Docstoc

Ch08 _EE_

Document Sample
Ch08 _EE_ Powered By Docstoc
					    8        Covalent Bonding

    8.1   Formation of Covalent Bonds
    8.2   Dative Covalent Bonds
    8.3   Bond Enthalpies
    8.4   Estimation of Average Bond Enthalpies using
          Data from Energetics



1
    8.5   Use of Average Bond Enthalpies to Estimate the
          Enthalpy Changes of Reactions
    8.6   Bond Enthalpies, Bond Lengths and Covalent
          Radii
    8.7   Shapes of Covalent Molecules and Polyatomic
          Ions
    8.8   Multiple Bonds
    8.9   Covalent Crystals




2
    8.1
           Formation of
          Covalent Bonds


3
Theory of covalent bond formation

    1. Lewis Model (A classical treatment)
    2. Valence Bond Theory (pp.31-35)
      (by Linus Pauling)
    3. Molecular Orbital Theory (pp.36-37)
      (By Robert Mulliken)



4
1. Lewis Model
A covalent bond is formed by sharing of valence
electrons between atoms of non-metals.
Reasons : 1.     They have small/no difference in
                 E.N.
          2.     They can achieve stable
                 electronic configurations.
                 (octet /duplet structures)
          3.     The total energy of the system is
                 lowered when electrons are
                 shared between the two nuclei.
5
                               nucleus
                               electron
                               Force between the
                               electron and nucleus.

                               Force along the bond
                               axis.
    A                 B
    If the electron is located between two
    nuclei, there is an attractive force holding
    the two nuclei together.


6
                                 BB’ > AA’
    A A’             B    B’
    If the electron is not located between two
    nuclei, there is a net force separating the
    two nuclei.


7
1. Lewis Model

    A Covalent bond is the result of shared
    valence-electron pair(s) positioned between
    two nuclei.

      Sharing of electron pair(s) between
      two nuclei.




8
Lewis(dot and cross) structure and Bond-line structure


• Electrons are represented by dots and/or crosses.

        H              H              H
                                      



                       




                                           
                                      
    H   C H         H CH          H C H
                                      

                       
        H              H              H

• Lewis structures basically obey octet rule.
• Hydrogen adopts duplet structrue.



9
 Lewis(dot and cross) structure and Bond-line structure

                                       H
         H

     H   C H                     H     C     H
         H
             Bond pairs                H



     H O H
                                 H     O      H

             Lone pairs      Only lone pairs are shown as
                          dots/crosses in bond-line structure

10
     O   O                 N   N



     O   O                N     N

     Double                Triple
      bond                 bond


              Multiple bonds


11
Rules for drawing Lewis structures
– A systematic way
 Using CO2, CO32 and HClO4 as examples

     1. Determine the arrangement of atoms
        within a molecules/polyatomic ion
       Central  least electronegative
       Terminal  H & F (most electronegative)




12
     CO2     O C O


     CO32   O C O
                 O

     HClO4   O        O
                 Cl
             O        O
                 H
13
2. Determine the total no. of valence electrons(V)
   in the species.
     For neutral molecule,
       V = sum of group numbers of atoms = m
     For anion, An,
       V=m+n
     For cation, Cn+,
       V=m–n



14
     Species   CO2      CO32   HClO4

               4+26   4+36+2 1+7+46
       V
                =16      =24     =32

       W


     W–V

15
3. Calculate the total no. of electrons that would
   be needed if each atom obeys the octet rule(W).
     2e for H
     8e for other atoms
4. The difference (W – V) gives the no. of
   electrons that have to be shared such that all
   atoms in the species can obey octet rule.




16
     Species     CO2       CO32     HClO4

               4+26      4+36+2 1+7+46
       V
                =16         =24     =32

                                     2+58
       W       38=24     48=32
                                      =42

                24-16      32-24      42-32
     W–V
               =4 pairs   =4 pairs   =5 pairs

17
5. Assign a single bond to each terminal atom in the
   species.




     CO2        O C O                 HClO4

                                   O          O
     CO32      O C O                  Cl
                    O             O           O
                                       H
18
5. Assign a single bond to each terminal atom in the
   species.
     If the resulting central atom has more than 4
     single bonds, rearrange the terminal atoms such
     that the central atom has 4 single bonds.
                                      H
          HClO4
                                      O
       O          O
           Cl                   O     Cl    O
      O           O
           H                          O
19
5. Assign a single bond to each terminal atom in the
   species.
     If the resulting central atom has more than 4
     single bonds, rearrange the terminal atoms such
     that the central atom has 4 single bonds.

     In SF6, W – V = 78 – (6+67) = 8
      4 single bonds are required
     However, S has to form 6 single bonds.
     Expansion of octet structure may happen for
     elements in Period 3 and beyond.

20
6. If bonding electrons remain, assign them in pair
   by making some of the bonds double or triple
   bonds.

     CO2      O      C     O
     Two bond pairs left  3 ways

              O      C     O

              O      C     O

              O      C     O

21
     CO32     O       C       O


                       O

     One bond pair left  3 ways


 O       C    O    O       C       O   O   C   O

        O                  O               O




22
             H
     HClO4

             O
                      No bond pair left
        O    Cl   O    Only one way

             O




23
7. Assign lone pairs to terminal atoms to give them
   octet.
     If any electrons still remain, assign them to the
     central atoms as lone pairs.

      CO2     No. of lone pairs left = (16 – 8)/2 = 4


                   O      C     O


      O      C      O          O      C     O


24
     CO32 No. of lone pairs left = (24 – 8)/2 = 8


                      O   C    O


                          O


       O     C    O                O    C    O

            O                           O


25
                        H
     HClO4

                        O

                 O      Cl     O

                        O

     No. of lone pairs left = (32 – 10)/2 = 11




26
8. Determine the formal charge on each atom.
     Formal charge is the charge an atom in a species
     would have if the bonding electrons are shared
     equally between the atoms.

     Formal charge of an atom in a species
                                      1
      Gp no. - no. of lone pair e s - (no. of bond pair e-s)
                                  -

                                      2

            Assume bond pairs are equally shared



27
     CO2
             O    C    O

                      1
           C  4 - 0 - 8  0
                      2
                    1
           O  64  4  0
                    2



28
     CO2                   
                  O    C    O


                            1
                 C  4  0  8  0
                            2
                            1
            Left O  6  6   2  -1
                            2
                            1
           Right O  6  2   6  1
                            2
29
     CO2                    
                  O    C    O

                            1
                 C  4  0  8  0
                            2
                            1
            Left O  6  2   6  1
                            2
                            1
           Right O  6  6   2  1
                            2
30
                                       
     O      C     O          O        C   O

     Separation of opposite formal charges
      Increase in potential energy of the species
      less stable Lewis structures

                  O      C       O

          Most stable Lewis structure

31
                               1
     CO3   2       C  4  0  8  0
                               2

                        O    C    O


                             O




      O         C   O              O     C   O

                O                        O


32
                          1
     CO32       O  64  4  0
                          2

                     O   C   O


                         O




      O      C   O            O     C   O

             O                      O


33
                            1
     CO32       O  6  6   2  1
                            2

                     O    C     O


                          O




      O      C   O               O      C   O

             O                          O


34
There is no separation of opposite formal charges
 All three structures are stable
                                    
                     O     C        O


                           O
                            
                                              
     O    C      O                  O     C     O

          O                               O
           
35
     CO32                        2

                     O    C   O


                          O

                     2                        2

      O      C   O            O        C   O

             O                         O


36
     HClO4                                1
                   H           H  1  0  2  0
                                          2
                                        1
                   O           O  64  4  0
                                        2

             O     Cl     O

                   O                        1
                                Cl  7  0   8  3
                                            2

                        1
             O  6  6   2  -1
                        2
37
     HClO4        H

                  O
                       
                   3+
             O    Cl    O

                  O
                  
Substantial separation of opposite formal charges
 Very unstable

38
     HClO4       H                       H

                 O                       O


             O   Cl    O          O     Cl     O


                 O                       O


Lone pair electrons must be rearranged to
minimize the separation of opposite formal charges.

39
     HClO4       H                     H

                 O                     O


             O   Cl   O          O     Cl    O


                 O                     O


Cl can expand the octet structure to accommodate
14 valence electrons.

40
     HClO4       H                    H

                 O                    O


             O   Cl    O        O     Cl     O


                 O                    O

                                  1
                      Cl  7  0   7  0
                                  2

41
     HClO4       H               H

                 O               O


             O   Cl   O      O   Cl   O


                 O               O

                               1
                      O  64  4  0
                               2
42
Breakdown of the Octet Rule (p.30)

 1. Valence shell expansion (Q.21(g), (h) (k) to (n))
     For elements from Period 3 and beyond can
     expand the octet by utilizing
     low-lying d-orbitals in bond formation.

     Max. no. of bond pairs = Group no.
     Gp 5       PF5
     Gp 6       SF6
     Gp 7       IF7

43
Breakdown of the Octet Rule (p.30)

2. Electron-deficient species (Q.21(o))

              F                     F


       F     B      F         F     B     F


     Dative bond (coordinate covalent bond) is a
     covalent bond in which the bond pair electrons are
     contributed solely by one of the bonding atoms.

44
Breakdown of the Octet Rule (p.30)

2. Electron-deficient species (Q.21(o))

             F                      F


       F     B      F         F     B     F


     The arrow(       )is pointing from the electron
     donor from the electron acceptor.


45
                        F
                              
                  F     B     F      unstable
                        

     Separation of opposite formal charges
     Not favourable for the most electronegative F
     to carry a positive formal charge




46
Q.21(d)
                                   

        O     O      O         O      O         O

     Separation of opposite formal charges is
     unavoidable.
              
         O       S       O     O     S     O

     Separation of opposite formal charges can be
     avoided by expansion of octet

47
 Q.21(a)
                 H                              H

                             +                 
         H       N           H             H    N   H


                 H                              H
                         
             H               Once formed, dative covalent
                             bond cannot be distinguished
     H       N       H       from normal covalent bond.
             H               The four N – H bonds are
                             identical.


48
 Q.21(a)
           H                           H

                   -                   
     H     B   H                   H   B   H

           H                           H
                               
                       H

               H       B   H

                       H



49
Breakdown of the Octet Rule (p.30)

 3. Odd-electron species
     Q.22                     
     N      O              N   O     Less stable


     Separation of opposite formal charges
     Positive formal charge on the more
     electronegative atom

50
Q.21(j)
                          

              O        N   O


             2NO2(g)  N2O4(l)

     O             O       O               O

         N     N                   N   N

     O             O           O           O
51
Q.21(p)
              Bond length   Bond strength

O         O    0.121 nm        strongest

          

O         O    0.133 nm

         

O         O    0.149 nm        weakest


52
Q.21(p)
               Bond length     Bond strength

O         O      0.133 nm         stronger

         
O         O      0.149 nm         weaker

Oxygen atoms have to be separated further apart
to minimize repulsion between negative formal
charges(lone pairs).
 Weaker bond
53
8.2 Dative Covalent Bonds (SB p.218)


Dative Covalent Bonds

      A dative covalent bond is formed by the
      overlapping of an empty orbital of an atom
      with an orbital occupied by a lone pair of
      electrons of another atom.
      Remarks
      (1) The atom that supplies the shared pair of electrons is
           known as the donor while the other atom involved in the
          dative covalent bond is known as the acceptor.

