Once again, welcome to MSB lecture series
on main group chemistry. In my previous lecture,
I was discussing about molecular orbital theory
and its utility in explaining bonding in simple
diatomic molecules such as H2 to second row
p block elements. I told you about how one
can use MO diagrams to explain bond strength,
and bond distance and all those thingsby just
looking into the number of electrons present
in bonding orbitals and number of electrons
present in anti bonding orbitals.
So, that will give you some magnitude of bond
order. Based on bond order, bond strength,
and bond enthalpy can be accessed. So, now
before I proceed further for hetero diatomic
molecules or poly atomic molecules let me
write MO diagram for F2 molecule.
Always remember to write arrow to show energy,
and here 2p and another 2p from another fluorine
atom, then we have 2s, 2s and then we have
1s we have 1s. So, here we will make bonding
and anti bonding σ, and σ*, we have two
electron here two electron here. So, this
shows the core electrons do not contribute
to the bonding. So, because here the bond
order is 0 that is the reason most of the
cases we ignore the writing MO diagram for
core electrons.
Now, σ and σ* again we have two electrons
each. So, now, what we have is σ, π, π
another σ. So, that is σ* here, it is π*,
and here it is π and it is σ . So now, we
have here five electrons. One , two like this
and now another four electrons will come here.
So, now, let us look into the bond order this
is F atomic orbital, and here atomic orbital
of F and MO for F2.
Let us look into the bond order: equals here,
these are cancelled, no point in counting
here. So, we shall count here ½ [6-4] = 1So,
between F and F there is a single bond. So,
we should be able to write MO diagram and
also depict the bond order here. So, we have
a single bond between two fluorine atoms in
F2.
So, when we write for a hetero diatomic molecule
that results in the formation of a polar covalent
compound, in these kind of polar covalent
compounds bonding molecular orbitals are closer
in energy to the atomic orbitals of the more
electronegative atom. And also the anti-bonding
molecular orbitals will be closer to the energy
of atomic orbitals of least electronegative
atom for example, I am taking two atoms of
the same element then if you take this one,
whether you consider bonding or anti bonding
molecular orbitals, they are at equidistant
from both the energy of both atomic orbitals
in the free state or isolated state.
But if I take something like this one with
low electronegativity higher in energy, and
other one with more electronegativity lower
in energy. So, when they are combined; obviously,
when we generate molecular orbitals the bonding
molecular orbitals will be closer to the one
which is more electronegative and; obviously,
anti bonding molecular orbitals will be closer
to the atomic orbitals of least electronegative
atom something like this.
For example, if I consider one here less electronegative,
another one is more electro negative atomic
orbital, and here atomic orbital and if I
constitute molecular orbital something like
this. So, this is the bonding molecular orbital
and this is the anti-bonding molecular orbital.
So, here you can see, this one would be much
closer to the energy of isolated atomic orbitals
having least electronnegativity. So, one should
remember always when we write a diatomic molecule
or for a covalent compound, we should consider
the electro negativity and accordingly we
should write their energy. So, that is the
reason I am telling you we should be extremely
careful in always mentioning this arrow, that
indicates the energy of the orbitals. Now
let me consider one such molecule that is
NO. We have N and O, N has little less electronegativity
compared to oxygen. Let me write the MO diagram.
.
So, here I am considering 2p atomic orbital
of nitrogen, and little lower in energy 2p
atomic orbital of oxygen. And then I am considering
here 2s, 2s. So, this will give, so, this
σ and σ* of 2s is completed. So, now, first
what we have is here π2p and then we have
σ2p, and then we have π2p* and σ2p*. So,
now we have here three electrons (2s)2 to
(2p)3 here we have (2s)2(2p)4.
So, like this now, we have to arrange this
seven electrons. So, this two are filled this
two are filled. So, we have one electron that
comes here it comes to the π* anti bonding
orbitals. So, let us look into this is MO
for NO. Let us look into the bond order bond
order equals we have six electrons here and
one electron here, so this is 2.5. So, bond
order of NO is 2.5 that is the reason NO has
a tendency to loss the loan electron present
in anti-bonding orbitals to get extra stability
once if you remove that one it becomes NO+
when it is NO+ we do not have any electron
in this is for NO.
