now in this hour actually about 15
minutes I will talk about equilibrium
and then I will solve some problems at
the end of chapters that I picked up in
the last half hour actually okay last
time remember we have started chemical
equilibrium and what we said that every
reaction actually reaches to in
equilibrium but depending on whether the
equilibrium is close to the products or
close to the reactants if it is too
close to the products that means most of
the 99.999% of the reactants are
converted to the products we simply say
that the reaction is complete right now
if on the other side if it is the
equilibrium is close to the reactant
side that means very very small amount
of the reactants is converted to the
products we simply say that the reaction
is not taken place but we should
remember that every reaction is
reversible and every reaction has simply
an equilibrium and equilibrium constants
now again let's go through very quickly
we have simply taken the decompositions
of dinitrogen tetroxide
to nitrogen dioxide and one is actually
colorless this is colorless this is
brown and when you start with the no.2
or you start with the only no.2 or if
you start only with n2o4 you reach to an
equilibrium equilibrium now the rate of
the reaction actually forward is given
like this for example when you start
with n2o4 dinitrogen tetroxide and rate
decreases as the time goes on this is
the forward rate expression rate low
expression now this is again
the formations of the n2o4 that means
the reverse rate of the reaction now
this is
Creason as the time goes on because the
concentrations of nitrogen dioxide
increases as it decomposes dinitrogen
tetroxide and it reaches to a point
where the forward rate and the reverse
rate are equal and we say this is the
equilibrium point and we say equilibrium
is achieved again this is as I said last
time dynamic equilibrium actually
dynamic equilibrium and dynamic
equilibrium is indicated in reactions if
he particularly emphasized would like to
emphasize on equilibrium instead of one
arrow we use two arrows this is means a
reversible reaction and the equilibrium
again now this is the rate of the
reactions simply like this and when we
simply reach the equilibrium point where
the forward and reverse reactions are
equal rate of the reverse and forward
reactions are equal then we have this
simply ratio of the product
concentration raised to the power 2
divided by the reactant concentration
molar concentrations would be equal to
KF over kr since KF and kr rate constant
and they are constant as long as the
temperature is constant that means that
this ratio of the molar concentrations
do not change if the if the if the if
temperature is not changing actually and
this is again remember it we have to
assume this imagine this in a closed
system in a closed system right not an
open system and this ratio of the rate
constants is called the rate equilibrium
constant of the reaction which is we
indicated as qaq as a matter of fact now
this is for example the reactions of
nitrogen with the hydrogen this is a
harbor yes
reaction fixations of simply nitrogen
with the hydrogen again this is a very
very important reaction industrially why
because it involves the preparation
productions of ammonia and ammonia is
used as a fertilizers that is converted
to other fertilisers and it is also used
to produce again some explosives as a
matter of fact and particularly this
product and the other products obtained
from the ammonia is used as fertilizers
the main fertilizer is synthetic
fertilizers is ammonia and ammonium
compounds of nitrogen actually the main
things here is as an element is nitrogen
because all the plants needs nitrogen
potassium actually and phosphorus in
some of them actually but nitrogen is
the main things that the plants really
need and if you give the nitrogen as its
compounds as a fertilizer and that is
good for the plant again now if I have
an equilibrium reaction like this and
then I can write down the equilibrium
constant expression this is called the
equilibrium constant expression for this
equilibrium reaction all in gaseous
state again now you see again we simply
take the concentrations of the product
molar concentrations raised to the power
two their coefficients in the balanced
chemical equations divided by the
concentrations of the reactants raised
to the power of their coefficients in
the balanced equations so here the
coefficient here is 1 so the power here
is 1 concentration at region 3 3 and to
2 and this is the equilibrium constant
you see again I have K sub C here this C
means that the equilibrium constant is
given in terms of molar concentrations
of the species reactant and product
we use simply the molar concentrations
now in general again if we consider an
equilibrium reaction like this the
equilibrium expression for the reaction
will be equal to K sub C actually
products and reactants what we should
