lecture 3: 123.101

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lecture 3: 123.101

  1. Unit One Parts 3 & 4:molecular bonding
  2. Unit OneParts 3&4 H O H3C Br O Br H CH3Locating electronsDescribing bonds PagesShape of molecules 34 & 43
  3. Unit One 3&4 if we know whereParts electrons are we can predict reactions and shape...they really are key to understanding chemistry H O H3C Br O Br H CH3Locating electronsDescribing bonds PagesShape of molecules 34 & 43
  4. Unit OneParts 3&4 H O H3C Br O Br H CH3 as I’ve taken the material out of order, I’ll give you someLocating electrons page numbersDescribing bonds PagesShape of molecules 35 & 45
  5. what are bonds?
  6. Na Cl Na Cl here we have an atom of sodium (Na) and an atom ofIonic bonds chlorine (Cl) Pg 34
  7. if we take one electron from Na and give it to Cl... Na Cl Na ClIonic bonds Pg 34
  8. Na Cl Na+ Cl- we get 2 charged species (cation = positive charge & anion = negative charge)Ionic bonds Pg 34
  9. Na+ Cl- NaCl opposite charges attract and give us an ionic bondIonic bonds Pg 34
  10. covalent bonds H H if we bring 2 atoms together and they... Pg 34
  11. covalent bonds H H share their 2 electrons we have a covalent bond Pg 34
  12. covalent bonds H H this is the bond we’ll be dealing with most often and is represented by the black line H H 2 electrons per bond Pg 34
  13. covalent bonds H H H H please remember 2 that this line is 2 electrons electrons per bond Pg 34
  14. chemistry apain these are just extremes
  15. reality is in the middle
  16. where do we find electrons?
  17. ONE DOES NOT SIMPLY
  18. Aufbau Principle THIS IS THE LONGVERSION...NOT THE VERSION I DO IN THE LECTURES lowest energy orbital
  19. Aufbau Principle don’t worry about the name...just that electrons like to have lowest energy possible...lowest energy orbital
  20. rather like many students...
  21. 1 18 1 H H He 2 13 14 15 16 17 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s1energy 2s 2px 2py 2pz hydrogen Pg 1s 43
  22. 1 18 1 H H He 2 13 14 15 16 17 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s1energy 2s 2px 2py 2pz just one electron hydrogen so in first orbital Pg 1s 43
  23. Pauli Exclusion Principle no two electrons are identical
  24. 1 18 2 H He He 2 13 14 15 16 17 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s2energy 2s 2px 2py 2pz helium Pg 1s 43
  25. 1 18 2 H He He 2 13 14 15 16 17 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s2energy 2s 2px 2py 2pz one electron has spin +½ (up) and the other spin –½ (down) helium Pg 1s 43
  26. 1 18 2 H He He 2 13 14 15 16 17 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s2energy 2s 2px 2py 2pz doesn’t matter what it means...just remember helium an electron can only be up or down Pg 1s 43
  27. 1 18 2 H He He 2 13 14 15 16 17 Li Be B C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s2energy 2s 2px 2py 2pz so can only ever have two electrons per orbital helium Pg 1s 43
  28. 1 18H He 2 13 14 15 16 173Li Be B C N O F NeLiNa Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 2px 2py 2pz 1s22s1 energy 2s lithium Pg 1s 43
  29. 1 18H He 2 13 14 15 16 173Li Be B C N O F NeLiNa Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 2px 2py 2pz 1s22s1 energy lithium obeys both rules...fill lowest orbital 2s first (until full) then fill next lowest) lithium Pg 1s 43
  30. 1 18H He 2 13 14 15 16one more ...adding 17 4 electron is easy...Li Be Be B C N O F NeNa Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 2px 2py 2pz 1s22s2 energy 2s beryllium Pg 1s 43
  31. 1 18H He 2 13 14 15 16 17 5Li Be B B C N O F NeNa Mg ...and another... Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 2px 2py 2pz 1s22s22p1 energy 2s boron Pg 1s 43
  32. 1 18H He 2 13 14 15 16 17 5Li Be B B C N O F NeNa Mg it could go in any of Al Si P S Cl Ar 3 4 5 6 7 8 9 10 x,11 y12 2pz, 2p 2p or they’re identical...wellK Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn they are As Se Br Kr energetically Ga Ge 2px 2py 2pz 1s22s22p1 energy 2s boron Pg 1s 43
  33. 1 18H He 2 13 14 15 16 17 5Li Be B B C N O F Ne but, where doesNa Mg the next (and most Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 important as itsK Ca Sccarbon) go?? Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Ti V Cr 2px 2py 2pz 1s22s22p1 energy 2s boron Pg 1s 43
  34. Hunds ruleelectrons as far apart as possible(de ge n er a t e o rb i tals) (as long as it doesn’t violate any of the previous rules!)
