Chemical Bonding
Chapter 8
Chemical Bonding & Structure
Molecular bonding and structure play
the central role in determining the
course of chemical reactions.
Bonds
Forces that hold groups of
atoms together and make
them function as a unit.
Bond Energy
-
It is the energy required to break a bond.
-
It gives us information about the strength
of a bonding interaction.
-
The stronger the bond, the higher the
bond energy.
Bond Energies
Bond breaking requires energy
(endothermic).
Bond formation releases energy
(exothermic).
Chemical Bonds
Chemical Bond
Ionic
Cation
Anion
Covalent
Molecule
Ionic Bonds
-
Formed from electrostatic attractions of
closely packed, oppositely charged ions.
-
Formed when an atom that easily loses
electrons (metal) reacts with one that has a
high electron affinity(nonmetal).
-
2Na(s) + Cl2(g) ----> 2Na+(aq) + 2Cl-(aq)
Figure 11.8: The structure of lithium fluoride
Covalent Bonding
Covalent bonds are formed by sharing
electrons between nuclei.
H. + .H ----> H-H
2 hydrogen atoms
hydrogen molecule
Figure 11.1: The formation of a bond between two
hydrogen atoms
Types of Covalent Bonds
Polar covalent bond -- covalent bond in which
the electrons are not shared equally
because one atom attracts them more
strongly than the other. A dipole moment
exists. HOH, HCl, & CO
Nonpolar covalent bond -- covalent bond in
which the electrons are shared equally
between both atoms. No dipole moment
exists. CO2, CH4, & Cl2
Electronegativity
The ability of an atom in a molecule
to attract shared electrons to itself.
As electronegativity increases, the
attraction for electrons increases.
Fluorine has the highest value at 4.0
and cesium and francium are lowest
at 0.7.
Increasing electronegativity
08_132
H
Decreasing electronegativity
2.1
Li
Be
B
C
N
O
F
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Na
Mg
Al
Si
P
S
Cl
0.9
1.2
1.5
1.8
2.1
2.5
3.0
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
0.8
1.0
1.3
1.5
1.6
1.6
1.5
1.8
1.9
1.9
1.9
1.6
1.6
1.8
2.0
2.4
2.8
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
0.8
1.0
1.2
1.4
1.6
1.8
1.9
2.2
2.2
2.2
1.9
1.7
1.7
1.8
1.9
2.1
2.5
Cs
Ba
La-Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
0.7
0.9
1.0-1.2
1.3
1.5
1.7
1.9
2.2
2.2
2.2
2.4
1.9
1.8
1.9
1.9
2.0
2.2
Fr
Ra
Ac
Th
Pa
U
Np-No
0.7
0.9
1.1
1.3
1.4
1.4
1.4-1.3
(a)
Increasing electronegativity
H
Decreasing electronegativity
2.1
Li
1.0
Na
0.9
K
B
Be
2.0
1.5
Al
Mg
1.2
Ca
Sc
Ti
V
Cr
Mn
Co
Ni
Cu
1.8
1.9
1.9
1.9
1.0
1.3
1.5
1.6
1.6
0.8
Y
Zr
Nb
Mo
Tc
Rh
Pd
Sr
Ru
Ag
Rb
1.6
1.8
1.9
2.2
2.2
2.2
1.9
W
Re
Os
Ir
Pt
1.7
1.9
2.2
2.2
2.2
0.8
Cs
1.0
Ba
1.2
1.4
La-Lu
Hf
Ta
1.5
1.5
Fe
0.7
0.9
1.0-1.2
1.3
Fr
Ra
Ac
Th
Pa
U
Np-No
1.1
1.3
1.4
1.4
1.4-1.3
0.7
0.9
Au
2.4
Zn
Si
P
1.5
1.8
2.1
Ga
Ge
As
4.0
3.5
3.0
2.5
F
O
N
C
S
2.5
Se
2.4
Cl
3.0
Br
2.8
1.6
1.8
2.0
Cd
In
Sn
Sb
1.7
1.7
1.8
1.9
2.1
Hg
Tl
Pb
Bi
Po
At
1.9
1.8
1.9
1.9
2.0
2.2
1.6
Te
(b)
Pauling Electronegativity Values
I
2.5
Electronegativity values for selected elements. See Figure 8.3
on page 334 in Zumdahl.
Homework #24
a. Rb < K < Na
b. Ga < B < O
c. Br < Cl < F
d. S < O < F
Three Possible Types of Bonds
Nonpolar Covalent
(Electrons equally
shared.)
Polar Covalent
(Electrons shared
unequally.)
Ionic
(Electrons are
transferred.)
Percent Ionic Character
xA  xB
100% 
% Ionic Character (IC) 
xA
where xA is the larger electronegativity and xB
is the smaller value.
Watch significant figures!!!
Ionic Bond
Polar Covalent
Nonpolar Covalent
% IC > 50 %
% IC 5 - 50 %
% IC < 5 %
Percent Ionic Character
What type of bonding & % ionic character
does KCl have? Ionic
xA  xB
100% 
% Ionic Character (IC) 
xA
3.0  0.8
100% 
% Ionic Character (IC) 
3.0
2.2
100%
% Ionic Character (IC) 
3.0
% Ionic Character (IC)  73%
Percent Ionic Character
What type of bonding & % ionic character
does HOH have? Polar covalent
xA  xB
100% 
% Ionic Character (IC) 
xA
3.5  2.1
100%
% Ionic Character (IC) 
3.5
1.4
100%
% Ionic Character (IC) 
3.5
% Ionic Character (IC)  40.%
Percent Ionic Character
What type of bonding & % ionic character
does N2 have? Nonpolar covalent
xA  xB
100% 
% Ionic Character (IC) 
xA
3.0  3.0
100% 
% Ionic Character (IC) 
3.0
0
100%
% Ionic Character (IC) 
3.0
% Ionic Character (IC)  0%
Polarity
A molecule, such as HF, that has a center
of positive charge and a center of negative
charge is said to be polar, or to have a
dipole moment.
H F
+
partial positive charge

