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CHAPTER 11
CHEMICAL BONDS: THE FORMATION
OF COMPOUNDS FROM ATOMS
SOLUTIONS TO REVIEW QUESTIONS
1.
smallest
Cl, Mg, Na, K, Rb largest.
2.
More energy is required for neon because it has a very stable outer electron structure
consisting of an octet of electrons in filled orbitals (noble gas electron structure). Sodium,
an alkali metal, has a relatively unstable outer electron structure with a single electron in
an unfilled orbital. The sodium electron is also farther away from the nucleus and is
shielded by more inner electrons than are neon outer electrons.
3.
When a third electron is removed from beryllium, it must come from a very stable
electron structure corresponding to that of the noble gas, helium. In addition, the third
electron must be removed from a +2 beryllium ion, which increases the difficulty of
removing it.
4.
The first ionization energy decreases from top to bottom because in the successive alkali
metals, the outermost electron is farther away from the nucleus and is more shielded from
the positive nucleus by additional electron energy levels.
5.
The first ionization energy decreases from top to bottom because the outermost electrons
in the successive noble gases are farther away from the nucleus and are more shielded by
additional inner electron energy levels.
6.
Barium and beryllium are in the same family. The electron to be removed from barium is,
however, located in an energy level farther away from the nucleus than is the energy level
holding the electron in beryllium. Hence, it requires less energy to remove the electron
from barium than to remove the electron from beryllium. Barium, therefore, has a lower
ionization energy than beryllium.
7.
The first electron removed from a sodium atom is the one valence electron, which is
shielded from most of its nuclear charge by the electrons of lower levels. To remove a
second electron from the sodium ion requires breaking into the noble gas structure. This
requires much more energy than that necessary to remove the first electron, because the
Na+ is already positive.
8.
(a)
(b)
(c)
K 7 Na
Na 7 Mg
O 7 F
(d)
(e)
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I 7 Br
Zr 7 Ti
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9.
The first element in each group has the smallest radius.
10.
Atomic size increases down a column since each successive element has an additional
energy level which contains electrons located farther from the nucleus.
11.
A Lewis structure is a representation of the bonding in a compound. Valence electrons are
responsible for bonding, therefore they are the only electrons that need to be shown in a
Lewis structure.
12.
Metals are less electronegative than nonmetals. Therefore, metals lose electrons more
easily than nonmetals. So, metals will transfer electrons to nonmetals leaving the metals
with a positive charge and the nonmetals with a negative charge.
13.
The noble gases are the most stable of all the elements because they have a complete
octet (8 electrons) in their valence level. When the elements in Groups 1A, 2A, 6A and
7A form ions, they want to establish a stable electron structure.
(a) The elements in Group 1A lose an electron (obtain a positive charge) in order to
achieve a noble gas configuration.
(b) The elements in Group 2A lose electrons (obtain a positive charge) in order to
achieve a noble gas configuration.
(c) The elements in Group 6A gain electrons (obtain a negative charge) in order to
achieve a noble gas configuration.
(d) The elements in Group 7A gain an electron (obtain a negative charge) in order to
achieve a noble gas configuration.
14.
A polar bond is a bond between two atoms with very different electronegativities. A polar
molecule is a dipole due to unequal electrical charge between bonded atoms resulting
from unequal sharing of electrons.
15.
The electron pair arrangement is the arrangement of both bonding and nonbonding
electrons in a Lewis structure. The molecular shape of a molecule is the three
dimensional arrangement of its’ atoms in space.
16.
A Lewis structure is a visual representation of the arrangement of atoms and electrons in
a molecule or an ion. It shows how the atoms in a molecule are bonded together.
17.
The more electronegative atom in the bond between two atoms will more strongly attract
electrons so it will have a partial negative charge ( -). The less electronegative atom will
have a partial positive charge (+) because the bonding pair of electrons has been pulled
away by the more electronegative atom.
18.
The dots in a Lewis structure represent nonbonding pairs of electrons. The lines represent
bonding pairs of electrons.
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19.
Group
20.
Lewis structures:
•
•
•
Cs–
Ba≠
Tl
–Pb
–Po
–At
≠Rn
ª≠
ª≠
ª≠
¶≠
¶≠
Each of these is a representative element and has the same number of electrons in its
outer energy level as its periodic group.
21.
