Matter and Materials

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Grade 11 Matter and Materials

Based on DocScientia

Atomic Bonds: Molecular structure
Intermolecular forces
Ideal gases and Thermal properties

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Matter and Materials

  1. 1. Matter and materials
  2. 2. Atomicbonds
  3. 3. Everything consists of of elements (except group 8) are not found alone. tsChemical en on p d om de c n re bobonds a s nit ru . ge lly ar ica fl m o e DocScientia p 14 ch
  4. 4. Chemicalbonds
  5. 5. with high Ep bond together, then existing bonds are broken. Octet rule: New bonds join molecules withChemicalbondinto lower Ep. Cℓ s DocScientia p 14
  6. 6. A chemical bond occurs when bond together to form a new substance with new properties and in so doing have anoble gas electron structure and a lower Ep.DocScientia p 14 Chemical bonds
  7. 7. A model describes an idea or thoughtBondingmodels DocScientia p 14
  8. 8. Covalent bond DocScientia p 14
  9. 9. DocScientia p 14
  10. 10. DocScientia p 14
  11. 11. DocScientia p 14
  12. 12. Between non-metals - Smallest e are sharedCovalent bond particle is a molecule. have half-filled orbitals that overlap to form a filled orbital - e negativity Polar or DocScientia p 14 must be the non-polar same or the bonds form diff. < 1,9
  13. 13. DocScientia p 15 Ionic bond
  14. 14. DocScientia p 15 Ionic bond
  15. 15. Ionic bond Between metal and non-metal - e are transferred Cations → electrostatic force/coulomb NON-METALS: force → anions High electrone- negativity > 2,1 ● METALS: e affinity ● Low ionisation cl i ● Accept e- art energy p anions t ● ● Donate e- l es al n DocScientia p 15 ● cation m o S i =
  16. 16. Negativity: Affinity:e- removed e- accepted,Energyrequired in Paymentpayment in energy
  17. 17. DocScientia p 15
  18. 18. DocScientia p 15 Metallic bond
  19. 19. Between metalsMetallic bond Low ionisation energy – form cations + core and  of delocalised e- Empty valence orbitals – e- move from one to the DocScientia p 15 next
  20. 20. DocScientia p 15
  21. 21. DocScientia p 15Valence electrons 4 n 3 +
  22. 22. Valence electrons e- T R A Outermost energy level #c nu orre mb s p er ond N of st the o shared S ele grou F E during a me p nt R reaction DocScientia p 15 R E D
  23. 23. Valency No sign # e- involved DocScientia p 16 in a reaction
  24. 24. DocScientia p 16 Lewis
  25. 25. Lewis Nucleus and core e : - represented by the atoms symbol Valence e ( ) are - N placed around the symbol placed, have all been Until they One at a DocScientia p 16 time Or all four sides are occupied.
  26. 26. DocScientia p 16 Lewis
  27. 27. DocScientia p 16 Lewis
  28. 28. DocScientia p 16 Lewis
  29. 29. DocScientia p 16 Lewis
  30. 30. Lewis Choose the central atom 1 2 Determine total # of valence e- in the molecule or ion Place the shared electron pairs between bonded atoms 3 4 Remaining valence electrons drawn as lone pairs, so each atom (except H) is surrounded by 8 e- With double/triple bonds, 5 DocScientia p 16 lone pairs will be less. Lewis structures
  31. 31. Lewis Determine the smallest electronegativity – middle. Rest go around 1 Valence electrons 2 Resonance structures Bonding electrons between atoms 3 DocScientia p 16 - Spread remaining e in octet around atoms 4
  32. 32. Lewis [ ] ? - ONO NO 3 3,0 3,5 5N O Resonance structures 18 O +1 This can happen to any of the oxygen neg. atoms. charge Only ALL 3 structures describes the actual bond. This is called resonance structures, and DocScientia p 16 when drawn, all three options should be done, with double arrows between to show the fact that the true structure is a mix of the three.