  35. Hunds rule makes sense as like charges always repel...electrons as far apart as possible(de ge n er a t e o rb i tals)
  36. 1 18 H He 2 13 14 15 16 17 6 Li Be B C C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s22s22p 12p 1 2px 2py 2pz x yenergy 2s 1s22s22p2 carbon Pg 1s 43
  37. 1 18 H He 2 13 14 15 16 17 6 Li Be B C C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s22s22p 12p 1 2px 2py 2pz could be 2pz, makes no x y difference...energy 2s 1s22s22p2 carbon Pg 1s 43
  38. thats a lot of electrons... luckily we don’t care about all them...
  39. all you have toremember is... ©jaci XIII@flickr
  40. 1 18H He 2 13 14 15 16 17 6Li Be B C C N O F NeNa Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s22s22p2 carbon atomic = number of number electrons Pg 45
  41. Valence electrons 1 18 H He 2 13 14 15 16 17 6 Li Be B C C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 1s22s22p 12p 1 2px 2py 2pz x yenergy 2s 1s22s22p2 carbon Pg 1s 43
  42. Valence electrons 1 18 H He 2 13 14 15 16 17 6 Li Be B C C N O F Ne Na Mg Al Si P S Cl Ar 3 4 5 6 7 8 9 10 11 12 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr only need consider high energy electrons or those on the outside called the 1s22s22p 12p 1 2px 2py 2pz valence electrons. x yenergy 2s 1s22s22p2 carbon Pg 1s 44
  43. C C if we consider the Bohr model of the atom, the 1s22s22p2 atomwhere we thinkplanet 2 one resembling a of an 2s22p with moons orbiting (or the solar system) N N 1s22s22p3 2s22p3group 1 2 13 14 15 16 17 18 H He Li Be B C N O F Ne Pg 44
  44. C C 1s22s22p2 2s22p2 then the valenceelectrons are those on the outer edge (like Neptune for young-upstarts or Pluto for us oldies) N N 1s22s22p3 2s22p3 group 1 2 13 14 15 16 17 18 H He Li Be B C N O F Ne Pg 44
  45. C C 1s22s22p2 2s22p2 then the valenceelectrons are those on the outer edge (like Neptune for young-upstarts or Pluto for us oldies) N N 1s22s22p3 2s22p3 group 1 2 13 14 15 16 17 18 H He Li Be B C N O F Ne Pg 44
  46. C C 1s22s22p2 2s22p2 N N 1s22s22p3 2s22p3 absolute rubbish...but moregroup 1 2 13 14 15 16 17 18 comprehendible! H He Li Be B C N O F Ne Pg 41
  47. C C 1s22s22p2 2s22p2 N N an easy we to remember the number of valence electrons is 1s22s22p3 2s22p3 to take group number...group 1 2 13 14 15 16 17 18 H He Li Be B C N O F Ne Pg 41
  48. C C 1s22s22p2 2s22p2 N N ...and ignore 1s22s22p3 2s22p3 first ‘1’ valenceelectrons 1 2 3 4 5 6 7 8 H He Li Be B C N O F Ne Pg 41
  49. C C 1s22s22p2 2s22p2 N N 1s22s22p3 2s22p3 valenceelectrons 1 2 3 4 5 6 7 8 so oxygen (group 16) has H He 6 valence Li Be B C N O F Ne electrons Pg 41
  50. what do valenceelectrons tell us?
