Molecular orbital theory1 by cuiliqing

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									                       MOLECULAR ORBITAL THEORY
This theory is based on quantum mechanics and it explains the properties of covalent
bond in terms of molecular orbitals (MO). These molecular orbitals (MO) are as a
result of interactions of atomic orbitals (AO) of the bonding atoms and are associated
with the entire molecule unlike in VB theory the overlapping atomic orbitals (AO)
are associated with only the individual atoms. This is to say that while in VB theory
we talk in terms hybridized AOs in MO theory we talk in terms of MOs
According to MO theory, when 1 AO overlaps with another 1 AO it creates two MO,
one which is a bonding MO (BMO) and the other is an anti-bonding (AMO).
Similarly 3 AO + 3 AO = 6 MO (3 BMO and 3 AMO). BMOs have lower energy and
therefore greater stability than its initial AOs from which it was formed while AMOs
have higher energy and lower stability that its initial AOs from which it was formed.
In a BMO, the electron density is greater between the nuclei of the bonding atoms, In
AMO the electron density is decreased to zero between the nuclei, so that placing
electrons in the BMO would yield a stable covalent bond while placing electrons in
the AMO would yield an unstable bond.
In terms of wave function, the formation of BMO corresponds to constructive
interferences of the wave function of the wave function of each AO, while AMO is
produced from destructive interferences of the wave function of each AO.




    BMO have energy lower that the individual AOs.
    AMO have energy higher than the individual AOs.
    Non bonding molecular orbitals (NMO) have the same energy as the AOs.
     Therefore an electron occupying an NMO does not affect the stability of the
     molecule.

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Overlapping of two s-orbitals
Sigma MO (whether bonding or anti bonding) have the electron density concentrated
symmetrically around a line (usually taken as the z-axis) between the nuclei of the
bonding atoms, therefore 2 electrons in a sigma MO form a sigma bond. These orbital
have cylindrical symmetry around the internuclear axis (z-axis). The interaction
between any two s-orbitals produces σ MOs




Overlapping of two p-orbitals
Two p-orbital can approach each other in 2 different ways, either end-to-end which
results in σ- MO and σ*- MO. This is usually the pz atomic orbital because lies along
the z-axis hence has a cylindrical symmetry. When two p orbitals overlap side-to-side
they produce π-MO and π*-MO. In the pi MO (bonding or anti bonding) the electron
density is concentrated above and below the line joining the 2 nuclei of the bonding
atoms. Therefore a pz orbital would not overlap with the pz-orbital of another atom but
not with its px- or py-orbitals.




             Change the 2px to 2pz
             for consistency




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Overlapping of an s-orbital to a p-orbital
This type of overlapping is only possible if the p-orbital lies on the z-axis such that
the overlapping with an s-orbital would be end-to-end resulting in σ MO (bonding
and anti bonding) as long as they are close in energy. Therefore sigma bonds are only
possible if the overlapping would produce a bond with cylindrical symmetry along the
z-axis, for example s and pz, but never with a px- or py- orbitals.

Overlapping of d-atomic orbitals
For d-block atoms an in              and                         there is a possibility of
forming sigma bonds with d-orbitals. The           orbitals can form d-d sigma MO since
it has cylindrical symmetry w.r.t. the internuclear axis (z-axis). It can also form sigma
MO with s-orbitals and pz-orbitals. The          - and       - orbitals look like p-orbitals
and therefore form π-MO with themselves or can interact with px- or py-orbitals
forming π-MO.          - and       - orbitals do not have the same symmetry as s or any
of the p orbitals or    ,  -and    orbital and cannot overlap with any but with
themselves forming delta MO δ-MO (bonding and anti bonding).

Rules governing molecular orbital electronic configuration

   1. The number of MO formed is equal to the sum of number of all the atomic
      orbitals involved.
   2. The filling of MO proceeds from the low energy orbitals to the high energy
      orbitals (from bonding to anti bonding).

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   3. It obeys Paulis Exclusion Principle –electrons fill in pair i.e. a maximum of 2
      electrons per MO.
   4. It obeys Hunds rule- electron will fill in MO of the same energy singly before
      pairing.
   5. The number of electrons in the MO is equal to the sum of all the electrons in
      the bonding atoms.
HOMONUCLEAR DIATOMIC MOLECULES OF THE FIRST PERIOD (H2 and
He2)
Homonuclear diatomic molecules are molecules that are made up of two atoms of the
same elements. Since it two atomic orbital overlapping (1s) and overlapping of s-
orbitals produces sigma molecular orbitals, we then have 2 molecular orbitals
to be filled by the electrons of this group.
Dihydrogen molecule (H2)

H – 1s1
H – 1s1
Total number of electrons = 2e
Overlapping of the two 1s orbital atomic orbitals =         2 molecular orbitals
Filling of these orbital =
We can determine the stability of the molecule can be from the bond order using




The bond order is 1 and therefore this molecule is stable and is a single bond.
There are no unpaired electrons so it is diamagnetic.




