Chemical Bonding I:
Basic Concepts
Chapter 9
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Valence electrons are the outer shell electrons of an
atom. The valence electrons are the electrons that
participate in chemical bonding.
Group
e- configuration
# of valence e-
1A
ns1
1
2A
ns2
2
3A
ns2np1
3
4A
ns2np2
4
5A
ns2np3
5
6A
ns2np4
6
7A
ns2np5
7
9.1
Lewis Dot Symbols for the Representative Elements &
Noble Gases
9.1
What are the correct Lewis dot symbols of
(A)F,F-, (B)S,S2-, (C) O,O2-, (D) N,N3-?
Group practice problem of chap9:Q1
Chemical Bonds
Two general types:
• Ionic bonds---- between ions of opposite charges
• Covalent bonds---- sharing of electrons
The Ionic Bond
Li+ F -
Li + F
1s22s1 1s22s22p5 1s2 1s22s22p6
[He] [Ne]
e- +
Li+ +
Li
Li+ + e-
F
F -
F -
Li+ F 9.2
The Octet Rule
Atoms tend to gain, lose, or share electrons until they
are surrounded by 8 valence electrons.
An octet = 8 electrons
= 4 pairs of electrons
= noble gas arrangement
= very stable
Exception: H = 2 electrons only
Example:
Na [Ne]3s1 loses 1 e----> Na+ [Ne]
Cl [Ne]3s23p5 gains 1 e- --> Cl- [Ar]
Electrostatic (Lattice) Energy
Lattice energy (E) is the energy required to completely separate
one mole of a solid ionic compound into gaseous ions.
Elattice
LiF(s)
Li+(g) + F-(g)
Q+Q E=k
r
Q+ is the charge on the cation
Q- is the charge on the anion
r is the distance between the ions
Lattice Energy
• Elattice increases as Charge
on ions increases
• Elattice decreases as inter-ionic
distances increase
(charge is much more important than distance)
cmpd
MgF2
MgO
LiF
LiCl
lattice energy
2957 Q= +2,-1
3938 Q= +2,-2
1036
r F- < r Cl853
9.3
Which of the following is a correct order of lattice energy?
(a)NaCl < AlCl3 < MgCl2
(b) LiF < LiCl < LiBr
(c) Na2O < MgO < Al2O3
(d) Ga2O3 < CaO < K2O
(e) LiCl < NaCl < KCl
Lattice Energy
• Elattice increases as Charge on ions increases
• Elattice decreases as inter-ionic distances increase
(charge is much more important than distance)
A covalent bond is a chemical bond in which two or more
electrons are shared by two atoms.
Why should two atoms share electrons?
F
+
7e-
F
F F
7e-
8e- 8e-
Lewis structure of F2
single covalent bond
lone pairs
F
F
lone pairs
single covalent bond
lone pairs
F F
lone pairs
9.4
Lewis structure of water
H
+
O +
H
single covalent bonds
H O H
or
H
O
H
2e-8e-2eDouble bond – two atoms share two pairs of electrons
O C O
double bonds
or
O
O
C
double bonds
8e- 8e- 8e-
Triple bond – two atoms share three pairs of electrons
triple bond
N N
8e-8e-
or
N
N
triple bond
9.4
Lengths of Covalent Bonds
Bond Lengths
Triple bond < Double Bond < Single Bond
Bond strength:
triple > double> single
9.4
Bond polarity
Polar covalent bond or polar bond is a covalent bond with
greater electron density around one of the two atoms
electron poor
region
H
electron rich
region
F
e- poor
H
d+
e- rich
F
d-
Electron density is pulled more tightly
around the F end of the molecule.
Nonpolar bond: two identical atoms form a bond & equally
share the bond’s electron pair
9.5
Electronegativity is the ability of an atom to attract
toward itself the electrons in a chemical bond.
Electron Affinity - measurable
X (g) + e-
X-(g)
Electronegativity - relative, F is highest
• the term used to describe the relative
attraction of an atom for the electrons in a
bond
• Atom of the bond with the greater
electronegativity will carry the partial
negative charge
9.5
Which of the following is a polar covalent bond?
(a)The HN bond in NH3.
(b) The SiSi bond in Cl3SiSiCl3.
(c) The CaF bond in CaF2.
(d) The OO bond in O2.
(e) The NN bond in H2NNH2.
Practice problem of chap9:Q4
The Electronegativities of Common Elements
9.5
Electronegativity:
• The most electronegative element: F (4.0)
• Next most electronegative element:
O (3.5)
• Next two most electronegative:
N (3.0) and Cl (3.0)
• Least electronegative element:
Cs (0.7)
Also
H (2.1)
C (2.5)
Na (0.9)
Classification of bonds by difference in electronegativity
Difference
Bond Type
0
Non polar Covalent
0 < and <2
Polar Covalent
2
Ionic
Increasing difference in electronegativity
Non polar
Covalent
share e-
partial transfer of e-
H-H
H-Cl
Polar Covalent
Ionic
transfer eNaCl 9.5
Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Non polar Covalent
Practice problem of chap9:Q11
9.5
Writing Lewis Structures
1. Draw skeletal structure of compound showing what
atoms are bonded to each other. Put least
electronegative element in the center. Arrange atoms &
connect with single bonds
– Single bond = 2 electrons
2. Count total number of valence e-. Add 1 for each
negative charge. Subtract 1 for each positive charge.
