Chemical Bonding in One Shot - JEE/NEET/Class 11th Boards || Victory Batch: transcript

come you all to the amazing platform of physics vala so today we are here on this platform english shivala and we are here to study the new chapter which is chemical bonding and the molecular structure we will be having the one short lecture of this chapter and we will be covering all the important topics related to this chapter so let's start the one shot lecture so first of all before starting this chapter how do you define the chemical bond as what do you understand by the chemical bond let's see for the definition it is defined as the force of interaction between two or more atoms molecules and the ions let's say i have a friend of mine okay we share a bond between each other but what is the basis of that bond we are having a force of attraction between each other that we are pulled towards each other right let's say we are the both atoms okay so this is a bond first of all

definition of a bond is this now the word comes that why do we form a bond why any of the atom the molecule or the ion any of the species why do they form a bond the answer is they want to get stabilized like if i'm coming into a class i'm the new student okay i'm just coming into the class and i will be sitting on a bench i will be very alone i will feel very lonely because i'm the new one i'm not having any friends but any of the person any of the student is coming towards me he's sitting beside me and he's talking to me so i will start feeling like i will feel calm i will feel yes i i'm not the new one i'm i'm also having friends i will feel happy because there are people who are coming to me who are talking to me i'm not lonely okay there are some exceptions some of the students like to sit lonely but that is a different case we are just having the normal a very small kid case okay so they will feel stable that

person that boy that student will feel stable because he has started making bond with someone so stability was defined in terms of that he's not he's not feeling lonely what about the atoms how do you define the stability of the atoms atoms are also forming bond why do they are why they are forming bonds because they want to get stabilized okay that string was having stability in terms of that he's not now feeling lonely if i talk about my stability like what is the stability of my in what parameters i define my stability i should have a good career i should have a good health everybody has their own significance of the stability so if we talk about the atoms molecules or ions they form bond to attain stability that is expressed in the terms of the octet right you want to form octet of electrons any of the atom wants to get the octet of electron what is octet of electrons getting the eight

electrons in the outermost shell by any of the method that eight electrons in the outermost shell is the stability definition for the atoms and that can be achieved by any of the method maybe you can achieve that by the shearing of electron by transfer of electrons or by any other method but the motif is to get the octet of electron done for that purpose only the atom forms the bond right or the molecule or the un molecule and we will discuss later first we are discussing about the atom right so why chemical bond is formed to achieve normal gas configuration to achieve minimum energy to achieve maximum stability see what happens is when you're getting the octet of electron you're getting the eight electrons in the outermost shell you are stable and if anything is stable it is having the very less amount of energy let's say you can remember this thing by the method like when you are very angry you are having a very high energy you are about

to like tordunga for dunga types right but when you are having a very calm mood your energy is also you are very calm you're very shy so there's a difference how you can learn that because if you're having a very high amount of energy you are very in the anger mode and you you're not stable at that moment you can do anything you can just slap someone you can just throw something so you're not stable now you're not in your senses so you're unstable right so you can remember in this respective manner that always your energy is inversely proportional to stability inversely proportional to stability right so why chemical bond is formed to have a normal gas configuration now what do you understand by this normal gas configuration you should have octet complete you should have octet complete you should have eight electrons in the outermost shell you should have eight

electron in the outermost shell you should have eight electrons in the outermost shell that you need to remember very nicely that can be achieved by any of the manner you can achieve by transfer of electron you can achieve this by transfer of electron and you can achieve this by sharing of electron transfer of electron sharing of electron you can achieve this by transfer of electrons and the sharing of electrons when you are having the transfer of electrons you are having the ionic bond when you are having the shearing of electron you are forming the covalent bond so that we will discuss further but first of all this is the basic meaning of why you are forming the chemical bond now whenever the bond is formed bond is formed between two atoms or maybe between two molecules also whenever the bond is formed between two atoms that can be called as the ionic bond when the transfer of electrons takes place from one atom to the other

atom one is the acceptor atom and other is the donor atom ionic bond right we will have a very deep understanding of the ionic bond further right now i'm just explaining you what all types of a bond is formed between two types of atoms right if you have a covalent bond you will be having the sharing of electrons whenever we have a covalent bonding shearing of electrons so we consider this as one electron one electron shedding like one atom is sharing it's one electron and other is also shading its other one electron let's say if i have an example let's say i have an example this is one and this is one so they are forming this type of a bond but what do you understand by the dative or a coordinate bond in just a very small uh explanation of the layman language dative bond is called as a bond when one is not sharing any electron and other is sharing both of the electrons right the bond is formed like for example

i have a bond over here between a and b b is having these two electron and a is not having any of the electron so the bond formed over here is like this these two electrons are part of the b also and the part of the a also these are shared equally between both of them a and b but a have not you know has not given any participation in the electron a have not given any electron that both of the electrons are of the b itself so i can call this as donation and acceptance donation and a is accepting electron donation and acceptance along with the sharing of electrons right so there is a donation okay this is a donor atom this is a donor atom and this a will be acceptor atom b is a donor atom a is the acceptor atom

okay one is known or one is acceptor so they are donating also and after donation and acceptance both are sharing these two electrons this is not the case that b has just given the electrons both of them to a and that is not the part of b now no he's giving also he is sharing also these two electrons are now the part of the b also and along with that the part of the a also so both are happy okay so you have to remember this thing that donation and acceptance along with the sharing of electrons donation and acceptance one is donor one is acceptor along with that sharing is also present so this is the definition of dative bond covalent bond and the ionic bond we will discuss all of these things in the deeper sense also what about between the molecules we can also have we can also have the bond between the molecules also right we can have a bond between molecules also so what kind of a bonds you can form between two molecules you can have the hydrogen bond and you can have the