  51. H 1 2H H4 3 O H the number of bonds
  52. the shape of molecules 109°
  53. ?how
  54. atoms are happy if they have a full valence shell... Ne 1s22s22p6noble gas
  55. Ne ...commonly this means 8 electrons 1s22s22p6noble gas
  56. 8electrons Ne full shell
  57. C1s22s22p2 4 bonds N1s22s22p3 3 bonds O1s22s22p4 2 bonds Pg 36
  58. Pg 45 C 4 valence electrons so for carbon to get to 8 it needs 4 more electrons1s22s22p2 Pg 45
  59. Pg 46 C or 4 new covalent bonds4 bonds Pg 46
  60. Cnitrogen has 5 valence1s22s22p2 electrons...so 4 bondsneeds 3 more... N 1s22s22p3 3 bonds O 1s22s22p4 2 bonds Pg 36
  61. N 5 valence electrons1s22s22p3 Pg 46
  62. so forms 3covalent bonds N 3 bonds Pg 46
  63. C 1s22s22p2 4 bonds oxygen needs 2more electrons soforms 2 covalent bonds N 1s22s22p3 3 bonds O 1s22s22p4 2 bonds Pg 36
  64. O 6 valence electrons1s22s22p4 Pg 46
  65. O2 bonds Pg 46
  66. C1s22s22p2 4 bonds hopefully, you can see 3 this is where those magic numbers in lecture one came N bonds from!1s22s22p3 O1s22s22p4 2 bonds Pg 34
  67. Pg8 36 H H C H HOctet rule: 8 valence electrons
  68. Pg8 37/46 H H H C N O H HOctet rule: 8 valence electrons
  69. Pg Lewis structures 37/46 Hydrofluoric acid HFH + F H F ≡ H F use octet rule to draw Methanol CH OH 3 the structure of stable molecules... H HC + O + 4H H C O H ≡H C O H H H
  70. Pg Lewis structures 41 Hydrofluoric acid HF H–F easy..H = 2 electrons (full s orbital) & F = 8...H + F H F ≡ H F Methanol CH3OH H HC + O + 4H H C O H ≡H C O H H H
  71. Pg Lewis structures 37/46 Lewis structure shows all valence electrons Hydrofluoric acid HF represented by our simple diagram H–FH + F H F ≡ H F Methanol CH3OH H HC + O + 4H H C O H ≡H C O H H H
  72. Pg Lewis structures 37/46 Hydrofluoric acid HFH + F H F ≡ works for more complex H F molecules Methanol CH3OH H HC + O + 4H H C O H ≡H C O H H H
  73. Pg Lewis structures 37/46 Hydrofluoric acid HFH + F H F ≡ H F Methanol CH3OH Note: it helps to leave lone pairs (of electrons) on diagram...this is where a lot of chemistry occurs... H HC + O + 4H H C O H ≡H C O H H H
  74. Acetone CH3COCH3 3 C + O + 6H how do we deal with more complex molecules? Pg 44
  75. Acetone CH3COCH3 3 C + O + 6H first draw all theatoms where you think O they might go... H H C H C C H H H Pg 44
  76. Acetone CH3COCH3 3 C + O + 6H now join all the atoms together...some of the atoms have full valence shells so wecan draw them in as on O the next slide... H C H C C H H H H Pg 44
  77. Acetone CH3COCH3 3 C + O + 6H the central C and O both have only 7 valence electrons... O H C H C C H H H H Pg 44
  78. Acetone CH3COCH3 3 C + O + 6H O ...but if they share 4electrons they both have H C H8 valence electrons...this C C gives us a double bond H H (alkene) H H O O C ≡ Pg H3C CH3 44
  79. –Borohydrideanion BH4 – what happens if we have a negative charge (anion)? Pg 44
  80. –Borohydrideanion BH4 – take the atoms as normal and...B + 3H + H Pg 44
  81. –Borohydrideanion BH4 – add electronB + 3H + H ...add an electron Pg 44
  82. –Borohydrideanion BH4 – add electron H HB + 3H + H H B H ≡ H B H H H Pg 44
  83. –Borohydrideanion BH4 – add electron does it matter which atom we give the electron to? H HB + 3H + H H B H ≡ H B H H H Pg 44
  84. –Borohydrideanion BH4 – add electron does it matter which atom we give the electron to? H HB + 3H + H H B H ≡ H B no! (but in this case H H H– makes more H chemical sense) Pg 44
