Ch. 9/10 Packet of Salmon
9.1: Valence Electrons: As we learned in Ch. 8, these are the outermost electrons, those that
occur after the noble gas in the noble gas notation. Usually, we are concerned only with the ein the highest principle E level (n). For example, we expect the chemistry of the element
bromine (configuration [Ar] 3d104s24p5 ) to be due to the seven e- in n = 4. When we number
the s- and p-blocks as groups 1A-8A, the group number is equal to the number of valence e-.
The 18 electrons represented by [Ar] in the example above are referred to as
electrons.
Lewis Dot Symbols (for atoms): This is simply a way to visualize the valence electrons of any
atom. We write the symbol of the atom in question, then draw the appropriate number of dots
(= # of valence e-). Top, bottom, left right; these are not important, but it is standard to place a
single dot on each side before doubling up. Sketch dot symbols for Li, N, Cl, O, B, and He in
the space below:
Chemical Bonds: are formed when valence electrons are redistributed between two or more
atoms. This forms an attractive force that holds the atoms together.
An ionic bond occurs when valence electron(s) is/are
to another. Created are positive ions (called
from one atom
) and negative ions (called
) which are held together by electrostatic attraction. A covalent bond is
formed by the
of valence electrons between atoms. The atoms are held
together by their mutual attraction for the shared electrons.
Energy and Chemical Bonds: Energy is always released when bonds are formed, energy is
always required when bonds are broken. Making an ionic compound from two oppositely
charged ions is an exothermic (H is negative, energy is released) process. We can predict the
degree of the energy change for an ion pairing compared to a different pairing using this
equation, which is similar to Coulomb’s Law:
E ion pair = C
(n +e)(n -e)
d
where n+e is the
positive charge on the cation, and n–e is the charge on the anion involved. d is the distance
between the ions—to determine this, we need to keep in mind the trends for ion size we

learned in Ch. 8.
• The
the charges, the greater the Eion pair
• The
the ion sizes, the greater the Eion pair
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Ch. 9/10 Packet of Salmon
• In general, the numerator (the charges) has a more profound effect on the size of Eion pair . The
distance is important when ranking compounds whose ions are of equal charge.
• Realize the equation above just gives the magnitude of Eion pair . The sign is always negative.
Practice: Place the following substances in order Eion pair , from most negative to least negative:
LiF, NaF, CaS, BeS, Fe2S3
Lattice Energy: is the energy change that occurs when (gaseous) ions combine to form one
mole of an ionic compound. Units: kJ/mol. This is an exothermic process, and Elattice thus
carries a
sign. We cannot calculate this directly, but since we have values for
ionization energies, electron affinities, and enthalpies of formations; we can calculate the
lattice energy as shown in A Closer Look, p. 380 in our text. Try problem 35 at the end of the
chapter for additional practice. With this in mind, why don’t ionic compounds such as KAr or
LiF2 exist?
GAP FOR 9.4 here…….
9.5 Bond Properties
Bond Order: This is the number of bonding e- shared by two atoms in a molecule.
For :C  O: , the carbon-oxygen bond order is 3. In :Ö=C=Ö: , the carbon-oxygen bond order

is 2. In


H  O H

, the hydrogen-oxygen bond order is one.
Of particular interest are compounds that have resonance. These Lewis structures show the
“asymmetrical” single and double bonds, therefore not all X-Y bonds are the same. These
compounds will have fractional bond orders.
The equation we use to determine bond order is
.
Bond order is something that requires a Lewis structure to determine. First sketch the
molecule, then determine bond order.
Ex 1: What is the S-O bond order in sulfur trioxide?
Ex. 2: What is the O-O bond order in ozone, O3?
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Bond Legnth: Is defined as the distance between nuclei of bonded atoms. As such, one important
factor in bond length is size of the bonded atoms. A second factor is the bond order between those
atoms. In general, the bond length
as bond order
See Table 9.8, p. 401 for approximate bond lengths.
Bond Energy: The Bond Dissociation Energy (D) is the enthalpy change (∆H) that occurs when a
molecule’s bond is broken (molecule should be in the gas phase).
The sign of D is
when bonds are broken, since energy must be
to break bonds.
Bond energies are found in Table 9.9 on p. 403.
In Chapter 6, we learned one way of calculating ∆H, that ∆H = ∑(∆H°f products) – ∑(∆H°f reactants)
Here we learn a second way to calculate ∆H: ∆H = ∑D(bonds broken) – ∑D(bonds formed)
Take, for example, the reaction C2H4 + Cl2  CH4Cl2 . If we wish to calculate ∆H for this
reaction, we can use the equation above, but first we must consider the Lewis sketch of the
molecules involved:
H H
H H
|
|
|
|
C  C + Cl – Cl 
 Cl – C – C – Cl
|
|
|
|
H H
H H
Looking at the ethene (C2H2) molecule compared to the molecule of the 1,2 dichloroethane
molecule that is our product, we can see that the double bond must be broken, as must the

