1: Shapes of the molecules and ions DateLesson OutcomesTask 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?

• 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

6

Shapes of molecules and ions

Finding out the

shape of a molecule is easy!

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 bonding pairs and lone

pairs.pairs.

ELECTRON PAIR REPULSION ELECTRON PAIR REPULSION THEORYTHEORY

“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.

AlBonds are further apart so repulsive forces are less

Bonds are closer together so repulsive forces are greater

All bonds are equally spaced out as far apart as possible

Al

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

REGULAR SHAPESREGULAR 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...

2 LINEAR 180º BeCl2

3 TRIGONAL PLANAR 120º BCl3

4 TETRAHEDRAL 109.5º CH4

5 TRIGONAL BIPYRAMIDAL 90º & 120º PCl5

6 OCTAHEDRAL 90º SF6

BOND BONDPAIRS SHAPE ANGLE(S) EXAMPLE

C

A covalent bond will repel another covalent bond

BERYLLIUM CHLORIDEBERYLLIUM CHLORIDE

Cl ClBe180°

BOND PAIRS 2

LONE PAIRS 0

BOND ANGLE...

SHAPE...

180°

LINEAR

ClBe Be ClCl

Beryllium - has two electrons to pair up

Chlorine - needs 1 electron for ‘octet’

Two covalent bonds are formed

Beryllium still has an incomplete shell

ADDING ANOTHER ATOM - ANIMATIONADDING ANOTHER ATOM - ANIMATION

B

BORON CHLORIDEBORON CHLORIDE

Cl

Cl

B120°

Cl

ClB

Cl

Cl

Cl

BOND PAIRS 3

LONE PAIRS 0

BOND ANGLE...

SHAPE...

120°

TRIGONAL PLANAR

Boron - has three electrons to pair up

Chlorine - needs 1 electron to complete ‘octet’

Three covalent bonds are formed; Boron still has an incomplete outer shell.

ADDING ANOTHER ATOM - ANIMATIONADDING ANOTHER ATOM - ANIMATION

METHANEMETHANE

BOND PAIRS 4

LONE PAIRS 0

BOND ANGLE...

SHAPE...

109.5°

TETRAHEDRAL

C H CH

H

H

H

109.5°

H H

C

H

H

Carbon - has four electrons to pair up

Hydrogen - 1 electron to complete shell

Four covalent bonds are formed

C and H now have complete shells

PHOSPHORUS(V) FLUORIDEPHOSPHORUS(V) FLUORIDE

FP

P

F

F

F

F

F

BOND PAIRS 5

LONE PAIRS 0

BOND ANGLE...

SHAPE...

120° & 90°

TRIGONAL BIPYRAMIDAL

120°

F

F

P

F

F

F

90°

Phosphorus - has five electrons to pair up

Fluorine - needs one electron to complete ‘octet’

Five covalent bonds are formed; phosphorus can make use of d orbitals to expand its ‘octet’

SULPHUR(VI) FLUORIDESULPHUR(VI) FLUORIDE

FS

BOND PAIRS 6

LONE PAIRS 0

BOND ANGLE...

SHAPE...

90°

OCTAHEDRAL

S

F

F

F

F

F

F

Sulphur - has six electrons to pair up

Fluorine - needs one electron to complete ‘octet’

Six covalent bonds are formed; sulphur can make use of d orbitals to expand its ‘octet’

F

F F

FS

F

F

90°

Pentagonal bipyramidal

IRREGULAR SHAPESIRREGULAR 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

OO O

As a result of the extra repulsion, bond angles tend to be slightly less as the bonds are squeezed together.