partial negative charge
Figure 11.2: Probability representations of the
electron sharing in HF
08_131
H
F

H




 H
F
F



 H
F

F

H
F
 H
H


F
H



 H
F
F

 H

(a)
F
(b)
The Effect of an electric field on hydrogen fluoride molecules.
08_133


+
H
O



H

(a)
(b)
Dipole Moment for the water molecule.
Polar Water Molecule
The polarity of water allows it to dissolve
ionic materials which are essential for life.
The polarity of the water molecule allows
water molecules to attract each other
strongly (hydrogen bonds). Because of
this fact water remains as a liquid at room
temperatures and allows the existence of
life as we know it.
08_134
3
+

N
H
H


H




(a)
(b)
Dipole moment for the ammonia molecule.
Homework #30
a.
C
O
b.PH is a pure covalent (nonpolar) bond since P and H have identical
electronegativities.
c.
H
d.
e.
Cl
Br
Te
Se
S
08_151
Nonpolar molecule--zero dipole moment.
Cation Size
Cations are always smaller than the
parent atom because they have lost an
entire electron shell. As well, the
number of protons is greater than the
number of electrons so the electrons
are held tighter.
Anion Size
Anions are always larger than the parent
atom because they have added
electrons which repel each other. As
well, the number of protons is less
than the number of electrons so they
are not held as tightly.
Relative sizes of some ions and their parent atoms.
Energy Changes & Chemical
Bonding
Most chemical reactions can be explained in
terms of the rearrangements of bonds.
• Energy must be added to break existing bonds
in the reacting substances.
• Energy is released when new bonds are formed
in the products.
• For the reactions of covalent compounds, the
net energy of the reaction, ∆Hrxn, = (sum of
energy added to break bonds (reactants) - (sum
of energy released when bonds form(products).
Energy Changes & Chemical
Bonding
Consider the reaction for the formation of
water from elemental hydrogen and
oxygen, 2H2(g) + O2(g) ---> 2H2O(g).
• First, write the Lewis dot structures for the reactants
•
and products, and determine which bonds must be
broken and which bonds must be formed.
Then use the average bond energy values found in
Table 8.4 p351 of the book. to calculate the energy
needed to break the bonds in the reactants and the
energy released when the bonds form in the products.
Energy Changes & Chemical
Bonding
2H2(g) + O2(g) ---> 2H2O(g).
• To break the H-H bond requires 432 kJ/mole H2(g).
•
•
Since there are two moles of H2(g) in the reaction, this
value is doubled to 864 kJ.
The oxygen atoms in the O2(g) are held together by a
double bond which requires 495 kJ/mole to break.
For each water molecule produced, there are two O-H
bonds formed. For the 2H2O(g), the total energy
released when four moles of O-H bonds form is: 4
moles O-H x 467 kJ/mole = 1868 kJ.
Energy Changes & Chemical
Bonding
2H2(g) + O2(g) ---> 2H2O(g).
• Therefore, the ∆Hrxn = (864 kJ to break H-H bonds
+ 495 kJ to break O=O bonds) – (1868 kJ released
when O-H bonds formed) = -509 kJ.
• The negative value indicates that the reaction
releases energy and is exothermic.
(Note that bond energies are average values, and
can vary depending upon what other elements
are bonded to the atoms which are being broken
apart.)
Homework #54
H
C
N+2 H
Bonds broken:
H
H
H
H
C
N
H
H
Bonds formed:
1 C ≡ N (891 kJ/mol)
1 C - N (305 kJ/mol)
2 H - H (432 kJ/mol)
2 C - H (413 kJ/mol)
2 N - H (391 kJ/mol)
ΔH = 891 kJ + 2(432 kJ) - [305 kJ + 2(413 kJ) + 2(391 kJ)] = -158 kJ
Part B
H
H
N
+2 F
N
H
F
4 H
F + N
H
Bonds broken:
Bonds formed:
1 N - N (160. kJ/mol)
4 H - F (565 kJ/mol)
4 N - H (391 kJ/mol)
1 N ≡ N (941 kJ/mol)
2 F - F (154 kJ/mol)
ΔH = 160. kJ + 4(391 kJ) + 2(154 kJ) - [4(565 kJ) + 941 kJ] = -1169 kJ
N
Energy Changes & Chemical
Bonding
For the formation of ionic compounds, the
∆Hrxn is based upon all of the energy
changes involved in the transfer of electrons
between the metal and nonmetal.
Consider the reaction for the formation of
sodium chloride from elemental sodium and
chlorine,
2Na(s) + Cl2(g) ---> 2NaCl(s).
Energy Changes & Chemical
Bonding
• Ionization energy must be added to completely
•
•
•
remove electrons from the metal in the gaseous state.
Since most metals are solids at room temperature,
energy must be added to vaporize the metal before
this ionization occurs. This energy value is called the
heat of formation, ∆Hºf.
