Chapter 3 Ionic Compounds

Chapter 3–1

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 a. Au+ = 79 protons, 78 electrons c. Sn2+ = 50 protons, 48 electrons b. Au3+ = 79 protons, 76 electrons d. Sn4+ = 50 protons, 46 electrons

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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 a. stannic = Sn4+ c. manganese ion = Mn2+ e. selenide = Se2– b. iodide = I– d. lead(II) = Pb2+

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

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. Fe(HSO3)3 = iron(III) sodium carbonate nitrate hydrogen sulfite, ferric bisulfite b. Ca(OH)2 = calcium d. Mn(CH3CO2)2 = manganese f. Mg3(PO4)2 = magnesium hydroxide acetate phosphate

3.26 Hydroxyapatite = Ca10(PO4)6(OH)2 Each Ca has a +2 charge; 10 Ca2+ = +20 3– Each PO4 has a –3 charge; 6 PO4 = –18 Each OH has a –1 charge; 2 OH– = –2 Total negative charge of –20 balances a total positive charge of +20.

Solutions to End-of-Chapter Problems

3.27 Use the criteria in Problem 3.1. a. CO2 = covalent b. H2SO4 = covalent c. KF = ionic d. CH5N = covalent

3.28 Use the criteria in Problem 3.1. a. C3H8 = covalent b. ClBr = covalent c. CuO = ionic d. CH4O = covalent

3.29 a. potassium (metal) and oxygen (nonmetal) = ionic b. sulfur and carbon (two nonmetals) = covalent c. two bromine atoms (two nonmetals) = covalent d. carbon and oxygen (two nonmetals) = covalent

3.30 a. carbon and hydrogen (two nonmetals) = covalent b. sodium (metal) and sulfur (nonmetal) = ionic c. hydrogen and oxygen (two nonmetals) = covalent d. magnesium (metal) and bromine (nonmetal) = ionic

3.31 Ionic bonds form between a metal and nonmetal because there is a transfer of electrons from the metal to the nonmetal. A metal gains a noble gas configuration of electrons by giving up electrons. A nonmetal can gain a noble gas configuration by gaining electrons.

3.32 No it is not proper to speak of sodium chloride molecules. Sodium chloride is an ionic compound because sodium is a metal and chlorine is a nonmetal. The compound is composed of ions. Molecules are compounds containing two or more atoms joined by covalent bonds.

3.33 a. four protons and two electrons c. 16 protons and 18 electrons e. 17 protons and 18 electrons = Be2+ = S2– = Cl– b. 22 protons and 20 electrons d. 13 protons and 10 electrons f. 20 protons and 18 electrons = Ti2+ = Al3+ = Ca2+

3.34 a. K+: 19 protons and c. Mn2+: 25 protons and e. Cs+: 55 protons and 18 electrons 23 electrons 54 electrons b. S2–: 16 protons and d. Fe2+: 26 protons and f. I–: 53 protons and 18 electrons 24 electrons 54 electrons

3.35 a. a period 2 element that forms a +2 cation = Be b. an ion from group 7A with 18 electrons = Cl– c. a cation from group 1A with 36 electrons = Rb+

3.36 a. a period 3 element that forms an ion with a –1 charge = Cl b. an ion from group 2A with 36 electrons = Sr2+ c. an ion from group 6A with 18 electrons = S2–

3.37 Elements in group 6A gain electrons to form anions because by gaining two electrons they have a filled valence shell.

3.38 Elements in group 2A lose electrons to form cations because by losing two electrons they have a filled valence shell.

3.39 a. sodium ion = Na+ c. manganese ion = Mn2+ e. stannic = Sn4+ 2– 3+ 2+ b. selenide = Se d. gold(III) = Au f. mercurous = Hg2

3.40 a. barium ion = Ba2+ c. oxide = O2– e. lead(IV) = Pb4+ b. iron(II) = Fe2+ d. ferrous = Fe2+ f. cobalt(III) = Co3+

3.41 The noble gas with the same number of electrons has the same electronic configuration as each ion.

a. O2– = Ne c. Al3+ = Ne e. F– = Ne b. Mg2+ = Ne d. S2– = Ar f. Be2+ = He

3.42 a. O2– F–, Na+ and Mg2+ have the same electronic configuration as neon. b. S2– Cl–, K+ and Ca2+ have the same electronic configuration as argon.

