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Chemistry 1000 (Fall 2013)
Problem Set #4: Metals and Ionic Compounds
Solutions
Answers to Questions in Silberberg (only those w/out answers at the back of the book)
2.51 The major isotope(s) of tellurium must contain more neutrons than the major isotope(s)
of iodine.
2.62
The total positive charge from the cations must be exactly equal in magnitude to the total
negative charge from the anions.
2.81
We use roman numerals in the names of ionic compounds containing transition metal
cations.
2.85
(a)
(b)
(c)
CsBr
BaS
CaF2
cesium bromide
barium sulfide
calcium fluoride
8.3
Main group elements which are metals tend to form cations. These elements can be
found toward the bottom left corner of the periodic table. (e.g. all Group 1 metals; Group
2 elements below Be; Group 13 elements below B; Sn; Pb)
Main group elements which are nonmetals close to the right side of the periodic table
tend to form anions. (e.g. all Group 17 elements, O, S, N)
8.16
Energy is released when cations and anions form ionic bonds. An ionic solid contains
many ionic bonds.
8.19
While energy is required to form S2- anions from S atoms and electrons (and also to form
K+ ions from K atoms, which release the electrons needed to form S2-), a large amount of
energy will be released when ionic bonds are formed between the K+ and S2- ions. This
energy released is greater than the energy required to form the initial ions.
8.21
(a)
Cs+
[Kr] 5s2 4d10 5p6 (same as Xe) (b) O2[He] 2s2 2p6 (same as Ne)
22
6
2+
S
[Ne] 3s 3p (same as Ar)
Sr
[Ar] 4s2 3d10 4p6 (same as Kr)
Cs2S
SrO
(c)
N3[He] 2s2 2p6 (same as Ne)
(d) Br[Ne] 3s2 3p6 (same as Ar)
2+
2
6
+
Mg
[He] 2s 2p (same as Ne)
Li
1s2 (same as He)
Mg3N2
LiBr
8.21 was assigned instead of 8.20 because 20d does not give the predicted ionic compound.
Instead of Rb2O, you actually get RbO2 (from the Rb+ and O2- ions). This is discussed in
Chapter 13.
13.16 The element in Group 1 has the largest atomic radius of all the elements in its period. As
such, it will also have the largest atomic volume (since Vatom = 4/3  r3). It follows that
the Group 1 element will require a greater volume of space per atom than any other
element in the same period in the solid phase.
The Group 1 element also has the smallest mass of all the elements in its period.
Since density is equal to mass divided by volume, the element with the smallest atomic
mass requiring the largest volume (per atom) will have the lowest density of all solids in
the period.
13.23 Both Group 1 metals and Group 2 metals react with increasingly vigor with water from
the top of the group to the bottom. A Group 2 metal, however, reacts less vigorously
with water than the Group 1 metal in the same period.
13.24
(a)
Li and Mg; Be and Al
(b)
Li and Mg
Both metals react with N2 to give the corresponding nitride. (Li is the only Group 1
metal that does so.)
Both metals react with O2 to give the corresponding oxide. (The other metals in Group 1
react with O2 to give more complex products.)
Be and Al
Both metals form oxoanions in strong base: [Be(OH)4]2– and [Al(OH)4]–
Both have amphoteric oxides (BeO and Al2O3)
Both form oxide coatings (passivation layers) so that they do not react with water
(c)
Li and Mg are similar in size – as are Li+ and Mg2+.
Be2+ and Al3+ are similar in charge density, both small and highly charged.
13.106
E photon 

hc

6.626  10
34 J
Hz
2.9979 10   10 nm
8 m
9
s
589.2nm 
1m
4 sig. fig.
E photon  3.371  1019 J
This is the energy for a single photon (and the answer to this question, as phrased).
To calculate the energy in J/mol, multiply by Avogadro’s number:
E  3.371  1019
J
photon
6.022141 10
23 photons
mol

