Unit 2 Review, pages 272–279
Knowledge
1. (b)
2. (d)
3. (c)
4. (d)
5. (c)
6. (a)
7. (d)
8. (d)
9. (c)
10. (d)
11. (d)
12. (d)
13. (c)
14. (d)
15. (c)
16. (d)
17. (a)
18. (a)
19. (b)
20. (d)
21. (c)
22. (d)
23. (c)
24. (b)
25. (b)
26. (d)
27. (b)
28. (c)
29. (d)
30. False. A proton is a positively charged subatomic particle.
31. True
32. True
33. False. The atomic number of an atom is the number of protons in the nucleus.
34. True
35. True
36. True
37. True
38. False. The highest energy state for an atom is its excited state.
39. True
40. True
41. True
42. False. An orbital is the region around the nucleus where there is a high probability of
finding an electron.
43. True
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Unit 2: Structure and Properties of Matter U2-11
44. False. The electron configuration for the element calcium is [Ar]4s2.
45. False. A magnesium atom has 2 valence electrons.
46. False. A lithium atom has 1 electron in its second energy level.
47. True
48. False. The bond between 2 chlorine atoms is covalent.
49. True
50. False. Ionic compounds are composed of a metal and a non-metal ion.
51. True
52. False. The Lewis structure for a water molecule has 2 lone pairs of electrons on the
central atom.
53. False. The valence shell electron-pair repulsion (VSEPR) theory describes how
electron pairs distribute themselves in space around an atom.
54. False. The polarity of a bond decreases as the electronegativity difference decreases.
55. False. The end-to-end overlap of orbitals forms a sigma bond.
56. False. Intramolecular forces occur within molecules.
57. (a) (i)
(b) (v)
(c) (x)
(d) (iv)
(e) (vii)
(f) (ix)
(g) (viii)
(h) (ii)
(i) (vi)
(j) (iii)
58. (a) (ii)
(b) (iv)
(c) (iii)
(d) (i)
Understanding
59. J.J. Thomson’s “blueberry muffin model” of the atom predicted that an atom was
made of negatively charged particles (electrons) in a sea of positively charged particles.
60. Robert Millikan’s contribution to atomic theory was the experimental measurement of
the charge of an electron. From the charge, and the charge-to-mass ratio determined by
Thomson, he was able to calculate the mass of an electron: 9.11 × 10–31 kg.
61. The results of Ernest Rutherford’s gold foil experiment differed from his expectations
because Rutherford expected all the alpha particles to pass through the foil, but even
though most particles went through, others were deflected or bounced back because
atoms are largely made of empty space with small positive nuclei that reflected back the
few particles that hit them.
62. The problem with the Rutherford model of the atom was that it described electrons as
moving in orbits around the nucleus. If this were the case, the electrons would lose
energy as they orbit and spiral into the nucleus. Since atoms are stable, we know that this
is not the case.
63. Isotopes of a given element share the same number of protons and electrons but have
different numbers of neutrons.
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Unit 2: Structure and Properties of Matter U2-12
64. The number of protons in a neutral atom is the same as the number of electrons.
65. Isotopes are forms of an element with the same numbers of protons and electrons as
one another, but different numbers of neutrons. While most isotopes are stable,
radioisotopes undergo radioactive decay and emit subatomic particles.
66. Thompson’s discovery of the electron disproved the part of Dalton’s theory that
stated that atoms cannot be subdivided.
67. Dalton’s theory that all atoms of a given element are identical in mass and other
properties is incorrect. The isotopes and ions that exist for each atom differ in mass and
other properties.
68. Neon has 10 protons. Therefore, the three isotopes of neon, with mass numbers 20,
21, and 22, have 10, 11, and 12 neutrons, respectively.
69. A cathode ray tube is an evacuated tube with an electrode on each end. A ray, or
stream of electrons, flows through the tube from the negative electrode to the positive
electrode.
70. The mass of the nucleus composes the majority of the mass of the atom.
71. Two phenomena that support the concept that light is a packet of energy are
blackbody radiation and the photoelectric effect.
72. The Bohr atomic model, although incorrect, was still a significant contribution to
atomic theory because it included the quantization of energy in atoms.
73. Scientists are able to determine what elements are present in distant stars by using
spectroscopic measurements of light from the stars and comparing them to known atomic
emission spectra. Each element has a unique pattern of lines in its atomic spectrum, much
like the unique groove pattern in a human fingerprint. The pattern of spectral lines can be
used to indentify elements in a distant star.
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Unit 2: Structure and Properties of Matter U2-13
74. An emission spectrum is a pattern produced when the electromagnetic radiation
resulting from electrons transitioning to a lower energy level is passed through a
spectrometer. The two types of emission spectra that can be produced are a continuous
spectrum (as from an electric arc in Figure 1a) and a line spectrum (as from hydrogen
gas in Figure 1b).
75. (a) From the observation that an atom emits a specific set of frequencies of light
when it is excited by thermal energy or electricity, scientists concluded that the electrons
in atoms can only exist at discrete energy levels, meaning that the energy of the electrons
is quantized.
