SMK SULTAN ISMAIL, JB NUR FATHIN SUHANA BT AYOB

Transcription

1 SMK SULTAN ISMAIL, JB NUR FATHIN SUHANA BT AYOB

2 4.4 Metallic bond

3 Formation of Metallic Bond Relationship between metallic bond & boiling point 4 1 Topic Overview 3 2 Properties of metal Factors affect the strength of metallic bond

4 LEARNING OUTCOMES At the end of the lesson student should be able to: (a) Explain the formation of metallic bond by using electron sea model. (b) Relate metallic bond to the properties of metal: i. malleability ii. ductility iii. electrical conductivity iv. thermal conductivity (c) Explain the factors that affect the strength of metallic bond. (d) Relate the strength of metallic bond to boiling point.

5 FORMATION OF METALLIC BOND Metal Are shiny solids at room temperature (except mercury), with characteristic high melting points and densities Metallic Bond Electrostatic attraction between the charged metallic ions and the sea or cloud of delocalised electrons.

6 METALLIC BONDS 1. It defined as the electrostatic attraction between the positively charged ions and the sea or cloud of delocalised electrons.

7 FORMATION OF METALLIC BOND The electron sea model Positive ions Delocalised valence electrons

8 METALLIC BONDING 1. Metal has an unique characteristics that is not shared by other substances (ionic or covalent) which is it can conduct electricity in the solid state. 2. Metals are hard, with high melting and boiling points. 3. It is ductile (melentur/ can be pulled into thin wires) and malleable (be knocked into shapes) 4. Metals are good conductor of heat. 5. X-ray crystallography shows that the atoms in metals are packed very tightly against one another in the crystal lattice. 6. The force that holds the atoms together in the solid state is called metallic bonds.

9 METALLIC BONDING 7. Sodium as an example. In the solid state, the sodium atoms are packed so close together that the 3s orbitals (containing one unpaired electron) of the atoms overlap with one another to form a giant molecular orbital. 8. The valence electron of each sodium atom is then free to move throughout the whole crystal structure and is no more bound to the outer shell of any sodium atom. Electrons are ionised and delocalised throughout the whole solid. 9. The attraction between the cations and the delocalised constitute the metallic bond. The delocalised electrons are referred to as a sea of delocalised mobile electrons.

10 METALLIC BONDING 10. When a electric potential is applied to the metal, the mobile electrons will move towards the positive terminal and conduct electricity (GOOD CONDUCTOR OF ELECTRICAL CONDUCTIVITY). 11. When one end of the metal is heated, the kinetic energy of the electrons increase. Then, it is transmitted to the other end by the delocalised mobile electrons. (GOOD CONDUCTOR OF HEAT) 12. Metals have low ionisation energy. The valence electron is easily excited to higher energy levels. When electrons returned to ground state, the excess energy is liberated as electromagnetic radiation, which makes solids shine. 13. When a piece of metal is knocked or pulled, the layers of cations simply slide over one another but are still held together by flexible glue of the delocalised electrons.

11 PROPERTIES OF METAL Malleability Ductility Is the ability of a metal to be hammered into shapes Is the ability of a metal to be drawn into wire. Whenever any stress is applied to the metals, one layer of atoms can slide over another without disrupting the metallic bonding. Metal ion Free electron

12 PROPERTIES OF METAL Electrical conductivity Good conductors of electric. When an electrical potential is applied to the metal, the mobile electrons will move towards the positive terminal and thus conducts electricity. Thermal conductivity Good conductors of heat. When one end of the metal is heated, the kinetic energy of the electrons increases. This is then transmitted to the other end by the system of delocalized mobile electrons.

13 PHYSICAL PROPERTIES OF METALS metals have high melting point high energy is required to overcome these strong electrostatic forces between the positive ions and the electron sea in the metallic bond e e e e e e e e e e e e e e e e e e Metallic bonds formed by the electrostatic forces exist between positive ions and the free moving electrons 13

14 METALLIC BOND ELECTRON SEA MODEL 1. All metal atoms in the sample contribute their valence electron to form an electron sea that is delocalized throughout the structure. 2. The valence electron (electron sea) are shared among all atoms in the substances 3. The mobile electron sea is acting as glue bonding nucleus (+ve charge) from repelling each others.