(a)
(b)
1A
E–
2A
E≠
3A
º≠
E
4A
º≠
–E
5A
–E
ªº≠
6A
•
–E
ª≠
7A
•≠
–E
¶
Elements with the highest electronegativities are found in the upper right hand
corner of the periodic table.
Elements with the lowest electronegativities are found in the lower left of the
periodic table.
22.
Valence electrons are the electrons found in the outermost energy level of an atom.
23.
By losing one electron, a potassium atom acquires a noble gas structure and becomes a
K + ion. To become a K 2+ ion requires the loss of a second electron and breaking into the
noble gas structure of the K + ion. This requires too much energy to generally occur.
24.
An aluminum ion has a +3 charge because it has lost 3 electrons in acquiring a noble gas
electron structure.
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CHAPTER 11
SOLUTIONS TO EXERCISES
1.
(a)
(b)
(c)
(d)
(e)
2.
(a)
(b)
(c)
(d)
(e)
A calcium atom is larger because it has electrons in the 4th shell, while a calcium
ion does not. In addition, Ca2+ ion has 20 protons and 18 electrons, creating a
charge imbalance and drawing the electrons in towards the nucleus.
A chloride ion is larger because it has one more electron than a chlorine atom in its
outer shell. Also, the ion has 17 protons and 18 electrons, creating a charge
imbalance, resulting in a lessening of the attraction of the electrons towards the
nucleus.
A magnesium ion is larger than an aluminum ion. Both ions will have 10 electrons
in their valence shells, but the aluminium ion will have a greater charge imbalance
since it has 13 protons and the magnesium ion has 12 protons. The charge
imbalance draws the electrons in more closely to the nucleus.
A sodium atom is larger than a silicon atom. Sodium and silicon are both in period
3. Going across a group, the radii of atoms decrease.
A bromide ion is larger than a potassium ion. The bromide ion has 35 protons and
36 electrons, creating a charge imbalance that results in a lessening of the attraction
of the electrons towards the nucleus. The potassium ion has 19 protons and 18
electrons, creating a charge imbalance that results in the electrons being drawn
more closely to the nucleus.
Fe2+ has one electron more than Fe3+, so will have a larger radius.
A potassium atom is larger than a potassium ion. They both have 19 protons. The
potassium atom has 19 electrons. The potassium ion has 18 electrons, creating a
charge imbalance that results in the electrons being drawn more closely to the
nucleus.
A chloride ion is larger than a sodium ion. The chloride ion has 17 protons and 18
electrons, creating a charge imbalance that results in a lessening of the attraction of
the electrons towards the nucleus. The sodium ion has 11 protons and 10 electrons,
creating a charge imbalance that results in the electrons being drawn more closely
to the nucleus.
A strontium atom is larger than an iodine atom. Strontium and iodine are both in
period 5. Going across a group, the radii of atoms decrease.
A rubidium ion is larger than a strontium ion. Both ions will have 36 electrons in
their valence shells, but the strontium ion will have a greater charge imbalance
since it has 38 protons and the rubidium ion has 37 protons. The charge imbalance
draws the electrons in more closely to the nucleus.
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- Chapter 11 -
3.
(a)
(b)
(c)
+
H
Na
H
O
F
N
(a)
(b)
(c)
+
H
Li
C
Cl
H
Cl
5.
(a)
ionic
(b)
covalent
(c)
covalent
(d)
ionic
6.
(a)
covalent
(b)
ionic
(c)
covalent
(d)
ionic
7.
Magnesium has an electron structure 1s22s22p 63s2, while the structure for chlorine is
1s22s22p 63s23p 5. When these two elements react with each other, each magnesium atom
loses its 3s2 electrons, one to each of two chlorine atoms. The resulting structures for both
magnesium and chlorine are noble gas configurations.
8.
(a)
9.
(a)
4.
(b)
10.
(a)
(b)
(d)
(e)
(f)
+
Pb
N
H
S
O
C
(d)
(e)
(f)
+
I
Mg
O
Br
H
F
F + 1 e- ¡ F-
(b)
Mg + F
+ F
MgF2
K + K
+ O
K 2O
Ca + O
Na + Br
Ca ¡ Ca2+ + 2 e -
CaO
NaBr
11.
Valence electrons:
H(1)
K(1)
Mg(2)
He(2)
Al(3)
12.
