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
b. CaF2 ionic
3.2
c. MgO ionic
d. Cl2 covalent
e. HF covalent
f. C2H6 covalent
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
b. H2 O compound, molecule
c. NaF compound
d. MgBr2 compound
e. F2 element, molecule
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.
3.5
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+
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
b. Se2– = 34 protons, 36 electrons
3.6
c. Zn2+ = 30 protons, 28 electrons
d. Fe3+ = 26 protons, 23 electrons
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
(group 2A): +2
b. iodine
(group 7A): –1
c. selenium
(group 6A): –2
d. rubidium
(group 1A): +1
3.7
a. Ne
3.8
b. Xe
c. Kr
d. Kr
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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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
b. barium (+2) and oxygen (–2) = BaO
3.12
c. magnesium (+2) and iodine (–1) =
2 I– anions are needed = MgI2
d. lithium (+1) and oxygen (–2) =
2 Li+ cations are needed = Li2O
a. In Na2S, there are twice as many Na+ cations (darker spheres) as there are S2– anions (lighter
spheres).
b. In MgCl2, there are twice as many Cl– anions (lighter spheres) as there are Mg2+ cations
(darker spheres).
a.
S2–
b.
Cl–
Mg2+
Na+
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.
3.15
a. S2–= sulfide
b. Cu += copper(I), cuprous
c. Cs+= cesium
d. Al3+= aluminum
e. Sn4+= tin(IV), stannic
a. stannic = Sn4+
b. iodide = I–
c. manganese ion = Mn2+
d. lead(II) = Pb2+
e. selenide = Se2–
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Chapter 3–3
3.16
Name the cation and then the anion.
a. NaF = sodium fluoride
b. MgO = magnesium oxide
3.17
c. SrBr2 = strontium bromide
d. Li2O = lithium oxide
e. TiO2 = titanium oxide
f. AlCl3 = aluminum chloride
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
Chromium has a variable charge, but here it must
have a +3 charge to balance the three chloride ions.
chromium(III) chloride, chromic chloride
d. PbO2
Lead has a variable charge, but here it must have a
+4 charge to balance the two oxide ions.
lead(IV) oxide
b. PbS
Lead has a variable charge, but here it must have a
+2 charge to balance the sulfide ion.
lead(II) sulfide
e. FeBr2
Iron has a variable charge, but here it must have a
+2 charge to balance the two bromide ions.
iron(II) bromide, ferrous bromide
c. SnF4
Tin has a variable charge, but here it must have a
+4 charge to balance the four fluoride ions.
tin(IV) fluoride, stannic fluoride
f. AuCl3
Gold has a variable charge, but here it must have a
+3 charge to balance the three chloride ions.
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
Fe2 O3= 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
Calcium is the cation (+2).
Bromide is the anion (–1).
CaBr2
c. ferric bromide
Iron (Fe) is the cation (+3).
Bromide is the anion (–1).
FeBr3
b. copper(I) iodide
Copper(I) is the cation (+1).
Iodide is the anion (–1).
CuI
d. magnesium sulfide
Magnesium is the cation (+2).
Sulfide is the anion (–2).
MgS
e. chromium(II) chloride
Chromium is the cation (+2).
Chloride is the anion (–1).
CrCl2
f. sodium oxide
Sodium is the cation (+1).
Oxide is the anion (–2).
Na2O
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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.
3.23
3.24
3.25
3.26
a. magnesium (+2)
MgSO4
c. nickel (+2)
NiSO4
b. sodium (+1)
Na2 SO4
d. aluminum (+3)
Al2(SO4)3
e. lithium (+1)
Li2SO4
Use Table 3.5 to determine the charge on the polyatomic ions.
a. sodium (+1) and bicarbonate
(–1): NaHCO3
c. ammonium (+1) and sulfate
(–2): (NH4)2SO4
e. calcium (+2) and bisulfate
(–1): Ca(HSO4)2
b. potassium (+1) and nitrate
(–1): KNO3
d. magnesium (+2) and
phosphate (–3): Mg3(PO4)2
f. barium (+2) and hydroxide
(–1): Ba(OH)2
a. OH– = KOH
b. NO2– = KNO2
c. SO42– = K2SO4
d. HSO3– = KHSO3
e. PO43– = K3PO4
f. CN– = KCN
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 =
sodium carbonate
c. Mg(NO3)2 = magnesium
nitrate
b. Ca(OH)2 = calcium
hydroxide
d. Mn(CH3 CO2)2 = manganese
acetate
e. Fe(HSO3)3 = iron(III)
hydrogen sulfite, ferric
bisulfite
f. Mg3(PO4)2 = magnesium
phosphate
Hydroxyapatite = Ca10(PO4)6(OH)2
Each Ca has a +2 charge; 10 Ca2+ = +20
Each PO4 has a –3 charge; 6 PO43– = –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
3.28
Use the criteria in Problem 3.1.