                    Molecular orbital energy or correlation diagram

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Dihelium molecule He2
He – 1s2
He – 1s2
Total number of electrons = 4e
Overlapping of the two 1s atomic orbital =          2 molecular orbitals
Filling of these orbital =
We can determine the stability of the molecule can be from the bond order using




The bond order is 0 and therefore this molecule is not stable and does not exist. It did
exist, it would be diamagnetic.
                                                    Pls note that this
                                                    should be He
                                                    and not H.




                               Molecular energy diagram of He2

HOMONUCLEAR DIATOMIC MOLECULE OF 2ND PERIOD ELEMENTS
(Li2, Be2, B2, C2, N2, O2, F2, Ne2).
In this group the general atomic are is 1s, 2s, 2px, 2py, 2pz (at most 5 atomic orbitals
from each atom) such that when two of atoms form a diatomic molecule the
interaction of the atomic orbital they produce at most 10 molecular
orbitals                                                             . These pi orbital in the
bracket are degenerate i.e. they are equal in energy                           and
       therefore electrons would fill in singly before pairing.
Diboron molecule B2
B – 1s2, 2s2, 2px1
                                                                                           34
B – 1s2, 2s2, 2px1
Total number of electrons = 10e
Overlapping of the two of 1s, 2s and 2px atomic orbital (total of 6 atomic orbital) = 6
molecular orbitals
Filling of these orbital =
We can determine the stability of the molecule can be from the bond order using




The bond order is 1 and therefore this molecule is stable and does exist. It has 2
unpaired electrons and therefore paramagnetic. In the molecular energy diagram the
         were not included because they are not valence shells but in test or exams
do include them.




Diflourine molecule F2
F – 1s2, 2s2, 2px2, 2py2, 2pz1
F – 1s2, 2s2, 2px2, 2py2, 2pz1
Total number of electrons = 18e
Overlapping of the two of 1s2, 2s2, 2px2, 2py2, 2pz1 (total of 10 atomic orbitals) = 10
molecular orbital
Note that here the       molecular orbital comes before the                 MO. This is
applicable for O2, F2, Ne2
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Filling of these orbital =
We can determine the stability of the molecule can be from the bond order using




The bond order is 1 and therefore this molecule is stable and exists. It is diamagnetic.




Note again that the            were not included.
 The reason why there is a change in the arrangement of the MO i.e. the position of
the      orbital is due to sharp decrease in the energy of the 2s orbital as you move
across the period which is most prominent from O to Ne (see diagram below). It
changes from
                                                           to




Energy level of the 2nd row of homonuclear diatomic molecules

                                                                                        36
            Molecules Li2, Be2, B2,C2 and N2           Molecules O2, F2 and Ne2 have π(2p)
            have π(2p) lower in energy than σ(2p)      higher in energy than σ(2p)
Taking a look at the molecular orbital electronic configuration of F2 the Highest
Occupied Molecular Orbital (HOMO) is the degenerate set                 while the
Lowest Unoccupied Molecular Orbital (LUMO) is         The HOMO and the LUMO
are known as FRONTIER ORBITALS, these orbital are used to explain a lot of
reaction mechanism, kinetic, structural and spectroscopic studies. Another term is
SOMO which is an ancroym.

Gerade and ungerade symmetry
Symmetry of orbitals and molecules is of great importance, 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 gerade (g)
or ungerade (u) (from German for even or odd). 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.
For molecules that possess a center of inversion are called centrosymmetric
molecules. Centrosymmetric molecules include:
     Homonuclear, X2
     Octahedral, EX6
     Square planar, EX4.
Non-centrosymmetric molecules include:
     Heteronuclear, XY
     Tetrahedral, EX4.



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A center of inversion is a symmetry operation that involves starting at an arbitrary
point in the molecule, travelling in a straight line to the centre of the molecule and
hen continuing an equal distance out on the other side of the centre. The test for
whether a MO is g or u is to find the possible center of inversion of the MO. If two
you start at a point a on the molecule and travel through its center and get to the
opposite side (in a straight line of equal distance) and you arrive to a same place a
then such a molecule is g but if you start with a and travelling you end up at b then its
u. a and b are opposite sign of the wave functions (phases)




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
Below is the molecular electronic configuration of F2 ,


To rewrite this based on symmetry would become



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Home Work
  1.     was formed by bombarding         with a fast moving electron. Confirm the
     existence of     and its magnetic properties. Draw the molecular orbital energy
     diagram of     and determine how many bonds it has.
  2. Write the electron configuration of                in terms of symmetry and
     determine their stability and magnetic properties.

