1: Shapes of the molecules and ions
Date
Lesson Outcomes
Task 1: I can demonstrate an understanding of the use of electron-pair
repulsion theory to interpret and predict the shapes of simple molecules and
ions
I can recall and explain the shapes of BeCl2, BCl3, CH4, NH3, NH4+, H2O, CO2,
gaseous PCl5 and SF6, and the simple organic molecules listed in Units 1 and 2
(GRADE C)
Task 2: I can apply the electron-pair repulsion theory to predict the shapes of
molecules and ions
I can show an understanding of the terms bond length and bond angle and
predict approximate bond angles in simple molecules and ions
(GRADE B)
Task 3: I can discuss the different structures formed by carbon atoms, including
graphite, diamond, fullerenes and carbon nanotubes, and the applications of
these, e.g. the potential to use nanotubes as vehicles to carry drugs into cells
(GRADE A)
BIG picture
• What skills will you be developing this lesson?
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ICT
Numeracy
Literacy
Team work
Self management
Creative thinking
Independent enquiry
Participation
Reflection
• How is this lesson relevant to every day life?
(WRL/CIT)
Shapes of Covalent Compounds
• What is the importance of
bonding in a compound?
• Bonding determines the physical
and chemical properties of the
compound.
• Covalent compounds form
molecules and ions whose
bonds are highly directional,
giving them a fixed shape which
determines the physical and
chemical properties of the
compound.
Shapes of Covalent Compounds
• Shape (structure) of
enzymes control the rate of
reaction in a biological
system.
• Example: organophosphates
are commonly used
insecticides. The molecules
are exactly the right shape to
interfere with an enzyme that
controls nerve impulses in an
insects body, so kiliing it.
Shapes of Ionic Compounds
Shapes of molecules and ions
All you need to do is to work out how many
electron pairs are involved in bonding and
then arrange them to produce the
minimum amount of repulsion between
them. You have to include both bonding
pairs and lone pairs.
Finding out
the shape of
a molecule is
easy!
6
ELECTRON PAIR REPULSION
THEORY
“THE SHAPE ADOPTED BY A SIMPLE MOLECULE OR ION IS
THAT WHICH KEEPS REPULSIVE FORCES TO A MINIMUM”
Molecules contain covalent
bonds. As covalent bonds
consist of a pair of
electrons, each bond will
repel other bonds.
Bonds will therefore push
each other as far apart as
possible to reduce the
repulsive forces.
Because the repulsions are
equal, the bonds will also be
equally spaced
Al
Bonds are closer
together so
repulsive forces
are greater
Bonds are
further apart so
repulsive forces
are less
Al
All bonds are
equally spaced
out as far
apart as
possible
REGULAR SHAPES
Molecules, or ions, possessing ONLY BOND
PAIRS of electrons fit into a set of
standard shapes. All the bond pair-bond
pair repulsions are equal.
All you need to do is to count up the
number of bond pairs and chose one of the
following examples...
BOND
PAIRS
SHAPE
BOND
ANGLE(S)
C
A covalent bond will
repel another
covalent bond
EXAMPLE
2
LINEAR
180º
BeCl2
3
TRIGONAL PLANAR
120º
BCl3
4
TETRAHEDRAL
109.5º
CH4
5
TRIGONAL BIPYRAMIDAL
90º & 120º
PCl5
6
OCTAHEDRAL
90º
SF6
BERYLLIUM CHLORIDE
Be
Cl
Cl
Be
Cl
Beryllium - has two electrons to pair up
Two covalent bonds are formed
Chlorine - needs 1 electron for ‘octet’
Beryllium still has an incomplete shell
BOND PAIRS
2
LONE PAIRS
0
180°
BOND ANGLE...
SHAPE...
180°
LINEAR
Cl
Be
Cl
ADDING ANOTHER ATOM - ANIMATION
BORON CHLORIDE
B
Cl
Cl
Cl
B
Boron - has three electrons to pair up
Cl
Chlorine - needs 1 electron to complete ‘octet’
Three covalent bonds are formed; Boron still
has an incomplete outer shell.