a. CO2 = covalent
b. H2SO4 = covalent
c. KF = ionic
d. CH5 N = covalent
Use the criteria in Problem 3.1.
a. C3H8 = covalent
b. ClBr = covalent
c. CuO = ionic
d. CH4 O = covalent
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Chapter 3–5
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
3.34
a. four protons and two electrons
= Be2+
b. 22 protons and 20 electrons
= Ti2+
c. 16 protons and 18 electrons
= S2–
d. 13 protons and 10 electrons
= Al3+
e. 17 protons and 18 electrons
= Cl–
f. 20 protons and 18 electrons
= Ca2+
a. K+: 19 protons and
18 electrons
b. S2–: 16 protons and
18 electrons
c. Mn2+: 25 protons and
23 electrons
d. Fe2+: 26 protons and
24 electrons
e. Cs+: 55 protons and
54 electrons
f. I–: 53 protons and
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+
b. selenide = Se2–
c. manganese ion = Mn2+
d. gold(III) = Au3+
e. stannic = Sn4+
f. mercurous = Hg22+
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Ionic Compounds 3–6
3.40
3.41
a. barium ion = Ba2+
b. iron(II) = Fe2+
e. lead(IV) = Pb4+
f. cobalt(III) = Co3+
The noble gas with the same number of electrons has the same electronic configuration as each
ion.
a. O2– = Ne
b. Mg2+ = Ne
3.42
c. oxide = O2–
d. ferrous = Fe2+
c. Al3+ = Ne
d. S2– = Ar
e. F– = Ne
f. Be2+ = He
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)
b. iodine = gain one electron (Xe)
c. sulfur = gain two electrons (Ar)
d. strontium = lose two electrons (Kr)
a. cesium = lose one electron (Xe)
b. barium = lose two electrons (Xe)
c. selenium = gain two electrons (Kr)
d. aluminum = lose three electrons (Ne)
3.44
3.45
Ions that contain an outer shell of eight electrons are likely to form.
a. S– No, only seven
electrons in outer shell
b. S2– Likely to form
3.46
c. S3– No, one electron in outer
shell
d. Na+ Likely to form
e. Na2+ No, only seven electrons
in outer shell
f. Na– No, two electrons in outer
shell
Ions that contain an outer shell of eight electrons are likely to form.
a. Mg+ No, one electron in
outer shell
b. Mg2+ Likely to form
c. Mg3+ No, only seven
electrons in outer shell
d. Cl + No, six electrons in
outer shell
e. Cl– Likely to form
f. Cl2– No, five electrons in outer
shell
3.47
Group
Number
1A
Number of Electrons
Gained or Lost
Lose 1
Charge
Example
a. X
Number of
Valence Electrons
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
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Chapter 3–7
3.48
Label each section of the periodic table.
