Localized Electron Model LE Model

1
Localized Electron Model(LE Model)
Lewis Structures
VSEPR Model
Hybridization
2
Lewis Structures
A Lewis Structure represents the locations of
bonds and unshared pairs of electrons in a
covalently bonded system of atoms (a molecule or
polyatomic ion).
All of the valence electrons (and only the
valence electrons) of all the atoms present are
included in the Lewis structure.
3
Steps for Writing Lewis Structures
Step One Count the total number of valence
electrons in the system.
4
Step 1
Example 1 Chloroform CHCl3
Number of valence electrons in each carbon
4
Number of valence electrons in each hydrogen
1
Number of valence electrons in each chlorine
7
Number of valence electrons for all chlorines
7 3 21
Total of valence electrons in the molecule
4 1 21 26
5
Step 1
Example 2 Nitrogen Triiodide NI3
Number of valence electrons in each nitrogen
5
Number of valence electrons in each iodine
7
Number of valence electrons for all iodines
7 3 21
Total of valence electrons in the molecule
5 21 26
6
Step 1
Example 3 Formaldehyde H2CO
Number of valence electrons in each oxygen
6
Number of valence electrons in each carbon
4
Number of valence electrons in each hydrogen
1
Number of valence electrons for all hydrogens
1 2 2
Total of valence electrons in the molecule
6 4 2 12
7
Steps for Writing Lewis Structures
Step One Count the total number of valence
electrons in the system.
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
The central atom, the one to which all the others
are bonded, is usually the first atom (other than
hydrogen) in the formula.
For the time being, when this statement is not
true, you will be told which atom is the central
atom.
A covalent single bond is represented by a dash.
Each single bond is composed of a pair of
electrons.
8
Step 2
Example 1 Chloroform CHCl3
The central atom is carbon. Bonds are drawn to
each of the other atoms present.
H

_
_
C
Cl
Cl

Cl
Subtract the electrons used from the total.
Total of valence electrons in the molecule
26
Total of electrons in 4 bonds
8
Total of valence electrons remaining unused
26 8 18
9
Step 2
Example 2 Nitrogen Triiodide NI3
The central atom is nitrogen. Bonds are drawn to
each of the other atoms present.
_
_
N
I
I

I
Subtract the electrons used from the total.
Total of valence electrons in the molecule
26
Total of electrons in 3 bonds
6
Total of valence electrons remaining unused
26 6 20
10
Step 2
Example 3 Formaldehyde H2CO
The central atom is carbon. Bonds are drawn to
each of the other atoms present.
_
_
C
H
H

O
Subtract the electrons used from the total.
Total of valence electrons in the molecule
12
Total of electrons in 3 bonds
6
Total of valence electrons remaining unused
12 6 6
11
Steps for Writing Lewis Structures
Step One Count the total number of valence
electrons in the system.
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Since the single bond represents two electrons,
each attached atom will need six more electrons
(three pairs).
The exception is hydrogen, which can only have
two electrons in its valence shell. No more
electrons will be added to a hydrogen.
12
Step 3
Example 1 Chloroform CHCl3
Add 3 pairs of electrons to each chlorine. None
are added to hydrogen.
H

..
..
_
_


C
Cl
Cl
..
..



Cl
..
Valence electrons remaining before this step
18
Total of valence electrons used in the step
18
Total of valence electrons remaining unused
18 18 0
13
Step 3
Example 2 Nitrogen Triiodide NI3
Add 3 pairs of electrons to each iodine.
..
..
_
_


N
I
I
..
..



I
..
Valence electrons remaining before this step
20
Total of valence electrons used in the step
18
Total of valence electrons remaining unused
20 18 2
14
Step 3
Example 3 Formaldehyde H2CO
Add 3 pairs of electrons to the oxygen. None are
added to hydrogens.
_
_
C
H
H



O
..
Valence electrons remaining before this step
6
Total of valence electrons used in the step
6
Total of valence electrons remaining unused
6 6 0
15
Steps for Writing Lewis Structures
Step One Count the total number of valence
electrons in the system.
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Step Four Place any remaining valence electrons
on the central atom as unshared pairs.
16
Step 4
Example 1 Chloroform CHCl3
Total of valence electrons remaining unused
0
H

..
..
_
_


C
Cl
Cl
..
..