Then energy is released when the nonmetal gains
electrons, as measured by the electron affinity value.
Energy is also released when the ions bond to form a
crystal lattice structure, called the lattice energy.
Ionic Compounds
2Na(s) + Cl2(g) ---> 2NaCl(s)
2Na(s) ---> 2Na(g)
add ∆Hºf = (108 kJ/mole Na x 2 moles Na) = 216 kJ
2Na(g) ---> 2Na1+(g) + 2e1add ionization energy = (496 kJ/mole Na x 2 moles Na) = 992 kJ
Cl2(g) ---> 2Cl(g)
add bond energy = 243 kJ/mole Cl2 x 1 mole Cl2) = 243 kJ
2Cl(g) + 2e1- ---> 2Cl1-(g)
release electron affinity = 349 kJ/mole Cl1- x 2 moles Cl1- = -698 kJ
2Na1+(g) + 2Cl1-(g) ---> 2NaCl(s)
release lattice energy = 788 kJ/mole NaCl x 2 moles NaCl = -1576kJ
∆Hrxn = 216 kJ + 992 kJ + 243 Kj + -698 kJ + -1576 kJ = -823 kJ
per mole of reaction
Homework #50
50.
Let us look at the complete cycle for Na2S.
2 Na(s) → 2 Na(g)
2 ΔHsub, Na = 2(109) kJ
2 Na(g) → 2 Na+(g) + 2 e
2 IE = 2(495) kJ
S(s) → S(g)
ΔHsub, S = 277 kJ
S(g) + e → S (g)
EA1 = -200. kJ
S-(g) + e → S2 (g)
EA2 = ?
2 Na+(g) + S2(g) → Na2S
LE = -2203 kJ
_____________________________________________________
2 Na(s) + S(s) → Na2S(s)
ΔH= -365 kJ
ΔH= 2 ΔH sub, Na + 2 IE + ΔHsubS + EA1 + EA2 + LE,
-365 = -918 + EA2,
EA2 = 553kJ
For each salt: ΔH= 2 ΔHsub, M + 2 IE + 277 - 200. + LE + EA2
# 50 continued
K2S: -381 = 2(90.) + 2(419) + 277 - 200. - 2052 + EA2,
EA2 = 576 kJ
Rb2S: -361 = 2(82) + 2(409) + 277 - 200. - 1949 + EA2,
EA2 = 529 kJ
Cs2S: -360. = 2(78) + 2(382) + 277 - 200. - 1850. + EA2,
EA2 = 493 kJ
We get values from 493 to 576 kJ.
The mean value is: = 538 kJ
Energy Changes & Chemical Bonding
The accepted values for the various energy changes that
occur during bonding reactions have been carefully
calculated using calorimeter experiments.
In your textbook,
• Ionization energies for some of the elements can be
•
•
•
•
found in Table 7.5 p.310 & Figure 7.6 p.310.
Electron affinity values are listed in Figure 7.7, page
313.
Lattice energies for ionic crystals are found in the
problems or given data
Values for the enthalpy of formation, ∆Hºf, are listed
in Figure Appendix 4 pp.A19-A22.
Values for the enthalpy of vaporization, ∆Hºvap, are
listed in The problems or given appendix.
Achieving Noble Gas Electron
Configurations (NGEC)
Two nonmetals react: They share
electrons to achieve NGEC.
A nonmetal and a representative group
metal react (ionic compound): The
valence orbitals of the metal are emptied
to achieve NGEC. The valence electron
configuration of the nonmetal achieves
NGEC.
Noble Gas Configuration
When a Group I, II, or III metal reacts with a
nonmetal to form a binary ionic
compound, the nonmetal gains electrons to
obtain the configuration of the next noble
gas. The metal loses electrons to gain the
configuration of the previous noble gas.
Na ----> Na+ + e- configuration of Ne
Cl + e- ----> Cl-
configuration of Ar
Lewis Structure
-
Shows how valence electrons are arranged
among atoms in a molecule.
-
Reflects central idea that stability of a
compound relates to noble gas electron
configuration.
-
Developed by G.N. Lewis in 1902.
Lewis Structures
Na. sodium atom
..
.S: sulfur atom
.
[Na]+ sodium ion
.. 2 
[ : S:] sulfide ion
..
Drawing Lewis structures
• Write the electron dot diagrams for each
element in the compound.
• Check the electronegativity difference
between the elements to determine if
electrons are transferred or shared.
• If the electronegativity difference > 1.67, the
reaction forms ions. Remove the electrons
from the metal and add them to the nonmetal.
Drawing Lewis Structures
Write the charges of the ions formed and use
coefficients to show how many of each ion
are needed to balance the overall charge.
+
2-
2Na , [ O ]
Ionic sodium oxide
Drawing Lewis structures
• If the electronegativity difference < 1.67, then
the atoms will share electrons.
• Position shared electron pairs between the two
atoms, and connect them with a single line to
represent a covalent bond.
• Place the extra pairs of electrons around atoms
until each has eight
• (Exception: For hydrogen or metallic
elements use only the valence electrons that
are available, so these atoms have less than an
octet.)
Drawing Lewis structures
• If an atom other than hydrogen or a metal has
less than eight electrons, move unshared pairs
to form multiple bonds.
• Add extra atoms, if needed, to obtain the
octets. Atoms with positive oxidation numbers
should be bonded to those with negative