BOND PAIRS
3
LONE PAIRS
0
Cl
120°
Cl
BOND ANGLE...
B
120°
Cl
SHAPE...
TRIGONAL PLANAR
ADDING ANOTHER ATOM - ANIMATION
METHANE
H
H
H
C
C
H
H
Carbon - has four electrons to pair up
Four covalent bonds are formed
Hydrogen - 1 electron to complete shell
C and H now have complete shells
BOND PAIRS
4
LONE PAIRS
0
H
109.5°
C
BOND ANGLE...
SHAPE...
109.5°
TETRAHEDRAL
H
H
H
PHOSPHORUS(V) FLUORIDE
P
F
F
F
F
P
Phosphorus - has five electrons to pair up
F
Fluorine - needs one electron to complete ‘octet’
F
Five covalent bonds are formed; phosphorus can
make use of d orbitals to expand its ‘octet’
BOND PAIRS
5
LONE PAIRS
0
F
90°
F
120°
BOND ANGLE...
SHAPE...
120° & 90°
TRIGONAL BIPYRAMIDAL
P
F
F
F
SULPHUR(VI) FLUORIDE
F
F
S
F
F
S
Sulphur - has six electrons to pair up
F
Fluorine - needs one electron to complete ‘octet’
F
Six covalent bonds are formed; sulphur can
make use of d orbitals to expand its ‘octet’
BOND PAIRS
6
LONE PAIRS
0
F
F
90°
F
F
S
BOND ANGLE...
SHAPE...
90°
OCTAHEDRAL
F
F
F
Pentagonal bipyramidal
IRREGULAR SHAPES
If a molecule, or ion, has lone pairs on the central atom, the shapes are slightly
distorted away from the regular shapes. This is because of the extra repulsion
caused by the lone pairs.
BOND PAIR - BOND PAIR < LONE PAIR - BOND PAIR < LONE PAIR - LONE PAIR
O
O
O
As a result of the extra repulsion, bond angles tend to
be slightly less as the bonds are squeezed together.
AMMONIA
H
N
H
H
N
H
BOND PAIRS
3
LONE PAIRS
1
TOTAL PAIRS
4
• Nitrogen has five electrons in its outer shell
• It cannot pair up all five - it is restricted to eight electrons in its outer shell
• It pairs up only three of its five electrons
• 3 covalent bonds are formed and a pair of non-bonded electrons is left
• As the total number of electron pairs is 4, the shape is BASED on a tetrahedron
AMMONIA
H
H
N
H
H
N
BOND PAIRS
3
LONE PAIRS
1
TOTAL PAIRS
4
• The shape is based on a tetrahedron but not all the repulsions are the same
• LP-BP REPULSIONS > BP-BP REPULSIONS
• The N-H bonds are pushed closer together
• Lone pairs are not included in the shape
N
H
H
N
N
H
H
H
H
H
107°
H
H
ANGLE... 107°
SHAPE... PYRAMIDAL
AMMONIA
H
N
H
H
N
H
BOND PAIRS
3
LONE PAIRS
1
TOTAL PAIRS
4
WATER
H
O
H
H
O
BOND PAIRS
2
LONE PAIRS
2
TOTAL PAIRS
4
• Oxygen has six electrons in its outer shell
• It cannot pair up all six - it is restricted to eight electrons in its outer shell
• It pairs up only two of its six electrons
• 2 covalent bonds are formed and 2 pairs of non-bonded electrons are left
• As the total number of electron pairs is 4, the shape is BASED on a tetrahedron
WATER
H
H
O
H
O
BOND PAIRS
2
LONE PAIRS
2
TOTAL PAIRS
4
• The shape is based on a tetrahedron but not all the repulsions are the same
• LP-LP REPULSIONS > LP-BP REPULSIONS > BP-BP REPULSIONS
• The O-H bonds are pushed even closer together
• Lone pairs are not included in the shape
O
H
O
O
H
H
H
H
104.5°
H
ANGLE... 104.5°
SHAPE... ANGULAR
Bent
XENON TETRAFLUORIDE
F
F
Xe
Xe
F
F
BOND PAIRS
4
LONE PAIRS
2
TOTAL PAIRS
6
F
• Xenon has eight electrons in its outer shell
• It pairs up four of its eight electrons
• 4 covalent bonds are formed and 2 pairs of non-bonded electrons are left
• As the total number of electron pairs is 6, the shape is BASED on an octahedron
XENON TETRAFLUORIDE
F
F
Xe
F
Xe
F
BOND PAIRS
4
LONE PAIRS
2
TOTAL PAIRS
6
F
• As the total number of electron pairs is 6, the shape is BASED on an octahedron
• There are two possible spatial arrangements for the lone pairs
• The preferred shape has the two lone pairs opposite each other
F
Xe
F
F
Xe
F
F
F
F
F
ANGLE... 90°
SHAPE... SQUARE PLANAR
CALCULATING THE SHAPE OF IONS
The shape of a complex ion is calculated in the same way a molecule by...
