Chapter 8 Problems

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Chapter 8 Problems

• 12, 16, 40, 46, 48, 50, 66, 70, 92, 102

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Electronegativity
Theres one more property of atoms thats
affected by the effective nuclear charge the
electronegativity.
Suppose we have two fluorine atoms bonded
together.
7
7
Whats the effective nuclear charge for each?
3
Electronegativity
electrons
This means that the electrons in the bond are
attracted to both of the atoms equally. Theyre
mostly located halfway down the bond.
4
Electronegativity
Now, suppose we have a carbon atom and a fluorine
atom.
C
F
4
7
electrons
Whats the effective nuclear charge for each?
This means that the electrons in the bond are
more attracted to the fluorine. Theyre located
closer to F.
We say that fluorine is more electronegative than
carbon.
5
Electronegativity
Finally, suppose we have a lithium atom and a
fluorine atom.
Li
F
1
7
electrons
Whats the effective nuclear charge for each?
The electrons in the bond are much more attracted
to the fluorine. Theyre located almost
completely on the F atom.
6
Bond Polarity
C
F
Li
F
F
F
electrons
electrons
electrons
When the electrons are in the exact center of a
bond, we say the bond is nonpolar.
When the electrons are closer to one end of a
bond, the bond is polar.
If the electrons are located completely on one
atom, the bond is an ionic bond. Otherwise, its
a covalent bond.
7
Electronegativity
How do we know whether a bond is nonpolar
covalent, polar covalent, or ionic?
It depends on the electronegativities of the two
elements.
Electronegativity is the ability of an atom to
attract the electrons in a bond.
8
Electronegativity
Electronegativity is usually higher as we go up
and to the right. Noble gases dont count,
because they usually dont form bonds.
Electronegativity
9
Electronegativity
If the difference between the electronegativities
of the two atoms in a bond is 0.5 or less, then
we have a nonpolar covalent bond.
If the difference is more than 0.5 but less than
1.75, it is a polar covalent bond.
If the difference is 1.75 or more, it is an ionic
bond.
10
Electronegativity
Place the following bonds in order, from most to
least polar, and tell what type of bond each
is CH PCl AgCl CaCl CS NBr AlCl
0.35 0.97 1.23 2.16 0.03 0.08 1.55
Electronegativities C 2.55 H 2.20 P
2.19 Cl 3.16 Ag 1.93 Ca 1.00 S 2.58 N
3.04 Br 2.96 Al 1.61
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Electronegativity
So the correct order is CS NBr CH PCl AgCl A
lCl CaCl
0.03 0.08 0.35 0.97 1.23 1.55 2.16
CS, NBr, and CH are nonpolar covalent. PCl
is polar covalent. So are AgCl and AlCl, which
is surprising! Only CaCl is ionic.
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Bonds
Molecules containing all ionic bonds usually
dissolve in water. Molecules containing all
covalent bonds often dont.
Molecules containing all ionic bonds usually have
higher melting and boiling points than molecules
containing all covalent bonds.
13
Lewis Dot Structures
We can get a more helpful picture of ionic and
covalent bonds by using Lewis dot structures.
In a Lewis dot structure, an atom is represented
by its element symbol, surrounded by dots
representing the valence electrons.
Examples
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Lewis Dot Structures
The rules for drawing Lewis dot structures are
simple
1. Only draw the valence electrons.
2. It doesnt matter which side of the atom gets
the electrons.
3. Dont pair the electrons until there is one
on each of the four sides.
4. We dont usually draw Lewis dot structures
for transition metals.
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Types of Bond
Lewis dot structures can help us understand the
difference between ionic and covalent bonds.
Suppose we bring a sodium atom and a chlorine
atom together to make NaCl

?
bond (bonds always contain two electrons)
Remember, because the difference in
electronegativities is so large, the electrons in
the bond are almost entirely on the chlorine.
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Ionic Bonds
Na
Cl-
In this bond, because the electronegativity is so
much higher for Cl, sodium basically loses its
valence electron, and chlorine gains it.
In other words, we end up with an Na ion and a
Cl- ion!
In an ionic bond, one or more electrons are
always transferred from one atom to another, so
we always get ions.
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The Octet Rule
Notice that, because it has gained an electron,
the chlorine now has 8 electrons in the valence
shell, and the sodium has none.
This is one example of a very important idea in
chemistry the Octet Rule
Atoms are most stable when they have valence
shells that are either completely full or
completely empty. Thus, atoms in Lewis dot
structures are usually surrounded by 8 electrons
or none.
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The Octet Rule
The Octet Rule helps to explain why some ions
have their charges
?
?
Al3
?
K
?
However, the Octet Rule has tons of exceptions.