  85. +Ammoniumcation NH4 + lose electron if we have a positive charge (cation) we do the opposite... Pg 44
  86. +Ammoniumcation NH4 + lose electron start with our normal atoms... N + 3H + H Pg 44
  87. +Ammoniumcation NH4 + lose electron N + 3H + H then remove an electron Pg 44
  88. +Ammoniumcation NH4 + lose electron H H N + 3H + H H N H ≡ H N H H H Pg 44
  89. where is the charge?is it on one atom?
  90. all over the molecule... No, its all over the molecule! But...
  91. but the truth isnt useful, so...
  92. formal charges localisecharge on an atom...
  93. this is ‘electron book- keeping’...we are just assigning charge to one atom to help explain chemistry...formal charges localisecharge on an atom...
  94. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons ...on an atom Pg 47
  95. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons ...according to the atoms position in the periodic table Pg 47
  96. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons...in lone pairs... Pg 47
  97. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons ...or the number of bonds to that atom Pg 47
  98. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons H H N + 3H + H H N H ≡ H N H H H cation N fc = 5-0-½(8)=+1 Pg 47
  99. formal number of number of ½ number charge = valence – unshared – of shared (fc) electrons electrons electrons H H N + 3H + H H N H ≡ H N H H Hno charge on H as: cation H = 1-0-½(2) = 0 N fc = 5-0-½(8)=+1 Pg 47
  100. formal number of number of number ofcharge = valence – unshared – bonds (fc) electrons electrons H H the simplified N + 3Hformula of bonds) + (just use number H H N H ≡ H N H H H cation N fc = 5-0-4=+1 Pg 47
  101. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons OO + O + O O O O ≡ O O O3 neutral ozone Pg 47
  102. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons OO + O + O O O O ≡ O O O3 neutral ozone lhs O; fc = 6-4-½(4)=0 Pg 47
  103. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons OO + O + O O O O ≡ O O O3 neutral ozone lhs O; fc = 6-4-½(4)=0 central O; fc = 6-2-½(6)=+1 rhs O; fc = 6-6-½(2)=-1 Pg 47
  104. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons OO + O + O O O O ≡ O O O3 neutral ozone lhs O; fc = 6-4-½(4)=0 central O; fc = 6-2-½(6)=+1 rhs O; fc = 6-6-½(2)=-1 Pg 47
  105. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons OO + O + O O O O ≡ O O ≡ O O O O3 neutral atoms formal ozone charges lhs O; fc = 6-4-½(4)=0 central O; fc = 6-2-½(6)=+1 rhs O; fc = 6-6-½(2)=-1 Pg 47
  106. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons OO + O + O O O O ≡ O O ≡ O O O ozone neutral as O3 neutral atoms formal + & – cancel each ozone charges other out lhs O; fc = 6-4-½(4)=0 central O; fc = 6-2-½(6)=+1 rhs O; fc = 6-6-½(2)=-1 Pg 47
  107. formal number of number of ½ numbercharge = valence – unshared – of shared (fc) electrons electrons electrons these charges explain why ozone is so reactive! OO + O + O O O O ≡ O O ≡ O O O O3 neutral atoms formal ozone charges lhs O; fc = 6-4-½(4)=0 central O; fc = 6-2-½(6)=+1 rhs O; fc = 6-6-½(2)=-1 Pg 47
  108. formal number of number of number ofcharge = valence – unshared – bonds (fc) electrons electrons OO + O + O O O O ≡ O O ≡ O O O O3 neutral atoms formal the simplified ozone charges formula (just use number of bonds) lhs O; fc = 6-4-2=0 central O; fc = 6-2-3=+1 rhs O; fc = 6-6-1=-1 Pg 47
  109. a bond O 2 H O is electrons
  110. a bond 2 is electrons
  111. simple model
  112. quantum model more accurate...
