“A” students work
(without solutions manual)
~ 10 problems/night.
Alanah Fitch
Flanner Hall 402
508-3119
afitch@luc.edu
Office Hours W – F 2-3 pm
Properties and Measurements
Property
Size
Volume
Weight
Temperature
1.66053873x10-24g
quantity
mole
Pressure
Energy, of electrons
Electronegativity
Unit
m
cm3
gram
Reference State
size of earth
m
mass of 1 cm3 water at specified Temp
(and Pressure)
oC, K
boiling, freezing of water (specified
Pressure)
amu
(mass of 1C-12 atom)/12
atomic mass of an element in grams
atm, mm Hg
earth’s atmosphere at sea level
energy of electron in a vacuum
F
Covalent bonding
Patterns in abundance suggest
a.
periodicity
b.
preferred electronic configuration of elements
Leading to the Second Rule: “Everybody wants to be “Like Mike”
a.
Ions: Groups 16 and 17 gain electrons; Groups 1 and 2 lose
b.
Other atoms share electrons to have eight electrons
= COVALENT BONDING
Rule 5:
Rule 4:
Rule 3:
Rule 2:
Rule 1:
There are no stupid questions
Slow me Down
Chemists are Lazy
Everybody wants to be a Noble Gas
Everybody wants a partner of the opposite charge
Covalent Bonding – getting to a noble gas electron configuration by
sharing electrons
1. Lewis structures and the Octet Rule
2. Valence bond theory
+
+
Repulsion of two hydrogen atoms with their
Proton core
Repulsion of two hydrogen atoms with their proton core
e
e
++
Repulsion is high close where
Protons see each other
+
e
e
Repulsion is low where
Electrons shield nucleus, and where
Electrons can be stabilized by both
Positive charges
Repulsive energy
+
Attractive energy
+
e
e
+
Atoms which are far apart
Do not even see each other
There is no energy, repulsive
Or attractive between the two
Repulsion of two hydrogen atoms with their proton core
e
e
++
Repulsion is high close where
Protons see each other
+ +
e
e
Repulsion is low where
Electrons shield nucleus, and where
Electrons can be stabilized by both
Positive charges
Repulsive energy
+
Attractive energy
e
e
+
Atoms which are far apart
Do not even see each other
There is no energy, repulsive
Or attractive between the two
Repulsion of two hydrogen atoms with their proton core
e
e
++
Repulsion is high close where
Protons see each other
+ +
e
e
Repulsion is low where
Electrons shield nucleus, and where
Electrons can be stabilized by both
Positive charges
Repulsive energy
+
Attractive energy
e
e
+
Electrons
are
Atoms which
arethe
farjelly
apart
Do not evenand
seepeanut
each other
There is no energy, repulsive
butter between
Or attractive between the two
the slices of
bread (protons)
Repulsion of two hydrogen atoms with their proton core
e
e
++
Repulsion is high close where
Protons see each other
+ +
e
e
Repulsion is low where
Electrons shield nucleus, and where
Electrons can be stabilized by both
Positive charges
Repulsive energy
+
Attractive energy
e
e
+
Atoms which are far apart
Do not even see each other
There is no energy, repulsive
Or attractive between the two
When the two hydrogen atoms are together, the electron configuration
On hydrogen Looks like?
When the two hydrogen atoms are together, the electron configuration
On hydrogen Looks like?
He
1s2
When the two hydrogen atoms are together, the electron configuration
On Hydrogen Looks like?
He
1s2
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2  [ He]
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
When the two hydrogen atoms are together, the electron configuration
Looks like?
1s2  [ He]
He
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
2
The “inner” shell 1s
Show in this diagram
electrons do not
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2  [ He]
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
Only the “outer-most” or valence shell electrons
Show in this Lewis Dot Structure
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2  [ He]
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
Only the “outer-most” or valence shell
electrons Show in this Lewis Dot
Structure
How many valence electrons?:
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2  [ He]
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
Only the “outer-most” or valence shell
electrons Show in this Lewis Dot
Structure
How many valence electrons?:
= last number in group
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2  [ He]
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
The shared pair of
Electrons = covalent
bond
Only the “outer-most” or valence shell
electrons Show in this Lewis Dot
Structure
When the two hydrogen atoms are together, the electron configuration
Looks like?
He
1s2  [ He]
When a hydrogen atom and a fluorine atom share electrons, the
Electron configuration on fluorine looks like?
1s2 2s2 2 p6  [ Ne]
The shared pair of
Electrons = covalent
bond
Only the “outer-most” or valence shell electrons
Show in this Lewis Dot Structure
The unshared pairs of electrons are “red hots”
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?

O 

When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?

O 

Valence electrons on hydrogen?
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?

O 

Valence electrons on hydrogen?
H
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?

  O  

  
H    O  H 
   


When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
  

  O   H    O  H 

   



1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
  

  O   H    O  H 

   



1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
  

  O   H    O  H 

   



1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Invoking Rule 2: Chemists are Lazy
the diagram above is too tedious to write out all the time
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
  

  O   H    O  H 

   



1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Invoking Rule 2: Chemists are Lazy
the diagram above is two tedious to write out all the time
Lewis dot structure for hydroxide
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
  

  O   H    O  H 

   



1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Invoking Rule 2: Chemists are Lazy
the diagram above is two tedious to write out all the time
Lewis dot structure for hydroxide
The single electron pair shared between the two bonded atoms
Is called a single bond
It is drawn as a line.
When a hydrogen atom and an oxygen atom share valence electrons plus an
Extra electron, the electron configuration on hydrogen and oxygen look like?
  

  O   H    O  H 

   



1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Invoking Rule 2: Chemists are Lazy
the diagram above is two tedious to write out all the time
Lewis dot structure for hydroxide
The single electron pair shaired between the two bonded atoms
Is called a single bond
It is drawn as a line.
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?

O 

When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?

O 

Valence electrons on hydrogen?
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?
Valence electrons on oxygen?

O 

Valence electrons on hydrogen?
H
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?




O   H   H   H  O  H
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Two shared electron pairs
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Two shared electron pairs
=
Two single bonds
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?
Valence shell of nitrogen?
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?
Valence shell of nitrogen?

N 

When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?

 N   H   H   H 


H  N  H

H
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?

 N   H   H   H 


H  O  H

H
1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?

 N   H   H   H 


H  O  H

H
1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
Three pairs of shared electrons = three single bonds
When two hydrogen atoms and an oxygen atom share valence electrons, the
electron configuration on hydrogen and oxygen look like?

O  

H   H 

H  O  H

1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When three hydrogen atoms and a nitrogen atom share valence electrons, the
electron configuration on hydrogen and nitrogen look like?

 N   H   H   H 


H  O  H

H
1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?
When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?
Valence shell of carbon?
When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?
Valence shell of carbon?

C 

When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C  H   H   C  H   H  


H

H

H  C   C  H
When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C  H   H   C  H   H  


1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
H

H

H  C   C  H
When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C  H   H   C  H   H  


1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
H

H

H  C   C  H
When four hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C  H   H   C  H   H  


1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
H

H

H  C   C  H
Two electron pairs shared is a
Double bond
When two hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?
When two hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C   H   C   H  


H  C    C  H
When two hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C   H   C   H  


1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
H  C    C  H
When two hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C   H   C   H  


1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
H  C    C  H
When two hydrogen atoms and two carbon atoms share valence electrons, the
electron configuration on hydrogen and carbon look like?


