Chem 2 AP Homework 9&10 Review:
Chapter 9: pg. 380-383 #74, 80, 81, 82, 86, 90, 96, 101, 104, 114, 117, 120;
Chapter 10: pg. 431-433 #71, 76, 85, 91, 92, 99, 101
Chapter 9
9.74
Which of the following are ionic compounds? Which are covalent compounds?
RbCl and KO2 are ionic compounds. Typically, ionic compounds are composed of a metal cation and a nonmetal anion.
PF5, BrF3, and CI4 are covalent compounds. Typically, covalent compounds are composed of two nonmetals.
9.80
Write three reasonable resonance structures for the azide ion, N3– in which the atoms are arranged as NNN. Show
formal charges.
−
9.81
N
+
N
N
−
N
+
N
N
2−
2−
N
N
The amide group plays an important role in determining the structure of proteins. Draw another resonance
structure for this group. Show formal charges.
O
N
O
C
N
H
9.82
+
N
C
H
Give an example of an ion or molecule containing Al that (a) obeys the octet rule, (b) has an expanded octet, and
(c) has an incomplete octet.
–
(a) An example of an aluminum species that satisfies the octet rule is the anion AlCl4 . The Lewis dot structure is
drawn in Problem 9.66.
3
(b) An example of an aluminum species containing an expanded octet is anion AlF6 . (How many pairs of electrons
surround the central atom?)
(c) An aluminum species that has an incomplete octet is the compound AlCl3. The dot structure is given in Problem
9.66.
3–
−
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
Cl
Cl
Al
Cl
Cl
Cl
Cl
Cl
9.86
Are the following statements true or false?
(a) Formal charges represent actual separation of charges? false
(b) ΔH°rxn can be estimated from the bond energies of reactants and products? true
(c) All second-period elements obey the octet rule in their compounds? false
(d) The resonance structures of a molecule can be separated from one another? False
9.90
Which of the following molecules has the shortest nitrogen-to-nitrogen bond? Explain. N2H4, N2O, N2, N2O4
Draw the Lewis structures:
H
N
N
H
H
H
N
N
N
N
O
N
N
O
N
N
O
O
O
O
Only N2 has a full triple bond (the resonance in N2O reduces the bond order slightly), so it has the shortest bond length.
2
Chem 2 Ap Ch. 9-10 Review
9.96
The triiodide ion (I3–) in which the I atoms are arranged in a straight line is stable, but the corresponding F3– ion
does not exist. Explain.
–
The central iodine atom in I3 has ten electrons surrounding it: two bonding pairs and three lone pairs. The central
iodine has an expanded octet. Elements in the second period such as fluorine cannot have an expanded octet as would be
–
required for F3 .
9.101
For each of the following organic molecules draw a Lewis structure in which the carbon atoms are bonded to each
other by single bonds? (a) C2H6, (b) C4H10, (c) C5H12. For (b) and (c) draw only structures in which each C atom is
bonded to no more than two other C atoms.
H
H
H
H
H
H
H
H
H
H
H
H
C
C
H
H
H
H
C
C
C
C
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
Other isomers of (b) and (c):
H
H
H
H
H
C
C
C
H
H
C
H
H
H
H
H
H
H
C
C
C
C
H
H
C
H
H
H
H
H
9.104
H
H
H
H
C
H
H
C
C
C
H
H
C
H
H
H
H
H
Calculate ΔH° for the reaction H2(g) + I2(g) → 2HI(g):
(a) Using Equation (9.3) of the text,
ΔH° = ∑BE(reactants) − ∑BE(products)
ΔH° = [(436.4 + 151.0) − 2(298.3)] = −9.2 kJ/mol
(b)
Using Equation (6.17) of the text,
ΔH ° = 2ΔH fo[HI( g )] − {ΔH fo[H2 ( g )] + ΔH fo[I2 ( g )]}
ΔH° = (2)(25.9 kJ/mol) − [(0) + (1)(61.0 kJ/mol)] = −9.2 kJ/mol
9.114
The N—O bond distance in nitric oxide is 115 pm, which is intermediate between a triple bond (106 pm) and a
double bond (120 pm).
(a) Draw two resonance structures for NO and comment on their relative importance.
N
O
N
O
The first structure is the most important. Both N and O have formal charges of zero. In the second structure, the
more electronegative oxygen atom has a formal charge of +1. Having a positive formal charge on an highly
electronegative atom is not favorable. In addition, both structures leave one atom with an incomplete octet. This
cannot be avoided due to the odd number of electrons.
(b) Is it possible to draw a resonance structure having a triple bond between the atoms?
N O It is not possible to draw a structure with a triple bond between N and O because that would place 9
electrons on the O atom. Any structure drawn with a triple bond will lead to an expanded octet. Elements in the
second row of the period table cannot exceed the octet rule.
9.117
Draw a Lewis structure for nitrogen pentoxide (N2O5) in which each N is bonded to three O atoms.