oxidation numbers.
• If extra electrons still remain, add them to the
central atom. All oxidation numbers should
add up to zero for a compound.
Lewis structures
Example CO2
Step 1
C-O-O
O-C-O
– Draw any possible structures
You may want to use lines for bonds.
Each line represents 2 electrons.
Lewis structures
Step 2
– Determine the total number of valence
electrons.
– CO2 1 carbon x 4 electrons
= 4
2 oxygen x 6 electrons = 12
Total electrons
= 16
Lewis structures
Try the C-O-O structure
C O O
No matter what you
try, there is no way
satisfy the octet for
all of the atoms.
Lewis structures
Step 3
Try to satisfy the octet rule for each atom
- all electrons must be in pairs
- make multiple bonds as required
O C O
This arrangement needs
too many electrons.
How about making some double bonds?
O=C=O
That works!
=
is a double bond,
the same as 4 electrons
Step 1
Ammonia, NH3
H
H N H
Step 3
H
Step 2
3 e- from H
5 e- from N
8 e- total
N has octet
H N H
H has 2 electrons
(all it can hold)
Lewis Structures
NO+
•
5 e- + 6 e- - 1 e- = 10 e-
•
[:N O:]+
•
Each atom has an octet and is satisfied.
Are there any other ways?
Yes, there is a simple calculation involving the
number of shared electrons and number of
bonding sites that tells you what kind of
bonding is going on.
Let S = N-A S= # of e¯ that need to be shared
N=needed e¯ (either 2 or 8)
A= available valence e¯
B=regions with e¯
Then S/2/B= kind of bonding
Caveat
•
Note this can be very helpful when it works
(when the octet rule is satisfied) and really
misleading when it doesn’t.
I personally never use it.
You might find it a good way to get started on
learning this material when you are in doubt
about resonance or number of double and
triple bonds.
Examples
S/2/B= 1
CCl4
S= 40-32=8
S/2/B= 8/2/4= 1
Single Bonds around C
Double bond example
S/2/B=2
CO2
Double bonds
S=24-16=8
S/2/B=8/2/2= 2
Double bonds around C
Greater Than 1= Resonance
S/2/B=fraction greater than 1
SO2
S=24-18=6
S/2/B= 6/2/2=3/2
3 e¯ pairs shared
between two bonding regions
Less Than one = Multiple
around central atom
S/2/B<1
IBr2
S=24-22
S/2/B=2/1/2=1/2
So need expanded for 5 e¯ pairs
Single, Double, & Triple Bonds
Single bonds -- one shared pair of
electrons.
Double bonds -- two shared pairs of
electrons.
Triple bonds -- three shared pairs of
electrons.
•Bond Strength = Triple > Double > Single
–For bonds between same atoms,
CN > C=N > C—N
–Though Double not 2x the strength of
Single and Triple not 3x the strength of
Single
•Bond Length = Single > Double > Triple
–For bonds between same atoms,
C—N > C=N > CN
Comments About the Octet Rule
-
2nd row elements C, N, O, F observe the
octet rule.
-
2nd row elements B and Be often have fewer
than 8 electrons around themselves - they are
very reactive.
-
3rd row and heavier elements CAN exceed
the octet rule using empty valence d orbitals.
-
When writing Lewis structures, satisfy octets
first, then place electrons around elements
having available d orbitals.
Four Failures of Lewis
Structures
Lewis Structures cannot adequately explain:
1. electron-deficient molecules.
2. the paramagnetism of oxygen and other
similar substances.
3. odd-electron molecules.
4. resonance.
Atoms with fewer than eight
electrons
Beryllium and boron will both form
compounds where they have less
than 8 electrons around them.
: :
..
:Cl Be Cl:
..
..
..
:F
.. B F:
..
:F:
..
Atoms with fewer than eight
electrons
Electron deficient. Species other than hydrogen and helium that
have fewer than 8 valence electrons.
They are typically very reactive species.
Coordinate covalent bond forms when N atom donates both
shared eF
H
F H
|
|
| |
F - B + :N - H
F-B-N-H
|
|
| |
F
H
F H
BF3 is called a Lewis acid because it accepts a pair of
electrons and NH3 is a Lewis base because it donates
a pair of electrons.
Atoms with more than eight
electrons
Except for species that contain hydrogen, this is
the most common type of exception.
For elements in the third period and beyond,
the d orbitals can become involved in
bonding.
Examples
5 electron pairs around P in PF5
5 electron pairs around S in SF4
6 electron pairs around S in SF6
An example: SF4
1. Write a possible
– arrangement.
F
F S F
F
2. Total the electrons.
• 6 from S, 4 x 7 from F
– total = 34
3. Spread the electrons
– around.
F
|
F - S- F
|
F
Homework # 68
a.
POCl3 has 5 + 6 + 3(7) = 32 valence electrons.
O
Cl
P
O
Cl
