Chapter 2 Review, pages 88–93
Knowledge
1. (c)
2. (c)
3. (d)
4. (d)
5. (a)
6. (b)
7. (b)
8. (b)
9. (d)
10. (b)
11. (d)
12. False. The equation below shows the formation of an anion.
13. False. Sodium chloride is an ionic compound.
14. True
15. True
16. False. When oxygen atoms and hydrogen atoms bond together to form water molecules,
electrons are shared between the atoms.
17. False. In molecules such as O2 and CO2, the atoms share two pairs of electrons, thus forming
double bonds.
18. True
19. False. A compound formed between potassium and fluorine would have ionic bonding.
20. True
21. False. The correct name for the compound (NH4)2SO3 is ammonium sulfite.
22. (a) (iii)
(b) (iv)
(c) (i)
(d) (vii)
(e) (ii)
(f) (v)
(g) (vi)
23. Sodium chloride, NaCl(s), is an example of an electrolyte because it forms a solution that
conducts electricity well.
24. An anion is a negatively charged ion and a cation is a positively charged ion.
25. The crystal lattice structure of sodium chloride arranges the sodium and chloride ions into a
cube shape. A sodium chloride crystal is cubic in shape, reflecting the crystal lattice structure.
26. What is special about the molecular structure of each of these elements is that they are
composed of two-atom molecules, so they are referred to as the diatomic elements.
27. If the electronegativity difference between two elements is 1.7, the two elements would form
an ionic compound.
28. Traditional and common names have been assigned to compounds throughout the ages
without regard to any standardized naming system. IUPAC names, on the other hand, follow
logical rules that give each compound a unique name. This allows chemists all over the world to
communicate the identities of chemical compounds without confusion.
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29. The zero-sum rule states that the total positive charge of the cations in a compound must
equal the total negative charge of the anions.
30. Three adverse health consequences of consuming too much sugar are increased chances of
developing type 2 diabetes, becoming obese, and developing tooth decay.
Understanding
31. When mixed with water, not all ionic compounds produce solutions that are good conductors
of electricity. Some ionic compounds dissolve only slightly in water and thus do not introduce
many ions into solution. It is also possible that only a tiny amount of a soluble ionic compound
has been added to water, again resulting in very few ions in solution. In either case, these ioniccompound–water mixtures will not conduct electricity well.
32. (a)
(b) X2Y5
33. Formation of ionic compound:
(a) Magnesium oxide:
(b) Aluminum sulfide:
(c) Lithium bromide:
34. (a) The number of protons in the nucleus of the sulfur atom does not change when it gains
electrons, so it remains a sulfur particle.
(b) When a sulfur atom gains 2 electrons it becomes a sulfur anion with a charge of –2.
35. Glucose, C6H12O6, cannot be represented with a formula unit of CH2O because it is a
molecular substance. Each molecule is composed of 6 carbon atoms, 12 hydrogen atoms, and 6
oxygen atoms bonded together. Formula units, which represent the simplest ratios of
combination, are reserved for ionic compounds because ionic compounds do not have a
molecular structure.
36. (a) Br2(l):
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(b) OF2(g):
(c) NF3(g):
(d) HCN(g):
(e) CS2(l):
(f) COCl2(g):
(g) NO+:
(h) PO33–:
(i) H2S(g):
(j) CCl4(l):
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37. The molecules in Question 36 that have double bonds are CS2 and COCl2; NO+ contains a
triple bond.
38. All of the entities in Question 36 contain polar covalent bonds except Br2 (non-polar
covalent) and NaF (ionic).
39. Electronegativity increases as atomic radius decreases because the positively charged nucleus
of a small atom can get much closer to bonding electrons of another atom than the nucleus of a
larger atom can. So, an atom with a small radius can exert a much stronger attractive force on
those electrons. The increased electrostatic attraction causes the atom to draw the bonding
electrons closer, thus giving the atom a higher electronegativity.
