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Chemistry 445 lecture 4 MO of diatomic molecules

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									    Chemistry 445.
      Lecture 4.



Molecular Orbital Theory
 of diatomic molecules
                    The non-existent He2 molecule
                          (bond order = 0)




                                                        BO = (2-2)/2 = 0




The MO diagram for the He2 molecule is similar to that for the H2 molecule,
but we see that the energy drop of the pair of electrons in the σ1s orbital is
negated because the other pair in the σ*1s rises in energy by an equal amount.
There is thus no net stabilization, and so the He2 molecule does not exist.
        The He2+ molecule/ion exists, bond order = ½

unpaired
electron so
is paramagnetic



                                                      BO = (2-1)/2 = ½




   The logic of the MO diagram suggests that if we remove an electron
   from the He2 molecule, we would obtain a stable [He2]+ cation, which
   is true in the gas-phase. This illustrates the power of the MO approach,
   since the Lewis dot diagram does not predict this.
              The Li2 molecule. Bond order = 1




                                                     BO = (2-0)/2 = 1

                                                    Bond energy for
                                                    Li2 = 110 kJ.mol-1,
                                                    Compared to 436
                                                    kJ.mol-1 for H2.



Note. In drawing up an MO diagram, only the valence shells are
considered, so for the diatomic molecules from Li2 to F2, the overlaps
of the pairs of 1s orbitals are ingnored. This is valid because these
are filled, and they make no net contribution to the bonding.
            The non-existent Be2 molecule,
                   bond order = 0




                                                BO = (2-2)/2 = 0




Here again, MO theory predicts that Be2 does not exist, which
Lewis dot diagrams do not predict.
                      σ and π bonding
 In the molecules we have considered so far, only σ overlaps have
 been of importance. In the formal definition, a σ–bond is one which
 lies along a rotational symmetry axis, (the rest of the molecule is
 ignored). A π–bond does not lie along a rotational axis (symmetry
 will be discussed later). In practical terms, a σ–bond lies along the
 bond connecting the two atoms, whereas a π–bond does not.
               z                   z

                    +
                                                                π*(pz)
  pz
                                                anti-bonding π* MO
orbitals      z                   z


                     +                                          π(pz)

                                                 bonding π MO
               O2 molecule, bond order = 2
                                      molecules with
                                      unpaired electrons
                                      are paramagnetic



The ability
to predict
the number                             BO = (6-2)/2 = 2
of unpaired
electrons in                              (disregardiing
molecules is                              overlap of
where MO                                  2s orbitals)
excels, and
Lewis-dot            O      O2    O
fails.
F2 molecule, bond order = 1
                        F2 has no unpaired
                        electrons, and so
                        is diamagnetic




                        BO = (6-4)/2 = 1
  Variation of the energies of the 2s and 2p
orbitals in crossing the periodic table from Li
              to F. (H&S Fig. 1.22)
      Energy levels of first-row homonuclear diatomic
                 molecules (H&S Fig 1.23)

                                 crossover point




Molecules Li2, Be2, B2,C2 and N2          Molecules O2, and F2 have π(2p)
have π(2p) lower in energy than σ(2p)     higher in energy than σ(2p)
 Be2 molecule, bond order = 0
(BO = 0, means does not exist)




                          BO = (2-2)/2 = 0
B2 molecule, bond order = 1




                         BO = (2-0)/2 = 1
N2 molecule, bond order = 3




                       diamagnetic


                           BO = (6-0)/2 = 3


       N       N2      N
C2 molecule, bond order = 2




                       diamagnetic


                          BO = (4-0)/2 = 2
                      Singlet oxygen (1O2)
                                                           BO = (6-2)/2 = 2

Singlet Oxygen
is an excited
state of the
ground state
triplet 3O2
molecule. It is
much more
reactive, and
will readily
attack organic
molecules.
                           O          O2        O
 The O2 molecule in its excited singlet state which is 25 kcal/mol in energy
 above the ground triplet state. Irradiation with IR light causes excitation to
 the singlet state, which can persist for hours because the spin-selection rule
 (see later) inhibits transitions that involve a change of spin state.
    Orbital parity – gerade (g) and ungerade (u)
Symmetry of orbitals and molecules is of great importance, and we
should be able to determine whether orbitals are gerade (g) or ungerade
(u) (from German for even or odd). This is because in the spectra of
inorganic compounds whether absorption of a photon to produce an
electronic transition can occur is determined by whether the two orbitals
involved are g or u. According to the Laporte selection rules, transitions
from gu and ug are allowed, but gg and uu are forbidden. An
orbital is g if it has a center of inversion, and u if it does not. So looking
at atomic orbitals, we see that s and d are g, while p orbitals are u: (see
next page for definition of center of sym.)




       s-orbital                p-orbital                   d-orbital
       gerade (g)              ungerade (u)                gerade (g)
            Orbital parity – gerade (g) and ungerade (u)
       Symmetry of orbitals and molecules is of great importance, and we
       should be able to determine whether orbitals are gerade (g) or
       ungerade (u) (from German for even or odd). This is because in the
       spectra of inorganic compounds whether absorption of a photon to
       produce an electronic transition is determined by whether the two
       orbitals involved are g or u. According to the Laporte selection rules,
       transitions from gu and ug are allowed, but gg and uu are
       forbidden. An orbital is g if it has a center of inversion, and u if it
       does not. So looking at atomic orbitals, we see that s and d are g,
       while p orbitals are u: (see next page for definition of center of sym.)

                                    a         not a           a
center of     a                               center of
inversion                                     inversion
  a=b                 b                     b    a≠b                   b

              s-orbital               p-orbital                d-orbital
              gerade (g)             ungerade (u)             gerade (g)
                  Parity (g or u) of molecular orbitals:
                   a                              a
 not a center                                                    center of
 of inversion                                                    Inversion
      a≠b                 b
                                                           b       a=b
 (sign of wave-
   function is
    opposite)
                   σ*(1s)u                        π*(2p)g
                   a                              a
                                                                  not a center
                                                                  of inversion
                          b                                           a≠b
  center of                                               b
  inversion
    a=b             σ(1s)g                            π(2p)u
The test for whether an MO is g or u is to find the possible center of inversion
of the MO. If two lines drawn out at 180o to each other from the center, and of
equal distances, strike identical points (a and b), then the orbital is g.
Energy levels of the N2 molecule




                             see if
                             you can
                             decide
                             which are g
                             or are u, and
                             bonding or
                             anti-bonding
      Energy levels of the N2 molecule
(calculated using semi-empirical MO theory)


                                   σ*2pu

                                   π*2pg

                                    σ2pg

                                   σ*2su

                                   π*2pu


                                    σ2sg
        Labeling molecular orbitals
                as g or u:
The following little table will help you to label molecular
orbitals as g or u. For σ-overlap, the bonding orbitals are
g, while the antibonding orbitals are u, while for π–
overlap the opposite is true:

                         bonding MO      anti-bonding MO


    σ-bonding                g                 u


    π-bonding                u                g
               To summarize:
• A bonding molecular orbital has overlap of the two
  atomic orbitals, and has no nodal plane. An anti-
  bonding orbital has a nodal plane between the two
  atoms forming the bond.
• g orbitals have even parity, and have a center of
  inversion. u orbitals have odd parity and have no
  center of inversion
• In drawing up an MO diagram, you should fully label
  all atomic orbitals and MO’s (indicate atomic orbital
  MO is derived from ( 1s, 2p, etc.), g or u, σ or π,
  bonding or non-bonding (*) ), indicate number of
  unpaired electrons, diamagnetic or paramagnetic,
  and bond order.

								
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