Cl
P
Cl
Cl
Cl
Skeletal
structure
Lewis
structure
This structure uses all 32 e while satisfying the octet rule for
all atoms. This is a valid Lewis structure.
SO42 =32 valence electrons.
2-
O
O
S
O
O
XeO4, 8 + 4(6) = 32 e
O
O
Xe
O
O
PO43, 5 + 4(6) + 3 = 32 e
3-
O
O
P
O
O
ClO4 has 7 + 4(6) + 1 = 32 valence electrons
O
O
Cl
O
O
-
b. NF3 has 5 + 3(7) = 26 valence electrons.
F
N
F
F
F
N
F
F
Skeletal
structure
Lewis
structure
c. SO32, 62-+ 3(6) + 2 = 26 e
O
S
O
O
PO33, 5 + 3(6) + 3 = 26 e
ClO3, 7 + 3(6) + 1 = 26 e
O
Cl
O
O
c.
ClO2 has 7 + 2(6) + 1 = 20 valence
O Cl
O
O Cl
O
SCl2, 6 + 2(7) = 20 e
Cl
S Cl
PCl2, 5 + 2(7) + 1 = 20 e
Cl
P Cl
d. Molecules ions that have the same number
of valence electrons and the same number
of atoms will have similar Lewis
structures.
Odd-Electron Molecules
NO2
• contains 17 electrons.
• cannot satisfy the octet rule.
• a more sophisticated model is neededthe molecular orbital model.
Species with an odd
total number of electrons
A very few species exist where the total
number of valence electrons is an odd
number.
This must mean that there is an unpaired
electron which is usually very reactive.
Radical - a species that has one or more
unpaired electrons.
They are believed to play significant roles
in aging and cancer.
Species with an odd
total number of electrons
Example - NO
Nitrogen monoxide is an example of a
compound with an odd number of electrons.
It is also known as nitric oxide.
It has a total of 11 valence electrons: six
from oxygen and 5 from nitrogen.
:
.
:N
O:
The best Lewis structure for NO is:
Resonance
Occurs when more than one valid Lewis
structure can be written for a particular
molecule.
These are resonance structures. The actual
structure is an average of the resonance
structures called a resonance hybrid.
Resonance structures
Sometimes we can have two or more equivalent
Lewis structures for a molecule.
O-S=O
O=S–O
They both - satisfy the octet rule
- have the same number of bonds
- have the same types of bonds
Which is right?
Resonance structures
They both are!
O -S=O
O =S - O
O
S
O
This results in an average of 1.5 bonds
between each S and O.
Resonance structures
Benzene, C6H6, is an example of a compound
for which resonance structures must be
written. At each corner of the hexagonal
ring, there is a carbon atom with a double
bond to one C and a single bond to another C
and to an H atom.
All of the bonds are the same length.
or
Stereochemistry
The study of the threedimensional arrangement
(molecular structure) of atoms or
groups of atoms within molecules
and the properties which follow
such arrangement.
Formal Charges
•
A bookkeeping system for electrons that is
used to predict which possible Lewis
structure is more likely.
•
They are used to show the approximate distribution
of electron density in a molecule or polyatomic
ion.
Rules for Formal Charges
1. Assign each atom half of the electrons in
each pair it shares.
2. Also give each atom all electrons from
unshared pairs it has.
3. Subtract the number of electrons assigned
to each atom from the number of valence
electrons for an atom of the element.
0
Formal Charges
0
0
-1
0
+1
O=C=O
O C=O
Structure 1
Structure 2
For
eachmost
oxygen
The
likely
For the single-bond oxygen
Lewis structures
are those
which
-
(6 e from unshared e + 1e from
have: assigned from
(4 electrons
bond) = 7 total
- from the
unshared
e- + 2 eobeying
Formalrule,
charge = 6 - 7 = -1
all atoms
the octet
bonds) = 6 total
the triple-bond
oxygen
all atoms with a formalFor
charge
of
zero,
or (2 e from unshared e + 3e from
Formal
6-6=0
the charge
most =electronegative
element
with the
bonds)
= 5 total
negative formal charge.Formal charge = 6 - 5 = +1
For carbon
For carbon
4 e- assigned from the bonds = 4 4 e- from the bonds = 4 total
Formal charge = 4 - 4 = 0
total
•
•
•
Another Example of Formal
0 Charges
0
+1
-1
C=O
For oxygen
Structure 1
(4 electrons assigned from
- + 2 e- from the
unshared
e
Although Structure
bonds) = 6 total
C=O
Structure 2
1 hasForalloxygen
atoms with a formal
- from unshared e- + 3e- from
(2
e
charge of zero, the carbon atom does not obtain
bonds) = 5 total
Formal
charge =Therefore,
6-6=0
an octet.
Structure
2 is=the
Formal charge
6 - 5most
= +1
all atoms obey the
For carbon
Forlikely
carbonLewis structure since
(2 electrons assigned from unshared
octet rule.
(2 electrons assigned from
unshared e- + 2 e- from the