(b) From the observation that a beam of electrons is deflected by an electric field toward
a positively charged plate, scientists concluded that electrons have a negative charge.
76. An atomic orbital is a region in space around an atom where there is a high
probability of finding an electron.
77. It impossible to determine both the exact location and the velocity of an electron
because any attempt to probe these tiny particles causes them to change position,
velocity, or both. Once you have measured either location or velocity, it is no longer
possible to accurately measure the other property.
78. The solutions to Schrödinger’s equation provide the energy and probability of
location of electrons in an atom.
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Unit 2: Structure and Properties of Matter U2-14
79. (a) The equation that shows the relationship between the principal energy level
number, n, and the number of sublevels (subshells), is
Number of subshells = n
Explanation: The secondary quantum number, l, represents subshells, and takes values
from 0 to n – 1. For example when n = l, l is 0, which represents one subshell
(s subshell). When n = 2, l is 0 or 1, resulting in two subshells (s and p). Therefore, the
total number of subshells at a given principal energy level will be equal to n, the principal
quantum number.
(b) The equation that shows the relationship between the principal energy level number,
n, and the number of orbitals, o, is
o = n2
For example, when n = 2, o = 4.
(c) The equation that shows the relationship between the principal energy level number,
n, and the maximum number of electrons in an energy level, e, is
e = 2n2
Explanation: e = o × 2 and o = n2, so e = 2n2.
For example, for n = 2, e = 8.
80. The set of quantum numbers n = 3, l = 3, and ml = 0 is not possible because l cannot
have a value of 3 when n has a value of 3. For every set of quantum numbers, l = 0 to
n – 1. Therefore, for n = 3, the maximum value for l is 2.
81. (a) The maximum number of electrons that can have the same n, l, ml, and ms
quantum numbers is 1.
(b) The maximum number of electrons that can have the same n, l, and ml quantum
numbers is 2.
(c) The maximum number of electrons that can have the same n and l quantum numbers
is 4n – 2 electrons, where n is the principal quantum number.
82. (a) The element with the electron configuration 1s22s22p63s23p63d104s24p6 is krypton,
since the atomic number of krypton is 36.
(b) The element with the electron configuration 1s22s22p63s2 is magnesium, since the
atomic number of magnesium is 12.
(c) The element with the electron configuration 1s22s22p63s23p63d104s24p64d75s1 is
ruthenium, since the atomic number of ruthenium is 44.
(d) The element with the electron configuration 1s22s22p63s23p6 is argon, since the
atomic number of argon is 18.
83. (a) An orbital representation of a 3p sublevel that shows a violation of Hund’s rule:
(b) An orbital representation of a 3p sublevel that shows a violation of the Pauli
exclusion principle:
(c) An orbital representation of a 3p sublevel that is a correct representation of a 3p
sublevel with 3 electrons:
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Unit 2: Structure and Properties of Matter U2-15
84. (a) The element that has 5 valence electrons in its third energy level is phosphorus.
(b) The noble gas with 2 valence electrons is helium.
(c) The non-metal with 1 valence electron is hydrogen.
(d) The semiconductor with 3 valence electrons is gallium.
85. (a) Aluminum has 3 valence electrons.
(b) Oxygen has 6 valence electrons.
(c) Helium has 2 valence electrons.
(d) Bromine has 7 valence electrons.
86. Answers may vary. Sample answer: Two atoms or ions other than Mg2+ and F– that
are isoelectronic are Na+ and O2–.
87. There are no transition metals in periods 1, 2, and 3 because the atoms in these
periods do not have their highest-energy electrons in d orbitals.
88. (a) The element with the highest electronegativity is fluorine, and its electron
configuration is 1s22s22p5.
(b) The non-metal element in Group 4A is carbon, and its electron configuration is
1s22s22p2.
(c) The transition element with the lowest atomic number is scandium, and its electron
configuration is 1s22s22p63s23p64s23d1.
(d) The noble gas that has 2 valence electrons is helium, and its electron configuration
is 1s2.
89. Two electrons are transferred from the beryllium atom to the sulfur atom when they
form an ionic bond.
90. A fluoride ion is larger than a fluorine atom because it has gained an electron.
91. (a) Calcium is more likely to form a cation.
(b) Chlorine is more likely to form an anion.
(c) Lithium is more likely to form a cation.
(d) Oxygen is more likely to form an anion.
92. In forming ionic bonds,
(a) sodium is most likely to lose 1 electron
(b) phosphorus is most likely to gain 3 electrons
(c) chlorine is most likely to gain 1 electron
(d) strontium is most likely to lose 2 electrons
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Unit 2: Structure and Properties of Matter U2-16
93. Answers may vary. Sample answer:
94. Ionic compounds are electrically neutral because the total positive charge of the
cation(s) balances the total negative charge of the anion(s) in each formula unit of the
compound.