      (2) Once formed, a dative covalent bond cannot be
          distinguished from a ‘normal’ covalent bond.
 54
8.2 Dative Covalent Bonds (SB p.218 – 219)


A. NH3BF3 molecule




 55
8.2 Dative Covalent Bonds (SB p.219)


B. Ammonium Ion (NH4+)




 56
8.2 Dative Covalent Bonds (SB p.219 – 220)


D. Aluminium Chloride Dimer (Al2Cl6)

                                             Al: relative small
                                             atomic size; high
                                             I.E.’s required to
                                             become a cation
                                             of +3 charge.


          AlCl3




 57
8.2 Dative Covalent Bonds (SB p.219 – 220)


D. Aluminium Chloride Dimer (Al2Cl6)
                                                  (a dimer of
  Why doesn’t Al form ionic compounds with Cl?    AlCl3)




                                             Check Point 8-2
 58
8.1 Formation of Covalent Bonds (SB p.213)


A. Electron Sharing in Covalent Bonds
          H H                            Attraction between
                                         oppositely charged nuclei
                                         and shared electrons
  Shared                                    electrostatic
                                         ( _____________ in nature)
  electrons
                             e-
                             e-              The shared electron
                                             pair spends most of the
                                             time between the two
                                             nuclei.

  Overlapping of atomic orbitals  covalent bond formation

 59
8.1 Formation of Covalent Bonds (SB p.213)

  A hydrogen molecule is achieved by
  partial overlapping of 1s orbitals




 60
8.1 Formation of Covalent Bonds (SB p.214)

Electron density map for covalent compounds
  There is
  substantial
  electron density
  at all points along
  the internuclear
  axis.
                                             Compare electron-
  Thus electrons                             density-map for
  are shared                                 ionic compounds:
  between the
  two atoms.




 61
8.1 Formation of Covalent Bonds (SB p.214)


B. Covalent Bonds in Elements
 •    Hydrogen molecule




                                   Dot and cross diagram




 62
8.1 Formation of Covalent Bonds (SB p.215)


 •    Chlorine molecule




 •    Oxygen molecule




 63
8.1 Formation of Covalent Bonds (SB p.215)


 •    Nitrogen molecule




 64
8.1 Formation of Covalent Bonds (SB p.216)


C. Covalent Bonds in Compounds




 65
8.1 Formation of Covalent Bonds (SB p.216)




 66
     VB theory vs MO theory

     VB : 1. Electrons distribute in localized bonds
          2. Only valence electrons are involved in bonding

     MO : 1. Electrons distribute in delocalized bonds
          2. All electrons are involved in “bonding”
            bonding and anti-bonding




67
                2                    2                    2

O     C    O          O     C    O         O      C     O
                                     
     O                      O                     O


VB : the real distribution of electrons in CO32 is described
     by a combination of three resonance structures in
     which all bonds are localized.




68
                                2




MO : The six electrons are delocalized over the entire
     structure as described by molecular orbitals.

               2                    2                  2

O    C     O         O     C    O         O    C     O
                                    
     O                    O                    O

69
 Four principles of molecular orbital theory

 1. The total number of molecular orbitals produced
    is equal to the total number of atomic orbitals
    contributed by the atoms that have combined.

     For H2, there are two modes of overlap between
     the two 1s orbitals
     1. Addition    ψσ1s  ψA,1s  ψB,1s
       Waves combine constructively (in phase)

     2. Subtraction ψσ1s*  ψA,1s  ψB,1s
        Waves combine destructively (out of phase)

70
 Antibonding orbital : the probability of finding the
 electrons between the two nuclei 
                                                               *
                                                               1s
  nuclei repel each other




        A ,1s       B ,1s     Head-on overlap

     Bonding orbital :  probability of finding electrons
     between the two nuclei.
                                                             1 s
      Nuclei are pulled together
71
     Electron density map of 1s molecular orbital of H2

                   A,1s   B ,1s
                   1s


72
     Head-on overlap




73
     Side-way overlap
     Build-up of electron cloud above and below
     the plane on which the bonding atoms lie.
74
 2. Energy : - Bonding MO < Parent AO < Antibonding MO

75
76
3. Electrons of molecules fill up the MOs in
   increasing order of energy according to the
   Pauli exclusion principle and Hund’s rule



                    1s*




                    1s



77
Q.26
There is no gain of stability when the AOs of two
helium atoms overlap.

                                  2 e involved in bonding
                                  2 e involved in antibonding
     He(A)   He2          He(B)
                   1s*           Overall : -
                                  No e involved in bonding

                                       Bond order = 0

                   1s

78
     8 e involved in bonding
     4 e involved in antibonding
     Overall : 4 e involved in bonding

                 O=O
             Bond order = 2

79
     10 e involved in bonding
     4 e involved in antibonding
     Overall : 6 e involved in bonding

                 NN
             Bond order = 3

80
4. AOs combine to give MOs most effectively
   when AOs are of similar energy.

Effectiveness of overlap : -
1s-1s > 1s-2s > 1s-2p > 1s-3s > 1s-3p > 1s-3d > …
The energy level diagram can be substantially
simplified by considering only the effective
overlaps. E.g. 1s-1s, 2s-2s, 2p-2p




81
     Only 1s-2px overlap is considered.




82
         Energetics of
     formation of covalent
            species


83
8.3 Bond Enthalpies (SB p.221)


Bond Dissoication Enthalpy
 Bond dissociation enthalpy is the enthalpy
 change for breaking one mole of a bond in a
 particular environment with the reactants
 and products in the gaseous state under
 standard conditions.
                                 B.D.E. / kJ mol1
 Cl – Cl(g)  Cl(g) + Cl(g)          242



 84
                                 B.D.E. / kJ mol1
Cl – Cl(g)      Cl(g) + Cl(g)        242
Br – Br(g)      Br(g) + Br(g)        193
I – I(g)        I(g) + I(g)          151
O = O(g)        O(g) + O(g)          498
N  N(g)        N(g) + N(g)          945




85
Q.27
                                     ΔHatm/ kJ mol1
                                       o


       1
       2   Cl2(g)  Cl(g)                 121

                                      B.D.E. / kJ mol1
Cl – Cl(g)  Cl(g) + Cl(g)                242

                          1
             ΔH o
                atm   =   2
                            B.D.E.




86
             Hatm
1/2Br2(l)                 Br(g)

     Hv                  1/2B.D.E.
              1/2Br2(g)

 Hv = enthalpy change of vapourization

 By Hess’s law,
 Hatm = Hv + 1/2B.D.E. > 1/2B.D.E.



87
             Hatm
1/2I2(s)                 I(g)

     Hs                 1/2B.D.E.
              1/2I2(g)

 Hs = enthalpy change of sublimation

 By Hess’s law,
 Hatm = Hs + 1/2B.D.E. > 1/2B.D.E.



88
Notes : -
1. B.D.E. values are always > 0
     Bond breaking processes are endothermic
2. B.D.E. values depend on the environment of the bond.

     CH3H(g)      CH3(g)   +   H(g)   B.D.E.   = 423 kJ mol1
     CH2H(g)      CH2(g)   +   H(g)   B.D.E.   = 480 kJ mol1
     CHH(g)       CH(g)    +   H(g)   B.D.E.   = 425 kJ mol1
     CH(g)        C(g)     +   H(g)   B.D.E.   = 334 kJ mol1




89
8.3 Bond Enthalpies (SB p.222)


Bond Enthalpy

      Bond enthalpy is the average of the bond
      dissociation enthalpies taken from a large
      number of species containing a particular
      chemical bond.




 90
     CH3H(g)         CH3(g)   +   H(g)   B.D.E.   = 423 kJ mol1
     CH2H(g)         CH2(g)   +   H(g)   B.D.E.   = 480 kJ mol1
     CHH(g)          CH(g)    +   H(g)   B.D.E.   = 425 kJ mol1
     CH(g)           C(g)     +   H(g)   B.D.E.   = 334 kJ mol1

 The average B.D.E. values of C – H bond in CH4
         1
          (423 480  425  334)kJ mol1  415.5 kJ mol1
         4

 Bond enthalpy of C – H bond = E(C – H) = 413 kJ mol1

 An average B.D.E. values from CH4, C2H6, C3H8…etc.




91
Estimation of B.E. from Hf and Hatm
                          o       o


                        4E(C – H)
          CH4(g)                         C(g) + 4H(g)

       ΔHf [CH4 (g)]
         o
                                              Hatm
                                                 o



                       C(graphite) + 2H2(g)

         ΔHf [CH4 (g)] + 4E(C – H)   Hatm
           o                             o



                        ΔHL [NaCl(s)]
                           o


          NaCl(s)                         Na+(g) + Cl(g)

       H [NaCl(s)]
          o
          f                               ΔHatm  1st IE  1st EA
                                             o


                                  1
                       Na(s) +      Cl (g)
                                  2 2
92
Example 1(a)

                          4E(C – H)
            CH4(g)                    C(g) + 4H(g)

         ΔHf [CH4 (g)]
           o
                                          Hatm
                                             o



                         C(graphite) + 2H2(g)

           ΔHf [CH4 (g)] + 4E(C – H)   Hatm
             o                             o




     4E(C – H) = [715 + 4  218  (75)] kJ mol1
                  = 1662 kJ mol1
      E(C – H) = 415.5 kJ mol1

93
                                C(g) + 4H(g)


                            4218 kJ mol1
Enthalpy / kJ mol1




                           C(g) + 2H2(g)


                                                   4E(C – H)
                            715 kJ   mol1

                            C(graphite) + 2H2(g)

                      75 kJ mol1
                                       CH4(g)


94
Example 1(b)
               ΔHatm[C2H6 (g)] = 6E(C – H) + E(C – C)
                 o


     C2H6(g)                                            2C(g) + 6H(g)

 ΔHatm[compound]
   o



     = the enthalpy change when one mole of the compound in
       gaseous state is broken down into its constituent atoms
       in gaseous state under standard conditions.
 ΔHatm[element]
   o


     = the enthalpy change when one mole of atoms of the
       element in gaseous state are formed from the
       element in its normal and most stable state under
       standard conditions.
                                 C(graphite)  C(g)

95
Example 1(b)
                ΔHatm[C2H6 (g)] = 6E(C – H) + E(C – C)
                  o


     C2H6(g)                                             2C(g) + 6H(g)


      ΔHf [C2H6 (g)]
        o
                                                 H     o
                                                         atm


                        2C(graphite) + 3H2(g)

         ΔHf [C2H6 (g)] + 6E(C – H) + E(C – C)  Hatm
           o                                        o




         E(C – C) = [2715 + 6218  (85) 6415.5] kJ mol1

                       = 330 kJ mol1



96
        E(C – C) in C2H6       E(C – C) in C2H6
        Estimated value       Experimental value
         330 kJ mol1            386 kJ mol1


     The discrepancy is due to the fact that
     the average B.D.E. values of C – H bonds in
     CH4 and C2H6 are not the same.




