1. Sample Problem 6.1 Ions Solution Study Check 6.1 Answer
a. Write the symbol and name for the ion that has 7 protons and 10 electrons.
b. Write the symbol and name for the ion that has 20 protons and 18 electrons.
Solution
a. The element with seven protons is nitrogen. In an ion of nitrogen with 10 electrons, the ionic charge would be 3–, [(7+) + (10–) = 3–]. The ion, written as N3–, is the nitride ion.
b. The element with 20 protons is calcium. In an ion of calcium with 18 electrons, the ionic charge would be 2+, [(20+) + (18–) = 2+]. The ion, written as Ca2+, is the calcium ion.
Study Check 6.1
How many protons and electrons are in each of the following ions?
a. S2– b. Al3+
Answer
a. 16 protons, 18 electrons b. 13 protons, 10 electrons

2. Sample Problem 6.2 Writing Symbols for Ions
Consider the elements aluminum and oxygen.
a. Identify each as a metal or a nonmetal.
b. State the number of valence electrons for each.
c. State the number of electrons that must be lost or gained for each to achieve an octet.
d. Write the symbol, including its ionic charge, and name for each resulting ion.
Solution
Study Check 6.2
Write the symbols for the ions formed by potassium and sulfur.
Answer
K+ and S2–

3. Sample Problem 6.3 Writing Formulas from Ionic Charges
Write the symbols for the ions, and the correct formula for the ionic compound formed when lithium and nitrogen react.
Solution
Lithium, which is a metal in Group 1A (1), forms Li+; nitrogen, which is a nonmetal in Group 5A (15), forms N3–. The charge of 3– is balanced by three Li+ ions.
3(1+) + 1(3–) = 0
Writing the cation (positive ion) first and the anion (negative ion) second gives the formula Li3N.
Study Check 6.3
Write the symbols for the ions, and the correct formula for the ionic compound that would form when calcium and oxygen react.
Answer
Ca2+, O2–, CaO

4. Sample Problem 6.4 Naming Ionic Compounds
Write the name for the ionic compound Mg3N2.
Solution
Step 1 Identify the cation and anion. The cation, Mg2+, is from Group 2A (2), and the anion, N3–, is from Group 5A (15).
Step 2 Name the cation by its element name. The cation Mg2+ is magnesium.
Step 3 Name the anion by using the first syllable of its element name followed by ide. The anion N3– is nitride.
Step 4 Write the name for the cation first and the name for the anion second. Mg3N2 magnesium nitride
Study Check 6.4
Name the compound Ga2S3.
Answer
gallium sulfide

5. Sample Problem 6.5 Naming Ionic Compounds with Variable Charge Metal Ions
Antifouling paint contains Cu2O, which prevents the growth of barnacles and algae on the bottoms of boats. What is the name of Cu2O?
Solution
Step 1 Determine the charge of the cation from the anion.

6. Sample Problem 6.5 Naming Ionic Compounds with Variable Charge Metal Ions
Continued
Step 2 Name the cation by its element name and use a Roman numeral in parentheses for the charge. copper(I)
Step 3 Name the anion by using the first syllable of its element name followed by ide. oxide
Step 4 Write the name for the cation first and the name for the anion second. copper(I) oxide
Study Check 6.5
Write the name for the compound with the formula Mn2S3.
Answer
manganese(III) sulfide

7. Sample Problem 6.6 Writing Formulas for Ionic Compounds
Write the correct formula for iron(III) chloride.
Solution
Step 1 Identify the cation and anion. Iron(III) chloride consists of a metal and a nonmetal, which means that it is an ionic compound.

8. Sample Problem 6.6 Writing Formulas for Ionic Compounds
Continued
Step 2 Balance the charges.
Step 3 Write the formula, cation first, using subscripts from the charge balance. FeCl3
Study Check 6.6
Write the correct formula for chromium(III) oxide.
Answer
Cr2O3

9. Sample Problem 6.7 Writing Formulas Containing Polyatomic Ions
An antacid called Amphojel contains aluminum hydroxide, which treats acid indigestion and heartburn. Write the formula for aluminum hydroxide.
Solution
Step 1 Identify the cation and polyatomic ion (anion).
Cation Polyatomic ion (anion)
aluminum hydroxide
Al3+ OH–
Step 2 Balance the charges.
Step 3 Write the formula, cation first, using the subscripts from charge balance. The formula for the compound is written by enclosing the formula of the hydroxide ion, OH–, in parentheses and writing the subscript 3 outside the right parenthesis.
Al(OH)3

10. Sample Problem 6.7 Writing Formulas Containing Polyatomic Ions
Continued
Study Check 6.7
Write the formula for a compound containing ammonium ions and phosphate ions.
Answer
(NH4)3PO4

11. Sample Problem 6.8 Naming Compounds Containing Polyatomic Ions
Name the following ionic compounds:
a. Cu(NO2)2 b. KClO3
Solution
Study Check 6.8
What is the name of Co3(PO4)2?
Answer
cobalt(II) phosphate

12. Sample Problem 6.9 Naming Molecular Compounds
Name the molecular compound NCl3.
Solution
Step 1 Name the first nonmetal by its element name. In NCl3, the first nonmetal (N) is nitrogen.
Step 2 Name the second nonmetal by using the first syllable of its element name followed by ide. The second nonmetal (Cl) is named chloride.
Step 3 Add prefixes to indicate the number of atoms (subscripts). Because there is one nitrogen atom, no prefix is needed. The subscript 3 for the Cl atoms is shown as the prefix tri. The name of NCl3 is nitrogen trichloride.

