TOPIC 8. Basic Concepts
of Chemical Bonding
| When
overhead transparencies are shown during lecture they will be noted in
boxes such as this one. If a similar illustration is in your text, its
figure or table number will also be included. |
1. Lewis Symbols and the "Octet" Rule
A. Valence Electrons
Recall that valence electrons
are those that are in the outer electron shell (the valence shell) of an
atom.
B. Lewis (Electron Dot) Symbols
Lewis symbols can be drawn for
elements showing their valence electrons. Here's an example showing the
Lewis symbols for the elements in row three of the Periodic Table.
C. The "Octet" Rule
Every noble gas has eight
electrons in its outer shell except for He which has only two. Most atoms
gain, lose or share electrons until they have the same number of outer-shell
electrons as one of the noble gases. There are exceptions to this
"octet" rule which will be described later.
D. Stability of Noble Gases
The noble gases are particularly
stable because of their electron structure. This can be seen in their:
1. High ionization
energies, DE > 0 -- they strongly
resist losing an electron
2. Low affinity
for electrons, DE > 0 -- they
strongly resist gaining an electron
3. Low (or lack of)
chemical reactivity
2. Ionic Bonding
A. Ionic Substances
1. Ionic substances are
usually found as solids, most of which have high melting points.
2. The atoms in ionic substances
are held together with ionic bonds.
B. Ionic bonds
Ionic bonds are found in ionic
compounds due to the electrostatic attractions between the positive(+) and
negative(-) ions.
C. Formula Units
Formulas of ionic compounds
represent the simplest ratio of ions that give an equal number of positive
and negative charges.
RECALL THAT THESE SIMPLE RATIOS OF
IONS ARE CALLED FORMULA UNITS:
NaCl, CaF2, Li2S
represent the formula units of sodium chloride, calcium fluoride and lithium
sulfide.
3. Formation of Binary Ionic Compounds
A. Formation of NaCl(s)
From Its Elements
Formation
of NaCl
Figure 9.4 Page 343 |
| Na(s) + ½Cl2(g) à
NaCl(s) |
DHf = -411
kJ |
| |
enthalpy of
formation of NaCl(s) |
B. The Steps That May Be
Involved
Recall that enthalpy is a state
function. Therefore, the enthalpy of formation, DHf,
of NaCl(s), can be found by adding up the terms 1 through 5 in the
figure below.
THE BORN-HABER CYCLE
We start by evaluating the five
terms:
1. Enthalpy of Formation of
Na(g): The Na(s) is converted to a gas.
Na(s) à
Na(g) DHf
= 108 kJ
enthalpy
of formation
of Na(g)
2. Enthalpy of Formation of
Cl(g): Diatomic Cl2(g) is converted to atomic
Cl(g).
½Cl2(g) à
Cl(g) DHf
= 122 kJ
enthalpy
of formation
of Cl(g)
3. Ionization Energy of Na(g):
The Na+ ion is formed.
Na(g) loses a valence
electron (leaving it with an octet of electrons)
Na(g) à
Na+(g) + e- DH
= 496 kJ
ionization
energy
(I) of Na
4. Electron Affinity of Cl(g):
The Cl- ion is formed.
Cl(g) gains a valence
electron (giving it an octet of electrons)
Cl(g) + e- à
Cl-(g) DH
= -349 kJ
electron
affinity (EA) of Cl
5. Lattice Energy of NaCl(s):
Finally the ions come together to form the compound.
The lattice energy
is the energy change that occurs when gaseous ions form an ionic solid.
|
Na+(g) +
Cl-(g) à NaCl(s)
|
DH = -788 kJ |
| |
lattice energy (U)
of NaCl(s) |
Lattice Energies (in kJ/mol)
for Some Ionic Compounds
Ionic
Compound |
Lattice
Energy |
| LiF |
-1047 |
| LiCl |
-834 |
| NaF |
-923 |
| NaCl |
-788 |
| KF |
-808 |
| KCl |
-701 |
| MgCl2 |
-2326 |
| MgO |
-3916 |
C. Adding These Steps Gives the
DHf of NaCl(s)
| DHf= |
DHf(Na) |
+ DHf(Cl) |
+ I(Na) |
+ EA(Cl) |
+ U(NaCl) |
| DHf= |
108 |
+ 122 |
+ 496 |
+ (-349) |
+ (-788) |
DHf=
-411 kJ <<< The enthalpy
of formation of NaCl(s)!