always remember here that the powers
here exponents are the coefficient of
this species in the balanced equations
right over here like this all of them
are the same way now we can also write
down the equilibrium expression in terms
of pressure and when we do that instead
of concentrations of the species we
write down the partial pressures of each
species in equilibrium but remember the
KP will have mean and if we are dealing
with gases reactions at least some of
the reactants and the products are gases
so instead of molar concentration we can
simply give the partial pressures of
each species and again in the
equilibrium expression here which is
called KP equilibrium expression we
raised the pressures of each species to
the exponents which is equal to the
coefficient in the balanced equations
again just the same form but just in one
case we have the molar concentration the
other case partial pressures of each
species there now there is certainly
relationship between KC and KP we can go
from KC to KP or from KP to KC what does
that means that simply means if we know
KC for an equilibrium reaction then we
can calculate KP for that reaction at a
given temperature now here how we do it
again we simply write down this equation
for gases PV is equal to NRT ideal gas
equations ideal gas law
now rearranging that we get simply V on
this side and over V RT now n over V is
actually number of moles of the gas per
unit volume liters again and that means
that the moles per liter
and it is equal to the molar
concentration of that species molar
concentrations of that species so if we
simply replace the molar concentrations
take the molar concentration here and
replace it with P over RT or vice versa
if you simply replace the P by molar
concentrations time RT then we have this
relationship which is simply says that
the KP would be equal to KC times our T
raised to the power Delta n what is
Delta n Delta n is the moles of gaseous
product minus moles of reactant products
again remember we only take into
consideration the gaseous reactant or
the product in the species we just take
the difference over here now if you just
say KC will be equal to KP RT minus
Delta and there and it would be the same
things so this means if you know one
equilibrium constant that music Leland
constant in partial pressures we can
convert it to the equilibrium constant
in molar concentrations but what we have
to know we have to know the temperature
right then this is the gas constant and
this is the temperature simply okay here
again
we give some values which simply says
that no matter where you start with what
concentration you start at the end
actually the ratio which is given an
equilibrium expression will always be
equal to the equilibrium constant and at
the same temperature again remember the
equilibrium constant will only depends
on the temperature on the temperature
when the temperature changes equilibrium
constant also changes so let's look at
here again as you can see the ratio of
no.2 two divided by this this is the
equilibrium expression for the
decomposition of the dinitrogen
tetroxide remains constant at this
temperature no matter what the initial
concentrations of no.2 and n2o4 n2o4 are
so and it doesn't matter where you start
and
what mixture of these you start when the
system reaches to the equilibrium at the
same temperature if the temperature is
the same that means that the ratio would
always be equal to the equilibrium
constant in this case K sub C if you
take those in molar concentrations like
this now magnitude of K this is what
really what the colibri inconstant tell
us right the value of the equilibrium
constant is very important for a
chemical reaction why because if the
equilibrium constant is very very large
right like 10 to the 20 10 to the 30 50
very large what does that means that
means almost all of the reactants are
converted to the products right now on
the other side if the equilibria
constant is very very small 10 to the
minus 50 for example right close to zero
in that case that means that the
reactants almost not converted to the
product at all so that we say that
reaction is not taking place so now what
the equilibrium constant value tells us
actually it tells us to what degree the
reactants are converted to the products
to reach to the equilibrium right to
reach the equilibrium if the equilibrium
constant is large then that means we are
on the product side if the equilibrium
constant is small that means we're in
the react
reactant side what does that means that
means the logical level constant means
at equilibrium we have products almost
all products very small amount of
reactants but almost nothing now on the
other side if the equilibrium constant
is too small that means at the
equilibrium then we have large amount of
reactants and very very small amount of
the products means the reaction in this
case is not taking place here we