J
E  2.030  105 mol
13.107 Be(s) + 2OH–(aq) + 2 H2O(l) → [Be(OH)4]2–(aq) + H2(g)
Zn(s) + 2OH–(aq) + 2 H2O(l) → [Zn(OH)4]2–(aq) + H2(g)
2 Al(s) + 2 OH–(aq) + 6 H2O(l) → 2 [Al(OH)4]–(aq) + 3 H2(g)
13.129
Step 1: Write a balanced chemical equation for this reaction
Cl2(g) + 2 NaOH(aq) → NaOCl(aq) + NaCl(aq) + 2 H2O(l)
Step 2: Calculate the mass of NaOCl to be produced
First, you must calculate the mass of 1000. L of bleach solution.
g
10001LmL   1.07  106 g
msolution  1000.L 1.07 mL
mNaOCl 
5.25%
100%
.25%
1.07  106 g   5.62  104 g
msolution  5100
%
Step 3: Calculate the moles of NaOCl to be produced
nNaOCl  5.62  104 g 
1mol
 755mol
74.4419g
Step 4: Calculate the moles of Cl2 to be consumed
nCl  755molNaOCl 
1molCl 2
 755molCl 2
1molNaOCl
Step 5: Calculate the volume of Cl2 to be consumed
2
PV  nRT
Pa m
755mol 8.3145 mol