  33. 33. Atoms with an empty orbital in the valence energy level can share a lone pair with another H atom/molecule. H HNH H H NH + H [ ] Dative covalentDocScientia p 22
  34. 34. VSEPR HA A E L E P Predict shapes of covalent L E E and radicals E U molecules C N L T I S L R C O R O I E L NDocScientia p 24 N
  35. 35. Theelectronpairs Bonding e-around thecentral Main Repel Lone pairsatom in a amoleculedetermine theshape of the molecule n g le c a u s es s e- arrange themselves as far apart as possibleDocScientia p 24
  36. 36. # electron pairs that surround anDocScientia p 24 Electron pair geometry
  37. 37. # that surround a central Also coordination numberDocScientia p 24 Molecular geometry
  38. 38. Repulsion strengths Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair Triple bond >double bond >single bondDocScientia p 24
  39. 39. Central atom with Two Electron Pairs There are two electron pairs in the valence shell of Beryllium. [1s2 2s2 ] Molecular geometry-Linear arrangement 180° H Be HDocScientia p 24
  40. 40. Central atom with Three Electron Pairs Three electron pairs in the valence shell of Boron. [1s2 2s2 2p1 ] Molecular geometry- Trigonal Planar arrangement F 120° B F FDocScientia p 24
  41. 41. Central atom with Four Electron Pairs Four electron pairs in the valence shell of Carbon. [1s2 2s2 2p2 ] Molecular geometry- Tetrahedral Bond angle -109.5⁰DocScientia p 24
  42. 42. Central atom with Five Electron Pairs Five electrons in the valence shell of Phosphorus. [1s2 2s2 2p6 3s2 3p3 ] Molecular geometry- Trigonal bipyramid Bond angle -120⁰ &90⁰DocScientia p 24
  43. 43. Central atom with Six Electron Pairs Six electrons in the valence shell of Sulfur. [1s2 2s2 2p6 3s2 3p4 ] Molecular geometry- Octahedral Bond angle-90⁰DocScientia p 24
  44. 44. Electronegativity The pull of an on a shared pair of electrons. Indication of Influenced by: bonding ability. ● Size of charge of # is on your nucleus periodic ● Size of table No units decreases increases Periodic tableDocScientia p 29
  45. 45. Difference predicts what type of bond will form between Bond Difference in electronegativity Non-polar covalent =0 Covalent and weak polar <1 Polar covalent >1 <2,1 Ionic – transfer of e- >2,1DocScientia p 29
  46. 46. Difference in electronegativity B What determines o polarity? n Shape d of the moleculeDocScientia p 30 s
  47. 47. Repulsive forces Attractive forces F O R C E SDocScientia p 36
  48. 48. Chemical bonds happen when two or more nuclei attract an e- W br hen poea b en k on ch tia /fo d t stores an l en rm s potential g erg , es y energy .DocScientia p 37
  49. 49. OB – bond length dissociation BM – bond energyPotential energy (kJ) B Distance between nuclei 0 N P M bonding Molecule most stable positionDocScientia p 37
  50. 50. Bond strength =measured by seeinghow much energy is necessaryto break the bond between two atoms.DocScientia p 38
  51. 51. Bond energy = the energy needed to break a bond.DocScientia p 38
  52. 52. Bond energy size weak Bond length strong Order: 1→2→3DocScientia p 38
  53. 53. Bond length=Distance between thetwo nuclei ofthe atoms bondingDocScientia p 38
  54. 54. Bond length Individual radii Bond order: The higher the order, the shorter the bond length.DocScientia p 38
  55. 55. Poly-atomic Diatomic molecules vibrate molecules Cannot absorb in different ways. move by stretching infrared Unequal stretch or and contracting Molecules that bend = equally. can dipole absorb/reflect moments infrared = No dipole greenhouse Infrared = moments gases. absorbedDocScientia p 39
  56. 56. Intermolecularforces
  57. 57. forcesDocScientia p 46 Intermolecular forces
  58. 58. Ionic and metallic bonds are strong -1 (400 – 4 000 kJ.mol ) usually found as solids.IM forces are mainlyfound between smallcovalent molecules.DocScientia p 46
  59. 59. Ion-dipole force 1 Ion-induced dipole 2Types force Van der Waals force DocScientia p 46 3
  60. 60. Types of van Dipole-dipole 1der waals force Dipole-induced dipole force 2 Induced dipole DocScientia p 46 forces (London) 3
  61. 61. Particles Type of bond Ions Coulomb forces Ion and polar molecule Ion-dipole Two polar molecules Dipole-dipole Ion and non-polar molecule Ion-induced dipole Polar and non-polar Dipole-induced dipole molecule Non-polar molecules London (dispersion) forcesDocScientia p 47
  62. 62. Ion-dipole forces Dipole approaches a positive or a negative ion. 240 pm 84 kJ DocScientia p 47
  63. 63. Ion-induced An ion approaching an atom ordipole forces molecule, it affects the electron cloud around the atom, causing a temporary dipole. + δ- δ+ DocScientia p 47
  64. 64. Dipole-dipole If two dipoles forces approach each other, they will turn so that their δ- oppositely charged ends 5 – 25 kJ.mol-1 will be closer. δ+ An attractive force will exist between these DocScientia p 47 dipoles.