(e) (b)
(a)
(e)
(f)
(f)
(e)
(e)
(d)
(c)
3.49
a. sulfate, SO42–
b. nitrite, NO2–
c. sulfide, S2–
3.50
a. carbonate, CO32–
b. sulfite, SO32–
c. nitrate, NO3–
3.51
3.52
3.53
3.54
a. sulfate = SO42–
b. ammonium = NH4 +
c. hydrogen carbonate = HCO3–
d. cyanide = CN–
a. acetate = CH3CO2–
b. bisulfite = HSO3–
c. dihydrogen phosphate = H2PO4–
d. hydronium = H3O+
a. OH– = 9 protons, 10 electrons
b. H3 O+ = 11 protons, 10 electrons
c. PO43– = 47 protons, 50 electrons
a. NH4+ = 11 protons, 10 electrons
b. CN– = 13 protons, 14 electrons
c. CO32– = 30 protons, 32 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
(S2–) = CaS
c. lithium (Li +) and iodine
(I–) = LiI
b. aluminum (Al3+) and
bromine (Br–) = AlBr3
d. nickel (Ni2+) and chlorine
(Cl–) = NiCl2
e. sodium (Na+) and selenium
(Se2–) = Na2Se
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Ionic Compounds 3–8
3.60
3.61
3.62
a. barium (Ba2+) and bromine
(Br–) = BaBr2
c. manganese (Mn2+) and
chlorine (Cl–) = MnCl2
b. aluminum (Al3+) and
sulfur (S2–) = Al2S3
d. zinc (Zn2+) and sulfur
(S2–) = ZnS
a. lithium (Li +) and nitrite
(NO2–) = LiNO2
c. sodium (Na+) and bisulfite
(HSO3–) = NaHSO3
b. calcium (Ca2+) and acetate
(CH3COO–) =
Ca(CH3COO)2
d. manganese (Mn2+) and
phosphate (PO43–) =
Mn3(PO4)2
a. potassium (K+) and
bicarbonate (HCO3–) =
KHCO3
c. lithium (Li +) and
carbonate (CO32–) =
Li2CO3
b. magnesium (Mg2+) and
nitrate (NO3–) =
Mg(NO3)2
d. potassium (K+) and
cyanide (CN–) = KCN
e. magnesium (Mg2+) and
fluorine (F–) = MgF2
e. magnesium (Mg2+) and
hydrogen sulfite (HSO3–) =
Mg(HSO3)2
e. ammonium (NH4+) and
phosphate (PO43–) =
(NH4)3PO4
3.63
X+
X2+
X3+
3.64
a. 2+
3.65
Y–
Y2–
Y3–
XY
XY2
XY3
X2 Y
XY
X2 Y3
X3 Y
X3 Y2
XY
b. 2+
c. 2+
d. 1+
Na+
Br–
NaBr
OH–
NaOH
HCO3–
NaHCO3
SO32–
Na2 SO3
PO43–
Na3 PO4
Co2+
CoBr2
Co(OH)2
Co(HCO3)2
CoSO3
Co3(PO4)2
3+
AlBr3
Al(OH)3
Al(HCO3)3
Al2(SO3)3
AlPO4
K+
I–
KI
CN–
KCN
NO3–
KNO3
SO42–
K2SO4
HPO42–
K2 HPO4
Mg2+
MgI2
Mg(CN)2
Mg(NO3)2
MgSO4
MgHPO4
CrI3
Cr(CN)3
Cr(NO3)3
Cr2(SO4)3
Cr2(HPO4)3
Al
3.66
Cr
3+
3.67
a. KHSO4
b. Ba(HSO4)2
c. Al(HSO4)3
d. Zn(HSO4)2
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Chapter 3–9
3.68
a. K2SO3
b. BaSO3
c. Al2(SO3)3
d. ZnSO3
a. Ba(CN)2
b. Ba3(PO4)2
c. BaHPO4
d. Ba(H2 PO4)2
a. Fe(CN)3
b. FePO4
c. Fe2(HPO4)3
d. Fe(H2 PO4)3
3.69
3.70
3.71
a. Na2O = sodium oxide
b. BaS = barium sulfide
c. PbS2 = lead(IV) sulfide
d. AgCl = silver chloride
e. CoBr2 = cobalt(II) bromide
f. RbBr = rubidium bromide
g. PbBr2 = lead(II) bromide
a. KF = potassium fluoride
b. ZnCl2 = zinc chloride
c. Cu2S = copper(I) sulfide
d. SnO = tin(II) oxide
e. AuBr3 = gold(III) bromide
f. Li2S = lithium sulfide
g. SnBr4 = tin(IV) bromide
a. FeCl2 = iron(II) chloride, ferrous chloride
b. FeBr3 = iron(III) bromide, ferric bromide
c. FeS = iron(II) sulfide, ferrous sulfide
d. Fe2S3 = iron(III) sulfide, ferric sulfide
a. CrCl2 = chromium(II) chloride,
chromous chloride
c. CrO= chromium(II) oxide, chromous oxide
b. CrBr3 = chromium(III) bromide,
chromic bromide
d. Cr2O3 = chromium(III) oxide,
chromic oxide
3.72
3.73
3.74
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
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Ionic Compounds 3–10
3.78
a. lithium sulfite and lithium sulfide = Li2SO3 and Li2S