40. Type of bond between elements:
(a) N and H: polar covalent (two non-metals, ΔEN = 0.8)
(b) Al and F: ionic (a metal and a non-metal, ΔEN = 2.4)
(c) N and O: polar covalent (two non-metals, ΔEN = 0.4)
(d) F and F: non-polar covalent (identical atoms, ΔEN = 0)
(e) Br and K: ionic (a metal and a non-metal, ΔEN = 2.2)
41. Chemical formula for ionic compound:
(a) Copper(II) oxide is CuO.
(b) Aluminum nitrate is Al(NO3)3.
(c) Manganese(II) chloride is MnCl2.
(d) Barium fluoride is BaF2.
(e) Lead(IV) oxide is PbO2.
(f) Iron(III) sulfate is Fe2(SO4)3.
42. Chemical formula for molecular compound:
(a) Carbon disulfide is CS2.
(b) Diarsenic trioxide is As2O3.
(c) Dichlorine monoxide is Cl2O.
(d) Diantimony pentoxide is Sb2O5.
43. (a) The IUPAC name for SrS(s) is strontium sulfide.
(b) The IUPAC name for (NH4)2SO4(s) is ammonium sulfate.
(c) The IUPAC name for SnF2(s) is tin(II) fluoride.
(d) The IUPAC name for FePO4(s) is iron(III) phosphate.
(e) The IUPAC name for Ca(OH)2(s) is calcium hydroxide.
(f) The IUPAC name for MgCO3(s) is magnesium carbonate.
44. (a) The IUPAC name for NF3(g) is nitrogen trifluoride.
(b) The IUPAC name for B2O3(s) is diboron trioxide.
(c) The IUPAC name for I2O5(s) is diodine pentoxide.
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(d) The IUPAC name for BrO(g) is bromine monoxide.
45. The chemical formula for the hydrated compound iron(III) nitrate nonahydrate is
Fe(NO3)3 •9H2O.
46. The name of the hydrated compound ZnCl2•6H2O(s) is zinc chloride hexahydrate.
47. A can of pop contains about 80 % of the recommended maximum quantity of added sugar.
48. The soft-drink industry uses mainly high fructose corn syrup (HFCS) as a sweetener instead
of sucrose because HFCS is sweeter than ordinary sucrose. This is because HFCS has been
processed to change some of its glucose to a much sweeter sugar, fructose. Sweetening
beverages with HFCS requires a smaller quantity of sweetener to produce the same degree of
sweet taste.
Analysis and Application
49. Toxicity can depend on the amount consumed. A toxic substance can often be consumed
safely in small amounts. For example, in relatively small amounts, Aspirin and vitamin A are
beneficial, but if taken in large amounts, they can become toxic.
50. A typical sports energy drink would conduct electricity well. Sports drinks typically contain
electrolytes, which dissolve as ions. The presence of ions in the drink would give it good
electrical conductivity.
51. This is how potassium bromide would dissolve in water.
The diagram shows that water molecules attract potassium and bromide ions away from the
crystal. They then surround the ions, preventing them from re-joining the crystal.
52. Aluminum oxide will have a higher melting point than sodium chloride. The ions in
aluminum oxide have greater charges than the ions in sodium chloride. As a result, the electrical
forces of attraction will be stronger between the ions in aluminum oxide. Stronger attraction, in
turn, will make it necessary to heat aluminum oxide to a higher temperature to break the ions free
of one another.
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53. (a) Sodium chloride must be liquid (molten) in a Downs cell, because sodium chloride is an
ionic compound, and ionic compounds can only conduct electricity in the liquid state.
(b) Sodium ions have a positive charge and must gain electrons in a Downs cell to become
atoms.
Chloride ions have a negative charge and must lose electrons in a Downs cell to become atoms.
54. If an unknown solid compound has a high melting point, is hard and/or brittle, and conducts
electricity in the liquid state, it is most likely ionic. If it has a low melting point, is soft or waxy,
and does not conduct electricity, it is likely molecular.
55. (a) An ionic compound will form between elements C and D. With 1 valence electron, C
must lie in column 1 of the periodic table. With 6 valence electrons, D must lie in column 16 of
the periodic table. These columns lie relatively far apart, so the difference in electronegativity
between C and D will likely exceed 1.7, particularly considering that C is metallic and D is nonmetallic. With a difference in electronegativity greater than 1.7, C and D will form an ionic bond.