bonds) = 4 total
Formal charge = 4 - 4 = 0
e- + 3 e- from the bonds) = 5 total
Formal charge = 4 - 5 = -1
Homework #82
82. For SO42, ClO4, PO43 and ClO3, only one of the
possible resonance structures is drawn.
a. Must have five bonds to P to minimize formal charge of
P. The best choice is to form a double bond to O since this
will give O a formal charge of zero and single bonds to Cl
for the same reason.
O
Cl
P
Cl
P, FC = 0
Cl
b.
Must form six bonds to S to minimize formal charge of S.
O
O S O
O
2-
S, FC = 0
c.Must form seven bonds to Cl to minimize formal
charge.
O
O
Cl
O
Cl, FC = O
O
d. Must form five bonds to P to minimize formal
charge.
3O
O
P
O
O
P, FC = 0
e. .
O
Cl
S
S, FC = 0
Cl, FC = 0
O, FC = 0
Cl
O
O
f. .
O
Xe
O
O
Xe, FC = 0
f. .
O Cl
O
Cl, FC = 0
O
h. We can’t. The following structure has a
zero formal charge for N: But N does not
expand its octet. We wouldn’t expect this
3resonance form to exist.
O
O N
O
O
VSEPR Model
Valence Shell Electron Pair
Repulsion -- The structure
around a given atom is
determined principally by
minimizing electron pair
repulsions.
Molecular Geometry
Parent Geometry is
Actual Geometry is the
electron pair
arrangement about
the central atom.
arrangement of atoms
about the central
atom.
•linear
•linear
•trigonal planar
•bent
•tetrahedral
•trigonal pyramid
08_142
Lone
pair
N
N
H
H
H
(a)
(b)
Lone pair of electrons on the ammonia molecule.
08_143
Lone pair
Bonding
pair
O
Bonding
pair
O
H
(a)
H
Lone pair
(b)
O
H
(c)
H
Lone pairs on the water molecule.
VSEPR
Two pairs of electrons are placed 180o apart -linear arrangement.
Three pairs of electrons are placed 120o apart
-- trigonal planar arrangement.
Four pairs of electrons are placed 109.5o apart
-- tetrahedral arrangement.
Double bonds and triple bonds count as one
effective pair of electrons.
Electron pair arrangement is the parent geometry. Molecular
structure is the actual geometry.
Parent & Actual Geometry
When every pair of electrons on the central
atom is shared with another atom, the
parent and actual geometry are the same.
When one or more pair of electron pairs
around a central atom are unshared(lone
pairs), the parent and actual geometry are
different.
VSEPR Model Summary
•
Determine the Lewis structure(s) for the
molecule.
•
For molecules with resonance structures, use any
of the structures to predict the molecular
structure.
•
Sum the electron pairs around the central atom to
determine the parent geometry.
•
The arrangement of the pairs is determined by
minimizing electron-pair repulsions.(Actual
Geometry)
VSEPR Model Summary
(Continued)
Lone pairs require more space
than bonding pairs since
they are tightly attracted
to only one nucleus. Lone
pairs produce slight
distortions of bond angles
less than 120o.
Linear - CO2
Trigonal planar, BCl3
Bent, H2O
Pyramidal, NH3
Tetrahedral, CH4
Molecular geometries
based
on
tetrahedral
H
Tetrahedral
C
H
O
H
H
H
Bent
H
Pyramidal
N
H
H
H
Bent and pyramidal are
actually tetrahedral but
some of the electron
pairs are not bonded.
Other geometries.
Other shapes are also observed.
Five bonds or lone electron pairs
Trigonal bipyramidal
Seesaw
T-shaped
Linear
Six bonds or lone electron pairs
Octahedral
Square pyramidal
Square planar
Trigonal bipyramidal
Square planar
Octahedral
Molecular geometry
H
H
C
H
H
C
H
H
As molecules get larger, the rules regarding
molecular geometry still hold.
Ethane
Tetrahedral shape around each carbon atom.
VSEPR shapes
Coordination
Number
Electron pairs General
Bonding Unshared
Formula
Shape
Linear
2
2
0
AB2
3
3
0
AB3
Trigonal planar
2
1
AB2
Bent
4
0
AB4
Tetrahedral
3
1
AB3
Trigonal
4
pyramidal
2
2
AB2
Bent
1
3
AB
Linear
VSEPR shapes
Coordination
Electron pairs
General
Number
Bonding Unshared
Formula
5
6
Shape
5
4
0
1
AB5
AB4
Trigonal Bipryramidal
3
2
AB3
T-shaped
2
3
AB2
Linear
6
0
AB6
Octahedral
5
1
AB5
Square pyramidal
4
2
AB4
Square Planar
Seesaw
Polar and nonpolar molecules
Most bonds between atoms of different elements in
a molecule are polar. That does not mean that
the molecule will be polar.
O=C=O
Electronegativities:
Oxygen = 3.5
Carbon = 2.5
Difference 1.0
(polar bond)
The electronegativity values
Show that the C-O bond would be polar with electrons
being pulled towards the oxygens. However, due to the
geometry, the pull happens in equal and opposite
directions.
Polar and nonpolar
molecules
For a molecule to be polar, the effects of bond