AMMONIAAMMONIA

HN NH H

HBOND 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

AMMONIAAMMONIA

ANGLE... 107°

SHAPE... PYRAMIDAL

HN NH H

HBOND PAIRS 3

LONE PAIRS 1

TOTAL PAIRS 4

H

H

N

H

H

H

N

H

107°H

H

N

H

• 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

AMMONIAAMMONIA

HN NH H

HBOND PAIRS 3

LONE PAIRS 1

TOTAL PAIRS 4

WATERWATER

H OH

HBOND PAIRS 2

LONE PAIRS 2

TOTAL PAIRS 4

O

• 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

ANGLE... 104.5°

SHAPE... ANGULAR

H

O

H

H OH

HBOND PAIRS 2

LONE PAIRS 2

TOTAL PAIRS 4

O

H

O

H

104.5°H

O

H

• 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

WATERWATER

Bent

XENON TETRAFLUORIDEXENON TETRAFLUORIDE

FXe Xe

F

F

F

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

BOND PAIRS 4

LONE PAIRS 2

TOTAL PAIRS 6

XENON TETRAFLUORIDEXENON TETRAFLUORIDE

FXe Xe

F

F

F

F

F

F F

FXe

F

F F

FXe

ANGLE... 90°

SHAPE... SQUARE PLANAR

• 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

BOND PAIRS 4

LONE PAIRS 2

TOTAL PAIRS 6

CALCULATING THE SHAPE OF IONSCALCULATING THE SHAPE OF IONSThe 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

SHAPES OF IONSSHAPES OF IONSDraw outer shell electrons of central atom N

EXAMPLEEXAMPLEEXAMPLEEXAMPLE

SHAPES OF IONSSHAPES OF IONS

N+ N

NH4+ NH2

-

Draw outer shell electrons of central atom

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

EXAMPLEEXAMPLEEXAMPLEEXAMPLE

SHAPES OF IONSSHAPES OF IONS

N+

H

H

H

H N+

N

H

H N

NH4+ NH2

-

Draw outer shell electrons of central atom

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

Pair up electrons in the usual way

EXAMPLEEXAMPLEEXAMPLEEXAMPLE

N

SHAPES OF IONSSHAPES OF IONS

N+

H

H

H

H N+

N

H

H N

NH4+ NH2

-

BOND PAIRS 4

LONE PAIRS 0

TETRADHEDRAL

H-N-H 109.5°

BOND PAIRS 2

LONE PAIRS 2

ANGULAR

H-N-H 104.5°

Draw outer shell electrons of central atom

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

Pair up electrons in the usual way

Work out shape and bond angle(s) from number of bond pairs and lone pairs.

EXAMPLEEXAMPLEEXAMPLEEXAMPLE

N

Ammonium ionExplain, in terms of electrons, how ammonia can react with hydrogen ions to form ammonium ions, NH4+. (2)

1. lone pair on N2. forms dative / co-ordinate bond with H+

SHAPES OF IONSSHAPES OF IONS

N BOND PAIRS 3 PYRAMIDAL

LONE PAIRS 1 H-N-H 107°

BOND PAIRS 4 TETRAHEDRAL

LONE PAIRS 0 H-N-H 109.5°

N

H

H

H

N+H

H

H

H N+

BOND PAIRS 2 ANGULAR

LONE PAIRS 2 H-N-H 104.5°N

H

H N

NH4+

NH2

-

NH3

REVIEWREVIEWREVIEWREVIEW

http://www.chemguide.co.uk/atoms/bonding/shapesdouble.html

MOLECULES WITH DOUBLE BONDSMOLECULES WITH DOUBLE BONDS

C O C OO

Carbon - needs four electrons to complete its shell

Oxygen - needs two electron to complete its shell

The atoms share two electrons

each to form two 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

MOLECULES WITH DOUBLE BONDSMOLECULES WITH DOUBLE BONDS

C O C OO

Carbon - needs four electrons to complete its shell

Oxygen - needs two electron to complete its shell

The atoms share two electrons

each to form two double bonds

DOUBLE BOND PAIRS 2

LONE PAIRS 0

BOND ANGLE...

SHAPE...

180°

LINEAR

O OC180°

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.

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

ELECTRON PAIR REPULSION THEORYELECTRON 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 repulsions2.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 .

37

RECAP

38

RECAP

ANSWERS ON NEXT PAGE

TEST QUESTIONSTEST QUESTIONS

For each of the following ions/molecules, state the number of bond pairsstate the number of lone pairsstate the bond angle(s)state, or draw, the shape

SiCl4

PCl6-

H2S

SiCl62-

PCl4+

BF3

TEST QUESTIONSTEST QUESTIONS

3 bp 0 lp 120º trigonal planar boron pairs up all 3 electrons inits outer shell

4 bp 0 lp 109.5º tetrahedral silicon pairs up all 4 electrons in

its outer shell 4 bp 0 lp 109.5º tetrahedral as ion is +, remove an electron

in the outer shell then pair up 6 bp 0 lp 90º octahedral as the ion is - , add one electron to

the 5 in the outer shell then pair up 6 bp 0 lp 90º octahedral as the ion is 2-, add two electrons

to the outer shell then pair up 2 bp 2 lp 92º angular sulphur pairs up 2 of its 6

electrons in its outer shell - 2 lone pairs are left

BF3

SiCl4

PCl6-

H2S

SiCl62-

PCl4+

ANSWERANSWER

For each of the following ions/molecules, state the number of bond pairsstate the number of lone pairsstate the bond angle(s)state, or draw, the shape

Predicting Molecular Geometry

Predict the following molecular geometries....