wonder walls forces we will discuss all of these things in a later of this chapter right if you have the bond between two metals that will be called as the metallic bond let's say if i have an example i have a element which is sodium and i write it as sodium solid so we cannot say that this is an atom of a sodium we will write it as a piece of a sodium in which so many sodium atoms are present so many sodium atoms are present so the bond between any of these two atoms is considered as the metallic bond so the bond between two metals will be considered as metallic bond bond between two are two non-metals let's say i have phosphorus nitrogen i have carbon nitrogen i have carbon oxygen all of the bonds formed between the two non-metals will be with the help of the sharing of electrons so that will be considered as the covalent bond bond between two

non-metals and bond between a metal and a non-metal is considered as ionic bond because metal and non-metal prefers to donate and accept electron one is electropositive and other is electronegative we will discuss this further but right now metal and a metal forms a metallic bond non-metallic non-metal forms a covalent bond metal and a non-metal will form a ionic point donation of electrons acceptance of electrons right so let's move forward first now we have to discuss the lewis theory and for that we have to discuss a very important thing which is known as the lewis symbol so what do you understand by this word which is called as a lewis symbol many of you are not knowing right now also that how do you define a lewis symbol as the lewis symbol is nothing just an element which is having electrons surrounding it now how do you define those electrons these electrons are actually

number of valence electrons number of valence electrons in an atom number of valence electrons in an atom i am not writing over here the valency i'm writing over here the valence electrons so let's say for example i have let's say for an example i have nitrogen so nitrogen is having electronic configuration as two comma five so the number of valence electrons are five so i have to write over here nitrogen with the five electrons surrounding so this is the lewis dot symbol of the nitrogen you can make for any of the element present just remember you have to surround that element with the number of valence electrons i'm not writing to i'm not telling you to write number of i'm not telling you to write the valency i am telling you to write the valence electrons right lewis symbol done after that what is lewis theory see first of all

there was when we were having the elements surrounding us they are forming the compounds also so what are the basis that some of the elements are combining and they are forming the compound okay we understand that two atoms are coming towards each other they are having the force of attraction between each other they are attracted towards each other because they think that if they are forming the bond with between each other they will have their octet complete they will be stable so now we have to understand what are the theories which were proposed in order for the bond formation right so first of all theory this is applicable only for the covalent compounds okay that means we are having over here the sharing of electrons now you people know that how electrons are shared how the bonds are formed you know all these things we are not going to discuss over here that how the sharing is taking place that we already know we have to understand that what was the

basis of the lewis theory so lewis theory said that every atom every atom forms a bond in order to in order to have octet of electrons octate is considered as stability okay but if i write an example also you can see the example also example if i have methane so methane over here is the stable compound because it is forming the bond with hydrogen

and as you can see this compound is stable as you can see this compound is considered as stable because the carbon is having the eight electrons so it is considered as stable but there was some of the limitations also attached with this respective theory that it's okay i understand that every atom wants to achieve the electrons that is octet but why still there are some compounds possible which are not having octet of electrons might be the case that electrons are less than eight surrounding them or more than eight surrounding them but still those compounds were also stable so this theory was not applicable to every of the compound possible but yes this was the basis the starting basis of every of the theories that is true that every atom wants to have octet of electron surrounding it but that but that's not the ultimatum so we have the limitations also of this respective theory so what

is the limitation first of all we have over here the hypervalence we see the superoctate or the expanded octet see these all three words are considered as same what do they mean they mean that there are no eight electrons surrounding they are having more than eight electrons surrounding for example i have over here for example i have over here let's say pcl5 i have if7 i have sf6 you know all these molecules they are not having eight electrons surrounding them they are having actually more than eight electrons surrounding i will make one of the example of sf6 you know that sulfur is having actually six electrons surrounding it because it is having the valence electrons as six if you're not remembering remembering the atomic number of the sulfur what you can do is if oxygen is having the six as a valence electron same will be applicable to all of the members of this family like sulfur

selenium tellurium polonium all of them will have 6 electrons in the valence shell right so let's talk about the fluorine now so fluorine we are having over here this is fluorine and we are forming a bond over here this is considered as fluorine so you can see very nicely over here you can see very nicely over here that the electrons surrounding the sulfur is not actually six these electrons are 12 so that is the case that is hypervalent superoctate expanded octate now what is the reason that other species also present hypovalent hypovalent is something which is having less than eight electrons for example i

have for hypovalent i have example as bf3 you can simply have the case of the bf3 also you know that in the case of the bf3 you are not having the octet of electrons you are less than octate you are less than octate you are two electrons less than octate you can simply see from the structure that the electrons surrounding the boron are only six there are only six electrons surrounding the boron so this is also not stable now this will be called as the hypovalene that is hypervalent this is hypovalent now we are also having some odd electrons pc odd electron species let's have an example first all the electrons pc we have an example over here let's say no2 if i make the structure of no2 i will be having like this the structure of no2 right and

this is oxygen and these two electrons are donated by the nitrogen to the oxygen right if you want to have i can have the color of the electrons of the oxygen as something different so you will be able to understand the electrons of oxygen very easily right so what is happening nitrogen is having two electrons shared with the oxygen two electrons donated to the left oxygen and one or electron is present at the as the odd electron so this respective molecule is having nitrogen is having around itself how many electrons just count and tell me one two three four five six seven total electrons nitrogen surrounding is seven so this is considered as odd electrons pc always remember that all the odd electron species are considered as paramagnetic always they are considered as paramagnetic it is being asked like what all species or what all examples in