The two most important are
1. Transition metals dont usually obey the
Octet Rule.
2. Hydrogen doesnt obey the Octet Rule.
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The Octet Rule
The Octet Rule is very helpful when were trying
to understand how molecules look.
Examples magnesium chloride
MgCl2
?
potassium nitride
K3N
?
20
Covalent Bonds
The situation is a little different for covalent
bonds. The electronegativity difference isnt as
large, so an electron isnt moved from one atom
to the other. Instead, each atom shares some
electrons.
the electrons in this bond are shared by both
atoms
Examples phosphorus trichloride
PCl3
?
ammonia
NH3
?
21
Lewis Dot Structures
Try writing Lewis dot structures for the
following compounds
carbon tetrabromide
CBr4
?
water
H2O
?
sodium hydroxide
NaOH
?
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Lewis Dot Structures
Try writing Lewis dot structures for the
following compound
nitric acid
HNO3
This ones more difficult. Is the hydrogen
attached to the nitrogen or to an oxygen? Are the
oxygens bonded to each other or to the nitrogen?
For larger compounds like this, theres a simple
method we can use to find out the Lewis dot
structure.
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Lewis Dot Structures
1. First, find the total number of valence
electrons in the compound. Remember to adjust
the number of electrons if the compound has a
charge!
Examples
HNO3
C2H2
NH4
CO32-
1
3 6 18
2 1 2
4 1 4
5
2 4 8
5
4
-1 electron
2 electrons
3 6 18
24 electrons
10 electrons
8 electrons
24 electrons
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Lewis Dot Structures
2. Sketch out a skeleton structure of the
molecule.
The central atom is usually obvious from the
formula. If there are three different elements,
oxygens are attached to the central atom and
hydrogens are attached to the oxygens.
HNO3
C2H2
NH4
CO32-
H-C-C-H
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Lewis Dot Structures
3. Add electrons to the atoms attached to the
central atom so that they obey the Octet Rule.
Remember, each bond you already drew contains two
electrons!
Note hydrogens dont obey the Octet Rule. They
only get a total of two electrons.
H-C-C-H
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Lewis Dot Structures
4. If there are any valence electrons left, use
them to try to make the central atom obey the
Octet Rule. You may run out of electrons before
this is possible!
H-C-C-H
24 electrons
10 electrons
8 electrons
24 electrons
-24
-6
-8
-24
4 remaining
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Lewis Dot Structures
5. Were out of electrons now! So now what can
we do for the central atoms that dont yet obey
the Octet Rule?
The central atom will steal an electron pair
from the outer atom that has the most electron
pairs. The stolen electrons are used to make
an extra bond.
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Lewis Dot Structures
H-CC-H
A double bond has formed!
If the central atom still doesnt obey the Octet
Rule, it can steal another pair of electrons.
This forms a triple bond.
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Lewis Dot Structures
6. Finally, if the structure represents an ion,
we place square brackets around it, and write the
charge as a superscript.
HNO3
C2H2
NH4
CO32-



2-
H-CC-H
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Resonance
Notice that when we drew the Lewis structures of
nitric acid and carbonate ion, we could have put
the double bonds in more than one place
HNO3
CO32-
In a real molecule, the double bond constantly
switches its position between each of the
possible locations. This phenomenon is called
resonance, and each of the structures are called
resonance structures.
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Lewis Structures
Sometimes determining a Lewis structure can be
very challenging! For example, try dinitrogen
monoxide.
N2O
16 valence electrons
The trouble is that we dont know whether the
central atom is nitrogen or oxygen! Lets start
drawing a Lewis structure for both.
N-N-O
N-O-N
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Lewis Structures
N-N-O
N-O-N
We need to make a double bond. But which atom
should we steal electrons from?
Lets try both possibilities (for each molecule).
The central atom still doesnt obey the Octet
Rule! Again, which atom should we steal
electrons from?
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Lewis Structures
How can we choose the correct structure?
These two are the same molecule!
So are these two.
And these two (look carefully).
To choose from among the five that remain, we
need to look at the formal charges.
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Formal Charges
The formal charge is the charge on an atom based
on the Lewis structure of the molecule.
Formal charge of an atom valence electrons
the atom has - of non-bonding electrons on
the atom - ½ the of electrons in the atoms
bonds
Example
Formal charges
H
1 0 ½ 2
0
O
6 4 ½ 4
0
non-bonding electrons
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Formal Charges
To choose the correct structure, we need to know
the formal charges of each atom.
