Chapter 3–1 Chapter 3 Ionic Compounds Solutions to In-Chapter Problems 3.1 The position of the elements in the periodic table determines the type of bonds they form. When a metal and nonmetal combine, as in (b) and (c), the bond is ionic. When two nonmetals combine, or when a metalloid bonds to a nonmetal, the bond is covalent. a. CO covalent c. MgO ionic e. HF covalent b. CaF2 ionic d. Cl2 covalent f. C2H6 covalent 3.2 An element is a pure substance that cannot be broken down into simpler substances by a chemical reaction. A compound is a pure substance formed by combining two or more elements together. A molecule is composed of atoms that are covalently bonded together. a. CO2 compound, molecule c. NaF compound e. F2 element, molecule b. H2O compound, molecule d. MgBr2 compound f. CaO compound 3.3 Vitamin C (C6H8O6) is likely to contain covalent bonds because it consists of the nonmetals C, H, and O. 3.4 The number of protons equals the atomic number. The charge is determined by comparing the number of protons and electrons. If the number of electrons is greater than the number of protons, the charge is negative. If the number of protons is greater than the number of electrons, the charge is positive. a. 19 protons and 18 electrons = K+ c. 35 protons and 36 electrons = Br– b. 7 protons and 10 electrons = N3– d. 23 protons and 21 electrons = V2+ 3.5 Use the identity of the element to determine the number of protons. The charge tells how many more or fewer electrons there are compared to the number of protons. A positive charge means more protons than electrons, while a negative charge means more electrons than protons. a. Ni2+ = 28 protons, 26 electrons c. Zn2+ = 30 protons, 28 electrons b. Se2– = 34 protons, 36 electrons d. Fe3+ = 26 protons, 23 electrons 3.6 Locate the element in the periodic table. A metal in groups 1A, 2A, or 3A forms a cation equal in charge to the group number. A nonmetal in groups 5A, 6A, or 7A forms an anion whose charge equals 8 – (the group number). a. magnesium b. iodine c. selenium d. rubidium (group 2A): +2 (group 7A): –1 (group 6A): –2 (group 1A): +1 3.7 a. Ne b. Xe c. Kr d. Kr 3.8 + 2+ a. Au = 79 protons, 78 electrons c. Sn = 50 protons, 48 electrons 3+ 4+ b. Au = 79 protons, 76 electrons d. Sn = 50 protons, 46 electrons © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Ionic Compounds 3–2 3.9 a. Mn = 25 protons, 25 electrons b. Mn2+ = 25 protons, 23 electrons c. 1s22s22p63s23p64s23d5 The two 4s2 valence electrons would be lost to form Mn2+. 3.10 Ionic compounds are composed of cations and anions. a. lithium (metal) and bromine (nonmetal): yes c. calcium and magnesium (two metals): no b. chlorine and oxygen (two nonmetals): no d. barium (metal) and chlorine (nonmetal): yes 3.11 •Identify the cation and the anion, and use the periodic table to determine the charges. •When ions of equal charge combine, one of each ion is needed. When ions of unequal charge combine, use the ionic charges to determine the relative number of each ion. •Write the formula with the cation first and then the anion, omitting charges, and using subscripts to indicate the number of each ion. a. sodium (+1) and bromine (–1) = NaBr c. magnesium (+2) and iodine (–1) = – 2 I anions are needed = MgI2 b. barium (+2) and oxygen (–2) = BaO d. lithium (+1) and oxygen (–2) = + 2 Li cations are needed = Li2O + 2– 3.12 a. In Na2S, there are twice as many Na cations (darker spheres) as there are S anions (lighter spheres). – 2+ b. In MgCl2, there are twice as many Cl anions (lighter spheres) as there are Mg cations (darker spheres). 