95. (a) Answers may vary. Sample answer: E is the central atom in a molecule with a
tetrahedral structure, and it is bonded to 4 H atoms, so C, Si, Ge, and Sn are possible
identities of element E.
(b) This molecule has a tetrahedral structure and bond angles of 109.5º.
96. A lone pair of electrons is unshared between bonding atoms, and exerts a stronger
repulsion effect than a bonded electron pair does.
97. Electron-pair repulsion is the nature of electron pairs to maximize the distance from
each other around a central atom.
98. The repulsion of electron pairs governs the position of bonding and lone pairs of
electrons around a central atom and thus determines the shape of the molecule.
99. Carbons 1 and 3 are trigonal planar, and carbons 2 and 4 are tetrahedral.
100. When the electronegativity difference between 2 atoms is very large, ionic bonds
form.
101. ENK = 0.8, ENBr = 2.8, ENNa = 0.9, ENO = 3.5, ENS = 2.5
In order of increasing electronegativity: K < Na < S < Br < O
102. (a) Of K and Ca, Ca has the highest electronegativity (ENK = 0.8, ENCa = 1.0).
(b) Of O and F, F has the highest electronegativity (ENO = 3.5, ENF = 4.0).
(c) Of S and Si, S has the highest electronegativity (ENS = 2.5, ENSi = 1.8).
(d) Of Ga and Ge, Ge has the highest electronegativity (ENGa = 1.6, ENGe = 1.8).
103. Given: ENH = ENP, , ENH < ENC
Required: order of decreasing polarity for P–H, O–H, N–H, F–H, C–H
Solution: Since ENH = ENP, ΔENP–H = 0.
Since ENH < ENC, ΔENC–H > 0.
Since electronegativity increases as you move right across the periodic table,
ΔENC–H < ΔENN–H < ΔENO–H < ΔENF–H.
In order of decreasing polarity: F–H > O–H > N–H > C–H > P–H
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Unit 2: Structure and Properties of Matter U2-17
104. (a) ΔEN Cl–Rb = EN Cl − EN Rb
= 3.0 – 0.8
ΔEN Cl–Rb = 2.2
ΔENCl–Rb > 1.7 , so an ionic bond forms between Rb and Cl.
(b) ΔENS–S = 0.0 , so ΔENS–S < 0.5
A non-polar covalent bond forms between S and S.
(c) ΔEN F–C = EN F − EN C
= 4.0 – 2.5
ΔEN F–C = 1.5
0.5 < ΔEN F–C < 1.7 , so a polar covalent bond forms between C and F.
(d) ΔENS–Ba = ENS − EN Ba
= 2.5 – 0.9
ΔENS–Ba = 1.6
0.5 < ΔENS–Ba < 1.7 , so a polar covalent bond forms between Ba and S. However, since
the electronegativity is so close to 1.7, this bond would have a significant degree of ionic
character as well.
(e) ΔEN N–P = EN N − EN P
= 3.0 – 2.1
ΔEN N–P = 0.9
0.5 < ΔEN N–P < 1.7 , so a polar covalent bond forms between N and P.
(f) ΔEN H–B = EN H − EN B
= 2.1 – 2.0
ΔEN H–B = 0.1
ΔENH–B < 0.5 , so a non-polar covalent bond forms between B and H.
105. Lewis structures and predictions of molecular structures:
Oxygen dichloride, OCl2
In the OCl2 molecule, the central O atom has 4 electron pairs, including 2 lone pairs, so
OCl2 has a bent structure. This is an asymmetrical structure, so OCl2 is polar.
Krypton difluoride, KrF2
In the KrF2 molecule, the central Kr atom has 5 electron pairs, including 3 lone pairs, so
KrF2 has a linear structure. The 2 bonds are identical, so KrF2 is non-polar.
Beryllium hydride, BeH2
In the BeH2 molecule, the central Be atom has 2 electron pairs and no lone pairs, so BeH2
has a linear structure. The 2 bonds are identical, so BeH2 is non-polar.
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Unit 2: Structure and Properties of Matter U2-18
Sulfur dioxide, SO2
In the SO2 molecule, the central S atom has 3 electron groups, including 1 lone pair, so
SO2 has a bent structure. This is an asymmetrical structure, so SO2 is polar.
Sulfur trioxide, SO3
In the SO3 molecule, the central S atom has 3 electron groups and no lone pairs, so SO3
has a trigonal planar structure. The 3 double bonds are identical, so SO3 is non-polar.
Nitrogen trifluoride, NF3
In the NF3 molecule, the central N atom has 4 electron pairs, including 1 lone pair, so
NF3 has a trigonal pyramidal structure. This is an asymmetrical structure, so NF3 is polar.
Iodine trifluoride, IF3
In the IF3 molecule, the central I atom has 5 electron pairs, including 2 lone pairs, so IF3
has a T-shaped structure. This is an asymmetrical structure, so IF3 is polar.
Tetrafluoromethane, CF4
In the CF4 molecule, the central C atom has 4 electron pairs and no lone pairs, so CF4 has
a tetrahedral structure. The 4 bonds are identical, so CF4 is non-polar.