3. Complete an octet for atoms bonded to the central atom
except hydrogen (H=2). Then add leftover electrons to
central atom. Electrons belonging to the central or
surrounding atoms must be shown as lone pairs if they
are not involved in bonding.
4. If the central atom has fewer than 8 electrons, try adding
double and triple bonds on central atom as needed.
9.6
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
F
N
F
F
9.6
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too few electrons, form double bond and re-check # of e-
O
C
O
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24
O
9.6
An atom’s formal charge is the difference between the
number of valence electrons in an isolated atom and the
number of electrons assigned to that atom in a Lewis
structure.
formal charge
on an atom in
a Lewis
structure
=
total number
total number
of valence
of nonbonding
electrons in electrons
the free atom
-
1
2
(
total number
of bonding
electrons
)
FC = [ Val. e-] -[(nonbonded e-) + 1/2(bonded e-)]
The sum of the formal charges of the atoms in a molecule
or ion must equal the charge on the molecule or ion.
• Remember: formal charges are NOT real charges
9.7
Practice
problem of
chap9:Q12
Resonance Structures
• Why necessary?
• Lewis structures are not adequate
• More than one Lewis Structure is needed to represent the real
molecule
A resonance structure is one of two or more Lewis structures
for a single molecule that cannot be represented accurately by
only one Lewis structure.
Example: experimentally, ozone has two identical bonds whereas
the Lewis Structure requires one single (longer) and one double
bond (shorter).
O
O
+
O
O
+
O
O
Resonance Structures:
• connect different resonance structures with <-->
• actual molecule is not depicted by resonance structures
• actual molecule does not go between resonance structures
• Molecules that can be depicted by resonance structures are
more stable than expected
Resonance in Benzene: C6H6
9.8
What are the resonance structures of the
carbonate (CO32-) ion?
-
O
C
O
O
-
O
C
O
O
-
-
O
Which of the following has resonance structure?
(a)NH4+
(b) CO2
(c) AlI3
(d) C6H6
(e) H2O
C
O
O
-
Practice problem of chap9:Q12, 7
9.8
Exceptions to the Octet Rule
The Incomplete Octet (too few electrons on central atom)
BeH2
BF3
B – 3e3F – 3x7e24e-
Be – 2e2H – 2x1e4e-
F
B
H
F
Be
H
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
F
9.9
Exceptions to the Octet Rule
Odd-Electron Molecules
Result: unpaired electron
N – 5eNO
O – 6eN
11e-
O
The Expanded Octet (too many e- on central atom; central atom with
principal quantum number n > 2)
SF6
S – 6e6F – 42e48e-
F
F
F
S
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
9.9
Which of the following is correct concerning the lone
pairs on the underlined atoms in compounds?
(a)ICl, 3
(b) H2S, 6
(c) CH4, 4
(d) CaH2, 2 (e) SCl2, 4
Which of the following obeys the octet rule?
(a)BF3
(b) SF4
(c) NO
(d) PF5
(e) NO3-
F
B
F
F
Practice problem of chap9:Q3,9
Bond energy
The enthalpy change required to break a particular bond in
one mole of gaseous molecules is the bond energy.
Bond Energy
H2 (g)
H (g) + H (g) DH0 = 436.4 kJ
Cl2 (g)
Cl (g) + Cl (g) DH0 = 242.7 kJ
HCl (g)
H (g) + Cl (g) DH0 = 431.9 kJ
O2 (g)
O (g) + O (g) DH0 = 498.7 kJ
O
O
N2 (g)
N (g) + N (g) DH0 = 941.4 kJ
N
N
Bond Energies
Single bond < Double bond < Triple bond
9.10
Bond Energies (BE) and Enthalpy changes in reactions
Imagine reaction proceeding by breaking all bonds in the
reactants and then using the gaseous atoms to form all the
bonds in the products.
DH0 = total energy input – total energy released
= SBE(reactants) – SBE(products)
Endothermic
reaction
Exothermic
reaction
9.10
Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g)
2HF (g)
DH0 = SBE(reactants) – SBE(products)
Type of
bonds broken
H
H
F
F
Type of
bonds formed
H
F
Number of
bonds broken
Bond energy
(kJ/mol)
Energy
change (kJ)
1
1
436.4
156.9
436.4
156.9
Number of
bonds formed
Bond energy
(kJ/mol)
Energy
change (kJ)
2
568.2
1136.4
DH0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
Practice problem of chap9:Q9
9.10