3.43 a. lithium = lose one electron (He) c. sulfur = gain two electrons (Ar) b. iodine = gain one electron (Xe) d. strontium = lose two electrons (Kr)

3.44 a. cesium = lose one electron (Xe) c. selenium = gain two electrons (Kr) b. barium = lose two electrons (Xe) d. aluminum = lose three electrons (Ne)

3.45 Ions that contain an outer shell of eight electrons are likely to form.

a. S– No, only seven c. S3– No, one electron in outer e. Na2+ No, only seven electrons electrons in outer shell shell in outer shell b. S2– Likely to form d. Na+ Likely to form f. Na– No, two electrons in outer shell

3.46 Ions that contain an outer shell of eight electrons are likely to form.

a. Mg+ No, one electron in c. Mg3+ No, only seven e. Cl– Likely to form outer shell electrons in outer shell b. Mg2+ Likely to form d. Cl+ No, six electrons in f. Cl2– No, five electrons in outer outer shell shell

3.47 Number of Group Number of Electrons Charge Example Valence Electrons Number Gained or Lost a. X 1 1A Lose 1 1+ Li b. Q 2 2A Lose 2 2+ Mg c. Z 6 6A Gain 2 2– S

d. A 7 7A Gain 1 1– Cl

3.48 Label each section of the periodic table.

(e) (b) (a) (f) (f) (e) (e) (e) (c) (d)

2– – 2– 3.49 a. sulfate, SO4 b. nitrite, NO2 c. sulfide, S

2– 2– – 3.50 a. carbonate, CO3 b. sulfite, SO3 c. nitrate, NO3

3.51 2– – a. sulfate = SO4 c. hydrogen carbonate = HCO3 + – b. ammonium = NH4 d. cyanide = CN

– – 3.52 a. acetate = CH3CO2 c. dihydrogen phosphate = H2PO4 – + b. bisulfite = HSO3 d. hydronium = H3O

3.53 – 3– a. OH = 9 protons, 10 electrons c. PO4 = 47 protons, 50 electrons + b. H3O = 11 protons, 10 electrons

3.54 + 2– a. NH4 = 11 protons, 10 electrons c. CO3 = 30 protons, 32 electrons b. CN– = 13 protons, 14 electrons

3.55 Transition metals have one or more d electrons. All of these electrons would have to be lost to follow the octet rule, and most transition metals do not lose that many electrons.

3.56 Yes, all isotopes of an element form the same type of ions. Electrons are gained or lost in the formation of an ion. Isotopes differ in the number of neutrons in the atom of an element.

3.57 Na donates an electron to F; then each atom has eight electrons in its outer shell, which follows the octet rule.

3.58 Li donates an electron to F; F– then has eight electrons, but Li+ only has two (same electron configuration as He).

3.59 a. calcium (Ca2+) and sulfur c. lithium (Li+) and iodine e. sodium (Na+) and selenium 2– – 2– (S ) = CaS (I ) = LiI (Se ) = Na2Se

b. aluminum (Al3+) and d. nickel (Ni2+) and chlorine – – bromine (Br ) = AlBr3 (Cl ) = NiCl2

3.60 a. barium (Ba2+) and bromine c. manganese (Mn2+) and e. magnesium (Mg2+) and – – – (Br ) = BaBr2 chlorine (Cl ) = MnCl2 fluorine (F ) = MgF2

b. aluminum (Al3+) and d. zinc (Zn2+) and sulfur 2– 2– sulfur (S ) = Al2S3 (S ) = ZnS

3.61 a. lithium (Li+) and nitrite c. sodium (Na+) and bisulfite e. magnesium (Mg2+) and – – – (NO2 ) = LiNO2 (HSO3 ) = NaHSO3 hydrogen sulfite (HSO3 ) = Mg(HSO3)2 b. calcium (Ca2+) and acetate d. manganese (Mn2+) and – 3– (CH3COO ) = phosphate (PO4 ) = Ca(CH3COO)2 Mn3(PO4)2

3.62 + + + a. potassium (K ) and c. lithium (Li ) and e. ammonium (NH4 ) and – 2– 3– bicarbonate (HCO3 ) = carbonate (CO3 ) = phosphate (PO4 ) = KHCO3 Li2CO3 (NH4)3PO4

b. magnesium (Mg2+) and d. potassium (K+) and – – nitrate (NO3 ) = cyanide (CN ) = KCN Mg(NO3)2