15 HOW THE MODEL EXPLAINS THE PROPERTIES OF METALS 1. Why metals are malleable and ductile? 2. Explanation : Because when a piece of metal performed by hammer, the metal ions (nuclei) slide past each other through the electron sea to new lattice positions So, the metal ions (nuclei) do not repel each other

16 HOW THE MODEL EXPLAINS THE PROPERTIES OF METALS 5. A : An external force applied to a piece of metal deforms the piece without breaking it 6. B : The force simply move metal ions (nuclei) past each other through the surrounding electron sea.

17 ELECTROSTATIC FORCE AND THE REASON IONIC COMPOUNDS CRACK 1. Why metals conduct heat and electricity well in both solid and liquid states? 2. Explanation : because the metal atoms have mobile electrons

18 FACTORS AFFECTING THE STRENGTH OF METALLIC BOND 1. Number of valence electron : No. of valence electrons increase, delocalized electron increase Attraction between nucleus and delocalized electron increase Strength metallic bond increase 2. Size of atoms : Size decrease Attraction between nucleus and delocalized electron increase Strength metallic bond increase

19 FACTORS AFFECTING THE STRENGTH OF METALLIC BOND 1. the strength of metallic bond number of valence electrons per atom metallic radius

20 FACTORS AFFECTING THE STRENGTH OF METALLIC BOND 3. Eg : the melting and boiling points of three metals are : Metal Sodium Magnesium Aluminium No. of valence electrons Melting point ( ) Boiling point ( ) Melting/boiling point increase in order Na<Mg<Al. Because : number of electrons per atom donated to the sea of delocalised electrons increases from 1 for sodium, 2 for magnesium, 3 for aluminium. So, the metallic bonding in Al is strongest. (factor no. 1)

21 FACTORS AFFECTING THE STRENGTH OF METALLIC BOND 4. Eg : The melting point beryllium and magenesium : Element Beryllium Magnesium No. of valence electrons 2 2 Atomic radius/nm Melting point ( ) The bigger size of the Mg atoms causes the attraction between the nucleus and the sea of delocalised electrons to be weaker. so, Mg has lower melting point.

22 FACTORS AFFECTING THE STRENGTH OF METALLIC BOND 5. However, the strength of metallic bonds also depend on the packing of atoms in the solid crystal. Element Magnesium Calcium No. of valence electrons 2 2 Atomic radius/nm Melting point ( ) Lattice structure Hexagon Face-centered cube Ca is larger than Mg. But, melting point is higher than Mg. This close-packed structure brings the calcium atom closer to one another compared to Mg.

23 Question : Explain the difference in the boiling point of the two metals given: Magnesium o C Aluminum o C 23

24 Answer The cationic size of Al is smaller compared to magnesium and its charge is higher (+3). Mg has two valence electrons Al has three valence electrons involved in the metallic bonding. The strength of metallic bond in Aluminium is greater than that of Magnesium Al has higher boiling point 24

25 The strength of metallic bond is directly proportional to the boiling point. The stronger metallic bond,the higher the boiling point. 25

26 4.5 INTERMOLECULAR FORCES

27 Type of Intermolecular forces 1 Exercises 4 Topic Outline 2 Factor influence the strength of van der waals forces 3 Effect of Hydrogen bond 27

28 LEARNING OUTCOMES At the end of the lesson student should be able to: (a) (b) Describe intermolecular forces: i. Van Der Waal Forces - dipole-dipole interactions - London forces or dispersion forces ii. Hydrogen bonding Explain the factors that influence the strength of van der Waals forces.

29 LEARNING OUTCOMES At the end of the lesson student should be able to: (c) (d) Explain the effects of hydrogen bonding on the following physical properties: i. boiling point ii. solubility iii. density of water compared to ice. Explain the relationship between: i. intermolecular forces and vapour pressure ii. vapour pressure and boiling point.