HETERONUCLEAR DIATOMIC MOLECULE (HF, LiF, CO, NO etc)
These molecules are made of 2 atoms of different elements and because of these
differences their contribution to the formation of the molecular orbital would not be
the same. This is unlike the case of homonuclear diatomic molecule their contribution
to the molecular orbital is the same. The differences arise as a result of the difference
in electronegativity
In drawing the molecular orbital diagram of heteronuclear diatomic molecules we
follow the procedure below
     Place the more electronegative atom on a lower level, while the less
       electronegative atom is place on a higher level. The larger the electronegativity
       difference the wide this gap between the combining species
atom          F             Li           H            C          N            O
χ             3.98          0.98         2.20         2.55       3.04         3.44
     Ignore inner orbital that are not non valence orbital since they are not likely
       participate in bonding like the 1s
     In drawing the MO diagram the more electronegative atom has the BMO
       closer to it while the less electronegative atom has the AMO closer to it.
     Only orbital of the similar energy and symmetry would overlap to produce
       MO.
     Because there are time when an s atomic orbital would overlap with a pz orbital
       to produce a sigma MO the labeling of the MO of heteronuclear diatomic
       molecules is 1σ, 2σ, 3σ etc or 1π, 2π etc

HF
H
F
Total number of valence electron 1+7 = 8

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Total number of atomic orbitals involved = 5
Therefore total number of molecular orbitals = 5

                        3




                 1s
             E
                  H
                                            1          2pz   2py   2px
                 AO

                                             2


                                             1               2s
                                             MO
                                                               F
                                                              AO

The Is (H) can interact with 2s(F) and 2pz(F) because they all have σ symmetry (can
overlap end-to-end). However due to the large energy difference in between the 2s(F)
and 1s(F) and therefore there would very little interaction between them such that the
σ MO formed is much more F character that it is considered a non-bonding σ MO
with no corresponding AMO. Nevertheless the interaction between the 1s(H) and
2pz(F) is strong because the energy difference between them is small and as such the σ
MO is formed which has both 1s(H) and 2pz(F)character.

The 2pz(F) and 2pz(F) are unaffected, as they have π-symmetry (side-to-side) and the
H-atom has no orbital of that symmetry. Therefore these orbitals not overlapping
with any other orbitals and as such do not have AMO but because of their symmetry
are labeled as π-MO and the electrons occupying them are non bonding. The
Molecular electronic configuration of HF is                      accounting for the 8
valence electrons.
H-H or F-F bonds are purely covalent because they have the same electronegativity
and share the bonding electrons equally. But in the case of H-F the bond pair of
electron are not shared equally rather due to high electronegativity of F the bond pair
have more F character and due to this polarization the bond has partial ionic character
(polar bond) hence is illustrated in the MO energy diagram to be more towards F than
central.
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Bond order = ½ (2-0) =1 the compound has a single polar covalent bond it is stable
and is diamagnetic

LiF
Li
F
Total number of valence electron 1+7 = 8
Total number of atomic orbitals involved = 5
Therefore total number of molecular orbitals = 5
This is similar to that of HF however the only difference is that the electronegitivity
difference between Li and F is greater than that between H and F. As a result the 2σ
MO would be so much more of 2pz(F) in character such that the bonding pair is
completely transferred to F atom as a lone pair making the bond an ionic bond rather
than covalent. In drawing the MO energy diagram the gap between Li and F in the
MO energy diagram would be wider than that of HF and the 1σ and 2σ closer to F
while the 3σ is much more closer to Li. The molecular electronic configuration is

                   3




             2s
             Li
             AO
         E




                                                           2pz    2py   2px
                                               1


                                                    2


                                                             2s
                                                    1
                                                    MO        F
                                                             AO

CO
C
O

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Total number of valence electron 4+6 = 10
Total number of atomic orbitals involved = 8
Therefore total number of molecular orbitals = 8
In this case the O is more electronegative than C however the difference is not much
so the would be polar covalent bond. In drawing the MO energy diagram the gap
between C and O would be smaller and it has 10 electrons like dinitrogen they are
isoelectronic when the 2p orbitals overlap, the π orbital would come before the σ
orbital. Since C and O both have2s and 2p orbitals they would overlap this way
2s(C)- 2s(O), 2px(C)-2px(O), 2py(C)-2py(O) and 2pz(C)-2pz(O). The 2pz orbital would
overlap to produce sigma MO because of their sigma symmetry while the 2px and 2py
would overlap to produce degenerate pi MO.
Molecular electronic configuration is
Bond order = ½(8-2) = 3. This compound has a triple bond it is stable and
diamagnetic
                             



                              
       2pz      2py   2px




                                               3
                                                               2pz    2py   2px
E


                                           1



                                         2

           2s
          C
          AO
                                                                 2s

                                                                 O
                                                                AO
                                          1
                                          MO



                                                                                  42
NO
N
O
Total number of valence electron 5+6 = 11
Total number of atomic orbitals involved = 8
Therefore total number of molecular orbitals = 8
This is similar to CO only that in this case because the total number electrons are
greater than 10 the 3σ would come before the 1π. Bond order = ½(8 - 3) = 2.5. this
molecule has mostly likely a radical it is also paramagnetic having an unpaired
electron. It is stable
                                 



                                  


           2pz     2py   2px

                                           1


                                                              2pz    2py   2px
    E                                      3




                                         2

              2s
             N
             AO
                                                                2s

                                                                 O
                                                                AO
                                          1
                                          MO

Home work
The halogen form compounds among themselves. One of these interhalogen
compounds is iodine monochloride, ICl, predict the molecular electronic
configuration and discuss the properties of this compound. Draw the MO energy
diagram also




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