlorine needs to gain only 1 electron to get an octet. Consequently, the compound that forms between Ca and Cl must have two chlorine atoms to every one calcium atom. Solution: The formula is therefore CaCl 2.

4. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.5 Writing Lewis Structures for Covalent Compounds 2. Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule. EXAMPLE 10.4 Write a Lewis structure for CO 2.Write a Lewis structure for CCl 4. Solution: Following the symmetry guideline, we write: O C O 1. Write the correct skeletal structure for the molecule. Solution: Following the symmetry guideline, we write: Total number of electrons for Lewis structure = Total number of electrons for Lewis structure =

5. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.5 Writing Lewis Structures for Covalent Compounds Continued 3. Distribute the electrons among the atoms, giving octets (or duets for hydrogen) to as many atoms as possible. Begin with the bonding electrons, and then proceed to lone pairs on terminal atoms, and finally to lone pairs on the central atom. EXAMPLE 10.4 4. If any atoms lack an octet, form double or triple bonds as necessary to give them octets. Since all of the atoms have octets, the Lewis structure is complete. Bonding electrons first. (8 of 32 electrons used) Lone pairs on terminal atoms next. (32 of 32 electrons used) Bonding electrons first. (4 of 16 electrons used) Lone pairs on terminal atoms next. (16 of 16 electrons used) Move lone pairs from the oxygen atoms to bonding regions to form double bonds.

6. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.5 Writing Lewis Structures for Covalent Compounds Continued EXAMPLE 10.4 SKILLBUILDER 10.4SKILLBUILDER 10.5 Write a Lewis structure for CO. Write a Lewis structure for H 2 CO. FOR MORE PRACTICE Example 10.15; Problems 49, 50, 51, 52, 53, 54.

7. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Begin by writing the skeletal structure. Since hydrogen atoms must be terminal, and following the guideline of symmetry, put the nitrogen atom in the middle surrounded by four hydrogen atoms. EXAMPLE 10.6 Writing Lewis Structures for Polyatomic Ions Write the Lewis structure for the NH 4 + ion. Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom and subtracting 1 for the positive charge. Next, place 2 electrons between each pair of atoms. Solution: Since the nitrogen atom has an octet and since all of the hydrogen atoms have duets, the placement of electrons is complete. Write the entire Lewis structure in brackets and write the charge of the ion in the upper right corner.

8. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.6 Writing Lewis Structures for Polyatomic Ions Write a Lewis structure for the ClO – ion. SKILLBUILDER 10.6 Writing Lewis Structures for Polyatomic Ions FOR MORE PRACTICE Problems 57bcd, 58abc, 59, 60. Continued

9. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Begin by writing the skeletal structure. Using the guideline of symmetry, make the three oxygen atoms terminal. EXAMPLE 10.7 Writing Resonance Structures Solution: O O N O Write a Lewis structure for the NO 3 – ions. Include resonance structures. Sum the valence electrons (adding 1 electron to account for the –1 charge) to determine the total number of electrons in the Lewis structure. Place 2 electrons between each pair of atoms. Distribute the remaining electrons, first to terminal atoms. Since there are no electrons remaining to complete the octet of the central atom, form a double bond by moving a lone pair from one of the oxygen atoms into the bonding region with nitrogen. Enclose the structure in brackets and write the charge at the upper right.

10. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Notice that you could have formed the double bond with either of the other two oxygen atoms. EXAMPLE 10.7 Writing Resonance Structures Write a Lewis structure for the NO 2 - ion. Include resonance structures. SKILLBUILDER 10.7 Writing Resonance Structures FOR MORE PRACTICE Example 10.16; Problems 57, 58, 59, 60. Continued Since the three Lewis structures are equally correct, write the three structures as resonance structures.

11. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.9 The central atom (P) has four electron groups. Predicting Geometry Using VSEPR 2. Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, and triple bonds each count as one group. EXAMPLE 10.8 The central atom (N) has three electron groups (the double bond counts as one group). Predict the electron and molecular geometry of PCl 3. Predict the electron and molecular geometry of the [NO 3 ] - ion. 1. Draw a Lewis structure for the molecule. 3. Determine the number of bonding groups and the number of lone pairs around the central atom. These should sum to the result from Step 2. Bonding groups include single bonds, double bonds, and triple bonds. Solution: PCl 3 has 26 electrons. Solution: [NO 3 ] – has 24 electrons. Three of the four electron groups around P are bonding groups, and one is a lone pair. All three of the electron groups around N are bonding groups.

12. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.9 Predicting Geometry Using VSEPR Continued EXAMPLE 10.8 SKILLBUILDER 10.8 Predict the molecular geometry of ClNO (N is the central atom). SKILLBUILDER 10.9 Predict the molecular geometry of The SO 3 2– ion. FOR MORE PRACTICE Example 10.17; Problems 67, 68, 71, 72, 75, 76. 4. Use Table 10.1 to determine the electron geometry and molecular geometry The electron geometry is tetrahedral (four electron groups), and the molecular geometry—the shape of the molecule—is trigonal pyramidal (four electron groups, three bonding groups, and one lone pair). The electron geometry is trigonal planar (three electron groups), and the molecular geometry—the shape of the molecule—is trigonal planar (three electron groups, three bonding groups, and no lone pairs).

13. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic Determine whether the bond formed between each of the following pairs of atoms is pure covalent, polar covalent, or ionic. (a) Sr and F (b) N and Cl (c) N and O Solution: (a) From Figure 10.2, we find the electronegativity of Sr (1.0) and of F (4.0). The electronegativity difference (ΔEN) is: ΔEN = 4.0 – 1.0 = 3.0 (b) From Figure 10.2, we find the electronegativity of N (3.0) and of Cl (3.0). The electronegativity difference (ΔEN) is: ΔEN = 3.0 – 3.0 = 0 Using Table 10.2, we classify this bond as pure covalent. (c) From Figure 10.2, we find the electronegativity of N (3.0) and of O (3.5). The electronegativity difference (ΔEN) is: ΔEN = 3.0 – 3.0 = 0 Using Table 10.2, we classify this bond as polar covalent.

14. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic Continued Determine whether the bond formed between each of the following pairs of atoms is pure covalent, polar covalent, or ionic. (a) I and I (b) Cs and Br (c) P and O SKILLBUILDER 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic FOR MORE PRACTICE Problems 83, 84.

15. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro Begin by drawing the Lewis structure of NH 3. Since N and H have different electronegativities, the bonds are polar EXAMPLE 10.11 Determining Whether a Molecule Is Polar FOR MORE PRACTICE Example 10.18; Problems 91, 92, 93, 94. Determine whether NH 3 is polar. The geometry of NH 3 is trigonal pyramidal (four electron groups, three bonding groups, one lone pair). Draw a three dimensional picture of NH 3 and imagine each bond as a rope that is being pulled. The pulls of the ropes do not cancel and the molecule is polar. Determine whether CH 4 is polar. SKILLBUILDER 10.11 Determining Whether a Molecule Is Polar Solution:

16. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.12 Lewis Structures for Elements What is the Lewis structure of sulfur? Solution : Since S is in Group 6A, it has 6 valence electrons. We draw these as dots surrounding its symbol, S.

17. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.13 Writing Lewis Structures for Ionic Compounds Write a Lewis structure for lithium bromide. Solution :

18. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.14 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound Use Lewis theory to predict the formula for the compound that forms between potassium and sulfur. Solution : The Lewis structures of K and S are: Potassium must lose 1 electron and sulfur must gain 2. Consequently, we need two potassium atoms to every sulfur atom. The Lewis structure is: The correct formula is K 2 S.

19. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.15 Writing Lewis Structures for Covalent Compounds Write a Lewis structure for CS 2. Solution : S C S

20. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.16 Writing Resonance Structures Write resonance structures for SeO 2. Solution : We can write a Lewis structure for SeO 2 by following the steps for writing covalent Lewis structures. We find that we can write two equally correct structures, so we draw them both as resonance structures.

21. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.17 Predicting the Shapes of Molecules Predict the geometry of SeO 2. Solution : The Lewis structure for SeO 2 (as we saw in Example 10.16) is composed of the following two resonance structures. Either of the resonance structures will give the same geometry. Total number of electron groups = 3 Number of bonding groups = 2 Number of lone pairs = 1 Electron geometry = Trigonal planar Molecular geometry = Bent

22. Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J. Tro EXAMPLE 10.18 Determining Whether a Molecule Is Polar Determine whether is SeO 2 polar. Solution : Se and O are nonmetals with different electronegativities (2.4 for Se and 3.5 for O). Therefore, the Se–O bonds are polar. As we saw in Example 10.17, the geometry of SeO 2 is bent. The polar bonds do not cancel but rather sum to give a net dipole moment. Therefore the molecule is polar.