Similarly, if you remove this electron, so,
it will be NO+. NO+ will be 6/2 it will be
3. So, NO bond is getting stabilized. That
is the reason NO ligand has a tendency to
act as a cationic ligand by eliminating the
electron present in the anti-bonding orbital;
that means, the the reactivity and properties
of molecules can be predicted simply by writing
MO diagrams and looking into the electronic
arrangement in the bonding and anti-bonding
orbitals. Ok.
So, this diagrams shows the relative energy
of various atomic orbitals, and when they
form diatomic molecules, especially I have
given here for second row elements, how that
is changing gradually. We can see, as we move
from left to the right of the periodic table
the, electronegativity steadily increasing,
as a result what happens, electrons are coming
more and more close to the nucleus, because
of the increased effective nuclear charge,
the orbitals are pulled more closer towards
the nucleus.
In this context the energy steadily drops,
you can see here, starting from Li to Ne energy
steadily dropping. At N2, it crosses here,
you can see here this σ will be lower in
energy compared to that is pz will be lower
in energy that, then this will be in higher
energy, this is what exactly happens that
is the reason I wrote two different diagrams
for molecules up to nitrogen, and nitrogen
onwards oxygen, fluorine, and neon and the
trend changes. σ2p will be lower in energy
compared to π2p. So, this, one should remember
andalso this diagram is there in most of the
textbooks, you can just have a look at it.
ok.
So, in this one just I have shown here the
electronegativity increases. As the orbitals
come more and more towards the nucleus there
is a possibility of mixing of atomic orbitals
even before the molecules being formed. So,
here without mixing MO patterns occurs as
expected here. I have shown MO diagrams with
mixing and without mixing. The first one is
here, without mixing is similar to what I
wrote for several molecules in my last lecture,
and also here due tothe s-p mixing the orbitals
shift occurs as shown here.
So, σ orbital will be higher in energy than
the πp orbitals, as a result you can come
across some sort of mixing and more refined
molecular orbitals. In fact, it shows mixing
of s as well as pz while the formation of
σ orbital. Similarly px, py do not mix with
s orbital. So, I will show you some more MO
diagrams where we can visualize how this mixing
occurs prior to the formation of molecular
orbitals. With this, so, let me write MO diagram
for water molecule. So, in water we have two
hydrogens having one s electron each, will
be combining with oxygen to form 2 O—H bonds.
From VSEPR theory as well as the valence bond
theory, we know the fact that oxygen in water
is tetrahedral with two lone pairs and two
bonded pairs, and the shape of water molecule
is bent or V shaped. Ok.
So, here I am considering 2p orbitals of oxygen,
and then here I am considering 2s orbitals
here, ligandlegend group orbitals: we are
considering here one and another one here,
and of course here we call this H2 bonding
and H2 anti-bonding. So, this is with a1 symmetry
this is with b2 symmetry and this are all
called Mulliken symbol at present you may
not be familiar with what is a1 what is b2,
how this Mulliken symbols are given for that
one should have the knowledge of group theory
perhaps, when you study group theory. And
when you go for the understanding of the spectroscopic
features of coordination compounds, I am sure
you will understand Mulliken symbols.
Right now I am not go into dig deep into the
Mulliken symbols; simply we should remember
here, we have Mulliken symbols for this one
this is for bonding H2 we have a1 and anti-bonding
we have b2. And now we are considering 2px
2py 2pz. So, for this also 2px is called as
b1 symmetry and 2py has b2 symmetry, and 2pz
has a1 symmetry. And of course here, s always
has a1 symmetry. And now we are considering
here, we have two electrons coming from two
hydrogen atoms, and now we have here six electrons
are coming from oxygen. Ok.
So, let me write them here. So this is a1
and this is b2 and this is a1 this is not
there, ignore this one, this is b1 and this
is b2 and this is a1 so, these two are anti-bonding.
So, here now I start filling, so total we
have two, this is not there. So, what we have
is 6+2 = 8, electron are there. We start filling:
two electrons are here, and two are here,
and two more are here and two more are here.