C   H   C   H  


1s2  [ He]
1s2 2 s2 2 p6  [ Ne]
H  C    C  H
Three electron pairs shared is a
Triple bond
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (last number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine
the only
number
ofone
valence
available
for distribution
after
H can
have
bondelectrons
because still
it can
share only
one
subtracting
electrons
for each single bond.
Electron. two
Poor
H.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
H
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting
two
electrons
for each
singledo
bond.
Halogens
have
lots
of electrons
but really
not like to share.
4. Determine
number of electrons required to complete the octet
Greedy the
halogens
a. AllHthey
getswant
onlyistwo
electrons
one
more to make up the Mike configuration
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.

6. Make up deficit of electrons by creating double
bonds


a.
C, N, O, S
F


Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double or triple bonds
a.
C, N, O, S
Rules for Writing Lewis Dot Structures
1. Count the number of valence electrons (lat number of group) of all atoms
a.
For an anion add the appropriate extra number of electrons
b.
For a cation subtract the appropriate extra number of electrons
2. Draw a molecular skeleton, joining by single bonds to the central atom.
a.
The central is usually the atom written first in the formula (N in NH4+, S
in SO2, and C in CCl4).
b.
The terminal atoms are usually H, O.
c.
Halogens are always terminal atoms.
3. Determine the number of valence electrons still available for distribution after
subtracting two electrons for each single bond.
4. Determine the number of electrons required to complete the octet
a.
H gets only two electrons
b.
Other exceptions to be noted below
5. Fill in the region required for the octet.
6. Make up deficit of electrons by creating double or triple bonds
a.
C, N, O, S
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) SO2
d) N2
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+C
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
C
Negative charge
Total electrons
=6
l= 7
=1
=14
O
+C
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+C
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
C
Negative charge
Total electrons
=6
l= 7
=1
=14
O
+C
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+C
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
Cl
Negative charge
Total electrons
=6
=7
=1
=14
O
+C
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+C
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
Cl
Negative charge
Total electrons
=6
=7
=1
=14
O
+C
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+C
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
Cl
Negative charge
Total electrons
=6
=7
=1
=14
O
+C
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+Cl
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
Cl
Negative charge
Total electrons
=6
=7
=1
=14
O
+C
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+Cl
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
Cl
Negative charge
Total electrons
=6
=7
=1
=14
O
+Cl
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol
c) N2
d) SO2
O
+Cl
+Negative charge
Total electrons
-1Single bond
6
7
1
14
-2
12
Hypochlorite?
Hypo – smallest number of oxygens
OClValence shell electrons?
O
Cl
Negative charge
Total electrons
=6
=7
=1
=14
O
+Cl
+Negative charge
Total electrons
-1Single bond
-2(6 electrons for O,Cl)
remaining



 O  Cl  
     

Skeleton
O  Cl


6
7
1
14
-2
12
12
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
O
+C
+4(H)
Negative charge
Total electrons
6
4
4
0
14
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
6
4
4
0
14
-10
4
Octets
Carbon has its octet
Hydrogen has its duet
Oxygen requires 4 more electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C O  H
H

H  C  O H

H
H
O
+C
+4(H)
Negative charge
Total electrons
-5single bonds
remaining
-octet for oxygen
remaining
6
4
4
0
14
-10
4
-4
0
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 8 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 4 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2N
Negative charge
Total electrons
10
0
10
Octets
Each nitrogen requires 6 more
2N
Negative charge
Total electrons
-1single bond
Remaining
Octet completion
Difference
10
0
10
-2
8
-12
-4
Skeleton
NN
2N
Negative charge
Total electrons
-1single bond
Remaining
10
0
10
-2
8
We are short 4 electrons for the octet,
The only way to get extra ones is to
Share four more electrons = triple
Bond.
NN
Place the remaining 4 electrons equally
On the two equal nitrogens


N  N 
Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



Draw Lewis structures of
a) Hypochlorite ion
b) Methyl alcohol, CH3OH
c) N2
d) SO2
Valence shell electrons?
2O
+1S
Negative charge
Total electrons
12
6
0
18
Skeleton, First atom in formula is central
OS O
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
12
6
0
18
-4
14
Octets
2O
+1S
Negative charge
Total electrons
-2(single bonds)
Remaining electrons
Octet for S
2(Octet for each O)
Deficit?
12
6
0
18
-4
14
-4
-12
-2
We are short 2 electrons for the octet,
The only way to get extra ones is to
Share two more electrons = double bond.
OS O
Place the remaining 14 electrons to fill
octets
  
 O S  O



We got this Lewis dot structure
OS O
No reason not to write instead

Which would lead to
 O




S O
O  S O



O  S  O 



Is there any reason for us to
Presume one of these is correct
And not the other?
No
  
 O S  O






O  S  O 


Grammar: double-headed arrow is used to separate resonance
structures
We got this Lewis dot structure
OS O
No reason not to write instead

Which would lead to
 O




S O
O  S O



O  S  O 



Is there any reason for us to
Presume one of these is correct
And not the other?
No
  
 O S  O






O  S  O 


Grammar: double-headed arrow is used to separate resonance
structures
We got this Lewis dot structure
OS O
No reason not to write instead

Which would lead to
 O




S O
O  S O



O  S  O 



Is there any reason for us to
Presume one of these is correct
And not the other?
No
  
 O S  O






O  S  O 


Grammar: double-headed arrow is used to separate resonance
structures
We got this Lewis dot structure
OS O
No reason not to write instead

Which would lead to
 O




S O
O  S O



O  S  O 



Is there any reason for us to
Presume one of these is correct
And not the other?
No
  
 O S  O






O  S  O 


Grammar: double-headed arrow is used to separate resonance
structures
We got this Lewis dot structure
OS O
No reason not to write instead

Which would lead to
 O




S O
O  S O



O  S  O 



Is there any reason for us to
Presume one of these is correct
And not the other?
No
  
 O S  O






O  S  O 


Grammar: double-headed arrow is used to separate resonance
structures
We got this Lewis dot structure
OS O
No reason not to write instead

Which would lead to
 O




S O
O  S O



O  S  O 



Is there any reason for us to
Presume one of these is correct
And not the other?
No
  
 O S  O






O  S  O 


Grammar: double-headed arrow is used to separate resonance
structures
Remember our Marshmallows?
No Clean Socks
NO3




N
5




O


O


3(O)
18
 




Charge
1
 
 
 
 
Total
24
 O N  O
O  N  O 



-single bonds
-6

 
 

 

Remaining
18




Octets (6x3 O +2 for N) 20





Deficit of 1 electron pair

       




O


 
Charge is distributed over
 
O
All three of the resonance
 O  N  O 

 

Forms
=
one
big
fat
marshmallow
O N  O







Remember our Marshmallows?
No Clean Socks
NO3




N
5




O


O


3(O)
18
 




Charge
1
 
 
 
 
Total
24
 O N  O
O  N  O 



-single bonds
-6

 
 

 

Remaining
18




Octets (6x3 O +2 for N) 20





Deficit of 1 electron pair

       




O


 
Charge is distributed over
 
O
All three of the resonance
 O  N  O 

 