  113. Atomic orbitalsits a quantum world... we’ve looked at a nice simple model so far...
  114. Atomic orbitalsits a quantum world... mathematicians and physicists have shown it’s a bit more complicated in ‘reality’
  115. but I dont like maths...so...heres some pretty pictures...
  116. 90%atomic orbital atomic orbital is the volume of space in which there is a 90% chance of finding an electron Pg 36
  117. 2 remember: aatomic orbital maximum of 2 electrons per orbital electrons Pg 37
  118. a 1s orbital is also a sphere...just a lot smaller2s Pg 38 Pic: Dr. Jonathan Gutow
  119. let’s ignore this nasty little effect of maths...2s Pg 38 Pic: Dr. Jonathan Gutow
  120. px z py z pz z y y y x x x2p Pg 38
  121. px z py z pz z y y y x x x2p each of the three 2p orbitals is dumbbell shaped... Pg 38
  122. px z py z pz z y y y x x x2p ...they are identical in all ways except... Pg 38
  123. px z py z pz z y y y x x x2p ...they point in different directions (hence the names) Pg 38
  124. px z py z pz z y y y this is one orbital (just has two different x x coloured areas) x2p Pg 34
  125. afraid?
  126. you will be...
  127. our simple Lewis model helps explain a lot of chemistry...especially reactions... what is a bond?
  128. what is a bond?...but it fails to explain such fundamental concepts as shape...
  129. ...actually, it can explain shape if we use VSEPR theory...but anyways, lets use those orbitals what is a bond?
  130. single (σ) bond H• + H• H Henergy here we have 2 hydrogen atoms (each with 1 electron in a 1s orbital) Pg H• 1s H• 1s 37
  131. single (σ) bond H• + H• H H σ* to form a covalent bond they mustenergy share their electrons... σ Pg H• 1s H–H H• 1s 35
  132. single (σ) bond H• + H• H H σ*energy ...this is achieved by combining the two atomic Pg orbitals to give... σ H• 1s H–H H• 1s 35
  133. single (σ) bond H• + H• H H σ* ...a new molecular orbital, a sigma σ orbital (or bond)energy σ Pg H• 1s H–H H• 1s 35
  134. single (σ) bond H• + H• H H ...this bonding σ* orbital is lower in energy than the atoms...so a bond will formenergy σ Pg H• 1s H–H H• 1s 35
  135. single (σ) bond H• + H• H H a consequence of the maths is we also get an anti-bonding sigma orbital (σ*)...2 orbitals σ* must give 2 new orbitalsenergy σ Pg H• 1s H–H H• 1s 37
  136. single (σ) bond H• + H• H H σ*energy ...but lets ignore this confusing little devil for the time being! σ Pg H• 1s H–H H• 1s 37
  137. single (σ) bondit is called a σ orbital as is symmetrical alongbond axis (you can rotate it like a cylinder and it doesn’t change) Pg H H 47
  138. single (σ) bond all bonds to H are sigma (as all are like a cylinder)...here we overlap 1s of H with 2p of C and get sigma bond)C• + H• C H Pg 37
  139. Pg single (σ) bond 38 σ*energy if we take two 2p orbitals and combine them head-to-head C• σ C• 2py C–C 2py
  140. Pg single (σ) bond 38 σ* ...we get a sigma σ bonding orbital...it is stillenergy like a cylinder... C• σ C• 2py C–C 2py
  141. Pg single (σ) bond 38 σ*energy ...this is the normal single bond we observe in alkanes etc. C• σ C• 2py C–C 2py