dependent on the K values and the action
is complete we call it so here the
expression it says when the K is much
larger than 1 the reaction favors the
product you know what that simply means
right
the products predominate at equilibrium
that means very large amount of products
if the K is small the reaction favors
reactors predominant at equilibrium so
again when we look at a chemical
equilibrium constant value for a
chemical reaction we can actually guess
to what degree the reactants will be
converted to the products reactants will
be converted to the products let's look
over here again the directions of the
chemical equilibrium and K the
equilibrium constant of the reaction in
the reverse reaction is reciprocal of
the equilibrium constant of the forward
reaction that means K C is equal to this
and equal equals to that at 100 degree
centigrade now if I simply write down
this reaction now this is just the
reverse of this reaction right 2 times
no.2 and dinitrogen tetroxide this is
the decomposition decompositions of
dinitrogen tetroxide this is the
formation of the dinitrogen tetroxide
starting from this these two are reverse
reaction if we write down the
equilibrium constant for this it would
be equal to that equilibrium constant
for this we go to that what is that
means that simply means that when we
reverse the reactions in fact the
equilibrium constant will be reversed
actually two like 1 over the value here
actually this value is equals to 1 over
this right 1 over that and that's equal
to this 1 so when we reverse the
equilibrium reactions the equilibrium
constant well values will be reversed to
but it would be like 1 over value now
when we let's see if I have that if I
simply multiply as you see an
equilibrium reaction by a coefficient in
this case we multiply by 2 and write
down the equilibrium expression for this
and this is equal to this again if you
compare these two what does that means
this is actually equal to the square of
Mount what does that means that means
that the equilibrium constant the new
equilibrium constant for this reaction
which is
multiplied by a coefficient two is the
square of this value if I multiply it by
three than the equilibrium expression
for that will be the cube of that right
reactions so in other words if I know
the equilibrium constant for this
reaction at a given temperature so I can
write down the equilibrium constant for
the reverse reaction I can write down
the equilibrium constant for the
reactions which is multiplied by a
constant any constant as a matter of
fact okay now here is again when the two
consecutive equilibrium a cure the
equation can be added to give a single
equilibrium the equilibrium constant of
the new reaction is the product of the
two constants this simply means that if
I have a reaction like this k1 is equal
to this right the equilibrium constant
for this reaction and second equilibrium
reaction k2 is equal to this when I
simply add up these two I get this
reaction over here now the equilibrium
constant of this reaction will be
directly obtained from the equilibrium
constant of the first and second
reaction that I added up side by side
and that simply means k3 equilibrium
constant for this reaction would be
equal to k1 times k2 so if I know the
values of k1 and k2 I can calculate the
value of k3 again now homogeneous versus
heterogeneous equilibrium now this
simply means that the phases of the
reactants and the product if all the
reactants and the products are in the
same phase we call them homogeneous and
if some are solid some are gas some are
solutions we call that heterogeneous
equilibria
let's see what it says here homogeneous
you could be a cure when all reactants
and the products are in the same phase
heterogeneous you could be a cure when
there's something in the equilibrium is
in a different phase the value used for
the concentrations of computer substance
is always 1 now this is little bit wrong
here
it's not right actually because this
this simply says that if I have solid
reactant or solid product right pure
solid reactant or solid product it says
that we assume the equilibrium constant
is a the the the concentration is equal
to one no it is not because what is the
concentration concentration is actually
the number of moles of the substance in
1 liter volume if the densities are
different actually the then the the
concentrations or the number of moles in
1 liter even its solid or liquid pure
liquid will be different right but when
we write down the equilibrium expression
as we will see it we don't really write
down the solids we only write down the
solutions or the gases as we have here
now that simply means like consider the
decomposition of calcium carbonate
solids to calcium oxide solid and carbon
dioxide and the equilibrium constant
here is written as KC is equal to the
equilibrium concentrations of carbon
dioxide or and KP is equal to this
directly this doesn't mean that the