  65. 65. Dipole-induced Polar molecules can induce a dipole forces temporary dipole in a non-polar molecule/atom. Usually a very weak force. δ- δ+ δ- δ+ DocScientia p 47
  66. 66. Induced dipole When 2 non-forces (London) The greater the polar atoms/ molecule, the greater molecules the attraction. Only approach, seen in the absence of there is a other forces. slight change in the electron cloud of both Ne Ne molecules or atoms. Temporary. DocScientia p 47
  67. 67. Hydrogen bonds H F F δ+ H δ- δ- H H δ+ δ+ F F δ- Hydrogen to a small atom with extremely high electronegativity. (N, O, F) Electrostatic force between δ- atom in the molecule and the H in the other. Very strong, but weaker DocScientia p 47 than covalent and ionic bonds.
  68. 68. nfluence of intermolecular forces onDocScientia p 55 In c re Phase as De e cr ea se In c re as e De cr ea se
  69. 69. Molecule size only affectsnfluence of intermolecular forces on van der Waals forces DECREASE F Cl Br I INCREASE Molecule size DocScientia p 55
  70. 70. nfluence of intermolecular forces onDocScientia p 55 Density Decreasse
  71. 71. nfluence of intermolecular forces on B.P. increase when molecular size increase. H2S is smaller than H2Se and so on. All these molecules have weak 100 H2O van der Waals forces. B.P. Of H2O is higher than expected. H2O have strong hydrogen bonds. H2Te H2Se -50 H2S B.P. DocScientia p 55
  72. 72. nfluence of intermolecular M.P. increase when molecular size increase. forces on HCl is smaller than HBr and so on. All these molecules have weak van der -25 HI Waals forces. M.P. Of HF is HF higher than expected. HBr HF have strong hydrogen bonds. -75 HCl M.P. DocScientia p 55
  73. 73. because IM forces weaken. Temp. increases, viscocity decreasesnfluence of intermolecular forces on Long, polar molecules: forces = greater, and viscosity is higher. Polarity = stronger attractive forces, and long chains become tangled. Viscosity Indication of resistance to flow. DocScientia p 55
  74. 74. nfluence of intermolecular forces on Degree of The more The expansion energy a particles depends particle has, separate, on material the IM force causing an type. weaken. increase in volume. Material expands on heating. Thermal expansion DocScientia p 55
  75. 75. nfluence of intermolecular forces on Covalent structures have no free e- and therefore are bad conductors. Exceptions: diamond and graphite. Thermal conductors DocScientia p 55
  76. 76. Microscopic properties ofwater. Covalent bondDocScientia p 70
  77. 77. Microscopic properties ofwater. Angular shapeDocScientia p 70
  78. 78. Microscopic properties ofwater. Hydrogen bondsDocScientia p 70
  79. 79. Microscopic properties ofwater. Greenhouse gasDocScientia p 70
  80. 80. How many water moleculesin 1 ℓ of water? 1 ℓ = 1000 g -1 M = 18,02 g.mol (H2O) n = m/M = 1000/18,02 = 55,5 mol DocScientia p 71
  81. 81. Cause of moderate climate on  P r o # of energy needed to p e change the temperature r t of 1 kg of a substance i Help organisms to by 1 °C e effectively regulate s body temp. of Specific heat capacity definition DocScientia p 71 water
  82. 82. P Water = high r latent heat o p eHeat absorbed/releasedr during phase changes. t i e Water releases heat slowly s when it cools down. ofLatent heatDocScientia p 71 water