30 TYPE OF INTERMOLECULAR FORCES (IMF) Hydrogen bonding Two type of IMF Van Der Waals forces London dispersion < dipole-dipole < H-bonding Increasing strength of intermolecular forces

31 INTERMOLECULAR FORCES Types of IMF Van Der Waals Forces Dipole-dipole forces/ permanent dipole forces London Forces/Dispersion forces/temporary dipoleinduced forces Hydrogen Bonds

32 INTERMOLECULAR FORCES Attraction between one molecule and a neighbouring molecule. Example :

33 TYPES OF INTERMOLECULAR FORCES 1. Van der Waals Force : Dipole-dipole forces (polar) HCI HCI Dispersion (london) force (Non-polar) 2. Hydrogen bond Ar Ar H 2 O CH 3 OH

34 DIPOLE-DIPOLE FORCES (PERMANENT DIPOLE FORCES) 1. Due to different EN between the atoms. 2. Attractive forces between polar covalent compounds. 3. The positive pole of one molecule attract the negative pole of another. (SEE figure 1.1) 4. If size and molar mass are almost same : Dipole moment (μ) increase, DD force increase, BP increase More polar molecule Stronger DD force More energy needed to separate them High BP

35 DIPOLE-DIPOLE FORCES Example : Dipole-dipole forces Dipole-dipole forces Chlorine is more electronegative, thus it has higher electron density Dipole-dipole forces; the partially positive end attracts the partially negative end Figure 1.1

36 DIPOLE-DIPOLE FORCES Substances RMM Dipole moment Boiling point ( ) CH 3 CH 2 CH CH 3 OCH CH 3 CN More polar molecule (high μ) Stronger DD force More energy needed to separate them High BP

37 DIPOLE-DIPOLE FORCES Example : Substance Molecular formula M r Dipole moment (D) Boiling point (K) Propane CH 3 CH 2 CH Acetaldehyde CH 3 COH Both propane and acetaldehyde have same relative molecular mass. However, acetaldehyde has boiling point than propane. Acetaldehyde is polar molecule and has dipole-dipole forces than propane.

38 DIPOLE-DIPOLE FORCES Example : Substance Molecular formula M r Dipole moment (D) Boiling point (K) Propane CH 3 CH 2 CH Acetaldehyde CH 3 COH Both propane and acetaldehyde have same relative molecular mass. However, acetaldehyde has high_ boiling point than propane. Acetaldehyde is _more_ polar molecule and has _stronger_ dipole-dipole forces than propane.

39 LONDON DISPERSION FORCES (TEMPORARY DIPOLE-INDUCED FORCES) Exist in all atoms and molecules Caused by temporary (instantaneous) of electron density in a molecules

40 LONDON DISPERSION FORCES (TEMPORARY DIPOLE-INDUCED FORCES) Occur between non polar molecules such as Cl 2, O 2, H 2, N 2 Attractive forces between non polar molecules is caused by the random movement of electrons

41 LONDON DISPERSION FORCES However, when these electrons revolve around the nucleus, At any instant, one side of the molecules has higher electron density than the other side. δ- δ+ Higher electron density Lower electron density It produces temporary dipoles

42 LONDON DISPERSION FORCES δ- δ+ Higher electron density Lower electron density It produces temporary dipoles Due to the random movement of the electrons, polarity of the temporary dipoles changes all the time.

43 LONDON DISPERSION FORCES (TEMPORARY DIPOLE-INDUCED FORCES) 1. For example, Neon (Z=10). Neon is a monatomic molecule. N 2 2. Since Neon is non-polar, the distributions of electrons in the atom, on the average is uniform. 3. However, the 10 electrons revolve around the nucleus of neon atom. 4. There is a high possibility that any given instant, the electron density might be higher in one part of the atom compared to the other parts.

44 LONDON DISPERSION FORCES (TEMPORARY DIPOLE-INDUCED FORCES) 5. This results to the formation of a temporary dipole of instantaneous dipole because it lasts for just a tiny fraction of time. 6. Then, the atom has a new instantaneous dipole. 7. temporary dipole can distort the electron clouds of neighboring atoms and giving rise to induced dipoles. 8. The attraction between induced dipoles and temporary dipoles produces intermolecular force.