(b) C and D will combine as follows:
(c) The chemical formula for the compound that will form is C2D.
(d) C2D is a formula unit.
56. (a) The difference in the structure of the two molecular forms of Mecoprop is that the two
forms are mirror images of one another that are impossible to superimpose.
(b) A “lock and key” model might explain the herbicidal effects of Mecoprop as follows: The
herbicide molecules must need to bind to specific receptors on cells of the weeds. Just as a key
made as a mirror image of the real key would not fit into and operate a lock, one of the mirrorimage forms of Mecoprop must not be able to fit into certain receptors on the plant cells.
57. (a) A Lewis structure that obeys the octet rule cannot be drawn for a nitric acid molecule
because no matter how the electrons in a nitrogen monoxide molecule are arranged, 1 electron
will remain unpaired in the Lewis structure, causing one of the atoms to have an odd number of
electrons in its valence shell. For example:
(b) The molecular structure of nitrogen monoxide accounts for its high reactivity and toxicity as
follows: Since one of the atoms in a nitrogen monoxide molecule does not have an octet of
electrons due to an unpaired electron, nitrogen monoxide molecules are unstable and, therefore,
very reactive. The high reactivity would contribute to the toxicity of nitrogen monoxide.
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58. (a) Lewis structure for BF4–:
(b) The tetrafluoroborate ion is more stable than boron trifluoride because, in the ion, the boron
atom is surrounded by 8 electrons: a stable octet, like that of a noble gas. This is a very stable
arrangement. A molecule of boron trifluoride, however, does not have a complete octet, so it is
reactive.
59. Structural formulas for three compounds with the molecular formula C3H8O:
60. (a) The hydrogen atoms are not likely to be connected to each other in a hydrazine molecule,
N2H4. Hydrogen atoms can only form one bond and consequently cannot be a “bridge” between
two other atoms. This means that hydrogen cannot be a central atom in a molecule. It will be the
nitrogen molecules that are connected.
(b) Lewis structure and structural formula for hydrazine:
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(c) Lewis structure and structural formula for two compounds with the formula C2H8N2:
61. Noble gas atoms (except for helium) have full valence shells in the form of an octet of
electrons, which is an extremely stable electron arrangement. In a noble gas compound, the noble
gas atom would have an overfilled valence shell. That such a compound could be prepared would
seem unlikely to chemists because a very stable octet of electrons would have to be disrupted.
For example, XeF2 would have a Lewis structure that places 10 electrons around the xenon atom,
as sketched below.
62. A polyatomic ionic compound contains both ionic and covalent bonds. Examples may vary.
Sample answer: For example, sodium sulfate has ionic bonds between sodium cations and sulfate
anions. However, it also has covalent bonds between the sulfur and oxygen atoms within the
sulfate anion.
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63. A beryllium atom has only 2 valence electrons, and each of these can join with an electron
from another atom to form a bonding pair of electrons. For example, 1 beryllium atom can
combine with 2 hydrogen atoms, and thus beryllium has a bonding capacity of two:
Even though an oxygen atom has 6 valence electrons, it has only 2 unpaired electrons in its
valence shell and can only form 2 single bonds to other atoms while obeying the octet rule. In a
water molecule, for example, an oxygen atom combines with 2 hydrogen atoms:
This illustrates that an oxygen atom has a bonding capacity of 2, just like a beryllium atom.
64. (a) The molecular formula for cholesterol is C27H46O.
(b) It is not surprising that cholesterol is a waxy solid at room temperature because a cholesterol
molecule is relatively large, and molecular compounds with large molecules are usually liquids
or solids at room temperature. In addition, it is not unusual for molecular compounds to be soft
and waxy.
(c) The advantage of representing molecules in a structural formula like that in Figure 4 is that it
shows only some of the hydrogen and carbon atoms in the molecule. If all of the hydrogen and
carbon atoms were shown, the structural formula would be very tedious to draw and the diagram
would be so cluttered that it might be difficult to interpret.