  142. Pg single (σ) bond 38 σ* this is one orbital NOT threeenergy C• σ C• 2py C–C 2py
  143. single (σ) bond the blue bit is the sigma orbital...ignore Pg the red orbitals for the time being... 35
  144. single bond or the simple C C version... THIS IS ALL YOU NEED TO KNOW σ (sigma) bond
  145. single bond σ (sigma) bond
  146. single bond σ (sigma) bond
  147. sp3 C an atom with 4 σbonds is called an sp3atom (as 1 x s and 3 x Pg p used in bonding) 38
  148. sp3 1 YOU NEED TO KNOW THIS 4 2 4points 3
  149. tetrahedral H Br C Br H 109° sp3 Pg 41
  150. tetrahedral H C Br Br H sp3 atoms are tetrahedral in shape (the bonds stay as far apart as possible) 109° sp3 Pg 41
  151. tetrahedral sp3maximum separation of four pointsmaximum separation of four valence electron pairs
  152. Pg double (σ + π) bonds 38 C C C=C π*energy C=C π two 2p orbitals can combine side-to-side carbon carbon 2pz C C 2pz
  153. Pg double (σ + π) bonds 38 C C C=C π*energy the new bond is a pi π bond C=C π carbon carbon 2pz C C 2pz
  154. Pg double (σ + π) bonds 38 C C C=C π*energy C=C π here we have a C–C σ bond and carbon a pi π bond carbon 2pz C C 2pz
  155. Pg double (σ + π) bonds 38 C C C=C π*energy C=C π the pi π bond is one orbital (with two bits to it) carbon carbon 2pz C C 2pz
  156. double (π) bond Pg 38
  157. double (π) bondit is called a pi π orbital as rotation around the C–Caxis causes a change (from red to blue) so no longer like a cylinder Pg 38
  158. double (π) bond remember: this is ONE orbital (just two different coloured halves) Pg 35
  159. double (π) bond we have an inner σ bond (the rod) and an outer πbond (the orbital) hence it is a double bond Pg 38
  160. norotation
  161. H3C CH3 CH3 CH3 CH3 the p bond prevents O H alkenes from rotating (the two bonds can’t twist passmultistep enzyme- each other)... light isomerisescatalysed reverse complexed process cis-retinal H3C CH3 CH3 CH3 O H CH3Pg38
  162. H3C CH3 CH3 CH3 CH3 this can effect O H shape of moleculemultistep enzyme- light isomerisescatalysed reverse complexed process cis-retinal H3C CH3 CH3 CH3 O H CH3Pg38
  163. H3C CH3 CH3 CH3 CH3 O H we must break π bond beforemultistep enzyme- light isomerisescatalysed reverse complexed alkene can rotate process cis-retinal H3C CH3 CH3 CH3 O H CH3Pg38
  164. H3C CH3 CH3 CH3 CH3 the change in O H shape initiates the visual cascade andmultistep enzyme- light isomerises our sightcatalysed reverse complexed process cis-retinal H3C CH3 CH3 CH3 O H CH3Pg38
  165. H3C CH3 CH3 CH3 CH3 O Hmultistep enzyme- light isomerisescatalysed reverse complexed process cis-retinal H3C CH3 CH3 CH3 O why do you think red path is easy H but blue hard? CH3Pg38
  166. double bond or the simple version... THIS IS ALL YOU NEED TO KNOW π (pi) bond
  167. double bond π (pi) bond
  168. sp2 an atom with three σ orbitals and one π orbital is called an sp2 atom (we only count the C orbitals used in making Pg s orbitals) 38
  169. sp2 1 3 3 2 1 double bond and 2 single bonds and wepoints have an sp2 atom
