Today we will try to focus on mainly there
are 18 electron count rules, and then some
of those oxidative addition, reductive elimination
if possible we will try to discuss in detail.
Before getting into the organometallic chemistry
codes I expect all of you to have some background
specifically you are expected to read about
structure and bonding, also you should have
some sort of understanding of counting electrons.
I will briefly go through the counting electrons,
well first of all organometallic chemistry
is a domain it is a huge domain right at this
point anything that metal is having electronegative
carbon attached with it, then usually it is
called organometallic complex.
The essential part of the organometallic chemistry
is one must have a metal and then we should
have a metal carbon bond with it.
Now in order to count the electrons of an
organometallic complex we usually follows
two method.
One is ionic and another is the covalent method.
In the ionic method we considered ligands
as donor of electrons, and in covalent method
we consider ligand as neutral species right.
Now for example, if we have a metal ligand
complex M-L.
M-L for covalent method we will be considering
metal and ligand in this configuration.
For example, if we have M-X for ionic method
we will be considering M+ and X-.
Now let us take one example and try to discuss
the electron count of it.
For the simplicity purpose I will be discussing
mainly the ionic method and that electron
count of the ionic method.
So covalent method those of you who are familiar
with and are comfortable with that there is
nothing wrong with it you can follow the covalent
method as well.
Mainly I will try to have the ionic method
as the discussion board for the electron count
because that makes things little bit easier.
For example, we have chromium complex, chromium
hexa carbonyl complex, so it is going to be
an octahedral complex as all of you know,
it is an octahedral complex.
Now if you see in the periodic table scandium,
titanium, vanadium, chromium it would be 6
electron for chromium and six of the cobalt's
are there sorry six of the carbon monoxides
are there.
So each of the carbon monoxide is giving two
electron so total electron is becoming six
plus 12 18.
So this is how usually we try to count the
electron.
For example, as we shown over here chromium
hex carbonyl complex, chromium having six
electrons and six of the carbon monoxide giving
two electrons each therefore we are having
six plus 12 total 18 electrons.
Let us try to take another example Platinum
tetrachloride 2- so the oxidation state of
platinum is platinum 2+ and therefore you
can count that electrons of it.
So platinum will be, you know OSIRPT osmium,
iridium, platinum showed overall 10 electrons.
And we will have then therefore on the platinum,
platinum 2+ will be having, you know eight
electrons right and four of the chlorine each
of them are having two electrons total of
eight electrons again.
So overall we will have 16 electrons is that
okay, this is what ionic method is where we
try to find out the oxidation state of each
and every metal involved into the complex.
In this case platinum is in + 2 oxidation
state each of the chloride will provide two
electrons to the complex and thereby we have
platinum 2+ total providing 8 electron into
the complex four chloride each giving two
electrons into the complex.
So therefore we have another 4 times 2 18
electron overall it is a 16 electrons complex.
Similarly something you can work out on your
own, cyclopentadienyl, ferrocenyl, dicarbonyl
chloride, if you count that electron you will
be able to find that it is a 18 electron complex
okay, that is something you can try to find
out on your own right.
So I think I would expect at this point that
you try to take some complex try to count
the electrons most often you will get you
know 16 or 18 electron complex it is possible
to have electron count other than 16 or 18
you just need to count it carefully now let
us look at a special case where we do see
metal bonding whatever the method you are
following either ionic method or coherent
method you're expected to add one electron
to the electron count and then come up with
the total electrons if you are having a metal
complex.
What is an example of an metal complex let
me show you one of them.
So this is a diagram complex bridged with
Cardinal terminal ligands are CP and carbon
you say it is a complex where we see that
iron is in plus 1 oxidation state right so
that electron counts for iron in plus 1 would
be totals giving you 7 each 7 electrons right
iron electron count for iron atom is B6s2
so total eight electrons RL is in plus 1 oxidation
state and therefore we have seven electrons
for iron 1 plus now if you look at carbon
monoxide it will be giving CO will be giving
two electrons as I tried to discuss iron plus
seven left one cyclopentadienyl this is in
ionic mode so CP minus that means 6 electron
bridging carbon monoxide will give you 2 electron
and then metal, metal bridge as we say should
give you one electron you should count 1electron
for metal-metal bond.