N
5 - 2 - ½ 6
0
N
5 - 4 - ½ 4
-1
N
5 - 6 - ½ 2
-2
N
5 - 0 - ½ 8
1
N
5 - 0 - ½ 8
1
N
5 - 0 - ½ 8
1
O
6 - 6 - ½ 2
-1
O
6 - 4 - ½ 4
0
O
6 - 2 - ½ 6
1
N
5 - 4 - ½ 4
-1
N
5 - 6 - ½ 2
-2
O
6 - 0 - ½ 8
2
O
6 - 0 - ½ 8
2
N
5 - 4 - ½ 4
-1
N
5 - 2 - ½ 6
0
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Formal Charges
N -2
N 0
N -1
N -2
N -1
O 2
N 1
N 1
N 1
O 2
N 0
O -1
O 0
O 1
N -1
Choose Lewis structures that have formal charges
closest to zero.
If theres still a choice, choose the Lewis
structure where the most electronegative atom has
the lowest charge.
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Bond Strength
As youd probably expect, triple bonds are
stronger than double bonds, and double bonds are
stronger than single bonds.
Also, ionic bonds are stronger than covalent
bonds.
Example
melting point 804C
melting point 148C
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Bond Strength
Exactly how strong are these bonds? Theres a
table in your book on page 330.
We can use this list to estimate the energy of a
chemical reaction!
For example, suppose we wanted to know the energy
of the reaction
CH4 2 O2 ? CO2 2 H2O
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Bond Enthalpy
CH4 2 O2 ? CO2 2 H2O
The approximate energy of the reaction is DH
Hbonds broken Hbonds formed
40
Bond Enthalpies
DH Hbonds broken Hbonds formed
In this reaction, we break 4 C-H bonds
2 OO bonds
we form 2 CO bonds
4 O-H bonds
DH 4 (413 kJ) 2 (495 kJ) 2 (799 kJ) 4
(463 kJ)
-808 kJ
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Bond Enthalpies
DH -808 kJ
CH4 2 O2 ? CO2 2 H2O
How close is this to reality? Lets compare it
to the real DH (you already know how to calculate
it!)
DH Hf(products) Hf(reactants)
DH -393.5 kJ 2 (-241.82 kJ) -74.8 kJ 2 (0
kJ)
-802 kJ
This is pretty good! But why isnt the number we
got from the bond enthalpies exact?
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Bond Strength
Notice that the bond enthalpies on page 330 are
all for covalent bonds. What if we want to know
the strength of the bond in NaCl?
Ionic bond strength is calculated differently.
We take advantage of the fact that ionic
compounds form crystals!
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Ionic Bond Strength
The strength of an ionic bond is, basically, the
energy it would take to take a crystal and break
it into the elements its made of.
NaCl (s) ? Na (s) ½ Cl2 (g) DH ?
Unfortunately, this is a difficult (impossible?)
reaction to perform. How do we calculate the
enthalpy of a reaction thats complex?
44
Ionic Bond Strength
NaCl (s) ? Na (s) ½ Cl2 (g) DH ?
We break the reaction into smaller steps, and add
the enthalpies of each step Hess Law!
What smaller steps should we use? First of
all, we can try the reaction where we destroy the
crystal!
NaCl (s) ? Na (g) Cl- (g) DH 788 kJ
The energy of the reverse of this reaction (i.e.,
-788 kJ), is called the lattice energy of the
crystal. Its the energy released when a crystal
is formed.
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Ionic Bond Strength
NaCl (s) ? Na (s) ½ Cl2 (g) DH ?
NaCl (s) ? Na (g) Cl- (g) DH 788 kJ
What other reactions can we use?
In Chapter 7, we saw reactions that had Na and
Cl- as products...
ionization energy
Na (g) ? Na (g) e- DH 496 kJ
Cl (g) e- ? Cl- (g) DH -349 kJ
electron affinity
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Ionic Bond Strength
NaCl (s) ? Na (s) ½ Cl2 (g) DH ?
NaCl (s) ? Na (g) Cl- (g) DH 788 kJ
Na (g) ? Na (g) e- DH 496 kJ
Cl (g) e- ? Cl- (g) DH -349 kJ
Na (g) e- ? Na (g) DH -496 kJ
Cl- (g) ? Cl (g) e- DH 349 kJ
Were going to want Na (g) and Cl- (g) to cancel
out with the previous reaction, so we need to
flip the last two equations.
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Ionic Bond Strength
NaCl (s) ? Na (s) ½ Cl2 (g) DH ?
NaCl (s) ? Na (g) Cl- (g) DH 788 kJ
Na (g) e- ? Na (g) DH -496 kJ
Cl- (g) ? Cl (g) e- DH 349 kJ
Were still not where we want to be if we stop
here, well have Na (g) and Cl (g) as our
products.
Appendix C to the rescue!
