Group
1A
2A
3A
4A
5A
6A
7A e configuration ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5
# of valence e -
1
2
3
4
5
6
7

Lewis Dot Symbols for the Representative Elements &
Noble Gases

What are the correct Lewis dot symbols of
(A)F,F , (B)S,S 2, (C) O,O 2, (D) N,N 3?

Chemical Bonds
Two general types:
• Ionic bonds---- between ions of opposite charges
• Covalent bonds---- sharing of electrons
Li + F Li + F
1s 2 2s 1
1s 2 2s 2 2p 5 1s 2
1s 2 2s 2 2p 6
[He] [Ne]
Li e
-
+ F
Li + + e
-
F
Li + + F
Li + F

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]3s 1 loses 1 e----> Na + [Ne]
Cl [Ne]3s 2 3p 5 gains 1 e- --> Cl [Ar]

Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
E lattice
LiF(s) Li + (g) + F (g)
E = k
Q
+ r
Q
-
Q
+ is the charge on the cation
Q
is the charge on the anion r is the distance between the ions cmpd lattice energy
MgF
MgO
2
2957 Q= +2,-1
3938 Q= +2,-2
LiF
LiCl
1036
853 r F < r Cl -

Which of the following is a correct order of lattice energy?
(a)NaCl < AlCl
3
< MgCl
2
(c) Na
2
O < MgO < Al
2
O
3
(e) LiCl < NaCl < KCl
(b) LiF < LiCl < LiBr
(d) Ga
2
O
3
< CaO < K
2
O
(c)
Lattice Energy
• E lattice
• E lattice increases as Charge on ions increases 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?
lone pairs
F
7e -
+ F
7e -
F F
8e 8e -
Lewis structure of F
2 single covalent bond lone pairs F F single covalent bond
F F lone pairs lone pairs

Lewis structure of water
H + O + H H O H
2e 8e 2e single covalent bonds or H O H
Double bond – two atoms share two pairs of electrons
O C O or O C O double bonds 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

Bond Lengths
Triple bond < Double Bond < Single Bond
Bond strength: triple > double> single

Bond polarity
Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms e poor e rich electron poor region electron rich region
H F
H F d
+ 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

X
( g)
+ e X
-
( g)
• 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

Which of the following is a polar covalent bond?
(a)The HN bond in NH
3
.
(b) The SiSi bond in Cl
3
SiSiCl
3
.
(c) The CaF bond in CaF
2
.
(d) The OO bond in O
2
.
(e) The NN bond in H
2
NNH
2
.
(a)
(b) nonpolar
(c) ionic
(d) nonpolar
(e) nonpolar

The Electronegativities of Common Elements

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
0 < and <2

2
Non polar Covalent
Polar Covalent
Ionic
Non polar
Covalent share e -
H-H
Increasing difference in electronegativity
Polar Covalent partial transfer of e -
H-Cl
Ionic transfer e -
NaCl

Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H
2
S; and the NN bond in H
2
NNH
2
.
Cs – 0.7
H – 2.1
N – 3.0
Cl – 3.0
S – 2.5
N – 3.0
3.0 – 0.7 = 2.3
Ionic
2.5 – 2.1 = 0.4
Polar Covalent
3.0 – 3.0 = 0 Non polar Covalent

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.

Write the Lewis structure of nitrogen trifluoride (NF
3
).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 )
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 the number of e valence e ?
in structure equal to number of
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
F N F
F

Write the Lewis structure of the carbonate ion (CO
3
2).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 )
-2 charge – 2e -
4 + (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

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 of valence electrons in the free atom
total number of nonbonding electrons
-
FC = [ Val. e ] -[(nonbonded e ) + 1/2(bonded e )]
1
2
( total number of bonding
) electrons
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

Write the free charge for:
FC = [ Val. e ] -[(nonbonded e ) + 1/2(bonded e )]
C: FC = 4 – 0 – ½ (4 + 2 + 2) = 0
O in C=O: FC = 6 – 4 – ½(4) = 0
O in C-O: FC = 6 – 6 – ½(2) = -1

Resonance Structures
• 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: C
6
H
6

What are the resonance structures of the carbonate (CO
3
2 -) ion?
-
O C O
-
O C O
-
O C
O O
-
Which of the following has resonance structure?
(a)NH
4
+
(d) C
6
H
6
(b) CO
(e) H
2
O
2
(c) Al I
3
O
-
O
(d) C
6
H
6

The Incomplete Octet (too few electrons on central atom)
BeH
2
Be – 2e -
2H – 2x1e -
4e -
H Be H
BF
3
B – 3e -
3F – 3x7e -
24e -
F B F
F
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24

Odd-Electron Molecules
Result: unpaired electron
NO
N – 5e -
O – 6e -
11e -
N O
The Expanded Octet (too many e- on central atom; central atom with principal quantum number n > 2)
SF
6
S – 6e -
6F – 42e -
48e -
F
F
S
F
F
F
F
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48

Which of the following is correct concerning the lone pairs on the underlined atoms in compounds?
(a)I Cl, 3 (b) H
2
S, 6 (c) CH
4
, 4
(d) CaH
2
, 2 (e) SCl
2
, 4
(a)
Which of the following obeys the octet rule?
(a)BF
(d) PF
3
5
(b) SF
4
(e) NO
3
-
(c) NO
F B F
F
(e)

Bond energy
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy .
H
2 ( g )
H
( g )
+ H
( g )
Bond Energy
D
H 0 = 436.4 kJ
Cl
2 ( g )
Cl
( g )
+ Cl
( g )
D
H 0 = 242.7 kJ
HCl
O
( g )
2 ( g )
H
( g )
+ Cl
( g )
D
H 0 = 431.9 kJ
O
( g )
+ O
( g )
D
H 0 = 498.7 kJ O O
N
2 ( g )
N
( g )
+ N
( g )
D
H 0 = 941.4 kJ N N
Bond Energies
Single bond < Double bond < Triple bond

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.
D
H 0 = total energy input – total energy released
=
S
BE(reactants) – S
BE(products)
Endothermic reaction
Exothermic reaction

Use bond energies to calculate the enthalpy change for:
H
2 ( g )
+ F
2 ( g )
2HF
( g )
D
H 0 =
S
BE(reactants) – S
BE(products)
Type of bonds broken
H H
F F
Type of bonds formed
H F
Number of bonds broken
1
1
Number of bonds formed
2
Bond energy
(kJ/mol)
436.4
156.9
Bond energy
(kJ/mol)
568.2
D
H 0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ
Energy change (kJ)
436.4
156.9
Energy change (kJ)
1136.4