• calculating the number of electrons in the outer shell of the central species
*
• pairing up electrons, making sure the outer shell maximum is not exceeded
• calculating the number of bond pairs and lone pairs
• using ELECTRON PAIR REPULSION THEORY to calculate shape and bond
angle(s)
* the number of electrons in the outer shell depends on the charge on the
ion
* if the ion is positive you remove as many electrons as there are positive
charges
* if the ion is negative you add as many electrons as there are negative
charges
e..g.
for PF6-
add one electron to the outer shell of P
for PCl4+
remove one electron from the outer shell of P
EXAMPLE
SHAPES OF IONS
Draw outer shell electrons of central atom
N
EXAMPLE
SHAPES OF IONS
Draw outer shell electrons of central atom
N
NH4+
For every positive charge on the ion,
remove an electron from the outer shell...
For every negative charge add an electron
to the outer shell...
for NH4+
remove 1 electron
for NH2-add 1 electron
N+
NH2N
SHAPES OF IONS
EXAMPLE
Draw outer shell electrons of central atom
N
NH2-
NH4+
For every positive charge on the ion,
remove an electron from the outer shell
N+
N
H
H
For every negative charge add an electron
to the outer shell..
for NH4+
for
NH2-add
remove 1 electron
1 electron
H
N+
H
Pair up electrons in the usual way
H
H
N
SHAPES OF IONS
EXAMPLE
Draw outer shell electrons of central atom
N
NH2-
NH4+
For every positive charge on the ion,
remove an electron from the outer shell
N+
N
H
H
For every negative charge add an electron
to the outer shell..
for NH4+
for
NH2-add
remove 1 electron
1 electron
H
N+
H
H
N
H
Pair up electrons in the usual way
Work out shape and bond angle(s) from
number of bond pairs and lone pairs.
BOND PAIRS
4
BOND PAIRS
2
LONE PAIRS
0
LONE PAIRS
2
TETRADHEDRAL
ANGULAR
H-N-H 109.5°
H-N-H 104.5°
Ammonium ion
Explain, in terms of electrons, how ammonia can react with
hydrogen ions to form ammonium ions, NH4+.
(2)
1. lone pair on N
2. forms dative / co-ordinate bond with H+
SHAPES OF IONS
REVIEW
H
NH3
N
H
N
BOND PAIRS
3
PYRAMIDAL
LONE PAIRS
1
H-N-H 107°
BOND PAIRS
4
TETRAHEDRAL
LONE PAIRS
0
H-N-H 109.5°
BOND PAIRS
2
ANGULAR
LONE PAIRS
2
H-N-H 104.5°
H
H
NH4
+
N+
H
N+
H
H
H
NH
2
N
H
N
http://www.chemguide.co.uk/atoms/bonding/shapesdouble.html
MOLECULES WITH DOUBLE BONDS
The shape of a compound with a double bond is calculated in the same way.