Selenium tetrafluoride, SeF4
In the SeF4 molecule, the central Se atom has 5 electron pairs, including 1 lone pair, so
SeF4 has a seesaw structure. This is an asymmetrical structure, so SeF4 is polar.
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Unit 2: Structure and Properties of Matter U2-19
Krypton tetrafluoride, KrF4
In the KrF4 molecule, the central Kr atom has 6 electron pairs, including 2 lone pairs, so
KrF4 has a square planar structure. This is a symmetrical structure, and the 4 bonds are
identical, so KrF4 is non-polar.
Iodine pentafluoride, IF5
In the IF5 molecule, the central I atom has 6 electron pairs, including 1 lone pair, so IF5
has a square pyramidal structure. This is an asymmetrical structure, so IF5 is polar.
Arsenic pentafluoride, AsF5
In the AsF5 molecule, the central As atom has 5 electron pairs and no lone pairs, so AsF5
has a trigonal bipyramidal structure. This is a symmetrical structure, and the 5 bonds are
identical, so AsF5 is non-polar.
OCl2, SO2, NF3, IF3, SeF4, IF5 are polar molecules
106. (a) (i) The atomic number of phosphorus is 15, so its electron configuration is
P: 1s22s22p63s23p3 or P: [Ne]3s23p3.
(ii) The atomic number of sulfur is 16, so its electron configuration is
S: 1s22s22p63s23p4 or S: [Ne]3s23p4.
(iii) The atomic number of bromine is 35, so its electron configuration is
Br: 1s22s22p63s23p64s23d104p5 or Br: [Ar] 4s23d104p5.
(iv) The atomic number of silicon is 14, so its electron configuration is
Si: 1s22s22p63s23p2 or Si: [Ne]3s23p2.
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Unit 2: Structure and Properties of Matter U2-20
(b) Orbital diagrams of valence electrons:
(i) Phosphorus:
(ii) Sulfur:
(iii) Bromine:
(iv) Silicon:
107. (a) P would use the 3p orbital to form PH3.
(b) S would use the 3p orbital to form H2S.
(c) Br would use the 4p orbital to form HBr.
(d) To form SiH4, Si would use the 3s and 3p orbitals to form four sp3 hybridized
orbitals.
(e) To form PH5, it is necessary for P to form 5 hybrid orbitals, one for each attached
hydrogen atom. To do this, 1 electron must be promoted from the 3s orbital to the 3d
orbital, to form 5 sp3d hybrids. So P would use the 3s, 3p, and 3d orbitals to form PH5.
108. (i) Reaction of boron and hydrogen:
(a) For boron, Z = 15, so its electron configuration is B: 1s22s22p1.
(b) Orbital diagram of valence electrons for boron:
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Unit 2: Structure and Properties of Matter U2-21
(c) Boron reacts with hydrogen to form boron dihydride, BH2 (2 B–H bonds and 1
lone pair) or borane, BH3 (3 B–H bonds); and so on. In each case, there are 3 groups
arranged around the B atom. To form these compounds of boron and hydrogen, an
electron must be promoted from the 2s orbital to the 2p orbital.
(d) Hybrid orbital for B:
(e) B: three half-empty sp2 hybrid orbitals; three orbitals available for bonding
(f) B would use an sp2 hybrid orbital to bond with H.
(g)
(ii) Reaction of nitrogen and hydrogen:
(a) For nitrogen, Z = 7, so its electron configuration is N: 1s22s22p3.
(b) Orbital diagram of valence electrons for nitrogen:
(c) Nitrogen reacts with hydrogen to form ammonia, NH3. Ammonia is trigonal
pyramidal, with 3 N–H bonds and 1 lone pair, so nitrogen requires 4 hybrid orbitals.
Therefore, the hybridization must be sp3. An electron is promoted from the 2s to the
2p orbital just prior to bonding, creating a pair of electrons in 2p (the lone pair).
(d) Hybrid orbital for N:
(e) N: three half-empty sp3 hybrid orbitals; three orbitals available for bonding
(f) N would use an sp3 hybrid orbital to bond with H.
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Unit 2: Structure and Properties of Matter U2-22
(g)
(iii) Reaction of carbon and hydrogen:
(a) For carbon, Z = 6, so its electron configuration is C: 1s22s22p2.
(b) Orbital diagram of valence electrons for carbon:
(c) Carbon reacts with hydrogen to form methane, CH4. The central C in CH4 forms 4
C–H bonds, so an atom will need to be promoted from 2s to 2p.
(d) Hybrid orbital for C:
(e) C: four half-empty sp3 hybrid orbitals; four orbitals available for bonding
(f) C would use an sp3 hybrid orbital to bond with H.
(g)
(iv) Reaction of fluorine and hydrogen:
(a) For fluorine , Z = 9, so its electron configuration is F: 1s22s22p5.