65. (a) Molecule formed by sulfur and chlorine:
(b) Molecule formed by arsenic and fluorine:
(c) Molecule formed by germanium and chlorine:
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(d) Molecule formed by chlorine and fluorine:
66. A Pauling electronegativity value measures the tendency of an atom to pull a pair of
electrons toward itself when bonded. Since helium, neon, and argon atoms do not bond to other
atoms, Pauling electronegativities cannot be assigned to them.
67. If the force of attraction between an atom’s nucleus and a free electron exceeds the repulsion
between the atom’s electrons and the free electron, first electron affinity will be positive. This
means energy is released when the atom captures a free electron. A greater first electron affinity
would mean that attractive forces exceed repulsive forces by a greater margin. If the attraction
between an atom and a free electron is large, the atom would very likely have a strong tendency
to pull electrons toward itself when bonded. Thus a large first electron affinity would likely mean
a high electronegativity.
68. (a) The electronegativity difference between hydrogen and sulfur is 0.4. This value is much
less than 1.7, so the bond between hydrogen and sulfur atoms will be covalent. An
electronegativity difference of 0.4 results in a bond that is slightly polar covalent.
(b) The electronegativity difference between calcium and fluorine is 3.0. This is much larger
than 1.7, so the bond between calcium and fluoride will be ionic bond.
(c) The electronegativity difference between sulfur and fluorine is 1.4. This value is less than 1.7,
so the bond between sulfur and hydrogen atoms will be polar covalent.
(d) The bonds and electronegativity differences in C6H12O6 are: carbon–hydrogen = 0.4, oxygen–
carbon = 0.8, oxygen–hydrogen = 1.2, and carbon–carbon = 0. Each of the first three values is
less than 1.7, so we would predict all of these bonds to be polar covalent in glucose, C6H12O6.
The electronegativity difference between carbon atoms is 0, so those bonds are non-polar
covalent.
69. Distribution of bonding electrons between atoms:
(a)
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(b)
(c)
70. Of nitrogen, N2(g), hydrogen fluoride, HF(g), and hydrogen bromide, HBr(g), hydrogen
fluoride would have bonds with the greatest δ+ and δ– charges. The δ+ and δ– charges result from
the electronegativity difference between the entities. The ΔEN values for the pairs of entities are:
nitrogen, 0; hydrogen and fluoride, 1.8; and hydrogen and bromine, 0.8. The bond in HF(g) is the
most polar, meaning HF(g) molecules would have the greatest δ+ and δ– charges.
71. Anhydrous copper(II) sulfate is white, whereas copper(II) sulfate pentahydrate is blue. When
the water molecules are driven out of a copper(II) sulfate pentahydrate sample by heating, the
colour will thus change from blue to white.
Evaluation
72. To facilitate tarnish removal from silver using aluminum foil immersed in hot water, I would
recommend dissolving washing soda, Na2CO3(s), in the water. Table sugar, C12H22O11(s), is a
molecular substance and will not release ions when dissolved in water. A sugar solution will not
be a good conductor of electricity. Washing soda, on the other hand, is an ionic compound.
When dissolved in water, it will form sodium cations and carbonate anions. The presence of
mobile ions will make the sodium carbonate solution a good conductor of electricity so the
tarnish removal reaction can proceed.
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73. The best model for a glycine molecule is structure (iii). The electron arrangement of the
nitrogen atom in structure (i) does not follow the octet rule, due to an underfilled valence shell on
the nitrogen atom and an overfilled shell on one of the oxygen atoms. In structure (ii), the carbon
atom that is bonded to the oxygen atoms has an overfilled valence shell and its electron
arrangement does not follow the octet rule. Also, structure (ii) has a total of 32 valence electrons
when it should have 30. Structure (iv) does have a total of 30 valence electrons as required, but
the carbon atom that is bonded to the nitrogen atom has an overfilled valence shell and its
electron arrangement does not follow the octet rule. Structure (iii) is the best model for glycine
because the nitrogen atom, oxygen atoms, and carbon atoms all have electron arrangements that
follow the octet rule and the total number of valence electrons is 30 as required.