[Notice that the ionization
energy is highly positive (+496) and would lead to a positive enthalpy of
formation if it were not offset by an even more highly negative lattice
energy (-788 kJ). If the ionization energy is too great a compound can not
be formed!]
Born-Haber
Cycle for MgO and NaF
See Figure 9.5 Page 345 |
D. Properties of Ions
An element exhibits
different properties than its ion:
- Na burns in air and reacts vigorously with water
- Cl2 is corrosive and poisonous
BUT...
... when their Na+
and Cl- ions combine they form NaCl, a substance that is
essential to life and stable in water.
4. Covalent Bonding
A. Molecular
Substances
1. Molecular
substances are found as gases, liquids, or solids with low
melting points.
2. The atoms in
molecular substances are held together with covalent bonds.
B. Covalent Bonds
Covalent bonds form
when both atoms in the bond have the same tendency to gain or lose
electrons. This is found in the bonding between nonmetals.
C. The
"Octet" Rule in Molecules
1. In molecules an
octet can be formed by sharing electrons - the covalent bond.
2. Each atom in a
covalently bonded molecule seeks to achieve the outer-shell noble gas
electron configuration: eight (or sometimes two) outer shell s and p
electrons around the atom.
3. Shared electrons
spend time in the region between the nuclei.
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EXAMPLE:
Notice that in the following cases the electrons
in the bond are counted twice -- once for each atom. (That's why we
call it sharing!)
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5. Bond Polarity and Electronegativity
A. Electronegativity
1. Electrons between
unlike atoms are not shared equally.
2. Electronegativity:
Measure of the attraction of an atom for the electrons in a bond. [Linus
Pauling (1901-1994)]
|
REVIEW
Ionization energy: A MEASURE OF
HOW STRONGLY AN ATOM HOLDS ON TO ONE OF ITS ELECTRONS
Electron affinity:
A MEASURE OF HOW STRONGLY AN ATOM ATTRACTS AN ADDITIONAL ELECTRON
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3. When the ionization
energy of an atom is very high, DE
>> 0, (it doesn’t give up an electron easily)...
... and its electron
affinity is very high, DE << 0,(it
strongly attracts an electron)...
...the atom will be
very electronegative!!
|
EXAMPLE:
The HF Molecule
- Fluorine has a high ionization energy
(1681 kJ/mole) and a high electron affinity (-328 kJ/mole) = high
electronegativity
- Hydrogen has a lower ionization
energy (1311 kJ/mole) and a lower electron affinity (-73 kJ/mole) = lower
electronegativity
- Therefore the electrons between the hydrogen
and the fluorine will be held more strongly by, and closer to the
fluorine.
H :F
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4. Electronegativity
Values
Relative
electronegativity values for the elements can be tabulated by considering
the covalent bond energy of the molecule (HF in the above example) and
those of the two diatomic molecules (H-H, and F-F) of the atoms involved.
(Bond energies will be discussed later in Topic 8.)
Pauling
Electronegativity Values
Similar to Figure 9.8 - Page 350 |
Electronegativity
Values of the Representative Elements
1A
(1) |
2A
(2) |
|
3A
(13) |
4A
(14) |
5A
(15) |
6A
(16) |
7A
(17) |
|
H
2.1
|
|
|
|
|
|
|
|
Li
1.0
|
Be
1.5
|
|
B
2.0
|
C
2.5
|
N
3.0
|
O
3.5
|
F
4.0
|
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Na
0.9
|
Mg
1.2
|
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Al
1.5
|
Si
1.8
|
P
2.1
|
S
2.5
|
Cl
3.0
|
|
K
0.8
|
Ca
1.0
|
|
Ga
1.6
|
Ge
1.8
|
As
2.0
|
Se
2.4
|
Br
2.8
|
|
Rb
0.8
|
Sr
1.0
|
|
In
1.7
|
Sn
1.8
|
Sb
1.9
|
Te
2.1
|
I
2.5
|
|
Cs
0.7
|
Ba
0.9
|
|
Tl
1.8
|
Pb
1.9
|
Bi
1.9
|
Po
2.0
|
At
2.2
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Note that generally
the
ELECTRONEGATIVITY INCREASES GOING TO THE RIGHT AND UP
the periodic table.