97
Q.28
         H   H   H   H

     H   C   C   C   C       H         4C(g) + 10H(g)
         H   H   H   H


     ΔHatm[C4H10 (g)]  5165 kJ mol -1 = 3E(C – C) +10E(C – H)
       o
                                                                 (1)

         H   H   H   H   H


     H   C   C   C   C   C       H        5C(g) + 12H(g)
         H   H   H   H   H



     ΔHatm[C5H12 (g)]  6337 kJ mol -1 = 4E(C – C) +12E(C – H)
       o
                                                                 (2)

     Solving (1) and (2)
     E(C – C) = 348 kJ mol1         E(C – H) = 412 kJ mol1
98
Use of bond enthalpies to estimate enthalpy changes of
reactions

Assumption : The bond enthalpies are transferable from
             one molecule to another.




99
      In each series,
      adjacent members
      differ by 1C & 2H
      Linear plot shows that
      bond enthalpies are
      transferable.




100
Use of bond enthalpies to estimate enthalpy changes of
reactions

 Assumption : The bond enthalpies are transferable from
              one molecule to another.

  ΔHreaction  E(bonds broken) - E(bonds formed)
    o




               Bond forming is always exothermic




101
Example 1

Given : E(ClCl) = 242 kJ mol1 E(CH) = 413 kJ mol1
        E(CCl) = 339 kJ mol1 E(HCl) = 431 kJ mol1

Estimate the H of the following reaction.
          CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)


Bond broken / B.E.(kJ mol1) Bond formed / B.E.(kJ mol1)

          C–H       413            C – Cl     339
          Cl – Cl   242            H – Cl     431

      ΔHreaction  (242 + 413 – 339 – 431) kJ
        o


                = 115 kJ
102
Q.29(a)

       N2H4(g) + 2F2(g)  N2(g) + 4HF(g)

Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
           N–N         163               NN     945
         4N – H       4390            4H – F   4565
          2F – F      2158

 ΔH     o
        reaction    (163+ 4390+2158– 945–4565) kJ
                   = 1166 kJ
      Highly exothermic, used as rocket fuel

103
Q.29(b)

      CH4(g) + H2O(g)  CO(g) + 3H2(g)

Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
        4C – H         4413               CO     1072
        2O – H         2463             3H – H    3436



 ΔH    o
       reaction      (4413 + 2463 – 1072 – 3436) kJ
                  = 198 kJ



104
Q.29(c)

      2O3(g)  3O2(g)

Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
        2O – O         2146             3O = O    3498
        2O = O         2498



 ΔH    o
       reaction      (2146 + 2498 – 3498) kJ
                  = 206 kJ



105
  What assumptions have been made ?
  1. Bond enthalpies are transferable.
  2. O3 has one O – O single bond and one O = O double
     bond

      O     O       O            O       O       O

  In fact, there is delocalization of electrons making the
  two O – O bonds identical with a bond order of 1.5




106
Q.29(d)

4CH3NHNH2(g) + 5N2O4(g)  4CO2(g) + 12H2O(g) + 9N2(g)

Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
      12C – H   12413              8C = O    8740
       4C – N   4305              24O – H    24463
       9N – N   9163               9N  N    9945
      12N – H   12390
      10N – O   10163
      10N = O   10594



107
Bond broken / B.E. (kJ mol1) Bond formed / B.E. (kJ mol1)
      12C – H        12413           8C = O   8740
        4C – N       4305           24O – H   24463
       9N – N        9163            9N  N   9945
      12N – H        12390
      10N – O        10163
      10N = O        10594



 ΔH   o
      reaction     (19893 – 25537) kJ
                 = 5644 kJ Experimental value = -5116 kJ
                               Energetics, Q.10
108
 Q.10
                                H
4CH3NHNH2(l) + 5N2O4(l)                 4CO2(g) + 9N2(g) + 12H2O(l)

                       5Hf[N2O4(l)]          4Hf[CO2(g)]
4Hf[CH3NHNH2(l)]
                                           12Hf[H2O(l)]
        4C(graphite) + 12H2(g) + 4N2(g) + 5N2(g) + 10O2(g)

      By Hess’s law,

      H = [4(393) + 12(286) - 4(+53) – 5(20)] kJ
          = 5116 kJ



109
  What assumptions have been made ?
  1. Bond enthalpies are transferable.
  2. N2O4 has two N – O single bonds and two N = O
     double bonds
                          O
                          O           O
                                      O
      O           O                            O             O

          N   N       
                              N
                              N   N
                                  N
                                                   N   N
                          O
                          O           O
                                      O
      O           O                             O           O



  In fact, there is delocalization of electrons making all
  four N – O bonds identical with a bond order of 1.5


110
Estimate the enthalpy change of the following reaction.

Given : ΔHfo [NH4ClO4 (s)] = 295 kJ mol1
                     ΔHreaction
                       o

      2NH4ClO4(s)                 N2(g) + Cl2(g) + 2O2(g) + 4H2O(g)


          2(295)                   2E(O=O) + 4E(H-H) 8E(O-H)


            N2(g) + Cl2(g) + 2O2(g) + 2O2(g) + 4H2(g)


  Hreaction = 2E(O=O) + 4E(H-H) 8E(O-H) 2(295)

            = 374 kJ


111
Estimate the lattice enthalpy of NH4ClO4.
                           ΔHL [NH4ClO4 (s)]
                             o

      NH4+(g) + ClO4(g)                       NH4ClO4(s)

  Given : ΔHfo [NH4ClO4 (s)] = 295 kJ mol1
            First IE of H = 1312 kJ mol1
            First EA of O = 141 kJ mol1
            E(Cl=O) = 533 kJ mol1
            E(Cl-O) = 209 kJ mol1
            E(Cl-Cl) = 242 kJ mol1



112
Estimate the lattice enthalpy of NH4ClO4.
                               ΔHL [NH4ClO4 (s)]
                                 o

      NH4+(g) + ClO4(g)                           NH4ClO4(s)

                         ½E(Cl-Cl)+2E(O=O)
                       -E(Cl-O)-3E(Cl=O) + 1st EA[O]   ΔHf
                                                         o


         ½E(NN)+2E(H-H)                               = 295
         -4E(N-H) +1st IE[H]



                        ½N2(g) + 2H2(g) + ½Cl2(g) + 2O2(g)

      ΔHL = -295-½E(NN)-2E(H-H)+4E(N-H)-1st IE[H]
        o


            - ½E(Cl-Cl)-2E(O=O)+ E(Cl-O)+3E(Cl=O)- 1st EA[O]
          = 556 kJ
113
 Relationship between bond strength and bond length

 Bond strength :
 - determined by B.D.E. or bond enthalpy
 - depends on the balance between
      attractive forces and repulsive forces within the
      molecule
      Attractive forces between bond pairs and bonding
      nuclei strengthen a covalent bond.

      Repulsive forces between lone pairs on adjacent
      bonding atoms weaken a covalent bond.
114
 Relationship between bond strength and bond length

 Bond length :
 - inter-nuclear distance between bonding atoms
 - determined by electron diffraction or
      X-ray diffraction




115
 Relationship between bond strength and bond length

 Bond length depends on
 1. the number of electron pairs involved in the
    covalent bond (bond order)
 2. the sizes (covalent radii) of the bonding atoms




116
Bond lengths listed in data books are average
values
 Bond lengths depend on environment.
Deviation of actual bond lengths from the average
ones are always small
 Bond lengths are transferable




117
          Molecule          Bond length / nm

          HO – H                0.0958
      H–O–O–H                   0.0960
                    O

      H     C                   0.0950

                O       H

          CH3O – H              0.0956


118
       Molecule     Bond length / nm

       diamond          0.15445

      CH3 – CH3         0.1536

      CH3 – CHF2        0.1540

                O

      H3C   C           0.1500

                H

119
       Molecule        Bond length / nm

       CH3 – H             0.1091

      CH3CH2 – H           0.1107

      CH2=CH – H           0.1087



              C    H       0.1084



120
 Relationship between bond strength and bond length

  1. Single Bond vs Multiple Bond (between atoms
     of the same kinds)
            Bond         CC       CC       CC
        Bond enthalpy
                         348       612       837
          (kJ mol-1)
       Bond length/nm    0.154    0.134     0.120

      Bond length : -
      - Since the bonding atoms are of the same kind,
        bond length only depends on bond order.

121
            Bond         CC       CC       CC
        Bond enthalpy
                         348       612       837
          (kJ mol-1)
       Bond length/nm    0.154    0.134     0.120

      Bond length : -
      - Since the bonding atoms are of the same kind,
        bond length only depends on bond order.
      - As the bond order increases, the
        electrostatic attraction between bond pairs
        and nuclei increases
       bonding atoms are drawn closer together
       shorter bond length
122
            Bond         CC       CC       CC
        Bond enthalpy
                         348       612       837
          (kJ mol-1)
       Bond length/nm    0.154    0.134     0.120

      Bond strength : -
      Since there is no lone pair on C,
       bond strength mainly depends on the
         attraction between bond pairs and nuclei
         which in turn depends on bond order.




123
          Bond            CC        CC     CC
      Bond enthalpy
                          348        612     837
        (kJ mol-1)
      Bond length/nm      0.154      0.134   0.120

                  bond      bond

              E(C=C) < 2E(C – C)

                  bond
      C – C  bond is stronger than C – C  bond


124
          Bond            CC        CC     CC
      Bond enthalpy
                          348        612     837
        (kJ mol-1)
      Bond length/nm      0.154      0.134   0.120

                  bond      bond

              E(CC) < 3E(C – C)

                 Two  bonds
      C – C  bond is stronger than C – C  bond


125
            Bond          NN      OO        FF
        Bond enthalpy
                          944       496       158
          (kJ mol-1)
       Bond length/nm    0.110     0.121     0.142

      Bond length : -
      NN < O=O < F-F (atomic size : N > O > F)
      - Bond order is the dominant factor




126
            Bond          NN     OO       FF
        Bond enthalpy
                          944         496   158
          (kJ mol-1)
       Bond length/nm     0.110   0.121     0.142

      Bond strength : -
       greatly from N2 to F2 because
      1. bond order  from N2 to F2
      2. repulsive forces between lone pairs on
         adjacent bonding atoms  from N2 to F2


127
  N          N     O       O        F       F


      Increasing repulsive forces between lone
      pairs on adjacent bonding atoms




128
2. Halogens

            Bond        FF     ClCl   BrBr   II
        Bond enthalpy
                        158     242     193      151
           (kJ mol-1)
       Bond length/nm   0.142   0.199   0.228   0.267

      Bond length : -
      Depends mainly on size of bonding atoms
      Shorter bond length doesn’t mean stronger bond.