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Unit 2: Structure and Properties of Matter U2-23
(b) Orbital diagram of valence electrons for fluorine:
(c) Fluorine reacts with hydrogen to form hydrogen fluoride, HF. The unpaired
electron in the 1s orbital of hydrogen will form a sigma bond with the unpaired
electron in the 2p orbital of fluorine. F does not need to promote any electrons to form
HF.
(d) Hybrid orbital for F: not applicable
(e) F: one half-empty 2p orbital; one orbital available for bonding
(f) F would use a 2p orbital to form a sigma bond with H.
(g)
109. (a) Be: sp hybridization; BeCl2: linear molecule
(b) S: sp3 hybridization; H2S: bent
(c) C: sp2 hybridization; H2CO: trigonal planar
(d) Si: sp3 hybridization; SiF4: tetrahedral
110. Of butane, CH3CH2CH2CH3, and 1-propanol, CH3CH2CH2OH, 1-propanol would
have the greater surface tension because of hydrogen bonding.
111. (a) Polar bonds are intramolecular and are created by the sharing of electrons
between atoms with differences in electronegativity. Hydrogen bonds are intermolecular
bonds and are a dipole–dipole interaction of slightly positive hydrogen atoms bonded to a
highly electronegative atom on a nearby molecule.
(b) Dipole–dipole interactions are a result of an electrostatic attraction between polar
molecules, and London dispersion forces are a result of temporary instantaneous dipole–
dipole interactions between non-polar molecules. Both are types of intermolecular forces
that attract molecules to each other.
112. (a) The reason that copper is the most common metal used in electrical wires is that,
like most metallic solids, copper conducts electricity well. Copper is also relatively
unreactive compared to most other metals.
(b) The reason that diamonds are used in industry as abrasives and drill tips is that
covalent network solids have the strongest intramolecular forces, resulting in extreme
hardness.
(c) The reason that liquid nitrogen (boiling point –196 °C) is used as a refrigerant is that
molecular solids have very weak intermolecular interactions and, consequently, low
melting and boiling points. Consequently, nitrogen readily absorbs thermal energy from
its environment to melt or boil. This loss of thermal energy keeps objects refrigerated.
113. Intramolecular forces are the chemical bonds between atoms in molecules, and
intermolecular forces are the interactions between molecules.
114. (a) (i) Van der Waals force is intermolecular.
(ii) London dispersion is intermolecular.
(iii) Dipole–dipole force is intermolecular.
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Unit 2: Structure and Properties of Matter U2-24
(iv) Ionic bonds are neither intramolecular nor intermolecular because they only occur
between ions, not between molecules.
(v) Hydrogen bonds are intermolecular.
(vi) Covalent bonds are intramolecular.
(b) From strongest to weakest: vi, iv, i (v, iii, ii) or covalent bonds, ionic
bonds, van der Waals (hydrogen bonds, dipole–dipole, London dispersion)
115. Molten KCl will conduct electricity even though solid potassium chloride does
not because when KCl melts the K+ and Cl– ions are separated and free to allow the
transfer of electrical charge.
116. (a) Answers may vary. Sample answer:
Characteristic
Electrostatic attraction
Elements involved
Bond Strength
Metallic bond
Y
only metals
variable. Some metals are
soft (e.g., the alkali metals)
while others are hard and
have extremely high
melting points (e.g., W)
Ionic bond
Y
metals and non-metals
Strong
(b) Answers may vary. Sample answer:
Characteristic
Formed by covalent bonding?
Strong bonding?
Bound by intramolecular forces?
Structure
Size
Network solid
Y
Y
Y
complex structure
Macroscopic
Covalent molecule
Y
Y
Y
single molecule
microscopic
117. (a) Gallium is a metal.
(b) The compound is ionic.
(c) Liquid tin(IV) chloride is molecular.
(d) Paradichlorobenzene is molecular.
(e) Mica is a covalent network.
(f) Boron is a covalent network.
(g) Hydrargyrum is a metal.
118. If titanium dioxide melts to produce a liquid that conducts electricity, it is an ionic
solid; if not, it is a covalent network solid.
119. (a) CO2 forms a molecular solid.
(b) SiO2 forms a covalent network.
(c) Si forms a covalent network.
(d) CH4 forms a molecular solid.
(e) Ru forms a metal.
(f) I2 forms a molecular solid.
(g) KBr forms an ionic solid.
(h) H2O forms a molecular solid.
(i) NaOH forms an ionic solid.
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Unit 2: Structure and Properties of Matter U2-25
(j) U forms a metal.
(k) CaCO3 forms an ionic solid.
(l) PH3 forms a molecular solid.
120. B2H6: molecular; SiO2: network; CsI: ionic; W: metallic
121. Lattice structures are like a scaffold network; at each connecting point of the
“scaffolding”(bonds) are the atoms that make up the lattice.
122. Ionic compounds are brittle because when a large enough force is applied it causes
the ion lattice to shift and then ions with like charges are forced to be close to each other,
and the resulting repulsion causes the lattice to break apart.
123. (a) Forming crystal lattices is a property of ionic compounds.
(b) Low melting and boiling points is a property of covalent compounds.