The following arrows when drawn on the periodic table show that trend:
|
àààààààààà
á
á
Electronegativity
á
á
|
5. The three most
electronegative atoms in decreasing order are:
F > O > N
B. Electronegativity and Bond
Polarity
1. Bond polarity is a way of
describing the sharing of electrons between atoms. The more unequal the
sharing of electrons in a bond, the greater the polarity of the bond.
2. The molecule HF has a polar
bond; its negative end is attracted to the positive electrode of an electric
field.
3. Bond polarity can be roughly
predicted by taking the difference between the electronegativities of the
atoms involved.
4. The more electronegative atom
in a bond has a partial negative charge (d-).
The other atom has a partial positive charge (d+).
This is shown by drawing an arrow pointing to the negative end with a plus
sign drawn at the positive end (+---->). This difference in polarity is
called a dipole.
A Dipole
+---->
H--Cl
d+ d-
(2.1) (3.0)
Difference in
electronegativity = 3.0 - 2.1 = 0.9
5. In a bond between two atoms,
the greater the difference in their electronegativities the more polar the
bond will be:
-
Small difference (< 0.5) = covalent bond
-
Intermediate difference (0.5 to 2.0) = polar
covalent bond
Covalent
and Polar Covalent Bonds
Figure 9.11 Page 353 |
Large difference (> 2.0) = ionic bond
a. Compare the bonds with
fluorine in the following molecules:
| |
F2
|
OF
|
HF
|
LiF
|
| Electronegativity
difference |
0
|
4.0 - 3.5
= 0.5
|
4.0 - 2.1
= 1.9
|
4.0 - 1.0
= 3.0
|
|
Type of bond
|
Covalent
|
Polar covalent
|
Polar covalent
|
Ionic
|
b. Bond polarities can be
compared using the periodic table.
|
EXAMPLES:
N-F is more polar than O-F
C-O is more polar than N-O
H-O is more polar than H-C
|
|
SAMPLE EXERCISE:
Using ONLY the Periodic Table determine which
of each of the following pairs of diatomic compounds is more polar.
1) B-Cl or C-Cl
2) P-F or P-Cl
3) Se-Cl or S-Br
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|
SAMPLE EXERCISE:
| The
Bonding Continuum-Template |
Place the following diatomic compounds and
elements in order of decreasing polarity (greatest polarity on the
left):
CF, LiF, H2, CN, Cl2,
CO, KF, FrF
| The
Bonding Continuum-Completed |
|
6. Localized Electron Bonding Model
The localized electron (LE) bonding model
is a simple method for describing bonding in molecules. It assumes that
molecules are composed of atoms that are bound together by sharing electrons
using the atoms' atomic orbitals. Electrons in the molecule that are not
involved in bonding are described as unshared electrons. In the LE model both
shared and unshared electrons are depicted as being localized in specific
regions in space. The LE model will be used in the following three instances:
- In order to draw simple Lewis structures of
molecules (done in the next Section).
- In order to predict the shapes of molecules using
VSEPR Theory (done in Topic 9 Molecular Geometry and Bonding Theories)
- In order to describe the types of atomic orbitals
that are used in molecules - Valence Bond Theory (also in Topic 9).
7. Drawing Lewis
Structures
Lewis structures help to describe
bonding. To draw them, we place eight electrons, in pairs, around each atom
involved in a bond (or two electrons for those that can achieve the electron
configuration of helium like H, Li, Be and B).
A. Single Bonds
1. Chlorine Molecule
[each Cl has 7 valence electrons (VE)].