129
2. Halogens

            Bond            FF   ClCl   BrBr   II
        Bond enthalpy
                            158   242     193      151
          (kJ mol-1)
       Bond length/nm     0.142   0.199   0.228   0.267

      Bond strength : -
      ClCl > BrBr > II
      It is because the attractive force between bond
      pair and nuclei increases as the size of the
      bonding atoms decreases.
130
2. Halogens

            Bond          FF     ClCl   BrBr   II
        Bond enthalpy
                          158     242     193      151
          (kJ mol-1)
       Bond length/nm     0.142   0.199   0.228   0.267

      Bond strength : -
      FF < ClCl > BrBr > II
      It is because F atoms are so small that the
      repulsion between lone pairs on adjacent F atoms
      becomes dominant
131
3. Hydrogen halides

            Bond        HF     HCl    HBr    HI
        Bond enthalpy
                        565     431     366     299
           (kJ mol-1)
       Bond length/nm   0.109   0.135   0.151   0.171

      Bond length : -
      Depends mainly on size of bonding atoms




132
3. Hydrogen halides

            Bond          HF     HCl    HBr    HI
        Bond enthalpy
                          565     431     366     299
          (kJ mol-1)
       Bond length/nm     0.109   0.135   0.151   0.171

      Bond strength : -
      HF > HCl > HBr > HI It is because
      1. the attraction between nuclei and bond pair 
         as the size of bonding atoms 
      2. there is no lone pair on H atom,
         thus no repulsion between lone pairs.
133
Q.30
          Bond         CH          CC    OO
      Bond enthalpy
                        415         612    496
         (kJ mol-1)
      Bond length/nm   0.109     0.154     0.121

  (a) Bond length : C–H < C=C
        because H atom is smaller than C atom
        Bond strength : C=C > C–H
        because more bond pairs are involved in C=C.

134
Q.30
          Bond         CH      CC       OO
      Bond enthalpy
                       415       612      496
        (kJ mol-1)
      Bond length/nm   0.109    0.154    0.121

  (b) Bond length : O=O < C=C
        because O atom is smaller than C atom
        Bond strength : O=O < C=C
        because there are repulsive forces between
        lone pairs on adjacent oxygen atoms.
135
Q.31

 NN < CC

 N is smaller than C




136
Q.31

 NN < CC < O=O
       CC has a greater
       bond order




137
Q.31

 NN < CC < O=O < N=N < C=C
          Atomic size : O < N < C




138
Q.31

 NN < CC < O=O < N=N < C=C < O–O

                    C=C has a greater
                    bond order




139
Q.31

 NN < CC < O=O < N=N < C=C < O–O < N–N < C–C
                          Atomic size : O < N < C




140
Covalent Radius
 •    Half the internuclear distance between two
      singly and covalently bonded atoms of the
      same kind in a molecule




141
  The covalent radii (in nm) of some elements

142
Van der Waals’ Radius
 •    Half the internuclear distance between two
      atoms in adjacent molecules which are not
      chemically bonded.



      I2



143
Atomic radius  from left to right across a period because
ENC experienced by bonding electrons  from left to right.


144
      Atomic radii of noble gases were obtained by
      extrapolation (smaller values)




145
                                        Calculated
                                          value

                                           0.031


                                           0.038


                                           0.071

                                           0.088

                                           0.108


  The covalent radii (in nm) of some elements

146
  Alternatively, van der Waals’ radii of noble gases
  are taken as the atomic radii (larger values).

                                             0.140


                                             0.154


                                             0.188

                                             0.202

                                             0.216


  The covalent radii (in nm) of some elements
                                           van der
                                         Waals’ radius
147
 Additivity rule of covalent radii




 Assumption : Electrons are equally shared between A and B
             Pure covalent bond

148
                       CBr in   CF in    CO in     CO in
          Bond
                        CBr4      CF4      CH3OH       CO2
       Experimental
                       0.1940    0.1320    0.1430     0.1160
         value/nm
      Estimated bond
                       0.1910    0.1480    0.1510     0.1275
        length/nm

       % deviation     -1.54%    12.12%    5.59%      9.91%


  Failure of additivity rule indicates formation of
  covalent bond with ionic character due to polarization of
  shared electron cloud to the more electronegative atom.

149
                       CBr in   CF in   CO in     CO in
          Bond
                        CBr4      CF4     CH3OH       CO2
       Experimental
                       0.1940    0.1320   0.1430     0.1160
         value/nm
      Estimated bond
                       0.1910    0.1480   0.1510     0.1275
        length/nm

       % deviation     -1.54%    12.12%    5.59%     9.91%


  Polarization of a covalent bond always results in the
  formation a stronger bond with shorter bond length.
                          +      
                         C        F
150
Benzene : Kekule structure               0.134 nm
                                   H


                        H          C         H
                              C         C
                                                 0.154 nm
                              C         C
                        H          C         H


                                   H

      In fact, all six carbon – carbon bonds are identical with
      a bond length equal to 0.139 nm which is intermediate
      between C – C bond length and C=C bond length


151
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)




(a) Predict the approximate bond lengths of
      Si – H, P – H, S – H and H – Cl from the following data:
               Bond             Bond length (nm)
               H–H                   0.074
              Si – Si                0.235
             P – P (P4)              0.221
             S – S (S4)              0.207
              Cl – Cl                0.199
      (Hint: Assume that covalent radii are additive.)
                                                           Answer
152
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)




                             0.235      0.074
(a) Bond length of Si – H =        nm        nm
                               2          2
                          = 0.154 5 nm
                              0.221      0.074
      Bond length of P – H =        nm        nm
                                2          2
                           = 0.147 5 nm
                              0.207      0.074
      Bond length of S – H =        nm        nm
                                2          2
                           = 0.140 5 nm
                               0.074      0.199
      Bond length of H – Cl =        nm        nm
                                 2          2
                            = 0.136 5 nm
153
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)




(b) The bond enthalpies of Si – H, P – H, S – H and H – Cl are
    given in the following table:

              Bond        Bond enthalpies (kJ mol–1)
              Si – H               +318
              P–H                  +322
              S–H                  +338
              Cl – H               +431
      Assume the actual bond lengths are very close to that
      calculated in (a), describe the relationship between bond
      length and bond enthalpy.
                                                         Answer
154
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)




  (b) The bond enthalpy of a covalent bond is related to the length. The
      larger the bond length, the weaker the attractive force between the
      two bonded atoms and the smaller is the bond enthalpy.




                                              Back




155
8.6 Bond Enthalpies, Bond Lengths and Covalent Radii (SB p.230)




         Why does the covalent radius of a given element
         change from one compound to another compound?

      The covalent radius of an atom is determined by the
      size of its bonding electron cloud which may vary due to
      the presence of different electron clouds and bonding
      atoms in different compounds.


                                               Back

156
      8.7
      Shapes of Covalent
         Molecules and
        Polyatomic Ions

157
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)


Valence Shell Electron Pair Repulsion
theory (VSEPR theory)
 Developed by Gillespie and Nyholm

 5 rules to predict the shapes of molecules
 and polyatomic ions




158
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)


 Rule 1
 The bond pairs and lone pairs in the valence
 shell arrange around the central atom in
 such a way as to minimize the electrostatic
 repulsion .
  Maximum angular separation between
    valence electron pairs around the
    central atom



159
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)

A. Molecules and Polyatomic Ions without
Lone Pair Electrons on the Central Atom
 •    Examples:
      1. Beryllium Chloride (BeCl2) Molecule
      2. Boron Trifluoride (BF3) Molecule
      3. Methane (CH4) Molecule
      4. Ammonium Ion (NH4+)
      5. Phosphorus Pentachloride (PCl5) Molecule
      6. Sulphur Hexafluoride (SF6) Molecule
160
  Step 1
  Draw the bond-line / Lewis structure

  Step 2
  Predict the shape using VSEPR theory




161
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)


      1. Beryllium Chloride Molecule (BeCl2)
      Electronic Diagram                    Shape in Diagram




                  Be         Cl
        Cl


                                               Bond angle
                                             = angle between
                                               2 bonds
        Shape in word


162
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.231)


      1. Beryllium Chloride Molecule (BeCl2)
      Electronic Diagram                    Shape in Diagram




                  Be         Cl
        Cl

                                        180 is the maximum
                                          possible angular
                                             separation
        Shape in word        Linear


163
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)


      2. Boron Trifluoride Molecule (BF3)
      Electronic Diagram                    Shape in Diagram



                  F




                  B

          F                F            120 is the maximum
                                          possible angular
                                             separation
164
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)


      2. Boron Trifluoride Molecule (BF3)
      Electronic Diagram                    Shape in Diagram



                  F




                  B

          F                F
                                               Shape in word

                                             Trigonal planar
165
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)


      3. Methane (CH4) Molecule
      Electronic Diagram                  Bond-line structure


                  H
                                                    H
                                                         90

                                          H         C          H
        H         C          H

                                                    H
                  H                      Not the maximum
                                         angular separation
166
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)


      3. Methane (CH4) Molecule
      Electronic Diagram                    Shape in Diagram


                  H




        H         C          H



                                               Shape in word
                  H
                                              Tetrahedral
167
          H


          C
                  H
      H
              H
                      bond in plane of the paper

                      bond coming out of the paper

                      bond going behind the paper

168
      The three-dimensional shape of a
      methane molecule can be represented as:




169
      mirror images of each other
               identical
170
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)


      4. Ammonium Molecule (NH4+)
      Electronic Diagram                    Shape in Diagram


                  H




        H         N          H



                                               Shape in word
                  H
                                              Tetrahedral
171
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)

      5. Phosphorus Pentachloride (PCl5) Molecule
      Electronic Diagram                  Bond-line structure


                  Cl                                Cl
        Cl                  Cl          Cl                     Cl
                  P
                                                    P       72
             Cl        Cl                    Cl           Cl
                                         Not the maximum
                                         angular separation
172
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)

      5. Phosphorus Pentachloride (PCl5) Molecule
      Electronic Diagram                    Shape in Diagram


                  Cl


        Cl                  Cl
                  P


             Cl        Cl
                                               Shape in word

                                        Trigonal bipyramidal
173
      stronger bond




       weaker bond

  Bond pairs occupying the axial positions have less angular
  separation.
  They experience stronger repulsion
  P – Cl bonds at axial position are weaker and have a
  greater bond length (214 pm).