3.63 Y– Y2– Y3– + X XY X2Y X3Y 2+ X XY2 XY X3Y2 3+ X XY3 X2Y3 XY

3.64 a. 2+ b. 2+ c. 2+ d. 1+

3.65 – – – 2– 3– Br OH HCO3 SO3 PO4 + Na NaBr NaOH NaHCO3 Na2SO3 Na3PO4 2+ Co CoBr2 Co(OH)2 Co(HCO3)2 CoSO3 Co3(PO4)2 3+ Al AlBr3 Al(OH)3 Al(HCO3)3 Al2(SO3)3 AlPO4

3.66 – – – 2– 2– I CN NO3 SO4 HPO4 + K KI KCN KNO3 K2SO4 K2HPO4 2+ Mg MgI2 Mg(CN)2 Mg(NO3)2 MgSO4 MgHPO4 3+ Cr CrI3 Cr(CN)3 Cr(NO3)3 Cr2(SO4)3 Cr2(HPO4)3

3.67 a. KHSO4 b. Ba(HSO4)2 c. Al(HSO4)3 d. Zn(HSO4)2

3.68 a. K2SO3 b. BaSO3 c. Al2(SO3)3 d. ZnSO3

3.69 a. Ba(CN)2 b. Ba3(PO4)2 c. BaHPO4 d. Ba(H2PO4)2

3.70 a. Fe(CN)3 b. FePO4 c. Fe2(HPO4)3 d. Fe(H2PO4)3

3.71 a. Na2O = sodium oxide e. CoBr2 = cobalt(II) bromide b. BaS = barium sulfide f. RbBr = rubidium bromide c. PbS2 = lead(IV) sulfide g. PbBr2 = lead(II) bromide d. AgCl = silver chloride

3.72 a. KF = potassium fluoride e. AuBr3 = gold(III) bromide b. ZnCl2 = zinc chloride f. Li2S = lithium sulfide c. Cu2S = copper(I) sulfide g. SnBr4 = tin(IV) bromide d. SnO = tin(II) oxide

3.73 a. FeCl2 = iron(II) chloride, ferrous chloride c. FeS = iron(II) sulfide, ferrous sulfide b. FeBr3 = iron(III) bromide, ferric bromide d. Fe2S3 = iron(III) sulfide, ferric sulfide

3.74

a. CrCl2 = chromium(II) chloride, c. CrO= chromium(II) oxide, chromous oxide chromous chloride

b. CrBr3 = chromium(III) bromide, d. Cr2O3 = chromium(III) oxide, chromic bromide chromic oxide

3.75 Copper cations can be 1+ or 2+, so the Roman numeral designation is required. Ca exists only as 2+.

CuBr2: copper(II) bromide or cupric bromide CaBr2: calcium bromide

3.76 Lead cations can be 2+ or 4+, so the Roman numeral designation is required. Zn exists only as 2+.

PbO: lead(II) oxide ZnO: zinc oxide

3.77 a. sodium sulfide and sodium sulfate = Na2S and Na2SO4 b. magnesium oxide and magnesium hydroxide = MgO and Mg(OH)2 c. magnesium sulfate and magnesium bisulfate = MgSO4 and Mg(HSO4)2

3.78

a. lithium sulfite and lithium sulfide = Li2SO3 and Li2S b. sodium carbonate and sodium hydrogen carbonate = Na2CO3 and NaHCO3 c. calcium phosphate and calcium dihydrogen phosphate = Ca3(PO4)2 and Ca(H2PO4)2

3.79 a. NH4Cl = ammonium c. Cu(NO3)2 = copper(II) nitrate, e. Fe(NO3)2 = iron(II) nitrate, chloride cupric nitrate ferrous nitrate

b. PbSO4 = lead(II) sulfate d. Ca(HCO3)2 = calcium bicarbonate, calcium hydrogen carbonate

3.80 a. (NH4)2SO4 = c. Cr(CH3CO2)3 = chromium(III) e. Ni3(PO4)2 = nickel(II) ammonium sulfate acetate, chromic acetate phosphate

b. NaH2PO4 = sodium d. Sn(HPO4)2 = tin(II) hydrogen dihydrogen phosphate phosphate, stannous hydrogen phosphate

3.81 a. magnesium carbonate = d. potassium hydrogen g. aluminum bicarbonate = MgCO3 phosphate = K2HPO4 Al(HCO3)3

b. nickel sulfate = NiSO4 e. gold(III) nitrate = h. chromous cyanide = Cr(CN)2 Au(NO3)3 c. copper(II) hydroxide = f. lithium phosphate = Cu(OH)2 Li3PO4