These 14 electrons are placed
around the atoms so that there are 8 around each Cl.
a. A shared pair
of localized electrons between two atoms is called a bond.
b. There are also three unshared
pairs of electrons localized on each chlorine. These are sometimes
called lone pairs
2. Hydrogen Chloride Molecule
[H has 1 VE; Cl has 7 VE].
These 8 electrons are placed around
the atoms so that there are 8 around the Cl and 2 around the H.
3. Hydrogen Molecule
[each H has 1 VE].
These 2 electrons are placed around
the atoms so that there are 2 around each H.
4. The pair of electrons in a
bond can be represented with a single line so that
H:H
can also be written as
H-H
5. When drawing Lewis structures
use the total number of valence electrons in each atom.
6. Molecules with more than two
atoms are treated the same way.
B. Combining Ability
Some common nonmetals typically
form a set number of bonds in neutral compounds when their formal charges
are zero. (Formal charges are described in a section below; until then
formal charges of all atoms that are given will be zero). These are called
the atoms' combining abilities. It will be helpful to know the ones in this
table. (Notice that in every case the Group # plus the Combining Ability add
up to 18, the Group # of the Noble Gases):
Atom
|
Group
# |
Combining Ability
|
Example
|
|
C
|
14 |
4 |
CH4
|
|
N
|
15 |
3 |
NH3
|
|
O
|
16 |
2 |
H2O
|
|
H, F, Cl, Br
|
17 |
1 |
HF
|
|
SAMPLE EXERCISE:
Draw the Lewis structures for the following
molecules:
1) CH4 (C has 4 VE; H has 1 VE)
2) NH3 (N has 5 VE)
(NH3 has one lone pair of
electrons.)
3) CCl4
(Each Cl has three lone pairs of
electrons.)
4) H2O
(H2O has two lone pairs of
electrons.)
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C. Multiple Bonds
Sometimes multiple bonds are
needed in order for eight electrons to be placed around EACH atom in a bond:
1. Double bonds:
|
EXAMPLE:
This is the Lewis structure for C2H4
H H
| |
H-C::C-H
Notice that four electrons are needed between
the two carbon atoms in order for there to be 8 electrons around each.
Since each pair of electrons in a bond can be
represented by a line we can draw two lines between the carbons:
H H
| |
H-C=C-H
This is called a DOUBLE BOND!
|
|
SAMPLE EXERCISE:
Draw the Lewis structures for the following
molecules:
1) H2CO
2) O2
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2. Triple bonds:
|
SAMPLE EXERCISE:
Draw the Lewis structures for the following
molecules (each one has a TRIPLE BOND):
1) C2H2
2) HCN
3) N2
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3. Localized Electrons:
Notice that in Lewis structures the electrons in double and triple bonds are
drawn as being localized between the atoms.
D. Polyatomic Ions
Lewis structures of polyatomic
ions are drawn the same way except that the total number of electrons must be
adjusted for the charge:
- take away one electron for each (+) charge,
- add one electron for each (-) charge.
|
EXAMPLE:
This is the Lewis structure for H3O+
(Oxygen has 6 VE. The three hydrogen atoms have 3 VE.
That's a total of 9 electrons. Now take one away for the
positive charge and that leaves us with 8 electrons around the
oxygen)
H
|
H-O:+
|
H
|
|
SAMPLE EXERCISE:
Draw the Lewis structures for the following
ions:
1) OH-
2) NO2-
3) NH4+
|
Note that a compound like NaOH has
- a covalent bond between O and H, and
- an ionic bond between Na and OH
8. Formal Charge
Formal charges can be used to
predict the arrangement of atoms in a molecule.
A. Finding the Formal Charge
To find the formal charges (FC)
on the atoms in a molecule or ion:
1. Start by drawing a Lewis
structure of the molecule or ion.
2. Assign a formal
charge on each atom using the formula:
FC = # of valence electrons
- (½ shared electrons +
unshared electrons)
|
EXAMPLE:
This is how the formal charges are assigned
to the nitrogen and hydrogen atoms in NH3.
Step 1: Draw the Lewis structure.