174
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.232)


      6. Sulphur Hexafluoride (SF6) Molecule
      Electronic Diagram                    Shape in Diagram


                 F
          F              F
                 S
         F              F
                F
                                               Shape in word
                                                 Octahedral
175
      Making balloon models



176
  Rule 2
  Double bond and triple bond are
  considered as one electron pair or one
  negative centre.




177
 Q.33     Linear

      O     C      O

            
      O     N      O


      H     C      N


178
 Q.33
           O

           S
                   Trigonal planar
      O        O

      Expansion of octet to form more bonds




179
 Q.33                                2

                       O   C    O


                           O


      All bonds are identical due to delocalization
      of electrons

                  2                 2                   2

 O       C    O        O    C    O        O    C      O
                                     
         O                  O                  O

180
                     2
 Q.33




        Trigonal planar




181
 Q.33




      All bonds are identical due to delocalization
      of electrons
                  Trigonal planar




182
 Q.33
          O
                   All bonds are identical due to
          S       delocalization of electrons
               O
               
      O
              O            Tetrahedral




183
 Q.33
                         O
                                      Tetrahedral (distorted)
              > 109.5
                         P       Cl
                Cl              Cl
                     < 109.5

      Repulsion between double bond and single bond
      > Repulsion between single bond and single bond




184
 Q.33                        +
                 H
        109.5
                 N
                         H
          H
                     H


           Tetrahedral




185
  Rule 3
  Repulsion between electron pairs
  follows the order : -

  lone pair–lone pair > lone pair–bond pair
                           > bond pair–bond
  pair




186
  Lone pairs are attracted by one nucleus
  only,
   they occupy larger space or
     they are more diffused,
   they produce more repulsion.

  Lone pairs are not considered in the shapes
  of molecules and polyatomic ions.

  Lone pairs are represented by


187
 Q.34(1)


      Cl   Sn   Cl         Sn
                     Cl          Cl

                     Trigonal planar
                     with respect to
                      electron pairs




188
 Q.34(1)

                        >120
                Sn              AX2E
           Cl             Cl
                <120


           Bent or V-shaped
            with respect to
                 atoms




189
 Q.34(2)



      H    N   H        N
                                H
                    H
                            H
           H
                     Tetrahedral
                   with respect to
                    electron pairs




190
 Q.34(2)


                       >109.5   AX3E
                 N
                           H
            H
                107   H

           Trigonal pyramidal
            with respect to
                 atoms



191
192
 Q.34(3)
                        H

      H   O   H         O
                    H

                    Tetrahedral
                  with respect to
                   electron pairs




193
 Q.34(3)
                       H
                           b
              104.5
      AX2E2            O           a > b > 104.5
                               a
                H
                       b

              Bent or V-shaped
               with respect to
                    atoms




194
195
 NH2                                 
                              H
            

      H     O
            N     H           N
                              O
                         H

          AX2E2           Tetrahedral
                      w.r.t. electron pairs

                      Bent or V-shaped
                        w.r.t. atoms


196
                 a                 b
           N                  P
                     H
      H                                H
          107   H       H
                             90   H

      Larger lone pair  stronger repulsion
      b > a > 109.5


197
 Q.35       180 > a > b
                                   180

  O     N     O            O       N      O


                               x
  O     N     O                     N
                           O        a     O
        y>x a>b

  O     N     O                Y
                                    N
                           O              O
                                    b
198
  Rule 4
  Single electron exerts less repulsive
  forces on other electrons than
  electron pair does.
  Repelling power : -
  lone pair > triple bond > double bond
  > single bond > single electron



199
 Q.36(1)                              -
                               Cl
           
      Cl   I   Cl              I

                               Cl

 Rule 5
 To minimize the great repulsion between
 lone pairs,
 (1) lone pairs should occupy positions with
     the largest angular separation.
     i.e. the equatorial position
200
 Q.36(1)
                   -
            Cl         Trigonal bipyramidal
                       w.r.t. electron pairs
            I
                            Linear
            Cl           w.r.t. atoms

           AX2E3


201
 Q.36(2)
                                      -
           Cl
                        Cl       Cl
      Cl   I    Cl           I
                        Cl       Cl
           Cl


 Rule 5
 To minimize the great repulsion between
 lone pairs,
 (2) lone pairs are located as far apart from
     one another as possible.
202
 Q.36(2)
                        -
                                 Octahedral
      Cl        Cl          w.r.t. electron pairs
            I
      Cl           Cl
                               Square planar
                                w.r.t. atoms

           AX4E2


203
 Q.37
          F               F
                              90
                                    F
      F   S   F           S         <120
                                    F
          F               F




                  F       S          F

                      F       F

204
 Q.37

                          Trigonal bipyramidal
                          w.r.t. electron pairs
      F       S       F
                                Seesaw
          F       F           w.r.t. atoms

          AX4E



205
 Q.37
                            Cl
                      90
      Cl   I    Cl   Cl     I


           Cl               Cl




206
 Q.37

            Cl
                   Trigonal bipyramidal
                   w.r.t. electron pairs
      Cl     I
                        T-shaped
            Cl         w.r.t. atoms

           AX3E2


207
 Q.37                              -
                               I
          
                        180
      I   I   I                I       AX2E3
                               I


                  Trigonal bipyramidal
                  w.r.t. electron pairs
                         Linear
                      w.r.t. atoms


208
 Q.37
              F                        F
                               < 90
                               F           F
      F       I       F                I       AX5E
                              F            F
          F       F


                               Octahedral
                          w.r.t. electron pairs
                           Square pyramidal
                             w.r.t. atoms

209
 Q.37
              O                O
                           F        F
      F       Xe       F       Xe
                           F        F
          F        F


      Lone pair and double bond must be located
      as far apart from each other as possible




210
 Q.37
              O                        O
                               < 90
                                F           F
      F       Xe       F               Xe       AX5E
                               F            F
          F        F


                                Octahedral
                           w.r.t. electron pairs
                            Square pyramidal
                              w.r.t. atoms

211
 Q.37
          F
                       F          F
      F   Xe   F            Xe    90   AX4E2
                       F          F
          F
                        Octahedral
                   w.r.t. electron pairs
                     Square planar
                      w.r.t. atoms

212
 Summary : -

      No. of negative
                        Pattern        Shape
          centre
            2            AX2           Linear

            3            AX3       Trigonal planar

            3            AX2E      Bent / V-shaped

            4            AX4         Tetrahedral

            4            AX3E     Trigonal pyramidal

            4           AX2E2      Bent / V-shaped

213
      No. of negative
                        Pattern         Shape
          centre
            5            AX5      Trigonal bipyramidal

            5            AX4E           Seesaw

            5           AX3E2          T-shaped

            5           AX2E3           Linear

            6            AX6          Octahedral

            6            AX5E      Square pyramidal

            6           AX4E2        Square planar

214
Hybridization and hydrid orbitals(by Linus Pauling)

 Problems identified : -

 For CH4
 Experimental findings : -
 1. Four identical C – H bonds
 2. Bond angle = 109.5




215
By VB Theory,
       2s                2p
  C                   

      Only 2 single bonds can be formed.

      Promotion of a 2s electron to a 2p
       orbital.
       2s            2p
  C*                       

216
        2s         2p                      1s
  C*                             4H    

       Two types of orbital overlaps : -
       2s–1s head-on overlap  stronger C – H bond
       2p–1s head-on overlap  weaker C – H bond




217
           2s         2p                     1s
  C*                                4H    

      The three 2p orbitals of carbon are at right
      angles, 90, and fail to match the tetrahedral
      angle of 109.5.

      The spherical 2s and 1s orbitals could overlap in
      any direction.




218
 Solutions to problems : -
 Atomic orbitals (2s, 2px, 2py, 2pz ) first mix to
 give four identical hybrid orbitals (sp3) arranged
 tetrahedrally.
                 2s          2p
           C*                   

                             sp3 hybridization


                                
                         sp3
219
 Solutions to problems : -
           sp3                         1s
                             4H     

                             sp3 – 1s
                             head-on overlaps


      four identical C – H single bonds arranged
      tetrahedrally.



220
                           sp3
                      hybridization




      Electron density of hybrid orbital is more
      concentrated in the bonding direction than
      s and p orbitals
       better overlap of orbitals and stronger bond.

221
                          sp3
                     hybridization




      Each sp3 hybrid orbital has
      25% s character and
      75% p character

222
223
Basic Concepts of the Theory of Hybridization

1.    Hybridization is a mathematical treatment of
      the wavefunctions of atomic orbitals to give
      new wavefunctions representing hybrid
      orbitals.

2. Only orbitals of similar energies can be mixed
   together to form stable hybrid orbitals.




224
 3. The number of hybrid orbitals produced equals
    the number of atomic orbitals mixed together.

 4. Hybridization is assigned only after the
    structure of a species is known.

      In other words, the theory of hybridization
      has no predicting power,
      it is just a backward statement to
      rationalize the experimental findings.




225
 Interpretation of shapes of molecules and
 polyatomic ions using the theory of hybridization

 1. BeCl2

      Experimental findings : -
      A linear molecule in gas state (over 520oC) with
      two identical Be – Cl bonds




226
       2s       2px
  Be   

                        excitation
       2s       2px
Be*            

                    sp hybridization

                           Two identical sp hybrid
                   
                           orbitals lying along the x-axis

227
                             sp
                        hybridization




      Each sp hybrid orbital has
      50% s character and
      50% p character




228
                sp                 3px
      Be*                  2Cl    

                          sp-3px
                          head-on overlaps


      Two identical Be – F  bonds along the
      x-axis with angular separation of 180




229
 Interpretation of shapes of molecules and
 polyatomic ions using the theory of hybridization

 2. BF3

      Experimental findings : -
      A trigonal planar molecule with three identical
      B – F bonds




230
       2s       2px 2py 2pz
  B           
                       excitation
       2s       2px 2py 2pz
  B*                 

                    sp2 hybridization

                           Three identical sp2 hybrid
                        orbitals lying on the x-y plane

231
                             sp2
                        hybridization




      Each sp2 hybrid orbital has
      1/3 s character and
      2/3 p character



232
               sp2                  2px
      B*                    3F    

                           sp2-2px
                           head-on overlaps


       Three identical B – F  bonds on the x-y
        plane with angular separation of 120




233
              2s      2px 2py 2pz
      B*                  


           The empty unhybridized 2pz
           orbital of B is perpendicular
           to the x-y plane




234
 Interpretation of shapes of molecules and
 polyatomic ions using the theory of hybridization

 4. NH3

      Experimental findings : -
      A trigonal pyramidal molecule with three
      identical N – H bonds




235
      2s    2px 2py 2pz
  N                      

                 sp3 hybridization


  N                  

      Four identical sp3 hybrid
      orbitals arranged tetrahedrally



236
           sp3
      hybridization




237
                 sp3                  1s
      N                    3H    

                            sp3-1s
                            head-on overlap


           Three identical N – H  bonds with
               angular separation of 107




238
             sp3
N              

                       Lone pair in sp3 hybrid
                       orbital




239
  Q.38

      H2O
                                 H
                    VSEPR
      H     O   H                O
                            H

                             Tetrahedral
                              sp3 hybridization
                            Backward statement !!!