  1kPa  16.9m 3
nRT
K 273.15K
V 

101.3kPa 
P
1000Pa
3
Note that 1 m3 = 1000 L, so this is 16,900 L (3 sig. fig.) of chlorine gas.
23.28 Addition of cryolite lowers the melting point of Al2O3. The electrolysis (which requires
that the Al2O3 be in liquid state) can therefore be performed at a lower temperature.
23.54
(a)
If they are not kept separate, the Cl2 and NaOH produced in the chlor-alkali process will
react to give NaOCl, NaCl and H2O.
Additional Practice Problems
1.
Name each of the following compounds:
(a)
CuI
copper(I) iodide
(b)
CuS
copper(II) sulfide
(c)
BaF2
barium fluoride
(d)
NiO
nickel(II) oxide
2.
Give the formula for each of the following compounds:
(a)
iron (III) chloride
FeCl3
(b)
aluminium oxide
Al2O3
(c)
cobalt (III) oxide
Co2O3
(d)
magnesium fluoride
MgF2
(e)
lithium sulfide
Li2S
(f)
copper(II) sulfide
CuS
3.
Suppose that you want to make 12 g of lithium oxide. What are the minimum masses of
lithium and oxygen you will need?
4 Li(s)
+
O2(g)
→
2 Li2O(s)
6.941 g/mol
31.9988 g/mol
29.8814 g/mol
n Li2O  12 g 
n Li  n Li2O 
1mol
 0.40mol
29.8814 g
4molLi
 0.80mol
2molLi2 O
m Li  0.80mol 
nO2  n Li2O 
2 significant figures for all three answers!
6.941g
 5.6 g
1mol
1molO2
 0.20mol
2molLi2 O
mO2  0.20mol 
31.9988 g
 6.4 g
1mol
4.
Write a balanced equation for each of the following reactions and name all products:
(a)
Na and Br2
2 Na(s) + Br2(l) → 2 NaBr(s)
sodium bromide
(b)
Li and N2
6 Li(s) + N2(g) → 2 Li3N(s)
lithium nitride
(c)
Li and O2
4 Li(s) + O2(g) → 2 Li2O(s)
lithium oxide
(d)
K and H2O
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)
or
2 K(s) + 2 H2O(l) → 2 K+(aq) + 2 OH-(aq) + H2(g)
potassium hydroxide and hydrogen gas
5.
The group 13 metals react readily with oxygen. This would seem to imply that
aluminium products should rust easily; however, they do not. Why don’t aluminium
products rust?
Pure aluminium reacts quickly with oxygen to form a thin layer of aluminium oxide on
its surface. This layer of aluminium oxide does not react further with oxygen. As such,
aluminium products do not rust. (This effect can be enhanced by “anodizing” the
aluminium which deposits a slightly thicker film of aluminium oxide on the surface of the
aluminium.)
6.
(a)
The bonding in BeCl2 is different from the bonding in MgCl2 or CaCl2. What is different
about BeCl2 and why is this the case?
Beryllium chloride is not an ionic compound (unlike MgCl2 or CaCl2). For an ionic
compound to exist, both the cation and anion must be relatively stable. A beryllium
cation (Be2+) would only have one shell of electrons (electron configuration 1s2). As
such, it would be very small (smaller than helium!) – too small to fully stabilize a +2
charge. So, BeCl2 is a covalent compound while MgCl2 and CaCl2 are ionic.
(b)
This is not the only case of beryllium behaving differently from the other alkaline earth
metals. Outside of Group 2, what element has the most similar chemical properties to
beryllium? This is an example of what kind of relationship?
The chemistry of beryllium and aluminium are often similar. This is an example of a
diagonal relationship.
7.
Write a balanced equation for each of the following reactions and name all products:
(a)
Ca and O2
2 Ca(s) + O2(g) → 2 CaO(s)
calcium oxide
(b)
Mg and H2O
Mg(s) + 2 H2O(l) → Mg(OH)2(s) + H2(g)
magnesium hydroxide and hydrogen gas
(c)
Ga and Br2(l)
2 Ga(s) + 3 Br2(l) → 2 GaBr3(s)
gallium bromide
(d)
CaCO3 and heat
CaCO3(s) → CaO(s) + CO2(g)
calcium oxide and carbon dioxide
(e)
MgCO3 and HCl(aq)
MgCO3(s) + 2 HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l)
magnesium chloride and carbon dioxide and water
or
MgCO3(s) + 2 H+(aq) → Mg2+(aq) + CO2(g) + H2O(l)
magnesium cation and carbon dioxide and water
8.
(a)
(b)
What is hard water?
Hard water is water containing Ca2+ and/or Mg2+ cations.
It may also contain other cations with large positive charges. e.g. Fe3+
Explain how/why water becomes hard.
Your answer should involve at least one balanced chemical equation.
Water dissolves minerals such as MgCO3 or CaCO3 as it passes through limestone, etc.
This is possible because CO2 from the air dissolves in water, making the water acidic:
H2O(l)
+
CO2(g)
H2CO3(aq)
The acidic water dissolves carbonates, making soluble bicarbonates:
CaCO3(s)
+
H2CO3(aq)
Ca(HCO3)2(aq)
(c)
(d)
9.
(a)
(b)
Why do deposits build up in pots and kettles which are regularly used to heat hard water?
Heating water reduces the solubility of gases such as CO2. This makes the water less acidic
and some of the soluble bicarbonates are converted back to carbonates and precipitate out.
This is an example of Le Châtelier’s principle as removal of CO2(g) drives both equilibria
shown in part (b) toward the reactant side.
What can be done to avoid these deposits? Explain the underlying chemistry.
Water can be softened. In other words, remove the Ca2+, Mg2+, etc. cations.
One way to do this is to use an ion exchange column in which anionic beads are initially
saturated with Na+ cations. When hard water passes through the ion exchange column, the
Ca2+/Mg2+/etc. cations are more strongly attracted to the anionic beads than the Na+ cations
were. So, the Ca2+/Mg2+/etc. cations adsorb to the beads and Na+ cations are released into
the water.
Since Na2CO3 is more soluble in water than CaCO3 or MgCO3, it does not precipitate the to
the same extent when the water is boiled and the CO2 evaporates out of solution.
Write a balanced chemical equation for the industrial production of sodium metal.
Include all states of matter.
2NaCl(l )  2Na(l )  Cl2(g)
Write a balanced chemical equation for the industrial production of sodium hydroxide.
Include all states of matter.
2NaCl(aq)  2H2O(l )  2NaOH(aq)  H2(g)  Cl2(g)
(c)
(d)
or 2Cl(aq)  2H2O(l )  2OH(aq)  H2(g)  Cl2(g)
These two industrial processes are based on the same method for making an otherwise
unfavourable reaction occur. What is this method?
electrolysis
Despite the fact that the two processes have reagents in common and employ the same
method, they lead to different products. Why?
Water is present in the production of sodium hydroxide.
When pure NaCl is electrolyzed the best electron acceptors are the Na+ cations, so neutral
Na is produced. When water is present, the best electron acceptors are the H+ cations, so
hydrogen gas is produced and the Na+ cations do not react.