the given question are paramagnetic you just have to follow a simple rule that is there any example present which is having the odd electrons surrounding so you can have other examples also you can have no you can have clo2 you can have clo3 and you are having no2 etc we are having many of them but right now i'm writing all these so odd electrons pc you are not having the octet of electron hypervalence pc you are having more than eight electrons hypovalence pc you are having the less than eight electrons so these are the examples present over here for the lewis theory and even we have to understand a very new word over here which is a pseudo-electronic configuration pseudo-electronic configuration is considered as it is considered as 18 electrons cations it is considered as 18 electrons cations let's say for example i have all of these zinc

plus two cadmium plus two mercury plus two and gallium plus three see i have zinc plus two over here cadmium plus 2 over here mercury plus 2 over here and you know gallium is coming after zinc only in the periodic table so all these electrons if you see they all have the 18 electron cations okay in the outermost shell so how do you define i if i write the electronic configuration of zinc zinc is what argon which is 3d 10 4s2 so if i remove the two electrons it will be argon 3d 10 remove 4s and what is the electronic configuration of argon 1s2 2s2 2p6 3s2 3p6 and now we have 3d 10 over here you can simply see from the electronic configuration that how many electrons are present in the outermost shell that

are considered as 18 electrons this is considered as 18 electrons so this will be called as the 18 electrons cation this is very very important you should know how to identify that same will be the case of the cadmium if you want to have the electronic configuration of cadmium you can write for the cadmium also what is the electronic application for the cadmium argon after that we have krypton then we have 4d10 and 5s2 argon krypton zenin xenon is having 5d10 and then we have 6s2 so you will remove two electrons you will again have the 18 electrons in the outermost shell so these will be also called as the pseudo electronic configuration even for the gallium also what you will called as gallium it is argon 3d 3d10 4s2 and 4p1 so if you will remove the three electrons it will be this is gallium okay this is the electronic configuration of the gallium so if you

write for the gallium plus 3 it will be argon 3d 10 only and that will be same as ring plus 2 and this will also be called as the pseudo electronic configuration that is the 18 electrons in the outer most shell okay so you have to remember all of these things present in the lewis theory why any of the atom is forming compound for the octet completion but still some of the compounds are possible which are not having the octet but still they are present around us then what is the basis of the stability that is governing that respective compound that we will discuss later but you should know these three these three are considered as the limitation these three are considered as the limitation okay you have to remember that what all of the limitations of lewis dot theory is present hypervalent hypovalent and the odd electrons pc everything is written on the blackboard and you have to remember everything everything is

very very important very very important okay so let's move forward now ionic bond now we have discussed the lewis theory we have understood symbol so many things now we have to move towards the first type of uh topic that is the ionic bond now what do you understand by the ionic bond bond formed by the transfer of electrons one is a donor atom other is the acceptor atom we know all these things now okay now what are the terms which can be asked in term of the in terms of the questions because knowing the bond is a different thing but what all words can be used to define that respective bond for the examination point of view that you need to remember okay it is a bond formed between the one electronegative atom and the electropositive atom see electronegative atom will attract the shared pair of electron towards itself electropositive will donate the electrons have tendency to donate the electrons right so these

two are the first cases donor must have less ionization enthalpy and acceptor must have very highly negative electron gain enthalpy so what do you understand by this let's say if i form a bond we know that there is a very famous ionic bond present over here let's say i have a nacl so you have n a two eight one you form any positive you get 2 comma 8 you have cl over here which is 2 comma 8 comma 7 and when you form cl negative over here you are having 2 comma 8 as what is happening this sodium is donating its electron to form sodium positive plus electron and chlorine is exactly accepting that electron only to form the cl negative ion this electron is moving here only so these two are coming towards each other this n a positive and c l negative is coming towards each other and there is a formation of a bond

over here this respective bond formation is known as the ionic bond electrostatic bond there is some amount of energy also released when these two come closer that is the lattice enthalpy we will discuss that later but first of all this is what is happening in the ionic bond now you know there is uh some of the terms associated with this see it is donating electrons so sodium over here is acting as a donor atom right cl is accepting electrons so cl over here will be called as acceptor atom it will be called as acceptor atom one is a donor atom other is the acceptor atom right so what they are saying is one should have a less ionization energy and acceptor must have a very high electron affinity c why they are saying this if you are donating electrons and you are you you require very high amount of energy to release that electrons that is

very you know that is very difficult because if this respective sodium cannot release electrons easily let's say it's a normal gas or it's a very stable electronic configuration so i have to supply very high amount of energy right and if the cl this atom cl is not having any tendency to accept electrons its electron affinity is very less let's say in the last class in the last one short lecture i have written nitrogen is having very very less electron affinity i say it is having the positive electron gain enthalpy alkaline earth metals they do not require any electrons normal gases they do not require any electrons so if i form the bond of the sodium with any of the atom which is not having any tendency to accept electrons how the bond formation will takes place it will be difficult now so the donor atom must have a very less ionization energy that it should remove electron very easily and the acceptor should accept electron very easily okay they are solid they are crystalline they are brittle these properties you have to

learn ionic compounds are non-directional in nature see they are having a 3d lattice i've written over here they are having a three dimensional lattice you have understood the chapter now the solid state right you know the chapter solid state in which we are having a three-dimensional lattice we are having the lattice points and so many other things so whenever you are forming a ionic compound you have a positive and negative charge over here present so they are having the three-dimensional lattice what they are having they are having the three dimensional lattice and the you know andy and the the bond present between these two positive and negative ions is considered as the ionic bond first of all you should know that why they are hard why they are brittle and even they have a very high melting point right they have a high conduction they have a high conductivity they have a high conductivity but in the