240
      2s     2px 2py 2pz
  O                

               sp3 hybridization
                                    1s
  O                      2H    

      Lone pairs             sp3-1s
                             head-on overlaps

           Two identical O – H  bonds with
             angular separation of 104.5
241
      CH3NH2


           H   H                   H
                                       H
                   VSEPR       N
       H   C   N
                               C
                                       H
           H   H           H
                                   H

                           Tetrahedral
                            sp3 hybridization




242
                         sp3 – 1s
                     head-on overlap

                                  H
                                       H
                           N
      Lone pair in                sp3 – sp3 head-on overlap
      sp3 orbital
                            C
                                       H
                     H
                                  H

                         sp3 – 1s
                     head-on overlap
243
 Hybridization Involving s, p, and d Atomic Orbitals

 PF5

 Experimental findings : -
 A trigonal bipyramidal molecule with 3 stronger
 and identical P – F bonds at equatorial positions
 and 2 weaker and identical P – F bonds at axial
 positions




244
           3s       3px 3py 3pz
  P                         
                           excitation
           3s       3px 3py 3pz                     3d
  P*                                  

                        sp3d hybridization

                                        Five identical sp3d hybrid
                                   orbitals in trigonal
                                        pyramidal arrangement

245
                         F

                                  F
                F        P
                                 F
                         F
      Five identical P – F sigma bonds formed by
              sp3d – 2p head-on overlaps



246
 Q.39

 SF6

 Experimental findings : -
 An octahedral molecule with 6 identical S – F bonds




247
           3s       3px 3py 3pz
  S                        
                           excitation
           3s       3px 3py 3pz                   3d
  S*                                     

                        sp3d2 hybridization

                                         Six identical sp3d2
                                   hybrid orbitals in
                                         octahedral arrangement

248
                         F
                  F              F
                         S
                 F               F
                         F

      Six identical S – F sigma bonds formed by
             sp3d2 – 2p head-on overlaps



249
      Rules for Identifying the
      Type of Hybridization of
      a Central Atom in a Given
      Species

      1. Work out the most
         stable Lewis structure
         of the species.
      2. Predict the molecular
         shape of the species
         with respect to
         electron pairs around
         the central atom using
         VSEPR theory.

250
      Rules for Identifying the
      Type of Hybridization of
      a Central Atom in a Given
      Species
      3. Match the predicted
         molecular shape with the
         type of hybridization
         according to the
         following table.




251
Q.40

 BH4                   
            H


            B               sp3
                    H
        H
                H




252
Q.40
                   
SF5       F
       F       F
           S       sp3d2
       F       F




253
Q.40
              Cl

 BCl3         B         sp2
         Cl        Cl


              F

  ClF3   F    Cl        sp3d

              F



254
Q.40         F

                  F
OSF4     O   S        sp3d
                  F
             F
                      4-
             O
XeO64
         O        O
             Xe
                           sp3d2
         O        O
             O


255
      Multiple Bonds



256
Ethene (CH2 = CH2) Molecule
Experimental findings :
A planar molecule with 4 identical C – H bonds and
one C  C bond. All bond angles are approximately
equal to 120o

      H   H             H            H
                            >120
               VSEPR
      C   C                 C    C   < 120

      H   H             H            H
                        trigonal planar
                        sp2 hybridized
257
       2s       2px 2py 2pz
  C                 
                       excitation
       2s       2px 2py 2pz
  C*                     

                    sp2 hybridization

                           Three identical sp2 hybrid
                        orbitals lying on the x-y plane

258
           sp2         2pz
                    
                             2pz orbital




          x- y plane


259
      sp2 – 1s head-on overlaps
       4 identical C – H  bonds




260
      sp2 – sp2 head-on overlap
       C – C  bond




261
      Side-way overlap




262
                                      E(C=C) < 2E(C–C)
                           kJ mol1    612     2(348)


      Head-on overlap is better than side-way overlap
       sigma bond is stronger than pi bond


263
      The 2pz orbitals cannot come closer to each
      other to have a better overlap since it will
      weaken the sigma bond.
264
 The electron cloud in a pi bond concentrates in regions
 above and below the nodal plane (x – y plane) where there
 is zero probability of finding the electrons.




265
      Orbital overlaps for C2H4




266
Ethyne (CH  CH) Molecule
  Experimental results :
  A linear molecule with two identical C – H bonds
  and one CC bond

                      VSEPR
      H   C   C   H              H    C   C    H

                                      Linear
                                  sp hybridized



267
      2s       2px 2py 2pz
  C                 
                       excitation

      2s       2px 2py 2pz
C*                      

                   sp hybridization

                           Two identical sp hybrid
                  
                           orbitals lying along the x-axis

268
      sp       2py   2pz
                           sp – sp head-on overlap
                  
                            C – C  bond




269
      sp       2py   2pz
                           sp – 1s head-on overlaps
                  
                            two identical C – H  bonds




270
      sp       2py   2pz
                           2py – 2py side-way overlap
                  
                            C – C  bond




271
      sp       2py   2pz
                           2pz – 2pz side-way overlap
                  
                            C – C  bond




272
                              Less energy is given out
                              when the second pi bond
                              is formed due to over-
                              crowded electrons
      Bond   B.E.(kJ mol1)
      C–C        348
                               264 first pi bond
      C=C         612
                               225 second pi bond
      CC        837

273
 Q.40

                VSEPR
 O      C   O           O     C      O
                            Linear
                        C : sp hybridized




274
 Q.40

                         VSEPR
 O         C        O              O     C      O
                                       Linear
      2s       2px 2py 2pz
                                   C : sp hybridized
 O                           O : sp2 hybridized

               sp2 hybridization

           sp2           2pz
                    

275
           sp       2py   2pz             sp2         2p
C                           2O               

    sp – sp2 head-on overlaps  Two identical C – O  bonds




276
           sp       2py   2pz             sp2   2py
C                            O         

    2py – 2py side-way overlap  C – O  bond




277
           sp       2py   2pz             sp2   2pz
C                            O         

    2pz – 2pz side-way overlap  C – O  bond




278
                  N          N
      Each N is surrounded by two negative centers
       N is sp hybridized
                                 2s    2px 2py 2pz
                             N                
                                      sp hybridization


                                  

279
       sp      2py   2pz            sp    2py   2pz
N                      N               

 sp – sp head-on overlap  N – N  bond




280
       sp       2py    2pz              sp   2py   2pz
N                        N                

 Lone pairs are in sp hybrid orbitals




281
       sp      2py   2pz            sp       2py   2pz
N                       N                 

 2py – 2py side-way overlap  N – N  bond




282
       sp      2py   2pz            sp       2py   2pz
N                       N                 

 2pz – 2pz side-way overlap  N – N  bond




283
 Benzene (C6H6)

 The simplest aromatic hydrocarbon
  Experimental Findings of Benzene : -
  A cyclic and planar structure with six identical
  C – H bonds and six identical C – C bonds.

        Bond       C–C        Benzene        C=C

  Bond length
                   1.54         1.39         1.34
     (nm)

      Bond order    1            1.5          2

284
 Benzene (C6H6)




              Electron density map
      Symmetrical distribution of electrons
285
Interpretation in Terms of Valence Bond Theory

 Step 1 : Most stable Lewis structure

           H                            H


      H           H          H              H




      H           H          H              H


           H                            H




286
  Q.42
              H                           H


      H               H           H               H




      H               H           H               H


              H                           H

      The real structure is the resonance hybrid of the
      two resonance structures.
       Each C – C bond has 50% single bond character
       and 50% double bond character.

287
      Kekulé structure(1865)



                          Thiele
                        structure
                          (1899)


288
 Step 2 : Type of hybridization of C

           H                           H


      H           H           H               H




      H           H           H               H


           H                           H

  Three negative centers around each C atom
   Trigonal planar
   sp2 hybridization
289
  Q.43 Sigma framework
                         sp2 – 1s head-on overlaps
                          six identical C – H  bonds




290
  Q.43 Sigma framework
                         sp2 – sp2 head-on overlaps
                          six identical C – C  bonds




291
  Q.43 Pi framework

      The six unhybridized 2pz orbitals are ⊥ to the
      molecular x – y plane




292
  Q.43 Pi framework

      2pz – 2pz side-way overlaps
       Three identical C – C pi bonds




293
294
  Problems Leading to a Modification of Valence
  Bond Theory
  1. More resonance structures can be drawn
     for benzene although they make little
     contribution to the real structure of benzene.




  Kekulé structure(1865)    Dewar structure(1867)
    More stable              Less stable
    More contribution to     Less contribution to
    the real structure       the real structure
295
  Q.44




                                           Least
               Less stable                 stable

      2pz orbitals are far apart from each other
       poor overlaps

296
  Problems Leading to a Modification of Valence
  Bond Theory

  2. All the resonance structures drawn are based
     on the assumption that bonding electrons are
     localized between two atoms.

      Better approach (MO theory) : -
      Delocalization of  electrons
      The six  electrons are shared by all six C atoms




297
      Delocalization of Pi Electrons – A simplified
      MO Theory




      The six 2pz orbitals overlap with one
      another to give a continuous Pi electron
      cloud above and below the molecular plane.