concentrations of the calcium oxide and
the constant concentrations of the
calcium carbonate is one no as it says
in the previous one previous slides
right now the concentrations of these do
not appear here because the
concentrations of solid calcium oxide is
constant what does that means that
simply literally means the number of
moles of calcium oxide in one liter
volume of solid calcium oxide right well
so that will always be constant it will
never change during the chemical
reaction and the same thing is also true
for the calcium carbonate the
concentrations of that solids again in 1
liter solid is number of moles would be
the same that that never changes so in
the equilibrium expression actually let
me just show you that
I can get this now the equilibrium
expression K sub C actually let's call
that K sub C prime would be equals to
the concentration of calcium oxide and
then concentration of carbon dioxide
divided by the concentrations of calcium
carbonate but now since these are
constant for the solids or if you have
pure water for example the
concentrations of water in pure water is
the number of moles of water in 1 liter
of water and that's equal to 56.66
something and it's constant it never
changes as long as it is U in the water
it is constant and this is also the same
the concentrations are not changing so
if these are not changing I can simply
combine them with this case of C let's
say times the concentrations of calcium
carbonate divided by the concentrations
of calcium oxide since these are all
constant that's constant so this is
equal to a new constant I call that K
sub C actually they case say K sub C is
equal to this and it's constant right
and that is equals to the carbon dioxide
concentrations of carbon dioxide so in
the previous slide as I told you which
says that the value the value used for
the concentrations of pure substance is
always 1 that's what we use but you
should know where it's coming from right
it's coming from because the
concentrations are constant so we
combine the constants here with this
constant obtain a new constant and that
one is K sub C which is equals to the
carbon dioxide same thing is also true
for the KP value 2 again yes I guess I
will stop it here let's say and solve
some problems from previous chapters or
the chapters that will be in
put it in the exam what is that not
tomorrow - Saturday right it's not
exactly okay oh let's see
I picked up some example problems and I
will go through those quickly if I can
simply says now one thing one one
problem I picked up from previous exams
here is the reactions given this is
about rate constant inaccurate remember
this a true means in active solutions in
water actually and plus bromide and plus
6h plus a cures in acidic solution
this is bromate ion actually and then
the Prolog it goes to three PR to liquid
and plus three H 2 O liquid this is the
reaction given in fact a chemical
reaction it says at a particular instant
in time at a particular instant in time
that means like in a reaction as the
reaction is taking place just one time
you know in an instant time the value of
value of minus Delta PR over delta T is
equals to two point two times ten to the
minus three mol per molarity per second
polarity per second so this is actually
the rate of decomposition or the rate of
of the bromide right rate of consumption
of the bromide is given at an instant
time reaction is going on whatever it is
it says what is the value what is the
value of Delta bromine molar
concentration delta T in the same time
that means it's within that in that
particular instant time right there at
that moment when this is equal to this
what is the value of that at actually
this simply means that in an instant
time for this reaction if the the rate
of consumptions of Rome I'd is equals to
two point two times ten to the minus
three it asks you what is the rate of
production of bromine at the same time
right bromine at the same time and it
can be calculated now remember what we
simply do to relate the consumption or
productions of the reactants and the
products we simply write down this the
consumption rate of this which is Delta
B r o2 o3 minus over delta t minus right
because this is the consumption it is
the equation would be equals to minus
one over or I should probably put the
minus on the other side one over five
minus Delta P R minus delta T right one
over five we just divided by five
because the rate of consumption of this
is five times of that in molar
concentration values right so we have to
divide it by five now if I related to
for example this one over here it will
be again one over six minor
delta h plus concentration divided by
delta t like this and then i can relate
it to the rate of production of the
products and that would be equals to
simply 1 over 3 and b delta B are 2
molar concentration and divided by delta
T and that would be also equals to