Na (s) ? Na (g) DH 108 kJ
½ Cl2 (g) ? Cl (g) DH 122 kJ
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Ionic Bond Strength
NaCl (s) ? Na (s) ½ Cl2 (g) DH ?
NaCl (s) ? Na (g) Cl- (g) DH 788 kJ
Na (g) e- ? Na (g) DH -496 kJ
Cl- (g) ? Cl (g) e- DH 349 kJ
Na (g) ? Na (s) DH -108 kJ
Cl (g) ? ½ Cl2 (g) DH -122 kJ
Na (s) ? Na (g) DH 108 kJ
½ Cl2 (g) ? Cl (g) DH 122 kJ
So DH 411 kJ
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Lattice Energy
Lets try another example. Whats the lattice
energy of TiO2 (s)?
Ti4 (g) 2 O2- (g) ? TiO2 (s) DH ?
Just like before, we must break this down into
smaller steps.
Start with ionization energies and electron
affinities.
Ti (g) ? Ti (g) e- DH 659 kJ
Ti (g) ? Ti2 (g) e- DH 1310 kJ
Ti2 (g) ? Ti3 (g) e- DH 2653 kJ
Ti3 (g) ? Ti4 (g) e- DH 4175 kJ
O (g) e- ? O- (g) DH -141 kJ
O- (g) e- ? O2- (g) DH 744 kJ
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Lattice Energy
Ti4 (g) 2 O2- (g) ? TiO2 (s) DH ?
Were going to want Ti4 (g) and O2- (g) to
reactant side, so we need to flip all these
equations.
Ti (g) e- ? Ti (g) DH -659 kJ
Ti2 (g) e- ? Ti (g) DH -1310 kJ
Ti3 (g) e- ? Ti2 (g) DH -2653 kJ
Ti4 (g) e- ? Ti3 (g) DH -4175 kJ
O- (g) ? O (g) e- DH 141 kJ
O2- (g) ? O- (g) e- DH -744 kJ
2 O- (g) ? 2 O (g) 2 e- DH 282 kJ
2 O2- (g) ? 2 O- (g) 2 e- DH -1488 kJ
Were also going to need 2 moles of O2- (g), so
we need to multiply the last two equations by two.
51
Lattice Energy
Ti4 (g) 2 O2- (g) ? TiO2 (s) DH ?
Ti (g) e- ? Ti (g) DH -659 kJ
Ti2 (g) e- ? Ti (g) DH -1310 kJ
Ti3 (g) e- ? Ti2 (g) DH -2653 kJ
Ti4 (g) e- ? Ti3 (g) DH -4175 kJ
2 O- (g) ? 2 O (g) 2 e- DH 282 kJ
2 O2- (g) ? 2 O- (g) 2 e- DH -1488 kJ
If we stop here, well have Ti (g) and 2 O (g) as
our products. Back to Appendix C!
Ti (s) ? Ti (g) DH 468 kJ
½ O2 (g) ? O (g) DH 248 kJ
Ti (s) O2 (g) ? TiO2 (s) DH -945 kJ
52
Lattice Energy
Ti4 (g) 2 O2- (g) ? TiO2 (s) DH ?
Ti (g) e- ? Ti (g) DH -659 kJ
Ti2 (g) e- ? Ti (g) DH -1310 kJ
Ti3 (g) e- ? Ti2 (g) DH -2653 kJ
Ti4 (g) e- ? Ti3 (g) DH -4175 kJ
2 O- (g) ? 2 O (g) 2 e- DH 282 kJ
2 O2- (g) ? 2 O- (g) 2 e- DH -1488 kJ
Ti (g) ? Ti (s) DH -468 kJ
O (g) ? ½ O2 (g) DH -248 kJ
Ti (s) O2 (g) ? TiO2 (s) DH -945 kJ
We need to flip two of the equations to get Ti
(g) and O (g) on the reactant side.
Finally, we need to multiply the second-to-last
equation by 2.
53
Lattice Energy
Ti4 (g) 2 O2- (g) ? TiO2 (s) DH ?
Ti (g) e- ? Ti (g) DH -659 kJ
Ti2 (g) e- ? Ti (g) DH -1310 kJ
Ti3 (g) e- ? Ti2 (g) DH -2653 kJ
Ti4 (g) e- ? Ti3 (g) DH -4175 kJ
2 O- (g) ? 2 O (g) 2 e- DH 282 kJ
2 O2- (g) ? 2 O- (g) 2 e- DH -1488 kJ
Ti (g) ? Ti (s) DH -468 kJ
2 O (g) ? O2 (g) DH -496 kJ
Ti (s) O2 (g) ? TiO2 (s) DH -945 kJ
So DH -11912 kJ
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Lattice Energy Activity