Valence electrons:
Si(4)
N(5)
P(5)
O(6)
Cl(7)
13.
Noble gas structures:
(a) Calcium atom, loses 2 e (b) Sulfur atom, gains 2 e (c) Helium, none
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- Chapter 11 -
14.
(a)
(b)
(c)
Chloride ion, none
Nitrogen atom, gain 3 e - or lose 5 e Potassium atom, lose 1 e -
15.
(a)
(b)
ionic, NaCl
molecular, CH 4
16.
(a)
(b)
(c)
(d)
Two oxygen atoms will form a nonpolar covalent compound. The formula is O2.
Hydrogen and bromine will form a polar covalent compound. The formula is HBr.
Oxygen and two hydrogen atoms will form a polar covalent compound. The
formula is H2O.
Two iodine atoms will form a nonpolar covalent compound. the formula is I2.
17.
(a)
(b)
NaH, Na 2O
CaH 2 , CaO
(c)
(d)
AlH 3 , Al 2O3
SnH 4 , SnO2
18.
(a)
(b)
SbH 3 , Sb2O3
H 2Se, SeO3
(c)
(d)
HCl, Cl 2O7
CH 4 , CO2
19.
Li 2SO4
Rb2SO4
Fr2SO4
lithium sulfate
rubidium sulfate
francium sulfate
K 2SO4
Cs 2SO4
potassium sulfate
cesium sulfate
20.
BeBr2
MgBr2
SrBr2
beryllium bromide
magnesium bromide
strontium bromide
BaBr2
RaBr2
barium bromide
radium bromide
21.
Lewis structures:
(c)
(d)
ionic, MgBr2
molecular, Br2
(e)
–
molecular, CO2
2–
(a)
Na
(b)
Br
(c)
22.
(a)
Ga
(b)
Ga3+
23.
(a)
(b)
covalent
ionic
(c)
(d)
ionic
covalent
24.
(a)
(b)
covalent
ionic
(c)
(d)
covalent
covalent
25.
(a)
ionic
(b)
covalent
(c)
covalent
26.
(a)
covalent
(b)
covalent
(c)
ionic
(c)
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O
Ca2+
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- Chapter 11 -
27.
(a) H H
28.
(a)
29.
O O
(b)
N
(b)
31.
S C S
H
H
H
N
(d)
H
3+
O S O
O
(a ) I
(b ) S
(c )
O N O
O
(c) H N H
H
(a ) Ba 2 +
(c )
32.
(d) Na
+
S H
H
(b ) Al
(d )
C
(e )
O C O
O
H
(d )
O C lO
O
(e )
O N O
O
2–
–
2–
O C O
O
I I
(c)
H H
(c) H C C H
H H
(a) Cl N Cl
Cl
(a)
(c) Cl Cl
(b) Br Br
(b) H O C O
O
H
30.
N
N
–
+
–
–
–
–
2–
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Cl
–
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33.
(a)
H 2O, polar
(b)
HBr, polar
(c)
CF4 , nonpolar
34.
(a)
F2 , nonpolar
(b)
CO2 , nonpolar
(c)
NH 3 , polar
35.
(a)
(b)
(c)
4 electron pairs, tetrahedral
4 electron pairs, tetrahedral
3 electron pairs, trigonal planar
36.
(a)
(b)
(c)
3 electron pairs, trigonal planar
4 electron pairs, tetrahedral
4 electron pairs, tetrahedral
37.
(a)
tetrahedral
(b)
pyramidal
(c)
tetrahedral
38.
(a)
tetrahedral
(b)
pyramidal
(c)
tetrahedral
39.
(a)
tetrahedral
(b)
pyramidal
(c)
bent
40.
(a)
tetrahedral
(b)
bent
(c)
bent
41.
Oxygen
42.
Potassium
HH
43.
N 2 H4
14 e–
NN
HN3 16 e– H – N = N = N
HH
44.
(a) NO2– 18 e–
–
O
N
O
bent
2–
O
(b) SO42– 32 e–
O
S
O
tetrahedral
Cl
pyramidal
O
(c) SOCl2 26 e–
Cl
S
O
(d) Cl2O
20 e–
Cl
O
Cl
bent
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45.
(a) C2H6 14 e–
H
(b) C2H4 12 e–
H
(c) C2H2 10 e–
H
H
H
C
C
H
H
C
C
H
H
C
C
H
H
H
46.