45 Example: London forces in Br 2 Br At any instant, the electron density might be higher on one side + - Br Br Temporary dipole molecule Br Electrons in a molecule move randomly about the nucleus + Br Br London forces - The temporary dipole molecule induce the neighboring atom to be partially polar 45

46 INSTANTANEOUS DIPOLE 1. Meaning : An atom or molecules with separation of charge produced by a momentary displacement of electrons from their normal distributions

47 INDUCED DIPOLE 1. Meaning : The separation of positive and negative charges in a neutral atom or a nonpolar molecule caused by the proximity of an ion or a polar molecule or instantaneous dipole.

48 DISPERSION (LONDON) FORCE 1. Results from instantaneous Dipole-induced dipole 2. The force are weak, but exist in any particles (non polar, polar molecule, ion, metal, etc) NOTES Van der Waals forces are weaker than covalent bonds or ionic bonds. Polar molecules have dipoledipole and temporary Dipole-induced dipole attraction

49 DD VS TEMPORARY DIPOLE??? 1. Temporary dipole-induced dipole also exist between polar molecules. Species Contribution to the van der Waals forces Dipole-dipole attraction Temporary dipole-induced dipole attraction Ar CO HCI Attractive forces Dipole-dipole Temporary dipole-induced dipole force Occurrence Between POLAR MOLECULES only Between all COVALENT MOLECULES

50 STRENGTH OF DISPERSIONS FORCE 1. Influenced by : Polarizability (size/molar mass) Molecular shape Molecular Polarity Factors which influence Van Der Waals forces Molecular Shape Molecular Size

51 FACTOR OF MOLAR MASS 1. Dispersion forces increase with number of electrons, which correlates closely with molar mass Molar mass increase Size increase Number of atoms increase Number of electrons increase Polarizibility increase Dispersion forces increase Boiling point increase

52 EXAMPLE : Neon Xenon Because of greater temporary dipoles, xenon molecules are stickier than neon molecules. Neon molecules will break away from each other at lower temperatures than xenon molecules. So, Neon has lower BP. This is why >> bigger molecules have HIGH BP than small ones. Bigger molecules have more electrons and more distance over which temporary dipoles can develop. So, bigger molecules are stickier.

53 MOLECULAR POLARITY Example : Compound Molecular formula Boiling point (K) Ethane CH 3 CH Fluoromethane F CH Both molecules having identical number of electrons, and almost similar molecular mass. However, fluoromethane has _high boiling point than ethane because it is a polar_molecules and has a dipole-dipole forces. Ethane is a non polar molecules and has a dispersion forces.

54 FACTOR OF MOLECULAR SHAPE 1. For non polar substance with the same polar mass: Surface area of molecules increase Contact between molecules increase Dispersion forces increase Boiling point increase POLAR D. F + D. D. F NON POLAR D. F Surface area depends on the shape of molecule

55 FACTOR OF MOLECULAR SHAPE 2. Boiling point: Spherical molecules make less contact each other than the cylindrical molecules Contact surface increase BP increase to break the surfaces (large surfaces) Contact surface decrease BP decrease (less electrons)

56 EXAMPLE : Long thin molecules can develop bigger temporary dipoles due to electron movement than short fat ones (same no. of electrons). Long thin molecules can lie closer together > the attractions are more effective if the molecules are really close. Example : BUTANE (arranged in single chain) and 2- METHYLPROPANE (shorter chain with a branch). Both have molecular formula, C 4 H 10. But, the atoms are arranged differently. Butane has HIGH BP because dispersion forces are greater and surface area larger, bigger temporary dipoles.

57 MOLECULAR SHAPE Example : Compound Molecular formula Boiling point ( C) Butane CH 3 CH 2 CH 2 CH methylpropane CH 3 CH 2 CH 3 I CH

58 MOLECULAR SHAPE Butane has a boiling point because it is a linear alkanes with surface area of contact and thus, dispersion forces. 2-methylpropane is a branched alkanes, more compact (spherical shape) and has a smaller area of contact. Hence, it has weaker dispersion forces.