Forms
=
one
big
fat
marshmallow
O N  O







Remember our Marshmallows?
No Clean Socks
NO3




N
5




O


O


3(O)
18
 




Charge
1
 
 
 
 
Total
24
 O N  O
O  N  O 



-single bonds
-6

 
 

 

Remaining
18




Octets (6x3 O +2 for N) 20





Deficit of 1 electron pair

       




O


 
Charge is distributed over
 
O
All three of the resonance
 O  N  O 

 

Forms
=
one
big
fat
marshmallow
O N  O







Remember our Marshmallows?
No Clean Socks
NO3




N
5




O


O


3(O)
18
 




Charge
1
 
 
 
 
Total
24
 O N  O
O  N  O 



-single bonds
-6

 
 

 

Remaining
18




Octets (6x3 O +2 for N) 20





Deficit of 1 electron pair

       




O


 
Charge is distributed over
 
O
All three of the resonance
 O  N  O 

 

Forms
=
one
big
fat
marshmallow
O N  O







Remember our Marshmallows?
No Clean Socks
NO3




N
5




O


O


3(O)
18
 




Charge
1
 
 
 
 
Total
24
 O N  O
O  N  O 



-single bonds
-6

 
 

 

Remaining
18




Octets (6x3 O +2 for N) 20





Deficit of 1 electron pair

       




O


 
Charge is distributed over
 
O
All three of the resonance
 O  N  O 

 

Forms
=
one
big
fat
marshmallow
O N  O







Remember our Marshmallows?
No Clean Socks
NO3-
O
O N  O

 






O
O






 


 









 O  N  O 
 O N  O




 
  
 
 

















O




Charge is distributed over
  



All three of the resonance
  O  N  O 

Forms = one big fat marshmallow    
  






N
3(O)
Charge
Total
-single bonds
Remaining
Octets (6x3 O +2 for N)
Deficit of 1 electron pair
5
18
1
24
-6
18
20


Remember our Marshmallows?
No Clean Socks
NO3




N
5




O


O


3(O)
18
 




Charge
1
 
 
 
 
Total
24
 O N  O
O  N  O 



-single bonds
-6

 
 

 

Remaining
18




Octets (6x3 O +2 for N) 20





Deficit of 1 electron pair

     




O


 
Charge is distributed over
 
O
All three of the resonance
 O  N  O 

 

Forms
=
one
big
fat
marshmallow
O N  O







Resonance
1. The “real” molecule is non of the three nitrates we drew but something
intermediate to the three.
2. Resonance can be “assumed” or “predicted” when there are equally plausible
Lewis dot structures.
3. Resonance forms differ only in the distribution of electrons and not in the
arrangement of atoms.
Resonance
1. The “real” molecule is none of the three nitrates we drew but something
intermediate to the three.
2. Resonance can be “assumed” or “predicted” when there are equally plausible
Lewis dot structures.
3. Resonance forms differ only in the distribution of electrons and not in the
arrangement of atoms.
Resonance
1. The “real” molecule is non of the three nitrates we drew but something
intermediate to the three.
2. Resonance can be “assumed” or “predicted” when there are equally plausible
Lewis dot structures.
3. Resonance forms differ only in the distribution of electrons and not in the
arrangement of atoms.
Resonance
1. The “real” molecule is non of the three nitrates we drew but something
intermediate to the three.
2. Resonance can be “assumed” or “predicted” when there are equally plausible
Lewis dot structures.
3. Resonance forms differ only in the distribution of electrons and not in the
arrangement of atoms.
Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the oxygen -3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the oxygen -3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the sulfur
-3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the sulfur
-3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the sulfur
-3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the sulfur
-3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the sulfur
-3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Write three resonance forms for SO3
Valence electrons
4(6)
Sulfur central atom
Three single bonds to the sulfur
-3(2)
Remaining electrons
2 electrons to complete S octet
3(6) electrons to complete O octets = 18
Deficit of two electrons = double bond

O
 

 O


24
-6
18
-2
-18
-2

O
 




S O
O S
O
 

 O 


 O



S  O 

Formal Charge helps determine the correct Lewis Dot Structure
Z

formal ch arg e  C f  X  Y  
2

X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure
Z = number of bonding e- shared by the atom in the Lewis structure
The correct Lewis dot structure is generally the one in which
a. The formal charges are as close to zero as possible
b. Negative charge is located on the more electronegative atom
Formal Charge helps determine the correct Lewis Dot Structure
Z

formal ch arg e  C f  X  Y  
2

X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure
Z = number of bonding e- shared by the atom in the Lewis structure
The correct Lewis dot structure is generally the one in which
a. The formal charges are as close to zero as possible
b. Negative charge is located on the more electronegative atom
Formal Charge helps determine the correct Lewis Dot Structure
Z

formal ch arg e  C f  X  Y  
2

X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure
Z = number of bonding e- shared by the atom in the Lewis structure
The correct Lewis dot structure is generally the one in which
a. The formal charges are as close to zero as possible
b. Negative charge is located on the more electronegative atom
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
H
6

formal ch arg e  CO  6  2    1
2


6

formal ch arg e  CC  4  2    11
2


H  C  O H
H
Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

6

formal ch arg e  CC  4  2    11
2

Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

6

formal ch arg e  CC  4  2    11
2

Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

6

formal ch arg e  CC  4  2    11
2

Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
Which is correct?
H

H  C  O H

H
2

formal ch arg e  CO  6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

 8
formal ch arg e  CC  4  0    0
 2
X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

8

formal ch arg e  C fC  4   0    0
2

X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

8

formal ch arg e  C fC  4   0    0
2

X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

8

formal ch arg e  C fC  4   0    0
2

X=number of valence e- in the free atom (last number of group)
Y = number of unshared e- owned by the atom in the Lewis structure formal ch arg e  C  X  Y  Z 
f



2
Z = number of bonding e- shared by the atom in the Lewis structure
formal ch arg e  C fO
H

H  C  O H

H
6

 6  2     1
2

formal ch arg e  C fC
6

 4  2     1
2

Which is correct?
H

H  C  O H

H
formal ch arg e  C fO
4

 6  4    0
2

8

formal ch arg e  C fC  4   0    0
2

The Octet Rule is a form of Rule #2- Everybody wants to be like a Noble gas
But: not everybody can share enough
Electrons to make up a perfect octet
H

 Be 


 B
The Octet Rule is a form of Rule #3- Everybody wants to be like a Noble gas
In same fashion: not everybody can share enough
Electrons to make up a perfect octet
These guys will have 1, 2, and 3 bonds only
H

 Be 


 B



 F  Be  F 




F
 

F



B  F 

The Octet Rule is a form of Rule #2- Everybody wants to be like a Noble gas
In same fashion: not everybody can share enough
Electrons to make up a perfect octet
This guy may end up “holding the bag”
Having an unpaired electron

N
 


Because he is
Not really
Strong enough
To always
Get the lion’s
Share of the
Electrons in
A covalent bond,
Particularly
With oxygen

 
N O

 

 O







NO


O  N  O 

The presence of these unpaired electrons on these gases
Gives rise to the many atmospheric reactions involved
In ozone destruction and formation of smog.
Para = paramour = love = similar orientation
Dia – diatribe = against = opposite orientation
The Octet Rule is a form of Rule #2- Everybody wants to be like a Noble gas
Some guys can take on more electrons because they
Make use of their d orbitals
3p
4s
3d
these guys
Have d orbitals
That allow them
To have more
Than 8 electrons