(c) Being brittle is a property of ionic compounds.
(d) Being soluble in water is a property of ionic compounds.
(e) Having low solubility in water is a property of covalent compounds.
(f) High melting and boiling points is a property of ionic compounds.
Analysis and Application
124. (a) How hydrogen and chlorine atoms will react to form a covalent bond:
(b) How 2 fluorine atoms will react to form a covalent bond:
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Unit 2: Structure and Properties of Matter U2-26
125. The photoelectric effect is best demonstrated with metal because metallic solids are
made of a “sea” of free-moving electrons around positive nuclei, in contrast to all the
other solids that do not have free electrons in their solid form. Since the electrons are
loosely held, they have low ionization energies. Therefore a photon has the right amount
of energy to free an electron from the metal surface. The free moving electrons in metal
are easier to activate with photons than attempting to free electrons in other solids, which
are held together more tightly and will not be able to escape the solid because the photons
cannot provide enough energy to free the electrons. (If a photon does not have enough
energy, an electron will not be freed no matter how many photons strike it.)
126. Water’s bent (V) shape helps to create its intense polarity, which gives it the ability
to form hydrogen bonds as well as create other strong dipole–dipole interactions. The
polarity of water and the resulting strong intermolecular forces give water its
characteristic high surface tension, high boiling point, and its abilities as a universal
solvent.
127. (a) Lewis structure for biacetyl, including hybridization of the carbon atoms:
Lewis structure for acetoin, including hybridization of the carbon atoms:
(b) In biacetyl, the C–C–O bond angles are all 120º. In acetoin, the C–C–O bond angle
around the double-bonded carbon is 120º, and the C–C–O bond angle around the singlebonded carbon is 109.5º.
(c) The biacetyl molecule has 11 sigma bonds and 2 pi bonds. The acetoin molecule has
13 sigma bonds and 1 pi bond.
128. (a) Bonding in HF:
There is 1 single bond in HF.
In H, there is 1 unpaired electron in 1s.
In F, there is 1 unpaired electron in 2p.
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Unit 2: Structure and Properties of Matter U2-27
When the hydrogen fluoride molecule forms, the 1s orbital in the hydrogen atom
overlaps with the 2p orbital of the fluorine atom to form a covalent bond (a sigma bond).
(b) Bonding in F2:
There is 1 single bond in F2.
In F, there is 1 unpaired electron in the 2p orbital.
When the fluorine molecule forms, the 2p orbitals of the 2 fluorine atoms overlap to form
a covalent bond (a sigma bond).
(c) Bonding in BCl3:
There are 3 identical single bonds in BCl3.
In B, there is one unpaired electron in 2p.
In each of the 3 Cl atoms, there is 1 unpaired electron in 3p.
In order for boron to form 3 identical covalent bonds, it must provide 3 single electrons in
3 separate, identical orbitals. An electron is promoted from the 2s to the 2p orbital just
prior to bonding:
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Unit 2: Structure and Properties of Matter U2-28
The s orbital and the 2 p orbitals become 3 sp2 hybrids.
The 3 atoms from the 3p orbitals of the Cl atoms overlap with boron’s 3 hybridized sp2
orbitals. For simplicity, only one of chlorine’s p orbitals is shown in the diagram below.
(d) Bonding in NF3:
There are 3 single bonds and 1 lone pair around the N atom.
In N, there is 1 lone pair in 2s and there are 3 unpaired electrons in 2p.
In each F, there is 1 unpaired electron in 2p.
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Unit 2: Structure and Properties of Matter U2-29
The central atom, N, has 3 single bonds and 1 lone electron pair, so nitrogen requires
4 hybrid orbitals. Therefore, the hybridization must be sp3. An electron is promoted from
the 2s to the 2p orbital just prior to bonding, creating a pair of electrons in 2p (the lone
pair).
The s orbital and the 3 p orbitals become 4 sp3 hybrids.
The 3 atoms from the three 2p orbitals of the F atoms overlap with 3 of nitrogen’s
hybridized sp3 orbitals, and the other sp3 orbital is occupied by nitrogen’s lone pair.
For simplicity, the fluorine atoms have not been shown in the diagram of nitrogen
trifluoride below.
(e) Bonding in C2Cl4:
Each C atom has 2 C–Cl bonds and 1 double C=C bond, so there are 3 groups of
electrons around each C.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Structure and Properties of Matter U2-30
In each C, there are 2 unpaired electrons in 2p.
In each Cl, there is 1 unpaired electron in 3p.
The central C atoms each have 3 electron groups, so carbon requires 3 hybrid orbitals.
Therefore, the hybridization must be sp2. An electron is promoted from the 2s to the 2p
orbital just prior to bonding:
Both carbon atoms are partially hybridized. The s orbital and 2 of the p orbitals become 3
sp2 hybrids.
The carbon atoms can now each form 3 sigma bonds using the sp2 orbitals: 1 sigma bond
forms with the other carbon atom and 2 sigma bonds form with chlorine atoms.