A double bond repels other bonds as if it was single e.g. carbon dioxide
C
O
O
C
O
Carbon - needs four electrons to complete its shell
The atoms share two electrons
Oxygen - needs two electron to complete its shell
each to form two double bonds
MOLECULES WITH DOUBLE BONDS
The shape of a compound with a double bond is calculated in the same way.
A double bond repels other bonds as if it was single e.g. carbon dioxide
C
O
O
C
O
Carbon - needs four electrons to complete its shell
The atoms share two electrons
Oxygen - needs two electron to complete its shell
each to form two double bonds
DOUBLE BOND PAIRS
2
LONE PAIRS
0
Double bonds behave exactly as single
bonds for repulsion purposes so the
shape will be the same as a molecule with
two single bonds and no lone pairs.
180°
O
C
O
BOND ANGLE... 180°
SHAPE... LINEAR
ELECTRON PAIR REPULSION THEORY
This theory can be used to interpret and predict the
shape of molecules. It is based on the following
ideas:
1. The electron pairs arrange themselves as far
apart from each other as possible in order to
minimise the repulsions
2. The repulsions between lone pairs is greater than
that between lone pair and a bond pair than that
between two bond pairs.
 The number of sigma bond pairs of electrons and
lone pairs in the molecule should be counted.
 Any pi bond pairs should be ignored when working
out the shape of the molecule .
RECAP
37
RECAP
38
TEST QUESTIONS
For each of the following ions/molecules,
state the number of bond pairs
state the number of lone pairs
state the bond angle(s)
state, or draw, the shape
BF3
SiCl4
PCl4+
PCl6-
SiCl62H2S
ANSWERS ON NEXT PAGE
TEST QUESTIONS
ANSWER
For each of the following ions/molecules,
state the number of bond pairs
state the number of lone pairs
state the bond angle(s)
state, or draw, the shape
BF3
3 bp
0 lp
120º
trigonal planar
boron pairs up all 3 electrons in
its outer shell
SiCl4
4 bp
0 lp
109.5º
tetrahedral
silicon pairs up all 4 electrons in
its outer shell
PCl4+
4 bp
0 lp
109.5º
tetrahedral
as ion is +, remove an electron
in the outer shell then pair up
PCl6-
6 bp
0 lp
90º
octahedral
as the ion is - , add one electron to
the 5 in the outer shell then pair up
SiCl62-
6 bp
0 lp
90º
octahedral
as the ion is 2-, add two electrons
to the outer shell then pair up
H2S
2 bp
2 lp
92º
angular
sulphur pairs up 2 of its 6
electrons in its outer shell 2 lone pairs are left
Predicting Molecular Geometry
Predict the following molecular geometries....
1. CH4 and PO43- (Ans. Tetrahedral)
(Ans. Square planar. Why not tetrahedral?)
2. XeF4
(Ans. Trigonal bipyramidal)
3. PCl5
4. BrF5
(Ans. Square pyramidal)
Questions
The Octet Rule is Often Violated
1. H, Be, B, Al violate the octet rule (< 8 valence electrons)
e.g. BeCl2, BH3, AlCl3
2. Nonmetals with a valence shell greater than n = 2
(e.g. P, Cl, Br, I, etc.)
–
May violate the octet rule when they are the CENTRAL
atom (e.g. ClF5 )
• How can they do this?
• Why doesn’t Fluorine violate the octet rule?
Lewis Structures for Organic
Compounds
1.
Alkanes: CnH2n+2
– Methane, Ethane, Propane, Butane, Pentane, Hexane
2. Alkenes: CnH2n have double bond(s)
– One double bond: Ethene (ethylene), Propene
(propylene)
3. Alcohols: CnH2n+1OH have hydroxyl group(s)
– methanol, ethanol
4. Carboxylic Acids: CnH2n+1COOH
have carboxyl group(s)
– Methanoic acid (formic acid), HCOOH
– Ethanoic acid (acetic acid, CH3COOH
Alkane
Butane
Click on the hyperlink to see: wire model, ball and stick model, space filling model and
other carbon compounds
Alkene
ETHANOL
Predicting Molecular
Shapes with More Than
One Central Atom
The tetrahedral
centers of ethanol.