molten state they have a high conductivity in the molten state and in the aqua state they have a very high conductivity in the molten state and the aqua state all these properties are associated all these properties are associated with the ionic bond because they have a three dimensional lattice we have to first break all of the bonds so that the melting point can be achieved so there is a very big lattice so we have to break so many number of bonds right three derivatives you have discussed about cesium chloride nacl zinc sulfide you know word zeit so many other compounds you have discussed in the solid state all of them are the ionic compounds and you know what type of a structure three dimensional structures they are having right okay so it is non-directional like you cannot say in the case of the ionic compounds that there is a specific direction of a bond let's say if i talk about the covalent bond i cannot make

methane like this i cannot make methane like this every bond has its own particular direction right every bond has own particular direction like this this is the way in which this methane is formed i cannot write methane like that i cannot write that is a incorrect representation of methane every bond has its own direction but that is not the case in the case of the ionic compound there is a nature which is associated with them that is the non-directional nature remember this thing very nicely they form three-dimensional lattice they have non-directional nature and all of the all of these things so these are the usual properties of the ionic compounds let's move forward see whenever there is a formation of an ionic bond there is some amount of energy released and maybe the energy will be required for the breaking of that ionic bond so we have to define now the lattice

enthalpy the definition is amount of energy evolved or absorbed when one mole of an ionic bond is formed from its constituent gaseous ions what they are saying is you were forming at that time n a positive and cl negative so you were having nacl so this bond formation was taking place whenever this bond is forming because both of them is getting stabilized and whenever there is a stability energy is released that energy released is considered as lattice enthalpy also the amount of energy required to break that respective bond you will require the same amount of energy let's say if the energy released over here is minus x so if you want to break this bond that same amount of energy you will require to

break the bond so lattice enthalpy is defined in terms of the making of the bond also and the breaking of the bond also but the sign will be opposite one will be positive and other will be negative now the factors on which lattice enthalpy depends see there is some sometimes there is very high amount of energy released sometimes there is very less amount of energy released so we have to now define what all factors are depending on the lattice enthalpy that they will define which of the case will have more energy release and which of the case will have less energy release so the factors will be charged on the ions radius of the ions and even the stability and the solubility also will be affected by the lattice enthalpy first we will discuss the these two factors okay this point is very very important this can be asked directly also that during the formation

of an ionic bond what is the major role of the energy which is the energy which is having the major role so you have to take the answer which will be the lattice enthalpy right so now we are having lattice enthalpy is directly proportional to product of the charges and inversely proportional to the radius of the ions you have to just remember these two okay factors and always the priority is given to the charge on the ions after that radius is considered okay let's say for an example i have sodium fluoride magnesium fluoride and the aluminium fluoride we have to first we have to first define the lattice enthalpy order mg f2 and alf 3. now what is happening first see the multiplication of the charges of these two ions present in the

molecule sodium is having plus one charge fluorine is having minus one with the sine of the mod also present magnesium is plus two fluorine is minus one with the mod aluminum is plus three and fluorine is minus one with the mode so this is having one this is two this is three so you don't even have to see the radius now because priority is given to the charge on the ion so answer will be very straight the lattice enthalpy of sodium fluoride will be the least one then we have for the mgf 2 and then we have for the alf 3 this will be the order of the lattice enthalpy now if i consider these two we have over here nacl na2s and then we have a three p these are the compounds present so if i

draw the q one q two over here this will be plus one minus one with the mod this will be plus one and minus two with the mod this will be plus 1 and minus 3 with the mode so answer will be 1 2 and 3 over here so i think you you are knowing the answer of this also [clears throat] nacl then we have na2s and then we have na3p so the answer of this will be this and answer of this will be this answer so everything is only depending on the product of the charge solely depending on the product of the charges so always have the priority on this factor if you want to have the example of the distance that is the radius so i can give that also let's say third case if i considered let's say i have

nacl kcl and let's say cesium chloride so if i write the product of the charges over here so i will have plus 1 and minus 1 plus 1 and minus 1 plus 1 and minus 1 obviously with the mod present so everybody is having over here the product of the charges as 1. so how will you now define the lattice enthalpy the product of the charges is same in all of the cases so what you have to do greater the lattice enthalpy less is the radius so answer will be in that case you will be having nacl rubidium chloride and then cesium chloride it means that when nacl will be formed more amount of energy will be released when the cesium chloride will be formed less amount of energy will be released so always remember priority is

always given priority is always given priority is always given to the charge on the ions i should write over here priority always so the charge on the ions after that radius of the iron if you're getting the proton of the q1 and q2 as same then you have to look into the respective radius of the ions okay perfect now bond have a cycle now we have to understand with the help of the bond haber cycle that how the formation of the ionic bond takes place right let's start let's say i want to form a compound magnesium is reacting with the f2 to form mgf2 solid this is the compound that you need to form now the first step

you will first take the magnesium it is in the solid state you will convert the magnesium in the gaseous state the energy released in this or required to convert this magnesium solid into the gaseous is the sublimation energy and that will be positive because you have to supply the energy for the formation after that after that let's say second process this magnesium gaseous will undergo ionization to form mg positive gaseous with the electron released then another electron will also be released because you want to have the mg plus two gaseous because you can see in the mgf2 magnesium is present in the plus two oxidation state so another electron will be evolved i will call it this as ionization one

and i will call this as ionization energy two first electron removal ie1 another electron removal ie2 right after that you have fluorine but you have fluorine as a molecule if you want to add electron into the fluorine you will require the single atom so first you have to break the f2 when you will break f2 it will be in the half f2 you can simply write as 2f atoms okay so the energy required for this will be your dissociation energy the energy required will be the dissociation energy after that i have another energy will be fourth one this fluorine gaseous atom will accept one electron to form f negative gaseous

another fluorine gaseous atom will accept other electron to form f negative gaseous now these two are called as electron gain enthalpy one and electron gain enthalpy one both are same because i have different two individual atoms i am not adding into the same i am adding into the different one okay now the [snorts] fifth one will be see this mg plus 2 iron will come near to the two f negative ions because in the medium i have a positively charged iron also negatively charged ions also they will come towards each other energy will be released they will form mg f2 solid and the energy released over here will be lattice enthalpy you know that whenever positive and negative charges are or the ions are coming towards each other