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Unit 2: Structure and Properties of Matter U2-31
Each carbon atom has 1 additional unpaired electron in a p orbital. These unpaired
electrons form a pi bond.
129. If you were designing a surfactant, you would want your surfactant to have a nonpolar end, such as a hydrocarbon chain, and a soluble polar end, such as a carboxylic
acid. The polar end of the surfactant orients toward the surface of the water, with which it
can interact through dipole–dipole forces. This lowers the surface tension of water by
interfering with interactions between water molecules. It also results in the non-polar end
facing away from the water. The non-polar end of the surfactant molecule can interact
with and dissolve dirt molecules. This is how adding soap to water assists cleaning.
130. The increased boiling point and decreased vapour pressure of hydrogen peroxide in
comparison to water are due to increased hydrogen bonding that takes place with H2O2 as
a result of the additional oxygen.
131. Answers may vary. Answers should include: Without hydrogen bonding, water
would not have its unusually high boiling point and would probably not exist as a liquid
in most places on Earth. Without liquid water, our planet would be different in countless
ways, particularly in relation to water as a solvent and its necessity for biological life.
132. If a small amount of indium impurity were added to pure selenium, a p-type
semiconductor would be formed.
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Unit 2: Structure and Properties of Matter U2-32
Evaluation
133. Answers may vary. Sample answers: Nanoparticles have medical applications in
drug delivery, imaging, and treatment of cancers. Widespread use of nanoparticles would
create a large influx of nanoparticles into the natural environment, but the results of this
have not been extensively researched. However, some research into zinc oxide
nanoparticles has shown these nanoparticles have toxic effects on human cells. It is
possible that large quantities of nanoparticles could adversely affect organisms in the
natural environment, and more research should be done to study their effects.
134. Answers may vary. Sample answer: The discoveries made by particle accelerators
are analogous to those made by J.J Thomson and Albert Einstein: experiments that
seemed costly and perhaps unjustified at the time led to pivotal discoveries through
which we transformed our everyday lives. As with earlier pivotal scientific
breakthroughs, the discovery of new subatomic particles has increased our understanding
of the evolution of the universe and how matter behaves in the universe, as well as having
the potential to develop new technologies.
135. Answers may vary. Students may evaluate the following types of bulbs:
Incandescent: Traditional light bulb used in households and industry for appliances,
advertising, flashlights, and more; Provides light by heating a filament composed of
tungsten in a vacuum sealed glass container.
Cost: Inexpensive
Environmental impact: Negative impact as a result of its light production inefficiency,
resulting in higher energy consumption and the increase in pollution release into the
environment
Advantages: Standard light bulb, cheap to produce, fitted for most current lighting
fixtures, used in most places in the world
Disadvantages: More than 90% of the energy produced by a bulb is heat, and the bulb
has a much shorter lifetime than most other bulbs
Compact fluorescent: Evacuated tube filled with gas and coated with fluorescent
material, current is run through the gas, which emits light and causes the coating to
fluoresce and produce light
Cost: Higher than incandescent, less expensive than most other alternatives
Environmental impact: Positive because of high energy efficiency leading to less
production of pollution than incandescent bulbs
Advantages: Much more energy efficient than incandescent bulbs, longer lifetime
Disadvantages: More costly to manufacture, some health issues in people with light
sensitivity associated with chronic use, the light produced, and characteristic “flicker”
associated with these light bulbs is not always as aesthetically pleasing as traditional light
bulbs
Copyright © 2012 Nelson Education Ltd.
Unit 2: Structure and Properties of Matter U2-33
Cold cathode: Works very similarly as a cathode ray tube, but instead of being
evacuated it is filled with a gas that will emit light with current put through it; Cold
cathode lamps are similar to fluorescent lamps but do not require heating to excite the gas
and emit light; Commonly used in neon lights and display lights in televisions and
computers
Cost: High
Environmental impact: Positive due to high efficiency of light production
Advantages: Very efficient lighting, long lifetime
Disadvantages: Range of brightness of light output is poor, manufacturing cost is higher
than traditional incandescent, range of practical uses is very narrow
High intensity discharge lamp (HID): Produces an electric arc in a tube filled with
gases such as mercury vapour and metal halides; Used mostly for lighting large areas,
such in outdoor or industrial settings
Cost: Much higher than traditional lighting
Environmental impact: Positive due to its high efficiency and high longevity
Advantages: Extremely energy efficient, long lasting, dimming capabilities, smaller size
and weight comparatively with other bulbs, flicker free
Disadvantages: More expensive to manufacture, require “warm-up” period before
emitting strong source of light, intensity of light produced narrows the situations where
their use is practical
Light emitting diode (LED): Passing current through a semiconductor diode results in
the emission of light; Use is as diverse as traditional incandescent lighting, and due to
their small size are frequently used in displays of electronic devices such as laptops,
televisions, mobile phones, cameras, and more
Cost: Highest of all light bulbs
Environmental impact: Most positive of all light bulbs due to its very high energy
efficiency and expansive lifespan
Advantages: Small, more efficient than both incandescent and fluorescent lights, longest
lasting, no flicker, easily dimmed
Disadvantages: Cost significantly more than most other bulbs, some fixtures built for
using traditional incandescent light bulbs cannot accommodate the switch to LED lights
Reflect on Your Learning
136. Answers may vary. Students will describe some of the strategies they have used to
understand the concepts Unit 2.