  170. trigonal planar 120° sp2 atoms are trigonal planar sp2 (flat and pointing to the corners of a triangle)...again, this is because the orbitalsPg try to be as far apart as possible41
  171. trigonal Pg 41planarsp 2maximum separation of three pointsmaximum separation of three valence electron pairs
  172. triple (σ + 2x π) bonds σ H C C H π (2pz + 2pz) σ a triple bond (like an π (2py + 2py)alkyne) is formed from oneσ bond and two π bonds (at right angles to each other due to the direct of the p orbitals that made them) σ π H C C π H Pg 39
  173. triple (σ + 2x π) bonds so...two p orbitals combine head-to-head to give a σ bond and two pairs of p orbitals combine side-to-side to give the two π orbitals (& there are only two π orbitals) σ H C C Hπ (2pz + 2pz) σ π (2py + 2py) σ π H C C π H Pg 39
  174. sp an atom with two σorbitals and two π orbitals is called an sp atom (as two orbitals made the basic σ skeleton) C Pg 39
  175. sp 1 2 2points
  176. linear an atom with two groups on it will be 180° sp Pg linear (a straight line) as the orbitals stay as far apart as possible 40
  177. linearspmaximum separation of two pointsmaximum separation of two valence electron pairs
  178. H3C H CO2H OH O O OCH3 H here is a real OH O molecule...we should be dynemicin A able to identify the types of atoms present...Pg40
  179. four groups attached so it must be sp3 and as those groups try to stay as far apart as possible it is tetrahedral H3C H CO2H OH O O OCH3 H OH O dynemicin A sp3 tetrahedralPg40
  180. ...only three groups so sp2 and flat, trigonal planar H3C H CO2H OH O O sp2 OCH3 trigonal H planar OH O dynemicin A sp3 tetrahedralPg40
  181. sp linear straight line, two groups must be sp H3C H and linear CO2H OH O O sp2 OCH3 trigonal H planar OH O dynemicin A sp3 tetrahedralPg40
  182. what is oxygen? H3C H CO2H OH O O OCH3 H OH O dynemicin APg40
  183. ...is it sp as what is attached to two oxygen? carbon atoms? H3C H CO2H OH O O OCH3 H OH O dynemicin APg40
  184. sp, sp2 or sp3?HO look at a simpler system...water, sp, H sp2 or sp3?
  185. sp, sp2 or sp3?HO H draw Lewis structure...
  186. sp, sp2 or sp3?HO H we have FOUR groups around O, two lone pairs & two H atoms. So it is...
  187. tetrahedral H O H sp3
  188. tetrahedral H O H sp3 that is why we draw water as a bent molecule...its shape is based on a tetrahedron...
  189. tetrahedral H O ...any atom with four atoms or lone pairs around it is sp3 with all that entails! H sp3
  190. sp, sp2or sp O 3? what kind of atom is the oxygen? CH H
  191. 1 double bond O ...and two lone pairs, so three groups around the C oxygen so it is...H H
  192. trigonalplanarH O C H sp2
  193. sp, sp2or sp3? N C H what kind of atom is the nitrogen?
  194. 1 triplebond N C H and one lone pair so two groups so it is...
  195. linear sp N C H
  196. what have ....we learnt? •e l e c t r o n s where they are •b o n d s what they areCourtesy: National Science Foundation •s h a p e
  197. ReadPages 36,41, 48-57 ©rachel_titiriga@flickr

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