F
 
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4

 F



Xe  F 

F
 


F
 
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4

 F



Xe  F 

F
 


F
 F
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4


XeFF
 F F 
Xe




FF

 


F
 F
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4


XeFF
 F F 
Xe




FF

 


F
 F
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4


XeFF
 F F 
Xe




FF

 

Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4
  
 F  
 F

  
  
F
 

Xe

F
F

Xe

F
 

  
   
F
 
F
  
  

F
 
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4

 F



Xe  F 

F
 


F
 
Draw the Lewis structure of XeF4
8+4(7)
=36 electrons
4bonds
= 8 electrons
Remainder
= 28 electrons
Octets: 4(6) for F = 24
Remainder to Xe = 4

 F



Xe  F 

F
 

Draw the Lewis structure of XeF2
Draw the Lewis structure of XeF2
8+2(7)
=22 electrons
Draw the Lewis structure of XeF2
8+2(7)
2bonds
=22 electrons
= 4 electrons
Draw the Lewis structure of XeF2
8+2(7)
2bonds
Remainder
=22 electrons
= 4 electrons
= 18 electrons
F  Xe  F
Draw the Lewis structure of XeF2
8+2(7)
=22 electrons
2bonds
= 4 electrons
Remainder
= 18 electrons
Octets: 2(6) for F = 12


 
 
 
 
F  Xe F


Draw the Lewis structure of XeF2
8+2(7)
=22 electrons
2bonds
= 4 electrons
Remainder
= 18 electrons
Octets: 2(6) for F = 12
Remainder to Xe = 6


 
 
 
 
F  Xe F


Draw the Lewis structure of XeF2
8+2(7)
=22 electrons
2bonds
= 4 electrons
Remainder
= 18 electrons
Octets: 2(6) for F = 12
Remainder to Xe = 6


 

 
 

 
F  Xe F


Can now predict SHAPE of molecule
determines: how two molecules orient themselves
for reaction together.
Can they dock? And actually do work
together in three dimensional space?
Valence
Shell
Electron
Pair
Repulsion
Valence electron pairs surrounding an atom repel
one another. Consequently, the orbitals
containing those electron pairs are oriented to be
as far apart as possible
Eel
d
kQ1Q2
E el 
d
8.99 x10 9 J  m
k
2
Coulomb 
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules


 F  Be  F 




Geometries of AX2-AX6 molecules


 F  Be  F 




Repulsion of valence
shell electrons pushes Fe
apart to a 180o orientation
Geometries of AX2-AX6 molecules


 F  Be  F 




Repulsion of valence
shell electrons pushes Fe
apart to a 180o orientation
Geometries of AX2-AX6 molecules


 F  Be  F 




Repulsion of valence
shell electrons pushes Fe
apart to a 180o orientation
Geometries of AX2-AX6 molecules


 F  Be  F 




Geometries of AX2-AX6 molecules


 F  Be  F 




Geometries of AX2-AX6 molecules


 F  Be  F 




Geometries of AX2-AX6 molecules

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
What
orientation would
put electrons as far
apart as possible?
120o
degrees apart
in a “circle”

F
 

F



B  F 

Geometries of AX2-AX6 molecules
Draw Lewis structures of
a) CH4
Valence shell electrons?
+C
+4(H)
Negative charge
Total electrons
4
4
0
8
Draw Lewis structures of
a) CH4
Valence shell electrons?
+C
+4(H)
Negative charge
Total electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
4
4
0
8
Draw Lewis structures of
a) CH4
Valence shell electrons?
+C
+4(H)
Negative charge
Total electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C H
H
4
4
0
8
Draw Lewis structures of
a) CH4
Valence shell electrons?
+C
+4(H)
Negative charge
Total electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C H
H
4
4
0
8
+C
+4(H)
Negative charge
Total electrons
-4single bonds
remaining
4
4
0
8
-8
0
Draw Lewis structures of
a) CH4
Valence shell electrons?
+C
+4(H)
Negative charge
Total electrons
Skeleton
Carbon is first in formula
Hydrogen is always terminal
H
H  C H
H
4
4
0
8
+C
+4(H)
Negative charge
Total electrons
-4single bonds
remaining
Octets
Carbon has its octet
Hydrogen has its duet
4
4
0
8
-8
0
Geometries of AX2-AX6 molecules
How can four
bonds be
organized in
3-D space to
be farthest apart?
H
H  C H
H
Geometries of AX2-AX6 molecules
How can four
bonds be
organized in
3-D space to
be farthest apart?
H
H  C H
H
Geometries of AX2-AX6 molecules
How can four
bonds be
organized in
3-D space to
be farthest apart?
H
H  C H
H
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
A pyramid is a space figure with a square base and 4 triangle-shaped sides.
(5 “faces”)
3
2
4
5
1
A tetrahedron is a space figure and 4 triangle shaped faces.
Dictionary: a four-sided solid; a
Triangular pyramid
3
2
1
4
A pyramid is a space figure with a square base and 4 triangle-shaped sides.
(5 “faces”)
Dictionary: Square base and sloping
Sides rising to an apex
3
2
5
4
1
A tetrahedron is a space figure and 4 triangle shaped faces.
Dictionary: a four-sided solid; a
Triangular pyramid
3
2
4
1
A pyramid is a space figure with a square base and 4 triangle-shaped sides.
(5 “faces”)
Dictionary 1: Square base and sloping
Sides rising to an apex
3
2
Dictionary 2: A solid figure with a
polygon base. The surface, or
lateral faces, are triangles having a
common vertex. in a regular pyramid
the base is a regular polygon and the
lateral faces are congruent triangles
4
5
1
A tetrahedron is a space figure and 4 triangle shaped faces.
Dictionary: a four-sided solid; a
Triangular pyramid
3
2
1
4
SIGNIFICANT AMBIGUITY
In nomenclature!!!!
Why tetrahedron and not this orientation?
1,190 o
Bond distance=1
4,1-
A
3,1-
2,1-
180 o
Why tetrahedron and not this orientation?
kQ1Q2
E el 
d
8.99 x10 9 J  m
k
2
Coulomb 
1,190 o
Bond distance=1
4,1-
A
2,1-
180 o
E el ,1,total
3,1-
 Q1Q2 Q1Q3 Q1Q4 
 k



d
d
d
13
14 
 12
Why tetrahedron and not this orientation?
kQ1Q2
E el 
d
8.99 x10 9 J  m
k
2
Coulomb 
1,190 o
Bond distance=1
A
4,1-
2,1-
180 o
E el ,1,total
3,1-
E el ,1,total
 1
1
1 
 k Q1Qi 



d
d
d
13
14 
 12
E el ,1,total
 1
1
1 




d
d
d
13
14 
 12
 Q1Q2 Q1Q3 Q1Q4 
 k



d
d
d
13
14 
 12
Since our example has all Q the same
Why tetrahedron and not this orientation?
E el ,1,total
1,1d14
d12
90 o
1
4,1-
A
d13
2,1-
E el ,1,total
3,1-
 1
1
1 




d
d
d
13
14 
 12
d12 
 A2   A1
2
2
 1 1 
1
1 
 1

 
 1914
.