Cl
..
No valence electrons remain unused in this
molecule. Proceed to step 5.
17
Step 4
Example 2 Nitrogen Triiodide NI3
Total of valence electrons remaining unused
2
The two electrons are placed as a pair on the
central atom.
..
..
..
_
_


N
I
I
..
..



I
..
18
Step 4
Example 3 Formaldehyde H2CO
Total of valence electrons remaining unused
0
_
_
C
H
H



O
..
No valence electrons remain unused in this
molecule. Proceed to step 5.
19
Steps for Writing Lewis Structures
Step One Count the total number of valence
electrons in the system.
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Step Four Place any remaining valence electrons
on the central atom as unshared pairs.
Step Five Count the number of valence electrons
on the central atom. If it is less than 8, form a
double or triple bond to complete the octet.
20
Step 5
Example 1 Chloroform CHCl3
The central atom, carbon, has 4 single bonds.
H

..
..
_
_


C
Cl
Cl
..
..



Cl
..
Since each is composed of two electrons, the
carbon has 8 valence electrons. This is a
complete Lewis structure.
21
Step 5
Example 2 Nitrogen Triiodide NI3
The central atom, nitrogen, has 3 single bonds
and one unshared pair.
..
..
..
_
_


N
I
I
..
..



I
..
Three bonds each containing two electrons plus
one unshared pair 8 electrons. This is a
complete Lewis structure.
22
Step 5
Example 3 Formaldehyde H2CO
The central atom, carbon, has 3 single bonds.
_
_
C
H
H



O
..
Each bond is composed of two electrons, and there
are no unshared pairs, so the carbon has only 6
valence electronsnot a complete octet. This is
not a complete Lewis structure.
This is not a complete Lewis structure.
23
Step 5
Example 3 (continued) Formaldehyde H2CO
So, a double bond is created between the carbon
and the oxygen by removing one of the oxygen
electron pairs and using it to form a bond.
_
_
_
_
C
H
H
C
H
H







O
O
..
..
Each single bond is composed of two electrons,
and the double bond is composed of two pairs, so
the carbon now has a complete octet.
This is a complete Lewis structure.
24
Special Note for Step 5
Remember that only a few atoms will form double
and triple bonds.
Double bonds can be formed by carbon (C),
nitrogen (N), oxygen (O), phosphorus (P), and
sulfur (S).
Triple bonds can be formed by carbon, nitrogen,
and oxygen.
25
Steps for Writing Lewis Structures
Step One Count the total number of valence
electrons in the system.
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Step Four Place any remaining valence electrons
on the central atom as unshared pairs.
Step Five Count the number of valence electrons
on the central atom. If it is less than 8, form a
double or triple bond to complete the octet.
26
Exceptions to the Octet Rule
Exception 1 Two elements are stable as central
atoms with fewer than eight valence electrons
beryllium (Be), which is stable with 4 electrons
(two single bonds), and boron (B), which is
stable with 6 (three single bonds). Memorize
these.
Exception 2 Atoms of elements in the third period
and higher can exceed the octet rule. There has
to be an unoccupied d sublevel in the valence
shell. Following the steps for drawing the Lewis
structures will automatically produce the correct
arrangement.
27
Steps for Writing Lewis Structures
Two more examples
28
Sulfur difluoride SF2
Step One Count the total number of valence
electrons in the system.
SF2
Number of valence electrons in sulfur
6
Number of valence electrons in each fluorine
7
Number of valence electrons for all fluorines
7 2 14
Total of valence electrons in the molecule
6 14 20
29
Sulfur difluoride SF2
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
The central atom is sulfur.
_
_
S
F
F
Total of valence electrons in the molecule
20
Total of electrons in 2 bonds
4
Total of valence electrons remaining unused
20 4 16
30
Sulfur difluoride SF2
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Add 3 pairs of electrons to each fluorine.
..
..
_
_


S
F
F
..
..
Valence electrons remaining before this step
16
Total of valence electrons used in the step
12
Total of valence electrons remaining unused
16 12 4
31
Sulfur difluoride SF2
Step Four Place any remaining valence electrons
on the central atom as unshared pairs.
Total of valence electrons remaining unused
4
The four electrons are placed as two pairs on the
central atom.
..
..
..
_
_