simply 1 over 3 Delta h2o divided by
delta T this is the relation we can
write down for every chemical reaction
what provided that we are given the
balanced chemical equation it should be
balanced chemical equation now what it
says here it says that the rate of
consumption is equal to that that means
that this value here is given right and
we are asked to calculate this so we can
simply pick up this one and this one
here and I'll calculate the rate of
production of bromine here just put in
the values so let's do that
again we have to be careful here
so I can simply write down one over five
minus Delta BR bromide over delta T like
this equals to 1 over 3 coefficient
there and then we have simply Delta er
to delta T so this value is given what
is that 2.2 times 10 to the minus 3 I'm
looking for this value here so I can
simply write it down as Delta BR to
bromine over delta T would be equals to
simply 3 over 5 - actually it's not -
right because this - is equal to that we
have to remember and that I can simply
put here 2.2 times 10 to the minus 3
that is the rate of consumption of
bromide ion bromide ion and this will
simply give me the result result is 1
point 32 times 10 to the minus 3 molar
per second so this is the answer
actually we can simply solve questions
like that once we have the balanced
chemical equation we relate the rate of
consumption or rate of production using
the coefficients with each other in fact
yes
okay now let's I have nine seven
that means chapter nine and problem
seven at the end of the chapter here a
molecule is given and that molecule let
me put it down here carbon double bond
carbon single bond carbon single bond
carbon triple bond carbon and single
bond carbon then we have the hydrogen's
yes let me see if I get then one two
three four five six carbon one two three
four five six carbon yes oh okay wait
wait yes that would be two hydrogen's
there one hydrogen here two hydrogen
there I guess this is correct
right now this is again very simple but
it says what is the hybridization at
each carbon atom in the molecule
hybridization in each carbon atom right
this is actually the Lewis structure of
that so can you tell me what is the
hybridization here sp2 yes sp2 now how
about this one that's also SP two right
three domains there sp2 now how about
this one sp3 this one is P that's pretty
good this one is also SP this one is sp3
all right so now what is the geometry
around this carbon if I consider this
one this one and that carbon trigonal
planar right trigonal planar and the
same thing is true here if I consider
this this and this again that's three
Gono planar that means they're all on
the same plane right they all on the
same plane and this one is different and
how about here this is linear there if I
simply find out the angles in fact then
these angles will will be 120 degrees
somewhere about 120 degrees and this
angle here simply this one is 180
degrees right that one is planar so we
we should know that it says how many
sigma bonds are there in the molecule
now
Sigma bonds is always present if the two
atoms are bonded to each other right so
one two three four five six seven eight
nine ten eleven twelve thirteen thirteen
Sigma bonds remember again the
symmetries these are symmetric with
respect to the axis bond axis now how
many PI bonds it says now again remember
if I have a double bond between two
carbon the second bond one is actually
the PI bond remember the PI bond is
simply like this in a plane actually so
in this case I have one here two three
here and that means three PI bonds right
now we have to remember this this is
very important again now when we have
these two pi bonds are in a plane which
are perpendicular to each other right
one PI bond is like this and the other
PI bond is like this perpendicular to
that and remember again and other things
when you simply form an sp2
hybridization there is one p orbital
here which is not involved in bonding
and that p orbital is always
perpendicular to this plane right so if
a PI bond is actually taking place
between these two over here that means
that that pi bond is in this plane here
right because I have a py orbital like
this here perpendicular to that p
orbital here perpendicular they have to
simply combine like this which is
perpendicular to the plane here to form
the PI bond in this case I have one like
that and I have one like that right so
that you should simply think about that
when you have
the pie bands in fact there are two pi
bonds between the two atoms
perpendicular to each other and then it
says identify the angles angles I guess
I guess you can identify that again it
is not going to be exactly 120 degrees
but if you just say in those cases about
like 120 degree use like about 120
degrees that's fine
or if you think that is less than 120
degrees you can just say less than
hundred 20 greater than but about 120
would be actually enough okay now
any questions