So, now, let me start connecting this one.
So, we have to connect now: the similar ones
a1 so, a1 and of course here also a1 is there
it should be connected. Now b1 is there. So,
b2 is from here we should connect to this
one and b2 is also here, it should be connected.
And now b1 we do not have any so, it remains
as non-bonding b1 the corresponding we do
not have. It remains as nonbonding. So, now,
b2 is connected here, and a1 is connected
here, and then a1 is connected here, and b2
is connected here. Ok.
So, now let us look into these diagrams here
this represents this, two electron here, and
these two electrons represent the presence
of two O—H bonds. And then these two electrons
and these two electrons represent the two
lone pairs present on oxygen atom in water
molecule. For example, if I write something
like this, I can simple correlate this one
to this one of this bond, and this one to
one of the other bond.
And then we have two lone pairs, so one of
this one will be this lone pair, and another
one represents this lone pair. So, this is
how you can clearly write the MO diagram for
even a triatomic molecule like water. Of course,
one can also write for poly atomic molecules,
that I will be considering in a few minutes
time. Hope you have understood how to write
the MO diagram for water molecule. We discussed
NO+ has a stronger NO bond, because the bond
order is 3, whereas in simple NO where due
to the presence of one electron in the anti-bonding
orbital it has a bond order of 2.5.
Now, let us consider another important molecules
such as CO. MO 
diagram for CO molecule. So, here we have
carbon and oxygen; oxygen being a more electronegative,
its energy both 2s and 2p energy will be relatively
lower compared to the energy of 2s and 2p.
So, we have to keep that in mind always. So,
here let me write 2s for C atomic orbitals
of carbon. And then I write here for 2p and
then relatively lower level I write for oxygen.
2s for oxygen atomic orbitals of oxygen here,
and here we have four electrons. In this case
we have six electrons. Ok.
So, first I am showing you the mixing. Now
both 2s and 2p they mix, and just remember
with which orbital 2s is mixing and what is
the nature of that bonding. whether it is
a σ bonding or π bonding. So, here let me
write relative energies: one here, one here
so here it 
is 1σ here 2σ and π2p and 3σ, π2p*and
3σ* . So, this is another way of representing.
So, now 
these two have to be connected here and then
this also to be connected, this one and this
also shows connection with both, and then
this also shows connection with this one whereas,
this one there is no connection from 2p, because
2s will not participate in π bonding, but
2s can correspond to this one. So, like this
let me start writing the electrons here, and
then here to four and then here we have a
total six electrons.
Now, the bond order: this is MO for CO. I
will show you again. So, 2p we have two electrons,
and 2p we have four electrons it is little
lower in energy, and then we have 2s orbital
and 2s orbital here; both have two electrons.
Now 2s has similarly for all σs interact
with 2s, it is head-to-head overlapping. So,
here all are connected, whereas π does not
connect with 2s. So, it shows only the interaction
of these 2p orbitals of carbon as well as
oxygen to form. Now we have a total of 8 electrons
are there they are filled here 2, 4, 6, 8.
So eight electrons are there; now let us look
into the bond order (6-0)/2 = 3. So, this
represents very clearly three bonds between
carbon and oxygen out of which one is σ and
two are π that can also be seen here. Same
information we got from the valence bond theory
and also when we look into Lewis dot structure.
This is how we wrote initially so here the
total number of ten electrons. So, here these
two are coming here and these two are coming
here in order to give octate for both. So
as a result we had this kind of situation.
So, here these two lone pairs on carbon makes
carbon monoxide, a soft Lewis base, and here
you can see triple bond is there same information
we are also getting from carbon monoxide MO
diagram.
So in my next lecture, so I would discuss
about more interesting molecules such as carbon
dioxide and BH3, BF3 and also hexacoordinated
sulfur compound, SF6, very interesting. And
also we can make an analogy of the geometry
and the structural information we get for
SF6 from MO diagram to what we have predicted
as octahedral geometry from the valence bond
theory using hybridization concept. Let us
consider those two things and whether any
difference is there or whether any abnormality
is there. We shall look into these things
in my next lecture until then have a pleasant
chemistry reading.
Thank you