13. Sample Problem 6.9 Naming Molecular Compounds
Continued
Study Check 6.9
Write the name for each of the following molecular compounds:
a. SiBr4 b. Br2O
Answer
a. silicon tetrabromide b. dibromine oxide

14. Sample Problem 6.10 Writing Formulas for Molecular Compounds
Write the formula for the molecular compound diboron trioxide.
Solution
Step 1 Write the symbols in the order of the elements in the name.
Step 2 Write any prefixes as subscripts. The prefix di in diboron indicates that there are two atoms of boron, shown as a subscript 2 in the formula. The prefix tri in trioxide indicates that there are three atoms of oxygen, shown as a subscript 3 in the formula.
B2O3

15. Sample Problem 6.10 Writing Formulas for Molecular Compounds
Continued
Study Check 6.10
Write the formula for the molecular compound iodine pentafluoride.
Answer
IF5

16. Sample Problem 6.11 Naming Ionic and Molecular Compounds
Identify each of the following compounds as ionic or molecular and give its name:
a. K3P b. NiSO4 c. SO3
Solution
a. K3P, consisting of a metal and a nonmetal, is an ionic compound. As a representative element in Group 1A (1), K forms the potassium ion, K+. Phosphorus, as a representative element in Group 5A (15), forms a phosphide ion, P3–. Writing the name of the cation followed by the name of the anion gives the name potassium phosphide.
b. NiSO4, consisting of a cation of a transition element and a polyatomic ion SO42–, is an ionic compound. As a transition element, Ni forms more than one type of ion. In this formula, the 2– charge of SO42– is balanced by one nickel ion, Ni2+. In the name, a Roman numeral written after the metal name, nickel(II), specifies the 2+ charge. The anion SO42– is a polyatomic ion named sulfate. The compound is named nickel(II) sulfate.
c. SO3 consists of two nonmetals, which indicates that it is a molecular compound. The first element S is sulfur (no prefix is needed). The second element O, oxide, has subscript 3, which requires a prefix tri in the name. The compound is named sulfur trioxide.
Study Check 6.11
What is the name of Fe(NO3)3?
Answer
iron(III) nitrate

17. Sample Problem 6.12 Drawing Lewis Structures
Draw the lewis structure for PCl3, phosphorus trichloride, used commercially to prepare insecticides and flame retardants.
Solution
Step 1 Determine the arrangement of atoms. In PCl3, the central atom is P because there is only one P atom.
Cl P Cl Cl
Step 2 Determine the total number of valence electrons. We use the group number to determine the number of valence electrons for each of the atoms in the molecule.
Step 3 Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be represented by a bond line.

18. Sample Problem 6.12 Drawing Lewis Structures
Continued
Step 4 Place the remaining electrons using single or multiple bonds to complete the octets. Six electrons (3 × 2 e–) are used to bond the central P atom to three Cl atoms. Twenty valence electrons are left.
26 valence e– – 6 bonding e– = 20 e– remaining
We use the remaining 20 electrons as lone pairs, which are placed around theouter Cl atoms and on the P atom, such that all the atoms have octets.
Study Check 6.12
Draw the Lewis structure for Cl2O.
Answer

19. Sample Problem 6.13 Drawing Lewis Structures for Polyatomic Ions
Sodium chlorite, NaClO2, is a component of mouthwashes, toothpastes, and contact lens cleaning solutions. Draw the Lewis structure for the chlorite ion, ClO2–.
Solution
Step 1 Determine the arrangement of atoms. For the polyatomic ion ClO2–, the central atom is Cl because there is only one Cl atom. For a polyatomic ion, the atoms and electrons are placed in brackets, and the charge is written outside to the upper right.
[O Cl O]–
Step 2 Determine the total number of valence electrons. We use the group numbers to determine the number of valence electrons for each of the atoms in the ion. Because the ion has a negative charge, one more electron is added to the valence electrons.