now another 9
problem 76 at the end of chapter it says
actually may just briefly right there
what it says the peroxide iron Oh - - -
has longer bond lengths than the
superoxide Oh - - iron so it says let's
say bond length of peroxide iron right
this is peroxide has longer bond length
than the super oxide that means longer
than bond length of superoxide which is
this super oxide iron and then question
is why explain it says and then it's
also B it says the magnetic properties
of no let me take this see see it says
all - 2 plus iron has stronger oxygen
oxygen bond than the o2 itself bond
strength of all - 2 plus ion is simply
greater than stronger actually bounced
on strength greater than the bond
strength
of oxygen o2 now considering the things
let me put the be there anyhow I guess
it says the magnetic property of b2 are
consistent with the PI to PI mo being
lower energy than the Sigma 2p mo for BR
b and i boron molecule and dye boron so
Sigma 2p is higher energy let's put that
here now 2 P PI 2p being lower energy
than the Sigma 2p is less energy no
greater energy then PI 2p orbitals
explain again all of these are why now
when we have questions like that the
only way to approach that is the
molecular orbital Theory approach mo
approach now we have simply covered only
the second row diatomic molecules
molecular orbitals of second law
diatomic molecules so the best way to
start a problem like this you just have
to do draw and molecular orbital energy
level diagram for simple second period
diatomic molecules which has simply 1s
2s and 2p electrons that's all so if I
do that let me this is the energy over
here I have 1s let's put that here and
then I have simply 2's here and then I
have P 2p orbitals and the same things
as over here these are all
diatomic molecules 2p and let's say this
is 2s and this is 1s usually we don't
simply consider the 1s it's the core
electrons interactions of those to form
the molecular orbitals will be very weak
but it is a good custom to keep because
you have to count the electrons you know
in the molecules or in the ion species
to calculate the bond order and so let's
say this is the Sigma 1 s and this is
Sigma star 1s and then I have simply
Sigma 2 s and this is Sigma star 2s and
then P orbitals this is again Sigma 2p
orbitals this is Sigma star you can call
that - pizzette it doesn't matter this
is Sigma star 2p now these two are pi
2px and 2py both are pi and this one
will be PI star 2p x and y you can call
that as again now to really say about
the bond length comparison of the bond
length we have to compare the bond order
remember the higher the bond order the
more stable the molecule or the ionic
species that means the stronger the bond
between them right the stronger the bond
stronger bond means
short-distance bond lengths short bond
lengths are represents strong bonds long
bonds represents weak bonds so we simply
have to calculate the all to let's say o
- - bond order right o to - pant order
now what we going to do here we will use
that we should always draw this diagram
somewhere put that to use it air you
know and I'll count the number of
electrons that I have in this species
the number of electrons here eighteen
right yeah that eighteen yes eight eight
sixteen this is two miners each oxygen
has eight electrons right and then two
oxygen 16 two more eighteen then I will
simply put the 18 electrons starting
from there just fill this out that
becomes 14 and then I will have simply
16 and 18 that is right now what is the
bond order 402 - I will simply write
down the number of electrons in bonding
molecular orbitals I can leave that up
again but you know if I have it here it
is probably better write down the number
of count the number of electrons in
bonding molecular orbitals 2 4 6 8 10
right so that means 10 - since total
number of electrons is 18 there's eight
electrons in the anti-bonding molecular
orbitals so that means minus 8 divided
by 2 this would be equals to simply 1
right
then we write down the calculate the
bond order for the oxygen simply super
oxide molecule o2 - and in that case Oh
- - how many electrons I have 17
electrons one
less one less and the bount order will
be equals to again I don't have to
really count once I I have this because
the electron that I will take will be
from the highest energy level I will
take one of that electrons what does
that means that means the number of
electrons in the anti-bonding electric
and anti-bonding molecular orbital would
be one less than the peroxide and that
means that the bond order 10 minus 7
divided by 2 which is equal to 1.5 so
which one has a stronger bond if i
consider both of these like this
the bunk again you know in one case it
is minus 1 the other case is 2 minus 1
which one has stronger bond karach side
or superoxide which one has stronger
bond superoxide right because bond order
is higher greater greater bond order
means more stable molecule that means
