1: Shapes of the molecules and ions

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?
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

Finding out the shape of a molecule is easy!
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!

ELECTRON PAIR REPULSION THEORY
“THE SHAPE ADOPTED BY A SIMPLE MOLECULE OR ION IS THAT WHICH KEEPS REPULSIVE FORCES TO A MINIMUM” Bonds are closer together so repulsive forces are greater Al Molecules contain covalent bonds. As covalent bonds consist of a pair of electrons, each bond will repel other bonds. Bonds are further apart so repulsive forces are less 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 All bonds are equally spaced out as far apart as possible

REGULAR SHAPES C A covalent bond will repel another covalent bond 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 º BeCl2 3 TRIGONAL PLANAR 120º BCl3 4 TETRAHEDRAL º CH4 5 TRIGONAL BIPYRAMIDAL 90º & 120º PCl5 6 OCTAHEDRAL 90º SF6 BOND BOND PAIRS SHAPE ANGLE(S) EXAMPLE

180° LINEAR BERYLLIUM CHLORIDE Cl Be Cl Be BOND PAIRS 2 LONE PAIRS 0
Beryllium - has two electrons to pair up Chlorine - needs 1 electron for ‘octet’ Two covalent bonds are formed Beryllium still has an incomplete shell BOND PAIRS 2 LONE PAIRS 0 Cl Be 180° BOND ANGLE... SHAPE... 180° LINEAR

ADDING ANOTHER ATOM - ANIMATION

120° TRIGONAL PLANAR BORON CHLORIDE B Cl Cl B BOND PAIRS 3
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. BOND PAIRS 3 LONE PAIRS 0 Cl 120° Cl B BOND ANGLE... SHAPE... 120° TRIGONAL PLANAR Cl

ADDING ANOTHER ATOM - ANIMATION

109.5° TETRAHEDRAL METHANE H C H C BOND PAIRS 4 LONE PAIRS 0 C 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 BOND PAIRS 4 LONE PAIRS 0 109.5° H C BOND ANGLE... SHAPE... 109.5° TETRAHEDRAL

120° & 90° TRIGONAL BIPYRAMIDAL PHOSPHORUS(V) FLUORIDE F F P P
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’ BOND PAIRS 5 LONE PAIRS 0 120° F P 90° BOND ANGLE... SHAPE... 120° & 90° TRIGONAL BIPYRAMIDAL

90° OCTAHEDRAL SULPHUR(VI) FLUORIDE S S BOND PAIRS 6 LONE PAIRS 0 F S
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’ BOND PAIRS 6 LONE PAIRS 0 F S 90° BOND ANGLE... SHAPE... 90° OCTAHEDRAL

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 N H N H Nitrogen has five electrons in its outer shell
BOND PAIRS 3 LONE PAIRS 1 TOTAL PAIRS 4 H 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 N H N BOND PAIRS 3 LONE PAIRS 1 TOTAL PAIRS 4 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 107° H N H N N H ANGLE ° SHAPE... PYRAMIDAL H H

AMMONIA N H N BOND PAIRS 3 LONE PAIRS 1 TOTAL PAIRS 4 H

WATER O H O H Oxygen has six electrons in its outer shell
BOND PAIRS 2 LONE PAIRS 2 TOTAL PAIRS 4 H 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 O H O BOND PAIRS 2 LONE PAIRS 2 TOTAL PAIRS 4 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 O 104.5° H O O H H ANGLE ° SHAPE... ANGULAR H H H

XENON TETRAFLUORIDE 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 Xe Xe F Xe F F Xe F F F 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 ANGLE ° 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

SHAPES OF IONS EXAMPLE N Draw outer shell electrons of central atom

SHAPES OF IONS NH2- NH4+ Draw outer shell electrons of central atom
EXAMPLE 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 NH4+ NH2- N+ N

EXAMPLE 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 N NH4+ NH2- N+ N H N+ H N

EXAMPLE 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. N NH4+ NH2- N+ N H N+ H N BOND PAIRS 4 LONE PAIRS TETRADHEDRAL H-N-H ° BOND PAIRS 2 LONE PAIRS ANGULAR H-N-H °

Explain, in terms of electrons, how ammonia can react with hydrogen ions to form ammonium ions, NH (2) Ammonium ion 1. lone pair on N 2. forms dative / co-ordinate bond with H+

SHAPES OF IONS NH3 NH4+ NH2-
REVIEW N H N NH3 BOND PAIRS PYRAMIDAL LONE PAIRS H-N-H 107° H N+ N+ NH4+ BOND PAIRS TETRAHEDRAL LONE PAIRS H-N-H ° H N N NH2- BOND PAIRS ANGULAR LONE PAIRS H-N-H °

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 O C O C O 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

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 O C O C O 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 O C 180° 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. BOND ANGLE... SHAPE... 180° 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: The electron pairs arrange themselves as far apart from each other as possible in order to minimise the repulsions 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 .