74. (a) The experimental data show that propene has two different carbon–carbon bond lengths,
one (151 pm) that agrees well with the typical length of a carbon–carbon single bond (154 pm)
and another (134 pm) that agrees with the typical length of a carbon–carbon double bond (134
pm). This is consistent with the Lewis structure for propene, which has one carbon–carbon single
bond and one carbon–carbon double bond. The experimental data show that benzene has carbon–
carbon bonds of equal length, 140 pm, and that this length is intermediate between that of a
typical carbon–carbon single bond and a carbon–carbon double bond. The Lewis structure for
benzene predicts single bonds with a length of about 154 pm alternating with double bonds
having a length of about 134 pm. Thus, the Lewis structure for benzene does not agree with the
experimental evidence.
(b) The carbon atoms in both Lewis structures in Figure 6 have octets of electrons as required by
the octet rule, so both structures satisfy the octet rule.
(c) The octet rule is not perfectly accurate in predicting the molecular structure of benzene. Even
though the Lewis structure for benzene follows the octet rule, the actual molecular structure of
benzene is inconsistent with the alternating double and single carbon–carbon bonds predicted
according to the octet rule. Benzene molecules appear to have the same type of bond all the way
around the ring, not alternating bond types. On the other hand, the actual molecular structure of
propene is consistent with a Lewis structure predicted on the basis of the octet rule.
75. The chemist should use fluorine and either xenon or radon in her research on synthesizing
molecular compounds composed of noble gas elements combined with other elements. The
chemist seeks the greatest possible movement of bonding electrons away from the noble gas
atoms, meaning she needs the most polar covalent bonds possible. This means that she should
choose pairs of elements with the greatest possible difference in electronegativity. Fluorine is the
most electronegative element in the periodic table and xenon and radon are the noble gases with
the lowest electronegativity values. These combinations of elements give the greatest possible
differences in electronegativity and thus the most polar covalent bonds.
76. (a) The chemist reasons that many ionic compounds dissolve in water to produce a solution
that conducts electricity, whereas the solutions of most molecular compounds do not conduct
electricity. Thus, if the solution of a dissolved compound conducts electricity, it is likely to be an
ionic compound. However, if the solution does not conduct electricity, then the compound is
more molecular in nature.
(b) A flaw in the chemist’s proposed method is that a test compound may not dissolve well in
water. Even if the compound were more ionic than covalent, it would release a limited number of
ions into the solution, resulting in low electrical conductivity. The observed electrical
conductivity would incorrectly suggest that the compound is more covalent than ionic.
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77. Answers may vary. Sample answer: I disagree with the clerk’s plan. Calcium chloride
dihydrate already contains water in its crystal structure and will not absorb and bind more water
molecules. Anhydrous calcium chloride would be a much better choice for a drying agent. The
anhydrous compound, lacking water, would readily absorb water vapour from the space inside
the package.
Reflect on Your Learning
78. Answers may vary. Look for students to cite an area of weakness and then develop a set of
specific strategies for improving their understanding. These strategies could include doing
Internet research or finding practice exercises on the Internet, reading related material in other
textbooks obtained from a library, or forming study groups with fellow students.
79. Answers may vary. Sample answer: I believe that science will never be able to explain
everything to the point that no exceptions, oddities, or questions remain. The variables in nature
are infinite. There will always be a level of understanding that lies beyond what we currently
know because the true picture, if there is one, is bigger than human minds can grasp.
80. Answers may vary. Sample answer: I would prefer to be an applied researcher. I appreciate
the importance of having curiosity about the way nature works and performing experiments for
the sole reason of acquiring knowledge for the sake of knowledge alone. However, I would be
really excited to put the discoveries of pure science into action in the service of people. I enjoy
working with others and would like to see my efforts make their lives better.
81. Answers may vary. Sample answer: I am now aware of the potential adverse health
consequences of consuming excessive quantities of table sugar. I do believe that most artificial
sweeteners are relatively safe, but still have my doubts about consuming artificial chemicals just
because I desire a sweet taste. I prefer to consume sucrose, but will do my best to limit my
intake, particularly by drinking less pop.