stronger bond between the atoms clear so
so what what I expect actually the bond
length of the o2 2 minus peroxide will
be greater than the superoxide why
because the bond order in peroxide
actually is less than the bond order in
superoxide clear that's how we explain
that now here for the bond strength of
o2 and all this over here if I simply
calculate that now let's calculate the
bond order for all true how many how
many electrons 16 electrons right in Oh
- and that would be equals to again we
will take 3 electrons from there
actually remember the oxygen molecule is
paramagnetic and the para magnetism of
oxygen molecule is
only be explained can only be explained
by the molecular orbital theory approach
with this approach otherwise you can't
really predict it from the Lewis or
valence band theories and so on
so I will simply take those two
electrons there leave their three
electrons only what does that means this
is going to be 10 minus 6 10 minus 6
divided by 2 equals to 2 bump order for
the all true is true now what is the
bond order of all - what is that 2 plus
now o - 2 plus has 14 electrons now
compared to this what does that means
that means I take all of those 4
electrons there and what happens if I
take those 4 electrons there then the
bond order will be calculated as 10
electrons again in bonding molecular
orbitals - 4 electrons in anti-bonding
molecular orbitals divided by 2 this is
equal to 3 right so you see again the
oxygen diatomic oxygen with two plus
charge has bound order of 3 which means
very very strong oxygen oxygen bond and
that means very very short oxygen oxygen
bond now for the boron molecule this
boron molecule that time how many
electrons 10 electrons there and that 10
electron if I simply count that 2 4 6 8
10
now boron molecule is paramagnetic that
means it has unpaired electrons if I use
this diagram boron molecule will be
diamagnetic
there will be no unpaired electrons but
remember what happens after the nitrogen
these two changes the Clay's the
energies of the PI and Sigma 2p reverses
now when that comes down over here then
the two electron will go
one here one there just like ones rule
right and then we have two unpaired
electron in Rome die
pour on molecule and it is poly magnetic
and that's what it says here actually in
over here this the para magnetism of be
two molecule proves that Sigma 2p is
less energy than the Sigma 2 pi and the
PI comes down pi is more energetic I
should probably not yes this is greater
means higher over here this goes over
here that goes down there that's what it
says actually yes question okay now what
I said is what this simply says that the
energy order Sigma 2 P is greater than 2
pi means this is higher this is lower
and that's here it's just to reverse
right it will be like this ok that means
that this will be about those two will
be down here now for the this B 2
molecules I have 10 electrons if I start
putting the two 10 electrons eight up to
here two more electrons if the next
level is this then I have to put two
electrons in the Sigma 2p which means
that it will be diamagnetic but if the
next level is this next one then I have
to put one electron there it will be
paramagnetic since the molecule is
paramagnetic mean is that it says that
Sigma 2p is higher energy than the Sigma
Pi 2p right by looking at the
paramagnetism we measured the
paramagnetism
or you find out simply that the molecule
is paramagnetic like oxygen molecule is
paramagnetic why because it has those
two
electrons here in the same direction
right unpaired electrons unpaired
electrons yes
now tomorrow afternoon I will be
available in my office to answer
questions if you have before the exam so
now another thing see let me let me just
say one more things remember
your book when he talks about the
hybridization of the atomic orbitals in
one atom it only goes up to sp3
hybridization right but we simply said
we have PCL 5 and we have SF 6 for
example we know the structure the
geometry here is trigonal bi-pyramidal
and there are five bonds bonded to the
phosphors here six bonds bond to the
phosphors and b number of electron pairs
around the phosphorous or electron
domains are octahedral they are simply
directed toward the corner of an
octahedron and the hybridization on the
sulfur here we said that sp3 now since
this sp3 will give you three hybrid
orbitals but you need five and when you
need five we simply said you just put
here D 2 SP 3 D 2 this one here SP SP 3
D only again four orbitals no 5 orbitals
5 hybrid orbitals 5 atomic orbitals 500
orbitals 6 atomic orbitals 6 hybrid
orbitals but your book says that the
hybridizations do not apply to these
molecules when you have more than 4
bonds but that's actually not true but
in the exam if there is a question about
this if you know that you're sure it's
an sp3 D or SP 3 D 2 you can write it
down or if you just say not applicable
or be accepted clear