59 Factors that influence the strength of the van der Waals forces. The molecular size/molecular mass Molecules with higher molar mass have stronger van der Waals forces as they tend to have more electrons involved in the London forces. Example: CH 4 has lower boiling point than C 2 H 6 Note: However if two molecules have similar molecular mass, the dipole-dipole interaction will be more dominant. Example: H 2 S has higher boiling point than CH 3 CH 3

60 DISPERSION VS DIPOLE-DIPOLE FORCES 1. In many cases, dispersion forces are comparable to an even greater than Dipole-dipole force between polar molecules. 2. Example : μ = 0 NON POLAR CCl 4 non polar Dispersion force (DF) Boiling point : 76.5 CHF 3 polar Dispersion force (DF) + D.D.F Boiling point : HCI HI HBr polar DF+ DDF Molar increase, D.F increase, BP increase

61 SUMMARY OF VAN DER WAALS FORCES 1. Dispersion (London) forces exist between all particles : Molar mass increase, strength increase All depends on molecular shape 2. Dipole-dipole forces (DDF) exist only in polar molecules Adds to the effect of Dispersion forces 3. When comparing substances of roughly comparable molar masses DD can effect properties such as melting and boiling point 4. When comparing substances of widely different molar masses Dispersion forces are usually more significant than DDF

62 EXAMPLE : 1. Arrange the following chlorine compounds in increasing order of their boiling points : Cl 2, CCl 4, CINO When comparing their strength of van der Waals forces, IMPORTANT to know their relative molecular mass of substances!!!

63 ANSWER (LOOK AT TEXT BOOK PGE 108): RMM of Cl 2 = 71, RMM of CCl 4 = 154, RMM of CINO = 65.5 Cl 2 and CCl 4 are NON POLAR molecules. But, CCl 4 has HIGH RMM. So, van der Waals forces of attraction in CCl 4 is higher. Then it has HIGH BP. Cl 2 < CCl 4 Cl 2 and CINO have similar RMM. But, CINO is a polar. So, CINO has stronger attraction. It has HIGH BP. Cl 2 < CINO CCl 4 and CINO has wide different of RMM. Molecules with higher RMM, has stronger attraction forces. CCl 4 has HIGH BP than CINO. CINO< CCl 4 Lastly : Cl 2 < CINO < CCl 4

64 HYDROGEN BOND

65 HYDROGEN BOND (FON) 1. It is special type of dipole-dipole forces between a hydrogen atom, which is covalently bonded to a small and highly electronegative atom (F, O, N) and the lone pair of electrons of another very electronegative atom. 2. HYDROGEN BOND is stronger than van der Waals forces but weaker than covalent bonds. A and B are F,O, N

66 HYDROGEN BOND H F H F Hydrogen bond

67 HYDROGEN BOND (FON) 1. The essential requirements for the formation of a hydrogen bond are : a) A hydrogen atom covalently bonded to a highly electronegative atom (F, O, N) b) An unshared pair of electrons on another electronegative atoms. A and B are F,O, N

68 HYDROGEN BOND Example : δ δ + δ H O H O H H H H N H N H δ δ + H δ H δ δ + H N H O H H H O H N H δ δ + H H δ δ δ H H F H N H δ + δ H δ + H N H F H δ δ

69 REMEMBER!! Hydrogen bond terbentuk apabila atom partially +ve bergabung dengan partially -ve yang ada lone pair

70 HYDROGEN BOND In ammonia molecule, the nitrogen atom has one lone pair of electrons. This means that each NH 3 can form one hydrogen bond. In water molecule, the oxygen has two lone pairs of electrons. This means that each oxygen atom in the H 2 O molecule can form two hydrogen bonds. Then, H 2 O molecule has two hydrogen atoms that can form hydrogen bonds with the other two H 2 O molecule.