Research
82. There are seven crystal systems: triclinic, monoclinic, orthorhombic, tetragonal, trigonal,
hexagonal, and cubic. Each system has a characteristic symmetry. Tetragonal crystals, for
example, have three axes that meet at 90o, with two axes of equal length and a third axis of
greater length. Each system also has characteristic lattice structures, such as simple and bodycentred for tetragonal crystals. Examples of each system are: copper(II) sulfate, CuSO4, blue
triclinic crystals; orthoclase, KAlSi3O8, whitish monoclinic crystals; topaz, Al2SiO4(FOH)2,
pinkish orthorhombic crystals; zircon, ZnSiO4, colourless or brownish tetragonal crystals;
corundum (rubies and sapphires), Al2O3, trigonal crystals with many different coloured varieties;
aquamarine, BeAlSiO2, pale blue hexagonal crystals; and potassium chloride, KCl, whitish cubic
crystals. Students’ presentations should include diagrams of the symmetry and associated lattice
structures for each crystal system and photographs of the example compounds or minerals they
find.
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83. In solid water, or ice, the water molecules align through hydrogen bonding into a structure
that is expanded compared to the spacing of the water molecules in the liquid state. The crystal
structure of ice at ordinary conditions is called “ice 1h” and can be modelled as two
interpenetrating lattices stacked in a closely packed hexagonal arrangement. Viewing a model of
this structure from the proper angle reveals hexagonal corridors. The expanded nature of ice 1h
compared to liquid water accounts for the lower density of ice. In addition, the hydrogen bonding
is strong enough to cause freezing water, which expands by about 9 % in forming ice, to break
filled containers and fracture rock. Students’ reports should focus on this information, but may
also include a description of other forms of ice that exist at very low temperatures and/or high
pressures. Students should include a diagram or model of the crystal lattice of ice. The diagram
should show three hexagonal rings of water molecules. Each ring is bonded to another in
different planes, with rings parallel to each other.
84. Formal charge is a system of assigning charges to atoms in Lewis structures in which each
atom’s formal charge equals its normal number of valence electrons minus the sum of its number
of lone pair electrons and one-half of the number of electrons in its bonds. Generally, the best
Lewis structure for a molecule or ion yields the lowest possible formal charges, places negative
formal charges on the most electronegative atoms, and does not result in adjacent atoms having
like formal charges. The formal charges must add up to equal the net charge on the molecule or
ion. Students’ reports should include this information as well as a specific example of using
formal charge to select between possible structures. Carbon dioxide, for instance, could be
represented as either of the two structures below:
Formal charge
Left oxygen atom
Carbon atom
Right oxygen atom
Total
Structure A
Structure B
6 – (3 + 2) = +1
4 – (4 + 0) = 0
6 – (1 + 6) = –1
1 + 0 + (–1) = 0
6 – (2 + 4) = 0
4 – (4 + 0) = 0
6 – (2 + 4) = 0
0+0+0=0
The third structure has the smallest number of formal charges (none), so you would choose this
as the better of the two Lewis structures.
85. Linus Pauling realized that the bond energy for two different atoms bonded together, say A–
B, is greater than the average of the bond energies for A–A and B–B. He attributed the difference
to bond polarity and developed an equation to calculate electronegativity values from sets of
bond energies. The Mulliken scale of electronegativities is based on an average of an element’s
electron affinity and ionization energy. The Allred–Rochow scale of electronegativities is based
on the electrostatic attraction an atom’s valence electrons experience due to the underlying
electrons and atomic nucleus. Students’ posters should include this information as well as the
equations used to calculate the values on each scale and the equations for converting Mulliken
and Allred–Rochow values to Pauling values.
86. D- and L-sugars are mirror images of one another. Gilbert Levin knew that the body only
metabolizes D-sugars. He hypothesized that L-sugars would taste sweet, but would not add
calories to foods. Taste tests proved that L-sugars have a sweet taste. Unfortunately, Levin could
find no way to produce them and make a profit. By accident, he found out that D-tagatose is
enough like L-fructose to have low caloric value, yet a pleasingly sweet taste. Moreover, Levin
found a way to produce D-tagatose profitably. Students’ reports should expand on this
information and include structural formulas for D-tagatose and L-fructose. Students should also
offer examples of food products that contain D-tagatose.
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