Now overall then if you count 7+ 29+6 15 +
2 + 1 so overall you have 18 electrons okay
so once again if you had a metal, metal bond
in an organometallic complex you are expected
to add one electron by either of the ionic
method or covalent method let us take another
example it is a rhodium complex.
I would like to discuss rhodium chloride bridge
complex it is a dinuclear complex terminal
ligands are ethylene now if you want you can
just you know split the molecule into half
what you see is a rhodium 1+ that means it
is going to have it is a d7s2 electronic configuration
total 9 electrons rhodium 1+ should be giving
8 electrons right everybody got it correct
8 electrons 2 of the ethylene each of them
are giving 2 electrons so total of 4 electrons
and chloride is giving you 2 electrons one
any one of them if you if you pickup.
So it is a 16 electron so far now as you know
this bridging chloride both of them in addition
of them being in negatively charged they will
also give you lone pair to the other rhodium
so another two electrons you have to consider
for overall electron so bridging electron
chloro bridge overall then you will have 10
14 + 2 16 electrons 16 electrons you will
have so this complex dirhodium dichloro bridge
and diethylene complex where you see that
the chloride is the bridging one each chloride
are attached with each of those rhodium therefore
total electrons coming from two chloride from
for any particular rhodium is going to be
2 + s2 total for rhodium is in +1oxidation
state that means it is having 8 electrons.
So 8 + 4 from the chloride and 4 from the
ethylene overall it is a 16 electron are you
all clear with this all right let us move
on for the next example now we can have yet
another example where we see that.
It is a once again it is a rhodium complex
dirhodium complex we have hydrogen or hydride
and chloride bridged and terminal ligands
being chloride and cyclopentadiene now if
you try to count that electron for rhodium
rhodium here we see that it is rhodium 3+
right now rhodium 3+ will rhodium is d7s2
so total 9 electrons rhodium 3 + will be 6
electrons CP star will again have 6 electrons
terminal chloride that will give you 2 electrons
now the bridging chloride for each of the
rhodium it will give two electrons and the
hydride H it will give you 2 electron.
Now this is a special case as you can see
the hydride will have overall 2 electron and
this is something called 2 electron 3 centered
bridging for each of the rhodium then you
have to count two electron for the hydride
region, so overall for each rhodium we have
two electrons from hydride two electrons from
chloride so two plus two for another two from
this chloride that is 6, 6 from the CP star
that is 12 rhodium is three plus that means
another 6 electron overall then you have a
18 electron complex.
So I hope it is becoming now clear that the
electron count of each of the metal complexes
can be done very carefully you can have ionic
method or covalent methods for counting that
electrons preferably fort his course I will
try to stick to the ionic method if any if
you are more comfortable with the covalent
method there is no problem with it you can
follow it as you know in the ionic method
will have a Cation and Anion formation.
And in the covalent bond will have a metal
and ligand as the ligand two electron donors
formation ligand will be deal as a neutral
stasis, once again most of the complex you
will find will have 18 electron some of them
will have 16 electron some rare cases you
might will have also 14 electron but whatever
it is you should remember few basic thing
that it is possible to have a two electron
three Center bond that is the last case we
have discussed.
Where hydride is bridge between the two rhodium
okay, so two electron are shared between the
three centers we if there is a metal-metal
bond then you have to add one electron more
to the total account and otherwise I guess
most of the example we have covered so far
I expect you to practice on these you know
electron count and you should be basically
able to count the electron in your almost
in your dream and none of the time you should
have a wrong calculation because this is going
to be very, very effective in terms of discussing
the fundamentals of some of the organ metallic
chemistry that we will be covering in this
course, okay.
Now as we are trying to discuss let us try
to see 16 electron complexes and 18 electron
complexes and the mode of reaction they undergo,
so we will discuss16 electron complexes and
18 electron complexes separately and try to
see how they are reacting is the reactivity
going to be similar for 16 electron and 18
electron complexes or are they in to differ
that is the major question we would like to
address.