The Lewis structure was found above to be:
H
|
H-N:
|
H
Step 2: Use the formula to find the
formal charges.
Nitrogen has 5 valence electrons.
In this molecule it is sharing 6 electrons.
It has 2 unshared electrons.
FCN = 5 - (½6 + 2)
= 0
Hydrogen has 1 valence electron.
In this molecule it is sharing 2 electrons.
It has 0 unshared electrons.
FCH = 1 - (½2 + 0)
= 0
|
|
SAMPLE EXERCISE:
Assign a formal charge to each atom in the
following molecules:
1) H2O
2) CH3NH2O (the
arrangement of the atoms will be given)
|
B. The Sum of the Formal
Charges on All the Atoms
- In a molecule the sum of the FC on all the
atoms must equal 0.
- In an ion the sum of the FC on all the
atoms must equal the charge on the ion.
C. Predicting the Arrangement
of Atoms in a Molecule
In a molecule with three or more
atoms the most stable arrangement of atoms can be predicted by calculating
the formal charges on the atoms in each arrangement. The one that gives the smallest
formal charges is the correct arrangement. The full set of rules
follows:
| Rules for Predicting the
Arrangement of Atoms Using Formal Charges (FC)
1. The arrangement where all FC are 0 is
preferred.
2. If no arrangement has all of its FC equal
to 0 then pick the one with the smallest FC's:
a. The one with the lowest number of FC not
equal to 0.
b. The one with one large FC rather than
many small FC's.
3. The FC's on adjacent atoms must have
opposite signs (or be 0).
4. If everything else is equal, pick the
arrangement where the (-) charge is on the most electronegative atom.
|
|
EXAMPLE:
What is the arrangement of the atoms in the
thiocyanate ion? Is it CNS-, CSN-, or NCS-?
Step 1: Draw the Lewis structures for
each of the three possibilities:
CNS-
CSN-
NCS-
Step 2: Determine the formal charges on
the atoms for each possibility:
CNS-
CSN-
NCS-
Step 3: Select the correct arrangement
using the rules given above.
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9. Resonance Structures
These are two correct Lewis
structures for the nitrite ion (NO2-):
RESONANCE STRUCTURES
The first one shows a double bond
between the N and the O below it and a single bond between the N and the O to
its right. The second structure has the double and single bonds switched.
Which one is the actual structure?
The answer is that the actual
structure is the average of these two resonance structures which
are drawn with a double-headed arrow between them. The average structure is
called the resonance hybrid and can be represented as follows:
RESONANCE HYBRID
This new representation of the
bonds is meant to indicate that the actual bond in this structure is part way
between a single bond and a double bond. Also notice that unshared electrons
can no longer be placed around individual atoms. Experiments have shown that
the N-O bonds in NO2- are identical.
Whenever more than one valid Lewis
structure can be drawn for a molecule or an ion a resonance hybrid of the
structure exists.
The necesity of drawing resonance
structures for some molecules and ions is a shortcoming in the localized
electron model. For example, in the NO2- case, the
resonance hybrid shows that a single bond exists between the N and each O and
that another two electrons are spread out over all three atoms. In Topic 9
you'll see how a Delocalized Electron model (Molecular Orbital
Theory) solves this problem!
|
SAMPLE EXERCISE:
SO32- and SO3
are composed of the same atoms. SO32- is an ion
that has two more electrons than SO3. Will the S-O bond
lengths in the two be the same? If not, which will have the shorter
S-O bond lengths?
1) Draw the Lewis structure of SO32-.
If there are other resonance structures draw them and the resonance
hybrid.
There is only one Lewis structure and no other
resonance structures in which sulfur has eight electrons. There's a
single bond between the sulfur and each of the oxygens. Experiment
shows that these bonds are all equal to 151 pm in length.
2) Draw the Lewis structure of SO3.
If there are other resonance structures draw them and the resonance
hybrid.
There are three resonance structures.