S
F
F
..
..
..
32
Sulfur difluoride SF2
Step Five Count the number of valence electrons
on the central atom. If it is less than 8, form a
double or triple bond to complete the octet.
The central atom, sulfur, has 2 single bonds
and two unshared pairs.
..
..
..
_
_


S
F
F
..
..
..
The central atom does have a complete octet, so
the Lewis structure is complete.
33
Thionyl Chloride SOCl2
Step One Count the total number of valence
electrons in the system.
SOCl2
Number of valence electrons in sulfur
6
Number of valence electrons in oxygen
6
Number of valence electrons in each chlorine
7
Number of valence electrons for all chlorines
7 2 14
Total of valence electrons in the molecule
6 6 14 26
34
Thionyl Chloride SOCl2
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
The central atom is sulfur.
_
_
S
Cl
Cl

O
Total of valence electrons in the molecule
26
Total of electrons in 3 bonds
6
Total of valence electrons remaining unused
26 6 20
35
Thionyl Chloride SOCl2
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Add 3 pairs of electrons to each chlorine.
..
..
_
_


S
Cl
Cl
..
..



O
..
Valence electrons remaining before this step
20
Total of valence electrons used in the step
18
Total of valence electrons remaining unused
20 18 2
36
Thionyl Chloride SOCl2
Step Four Place any remaining valence electrons
on the central atom as unshared pairs.
Total of valence electrons remaining unused
2
The two electrons are placed as a pair on the
central atom.
..
..
..
_
_


S
Cl
Cl
..
..



O
..
37
Thionyl Chloride SOCl2
Step Five Count the number of valence electrons
on the central atom. If it is less than 8, form a
double or triple bond to complete the octet.
The central atom, sulfur, has 3 bonds
and one unshared pair.
..
..
..
_
_


S
Cl
Cl
..
..



O
..
The central atom does have a complete octet, so
the Lewis structure is complete.
38
Phosgene COCl2
Step One Count the total number of valence
electrons in the system.
COCl2
Number of valence electrons in carbon
4
Number of valence electrons in oxygen
6
Number of valence electrons in each chlorine
7
Number of valence electrons for all chlorines
7 2 14
Total of valence electrons in the molecule
4 6 14 24
39
Phosgene COCl2
Step Two Identify the central atom and draw a
single bond from it to each of the attached
atoms. Subtract the electrons used (2 per bond)
from the total number of valence electrons.
The central atom is carbon.
_
_
C
Cl
Cl

O
Total of valence electrons in the molecule
24
Total of electrons in 3 bonds
6
Total of valence electrons remaining unused
24 6 18
40
Phosgene COCl2
Step Three Complete the valence shell of each
attached atom. Subtract the electrons used.
Add 3 pairs of electrons to each chlorine and the
oxygen.
..
..
_
_


C
Cl
Cl
..
..



O
..
Valence electrons remaining before this step
18
Total of valence electrons used in the step
18
Total of valence electrons remaining unused
18 18 0
41
Phosgene COCl2
Step Four Place any remaining valence electrons
on the central atom as unshared pairs.
Total of valence electrons remaining unused
0
Go to step five.
..
..
_
_


C
Cl
Cl
..
..



O
..
42
Phosgene COCl2
Step Five Count the number of valence electrons
on the central atom. If it is less than 8, form a
double or triple bond to complete the octet.
The central atom, sulfur, has 3 bonds and no
unshared pairs
..
..
_
_


C
Cl
Cl
..
..



O
..
The central atom does not have a complete octet,
so the Lewis structure is not complete.
43
Phosgene COCl2
Step Five Count the number of valence electrons
on the central atom. If it is less than 8, form a
double or triple bond to complete the octet.
So, a double bond is created between the carbon
and the oxygen by removing one of the oxygen
electron pairs and using it to form a bond.
..
..
..
..
_
_
_
_




C
Cl
Cl
C
Cl
Cl
..
..
..
..







O
O
..
..
Each single bond is composed of two electrons,
and the double bond is composed of two pairs, so
the carbon now has a complete octet.
This is a complete Lewis structure.