16 electron complex I think a moment ago we
have discussed some of them 16 electron complexes,
now let us take a simplified example ML4 and
assume that this is 16 electron ligand is
each of the ligand are having 2 electron let
and you know let us give an example this nickel
complex where we have nickel phosphine, aryl
fluoride and triethyl phosphine two of them
now if you try to count the electron for these
complexes.
So what we will find one two three four, four
of the ligand that means eight electrons four
ligand to each nickel is in plus 2 oxidation
state that means it is a eight lecture on
count again for it so nickel is d8 s2 total
10 electrons nickel is in plus 2 oxidation
state eight electrons total them and ate from
the four other ligand total it is a 16 electron
complex, usually if it is reacting with another
ligand these 16electron complex reacting with
another ligand.
And the rate constant is K this is a slow
process what we see usually we do see that
a 18 electron species formation if L prime
is reacting with ML 4 we get ML 4 L prime
complex 16 electron complex reacting with
two electron donor ligand it is becoming 18
electron complex the next step will be faster
it is a first reaction and it will undergo
ligand exchange overall will then have ML3
L’ and L will come out of the system.
So a 16 electron complex will undergo associative
mechanism this is the one where ligand is
getting associated with the 16 electron complex
becoming or giving rise to an 18 electron
complex this 18 electron complex then will
undergo first dissociation you have a ligand
exchange and will give rise to ML3 L’along
with formation of the ligand L. Now let me
give you a practical example that the same
one will use the nickel complex,
Where we will see that nickel is tetra coordinated,
of course it is a square planar complex the
same one which we were discussing a moment
ago, it is a 16electron complex by now hopefully
you will be able to count the electrons very
quickly, it reacts with let's a period in,
that is the exogenous ligand, overall now
from Petra coordinate we have a Penta coordinated
nickel complex pyridine is associated with
the nickel, along with the other ligand right.
So 16 electron complex going to 18 electron
complex, and from here on chloride will come
out okay, and we will have a tetra dentate
or tetra coordinated nickel complex back,
where it will be again square planar, along
with formation of the fluoride from the system,
just to clarify once again this is a 16 electron
complex coming, to give you 18 electron complex
while reacting with pyridine, and chloride
is coming out from the system to give you
back a 16electron complex.
So therefore it is to simplify usually 16
electron complexes, undergo associative mechanism
to give you 18 electron intermediate, which
will then rapidly be shows here to give you
back 16 electron complexes.
Now for this reaction the one we have just
discussed where pyridine is exchanging with
the chloride, how can of course you can do
some experiment to prove that this is what
is the trend, now if you are looking at the
aryl group, this aryl group if you substitute
the aryl in one case if you have if you have
toile, okay in another case if you have messy
tile, the expected pattern of reactivity or
the observed pour pattern of reactivity is
aryl if it is toile, such as if it is aryl
messy tile, the rate constant varies six thousand
times.
So the air I look one when it is toile line
that means less technically demanding, that
is the one which will react very fast compared
to the one with messy tile, which spherically
demanding or bulky, and thereby will be preventing
the pyridine coordination this is the one
which is the slow step, period in coordination
will be hindered if this aryl, messy tile
one that means ethically demanding one the
reaction will be slow.
If the iron is less technically demanding,
then this step will be faster and then their
pie depending on the aryl whether it is a
toile or messy tile, we can have a significant
difference in the rate constant and do it
constant if you compare the toile one in six
thousand times faster compared to the messy
tile one.
So a deuce just see these complexes nickel,
complexes underwent an associative mechanism
we're starting nickel complex one 16 electron,
and then the ligand came in give rise to 18
electron complex which then rapidly dissociate
to give you the 16 electron complex, overall
it is a 16 electron complex undergoing an
associative mechanism, to give you an18 electron
complex which then dissociates to give you
16 electron compact.
So 16 electron complex starting out resultant
is also a 16 electron complex in between you
have an 18electron complex.
If you try to draw them, you know the energy
profile of this reaction it should somewhat
looks like, this is the first step which is
the slow step, and then the next step is the
very first one first one to give you the product
okay with this we'll meet again shown in the
next class, thank you all and I expect you
will be studying on the electron count mainly,
and look at the 16electron complexes as well,
I hope all of you will be able to do the electron
count very easily, so that we can discuss
at the same space have a nice one see you
bye.
