1. No category

Chem 2 AP Homework 9&10 Review: pg. 380-383 #74, 80, 81, 85, 86, 90, 93, 96, 101, 104, 114, 117, 120; pg. 431-433
#71, 74, 76, 85, 91, 92, 93, 99, 101, 102
Chapter 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.
80
Write three reasonable resonance structures for the azide ion, N3– in which the atoms are arranged as
NNN. Show formal charges.
−
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
85
+
N
C
H
Draw reasonable resonance structures for the following ions (a) HSO4–, (b) PO43–, (c) HSO3–, (d) SO32–.
O
H
O
S
O
+
O
−
H
+
O
S
O−
O
O
(b)
−
O
P
O
O
−
−
O
O−
−
O
O
H
O
O
S+ O
−
H
O
O
−
−
P
−
O
O
−
−
O
O
−
2+
S
O
−
O−
−
+
P
O
−
O−
There are two more equivalent resonance structures to the first structure above.
O
O
H
O
S
O
−
H
O
−
O
S
O
O
O
−
−
O
S
−
O
O
H
O
−
+
S
O
−
−
S+ O
−
(d)
There are two more equivalent resonance structures to the first structure.
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
(c)
(a)
2
90
N 2O 4
Chem 2 Ap Ch. 9-10 Review
Which of the following molecules has the shortest nitrogen-to-nitrogen bond? Explain. N2H4, N2O, N2,
Draw the Lewis structures:
H
N
N
H
H
N
H
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.
93
The following species have been detected in interstellar space: (a) CH, (b) OH, (c) C2, (d) HNC, (e)
HCO. Draw Lewis structures for these species and indicate whether they are diamagnetic or
paramagnetic.
H paramagnetic
(a) C
(d) H
+
-
C diamagnetic
N
H paramagnetic
O
(b)
(e) H
(c)
C
C
diamagnetic
O paramagnetic
C
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–.
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
C
C
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
Note that for C4H10 and C5H12 there are a number of structural isomers.
104
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 !f [HI( g )] − {ΔH !f [H 2 ( g )] + ΔH !f [I 2 ( g )]}
ΔH° = (2)(25.9 kJ/mol) − [(0) + (1)(61.0 kJ/mol)] = −9.2 kJ/mol
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
3
Chem 2 Ap Ch. 9-10 Review
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.
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
120
O
N
O
O
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: (694 kJ/mol )/ 2 = 347 kJ / mol
Chapter 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
P
Cl
Cl
F
Cl
F
Xe
F
F
4
Chem 2 Ap Ch. 9-10 Review
Cl
Cl
Cl
F
Cl
P
F
Cl
AB5
trigonal bipyramid
µ=0
AB6
octahedral
µ=0
F
S
F
74
F
F
Antimony pentafluoride, SbF5, reacts with XeF4 and XeF6 to form ionic compounds, XeF3+SbF6– and
XeF5+SbF6–. Describe the geometries of the cations and anion in these two compounds
F
+
F
Xe
F
Xe
+
AB3E2
T-shaped
76
F
F
F
F
−
F
Sb
F F
F
F
AB5E
Square Pyramidal
F
F
AB6
Octahedral
Predict the bond angles for the following molecules:
Cl
Cl
Be
Cl
(b)
Cl
Cl
(f)
H
(c)
Cl
C
H
H
Cl
Hg
H
(d)
Hg
Cl
(e)
H
O
(g)
Cl
C
Cl
Cl
H
O
Sn
Cl
B
Cl
(a)
Cl
H
H
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)
85
SnCl2:
AB2E type, roughly 120° (bent).
(g) H2O2:
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).
Does the following molecule have a dipole moment?
H
Cl
C
C
H
Consider the overlap of the 2p orbitals on each carbon atom.
C
Cl
5
Chem 2 Ap Ch. 9-10 Review
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:
91
Write the ground-state electron configuration for B2 Is the molecule diamagnetic or paramagnetic?
The orbital diagram (from page 422), just considering valence electrons, is:
B2
*
σ 2 px
π 2* p y , π 2* pz
σ 2 px
It
π 2 p y , π 2 pz
↑
σ 2* s
σ2s
92
↑↓
↑↓
What are the hydribidization states of the C and N atoms in this molecule?
NH2
The carbons are in sp2 hybridization states. The nitrogens are in the sp3 hybridization state,
except for the ring nitrogen double-bonded to a carbon that is sp2 hybridized. The oxygen
H
C
atom is sp2 hybridized.
N
C
C
O
93
is paramagnetic.
↑
C
N
H
H
Use molecular orbital theory to explain the difference between the bond energies of F2 and F2–.
The orbital diagrams (valence electrons only) are shown below:
F2
F 2–
*
*
↑
σ 2 px
σ 2 px
π 2* p y , π 2* pz
↑↓
↑↓
π 2* p y , π 2* pz
↑↓
↑↓
π 2 p y , π 2 pz
↑↓
↑↓
π 2 p y , π 2 pz
↑↓
↑↓
σ 2 px
↑↓
σ 2 px
↑↓
σ 2* s
σ2s
↑↓
↑↓
σ 2* s
σ2s
↑↓
↑↓
Since F2− has an extra electron in σ 2 p x , it only has a bond order of ½, compared with a bond order of one for F2.
*
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.
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.
6
Chem 2 Ap Ch. 9-10 Review
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
(b) Describe the hybridization state of Al in AlCl3 and Al2Cl6.
The hybridization of Al in AlCl3 is sp2. The molecule is trigonal planar. The hybridization of Al in Al2Cl6 is
sp3.
(c) Sketch the geometry of the dimer.
Cl
Cl
Al
Cl
Cl
Al
Cl
Cl
(d) Do these molecules possess a dipole moment?
All of the Al—Cl dipoles cancel; the molecules do not possess a dipole moment.
102
Assume that the third-period element phosphorous forms a diatomic molecule, P2, in an analogous way
as nitrogen does to form N2.
(a) Write the electronic configuration for P2. Use [Ne2] to represent the electron configuration for the
first two periods.
Following molecular orbital diagram for N2 shown in Table 10.5 of the text, and replacing the 2nd energy
level with the 3rd, we get.
P2
*
σ 3 px
π3* p y , π3* pz
σ 3 px
π 3 p y , π 3 pz
↑↓
↑↓
↑↓
↑↓
σ
↑↓
σ 3s
So the electron configuration for P2 becomes
[Ne 2 ](σ 3s )2 (σ 3s* )2 ( π3 p y )2 ( π3 pz )2 (σ 3 p x )2
*
3s
(b) Calculate its bond order
Past the Ne2 core configuration, there are 8 bonding electrons and 2 antibonding electrons. The bond order
is:
bond order = ½(8 − 2) = 3
(c) What are its magnetic properties (diamagnetic or paramagnetic)?
All the electrons in the electronic configuration are paired. P2 is diamagnetic.