TEST QUESTIONS BF3 SiCl4 PCl4+ PCl6- SiCl62- H2S
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- SiCl62- H2S ANSWERS ON NEXT PAGE

TEST QUESTIONS BF3 SiCl4 PCl4+ PCl6- SiCl62- H2S
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 4 bp 0 lp 109.5º tetrahedral silicon pairs up all 4 electrons in 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 electrons in its outer shell - 2 lone pairs are left SiCl4 PCl4+ PCl6- SiCl62- H2S

Predicting Molecular Geometry
Predict the following molecular geometries.... CH4 and PO43- XeF4 PCl5 BrF5 (Ans. Tetrahedral) (Ans. Square planar. Why not tetrahedral?) (Ans. Trigonal bipyramidal) (Ans. Square pyramidal)

Questions

The Octet Rule is Often Violated
H, Be, B, Al violate the octet rule (< 8 valence electrons) e.g. BeCl2, BH3, AlCl3 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? 1. They have dative bond. AlCl3 exists are dimer.

Lewis Structures for Organic Compounds
Alkanes: CnH2n+2 Methane, Ethane, Propane, Butane, Pentane, Hexane Alkenes: CnH2n have double bond(s) One double bond: Ethene (ethylene), Propene (propylene) Alcohols: CnH2n+1OH have hydroxyl group(s) methanol, ethanol 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 The tetrahedral centers of ethanol.
Predicting Molecular Shapes with More Than One Central Atom The tetrahedral centers of ethanol.

Halogenoalkane

Aldehyde

Ketone

SAMPLE PROBLEM 10.9 Predicting Molecular Shapes with More Than One Central Atom 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. SOLUTION: tetrahedral tetrahedral trigonal planar >1200 <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. (i) CH3– (2) (ii) CH (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 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

Answers

How can I improve on task 2?
Task 2: Review Go back to your lesson outcome grid and fill out the ‘How I did’ and the ‘Targets’ column. Lesson Outcomes How I did Targets Task 2: Grade B Met? Partly met? Not met? 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 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. is very hard. This is again due to the need to break very strong covalent bonds operating in 3-dimensions. doesn't conduct electricity. All the electrons are held tightly between the atoms, and aren't free to move. 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
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.

How can I improve on task 3?
Task 3: Review Go back to your lesson outcome grid and fill out the ‘How I did’ and the ‘Targets’ column. Lesson Outcomes How I did Targets Task 3: Grade A/A* Met? Partly met? Not met? 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 BeH2 LINEAR 180°

3 total 3 bond 0 lone B A BF3 TRIGONAL PLANAR 120°

3 total 2 bond 1 lone SO2 BENT <120°

4 total 4 bond 0 lone B A CH4 TETRAHEDRAL 109.5°

4 total 3 bond 1 lone NH3 TRIGONAL PYRAMIDAL 107°

4 total 2 bond 2 lone H2O BENT 104.5°

Be Ba 5 total 5 bond 0 lone PCl5 TRIGONAL BIPYRAMIDAL 120°/90°

6 total 6 bond 0 lone B A SF6 OCTAHEDRAL 90°

Examples F P F F PF3 4 total 3 bond 1 lone TRIGONAL PYRAMIDAL 107°

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
Shared Electron Pairs Unshared Electron Pairs Type Formula Ideal Geometry Observed Molecular Shape Examples AB2 AB2E AB2E2 AB2E3 AB3 AB3E AB3E2 AB4 AB4E AB4E2 AB5 AB5E AB6 2 3 4 5 6 1 2 3 Linear Trigonal planar Tetrahedral Trigonal bipyramidal Triangular bipyramidal Octahedral Linear Angular, or bent Trigonal planar Triangular pyramidal T-shaped Tetrahedral Irregular tetrahedral (or “see-saw”) Square planar Triangular bipyramidal Square pyramidal Octahedral 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

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