2
 2 2
2
E el ,1,total
 1
1
1 




d
d
d
13
14 
 12
1
d12
A
4
2
3
E el ,1,total
 1
1
1 




d
d
d
13
14 
 12
14
1
2
14
A1  1
60o
2
4
A
A4  1
1
d12
A
4
2
3
E el ,1,total
opp
sin  
hyp
 1
1
1 




d
d
d
13
14 
 12
14
sin60 o  
1
A1  1
1732
.
 14
60o
2
4
2
1
2 sin60 o   14
2
14
14
A
A4  1
1
d12
A
4
2
3
E el ,1,total
opp
sin  
hyp
 1
1
1 




d
d
d
13
14 
 12
14
sin60 o  
1
A1  1
1732
.
 14
60o
2
4
2
1
2 sin60 o   14
2
14
14
A
A4  1
1
E el ,1,total
1
1 
 1



 1732
.

.
1732
.
1732
.
 1732

d12
A
4
2
3
Other arrangement = 1.914
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
Triangular
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
Triangular
pyramid
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
Triangular
bipyramid
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
How can five bonds
be arranged in space
to be as far apart as
possible?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
How can these six guys
best position themselves
away from each other?
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
Geometries of AX2-AX6 molecules
A = central atom
X = terminal atoms
Some of the molecules we constructed using Lewis Dot structures
had
UNSHARED PAIRS of electrons on the CENTRAL ATOM
What effect will this have on the geometry?.

 O







S O


O  S  O 





O   H   H   H  O  H
 

 
N   H   H   H   H  N  H
 

H
Some of the molecules we constructed using Lewis Dot structures
had
UNSHARED PAIRS of electrons on the CENTRAL ATOM
What effect will this have on the geometry?.
  
 O S  O






O  S  O 






O   H   H   H  O  H
 

Unshared electron pairs orient
themselves pretty much the same
as single bonds.
 
N   H   H   H   H  N  H
 

H
Some of the molecules we constructed using Lewis Dot structures
had
UNSHARED PAIRS of electrons on the CENTRAL ATOM
What effect will this have on the geometry?.
  
 O S  O






O  S  O 






O   H   H   H  O  H
 

Unshared electron pairs orient
themselves pretty much the same
as single bonds.
 
N   H   H   H   H  N  H
 

H
The observed molecular
geometry (invisible
electrons) is very
different
Some of the molecules we constructed using Lewis Dot structures
had
UNSHARED PAIRS of electrons on the CENTRAL ATOM
What effect will this have on the geometry?.

 O







S O


O  S  O 





O   H   H   H  O  H
 

AX2E
Unshared electron pairs orient
themselves pretty much the same
as single bonds.
 
N   H   H   H   H  N  H
 

H
The observed molecular
geometry (invisible
electrons) is very
different
  
 O S  O






O  S  O 


AX2E
This geometry, with respect to electron pairs and bonds,
is triangular planar (three guys trying to get out of each others way)
  
 O S  O






O  S  O 


AX2E
This geometry, with respect to electron pairs and bonds,
is triangular planar (three guys trying to get out of each others way)
  
 O S  O






O  S  O 


AX2E
This geometry, with respect to electron pairs and bonds,
is triangular planar (three guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from triangular planar
Actual degrees observed
is slightly less than 120o because
unshared electron pair expands
  
 O S  O






O  S  O 


AX2E
This geometry, with respect to electron pairs and bonds,
is triangular planar (three guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from triangular planar
Actual degrees observed
is slightly less than 120o because
unshared electron pair expands
Molecular Geometry is “bent”
  
 O S  O






O  S  O 


AX2E
This geometry, with respect to electron pairs and bonds,
is triangular planar (three guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from triangular planar
Actual degrees observed
is slightly less than 120o because
unshared electron pair expands
Molecular Geometry is “bent”
Some of the molecules we constructed using Lewis Dot structures
had
UNSHARED PAIRS of electrons on the CENTRAL ATOM
What effect will this have on the geometry?.
  
 O S  O






O  S  O 






O   H   H   H  O  H
 

Unshared electron pairs orient
themselves pretty much the same
as single bonds.
 
N   H   H   H   H  N  H
 

H
AX3E
The observed molecular
geometry (invisible
electrons) is very
different
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from tetrahedral
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from tetrahedral
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from tetrahedral
Triangular
Pyramid
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from tetrahedral
Triangular
Pyramid
 
H  N  H AX E
3
H
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But one of the “terminal atoms” is missing so the
molecular geometry differs from tetrahedral
Triangular
Pyramid
Explains why the electron pair on
ammonia is a “red hot”
Effect of lone pairs on substituents



 



F N F
       


 F

102.3o
F
F
F
 


 
H N H
H
107.2o
Effect of F is NOT by geometry of it’s lone pairs
BUT
By it’s electronegativity which pulls electrons along the bond, lowers
Density of electrons in the bondings area
Allows N lone pair to expand



 



F N F
       


 F

 


 
H N H
H
Lone pair expands
102.3o
F
F
F
107.2o
Some of the molecules we constructed using Lewis Dot structures
had
UNSHARED PAIRS of electrons on the CENTRAL ATOM
What effect will this have on the geometry?.
AX2E2




O   H   H   H  O  H
 

Unshared electron pairs orient
themselves pretty much the same
as single bonds.
 
N   H   H   H   H  N  H
 

H
The observed molecular
geometry (invisible
electrons) is very
different


 


H O H
 


 


H O H
 
AX2E2


 


H O H
 
AX2E2


 


H O H
 
AX2E2
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)


 


H O H
 
AX2E2
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But two of the “terminal atoms” are missing so the
molecular geometry differs from tetrahedral


 


H O H
 
AX2E2
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But two of the “terminal atoms” are missing so the
molecular geometry differs from tetrahedral


 


H O H
 
AX2E2
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But two of the “terminal atoms” are missing so the
molecular geometry differs from tetrahedral


 


H O H
 
AX2E2
This geometry, with respect to electron pairs and bonds,
is tetrahedral (four guys trying to get out of each others way)
But two of the “terminal atoms” are missing so the
molecular geometry differs from tetrahedral
This shape is “bent”
Both are bent,
but the angle is different.
Depends upon the number
of valence shell electron pairs
AX2E
  
 O S  O



 
AX2E2
H  O  H
 
Both are bent,
but the angle is different.
Depends upon the number
of valence shell electron pairs
  
 O S  O



 
H  O  H
 
Geometries of molecules with expanded octets (Fig 7.8)
5 ELECTRON PAIRS
Geometries of molecules with expanded octets (Fig 7.8)
5 ELECTRON PAIRS
This geometry, with respect to electron pairs and bonds,
is triangular bipyramidal (five guys trying to get out of each
others way)
Geometries of molecules with expanded octets (Fig 7.8)
5 ELECTRON PAIRS
This geometry, with respect to electron pairs and bonds,
is triangular bipyramidal (five guys trying to get out of each
others way)


 

 
F  Xe F
 

 


AX2E3
Geometries of molecules with expanded octets (Fig 7.8)
5 ELECTRON PAIRS
This geometry, with respect to electron pairs and bonds,
is triangular bipyramidal (five guys trying to get out of each
others way)
The molecular
geometry will reflect
 