298
      In fact, the combination of six 2pz orbitals
      gives rise to six molecular orbitals.




299
      Anti-bonding
        orbitals

           
                     2pz

        Bonding
        orbitals



300
      The most stable bonding molecular orbital
      which can accommodate only two electrons



301
Simple Rules for Recognizing Delocalization
of  Electrons
 1. Two or more resonance structures (with the
    same/similar stability) can be drawn to
    represent  bond arrangement of a species




302
Simple Rules for Recognizing Delocalization
of  Electrons

2. There are more than two adjacent atoms
   having p orbitals that are parallel to one
   another.
                     p orbitals parallel to one another
                      Maximum side-way overlap

                     p orbitals ⊥ to one another
                      Minimum side-way overlap


303
  Q.45

  O       C     O     O   C   O       O   C   O



       stable                 unstable

      Only ONE stable structure can be drawn
       No significant delocalization of pi electrons




304
  Q.45




      There are only TWO adjacent p orbitals
      parallel to each other : 2py // 2py ; 2pz // 2pz
       No significant delocalization of pi electrons

305
  Using Curly Arrows to illustrate the
  Delocalization of  Electrons (p.60)




      Rule 1 : A curly arrow represents the movement
                   of an electron pair (pi electron pair or
               lone pair)


306
  Using Curly Arrows to illustrate the
  Delocalization of  Electrons (p.60)




      Rule 2 : If an electron pair moves in on a new
               atom, another pair must leave that atom
               so that the octet rule is observed.


307
 Driving Force for Delocalization of Pi Electrons

  1. There is more space for electrons to distribute within
     the molecule.
     Repulsion between electrons is minimized.

  2. Electrons can be attracted by more than two nuclei at
     the same time.




308
  Evidence for the Stabilization of Benzene by
  Delocalization of Pi Electrons

      Q.46
                                                    H


               + H–H
                                                    H

        ΔHo = 120 kJ mol1
              E(C=C) + E(H–H) – E(C–C) – 2E(C–H)




309
                                                      H


               + H–H
                                                      H

      ΔHo = 120 kJ mol1
              E(C=C) + E(H–H) – E(C–C) – 2E(C–H)
                                            H

                                      H           H


                + 3H – H
                                      H           H

                                            H

      ΔH  3E(C=C) + 3E(H–H) – 3E(C–C) – 6E(C–H)
         o


             = 3(120) kJ mol1 = 360 kJ mol1

310
                       + 3H2(g)       Delocalization enthalpy
      Enthalpy




                      H = 152 kJ mol1
                    + 3H2(g)

                                             H = 360 kJ mol1

                 H = 208 kJ mol1




                       Reaction coordinate
311
 More examples of delocalization of pi electrons
  NO3
  Rule 1 : -

       O            O                O

           N   O        N   O            N   O

       O            O                O




312
 More examples of delocalization of pi electrons

         O                  O                   O

             N   O              N   O               N   O

         O                  O                   O


      N and O are sp2 hybridized such that
      the 2pz orbital of N and the 2pz orbitals of adjacent O
      atoms are parallel to one another
       maximum side-way overlap




313
 More examples of delocalization of pi electrons
  Rule 2 : -

              O               O                       O

                  N   O           N       O               N       O

              O               O                       O


                                                
      O                 O                     O                  
                 N   O               N       O              N       O
                                                  
          O                   O                       O
                                                     



314
 Q.47

  HNO3
          O                        O

          N        H               N        H
      O       O               O         O



          Structures with equal stability


315
  The ionizable hydrogen atom must bond to the
  more electronegative oxygen
              O

      H   O    S   O       H
                                       O
              O

          Hydroxyl group       H   O   N     O
              O

H         O   S    O       H



316
          H
                                 O
          O
                   H     O       C   O       H

      O   Cl   O
                                 O
          O
                   H3C       C

                                 O       H

317
            O                        O

            N         H              N       H
      O          O              O        O




                                     O
          Unstable due to greater
          separation of opposite     N       H
          formal charges         O       O

318
                             Bond order = 1.5
                              0.122 nm
                130
         0.122 nm
Bond order = 1.5

                       0.141 nm

                   Bond order = 1


319
  NO3 Structures with equal stability


      O              O               O

          N   O          N   O           N   O

      O              O               O




320
  NO3
         All ∠ONO = 120
         All N – O bonds are identical with
         a bond length of 0.121 nm and
                           1
         a bond order of 1
                           3




321
 CH3COO Structures with equal stability

                O                          O

      H3C   C                 H3C     C

                O                          O



                      Two identical C – O bonds
                      with a bond order of 1.5


322
 CO32
          O                  O                  O


          C                  C                  C
      O         O        O        O        O          O

              Structures with equal stability
          All C – O bonds are identical with a bond
                     1
          order of 1
                     3

323
      O3




            Structures with equal stability


                    ½          ½

      All O – O bonds are identical with a bond
      order of 1.5

324
  NO2




          Structures with equal stability


                  ½            ½


      All N – O bonds are identical with a bond
      order of 1.5
325
      Covalent Crystals



326
8.9 Covalent Crystals (SB p.240)


      Covalent Crystals
      1. Simple molecular structures
      2. Molecular structures (e.g. C60)
      3. Giant covalent structures




327
8.9 Covalent Crystals (SB p.240)


      Substances with Simple Molecular
      Structures
      •   Consist of discrete molecules held
          together by weak intermolecular forces
      •   Atoms in a molecule are held together by
          strong covalent bonds
      •   Examples: H2 , O2 , H2O, CO2, I2

                               p.82
328
8.9 Covalent Crystals (SB p.240)


      Substances with Giant Covalent
      Structures
      •   Consist of millions of atoms bonded
          covalently together to give a
          3-dimensional network
      •   No simple molecules present
      •   Examples: diamond, graphite and quartz
          (silicon(IV) oxide)


329
8.9 Covalent Crystals (SB p.240)

  1. Diamond




                          A diamond crystal

330
8.9 Covalent Crystals (SB p.241)




                     The structure of diamond


331
 Similar to the structure of ZnS, p.18)

 Unit cell           Each C atom is tetrahedrally
                     bonded to four other C
                     atoms
                     All the C – C bonds are
                     strong and identical
                     Bond angle = 109.5

                     Bond length = 0.154 nm
                     Bond order = 1

332
8.9 Covalent Crystals (SB p.241)

 2. Graphite




                              Graphite

333
      Each C atom is covalently bonded to 3 other C
      atoms in the same layer.
      A network of coplanar hexagons is formed
334
      Weak van der Waals’ forces hold the layers
      together

335
                     A

                     B

                     A
      3D structure

336
  Q.48

 Why is the C – C bond length of graphite(0.142 nm)
 shorter than that of diamond (0.154 nm) ?

337
C : sp2 hybridized


  Firstly,
  s character : sp2 > sp3
  sp2-sp2 overlap is better than sp3-sp3 overlap
  Strength of  bond : graphite > diamond
338
Secondly,
Each C atom has a half filled 2pz orbital ⊥ to the plane.
Side-way overlaps of 2pz orbitals give extensively
delocalized  bonds.
339
      C=C > C–C > C-C
      1.34 nm   1.42 nm   1.54 nm
340
8.9 Covalent Crystals (SB p.242)


      Comparison of the properties of
      diamond and graphite
           Property                   Diamond        Graphite
  Density (g cm-3)                 3.51           2.27
  Hardness                         10 (hardest)   < 1 (very soft)
  Melting point (C)               3 827          3 652 (sublime)
  Colour                           Colourless     Shiny black
  Electrical conductivity          None           High

Diamond : extremely hard
                 used in drills and cutting tools
341
8.9 Covalent Crystals (SB p.242)


      Comparison of the properties of
      diamond and graphite
           Property                   Diamond        Graphite
  Density (g cm-3)                 3.51           2.27
  Hardness                         10 (hardest)   < 1 (very soft)
  Melting point (C)               3 827          3 652 (sublime)
  Colour                           Colourless     Shiny black
  Electrical conductivity          None           High

Graphite : soft and slippery
                 used as pencil ‘lead’ and lubricant
342
8.9 Covalent Crystals (SB p.242)


      Comparison of the properties of
      diamond and graphite
           Property                   Diamond        Graphite
  Density (g cm-3)                 3.51           2.27
  Hardness                         10 (hardest)   < 1 (very soft)
  Melting point (C)               3 827          3 652 (sublime)
  Colour                           Colourless     Shiny black
  Electrical conductivity          None           High

Graphite : high electrical conductivity
                 Used as electrode
343
Q.49

The layers in graphite are held by weak van der
Waals’ forces.
The layers can slip over each other when a force is
applied to them.
Graphite is slippery and can be used to make
lubricant




344
Q.49

The high electrical conductivity of graphite is due
to the 2pz electrons which delocalize extensively
throughout the whole structure.
Graphite is used to make electrodes.




345
8.9 Covalent Crystals (SB p.242)


      Comparison of the properties of
      diamond and graphite
               Property               Diamond        Graphite
  Density (g cm-3)                 3.51           2.27
  Hardness                         10 (hardest)   < 1 (very soft)
  Melting point (C)               3 827          3 652 (sublime)
  Colour                           Colourless     Shiny black
  Electrical conductivity          None           High

  Density : diamond > graphite
      C.N. :         4                  3
346
8.9 Covalent Crystals (SB p.242)

3. Quartz
                                   6B




347
8.9 Covalent Crystals (SB p.242)




                               Quartz

348
      No. of Si atoms = 8(⅛) + 6(½) + 4 = 8
      No. of O atoms = 16




                                 Si : O = 1 : 2




349
      The END




350
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)




             Example 8-7A                      Example 8-7B




                                Check Point 8-7



351
8.1 Formation of Covalent Bonds (SB p.218)




(a) How many lone pair and bond pair electrons are present
    in NH3 and H2O molecules respectively?
                                                           Answer
 (a) Ammonia has one lone pair and three bond pairs of electrons.
      Water has two lone pairs and two bond pairs of electrons.




352
8.1 Formation of Covalent Bonds (SB p.218)




(b) Nitrogen can only form one chloride, NCl3, while
    phosphorus can form two chlorides, PCl3 and PCl5.
    Explain briefly.
                                                           Answer
 (b) The electronic configuration of nitrogen is 1s22s22px12py12pz1. Its
     outermost shell electrons are filled in the second quantum shell.
     There are no lowlying d orbitals available for nitrogen to expand octet.
     It has a maximum of three half-filled p orbitals to form three bonds,
     i.e. NCl3.




353
8.1 Formation of Covalent Bonds (SB p.218)




 (b) The ground state electronic configuration of phosphorus is
     1s22s22p63s23px13py13pz1. It has three half-filled p obitals for bond
     formation. Thus, three chlorine atoms can form bonds with it to give
     PCl3.
      After promoting one 3s electron to the low-lying d orbitals, the excited
      state electronic configuration of phosphorus becomes
      1s22s22p63s13px13py13pz13d1. It now has five half-filled orbitals
      available for bond formation. Therefore, five chloride atoms can form
      bonds with it to give PCl5.




354
8.1 Formation of Covalent Bonds (SB p.218)




 (b) Phosphorus has low-lying d orbitals which allow it to expand octet
     (contain more than eight outermost shell electrons) whereas nitrogen
     has not.




