2.2 CHEM HL

• Resonance hybrid structures can be used to represent molecules for which there are more than one valid Lewis formula in terms of electron distribution
• Resonance structures of the nitrate (V) ion


  Three structures possible

  
• Resonance hybrid nitrate (V) ion

  • The criteria for forming resonance hybrids structures is that molecules must have a double bond (pi bond) that is capable of migrating from one part of a molecule to another
  • This usually arises when there are adjacent atoms with equal electronegativity and lone pairs of electrons that can re-arrange themselves and allow the double bonds to be in different positions

  
• Resonance Hybrids Table

  Carboxylate ion, RCOO-

  
• Ozone, O3

  
• Benzene, C6H6

  
• Carbonate ion, CO32-

  
• Structure of Benzene

  The structure consists of 6 carbon atoms in a hexagonal ring, with alternating single and double carbon-carbon bonds

  • This suggests that benzene should react in the same way that an unsaturated alkene does
• Remaining p orbitals


  1. Overlap laterally with p orbitals of neighbouring carbon atoms

  2. Form a π system
• Extensive sideways overlap of p orbitals


  Results in the electrons being delocalised and able to freely spread over the entire ring causing a π system
• π system


  • Made up of two ring-shaped clouds of electron density - one above the plane and one below it
• Benzene and other aromatic compounds


  • Regular and planar compounds with bond angles of 120o
• Delocalisation of electrons


  All of the carbon-carbon bonds in these compounds are identical and have both single and double bond character
• The bonds all being the same length is evidence for the delocalised ring structure of benzene
• Evidence for delocalisation

  • This evidence of the bonding in benzene is provided by data from:
  • Enthalpy changes of hydrogenation
  • Carbon-carbon bond lengths from X-ray diffraction
  • Saturation tests
  • Infrared spectroscopy
• Hydrogenation of cyclohexene


  • Each molecule has one C=C double bond
  • The enthalpy change for the reaction of cyclohexene is -120 kJ mol-1
• Hydrogenation of cyclohexene


  1. C6H10 + H2 → C6H12

  2. ΔHꝋ = -120 kJ mol-1
• Hydrogenation of benzene


  • The Kekulé structure of benzene as cyclohexa-1,3,5-triene has three double C=C bonds
  • It would be expected that the enthalpy change for the hydrogenation of this structure would be three times the enthalpy change for the one C=C bond in cyclohexene
• Hydrogenation of benzene


  1. C6H6 + 3H2 → C6H12

  2. ΔHꝋ = 3 x -120 kJ mol-1 = -360 kJ mol-1
• When benzene is reacted with hydrogen, the enthalpy change obtained is actually far less exothermic, ΔHꝋ = -208 kJ mol-1
• The actual structure of benzene is more stable than the theoretical Kekulé model


  152 kJ mol-1 less energy is produced than expected
• Carbon-carbon bond lengths

  • Cyclohexene contains two different carbon-carbon bonds 
  • The single carbon-carbon bond (C-C) has a bond length of 154 pm
  • The double carbon-carbon bond (C=C) has a bond length of 134 pm
  • The Kekulé structure of benzene as cyclohexa-1,3,5-triene has three single C-C and three double C=C bonds
  • It would be expected that benzene would have an equal mixture of bonds with lengths of 134pm and 154 pm
  • All of the carbon-carbon bond lengths are 140 pm suggesting that they are all the same and also intermediate of the single C-C and double C=C bonds
• Saturation tests

  • Cyclohexene will decolourise bromine water as an electrophilic addition reaction takes place
  • The Kekulé structure of benzene as cyclohexa-1,3,5-triene has three double C=C bonds
  • It would, therefore, be expected that benzene would easily decolourise bromine water
  • Benzene does not decolourise bromine water suggesting that there are no double C=C bonds
• Infrared spectroscopy

  • Cyclohexene shows a peak in the range of 1620 - 1680 cm-1 for the double C=C bond within its structure
  • The Kekulé structure of benzene as cyclohexa-1,3,5-triene has three double C=C bonds
  • It would, therefore, be expected to also show a peak at 1620 - 1680 cm-1 for the double C=C bonds
  • Benzene does not show a peak in this range for the double C=C bonds, instead, peaks are seen at around 1450, 1500 and 1580 cm-1 which are characteristic of double C=C bonds in arenes
• Expansion of the octet