 

the fact that there are


AX2E3 three sites not


 

 
occupied by terminal
atoms
F  Xe F
Geometries of molecules with expanded octets (Fig. 7.8)
5 ELECTRON PAIRS
Triangular bipyramid shape for electron pairs
AX2E3
Geometries of molecules with expanded octets (Fig. 7.8)
5 ELECTRON PAIRS
This molecule will have two terminal atom
positions occupied by electrons.
What will be the molecular geometry?
AX2E3
Comparing where the non-bonded electron pair will go
Variations: Axial versus Equatorial orientation
axial
equatorial
90o
120o
E
E
E = non-bonding electron pair
S = bonded electron pair
E Total ,1   E  E  equatorial  2( E  S ) equatorial  4 E  S  axial  2 S  S  axial
E Total ,2  3( S  S ) equatorial  6 E  S  axial
If we put the E at the axial orientation they
minimize E-E repulsion; increase E-S repulsion
If we put E at the equatorial orientation
E-E repulsion exists, but we decrease the E-S repulsion
E
Minimizes impact
Of E on S
E
Geometries of molecules with expanded octets (Fig. 7.8)
5 ELECTRON PAIRS
AX2E3
Geometries of molecules with expanded octets (Fig. 7.8)
Geometries of molecules with expanded octets (Fig. 7.8)

F
 

 F



Xe  F 

F
 

Geometries of molecules with expanded octets (Fig. 7.8)

F
 
AX4E2

 F



Xe  F 

F
 

Geometries of molecules with expanded octets (Fig. 7.8)

F
 
AX4E2
This geometry, with respect
to electron pairs and bonds,
is octahedral (six guys
trying to get out of each
others way)

 F



Xe  F 

F
 

Geometries of molecules with expanded octets (Fig. 7.8)

F
 
AX4E2
This geometry, with respect
to electron pairs and bonds,
is octahedral (six guys
trying to get out of each
others way)

 F


The molecular geometry has two of the terminal atom
positions occupied by unshared electron pairs

Xe  F 

F
 

Geometries of molecules with expanded octets (Fig. 7.8)

F
 

 F



Xe  F 

F
 

Why electron configuration is important:
controls shape of molecule
dictates 3D interaction of molecules
Square planar lets it slide into the DNA grove
Geometries of molecules with expanded octets (Fig. 7.8)
Multiple Bonding
• Has no effect upon geometry:
– BF3 and SO3 : both AX3 , same geometry


F
 

F



O
 

B F




 O


O
 

S O

O
 




O  S  O 

 O



S  O 

Compare Molecular Geometries
for BeF2 and CO2
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2


 F  Be  F 




Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
4
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
4
12
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
Negative charge
Total electrons
4
12
0
16
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
Negative charge
Total electrons
4
12
0
16
Skeleton
Carbon is first in formula= central atom
O C O
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
Negative charge
Total electrons
4
12
0
16
Skeleton
Carbon is first in formula= central atom
O C O
+C
+2(O)
Negative charge
Total electrons
-2single bonds
remaining
4
12
0
16
-4
12
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
Negative charge
Total electrons
4
12
0
16
Skeleton
Carbon is first in formula= central atom
O C O
+C
+2(O)
Negative charge
Total electrons
-2single bonds
remaining
4
12
0
16
-4
12
Octets
Carbon needs 4 electrons
O each need 6
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
Negative charge
Total electrons
4
12
0
16
Skeleton
Carbon is first in formula= central atom
O C O
+C
+2(O)
Negative charge
Total electrons
-2single bonds
remaining
4
4
0
16
-4
12
Octets
Carbon needs 4 electrons
O each need 6
+C
+2(O)
Negative charge
Total electrons
-2single bonds
remaining
e required for octets
deficit = multiple bonds
4
12
0
16
-4
12
16
4
Compare Molecular Geometries
for BeF2 and CO2
1. We already did BeF2
Valence shell electrons for CO2?
+C
+2(O)
Negative charge
Total electrons
4
12
0
16
Skeleton
Carbon is first in formula= central atom
O C O




O C O
+C
+2(O)
Negative charge
Total electrons
-2single bonds
remaining
4
4
0
16
-4
12
Octets
Carbon needs 4 electrons
O each need 6
+C
+2(O)
Negative charge
Total electrons
-2single bonds
remaining
e required for octets
deficit = multiple bonds
4
4
0
16
-4
12
16
4
Multiple Bonding
• Has no effect upon geometry:
– BF3 and SO3 : both AX3 , same geometry
– BeF2 and CO2 : both AX2, both linear


 F  Be  F 








O C O
NO CENTRAL ATOM
H
 


H C 
H
 


 C H
NO CENTRAL ATOM
H
 


H C 
H
 


 C H
Consider each carbon separately
NO CENTRAL ATOM
H
 


H C 
H
 


 C H
Consider each carbon separately
AX3
NO CENTRAL ATOM
H
 


H C 
H
 


 C H
Consider each carbon separately
AX3
Geometry around
the carbon = Triangular Planar
NO CENTRAL ATOM
H
 


H C 
H
 


 C H
Consider each carbon separately
AX3
Geometry around
the carbon = Triangular Planar
NO CENTRAL ATOM


H C
  
  


C H
NO CENTRAL ATOM


H C
  
  


C H
Consider each carbon separately
NO CENTRAL ATOM


H C
  
  


C H
Consider each carbon separately
AX2
NO CENTRAL ATOM


H C
  
  


C H
Consider each carbon separately
AX2
Geometry around the carbon =Linear
NO CENTRAL ATOM


H C
  
  


C H
Consider each carbon separately
AX2
Geometry around the carbon =Linear
Polarity
•
Bond Polarity
•
Molecular Polarity
– Diatomic molecules
– Polyatomic molecules
Bond Polarity
•All bonds are polar unless the two atoms
joined are identical
•(H–H).
•Extent of polarity depends upon difference
in electronegativity.
Decreasing size
Electronegativity is a measure of
Redhotness, scaled to a maxium of 4
What order of “redhotness” would we give these guys? Ability to attract more electrons.
Remember, “redhotness” is related to charge/volume
Smaller guys at top right
Bond Polarity
•All bonds are polar unless the two atoms
joined are identical
•(H–H).
•Extent of polarity depends upon difference
in electronegativity.
–H–H
–H–C
–H–F
Bond Polarity
•All bonds are polar unless the two atoms
joined are identical
•(H–H).
•Extent of polarity depends upon difference
in electronegativity.
–H–H
–H–C
–H–F
E.N. = 0
nonpolar
Bond Polarity
•All bonds are polar unless the two atoms
joined are identical
•(H–H).
•Extent of polarity depends upon difference in
electronegativity.
–H–H
–H–C
–H–F
E.N. = 0
E.N.
nonpolar
Bond Polarity
•All bonds are polar unless the two atoms
joined are identical
•(H–H).
•Extent of polarity depends upon difference in
electronegativity.
–H–H
–H–C
–H–F
E.N. = 0
nonpolar
E.N. = 2.5-2.2= 0.3 slightly polar
Bond Polarity
•All bonds are polar unless the two atoms joined
are identical
•(H–H).
•Extent of polarity depends upon difference in
electronegativity.
–H–H
–H–C
–H–F
E.N. = 0
nonpolar
E.N. = 2.5-2.2=0.3 slightly polar
E.N. =
Bond Polarity
•All bonds are polar unless the two atoms joined
are identical
•(H–H).
•Extent of polarity depends upon difference in
electronegativity.
–H–H
–H–C
–H–F
E.N. = 0
nonpolar
E.N. = 2.5-2.2=0.3 slightly polar
E.N. = 4-2.2=1.8 strongly polar
Molecular Polarity
•Diatomic molecules: polar if atoms differ
–H–Cl
– polar
Cl–Cl
nonpolar
•HCl molecules line up in an electric field,
Cl2 molecules don’t.
Polar molecules line up in an electric field
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
• F  Be  F
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
• F  Be  F
Arrow indicates the direction in which electrons
are biased - the negative pole
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N. = 4-1.6=2.4
• F  Be  F
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N. = 4-1.6=2.4
• F  Be  F
Vectors cancel each other
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N. = 4-1.6=2.4
• F  Be  F
• compound nonpolar
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
O
H
H
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
H
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
H
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
H
Net charge direction
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
polar
H
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
polar
H
CH4
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
polar
H
Bonds are weak
polar
E.N. = 2.5-2.2=0.3
CH4
H
C
H
H
H
H
C
H
H
H
Non polar
molecule
H
C
H
H
H
Molecular Polarity (cont.)
•Polyatomic molecules
–even though bonds are polar, molecule may be
nonpolar if bonds are symmetrically arranged:
Bonds are polar
E.N.==4-1.6=2.4
4-1.6=2.4
E.N.
• F  Be  F
•nonpolar
Bonds are polar
E.N. = 3.5-2.2=1.3
O
H
polar
H
Bonds are weak
polar
E.N. = 2.5-2.2=0.3
CH4
nonpolar
Non-polar
polar
polar
Non polar
Very polar bonds (Cl = 3.2; C=2.5)
The “fly in the ointment”
Our Valence Shell Electron Repulsion Model
appears to conflict with our knowledge of s, p, d
orbitals and how they are filled
Solution: Invoke Concept of Hybridization