energy will be released and that energy will be called as lattice enthalpy okay fine perfect now for this whole process okay we are calculating the enthalpy of formation of this we are calculating over here enthalpy of formation of mg f2 we are calculating enthalpy of formation of mgf2 so what we need to do is we need to add all of these energies okay so what all energies are present over here we have sublimation energy we have ionization energies we have dissociation energy we have two times we have to multiply the electron gain enthalpy and then we have to plus the lattice enthalpy also so we are adding all of these energy now we have to cancel them

okay now add all of these when you will add all of these left hand side will be added to the left hand side right hand side will be added to the right hand side okay so let's cancel what all terms will be cancelled okay i'm taking the pink color stay on the whole of the left side is there any mg solid present only one in the right hand side is there any present no so this will not be cancelled magnesium gaseous any other no magnesium gases mg positive mg positive cancel these two okay mg gaseous will be cancelled with the mg gaseous over here this f2 gaseous is not cancelled anywhere these two fluorine atoms are cancelled

with these two fluorine atoms okay these two electrons are cancelled by these two electrons okay after that these two f negative atoms are cancelled by these two f negative atoms and this mg plus two will be cancelled by this mg plus 2. this is very very important you need to understand every of the word of this bond haber cycle just have a very nice concentration in this respective topic okay so you are left with the f2 you are left with the magnesium solid and you are left with the mg f2 okay whatever is left just right you are left with the magnesium gaseous not magnesium gaseous you are left with the magnesium solid f2 gaseous and you are going to form mgf2 solid and

here you have only started your respective reaction so this is the process through which ionic bond formation takes place now what type of a question can be asked on this see what they will what they will do they will give you the values of all of the energies okay sublimation ionization electron again enthalpy lattice enthalpy and they will ask you the enthalpy of emission you have to just put and just add all of these you will get the answer of the enthalpy of formation sometimes it can happen and the help of a mission is given they will ask you the lattice enthalpy so just minus mgf2 just take it over here let us enter over here and get your answer so what you have to remember for the numerical point of view this respective and this is the method through which you have to understand what is happening [snorts] adding all steps

desired equation is achieved [snorts] just go through this i'm having a water one minute break just go through this this is very very important very very important it is just go through this now let's move forward now solubility of the ionic compounds

i am giving you in the terms of the very short layman language just follow this and solve your questions although this topic is very deep but i will tell you in a very short form you will understand and you will be able to solve the questions now see i have a compound which is a b solid it's a ionic compound it's a ionic compound i dissolve this compound into the water now there might be a case this is soluble there might be a case it is not soluble first of all ionic compounds are soluble in the polar solvent okay why because you know there is a concept we call it as like dissolves like okay this is followed by a concept which is like

dissolves like as ionic compounds itself are polar so they are soluble in the polar solvents like water ethanol okay what are the what is the definition of a polar solvent the solvent which is having a very high dielectric constant the polar solvents are having very high dielectric constant now what is dielectric constant okay see every solvent is having a property so that when any ionic compound is dissolved into that respective solvent it can dissociate that compound into the positive and negative ions effectively if you have a very high dielectric constant that means you can separate the positive and negative ions very easily and if you have a very less dielectric constant you are not able to do that process effectively so so for that solubility will be less okay so solvents having die high dielectric constant are considered as

more polar you know now the meaning of the dielectric constant okay now first [snorts] of all we have to break this compound into a positive ion and b negative ion and the energy used in this will be lattice energy you have to first break this compound and the energy required to break this compound will be the largest energy now this a positive ion will be added into the water and it will be surrounded by the water molecule now this will be in the aqueous form that means now it is hydrated it means that this a positive will have water molecules around it okay this will have water molecules surrounding it there will be very high number of

water molecules surrounding it and why they are surrounding it what is the force that is you know attracting these two this a is a positive one and oxygen is having the del negative charge so we have a force over here if you know that this is known as iron dipole interaction we call it as iron dipole interaction why because this is the ion this oxygen is having polarity as del negative this molecule itself is a polar molecule having a dipole one is still positive and other is del negative so the force present over here is called as is called as iron dipole interaction okay perfect similarly the case will be with the b negative ion similarly the case will be with the b negative ion it will give you b negative aqueous okay

b negative is present over here and water molecules are surrounding it so many water molecules will surround like this okay same this will be having the interaction as ion dipole itself this is also iron dipole b negative is the iron and water is having the polarity one is having del positive one is having del negative so the molecule is polar so these two ions in the aqueous form are now hydrated we are having the a positive hydrated we are having the b negative hydrated so the energy released over here will be the enthalpy of hydration of a positive and the energy released over here will be

enthalpy of hydration of b negative now what are the conditions which will decide that ionic bond will be soluble or not so understand this these two factors itself if lattice enthalpy is greater than hydration enthalpy your compound is not soluble or you can simply say less soluble and if lattice enthalpy and hydration enthalpy is having these type of relationship that means compound is soluble you can simply say compound is highly soluble the compound will be highly soluble so just understand these two points what is lattice enthalpy how it is used over here what is hydration

enthalpy over here what we have to do we will having we will be having some of the value of the lattice enthalpy we will be having some of the value of the hydration enthalpy of the positive line some of the negative ion we will add these two will have the comparison of the magnitude of the both which will be having the greater value will decide whether the compound will be soluble or not soluble this is the basis of that selection of the solubility okay now now we have to discuss before this we need to discuss one more important thing see we have understood that how we can decide that the compound will be soluble in the water or not but what exactly happens when some of the compound or the iron enters into the water what all situations will take place so let's see let's say i have dissolve nacl in water