Halogenoalkane
Aldehyde
Ketone
SAMPLE PROBLEM 10.9
PROBLEM:
PLAN:
Predicting Molecular Shapes with More Than
One Central Atom
Determine the shape around each of the central atoms in
acetone, (CH3)2C=O.
Find the shape of one atom at a time after writing the Lewis
structure.
SOLUTION:
tetrahedral
H
H C
H
O
C
tetrahedral
H
C H
H
trigonal planar
O
H
H
C
C
C
H
HH
>1200
H
<1200
Questions
1. The following ions may be formed as intermediates in
chemical reactions. Their shapes can be predicted in the
usual way. Draw a clear diagram of each, indicating values
for the bond angles.
Pyramidal diagram (1), angle < 109° (1)
(i)
CH3–
(2)
(ii)
CH3+
(2)
Trigonal planar (1), 120° (1)
2. Outline the basic principle for predicting the shape of a
simple molecule.
(5)
1.pairs of electrons
2.around a central atom
3. repel each other
4.arranging themselves as far apart as possible
5. to minimise repulsion or reach lowest energy state
Questions
Questions
Answers
Task 2: Review
Go back to your lesson outcome grid and fill out
the ‘How I did’ and the ‘Targets’ column.
Lesson Outcomes
Task 2:
Grade B
How I did
Met?
Partly met?
Not met?
Targets
How can I improve
on task 2?
Allotropy Task 3
Diamond
• Pure Diamond is
composed entirely of
interlocking carbon
atoms, each of which is
covalently bonded to its
four nearest neighboring
carbon atoms.
• Due to the strong C-C
bonds and interlocked
crystal structure,
Diamond is the hardest
known substance.
How to draw the structure of diamond?
 Properties of Diamond
1. has a very high melting point (almost 4000°C). Very strong carbon-carbon
covalent bonds have to be broken throughout the structure before melting
occurs.
2. is very hard. This is again due to the need to break very strong covalent bonds
operating in 3-dimensions.
3. doesn't conduct electricity. All the electrons are held tightly between the atoms,
and aren't free to move.
4. is insoluble in water and organic solvents. There are no possible attractions
which could occur between solvent molecules and carbon atoms which could
outweigh the attractions between the covalently bound carbon atoms.
Graphite
335 pm
142 pm
Properties of Graphite
1.
2.
3.
4.
5.
has a high melting point, similar to that of diamond. In order to melt
graphite, it isn't enough to loosen one sheet from another. You have to
break the covalent bonding throughout the whole structure.
has a soft, slippery feel, and is used in pencils and as a dry lubricant for
things like locks. You can think of graphite rather like a pack of cards each card is strong, but the cards will slide over each other, or even fall
off the pack altogether. When you use a pencil, sheets are rubbed off
and stick to the paper.
has a lower density than diamond. This is because of the relatively large
amount of space that is "wasted" between the sheets.
is insoluble in water and organic solvents - for the same reason that
diamond is insoluble. Attractions between solvent molecules and carbon
atoms will never be strong enough to overcome the strong covalent
bonds in graphite.
conducts electricity. The delocalised electrons are free to move
throughout the sheets. If a piece of graphite is connected into a circuit,
electrons can fall off one end of the sheet and be replaced with new ones
at the other end.
Other forms of Carbon
Amorphous Carbon: Non Crystalline form of
Carbon
Fullerenes
Nanotubes
Task 3: (Grade A)
Answers
1 In diamond, all the carbon atoms are bonded together by
strong covalent bonds. The diamond is hard because it
is difficult to split the structure. All of the electrons are
fixed in bonds. In graphite, there is strong bonding within
a 2-D layer but the forces between the layers are very
weak so layers can slide over each other. There are free
electrons within the structure that can move and conduct
electricity.
2 Fullerenes are molecular but diamond and graphite are
giant structures of atoms (macromolecular).
3 There has been insufficient long-term research of the
penetration of the skin by nanoparticles.
Task 3: Review
Go back to your lesson outcome grid and fill out
the ‘How I did’ and the ‘Targets’ column.