polarity must not cancel out.
One way is to have a geometry that is not
symmetrical
Olike in water.Electronegativity
H
H
difference = 1.3
Here, the effects of the polar bonds do not cancel,
so the molecule is polar.
Polar and nonpolar molecules
Polarity is an important property of molecules.
It affects physical properties such as melting
point, boiling point and solubility.
Chemical properties also depend on polarity.
Dipole moment, µ, is a quantitative measure of
the polarity of a molecule.
Dipole moment
This property can be measured by placing
molecules in an electrical field. Polar molecules
will align when the field is on. Nonpolar
molecules will not.
+
-
+
-
Polar and nonpolar molecules
A molecule is nonpolar if the central atom is
symmetrically substituted by identical
atoms.
CO2, CH4 , CCl4
A molecule will be polar if the geometry is
not symmetrical.
H O, NH , CH Cl
Geometry and polar
molecules
For a molecule to be polar
– - must have polar bonds
– - must have the proper geometry
• CH4
• CH3Clpolar
–
• CHCl3 polar
• CCl4
non-polar
CH2Cl2
non-polar
polar
Bonding theory
Two methods of approximation are used to
describe bonding between atoms.
Valence bond method
Bonds are assumed to be formed by overlap of
atomic orbitals
Molecular orbital method
Valence bond method
According to this model, the H-H bond forms as
a result of the overlap of the 1s orbitals from
each atom. The bonding pair held directly
between both nuclei and is called a sigma
(Ó) bond.
74 pm
Valence bond method
Multiple bonds are formed by the side-to-side overlap of
orbitals. The bonding pair is held above and below the
two nuclei and is called a pi (π) bond.
π bond
C2H4
H
H
Ó bond
C
H
C
H
π overlap
Valence bond method
Hybrid orbitals are need to account for the
geometry that we observe for many
molecules.
Example - Carbon
Outer electron configuration of 2s2 2px1 2py1
We know that carbon will form four
equivalent bonds - CH4, CH2Cl2 , CCl4.
Hybridization
To explain why carbon forms four identical single
bonds, we assume the the original orbitals will
blend together.
2p
energy
2sp3
2s
Unhybridized
Hybridized
Hybridization
In the case of a carbon that has 4 single bonds, all
of the orbitals are hybrids.
sp3
+3
1
4
25% s and 75% p character
s
p
sp3
Ethane, CH3CH3
sp3
hybrids
Ó bond - formed
Ó bond
1s orbital
from H
by an endwise
(head-on) overlap.
Molecules are
able to rotate
around single
bonds.
Ethane , CH3CH3
Rotation of single bond
sp2 hybrid orbitals
To account for double bonds, a second type of
hybrid orbital must be pictured. An sp2
hybrid is produced by combining one s and
2 p orbitals. One p orbital remains.
2p
energy
2p
2sp2
2s
Unhybridized
Hybridized
sp2 hybrid orbitals
The unhybridized p orbitals are able to overlap,
resulting in the formation of a second bond π bond.
C
C
A π bond is a
sideways overlap
that occurs both
above and below the
plane of the molecule
Parts of the molecule
are no longer able to
rotate about the bond.
Ethene
Bonding in ethene
1s orbital
π overlap
sp2
hybrids
H
H
C
Ó bond
H
C
H
π overlap
sp2
hybrids
π bond
Bonding in ethene
sp
hybrid
orbital
Forming a triple bond is also possible.
This
requires that two p orbitals remain
unhybridized.
2p
energy
2p
2sp
2s
Unhybridized
Hybridized
sp hybrid orbital
Now two p orbitals are available to
form π bonds.
C
C
Ethyne
Bonding in ethyne
sp hybrid
π overlaps
Bonding in ethyne
Other hybrid orbitals
d orbitals can also be involved in the formation of
hybrid orbitals.
Hybrid
Shape
sp
Linear
sp2
Trigonal planar
sp3
Tetrahedral
sp3d
Trigonal bipyramidal
sp3d 2
Octahedral
Molecular Orbital Method
When atomic orbitals combine to form
molecular orbitals, the number of
molecular orbitals formed must equal the
number of atomic orbitals mathematically
combined.
Example - H2
Two 1s orbitals will combine forming two
H2 molecular orbital diagram
energy
Ó*1s
1s
H
1s
H
Ó 1s
H2
Orbital shapes
Molecular orbitals
When two atomic orbitals combine, three types
of molecular orbitals are produced.
Bonding orbital - Ó or π
The energy is lower than the atomic orbitals
and the electron density overlaps.
Antibonding orbital - Ó* or π*
The energy is higher than the atomic orbitals
MO diagram of helium
energy
Ó*1s
1s
He
1s
He
Ó1s
He2
If we develop a
diagram for helium
we see that both
a bonding and
antibonding
orbital will be
filled.
The result is that
it is no more stable
than the unbonded