20. Sample Problem 6.13 Drawing Lewis Structures for Polyatomic Ions
Continued
Step 3 Attach each bonded atom to the central atom with a pair of electrons. Each bonding pair can also be represented by a line, which indicates a single bond.
Step 4 Place the remaining electrons using single or multiple bonds to complete the octets. Four electrons are used to bond the O atoms to the central Cl atom, which leaves 16 valence electrons. Of these, 12 electrons are drawn as lone pairs to complete the octets of the O atoms.
The remaining four electrons are placed as two lone pairs on the central Cl atom.
Study Check 6.13
Draw the Lewis structure for the polyatomic ion NH2–.
Answer

21. Sample Problem 6.14 Drawing Lewis Structures with Multiple Bonds
Draw the Lewis structure for carbon dioxide, CO2, in which the central atom is C.
Solution
Step 1 Determine the arrangement of atoms. O C O
Step 2 Determine the total number of valence electrons. Using the group number to determine valence electrons, the carbon atom has four valence electrons, and each oxygen atom has six valence electrons, which gives a total of 16 valence electrons.
Step 3 Attach each bonded atom to the central atom with a pair of electrons.
O : C : O or O—C—O
Step 4 Place the remaining electrons to complete octets. Because we used four valence electrons to attach the C atom to two O atoms, there are 12 valence electrons remaining.
16 valence e– – 4 bonding e– = 12 e– remaining

22. Sample Problem 6.14 Drawing Lewis Structures with Multiple Bonds
Continued
The 12 remaining electrons are placed as six lone pairs of electrons on the outside O atoms. However, this does not complete the octet of the C atom.
To obtain an octet, the C atom must share lone pairs of electrons from each of the O atoms. When two bonding pairs occur between atoms, it is known as a double bond.
Study Check 6.14
Draw the Lewis structure for HCN, which has a triple bond.
Answer

23. Sample Problem 6.15 Bond Polarity
Using electronegativity values, classify each of the following bonds as nonpolar covalent, polar covalent, or ionic:
O—K, Cl—As, N—N, P—Br
Solution
For each bond, we obtain the electronegativity values and calculate the difference in electronegativity.
Study Check 6.15
Using electronegativity values, classify each of the following bonds as nonpolar covalent, polar covalent, or ionic:
a. P—Cl b. Br—Br c. Na—O
Answer
a. polar covalent (0.9) b. nonpolar covalent (0.0) c. ionic (2.6)

24. Sample Problem 6.16 Shapes of Molecules
Predict the shape of a molecule of SiCl4.
Solution
Step 1 Draw the Lewis structure. Using 32 e–, we draw the bonds and lone pairs for the Lewis structure of SiCl4.

25. Sample Problem 6.16 Shapes of Molecules
Continued
Step 2 Arrange the electron groups around the central atom to minimize repulsion. To minimize repulsion, the electron-group geometry would be tetrahedral.
Step 3 Use the atoms bonded to the central atom to determine the shape. Because the central Si atom has four bonded pairs and no lone pairs of electrons, the SiCl4 molecule has a tetrahedral shape.
Study Check 6.16
Predict the shape of SCl2.
Answer
The central atom S has four electron groups: two bonded atoms and two lone pairs of electrons. The shape of SCl2 is bent, 109°.

26. Sample Problem 6.17 Predicting Shape of an Ion
Use VSEPR theory to predict the shape of the polyatomic ion NO3–.
Solution
Step 1 Draw the Lewis structure. The polyatomic ion, NO3–, contains three electron groups (two single bonds between the central N atom and O atoms, and one double bond between N and O). Note that the double bond can be drawn to any of the O atoms.
Step 2 Arrange the electron groups around the central atom to minimize repulsion. To minimize repulsion, three electron groups would have a trigonal planar geometry.
Step 3 Use the atoms bonded to the central atom to determine the shape. Because NO3– has three bonded atoms, it has a trigonal planar shape.
Study Check 6.17
Use VSEPR theory to predict the shape of ClO2–.
Answer
With two bonded atoms and two lone pairs of electrons, the shape of ClO2– is bent with a bond angle of 109°.

27. Sample Problem 6.18 Polarity of Molecules
Determine whether a molecule of OF2 is polar or nonpolar.
Solution
Step 1 Determine if the bonds are polar covalent or nonpolar covalent. From Figure 6.6, F and O have an electronegativity difference of 0.5 (4.0 – 3.5 = 0.5), which makes each of the O—F bonds polar covalent.

28. Sample Problem 6.18 Polarity of Molecules
Continued
Step 2 If the bonds are polar covalent, draw the Lewis structure and determine if the dipoles cancel. The Lewis structure for OF2 has four electron groups and two bonded atoms. The molecule has a bent shape in which the dipoles of the O—F bonds do not cancel. The OF2 molecule would be polar.
Study Check 6.18
Would a PCl3 molecule be polar or nonpolar?
Answer
polar

29. Sample Problem 6.19 Attractive Forces Between Particles
Indicate the major type of molecular interaction—dipole–dipole attractions, hydrogen bonding, or dispersion forces—expected of each of the following:
a. HF b. Br2 c. PCl3
Solution
a. HF is a polar molecule that interacts with other HF molecules by hydrogen bonding.
b. Br2 is nonpolar; only dispersion forces provide attractive forces.
c. The polarity of the PCl3 molecules provides dipole–dipole attractions.
Study Check 6.19
Why is the melting point of H2S lower than that of H2O?
Answer
The attractive forces between H2S molecules are dipole–dipole attractions, whereas the attractive forces between H2O molecules are hydrogen bonds, which are stronger and require more energy to break.