3.82 a. copper(I) sulfite = d. lead(IV) carbonate = g. ammonium cyanide = Cu2SO3 Pb(CO3)2 NH4CN

b. aluminum nitrate = e. zinc hydrogen phosphate h. iron(II) nitrate = Fe(NO3)2 Al(NO3)3 = ZnHPO4 c. tin(II) acetate = f. magnesium dihydrogen Sn(CH3CO2)2 phosphate = Mg(H2PO4)2

3.83 – – 3– a. OH = Pb(OH)4 c. HCO3 = Pb(HCO3)4 e. PO4 = Pb3(PO4)4 lead(IV) hydroxide lead(IV) bicarbonate lead(IV) phosphate 2– – – b. SO4 = Pb(SO4)2 d. NO3 = Pb(NO3)4 f. CH3CO2 = Pb(CH3CO2)4 lead(IV) sulfate lead(IV) nitrate lead(IV) acetate

3.84 – – 3– a. OH = Fe(OH)3 c. HPO4 = Fe(HPO4)3 e. PO4 = FePO4 iron(III) hydroxide iron(III) hydrogen iron(III) phosphate phosphate 2– – – b. CO3 = Fe2(CO3)3 d. NO2 = Fe(NO2)3 f. CH3CO2 = Fe(CH3CO2)3 iron(III) carbonate iron(III) nitrite iron(III) acetate

3.85 The false statements are corrected.

a. True: Ionic compounds have high melting points. b. False: Ionic compounds are solids at room temperature. c. False: Most ionic compounds are soluble in water. d. False: Ionic solids exist as crystalline lattices with the ions arranged to maximize the electrostatic interactions of anions and cations.

3.86 The false statements are corrected.

a. True: Ionic compounds have high boiling points. b. False: The ions in a crystal lattice are arranged to maximize the electrostatic interactions of anions and cations. c. True: When an ionic compound dissolves in water, the solution conducts electricity. d. False: In an ionic crystal, ions having like charges are surrounded by ions of the opposite charge. Therefore, ions of opposite charge are arranged close together.

3.87 NaCl has a higher melting point than CH4 or H2SO4 because it is an ionic compound, whereas the other two compounds are covalent. Ionic solids have higher melting points.

3.88 Cl2 has the lowest boiling point. Cl2 is a covalent compound, whereas KI and LiF are ionic compounds. Covalent compounds have lower boiling points.

3.89 a. A neutral zinc atom has 30 protons and 30 electrons. b. The Zn2+ cation has 30 protons and 28 electrons. c. The electronic configuration of zinc: 1s22s22p63s23p64s23d10 The 4s2 electrons are lost to form Zn2+.

3.90 a. A neutral copper atom has 29 protons and 29 electrons. b. The Cu+ cation has 29 protons and 28 electrons. c. The Cu2+ cation has 29 protons and 27 electrons. d. The formula of zinc acetate is Zn(CH3CO2)2.

3.91 Cation a. Number of b. Number of c. Noble d. Role Protons Electrons Gas Na+ 11 10 Ne Major cation in extracellular fluids and blood; maintains blood volume and blood pressure K+ 19 18 Ar Major intracellular cation Ca2+ 20 18 Ar Major cation in solid tissues like bone and teeth; required for normal muscle contraction and nerve function Mg2+ 12 10 Ne Required for normal muscle contraction and nerve function

3.92 Calcium carbonate is insoluble in water as indicated by the photo in Section 3.6C. The shells of oysters and other mollusks are composed largely of calcium carbonate.

+ – 3.93 silver (Ag ) nitrate (NO3 ) = AgNO3

+ 2– 3.94 ammonium (NH4 ) carbonate (CO3 ) = (NH4)2CO3

3.95 CaSO3 = calcium sulfite

3.96 a. CdS = cadmium sulfide c. Cr2O3 = chromium(III) oxide b. TiO2 = titanium(IV) oxide d. Mn3(PO4)2 = manganese(II) phosphate

+ – 3.97 ammonium (NH4 ) nitrate (NO3 ) = NH4NO3

+ 3– 3.98 sodium (Na ) phosphate (PO4 ) = Na3PO4

3.99 a. magnesium oxide (MgO) and potassium iodide (KI) b. CaHPO4 = calcium hydrogen phosphate c. FePO4 = iron(III) phosphate, ferric phosphate d. sodium selenite = Na2SeO3 e. The name chromium chloride is ambiguous. Without a designation as chromium(II) or chromium(III), it’s impossible to know the ratio of chromium cations to chloride anions.

3.100 potassium dichromate = K2Cr2O7 potassium permanganate = KMnO4