1. CH4 and PO43-

2. XeF4

3. PCl5

4. BrF5

(Ans. Tetrahedral)

(Ans. Square planar. Why not tetrahedral?)

(Ans. Trigonal bipyramidal)

(Ans. Square pyramidal)

QuestionsQuestions

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, ethanol4. 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

The tetrahedral centers of ethanol.

Predicting Molecular Shapes with More Than One Central Atom

ETHANOL

Halogenoalkane

Aldehyde

Ketone

SAMPLE PROBLEM 10.9 Predicting Molecular Shapes with More Than One Central Atom

SOLUTION:

PROBLEM: Determine the shape around each of the central atoms in acetone, (CH3)2C=O.

PLAN: Find the shape of one atom at a time after writing the Lewis structure.

C C C

OH

H

H

HH

H

tetrahedral tetrahedral

trigonal planar

C

O

HC

HHH

CH

H>1200

<1200

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.(i) CH3– (2)(ii) CH3+ (2)

Pyramidal diagram (1), angle < 109° (1)

Trigonal planar (1), 120° (1)

2. Outline the basic principle for predicting the shape of a simple molecule. (5)

1.pairs of electrons2.around a central atom3. repel each other4.arranging themselves as far apart as possible5. to minimise repulsion or reach lowest energy state

Questions

Questions

Questions

Answers

Task 2: Review

Lesson Outcomes How I did Targets

Task 2:

Grade B

Met?

Partly met?

Not met?

How can I improve on task 2?

Go back to your lesson outcome grid and fill out the ‘How I did’ and the ‘Targets’ column.

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

142 pm

335 pm

Properties of Graphite1. 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.

2. 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.

3. has a lower density than diamond. This is because of the relatively large amount of space that is "wasted" between the sheets.

4. 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.

5. 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

Lesson Outcomes How I did Targets

Task 3:

Grade A/A*

Met?

Partly met?

Not met?

How can I improve on task 3?

Go back to your lesson outcome grid and fill out the ‘How I did’ and the ‘Targets’ column.

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

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.

Review of lesson

Common Molecular Shapes

2 total

2 bond

0 lone

LINEAR180°BeH2

B BA

3 total

3 bond

0 lone

TRIGONAL PLANAR

120°

BF3

Common Molecular Shapes

B

B

A

B

Common Molecular Shapes

3 total

2 bond

1 lone

BENT

<120°

SO2

4 total

4 bond

0 lone

TETRAHEDRAL

109.5°

CH4

Common Molecular Shapes

B

A

BB

B

4 total

3 bond

1 lone

TRIGONAL PYRAMIDAL

107°

NH3

Common Molecular Shapes

4 total

2 bond

2 lone

BENT

104.5°

H2O

Common Molecular Shapes

5 total

5 bond

0 lone

TRIGONAL BIPYRAMIDAL

120°/90°PCl5

Common Molecular Shapes

A Be

Be

Be

Ba

Ba

6 total

6 bond

0 lone

OCTAHEDRAL

90°

SF6

Common Molecular Shapes

B

B

B

B

B

BA

• PF3

4 total

3 bond

1 lone

TRIGONAL PYRAMIDAL

107°

F P FF

Examples

144-145 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)

146-149 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)

146-149 I can apply the electron-pair repulsion theory to predict the shapes of molecules and ions (2.3c)

146-149 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)

150-151 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

AB2

AB2EAB2E2

AB2E3

AB3

AB3E

AB3E2

AB4

AB4E

AB4E2

AB5

AB5EAB6

222233

34

4

45

56

012301

20

1

20

10

LinearTrigonal planarTetrahedralTrigonal bipyramidalTrigonal planarTetrahedral

Triangular bipyramidalTetrahedral

Triangular bipyramidal

OctahedralTriangular bipyramidal

OctahedralOctahedral

LinearAngular, or bentAngular, or bentLinearTrigonal planarTriangular pyramidal

T-shapedTetrahedral

Irregular tetrahedral (or “see-saw”)Square planarTriangular bipyramidal

Square pyramidalOctahedral

CdBr2

SnCl2, PbI2

OH2, OF2, SCl2, TeI2

XeF2

BCl3, BF3, GaI3

NH3, NF3, PCl3, AsBr3

ClF3, BrF3

CH4, SiCl4, SnBr4, ZrI4

SF4, SeCl4, TeBr4

XeF4

PF5, PCl5(g), SbF5

ClF3, BrF3, IF5

SF6, SeF6, Te(OH)6, MoF6

TypeFormula

Shared Electron

Pairs

Unshared Electron

Pairs

IdealGeometry

ObservedMolecular Shape Examples