(a)
Hg
(b)
Be
(c)
N
47.
(a)
Zn
(b)
Be
(c)
N
48.
(1)
(2)
Fluorine is missing one electron from its 2p level while neon has a full energy level.
Fluorine’s valence electrons are closer to the nucleus. The attraction between the
electrons and the positive nucleus is greater.
49.
Lithium has a +1 charge after the first electron is removed. It takes more energy to
overcome that charge and to remove another electron than to remove a single electron
from an uncharged He atom.
50.
Yes. Each of these elements have an ns1 electron and they could lose that electron in the
same way elements in Group 1A do. They would then form +1 ions and ionic
compounds such as CuCl, AgCl, and AuCl.
51.
SnBr2 , GeBr2.
52.
The bond between sodium and chlorine is ionic. An electron has been transferred from a
sodium atom to a chlorine atom. The substance is composed of ions not molecules. Use
of the word molecule implies covalent bonding.
53.
A covalent bond results from the sharing of a pair of electrons between two atoms, while
an ionic bond involves the transfer of one or more electrons from one atom to another.
54.
This structure shown is incorrect since the bond is ionic. It should be represented as:
+
Na
2–
O
(d)
+
Na
55.
The four most electronegative elements are F, O, N, Cl.
56.
highest
F, O, S, H, Mg, Cs
lowest.
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Fr
(e)
Au
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57.
It is possible for a molecule to be nonpolar even though it contains polar bonds. If the
molecule is symmetrical, the polarities of the bonds will cancel (in a manner similar to
a positive and negative number of the same size) resulting in a nonpolar molecule.
An example is CO2 which is linear and nonpolar.
58.
Both molecules contain polar bonds. CO2 is symmetrical about the C atom, so the
polarities cancel. In CO, there is only one polar bond, therefore the molecule is polar.
59.
(a)
105°
60.
(a)
Both use the p orbitals for bonding. B uses one s and two p orbitals while N uses
one s and three p orbitals for bonding.
BF3 is trigonal planar while NF3 is pyramidal.
BF3 no lone pairs
NF3 one lone pair
BF3 has 3 very polar covalent bonds. NF3 has 3 polar covalent bonds
(b)
(c)
(d)
(b)
107°
(c)
109.5°
(d)
109.5°
61.
Fluorine’s electronegativity is greater than any other element. Ionic bonds form between
atoms of widely different electronegativities. Therefore, Fr–F, Cs–F, Rb–F or K–F would
be ionic substances with the greatest electronegativity difference.
62.
Each element in a particular column has the same number of valence electrons and
therefore the same Lewis structure.
63.
S
1.40 g
= 0.0437 mol
g
32.07
mol
2.10 g
= 0.131 mol
O
g
16.00
mol
Empirical formula is SO3
0.0437
= 1.00
0.0437
0.131
= 3.00
0.0437
O
O
64.
S
O
We need to know the molecular formula before we can draw the Lewis structure. From
the data, determine the empirical and then the molecular formula.
14.5 g
1.21
= 1.00
= 1.21 mol
C
g
1.21
12.01
mol
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85.5 g
2.41
= 2.01
= 2.41 mol
g
1.21
35.45
mol
CCl 2 is the empirical formula
empirical mass = 1(12.01) + 2(35.45) = 82.91
166
= 2.00
82.91
Therefore, the molecular formula is (CCl 2)2 or C2Cl 4
Cl
Cl
Cl
C
C
Cl
65.
(a)
Cl
incorrect, correct structure is:
O
C
(d)
incorrect, correct structure is:
–
O
O
N
O
O
(b)
correct
(e)
incorrect, correct structure is:
H
(c)
correct
(f)
C
N
incorrect, correct structure is:
2–
O
O
S
O
O
66.
67.
(25 g Li) a
1 mol
520 kJ
ba
b = 1.9 * 103 kJ
6.941 g
mol
Removing the first electron from 1 mole of sodium atoms requires 496 kJ. To remove a
second electron from 1 mole of sodium atoms requires 4,565 kJ. The conversion is:
(15 mol Na) a
496 KJ
b = 7.4 * 103 kJ
mol
(15 mol Na) a
4565 KJ
b = 6.8 * 104 kJ
mol
(7.4 * 103 kJ) + (6.8 * 104 kJ) = 7.5 * 104 kJ
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