  83. 83. M.P. B.P. P r Large amounts of energy required o to break HB. Essential for life p Points are on  – otherwise e therefore r H2O would be in t higher than the gaseous state. i expected. e s Strong hydrogen bonds ofDocScientia p 72 water
  84. 84. D P re o pn es r ti i et sy of waterDocScientia p 72
  85. 85. Adhesion and cohesion P r Forces between two different o types of molecules p e r t i Forces between the same type of molecule. e s ofDocScientia p 73 water
  86. 86. Surface action P r o Due to the p cohesive forces e of molecules r on the surface of t a volume i e of water. s ofDocScientia p 73 water
  87. 87. Capillary action P r Tendency to rise in a tube o as a result of surface p tension – i.e. adhesive e properties of water. r t i e s ofDocScientia p 73 water
  88. 88. Idealgasesand thermalpropertiesDocScientia p 85
  89. 89. v and EK Small = different particles KMT for individual Continuous motion particles. Avg. EK = Empty spaces constant between particles if temp = forces constantlastic collisions Motion DocScientia p 85 Of particles
  90. 90. Condensation of gases Gases fill the KMT whole container B R M O O W T N I I O A N NDi f f u s i o nDocScientia p 85 Explains
  91. 91. DocScientia p 85
  92. 92. Receive energy Temperature and avg. EK Individual T α EK EK Avg. E KDocScientia p 86 and temperature
  93. 93. 2 EK = ½ mv T α EK 2 T α ½ mv Avg. E KDocScientia p 86 and temperature
  94. 94. DocScientia p 86
  95. 95. DocScientia p 86
  96. 96. Properties:➔Particles = identical in Real gas: every way Approach ideal➔Only occupies volume gas behaviour due to motion of when: particles; ➢ Low particles themselves temperatures = no volume ➢ High pressure ➔No forces ➔Collisions are perfectly elastic Ideal gas DocScientia p 86 model
  97. 97. Low temperature: ➔EK decrease High pressure: ➔Collisions ➢ Particles own volume decrease contributes to total volume of the gas, = larger.➔Pressure = lower ➢ Larger particles = ➔Move closer stronger intermolecular forces ➢ Liquefaction occurs – gas together experiences high pressure➔Attractive forces under critical temp. increase – could cause condensation Ideal gas DocScientia p 86 model
  98. 98. pV Ideal gasDocScientia p 86 p model
  99. 99. VDocScientia p 86 P
  100. 100. V 1/pDocScientia p 86
  101. 101. pDocScientia p 86 1/V
  102. 102. VDocScientia p 86 T
  103. 103. For comparisons, temp. and pressurehave to be identical – STP V&p 0°C/273K, and 101,3 kPaDocScientia p 91
  104. 104. Pressure = # collisions against a Fcontainer per unit time. p= A -2 1 Pa = 1 N.mDocScientia p 91
  105. 105. DocScientia p 91
  106. 106. Temperature =constant.Avg. EK = same Volume (cm3) V o l um ecollisions pVPressure 1/p (kPa-1) Pressure (kPa)DocScientia p 94
  107. 107. The volume of anenclosed mass of gas isinversely proportional tothe pressure of the gas,provided the temperature DocScientia p 94 Boyles Law
  108. 108. V α1/p More than one set: pV=k p1V1 = p2V2DocScientia p 94
  109. 109. Volume (cm3) High temp.Diff. temperatures Med. temp. Low temp. 1/p (kPa-1) DocScientia p 94
  110. 110. Just clickDocScientia p 94
  111. 111. Volume and temp. 3 V (cm ) -273 °C DocScientia p101 T (°C)
  112. 112. Volume and temp. VαT V = kT V1 = V 2 K T =T 1 2 DocScientia p 101