Cl-Cl single bond. On the products side, we are forming a C-C single bond, and 2 Cl-C single
bonds. We look to table 9.9 for the associated values of D.
Bonds broken
Bonds formed
C=C (x1)
602 kJ
C-C (x1)
346 kJ
Cl-Cl (x1)
242 kJ
C-Cl (x2)
339 (x2) = 678 kJ
∑ D (bonds broken) = 844 kJ
∑ D (bonds formed) = 1024 kJ
Then, ∆Hrxn = 844 kJ – 1024 kJ = –180 kJ
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•Just like in Chapter 6, if there are coefficients of the balanced chemical equation, this means
we need to multiply by that coefficient.
Example 2:
2 H2 + O2  2 H2O
What is ∆Hrxn?
H–H + H–H + O=O  H–O–H + H–O–H
Bonds broken:
Bonds formed:
2 x H–H = 2 x 436 kJ
= 872 kJ
1 x O=O = 1 x 498 kJ
= 498 kJ
4 x O–H = 4 x 463 kJ
= 1852 kJ
∑ = 1370 kJ
∆Hrxn = ∑ D(bonds broken) – ∑ D(bonds formed) = 1370 kJ – 1852 kJ = –482 kJ
Had we used the Chapter 6 equation, we would have calculated:
∆Hrxn = ∑(∆H°f products) – ∑(∆H°f reactants) = [2(-241.8)] – [2(0) + 0] = –483.6 kJ
Pretty good agreement!
Bond Polarity/Electronegativity:
We say that electrons are shared in a covalent bond, but the only time they are shared equally is
when two identical atoms are bonded as in F2 or H2. When different atoms are bonded
covalently, we call the bond a polar covalent bond. This means that the electrons are shared
unequally. Electronegativity allows us to understand this. The electronegativity () of an
element is a measure of an atom’s ability to attract electrons towards itself when forming a
bond. This unequal sharing has the effect of creating a partially positive (∂+) end and a
partially negative (∂-) end of the bond. The more electronegative element in the bond is
considered ∂-, the less electronegative element is ∂+
∂+
∂–
A
B
Which atom above has higher electronegativity, A or B?
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Ch. 9/10 Packet of Salmon
• The trend for electronegativity is the same as for electron affinity. It is useful to remember the
4 elements with highest electronegativity: F > O > N > Cl (“Phonecall” is a useful mnemonic)
Vocab:
Polar covalent bond = polar bond = dipolar = bond with unequally shared e-.
Nonpolar covalent bond = nonpolar bond = bond with equally shared e-.
Realize that bond polarity is a spectrum: The greater ∆ is between the two bonded atoms, the
more polar the bond. At one end of this spectrum are nonpolar bonds, where the
electronegativity difference between the two bonded atoms is zero and the e- are shared
equally. A carbon-hydrogen bond (C = 2.5, H = 2.1) has ∆ = 0.4; this is a weakly polar
bond. A bond between fluorine( = 4.0) and hydrogen ( = 2.1) has ∆ of 1.9; this is a strongly
polar bond.
Use your periodic table to answer the following:
For each, draw an arrow showing direction of polarity, and label the atoms ∂+ and ∂- Then,
place the bonds in order of increasing polarity.
C—F
C—H
C—O
C—C
C—N
Formal Charge: The formal charge of an atom in a molecule is determined using the equation:
Formal charge = group # of atom – [number of lone pair electrons + ½ (# of bonding electrons)]
I like to think of the number in the brackets above as “all the e- the atom has to itself + ½ the eit shares.
The sum of formal charges of all atoms in a molecule or ion must equal the charge on the
molecule or ion.
Again, you must have a Lewis sketch to assign formal charges to atoms in a molecule or ion.
Sketch the following molecules and ions, and assign formal charges to each atom.
a) NO31-
b) HCN
c) CO2
d) NH41+
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The electroneutrality principle combines ideas of formal charge and electronegativity. It says
that the most likely structure for a molecule is the one where the formal charges are closest to
zero. When a negative charge exists, it should be placed on the most electronegative atom.
Here are two possible resonance structures for CO2:
••