okay we are dissolving nacl in water i have a beaker over here i'm adding nacl over here this is water you know that whenever any ion enters the solvent enters the water solvent what will happen it will [snorts] break into any positive ion and cl negative ion and both will be in the hydrated form now what do you understand by this word hydrated you know that any positive and cl negative will be surrounded by will be surrounded by many of the water molecules they will be surrounded by many of the water molecules so this is the meaning of the hydration first but you know there can be a question related to this that how you will decide that one will be less soluble and other will be more soluble so we have to see another factor over here see

if i ask you one compound is nacl other is kcl and other is cesium chloride let's say i have these three compound and you have to tell me the solubility order of these three we have to tell that what is the solubility order of these three what is the hydration order of although sometimes these two words are used simultaneously sometimes they are not okay but right now we are just talking about the solubility and hydration in the same sense okay okay now this can be decided by only one value which is called as hydration now on what factors the hydration depends hydration depends on charge by radius ratio if you have a greater charge by radius ratio you will be more hydrated you will be having the more hydration more

hydration energy will be released let's say i have the comparison see cl negative is same in all of these three so answer will not be depending on the cl negative ion answer will be depending on the n a positive k positive and the cesium positive okay so you have any positive k positive and cesium positive present over here okay so if i calculate the charge by radius ratio this will be also having plus one plus one and plus one both are having charges as plus one so radius i'm not knowing the value of the radius but i can hypothetically write over here as one two three because i know they have these type of sizes one is less and one is more so you will find out that charge by radius ratio will be higher for the sodium positive now what do you

understand by this it means that greater is the hydration i can write over here one more thing i can add one more thing over here more water molecules more water molecules will surround okay now if i write the picture over here this will be like this any positive cesium positive and k positive okay so if i surround the water molecules the case will be like this the case will be like this so many water molecules will surround the any positive ion what about potassium positive potassium positive will be like this

and what about cesium positive now this is cesium positive so which of the following is having greater number of water molecules attached to it obviously any positive it is more hydrated if i talk about the ionic radius and if i talk about the hydrated radius these are the two different things okay this is this pink is considered as ionic radius radius ionic this is radius ionic this is radius ionic and if i talk about the hydrated radius if i talk about the hydrated radius that will be this much that will be this much that will be this much so greater is the hydrated radii greater is the hydration and that is decided by the charge by radius ratio greater is the

charge by radius ratio more you have the tendency to attract the water molecules towards yourself more you will be hydrated more you will be having the hydrated radii and more water molecules will surround everything is related with each other okay now what is happening over here another important thing you have understood that you have understood that yes ma'am hydration will be highest for sodium positive then k positive and then and then mam for the cesium positive because sodium positive is having the higher charge by radius ratio followed by the k positive and then the cesium positive very true so now there is a question which is related to this which is called as ionic mobility what do you understand by ionic mobility the movement how easily you can move inside the solution

ionic mobility now i ask you a very simple question if someone is very much in the size so will it be able to move easily no no it will move in a very difficult manner like this but if someone is having a very small size so they will be moving very very fast right [snorts] so this is the case here itself also now ionic mobility is directly proportional to inversely proportional to hydration or you can simply say hydrated radiator the hydrated radii less will be the movement or the ionic mobility so the ionic mobility will be highest for cesium then k positive and then sodium positive do not remember this thing do not forget this thing this is asked in the electrochemistry chapter also because you know that chemical bonding is the

basis so many other concepts are also the basis of this chapter okay so this is also asked in the uh electrolytic cell the conductance part there we have the application of the ionic mobility see we are discussing all of these things in the solution form right in the solution form ionic [snorts] mobility is simply depending upon the ability of an iron to move if you have a greater size you cannot move easily but in the water you are having a greater size okay in the water you are having the greater size if i ask this question the same question in the gaseous state if i ask the same question in the gaseous state what will be the answer in the gaseous state size of iron

size of iron will be cesium positive then potassium positive and then sodium positive so if you defined define here the ionic mobility yes that will be like this only because in the gaseous form we are never talking about the hydration or all of these terms so sometimes the question can be confusing so i have made a boundary over here in the solution state and in the gaseous state answers are different okay very very important hydration ionic mobility and all these things and even i am writing a order of a very important order over here always remember that h positive and o h negative are having very high

are having very high ionic mobility they are always having very high ionic mobility they are always having very ionic mobility always remember you have to see the order after this whether hydration no hydration these will always have to be considered in the first okay there's a reason related to that that with the help of the hydrogen bonding they are able to you know get into that electrode very easily but yes we are talking about in the solution form okay in the solution from this question can be asked in the electrochemical electrochemistry also so let's say if i ask you a question there is k positive there is lithium positive there is h positive so how will you define the let's say the ionic mobility order in the aqueous solution always you have to write h positive first always there is no exception k positive and l i positive

smaller size great hydration less will be the mobility so answer will be k positive and the lithium positive so don't forget this thing that h positive and oh negative are having exceptionally high ionic mobility don't forget this thing okay let's move forward we are done with the solubility uh no other any other topic is required for the solubility just follow this and you will be able to solve your questions effectively now if you have a greater ionic mobility you have a greater conductance also because if your a if the ions are able to move in a very high speed so they will reach to their respective electrodes in a very short span of time so they will have a greater conductance so you can have a add-on factor in your notes also this is all related to the electrochemistry but it is the basis of the chemical bonding okay now the next thing is in which there's a question in which of the following hf is