b. sodium carbonate and sodium hydrogen carbonate = Na2 CO3 and NaHCO3
c. calcium phosphate and calcium dihydrogen phosphate = Ca3(PO4)2 and Ca(H2PO4)2
3.79
a. NH4 Cl = ammonium
chloride
c. Cu(NO3)2 = copper(II) nitrate,
cupric nitrate
b. PbSO4 = lead(II) sulfate
d. Ca(HCO3)2 = calcium
bicarbonate, calcium
hydrogen carbonate
a. (NH4)2SO4 =
ammonium sulfate
c. Cr(CH3CO2)3 = chromium(III)
acetate, chromic acetate
b. NaH2PO4 = sodium
dihydrogen phosphate
d. Sn(HPO4)2 = tin(II) hydrogen
phosphate, stannous hydrogen
phosphate
e. Fe(NO3)2 = iron(II) nitrate,
ferrous nitrate
3.80
e. Ni3(PO4)2 = nickel(II)
phosphate
3.81
a. magnesium carbonate =
MgCO3
b. nickel sulfate = NiSO4
g. aluminum bicarbonate =
Al(HCO3)3
h. chromous cyanide = Cr(CN)2
c. copper(II) hydroxide =
Cu(OH)2
d. potassium hydrogen
phosphate = K2 HPO4
e. gold(III) nitrate =
Au(NO3)3
f. lithium phosphate =
Li3PO4
a. copper(I) sulfite =
Cu2SO3
d. lead(IV) carbonate =
Pb(CO3)2
g. ammonium cyanide =
NH4CN
b. aluminum nitrate =
Al(NO3)3
c. tin(II) acetate =
Sn(CH3CO2)2
e. zinc hydrogen phosphate
= ZnHPO4
f. magnesium dihydrogen
phosphate = Mg(H2PO4)2
h. iron(II) nitrate = Fe(NO3)2
a. OH– = Pb(OH)4
lead(IV) hydroxide
b. SO42– = Pb(SO4)2
lead(IV) sulfate
c. HCO3– = Pb(HCO3)4
lead(IV) bicarbonate
d. NO3– = Pb(NO3)4
lead(IV) nitrate
e. PO43– = Pb3(PO4)4
lead(IV) phosphate
f. CH3CO2– = Pb(CH3CO2)4
lead(IV) acetate
a. OH– = Fe(OH)3
iron(III) hydroxide
c. HPO4– = Fe(HPO4)3
iron(III) hydrogen
phosphate
d. NO2– = Fe(NO2)3
iron(III) nitrite
e. PO43– = FePO4
iron(III) phosphate
3.82
3.83
3.84
b. CO32– = Fe2(CO3)3
iron(III) carbonate
f. CH3CO2– = Fe(CH3CO2)3
iron(III) acetate
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Chapter 3–11
3.85
The false statements are corrected.
a.
b.
c.
d.
3.86
True: Ionic compounds have high melting points.
False: Ionic compounds are solids at room temperature.
False: Most ionic compounds are soluble in water.
False: Ionic solids exist as crystalline lattices with the ions arranged to maximize the
electrostatic interactions of anions and cations.
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.
b.
c.
d.
A neutral copper atom has 29 protons and 29 electrons.
The Cu+ cation has 29 protons and 28 electrons.
The Cu2+ cation has 29 protons and 27 electrons.
The formula of zinc acetate is Zn(CH3 CO2)2.
3.91
Cation a. Number of
Protons
Na+
11
K+
b. Number of
Electrons
10
c. Noble
Gas
Ne
d. Role
Major cation in extracellular fluids and blood;
maintains blood volume and blood pressure
19
18
Ar
Major intracellular cation
2+
20
18
Ar
Mg2+
12
10
Ne
Major cation in solid tissues like bone and
teeth; required for normal muscle contraction
and nerve function
Required for normal muscle contraction and
nerve function
Ca
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
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Ionic Compounds 3–12
3.94
ammonium (NH4 +) carbonate (CO32–) = (NH4)2CO3
3.95
CaSO3 = calcium sulfite
3.96
a. CdS = cadmium sulfide
b. TiO2 = titanium(IV) oxide
3.97
ammonium (NH4 +) nitrate (NO3–) = NH4NO3
3.98
sodium (Na+) phosphate (PO43–) = Na3PO4
c. Cr2O3 = chromium(III) oxide
d. Mn3(PO4)2 = manganese(II) phosphate
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
© 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.