137. Answers may vary. Students will explain what information in Unit 2 they found
most and least interesting, and why.
138. Answers may vary. Students will identify any misconceptions they had about atomic
structure and bonding before beginning Unit 2, explain how they came about, and discuss
how working through Unit 2 helped them to identify and correct their misconceptions.
139. Answers may vary. Students will suggest ways to explain what they learned in
Unit 2 to a friend or family member who does not have any chemistry training.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Structure and Properties of Matter U2-34
Research
140. (a) Disposable diapers utilize an absorbent polymer crystal called sodium
polyacrylate to absorb liquid.
(b) The widespread use of disposable diapers has increased the output of the absorbent
polymer used in these diapers, generated a large amount of waste in landfills, and
released large amounts of effluents from the diaper manufacturing process.
141. (a) Silicon dioxide, SiO2, is a covalent compound.
(b) Glass is brittle, has a high melting point, does not conduct electricity as a solid, and is
insoluble in water. Most of its characteristics could be either ionic or covalent and though
glass has a lower melting point than most covalent network solids; its lack of solubility in
water demonstrates that it is a covalent material.
142. Answers may vary. Technologies based on the principles of atomic emission that
can be discussed include: Laser, astronomical spectroscopy, instruments used in
analytical spectroscopy such as: IR spectroscopy; fluorescence spectroscopy; and Raman
spectroscopy, and others.
Answers should include a description of the technology and how it utilizes the principles
of atomic emission; how the technology is used; the benefits the technology provides for
society; and, the current economic impacts of the technology (i.e., how much of it is used
in the world, how much is the industry it is used in worth, future growth of the
technology, and industry it is involved in).
143. Answers may vary. Fireworks are composed of an oxygen producer, fuel, a binder,
and a colour producer. The processes used to create colored light in fireworks are
incandescence (heat) and luminescence. Answers should include a description of oxygen
producers, fuels, and binding materials; a description of incandescence and luminescence;
a few examples of the colours that different metal salts emit; and how the composition of
each firework is designed based on the desired timing and colours to be emitted.
144. Answers may vary. Timelines should include:
• two distinct and important contributors to atomic theory from each discipline
• a description of the contribution of each individual to the field of atomic theory
• a description of how the contribution of each individual played a role in future
endeavours and discoveries of atomic theory
The timeline should include the date—or range of dates for less recent contributors—that
each contribution to atomic theory was made, with dates in chronological order.
145. (a) Mercury, silver, tin, and copper are used in dental amalgams. Amalgam is used
in dentistry for its low cost, strength, durability, and moldability.
(b) Answers may vary. Some metals used in coins include: copper, tin, gold, nickel, and
zinc. These metals are non-toxic, non-reactive, corrosion-resistant, and imprintable but
not too weak to be bent out of shape or broken.
Copyright © 2012 Nelson Education Ltd.
Unit 2: Structure and Properties of Matter U2-35
146. Answers may vary. Answers should include the following information: Peptide
bonds are responsible for linking amino acids together to form the peptide sequence. The
three-dimensional structure of keratin is governed by hydrogen bonds within the aminoacid backbone, which give the protein an alpha-helix structure. Hydrophobic interactions
between two individual keratins cause them to twist around each other, forming a coiledcoil dimer. Multiple dimers are cross-linked by disulfide bonds and form higher-order
filaments.
Perming applies chemicals (reducing agents) to hair, which break the disulfide bonds
between amino acid chains. An oxidizing agent is then applied to re-establish disulfide
bonding to set the hair in a new formation.
Colouring is a multi-step process. It first involves the removal of natural hair colour by
breaking down pigments with oxidizing agents such as hydrogen peroxide. During the
dying process, a dye precursor such as 1,4-diaminobenzene is oxidized and then
combines with a coupling compound that will produce the desired colour. The resulting
combination of the two compounds is large enough to attach to the keratin strands and
colour the hair.
147. Answers may vary. Some possible isotopes to research are: 123I, 201Tl, 67Ga, 111In,
133
Xe, 18F. Answers should include:
• the storage measures used for the chosen isotope, as outlined by the Canadian Nuclear
Safety Commission
• the amount of radiation the chosen isotope emits and how this is minimized for patients
• the duration of the isotope in the body, how it distributes in the body, and how this is
taken into account for dosage
• the maximum allowance of the chosen isotope a patient can receive
• occupational health and safety measures for the chosen isotope, as outlined by the
Canadian Nuclear Safety Commission
• the disposal procedures and requirements, as outlined by the Canadian Nuclear Safety
Commission
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Unit 2: Structure and Properties of Matter U2-36