 F  Be  F 




Repulsion of valence
shell electrons pushes Fe
apart to a 180o orientation
VSEPR
AX2
Linear


 F  Be  F 




VSEPR
AX2
Linear
VSEPR model
suggests that once
Be bonds to F the orbitals
are “equivalent” and therefore are equidistant from each other.


 F  Be  F 




VSEPR
AX2
Linear
VSEPR model
suggests that once
Be bonds to F the orbitals
are “equivalent” and therefore are equidistant from each other.
But the electron orbital diagram suggests otherwise.


 F  Be  F 




VSEPR
AX2
Linear
VSEPR model
suggests that once
Be bonds to F the orbitals
are “equivalent” and therefore are equidistant from each other.
But the electron orbital diagram suggests otherwise.
Orbital diagram
Of isolated atoms
F and Be
1s
()
2s
()
2p
() () ()
Be ()
()
  
F
F ()
()
() () ()


 F  Be  F 




VSEPR
AX2
Linear
VSEPR model
suggests that once
Be bonds to F the orbitals
are “equivalent” and therefore are equidistant from each other.
But the electron orbital diagram suggests otherwise.
Orbital diagram
Of isolated atoms
F and Be
1s
()
2s
()
2p
() () ()
Be ()
()
  
F
Diagram suggests that Be
has paired electrons and
F ()
wouldn’t even make bonds,
much less two equivalent bonds
()
() () ()


 F  Be  F 




1s
()
2s
()
2p
() () ()
Be ()
()
  
F
Orbital diagram
Of isolated atoms
F and Be
F ()
()
() () ()
Be has a “full” orbital so “shouldn’t share” electrons?


 F  Be  F 




Orbital diagram
Of isolated atoms
F and Be
1s
()
2s
()
2p
() () ()
Be ()
()
  
F
F ()
()
() () ()
Be has a “full” orbital so “shouldn’t share” electrons?
So what it does is it “mixes” the “2s” with one “2p” orbital to make
two “equivalent” orbitals.


 F  Be  F 




2s
()
2p
() () ()
Be ()
()
  
F
Orbital diagram
Of isolated atoms
F and Be
F ()
F
Orbital diagram
Of isolated
F and hybridized
Be atoms
1s
()
1s
()
Be ()
F ()
()
2s
()

()
() () ()
2p
() () ()
sp
    
() () ()
Formation of sp hybrid orbitals
[


 F  Be  F 




2s
()
2p
() () ()
Be ()
()
  
F
Orbital diagram
Of isolated atoms
F and Be
Once Be has “created” two
identical sp orbitals out of the
2s and one 2p orbital it can
now accept fluorine’s electrons
AND orient electrons
equally as predicted by VSEPR
model
1s
()
F ()
F
1s
()
Be ()
F ()
2s
()

()
()
() () ()
2p
() () ()
sp
    
() () ()
Share
electrons
Formation of Hybrid Orbitals
•s orbital + p orbital  two sp hybrids
–The number of hybrid orbitals formed is equal
to the number of atomic orbitals mixed
–Energies of hybrid orbitals are intermediate to
those of the atomic orbitals from which they are
formed


 F  Be  F 




2s
()
2p
() () ()
Be ()
()
  
F
Orbital diagram
Of isolated atoms
F and Be
F ()
F
Orbital diagram
Of F in BeF2
1s
()
1s
()
Be ()
F ()
()
2s
()

()
() () ()
2p
() () ()
sp
    
() () ()
Shared
electrons

F
 

F



B  F 

VSEPR
AX3
Triangular Planar

F
 

F



B  F 

Orbital diagram
Of isolated atoms
F and B
2p
(  ) (  ) ( )
F
1s
(  )
2s
(  )
B
(  )
(  )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )
 
 
Boron needs to have three of its electrons shared
with the three F electrons.

F
 

F



B  F 

Orbital diagram
Of isolated atoms
F and B
2p
(  ) (  ) ( )
F
1s
(  )
2s
(  )
B
(  )
(  )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )
 
 
Boron needs to have three of its electrons shared
with the three F electrons.
This means it needs to create three equivalent orbitals

F
 

F



B  F 

Orbital diagram
Of isolated atoms
F and B
Orbital diagram
of isolated F atoms
and isolated
hybridized B atoms
2p
(  ) (  ) ( )
F
1s
(  )
2s
(  )
B
(  )
(  )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )
F
1s
(  )
 
 
2p
(  ) (  ) ( )
2s
(  )
sp2
 
   
B
(  )
( )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )

F
 

F



B  F 

Orbital diagram
Of isolated atoms
F and B
Orbital diagram
of isolated F atoms
and isolated
hybridized B atoms
Share electrons
2p
(  ) (  ) ( )
F
1s
(  )
2s
(  )
B
(  )
(  )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )
F
1s
(  )
 
 
2p
(  ) (  ) ( )
2s
(  )
sp2
 
   
B
(  )
( )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )

F
 

F



B  F 

Orbital diagram
Of isolated atoms
F and B
Orbital diagram
of bonded F and B
2p
(  ) (  ) ( )
F
1s
(  )
2s
(  )
B
(  )
(  )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )
F
1s
(  )
 
 
2p
(  ) (  ) ( )
2s
(  )
sp2
 
   
B
(  )
( )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )
H
H  C H
H
VSEPR
AX4
Tetrahedral
2s
2p
H
1s
( )
H  C H
C
(  )
(  )
     