negative due to the lattice enthalpy or most contribution in bond hyper cycle is from what they are telling you is this is not hf over here this is enthalpy of formation over here they are saying that enthalpy of formation is negative due to the lattice enthalpies what they're saying is that lattice enthalpy is very high over here and high means that it is highly negative over here so what will be the case in which case the lattice enthalpy will be the highest one you know that it all depends upon the q1 q2 and the r you know that lattice enthalpy depends on q1 q2 that is divided by the r this is the uh formula this is plus 1 into minus 1 this is plus 2 into minus 2 this is plus 1 into minus 1 this is plus 1 into minus 1 so we are having the answer as mgo this will be having the highest product highest contribution due

to the lattice enthalpy because of the very high charges okay now you know that ionic compounds are having this property that they follow this isomorphism okay this is a property which is followed by ionic compounds ionic compounds show rna compounds show iso isomorphism ionic compounds show isomorphism so how do you differentiate or how do you identify that these are isomorphous so let's have a look equal number of total ions must be there number of electrons in both cations and anions should be equal and these are the examples okay nacl and mgs na positive and mg plus two will have the same number of electrons cl negative and s minus 2 will be having the same number of electrons as both are isoelectronic and we know what is

isoelectronic same is the case of the nano 3 and ca co3 if you calculate the no3 negative total co2 negative total any positive calcium plus two you will have the same thing okay now the next thing when same general formula and equal number of water of crystallization okay so in this case zinc sulfate magnesium sulfate iron sulfate all of these are also considered as isomorphous so how you will define this is the way okay remember the two forms of the isomorphism let's move forward alums alums are also actually the fused souls there are so many songs which are fused together with some water molecules attached to them so ionic compounds also form the alums okay salts basically are the ionic compounds and that is the alums the fused salt actually so general formula is m2 so4 dot m2 so4 dot 24 h2o okay

lithium does not form alums you have to remember this thing in alum metal ion is in the hydrated form after that aqua solution is always acidic strong acid plus strong bases and strong acid plus weak base is always present all alums are isomorphous i have explained you isomorphous in the previous page k2 so4 that dot aluminium sulphate can be present that is the one case other is ammonium sulfate or chromium sulfate dot 24 h2 is present that you can call as potash alan this is the chromium alum okay so you have to remember just one important thing over here that lithium does not form any alum and this is the formula and in the very high number of cases these are mostly acidic okay this is the concept related to the alums okay this is also the property shown shown by the ionic compounds

ionic compounds form alums ionic compound form allen see it's a very big chapter i'm just considering the most important points over here that can be asked in the let's move towards the covalent compounds covalent compounds as you all know it is formed by the shearing of electrons two atoms will be there they will share one of the electrons of each and there will be the bond formation that will be called as the covalent bonds we know all of these things now we have to understand other important properties related to the examination point of view sharing occurs equally between two atoms polar covalent bonds are present and non-polar covalent bonds are present okay so let's discuss all of the all of the points over here these are formed by the sharing of electrons let's say we have so many examples we have h2 over here we have o2 over here n2 over here

etc we have so many compounds like when we start the class 11th we first of all we understand that how we can make the covalent compounds with the help of the lewis dot structures and we know all these things if you want to make the o2 we can simply just have a look on the o2 that how it is forming it is forming with the help of the sharing of electron just for an example just for uh the sake of the examples i have just made one of them okay sharing occurs equally between two atoms okay like one atom is sharing electrons and other is sharing the electrons if you have unequal sharing then that might uh you know move in the form of the in the way of the dative bond so there will be the equal sharing of electrons but sometimes there are exceptions also see covalent bond can be one electron one electron sharing that means one electron electron means that both of the atoms are sharing same number of electrons like same number of electrons are shared same

number of electrons are shared same number of electrons are shared but if there is like there are sometimes two center three electron bond there are two center four electron bonds okay so these type of bonds are not having the case like there are only i should change it i change this as first of all this is three center four electron bond sometimes there is three center two electron bond you have studied the case of the banana bond in the boron chemistry there we study these types of bonds also and they are also considered but not the same number of electrons are shared by the each atom not same number of electrons are shared by

each atom so you have to understand that not every of the atom is sharing the same number of electrons to form a to form a bond so covalent bond can be one electron electron sharing also and some other number of electrons sharing also okay the bond can be polar covalent bond and a nonpolar covalent bond so how do you define these as whenever they are writing as a polar covalent bond whenever they are writing as a polar covalent bond that means you are having like clf you are having hf you are having hcl all these are examples of the polar covalent bonds how do you define them as they are having difference of electronegativity as non-zero there is no difference of the electronegativity present so there is a like if i write the electronegativity of fluorine and chlorine fluorine is having a greater electronegativity as compared to the chlorine so there will always there is always difference of electronegativity

present so difference of electronegativity will not be zero but in the case of the non-polar covalent bond in the case of the non-polar covalent bond but in the case of the non-polar covalent bond there is a polar covalent bond so in that case difference of electronegativity is zero let's say i have cl2 i have h2 i have f2 all these respective molecules are considered as polar not polar non-polar covalent bond because there is no difference of electronegativity what should i write like in this case i can write over here i can write for the clf i can write del positive over here del negative over here i can write this polarity i can define this polarity but what about cl2 how should i write del positive and del negative over here so they are non polar covalent bonds they will be considered as non polar covalent bonds so this much

is required first of all for the covalent bonds after that we have to understand the meaning of this word which is called as covalency now what do you understand by the covalent see covalency is defined as covalency is defined as like how many bonds a respective atom can make okay how many maximum number of bonds a respective atom can make that will be defined as a covalency but in the case of a molecule it is decided by what number of bonds that respective atom in that respective state is forming like first i should explain this like like i say i have a molecule like this aluminium h2o 6 plus 3 i have a molecule like this so what will be the covalency of aluminum over here what will be the covalency of aluminium over here that will be six that means that six water molecules are surrounding it six bonds it is forming