form -- it will not
bond
Molecular orbital bonding
For a molecule to be stable, you must have more
electrons in bonding orbitals than in antibonding
orbitals.
The bonded form will be at a lower energy so will
be more stable.
#bonds formed=bonding e- -antibonding e-
MO π*diagram for O2
2pz
π*2px π*2py
2px 2py 2pz
π 2px
π 2py
2px 2py 2pz
π2pz
2s
Ó*2s
Ó 2s
2s
Ó*1s
1s
Ó1s
1s
#bonds formed = 10 bonding e- - 6 antibonding e2
=2
π * 2px
MO diagram for NO
π *2py π *2pz
2px 2py 2pz
π 2py
π 2pz
2px 2py 2pz
π 2px
2s
Ó *2s
Ó 2s
1s
N
2s
Ó *1s
Ó 1s
1s
NO
O
# bonds = 10 bonding e- - 5 antibonding e- = 2.5
2
Delocalized electrons
MO diagrams for polyatomic species are often
simplified by assuming that all Ó and some
π orbitals are localized -- shared between
two specific atoms.
Resonance structures require that electrons
in some π orbitals be pictured as delocalized.
Delocalized - free to move around three or
more atoms.
Delocalized electrons
Benzene is a good example of delocalized
electrons.
We know that the bonding between carbons
has an order of 1.5 and that all of the bonds
are equal.
=
Aromatic hydrocarbons
H
H
H
H
H
H
p orbitals overlap
sidewise all around
the ring. No localized
double bonds.
Band theory of bonding in
solids
This is an extension of delocalized orbitals.
Each atom interacts with all of the others in the
crystal, resulting in an enormous number of
‘molecular orbitals.’
3s
Na
3s
Na
Band
Band theory of bonding in
solids
A group of very closely spaced energy
levels.
Energy gap
The difference in energy between the
bonding and antibonding orbitals.
Band theory of bonding in solids
Conductor - A material with a partially filled
energy band.
Insulator - The highest occupied band is filled
or almost completely filled. The forbidden
band just above the highest filled is wide.
Semiconductor - The gap between the highest
filled band and the next higher permitted
band is relatively narrow.
Band theory of bonding in solids
Energy
Empty
Forbidden, wide
Filled
Energy
insulator
Empty
Forbidden, narrow
Filled
Energy
semiconductor
No
Forbidden
conductor
Intermolecular Attractions
What type of attractive
forces hold molecules
together in liquids and
solids?
Intermolecular Forces
Intermolecular forces - are forces between
the molecules of molecular substance
that tend to be weaker than normal
covalent bonds
There are several types of intermolecular
bonds which we will discuss
London forces
Temporary dipole attractions that exist
between nonpolar molecules - also called
the dispersion forces.
Results from an unbalanced distribution of
valence electrons, resulting in a
temporary dipole.
A relatively weak force that increases with
Dipole-Dipole Interactions
Dipole-dipole interactions are weak
intermolecular forces that can occur
between polar molecules.
Polar molecules attract one another when a
positive side (∂+) and a negative side (∂) of two polar molecules are near one
another.
Dipole-dipole
attractions
The greater the percent ionic character, the stronger
the dipole-dipole forces.
∂+ and ∂- ends are attracted to each other.
-
+
H
Cl
Cl
Cl
+
H
-
+
H
Cl
+
H
Cl
Cl
+
H
solid
-
+
H
Cl
+
H
Cl
Cl
+
H
+
H
liquid
Hydrogen Bonding
Are intermolecular forces resulting from a
specific interaction between a hydrogen
atom in one molecule and a fluorine,
oxygen, chlorine or nitrogen atom in
another molecule
The hydrogen attempts to form a coordinate
covalent bond with an extra pair of
electrons on a neighboring atom.
Hydrogen Bonding
The small sizes of the elements involved and the
large electronegativity differences result in
large ∂+ and ∂- values.
Hydrogen bonds are represented using a dashed
line.
Hydrogen bonding
The hydrogens of
one water molecule
interact with the
oxygen on other
water molecules.
Ion-Dipole Attractions
When ionic compounds dissolve in water,
the negative ions are attracted to the
oxygen atoms, and the positive ions are
attracted to the hydrogen atoms.
The degree of dissociation depends on the
strength of these ion-dipole attractions
Intermolecular forces
For molecules to form liquids and solids, there
must be attractions between the them.
Intermolecular attractive forces
• ion-dipole attraction
• dipole-dipole attraction including hydrogen bonding
• London (dispersion) forces
Relative strength
Dissolving covalent compounds
Covalent compounds do not dissociate
(separate into ions) when they dissolve.