  • Elements in period 3 and above have the possibility of having more than eight electrons in their valence shell
  • This is because there is a d-subshell present which can accommodate additional pairs of electrons
  • This is known as the expansion of the octet
  • The concept explains why structures such as PCl5 and SF6 exist, which have 5 and 6 bonding pairs of electrons respectively, around the central atom
• Phosphorus pentachloride, PCl5

  • An example of a molecule with five bonding electron pairs is phosphorus pentachloride, PCl5
  • The total number of valence electrons is = P + 5Cl = 5 + (5 x 7) = 40
  • The number of bonding pairs is 5, which accounts for 10 electrons
  • The remaining 30 electrons would be 15 lone pairs, so that each Cl has 3 lone pairs
  • The completed Lewis formula looks like this:

  
• Sulfur tetrafluoride, SF4

  • The total number of valence electrons is = S + 4F = 6 + (4 x 7) = 34
  • The number of bonding pairs is 4, which accounts for 8 electrons
  • The remaining 26 electrons would be 13 lone pairs
  • Fluorine cannot expand the octet so each fluorine would accommodate 3 lone pairs, accounting for 24 electrons, leaving 1 lone pair on the sulfur (sulfur has expanded the octet)
  • The completed Lewis formula looks like this:

  
• Chlorine trifluoride, ClF3

  • The total number of valence electrons is = Cl + 3F = 7 + (3 x 7) = 28
  • The number of bonding pairs is 3, which accounts for 6 electrons
  • The remaining 22 electrons would be 11 lone pairs
  • Fluorine cannot expand the octet so each fluorine would accommodate 3 lone pairs, accounting for 18 electrons, leaving 2 lone pairs on the chlorine
  • The completed Lewis formula looks like this:

  
• Triiodide ion, l3-

  • The total number of valence electrons is = 3I + the negative charge = (3 x 7) + 1 = 22
  • The number of bonding pairs is 2, which accounts for 4 electrons
  • The remaining 18 electrons would be 9 lone pairs
  • Iodine would accommodate 3 lone pairs, accounting for 12 electrons, leaving 3 lone pairs on the central iodine
  • The completed Lewis formula looks like this:

  
• Sulfur hexafluoride, SF6

  • An example of a molecule with six bonding electron pairs is sulfur hexafluoride, SF6
  • The total number of valence electrons is = S + 6F = 6 + (6 x 7) = 48
  • The number of bonding pairs is 6, which accounts for 12 electrons
  • The remaining 36 electrons would be 18 lone pairs, so that each F has 3 lone pairs, accounting for all electrons and no lone pairs
  • The completed Lewis formula looks like this:

  
• Bromine pentafluoride, BrF5

  • The total number of valence electrons is = Br + 5F = 7 + (5 x 7) = 42
  • The number of bonding pairs is 5, which accounts for 10 electrons
  • The remaining 32 electrons would be 16 lone pairs
  • Fluorine cannot expand the octet so each fluorine would accommodate 3 lone pairs, accounting for 30 electrons, leaving 1 lone pair on the bromine
  • The completed Lewis formula looks like this:

  
• Xenon tetrafluoride, XeF4

  • The total number of valence electrons is = Xe + 4F = 8 + (4 x 7) = 36
  • The number of bonding pairs is 4, which accounts for 8 electrons
  • The remaining 28 electrons would be 14 lone pairs
  • Each fluorine would accommodate 3 lone pairs, accounting for 24 electrons, leaving 2 lone pairs on the xenon
  • The completed Lewis formula looks like this:

  
• • Molecular geometry refers to the shape of the molecules based on the relative orientation of the atoms
• • Electron domain geometry refers to the relative orientation of all the bonding and lone pairs of electrons
• • PCl5  is a symmetrical molecule so the electron cloud charge is evenly spread
  • This means that it will be a non-polar molecule as any dipoles from the P–Cl bonds would be cancelled out
• • SF4 and ClF3 are asymmetrical molecules having one or two lone pairs on one side of the central axis making the overall molecule polar
• • SF6 is a symmetrical molecule so the electron cloud charge is evenly spread with 90o between the bonds
  • This means that it will be a non-polar molecule as any dipoles from the S–F bonds would be cancelled out
• • XeF4 is also non-polar despite having two lone pairs
  • The bonding pairs are at 90o to the plane and the lone pairs are at 180o
  • The lone pairs are arranged above and below the square plane resulting in an even distribution of electron cloud charge
• • BrF5 is asymmetrical having a lone pair at the base of the pyramid making the overall molecule polar
• • Formal charge is described as the charge assigned to an atom in a molecule, assuming that all the electrons in the bonds are shared equally between atoms, regardless of differences in electronegativity