                                                 Back




355
8.2 Dative Covalent Bonds (SB p.220)




(a) Draw a “dot and cross” diagram for the product formed in
    the reaction between an ammonia molecule and a
    hydrogen chloride molecule.
                                               Answer
 (a)




356
8.2 Dative Covalent Bonds (SB p.220)




(b) There is a dative covalent bond present in a HNO3
    molecule. Draw a “dot and cross” diagram of the
    molecule.
                                               Answer
 (b)




357
8.2 Dative Covalent Bonds (SB p.220)




(c) State the major difference between an ordinary and a
    dative covalent bond.
                                                        Answer
(c) A dative covalent bond is covalent bond in which the shared pair of
    electrons is supplied by only one of the bonded atoms, whereas
    electrons in an ordinary covalent bond come from both bonded atoms.




                                               Back




358
8.3 Bond Enthalpies (SB p.222)




       Why do two atoms bond together? How does covalent
        bond strength compare with ionic bond strength?
                                                                 Answer
      They are of similar strength. For example, the lattice enthalpy of NaCl
      is 771 kJ mol–1 while the H–H bond enthaly is 436 kJ mol–1. It is a
      misconception that ionic bond must be stronger (or weaker) than
      covalent bond.


                                                     Back

359
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)



(a) Referring to Table 8-2 on page 222, calculate the enthalpy
    change for the following reactions and state whether the
    reactions are endothermic or exothermic.
    (i) Reaction between nitrogen and hydrogen.
        N2(g) + 3H2(g) → 2NH3( g)
                                                        Answer
(a) (i)


          Sum of average bond enthalpies of reactants
          = E(N  N) + 3 E(H – H)
          = [+944 + 3  (+436)] kJ mol–1
          = +2 252 kJ mol–1
360
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)




(a) (i) Sum of average bond enthalpies of products
       = 6 E(N – H)
       = 6  (+388) kJ mol–1
       = +2 238 kJ mol–1
       H = [+2 252 – (+2 328)] kJ mol–1
          = –76 kJ mol–1
        The reaction is exothermic.




361
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)



(a) (ii) Reaction between hydrogen and chlorine.
         H2(g) + Cl2(g) → 2HCl(g)
                                                       Answer
(a) (ii) H – H + Cl – Cl → 2H – Cl
        Sum of average bond enthalpies of reactants
        = E(H – H) + E(Cl – Cl)
        = (+436 + 242) kJ mol–1
        = +678 kJ mol–1


                                                Back

362
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)




(a) (ii) Sum of average bond enthalpies of products
       = 2 E(N – Cl)
       = 2  (+431) kJ mol–1
       = +862 kJ mol–1
       H = [+678 – (+862)] kJ mol–1
          = –184 kJ mol–1
        The reaction is exothermic.




363
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)



(a) (iii) Complete combustion of hydrogen.
                                                          Answer
(a) (iii)


            Sum of average bond enthalpies of reactants
            = 2 E(H – H) + E(O = O)
            = [2  (+436) + 496] kJ mol–1
            = +1 368 kJ mol–1




364
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)




(a) (iii) Sum of average bond enthalpies of products
       = 4 E(O – H)
       = 4  (+463) kJ mol–1
       = +1 852 kJ mol–1
       H = [+1 368 – (+1 852)] kJ mol–1
           = –484 kJ mol–1
        The reaction is exothermic.




365
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)



(a) (iv) Complete combustion of ethanol.

(a) (iv)
                                                             Answer




           Sum of average bond enthalpies of reactants
           = E(C – C) + E(C – O) + E(O – H) + 5 E(C – H) + 3 E(O = O)
           = [+348 + 360 + 463 + 5  (+412) + 3  (+496)] kJ mol–1
           = +4 719 kJ mol–1


366
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)




(a) (iv) Sum of average bond enthalpies of products
       = 4 E(C = O) + 6 E(O – H)
       = [4  (+743) + 6  (+463)] kJ mol–1
       = +5 750 kJ mol–1
       H = [+4 719 – (+5 750)] kJ mol–1
          = –1031 kJ mol–1
        The reaction is exothermic.




367
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)



(a) (v) Complete combustion of octane.
                                                          Answer
(a) (v)




          Sum of average bond enthalpies of reactants
          = 14 E(C – C) + 36 E(C – H) + 25 E(O = O)
          = [14  (+348) + 36  (+412) + 25  (+496)] kJ mol–1
          = +32 104 kJ mol–1
368
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)




(a) (iv) Sum of average bond enthalpies of products
       = 32 E(C = O) + 36 E(O – H)
       = [32  (+743) + 36  (+463)] kJ mol–1
       = +40 444 kJ mol–1
       H = [+32 104 – (+40 444)] kJ mol–1
          = –8 340 kJ mol–1
        The reaction is exothermic.




369
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)



(b) Calculate the enthalpy change for the reaction
      CH4(g) + H2O(g) → CO(g) + 3H2(g)
      using the following bond enthalpies.
      E(C – H in CH4) = +435 kJ mol–1
      E(C  O in CO) = +1 078 kJ mol–1
      E(H – H in H2) = +436 kJ mol–1
      E(H – O in H2O) = +464 kJ mol–1
                                                      Answer

370
8.5 Use of Average Bond Enthalpies to Estimate the Enthalpy Changes
of Reactions (SB p.227)

                                                Back
(b) CH4(g) + H2O → CO(g) + 3H2(g)
      Sum of average bond enthalpies of reactants
      = 4 E(C – H) + 2 E(O – H)
      = [4  (+435) + 2  (+464)] kJ mol–1
      = +2 668 kJ mol–1
      Sum of average bond enthalpies of products
      = E(C  O) + 3 E(H – H)
      = [+1 078 + 3  (+436)] kJ mol–1
      = +2 386 kJ mol–1
      H = [+2 668 – (+2 386)] kJ mol–1 = +282 kJ mol–1

371
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237)



(a) Explain why a molecule of CCl4 is tetrahedral, but a
    molecule of NCl3 is trigonal pyramidal in shape.
                                                              Answer
(a) In a CCl4 molecule, there are four bond pairs of electrons on the central
    carbon atom. The bond pairs have to stay as far away as possible. They
    take up the shape of a tetrahedron and thus the molecule is tetrahedral
    in shape. The four electron pairs in a NCl3 molecule take up the shape
    of a tetrahedron as well. However, one of the electron pairs is a lone
    pair and the other three are bond pairs. The shape of a NCl3 molecule is
    thus trigonal pyramidal.




372
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237)



(b) Deduce the shape of a molecule of BCl3.
                                                          Answer
(b) A BCl3 molecule has six outermost shell electrons around the central
    boron atom, forming three bond pairs. The shape of the BCl3 molecule
    is thus trigonal planar.




373
7.5 Ionic Radii (SB p.208)



(c) Draw the structures of molecules of XeF2, XeF4 and XeF6
    where Xe is a noble gas element with eight electrons in
    its outermost shell.
                                               Answer
(c)




                                        Back

374
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)



The following data refer to the molecules NH3, H2O and HF.

      Molecule       Bond length (nm)              Bond angle
        NH3                  0.101                     107 
        H2O                  0.096                   104.5 

        HF                   0.092                       –




375
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)



(a) Briefly explain the variation in bond length.
                                                              Answer
(a) The bond lengths of the three molecules decrease as follows:
            H–N              H–O              H–F
          0.101 nm         0.096 nm         0.092 nm


      The atomic radius of H is the same in the three molecules, so the bond
      lengths of the molecules depend on the size of the N, O and F atoms.
      N, O and F are in the same period in the Periodic Table. Since atomic
      sizes decrease across a period owing to the increase in effective
      nuclear charge, the bond lengths of the three molecules decrease
      accordingly.



376
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)



(b) Explain why the bond angle of H2O is less than that of
    NH3.
                                                                Answer
(b) This can be explained by the valence shell electron pair repulsion theory.
      The central oxygen atom in H2O has two lone pairs and two bond pairs
      of electrons while the central nitrogen atom in NH3 has one lone pair
      and three bond pairs of electrons.
      The electrostatic repulsion between electron pairs decreases in this
      order:
      lone pair and lone pair > lone pair and bond pair > bond pair and bond
      pair
      Thus, the bond angle of H2O is less than that of NH3.


377
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.237 – 238)



(c) Match the following bond enthalpies to the bonds in the
    above three molecules:
       +562 kJ mol–l, +388 kJ mol–l, +463 kJ mol–l
                                                             Answer
(c) The bond enthalpies are:
            H–N             H–O             H–F
        +388 kJ mol–l   +463 kJ mol–l   +562 kJ mol–l


      The bond enthalpies increase as shown owing to the decrease in bond
      length and increase in polarity of bonds.

                                                  Back

378
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.238)




What are the shapes of a H2S molecule and a H3O+ ion?
Explain their shapes in terms of the valence shell electron
pair repulsion theory.
                                                             Answer
H2S molecule is V-shaped. In H2S molecule, there are two bond pairs and
two lone pairs of electrons in the outermost shell of the central sulphur atom.
All three types of electrostatic repulsion (lone pair – lone pair, lone pair –
bond pair, bond pair – bond pair) are present. The two lone pairs will stay
the furthest apart and the separation between the lone pair and a bond will
be greater that that between the two bond pairs. Therefore, the H – S – H
bond angle in the H2S molecule is about 104.5 instead of 109.5 in
tetrahedron.


379
8.6 Shapes of Covalent Molecules and Polyatomic Ions (SB p.238)




H3O+ ion has a trigonal pyramidal shape. In H3O+ ion, the central oxygen
atom forms two covalent bonds with two hydrogen atoms respectively. Also,
one dative covalent bond is formed between the oxygen atom and the
remaining hydrogen ion. We can regard the central oxygen atom has three
bond pairs and one lone pair of electrons. According to the valence shell
electron pair repulsion theory, the lone pair will stay further away from the
three bond pairs. The three bond pairs are in turn compressed closer
together. Thus, the H – O – H bond angles in the H3O+ ion are about 107
instead of 109.5 in tetrahedron.




                                                   Back

380
8.8 Multiple Bonds (SB p.240)




(a) Does sulphur obey the octet rule in forming a SO2
    molecule? Explain your answer.
                                                              Answer
(a) In the formation of SO2 molecule, sulphur does not obey the octet rule
    because sulphur has 10 electrons in its outermost shell.




381
8.8 Multiple Bonds (SB p.240)

                                                 Back


(b) Draw a “dot and cross” diagram of the hydrogen cyanide
    molecule (HCN). Describe and explain the shape of the
    molecule.
                                                     Answer
(b)




      HCN molecules has a linear shape as the central carbon atom does not
      have any lone pair electrons. In order to minimize electrostatic
      repulsion, the two electron clouds of the central carbon atom are
      separated at a maximum with bond angles of 180.


382

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:60
posted:8/19/2012
language:English
pages:382