••
••

O = C = O 
 O  C – O :
0
0
•
•
0
••
1+
0
1-
We see that the sketch on the left is the more likely.
VSEPR: Valence Shell Electron Pair Repulsion Theory—This theory says that sets of electrons
around a central atom in a molecule will move as far away from each other as possible, as the
sets repel each other due to their negative charge.
A “set” of electrons is not a pair of electrons! One set of electrons is either: one lone pair or any
bond. A single bond is one set of e-. A double bond is one set of e-. A triple bond is one set of e-.
How many sets of electrons are about the central atom of each molecule below?

H—C—H
||
:O:
3
H  O H

4
:Ö=C=Ö:
3
H
|
H—C—H
|
H
4
Cl
Cl
:I:
Cl
Cl
6
You must have memorized the shapes & bond angles that result from 2, 3, 4, 5, and 6 sets of e6
Ch. 9/10 Packet of Salmon
Sets e-
Shape name
Angle between e-
2
Linear
180
3
Trigonal planar
120
4
5
6
Tetrahedral
Sketch
109
Trigonal bipyramid
120 & 90
Octahedral
90
The shape of the electron sets is called “electron geometry.”
Electron geometry does not care whether a given set of e- is a lone pair or a shared pair. To write the
molecular geometry (a molecule’s shape), we need to account for both the electron geometry and which
of those sets have atoms bonded to them. The shape of the molecule only reflects the position of atoms in
space. How those atoms are positioned is, of course, dictated by the number of sets of e- around the
central atom.
Sets of Atoms bonded Electron
to center atom geom.
e-
2
3
3
4
4
4
Molecular geom.
(shape)
Bond angles
Example
2
Linear
Linear
180
CO2
3
Trig. Planar
Trig. Planar
120
BF3, CH2O
2
Trig. Planar
Bent
120
SO2, O3
4
Tetrahedral
Tetrahedral
109
CH4, CX4
3
Tetrahedral
Trigonal pyramid
107
NH3, PCl3
2
Tetrahedral
Bent
105
H2O, SF2
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5
5
5
5
6
6
6
5
Trigonal bipyramid
Trig. bipyramid
90 & 120
PCl5
4
Trigonal bipyramid
See-saw
90 & 120
SF4
3
Trigonal bipyramid
T-shaped
90
ICl3
2
Trigonal bipyramid
Linear
180
XeF2
6
Octahedral
Octahedral
90
SF6
5
Octahedral
Square pyramid
90
ClF5
4
Octahedral
Square planar
90
XeF4
Sketch each molecule, and give its a) electron geometry b) shape of molecule c) bond angles
1. PCl3
4. SeCl4
2. SO2
5. AsCl5
3. BrF3
6. SO42–
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Polar Molecules:
We’ve talked about the “like dissolves like” concept before—polar substance dissolve in polar
solvents; Nonpolar substances dissolve in nonpolar solvents. However, the polarity this concept refers
to is not the polarity of bonds; it is the polarity of molecules. Therefore, we must be able to ascertain
the polarity of an entire molecule.
We’ve seen how to show the polarity of bonds using arrows that point in the direction of edistribution. A molecule that is polar will have a net polarity. In other words, when looking at the
polarities of all the bonds of the molecule, there is an overall inequality of e- distribution in the
molecule. For you to be able to determine the polarity of a molecule, you msut consider 3 things:
1. The presence of polar bonds. In order to have a net polarity in a molecule, there must be
bonds that share e- unequally. Any molecule that has only nonpolar bonds must be
nonpolar.
2. The shape of the molecule. Certain shapes lend themselves to creating nonpolar
molecules, even when there are polar bonds present in the molecule. Other shapes, when
created with polar bonds, must lead to polar molecules. You must be able to recognize
both of these scenarios.
3. The magnitude of the bond polarities. Although certain shapes can give rise to nonpolar