H
H
( )
H
( )
H
( )
H
Orbital diagram
Of isolated atoms
H and C
2s
2p
(  )
     
2s
2p
H
1s
( )
C
(  )
( )
H
( )
H
( )
H
( )
H
H
1s
( )
H  C H
C
(  )
H
H
( )
H
( )
H
( )
Orbital diagram
Of isolated atoms
H and C
In order to accept
electrons for bonds
from H carbon
needs to create
four identical
and energetically
equivalent bonds
sp3
     
2s
2p
(  )
     
2s
2p
H
1s
( )
C
(  )
( )
H
( )
H
( )
H
( )
H
H
1s
( )
H  C H
C
(  )
H
H
( )
H
( )
H
( )
Orbital diagram
Of isolated atoms
H and C
In order to accept
electrons for bonds
from H carbon
needs to create
four identical
and energetically
equivalent bonds
sp3
     
2s
2p
(  )
     
2s
2p
H
1s
( )
C
(  )
( )
H
( )
H
( )
H
( )
H
H
1s
( )
H  C H
C
(  )
H
H
( )
H
( )
H
( )
Orbital diagram
Of isolated atoms
H and C
In order to accept
electrons for bonds
from H carbon
needs to create
four identical
and energetically
equivalent bonds
sp3
     
So far we have considered
sp


 F  Be  F 




2 orbitals
VSEPR
AX2
Linear

sp2
3 orbitals
F
 

F



B  F 

VSEPR
AX3
Triangular Planar
H
sp3
H  C H
4 orbitals
H
VSEPR
AX4
Tetrahedral
So far we have considered
sp


 F  Be  F 




2 orbitals
VSEPR
AX2
Linear

sp2
3 orbitals
F
 

F



B  F 

VSEPR
AX3
Triangular Planar
H
sp3
H  C H
4 orbitals
H
What about the guys with expanded octets?
VSEPR
AX4
Tetrahedral
VSEPR
AX5
Triangular bipyramid
3p
4s
3d
these guys
Have d orbitals
That allow them
To have more
Than 8 electrons
Orbital diagrams
of isolated Cl and P
atoms
Cl

10
Ne

3s
(  )
P

10
Ne

(  )
Cl

10
Ne

Cl

10
Ne
Cl

10
Cl

10
VSEPR
AX5
Triangular
bipyrimad
3p
(  ) (  ) ( )
 
3d
        
   
        
(  )
(  ) (  ) ( )
        

(  )
(  ) (  ) ( )
        
Ne

(  )
(  ) (  ) ( )
        
Ne

(  )
(  ) (  ) ( )
        
Orbital diagrams
of isolated Cl and P
atoms
Cl

10
Ne

3s
(  )
P

10
Ne

(  )
Cl

10
Ne

Cl

10
Ne
Cl

10
Cl

10
VSEPR
AX5
Triangular
bipyramid
3p
(  ) (  ) ( )
 
3d
        
   
        
(  )
(  ) (  ) ( )
        

(  )
(  ) (  ) ( )
        
Ne

(  )
(  ) (  ) ( )
        
Ne

(  )
(  ) (  ) ( )
        
Orbital diagrams
of isolated Cl and P
atoms
Cl

10
Ne

3s
(  )
P

10
Ne

( )
Cl

10
Ne

Cl

10
Ne
Cl

10
Cl

10
VSEPR
AX5
Triangular
bipyramid
3p
(   ) (  ) ( )
 
sp3d
3d
        
   
         
(  )
(   ) (  ) ( )
        

(  )
(   ) (  ) ( )
        
Ne

(  )
(   ) (  ) ( )
        
Ne

(  )
(   ) (  ) ( )
        
Orbital diagrams
of isolated Cl and P
atoms
Cl

10
Ne

3s
(  )
P

10
Ne

( )
Cl

10
Ne

Cl

10
Ne
Cl

10
Cl

10
VSEPR
AX5
triangular
bipyramid
3p
(   ) (  ) ( )
 
sp3d
3d
        
   
         
(  )
(   ) (  ) ( )
        

(  )
(   ) (  ) ( )
        
Ne

(  )
(   ) (  ) ( )
        
Ne

(  )
(   ) (  ) ( )
        
VSEPR
AX2
AX3
AX4
AX5
AX6
Nothing new here – same as we got with VSEPR


 F  Be  F 




F
1s
()
Be ()
F ()
2s
()

()
2p
() () ()
sp
    
() () ()
What does the orbital between the Be atomic “sp” and
the F “p” electrons “look” like?
sp


 F  Be  F 




F
1s
()
Be ()
F ()
2s
()

()
2p
() () ()
sp
    
() () ()
What does the orbital between the Be atomic “sp” and
the F “p” electrons “look” like?
What does the orbital between the Be atomic “sp” and
the F “p” electrons “look” like?
Whenever we have a “single” bond we can assume that
it has the sigma shape, resulting from hybridization between
atomic orbitals
Sigma bond
Single bond

Whenever we have a “single” bond we can assume that
it has the sigma shape, resulting from hybridization between
atomic orbitals
Sigma bond
Single bond

For double and triple bonds, we do not need to create
more equivalent bonds which can be moved as far apart
as predicted by Valence Shell Electron Pair Repulsion.
Whenever we have a “single” bond we can assume that
it has the sigma shape, resulting from hybridization between
atomic orbitals
Sigma bond
Single bond

For double and triple bonds, we do not need to create
more equivalent bonds which can be moved as far apart
as predicted by Valence Shell Electron Pair Repulsion.
We need to simply create additional bonds within the
shape predicted by VSEPR
Whenever we have a “single” bond we can assume that
it has the sigma shape, resulting from hybridization between
atomic orbitals
Sigma bond
Single bond

For double and triple bonds, we do not need to create
more equivalent bonds which can be moved as far apart
as predicted by Valence Shell Electron Pair Repulsion.
We need to simply create additional bonds within the
shape predicted by VSEPR
Pi bond
Double bond around single bond



 F  Be  F 




F
1s
()
Be ()
F ()
2s
()

()
2p
() () ()
sp
    
() () ()
What does the orbital between the Be atomic “sp” and
the F “p” electrons “look” like?
What does BeF3 Look Like?
1s
(  )
F
2p
(  ) (  ) ( )
2s
(  )
sp2
 
   
B
(  )
( )
F
(  )
(  )
(  ) (  ) ( )
F
(  )
(  )
(  ) (  ) ( )

sp2
3 orbitals
F
 

F



B  F 

VSEPR
AX3
Triangular Planar
H
H  C H
H
 
H  N  H
H
What does ethylene Look Like?
H
H  C 
 


H
C  H
 
AX3
Geometry around
the carbon = Triangular Planar
H
H  C 
 


H
C  H
 
AX3
Geometry around
the carbon = Triangular Planar
What does acetylene look like?
H
H  C 
 


H
C  H
 
AX3
Geometry around
the carbon = Triangular Planar
H  C 
 
 
C  H
AX2
Geometry around the carbon =Linear
H
H  C 
 


H
C  H
 
AX3
Geometry around
the carbon = Triangular Planar
H  C 
 
 
C  H
AX2
Geometry around the carbon =Linear
Delocalization/Resonance Structures
Nitrate is a marshmallow
Poor Nitrate
kQ1Q2
E el 
r1  r2
1 charge
Large radius
Low
Charge
density