71 Other examples: NH 3 liquid.. N H 2 O Hydrogen intermolecular bond.. N Hydrogen intermolecular bond O O Covalent bond O O Hydrogen intermolecular bond 71

72 Consider ethanol, CH 3 OH CH 3 OH CH 3 OH and methane Hydrogen bond C Not a hydrogen bond in O C H is not bonded to either F, O or N O C 72

73 Example: H 2 O H O H covalent bond hydrogen bond H H O H O H O H H H O H 73

74 74

75 EXAMPLE From these compounds, select those who can form hydrogen bonds : HCI, HF, CH 4, H 2 S, H 2 O, CH 3 NH 2, CH 3 COOH, CH 3 CONH 2, PH 3

76 ANSWER : From these compounds, select those who can form hydrogen bonds : HCI, HF, CH 4, H 2 S, H 2 O, CH 3 NH 2, CH 3 COOH, CH 3 CONH 2, PH 3 HF, H 2 O, CH 3 NH 2,CH 3 COOH, and CH 3 CONH 2, can form hydrogen bonds Because : each of these molecules has a hydrogen atom which is bonded covalently to F, O, and N.

77 H BONDING AND DENSITY OF H 2 O 1. As results of geometric arrangement of H bonds in H 2 O, ice has an open, hexagonally shaped crystal structure. 2. When water freeze, Volume increase, density decrease Ice floats in water δ = m v meningkat) (bila terapung, density kurang, so, volume

78 EFFECT OF HYDROGEN BONDING ON DENSITY H 2 O is unusual in its ability to form an extensive hydrogen bonding network. Density of ice In ice, water molecules form tetrahedral arrangement to maximise the amount of hydrogen bonding between them. Hence, gives rise to an open structure. This arrangement of molecules has greater volume (is less dense) than liquid water, thus water expands when frozen. So, ice is less dense than water

79 EFFECT OF HYDROGEN BONDING ON DENSITY The arrangement of water molecules in ice H O H H H O H O H H H H H H O H H O H H O H H 79

80 Density Ice (solid H 2 O) has lower density compared to its liquid. Refer to the structure of ice

81 Ice form tetrahedral arrangement Hydrogen bond takes one of the tetrahedral orientation and occupy some space

82

83 HYDROGEN BONDING IN ICE AND WATER 1. Open structures of ice accounts for the fact that ice is less dense than water at 0. When ice melts, its volume decreases. 2. When ice melts, some of the hydrogen bonds break and the water molecules are packed closer together. Hence, water has higher density than ice. 3. Ice melts on heating. When further heating, more hydrogen bonds are broken as more water molecules are produced.

84 HYDROGEN BONDING IN ICE AND WATER 4. Density of water increases with the temperature. 5. At the same time, water expands as it is being heated and causes its density to decrease. 6. Beyond 4, the density of water decreases with increasing temperature.

85 EFFECT OF HYDROGEN BONDING ON DENSITY When ice melts, some of the hydrogen bonds are broken. Density of water This allows the water molecules to be more compactly arranged and decrease in volume. Therefore, water has a higher density than ice. 85

86 86

87 EFFECT OF HYDROGEN BONDING ON DENSITY H O H H O H H O H H O H H O H The arrangement of water molecules in liquid phase

88 SOLUBILITY AND INTERMOLECULAR FORCES 1. Substances with similar types of intermolecular forces dissolve in each other 2. Example : Solid NaCl Ionic Ionic bond Dispersion force Water Polar H bond Dispersion force So, NaCl dissolves in water LIKE dissolves LIKE Rule

89 SOLUBILITY AND INTERMOLECULAR FORCES 3. Example : CCl 4 Non polar Dispersion force C 6 H 6 Non polar Dispersion force So, CCl 4 dissolves in C 6 H 6 (benzene) Sebab ada benda yang sama H 2 O polar H bond D.D.F Dispersion force CCl 4 Non polar Dispersion force So, CCl 4 do not dissolves in H 2 O

90 THE EFFECTS OF HYDROGEN BONDING Boiling point The effects on physical properties Solubility Density

91 EFFECT OF HYDROGEN BONDING ON BOILING POINT Figure: The boiling point of the hydrides of element in group 14 to

92 EFFECT OF HYDROGEN BONDING ON BOILING POINT 1. Based on the graph, the hydrides of Group 14 elements (CH 4, SiH 4, GeH 4, SnH 4 ) display normal behavior. 2. BP increases regularly when the RMM increases. 3. This is due to the increasing strength of the van der Waals forces of attraction as the molecular size increases from CH 4 to SnH 4.