2– a. S b. Cl– 2+ Na+ Mg 3.13 Zinc forms Zn2+ and oxygen forms O2–; thus, zinc oxide = ZnO. 3.14 When a metal forms more than one cation, the cations are named by one of two methods. Method [1]: Follow the name of the cation by a Roman numeral in parentheses to indicate its charge. Method [2]: Use the suffix -ous for the cation with the lower charge, and the suffix -ic for the cation with the higher charge. These suffixes are often added to the Latin names of the elements. Anions are named by replacing the ending of the element name by the suffix -ide. a. S2–= sulfide c. Cs+= cesium e. Sn4+= tin(IV), stannic b. Cu+= copper(I), cuprous d. Al3+= aluminum 3.15 4+ 2+ 2– a. stannic = Sn c. manganese ion = Mn e. selenide = Se – 2+ b. iodide = I d. lead(II) = Pb © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Chapter 3–3 3.16 Name the cation and then the anion. a. NaF = sodium fluoride c. SrBr2 = strontium bromide e. TiO2 = titanium oxide b. MgO = magnesium oxide d. Li2O = lithium oxide f. AlCl3 = aluminum chloride 3.17 First determine if the cation has fixed or variable charge. To name an ionic compound that contains a cation that always has the same charge, name the cation and then the anion (using the suffix -ide). When the metal has variable charge, use the overall anion charge to determine the charge on the cation. Then name the cation (using a Roman numeral or the suffix -ous or -ic), followed by the anion. a. CrCl3 d. PbO2 Chromium has a variable charge, but here it must Lead has a variable charge, but here it must have a have a +3 charge to balance the three chloride ions. +4 charge to balance the two oxide ions. chromium(III) chloride, chromic chloride lead(IV) oxide b. PbS e. FeBr2 Lead has a variable charge, but here it must have a Iron has a variable charge, but here it must have a +2 charge to balance the sulfide ion. +2 charge to balance the two bromide ions. lead(II) sulfide iron(II) bromide, ferrous bromide c. SnF4 f. AuCl3 Tin has a variable charge, but here it must have a Gold has a variable charge, but here it must have a +4 charge to balance the four fluoride ions. +3 charge to balance the three chloride ions. tin(IV) fluoride, stannic fluoride gold(III) chloride 3.18 a. Cu2O = copper(I) oxide, cuprous oxide c. CuCl = copper(I) chloride, cuprous chloride b. CuO = copper(II) oxide, cupric oxide d. CuCl2 = copper(II) chloride, cupric chloride 3.19 Fe2O3= iron(III) oxide, ferric oxide 3.20 Identify the cation and the anion and determine their charges. Balance the charges. Write the formula with the cation first, and use subscripts to show the number of each ion needed to have zero overall charge. a. calcium bromide c. ferric bromide Calcium is the cation (+2). Iron (Fe) is the cation (+3). Bromide is the anion (–1). Bromide is the anion (–1). CaBr2 FeBr3 b. copper(I) iodide d. magnesium sulfide Copper(I) is the cation (+1). Magnesium is the cation (+2). Iodide is the anion (–1). Sulfide is the anion (–2). CuI MgS e. chromium(II) chloride f. sodium oxide Chromium is the cation (+2). Sodium is the cation (+1). Chloride is the anion (–1). Oxide is the anion (–2). CrCl2 Na2O © 2013 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. Ionic Compounds 3–4 3.21 Ionic compounds have high melting points and high boiling points. They usually dissolve in water. Their solutions conduct electricity and they form crystalline solids. 3.22 Write the formula formed from polyatomic ions with the cation first and then the anion, omitting charges. Use parentheses around polyatomic ions when more than one appears in the formula, and use subscripts to indicate the number of each ion. a. magnesium (+2) c. nickel (+2) e. lithium (+1) MgSO4 NiSO4 Li2SO4 b. sodium (+1) d. aluminum (+3) Na2SO4 Al2(SO4)3 3.23 Use Table 3.5 to determine the charge on the polyatomic ions. a. sodium (+1) and bicarbonate c. ammonium (+1) and sulfate e. calcium (+2) and bisulfate (–1): NaHCO3 (–2): (NH4)2SO4 (–1): Ca(HSO4)2 b. potassium (+1) and nitrate d. magnesium (+2) and f. barium (+2) and hydroxide (–1): KNO3 phosphate (–3): Mg3(PO4)2 (–1): Ba(OH)2 3.24 – 2– 3– a. OH = KOH c. SO4 = K2SO4 e. PO4 = K3PO4 – – – b. NO2 = KNO2 d. HSO3 = KHSO3 f. CN = KCN 3.25 First determine if the cation has fixed or variable charge. To name an ionic compound that contains a cation that always has the same charge, name the cation and then the anion. When the metal has variable charge, use the overall anion charge to determine the charge on the cation. Then name the cation (using a Roman numeral or the suffix -ous or -ic), followed by the anion. a. Na2CO3 = c. Mg(NO3)2 = magnesium e.