molecules, this will only happen when the magnitudes of all polar bonds are equal.
Predicting molecular polarity is essentially a question of vector addition, where the
vectors we add are the polarity arrows. The magnitude of a bond’s polarity is equal to the
magnitude of the vector.
• A molecule will be nonpolar in two cases:
1. Its bonds are all nonpolar OR
2. The bond polarities cancel each other out due to shape. (The sum of the
polarity vectors is zero)
• A molecule will be polar when:
*
1. It has polar bonds AND
2. These polarities are not cancelled due to shape
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You cannot comment on the polarity of a molecule without considering all of
* things! This means that you must sketch each before deciding if a
these
molecule will be polar or nonpolar.
Which molecular shapes can lead to a nonpolar molecule, even with polar bonds present?
2/2 linear
(or 5/2)
3/3 trig. planar
4/4 tetrahedral
5/5 trig. bipyramid
6/6 octrahedral
6/4 square planar
What is the other requirement so that molecules with these shapes are nonpolar?
All ‘outer’ atoms are the same
Which shapes, when polar bonds are present, must give a polar molecule?
Bent
T-shaped
Trig. Pyramid
square pyramid
See-saw
What is the rule for molecule polarity for diatomic molecules?
If the bond is polar, so is the molecule (HCl, HF, NO). If the bond is nonpolar, so
is the molecule (H2, N2, Br2)
Are the following molecules polar or nonpolar?
(Psst… Sketchit!)
1. BCl3
2. SO2
3. O3
4. PCl3
5. PCl5
6. BrF3
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7. TeCl4
8. SF6
9. CH4
10. CH3I
11. AsF4Cl
12. CO2
Chapter 10: Valence bond theory
In Ch. 7 and 8, we learned that electrons are generally located in atomic orbitals. In Ch. 9, we
learned VSEPR theory, which states that sets of electrons about an atom wish to get as far away
from other sets as possible. We then learned the bond angles and geometries that make this
possible. In valence bond theory, we see that the above ideas have requirements that can’t be met
at the same time: How can a carbon atom in a molecule of CH4 share electrons with tetrahedral
geometry/109.5° bond angles, when its electrons are located in atomic s and p orbitals?
To address this, there is the theory of hybridized orbitals, where the atomic orbitals combine to
attain the needed properties. Let’s continue looking at CH4 as an example to illustrate:
Carbon’s four valence electrons are in the 2s and 2p sublevels:


px py
pz



sp3 sp3


sp3
sp3
s
In order to create the four identical molecular orbitals at 109.5°, all four of the above orbitals are
used: the four (s, p, p, p) atomic orbitals are hybridized into four new hybrid orbitals, which are
named sp3. The four valence electrons are redistributed among these new orbitals as shown.
These orbitals have the needed geometry, and the electrons are unpaired and ready to be shared
with the electrons from the hydrogen atoms to form the CH4 molecule.
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The CH2O molecule has a similar challenge: Sketching the molecule, we see that the carbon
atom has 3 sets around it, necessitating a 120° bond angle. How is this achieved?



px py
pz


sp2 sp2

p

sp2
s
Here, three of the atomic orbitals (s, p, p) are hybridized into three sp2 orbitals; carbon’s fourth
valence electron is in the one remaining unhybridized p orbital.
Finally, in the CO2 molecule, there are two sets of electrons around the carbon atom. This
requires a 180° angle. Let’s look at how this geometry is achieved:


px
py
pz


sp
sp


p
p

s
In this example, two atomic orbitals (s, p) are hybridized into two sp2 orbitals; the two other
valence electrons are in the unhybridized atomic p orbitals.
In order to attain the needed geometry for forming expanded octets, either 5 or 6 atomic orbitals
are needed in order to make the 5 or 6 hybridized orbitals with appropriate trigonal bipyramid or
octahedral shapes. We can see that we’ll need more orbitals than are available in the s and p
orbitals; therefore d orbitals must be used. This explains why only atoms from the 3 period and
beyond can host expanded octets; only they have d sublevels.
For the PCl5 molecule:
dxy dxz dyz dz2 dx2-y2
  
px py

pz


3

3

3

3
sp d sp d sp d sp d sp3d
-y2
s
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So, 5 atomic orbitals (s, p, p, p, d) are used to create the 5 hybridized sp3d orbitals that have
trigonal bipyramid geometry. (The unhybridized d orbitals are of no interest to us.)
OK, your turn: Draw the way in which atomic orbitals are hybridized on the sulfur atom in a
molecule of SF6.
We see that
hybridized
atomic orbitals (
) are used to create the
orbitals. The geometry of this is
.
Here’s the Cliff’s Notes® version of the above. Make sure you know it cold.
Sets of e-
e- geometry
angles
hybridization
2
Linear
180
sp
3
Trig pyr.
120
sp2
4
Tetrahedral
109
sp3
5
Trig bipyramid
90/120
sp3d
6
Octahedral
90
sp3d2
Let’s reinforce the importance of being able to draw Lewis structures. Your ability to flawlessly
create these impacts your ability to name: the electron geometry, the molecular geometry, the
bond angles, the presence of resonance structures, the bond order, the formal charge of atoms,
the polarity of the molecule, and the hybridization about different atoms in the molecule. You
also must sketch molecules correctly in order to calculate the energy change of a reaction using
bond energies.
What of the electrons that are left over in unhybridized atomic p orbitals, as in the CH2O and
CO2 molecules above? It turns out that these are used in forming multiple bonds. Let’s look at
the CO2 molecule as an example.
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Looking at the structure, we see that there are 2 sets
of e- around the carbon atom, and 3 sets of e- around
each oxygen atom. VSEPR dictates a 180° bond
angle, and suggests that the sets about oxygens
should be arrayed at 120° in trigonal planar fashion.
Reviewing our work with CO2 above, we saw that 2 of carbon’s four valence e- were found
singly in the 2 sp hybrid orbitals, and that the other 2 were left in the two unhybridized atomic p
orbitals. Knowing VSEPR, it makes sense that these p orbitals should be arranged in such a way
that they are far from the sp orbitals. This is done by having the two atomic p orbitals at 90° from
the sp orbitals, and from each other:
sp2
2
sp

π bond

sp2
O


sp
sp
sp1
C
py
 bond

2

π bond

sp' 

 sp2
O

 bond
px
px
px

sp2
Each oxygen atom has three sets of e-, and thus needs sp2 hybridization. With 6 valence e-, what
we get for each oxygen atom is this:

s
+



px
py
pz




sp2
sp2
sp2
+

p
The unpaired electron in the sp2 orbital pairs with an an unpaired electron in carbon’s sp orbital.
This happens twice, once for each oxygen atom, thus pairing both the unpaired sp electrons of
carbon.
These shared pairs occur in between the nuclei of the bonding atoms; such a bond is called a
sigma () bond.
Additionally, another bond is formed—This bond occurs between the unpaired electrons that are
found in the unhybridized atomic p orbitals. This bond, which occurs above/below the region of
the  bond, creates the double bond that exists between each carbon and oxygen atom in the CO2
molecule. Such bonds between e- in atomic p orbitals are called pi (π) orbitals.
In general, the first bond between two atoms is always a  bond, any second or third bonds are π
bonds. For example, in a molecule of CO, the first bond is , the other two bonds are both π.
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A fairly typical question asked is “what is the total number of sigma and pi bonds in a molecule
of XXX?
How many sigma and pi bonds are on the
amino acid molecules at left? (Be careful,
bonds to hydrogen atoms are not always
shown in organic chemistry!)
Tyr: _____  bonds, _____ π bonds
Asp: _____  bonds, _____ π bonds
Val: _____  bonds, _____ π bonds
Sketch the following molecules, and determine the hybridization about the center atom for each.
IF4-1
PCl3
BF3
H2O
XeF2
HCN
15