The resonance hybrid shows a molecule with
S-O bonds that are equal to each other. We expect their lengths to
be between that of a single bond and a double bond (shorter than a
single bond). Experiment shows that all three bonds are 142 pm in
length, shorter than the single bonds in SO3
|
10. Exceptions to the Octet Rule
A. More Than Eight Electrons
Around an Atom
Sometimes MORE than 8
electrons may be found around atoms that are beyond the second row of the
periodic table. In these cases 10 or 12 electrons are present.
|
EXAMPLES:
|
In these cases the central atom
also uses electrons in one or two d orbitals.
|
SAMPLE EXERCISE:
Draw the Lewis structure for ICl4-
|
B. Less Than Eight Electrons
Around an Atom
Sometimes LESS than 8
electrons can be found around atoms that have too few valence electrons.
|
EXAMPLE:
In BH3 the boron atom bonds to 3
hydrogen atoms. Since boron has only three electrons to share, it can
have only 6 electrons around it after each of the three hydrogens
shares one electron with it.
|
C. Odd Number of Electrons
Around an Atom
Some molecules and ions have an
odd number of valence electrons like NO (11 valence electrons) and NO2 (17
valence electrons). Obviously all of the electrons in these molecules (called
free radicals) can not be paired! Only a few common molecules fit this
description.
11. Strengths of Covalent Bonds
A. Energy is Needed to Break a
Bond.
DX-Y is the bond
enthalpy (or bond dissociation energy) of a molecule
XY.
XY(g) à
X(g) + Y(g) DX-Y = DHo298
|
EXAMPLE:
H2(g) à
2 H(g) DH-H = 2DHof[H(g)]
= 436 kJ
NOTE the enthalpy
of formation of H(g):
½ H2(g)
à H·(g) DHof
= 218 kJ/mol
|
B. Average Bond Enthalpies
| Here's a table of Average
Bond Enthalpies that gives the enthalpy needed to break
various bonds. The enthalpy gained when a bond is formed is MINUS
the value shown. |
C. Estimating the Enthalpy of
Reaction from Bond Enthalpies
The enthalpy of a reaction, DHorxn,
can be estimated by considering the enthalpy needed to break the bonds in
the reactants, and the enthalpy gained by forming new bonds in the products:
|
EXAMPLE:
Estimate the DHorxn
for the following reaction:
H2(g) + Cl2(g)
à 2HCl(g) DHorxn
= ?
Step 1: Find the bond enthalpies of the
reactants and the products from the table remembering that when bonds
are formed the values must be subtracted.
break these bonds form
these bonds
H-H + Cl-Cl à
H-Cl + H-Cl
DX-Y = 436 kJ
243 kJ
-432 kJ -432 kJ
Step 2: Add up the bond enthalpies.
+679 kJ needed to break the bonds
-864 kJ gained when bonds are formed
DHorxn
= +679 kJ -864 kJ = -185 kJ
Therefore:
H2(g) + Cl2(g) à
2HCl(g) DHorxn
= -185 kJ
NOTE: This is twice
the DHof of HCl
2 x (-92.307 kJ) =
-184.614 kJ
|
|
SAMPLE EXERCISE:
Estimate the DHorxn
for the following combustion reaction:
2H3C-CH3(g) + 7O2(g)
à 4CO2(g) + 6H2O(g)
Step 1: Rewrite the equation to show all
of the bonds (the Lewis structures of O2 and CO2
show the double bonds).
H H
| |
2H-C-C-H(g) + 7O=O(g) à
4O=C=O(g) + 6H-O-H(g)
| |
H H
Step 2: Find the enthalpy needed to
break all of the bonds.
12 C-H bonds =
12 X
=
2 C-C
bonds = 2 X
=
7 O=O
bonds = 7 X
=
TOTAL ENTHALPY NEEDED =
Step 3: Find the enthalpy gained when
the new bonds are formed.
8 C=O
bonds = 8 X
=
12 H-O bonds =
12 X
=
TOTAL ENTHALPY GAINED =
Step 4: Add these up.
DHorxn
= Enthalpy Needed + Enthalpy Gained
(DHorxn
= -2340 kJ)
|
D. Multiple Bonds, Bond
Enthalpy and Bond Length
As the number of bonds between two
atoms increases, the bond strength increases and the bond length decreases.
HW