  113. 113. The volume of a fixed mass of gas is directly proportional to the temperature of the gas,Charles Law provided to pressure remains constant. DocScientia p 101
  114. 114. Charles Law DocScientia p 101
  115. 115. Temp. and pressureDocScientia p102
  116. 116. Pressure of a constant volumeTemp. and pressure of gas with a fixed mass is directly proportional to the absolute temperature. DocScientia p 102
  117. 117. p (kPa)Temp. and pressure Absolute zero Extrapolation -273 °C DocScientia p102 T (°C)
  118. 118. Absolute zero The absolute zero is the lowest possible temperature that any substance can ever reach. DocScientia p 102
  119. 119. Because absolute zero is -273°C, a scale was created whereTemp. and pressure absolute zero is actually zero. Kelvin scale (K) Pressure (kPa) Temperature (K) DocScientia p102
  120. 120. Guy-Lussacs Law Temperature in K is directly proportional to the pressure of an enclosed mass of gas, provided the volume remains constant. DocScientia p103
  121. 121. Represented by a tKelvin temp scale -273 0 100 °C 0 273 373 K T = t + 273 Represented t = T - 273 by a T DocScientia p103
  122. 122. Factors that determine the pressure of a gas: # of collisions Intensity of collisions Pressure, Volume and temperatureDocScientia p106
  123. 123. Boyles Law: p 1V 1 = p 2V 2 Pressure, Volume and temperatureDocScientia p106
  124. 124. Charles Law: V1 = V 2 T1 T2 Pressure, Volume and temperatureDocScientia p106
  125. 125. Guy-Lussacs Law: p1 = p 2 T1 T2 Pressure, Volume and temperatureDocScientia p106
  126. 126. p 1V 1 = p 2V 2 T1 T2 Pressure, Volume and temperatureDocScientia p106
  127. 127. pV α T pV = kT Ideal gas LawDocScientia p106
  128. 128. k depends on the # of molecules k can therefore be substituted with n (the # of moles) and R, which is the general gas constant. pV = nRT Ideal gas LawDocScientia p106
  129. 129. mol K pV = nRT Pa 3 m 8,31 Ideal gas LawDocScientia p106
  130. 130. Textbook: DocScientia, Grade 11 workbook, 2013Images: attempt has been made to acknowledge allsources, if an images source could not be found, it wasacknowledged as such.Slide 1 – llnl.gov Slide 2 – flickr.comSlide 4 – CAIROO software Slide 7 – reference.comSlide 8 – source unknown Slide 9 to 11 – CAIROO softwareSlide 12 a – 123rf.com Slide 12 b – source unknownSlide 12 c – equipmentexplained.com Slide 13 – tumblr
  131. 131. Slide 14 – tumblrSlide 17 – CAIROO software Slide 18 – launch.tased.edu.auSlide 20 – CAIROO sorftware Slide 23 – eklavya.orgSlide 24 – chemistryland.com Slide 26 - 29 – Lily KotzeSlide 41 - 43 – worldofteaching.com Slide 56 – swarooproy.deviantartSlide 57 – reference.com Slide 62 – source unknownSlide 64 - 65 – source unknown Slide 68 a - b – source unknownSlide 70 – source unknown Slide 73 – soundcloudSlide 74 - 75 – scienceclarified Slide 76 – CAIROO software
  132. 132. Slide 77 – CAIROO software Slide 79 – google imagesSlide 85 – milkywayscientists Slide 86 – google imagesSlide 87 – CAIROO software Slide 88 – CAIROO softwareSlide 91 – TutorVista.com Slide 94 - 95 – CAIROO softwareSlide 105 – hip2b2 Slide 110 – mindsetlearnSlide 114 – CAIROO software Slide 115 – CAIROO software

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