One resonance form showing formal charges is:
O
N
O
O
N
O
O
3
Chem 2 Ap Ch. 9-10 Review
9.120
Experiments show that it takes 1656 kJ/mol to break all the bonds in methane (CH4) and 4006 kJ/mol to break all
the bonds in propane (C3H8). Based on these data, calculate the average bond energy of the C—C bond.
There are four C−H bonds in CH4, so the average bond energy of a C−H bond is:
1656 kJ/mol
= 414 kJ/mol
4
The Lewis structure of propane is shown below. There are eight C−H bonds and two C−C bonds.
H H H
H
C
C
C
H H
H
H
8(C−H) + 2(C−C) = 4006 kJ/mol
8(414 kJ/mol) + 2(C−C) = 4006 kJ/mol
2(C−C) = 694 kJ/mol
So, the average bond energy of a C−C bond is: (694kJ/mol) / 2 = 347 kJ / mol
Chapter 10
10.71 Sketch the bond moments and resultant dipole moments for the following molecules:
Dot structure
Label
Shape
Bond dipole
Resultant
dipole moment
O
H
H
AB2E2
bent
µ>0
AB3E
trigonal
pyramidal
µ>0
AB4E2
square planar
µ=0
AB5
trigonal bipyramid
µ=0
AB6
octahedral
µ=0
P
Cl
Cl
Cl
F
F
Xe
F
F
Cl
Cl
Cl
F
Cl
P
F
Cl
F
S
F
F
F
4
10.76
Chem 2 Ap Ch. 9-10 Review
Predict the bond angles for the following molecules:
Cl
Cl
Be
Cl
(b)
Cl
Cl
Cl
Cl
H
C
Cl
(c)
Hg
Cl
C
H
H
H
(d)
Hg
Cl
(e)
H
O
H
H
(g)
(f)
Cl
H
O
Sn
Cl
B
Cl
(a)
Cl
Sn
(h)
H
(a)
BeCl2:
AB2 type, 180° (linear).
(b)
BCl3:
AB3 type, 120° (trigonal planar).
(c)
CCl4:
AB4 type, 109.5° (tetrahedral).
(d)
CH3Cl: AB4 type, 109.5° (tetrahedral with a possible slight distortion resulting from the different sizes of the
chlorine and hydrogen atoms).
(e)
Hg2Cl2: Each mercury atom is of the AB2 type. The entire molecule is linear, 180° bond angles.
(f)
SnCl2:
AB2E type, roughly 120° (bent).
(g)
H 2O 2:
The atom arrangement is HOOH. Each oxygen atom is of the AB2E2 type and the H−O−O angles will be
roughly 109.5°.
(h)
SnH4:
AB4 type, 109.5° (tetrahedral).
10.82
F
N
F
F
N
N
N
F
(a) The hybridization around each N is sp2 (electron pair arrangement is trigonal planar.) . It has the electron
2
arrangement of three sp hybrid orbitals.
(b)
The two N−F bond moments in the left-hand structure cancel. The sum or resultant dipole moment will be zero.
The structure on the right has a dipole moment because the two N−F bond moments do not cancel each other out
and so the molecule has a net dipole moment.
10.85
Does the following molecule have a dipole moment?
Cl
H
C
C
C
H
Consider the overlap of the 2p orbitals on each carbon atom.
Cl
H
H
Cl
C
C
C
Cl
The geometric planes containing the CHCl groups at each end of the molecule are mutually perpendicular. This is
because the two carbon-carbon double bonds must use different 2p orbitals on the middle carbon, and these two 2p
orbitals are perpendicular. This means that the two chlorine atoms can be considered to be on one side of the molecule
and the two hydrogen atoms on the other. The molecule has a dipole moment. Look at the 3-D drawing:
5
Chem 2 Ap Ch. 9-10 Review
10.92
What are the hydribidization states of the C and N atoms in this molecule?
2
3
NH2
The carbons are in sp hybridization states. The nitrogens are in the sp hybridization state, except
2
2
H for the ring nitrogen double-bonded to a carbon that is sp hybridized. The oxygen atom is sp
C
hybridized.
N
C
C
O
C
N
H
H
10.99
Only one of the following two molecules containing C and H atoms exists. Which one is it?
In the structure on the left, the bonds are too strained (bond angles too small). This molecule
does not exist. The structure on the right does exist, although it is highly reactive. This
molecule is called benzyne.
10.101 Aluminum trichloride (AlCl3) is an electron-deficient molecule. It has a tendency to form a dimer (a molecule
made of two AlCl3 units):
AlCl3 + AlCl3 → Al2Cl6
(a) Draw a Lewis structure for the dimer.
Cl
Cl
Al
Cl
(b)
(c)
Cl
Al
Cl
Cl
Describe the hybridization state of Al in AlCl3 and Al2Cl6.
2
3
The hybridization of Al in AlCl3 is sp . The molecule is trigonal planar. The hybridization of Al in Al2Cl6 is sp .
Sketch the geometry of the dimer.
Cl
Cl
Al
Cl
(d)
Cl
Al
Cl
Cl
Do these molecules possess a dipole moment?
All of the Al—Cl dipoles cancel; the molecules do not possess a dipole moment.