Brad Collins, DTCC
Chemical Structures
(Part 1)
Chapter 10
Some images Copyright © The McGraw-Hill Companies, Inc.
Covalent Bonds
• Covalent bonds involve the sharing of electrons between two
atoms.
• Bonds form through overlap or combining of orbitals.
• Bond length is the distance between the 2 nuclei of atoms
sharing electrons.
• Atoms give, take, or share electrons to achieve a stable
electron arrangement - octet rule
• Metals donate electrons to non-metals - ionic
• Non-metals share electrons in bonds - covalent
10.1
Properties of Covalent Bonds
• Two types of attractive forces in covalent compounds:
• Bond energy - holds atoms together in the molecule
• Intermolecular Attraction - between molecules, weak
Covalent compounds tend to be gases, liquids or lowmelting solids.
Most covalent compounds are insoluble in water. The
ones that do dissolve, are nonelectrolytes.
Molten covalent compounds do not conduct electricity.
10.1
Review
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
10.1
9.1
Lewis Structures
• Structural formulas give
information about the structure of
a molecule.
• Lewis structures are a type of
structural formula.
• Lewis structures represent
valence electrons as dots
10.1
10.1
Lewis Structures
Terminology
10.1
Lewis Structures
Terminology
Lone pairs
Bonding pairs
Lone pairs
Bonding pairs
10.1
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
lone pairs
F
F
lone pairs
single covalent bond
10.1
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
or
O
O
C
double bonds
double
8e- 8e-bonds
8e-
Triple bond – two atoms share three pairs of electrons
N N
8e- bond
triple
8e-
or
N
N
triple bond
10.1
Bond Order
• Describes the bonding between 2 atoms
• First Order - Single bond
• Second Order - Double bond
• Third Order - Triple bond
10.1
Lengths of Covalent Bonds
Bond Length depends on ion size and bond type.
Bond
Type
Bond
Length
(pm)
C-C
154
C=C
133
C≡C
120
C-N
143
C=N
138
C≡N
116
Bond Lengths
Triple bond < Double Bond < Single Bond
10.1
Writing Lewis Structures
For an inorganic compound of type AXn where A is a
central atom bonded to n number of outer atoms (X).
1. Count total number of valence e-. Add 1 for each negative charge.
Subtract 1 for each positive charge.
2. Bond all of the outer atoms to the central atom using single bonds.
Put least electronegative element in the center.
3. Complete an octet for all atoms except hydrogen starting with the
outside atoms and leaving the central atom for last. Add electrons as
lone pairs.
4. If the central atom still needs 8 electrons, remove lone pair electrons
from an outer atom and replace its single bond with a double bond.
If the central atom still doesn’t have 8 electrons, form another double
bond with another outer atom, or form a triple bond.
10.2
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 F atoms. If electrons remain complete an octet
for N atom.
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
10.2
Resonance Structures
• Resonance structures arise when a molecule
can have more than one valid Lewis structure.
• Examples: ozone, nitrate ion, carbonate ion
O
O
+
O
-
-
O
+
O
O
10.3
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 O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 9 lone pairs (9x2) = 24 valence electrons
Step 5 - Too few electrons to make octet for C so, form double bond
and re-check # of e2 single bonds (2x2) = 4
1 double bond = 4
O
C
O
8 lone pairs (8x2) =
Total =
O
10.3
Major Structures
• In some cases, the resonance structures look
different from each other. Example: Thiocyanate
• In these cases we must use other concepts to
designate a ‘major (favored) structure’
• Electronegativity
• Formal charge
10.3
Electronegativity is the ability of an atom to attract
toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e-
X-(g)
Electronegativity - relative, F is highest
10.3
10.3
Electronegativity increases from left to right across a period as metallic character
decreases.
Within a group electronegativity decreases as atomic number (and metallic
character) increases.
10.3
Transition metals do not follow these trends.
Two possible skeletal structures of formaldehyde (CH2O)
H
C
O
H
H
C
H
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
-
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.
10.3
H
-1
+1
C
O
formal charge
on an atom in
a Lewis
structure
H
=
C – 4 eO – 6 e2H – 2x1 e12 e-
total number
of valence
electrons in
the free atom
-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
of nonbonding
electrons
formal charge
on C
= 4 -2 - ½ x 6 = -1
formal charge
on O
= 6 -2 - ½ x 6 = +1
-
1
2
(
)
total number
of bonding
electrons
10.3
H
H
0
C
formal charge
on an atom in
a Lewis
structure
0
O
=
C – 4 eO – 6 e2H – 2x1 e12 etotal number
of valence
electrons in
the free atom
-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
total number
of nonbonding
electrons
formal charge
on C
= 4 - 0 -½ x 8 = 0
formal charge
on O
= 6 -4 - ½ x 4 = 0
-
1
2
(
)
total number
of bonding
electrons
10.3
Determining the Major Structure
1. For neutral molecules, a Lewis structure in which there are no formal
charges is preferable to one in which formal charges are present.
2. Lewis structures with large formal charges are less plausible than
those with small formal charges.
3. Among Lewis structures having similar distributions of formal charges,
the most plausible structure is the one in which negative formal
charges are placed on the more electronegative atoms.
Which is the most likely Lewis structure for CH2O?
H
-1
+1
C
O
H
H
H
0
C
0
O
10.3
Exceptions to the Octet Rule
The Incomplete Octet
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
10.4
Exceptions to the Octet Rule
Odd-Electron Molecules
NO
N – 5eO – 6e11e-
N
O
The Expanded Octet (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
1 e– in 3s, 3 e– in 3p and 2
e– in 3d orbitals
10.4
10.4
10.4
10.4