Lesson Outcomes
Task 3:
Grade A/A*
How I did
Met?
Partly met?
Not met?
Targets
How can I improve
on task 3?
Homework
• Homework task: Read page 152 and 153 and
use AUTOLOGY to explain the applications of
these, e.g. the potential to use nanotubes as
vehicles to carry drugs into cells.
• Due date: NEXT LESSON
Review of lesson
The following slides will give you information about
the bonds of the molecule.
You need to identify the name of the shape and give an
example of the molecule.
Common Molecular Shapes
2 total
2 bond
0 lone
B
A
B
LINEAR
BeH2
180°
Common Molecular Shapes
3 total
3 bond
0 lone
B
A
B
BF3
B
TRIGONAL PLANAR
120°
Common Molecular Shapes
3 total
2 bond
1 lone
SO2
BENT
<120°
Common Molecular Shapes
4 total
4 bond
0 lone
B
A
B
B
B
CH4
TETRAHEDRAL
109.5°
Common Molecular Shapes
4 total
3 bond
1 lone
NH3
TRIGONAL PYRAMIDAL
107°
Common Molecular Shapes
4 total
2 bond
2 lone
H2O
BENT
104.5°
Common Molecular Shapes
Ba
5 total
5 bond
0 lone
Be
Be
Be
Ba
PCl5
TRIGONAL
BIPYRAMIDAL
120°/90°
Common Molecular Shapes
6 total
6 bond
0 lone
B
B
A
B
B
B
B
SF6
OCTAHEDRAL
90°
Examples
• PF3
4 total
3 bond
1 lone
F P F
F
TRIGONAL
PYRAMIDAL
107°
144-145
146-149
146-149
146-149
150-151
I can demonstrate an understanding of the
use of electron-pair repulsion theory to
interpret and predict the shapes of simple
molecules and ions (2.3a)
I can recall and explain the shapes of BeCl2,
BCl3, CH4, NH3, NH4+, H2O, CO2, gaseous
PCl5 and SF6, and the simple organic
molecules listed in Units 1 and 2 (2.3b)
I can apply the electron-pair repulsion theory
to predict the shapes of molecules and ions
(2.3c)
I can show an understanding of the terms
bond length and bond angle and predict
approximate bond angles in simple molecules
and ions (2.3d)
I can discuss the different structures formed
by carbon atoms, including graphite,
diamond, fullerenes and carbon nanotubes,
and the applications of these, e.g. the
potential to use nanotubes as vehicles to
carry drugs into cells (2.3e)
Geometry of Covalent Molecules ABn, and ABnEm
Type
Formula
Shared
Electron
Pairs
Unshared
Electron
Pairs
AB2
AB2E
AB2E2
AB2E3
AB3
AB3E
2
2
2
2
3
3
0
1
2
3
0
1
Linear
Trigonal planar
Tetrahedral
Trigonal bipyramidal
Trigonal planar
Tetrahedral
Linear
Angular, or bent
Angular, or bent
Linear
Trigonal planar
Triangular pyramidal
CdBr2
SnCl2, PbI2
OH2, OF2, SCl2, TeI2
XeF2
BCl3, BF3, GaI3
NH3, NF3, PCl3, AsBr3
AB3E2
AB4
3
4
2
0
Triangular bipyramidal
Tetrahedral
T-shaped
Tetrahedral
ClF3, BrF3
CH4, SiCl4, SnBr4, ZrI4
AB4E
4
1
Triangular bipyramidal
SF4, SeCl4, TeBr4
AB4E2
AB5
4
5
2
0
Octahedral
Triangular bipyramidal
Irregular tetrahedral
(or “see-saw”)
Square planar
Triangular bipyramidal
AB5E
AB6
5
6
1
0
Octahedral
Octahedral
Square pyramidal
Octahedral
ClF3, BrF3, IF5
SF6, SeF6, Te(OH)6,
MoF6
Ideal
Geometry
Observed
Molecular Shape
Examples
XeF4
PF5, PCl5(g), SbF5