93 EFFECT OF HYDROGEN BONDING ON BOILING POINT 1. The exceptions of NH 3, NF and H 2 O, the BP of hydrides of Group 15,16 and 17 elements also increase with the increase of RMM. 2. RMM increases, van der Waals forces increase and BP increases. 3. NH 3, NF and H 2 O, are high BP due to the existence of hydrogen bonds which are stronger than van der Waals forces. 4. BP of NF is higher than NH 3. This is due to flourine atom is more electronegative than N.

94 EFFECT OF HYDROGEN BONDING ON BOILING POINT 5. BP of H 2 O is higher than BP of HF, although F is more electronegative than O. 6. This is because each oxygen atom in the water molecule is able to form two hydrogen bonds with two other water molecules. 7. Flourine atom in the HF molecule can form only one hydrogen bond with another HF molecule.

95 Properties of compounds with Hydrogen intermolecular forces Boiling point Have relatively high boiling point than compounds having dipole-dipole forces or London forces - the Hydrogen bond is the strongest attraction force compared to the dipoledipole or the London forces. 95

96 Solubility A. Dissolve in polar solvent The molecules that posses Hydrogen bonds are highly polar. They may form interaction with any polar molecules that act as solvent. B. Dissolve in any solvent that can form Hydrogen bonds 96

97 Example NH 3 dissolves in water because it can form Hydrogen intermolecular bond with water. O Hydrogen bond.. N.. N 97

98 Problem: Explain the trend of boiling point given by the graph below: T/ o C HF HBr HI HCl Molecular mass 98

99 Answer HF can form hydrogen bonds between molecules while HCl, HBr and HI have van der Waals forces acting between molecules. Hydrogen intermolecular bond is stronger that the van der Waals forces. More energy is required to break the Hydrogen bond. Boiling point increases from HCl to HI. The strength of van der Waals forces increases with molecular mass. Since molecular mass increases from HCl to HI, thus the boiling point will also increase in the same pattern. 99

100 The effect of Hydrogen bond on water molecules The density of water is relatively high compared to other molecules with similar molar mass. Reason: Hydrogen intermolecular bonds are stronger than the dipole-dipole or the London forces. Thus the water molecules are drawn closer to one another and occupy a smaller volume. 100

101 Example : Explain the trend of boiling points given below: The order of the increase in boiling point is: H 2 O > HF > NH 3 > CH 4 101

102 Answer: by looking at the polarity of the bond, we have (Order of polarity: HF > H 2 O > NH 3 ) but H 2 O has the highest boiling point. For H 2 O, the number of hydrogen bonds per molecule affects the boiling point. Each water molecule can form 4 hydrogen bonds with other water molecules. More energy is required to break the 4 Hydrogen bonds. HF has higher boiling point than NH 3 because F is more electronegative than Nitrogen. CH 4 is the lowest - it is a non polar compound and has weak van der Waals forces acting between molecules. 102

103 INTERMOLECULAR AND INTRAMOLECULAR HYDROGEN BONDING

104 INTERMOLECULAR HYDROGEN BONDING Hydrogen bonds that occur between two different molecules are called INTERMOLECULAR HYDROGEN BONDS. Eg : Hydrogen bonds between molecules of HF, NH 3, H 2 O, and CH 3 COOH. Intermolecular hyrogen bonding can lead to the formation of chains, rings or three dimensional structures. 104

105 INTRAMOLECULAR HYDROGEN BONDING Hydrogen bonds that occur within the same molecules are called INTRAMOLECULAR HYDROGEN BONDS. Eg : 4-nitrophenol has a higher BP than 2-nitophenol. Because 4-nitrophenol has intermolecular hydrogen bonds and 2- nitrophenol has intramolecular hydrogen bonds. 2-nitrophenol, -OH group and the -NO 2 group are on adjacent carbon atoms and close to one another. INTRAMOLECULAR occur. 4-nitrophenol, -OH group and the -NO 2 group are far apart. Instead, the form intermolecular hydrogen bonds (see page 114) 105