with the water molecule so the covalency will be six but what in the case of the elemental form so for any of the second period elements the maximum covalency that you can achieve that maximum covalency that you can achieve will be the maximum covalency that you can achieve that you can achieve is 4 not more than that because what is the reason what is the reason for that because you know for the second period elements for the second period elements the electrons are filled in what shell let's say i start with let's say i start with lithium ion okay so lithium is having the filling of the 2s one and then we move from lithium to neon okay that is having its 2p6 completely filled so you are filling 2s and then 2p so you are having 2s the one orbital and then you're having the two p

other three orbital so total number of orbitals involved over here are four you cannot make any more bonds uh greater than four present over here so always remember this for the second period elements always every atom can have maximum covalency as for not more than that you can form a bond you can form form a bond from here here here and here what after that you cannot form any bond after this now i'm just explaining you in the very you know very layman language i can explain you in a very deep manner also but this much is required just follow this okay you will be able to solve many questions all of the questions by following this because you are not having any other orbital present you cannot form any other bond but what about the third period see for the third period now we cannot define simply we cannot simply define what should be the covalency some of the atoms can have more than uh their required capacity sometimes less than the required capacity so even it is never asked that what should be the covalency for the third period

elements because it is not completely defined it is not completely defined so you can just write a dash over here not required but for the second period elements it is always asked that what can be the maximum covalency what maximum number of a bond a particular atom can make in the second period element so that you are required to remember and i have told you the reason okay just remember this thing after that see since you're making the covalent bonds you're making the covalent compounds there are some charges present on the atoms whenever you make a bond like if i if i say for an example if i say for an example you are having a n3 negative ion you are having n3 negative ion how do you write this n3 negative how do you get to know that i should write negative over here so first of all i have to follow a simple rule i have to follow a simple rule a simple trick over here that is that you need to remember is for the second period elements for the

second period elements always remember for the second period elements always remember if you have bonds if you have bonds equal to valency if you have bonds greater than greater than your valency and if you have bonds less than your valency there are these three cases present okay not other not other than these any case will be present so if you have a bond equal to valency you will be having a no charge bonds greater than valency positive one less than valency negative so these all charges will be present these all charges will be present one two three these three cases will be present so first of all let's say i have an example which is n3 negative okay so one nitrogen over here

one nitrogen over here and one nitrogen over here okay let's say i make two bonds over here and two bonds over here so what should be the charge that should be present nitrogen over here is forming four bonds so it should be having one plus one charge present because the bonds is greater than valency nitrogen is having valency as minus three so three we should take the one more extra bond is formed so we will write over here plus one nitrogen is having valency three and it is forming only two bonds over here so i will write minus one over here and minus one over here just calculate just cross plus one and minus one you will be left with n three minus okay so this is how you have to make the n3 minus now let's say i have any other compound but before that we need to understand one more important thing which is

which is your dative bond you need to understand the dative bond also because after that you will be under you will be able to calculate the formal charges more effectively see what do you understand by the dative bond whenever there is a zero two electron sharing whenever there is a zero two electron sharing always there will be the formation of a dative bond one atom is sharing no electron and other atom is sharing both of the electrons so there will be dative bond formation okay always dative bond and even we call it as coordinate bond even we call it as coordinate bond dative bond we can call or coordinate bond we can call okay after that what is happening is let's say if you have a donor atoms since you are donating electrons donor atom and other you are having as the acceptor atom

so always remember in terms of the formal charges or in terms of the charges whenever there is a donor atom it will always acquire the positive charge and this will always acquire the negative charge always one atom will have the positive and the factor will have the negative charge okay in this respective bond there is always i have already explained you that there will be donation also plus there will be sharing also there will be donation of electrons plus there will be sharing of electrons both of the things will be present okay hello then after that donation plus sharing both now let's uh move towards the example first right let's move towards the example first let's say i want to make ammonium ion i want to make ammonium ion this molecule is having the coordinate bond now what is happening you are having ammonia molecule present and h positive is coming and h positive is saying to ammonia that you are having complete octet of electron and what what problem

you are having you are having complete octet of electron you are having these three bonds which are giving you six electrons and there is one lone pair present so you do one thing see i have i'm having one s0 i'm not having even a single electron present just donate me this electron i'm not telling you to just give me these electron and just and just go away these electrons will be the part of you also and me also so there will be the formation of the coordinate bond so he will say yes okay i can do this so this type of a structure will be present this type of a structure will be present first of all i'm making this like this and in terms of a charge if you want to make this bond so this is not the way that in terms of a charge we will be making then the bond formation will be like this now since nitrogen is a donor atom it will acquire the positive charge since hydrogen is the acceptor atom it

should acquire the negative charge but it but it is already positive so positive negative will make it neutral so this is how you will make the compound this is how you will make the compound both of the compounds are correct one of them is showing the arrow and other is showing the bond donor is positive acceptor is negative okay i hope this is clear to everyone i hope this is clear to everyone okay i have given you only the one example i am telling you other examples also let's say i have this co a very famous carbonyl molecule carbon monoxide molecule so this is the structure of this molecule first of all they are sharing two two electrons oxygen and after that these two electrons which are present in between are given by oxygen itself that means these two electrons are given to the carbon for the shearing of electrons because you

know obviously both of them in this case will have octet of electrons in this case both of them will have the octet of electrons both of them will have the octet of electrons both of them okay so even if you want to make in some other form you can make in some other form also that will be carbon double bond o now positive and then negative this will be the structure of carbon monoxide oxygen in the donor atom positive carbon is accepted atom negative so this is the structure of carbon monoxide i know i hope you have understood that what you have to do now i'm writing you the examples of the dative bond see you can have the dative bond in the hno3 molecule i think this color is visible hno3 molecule and after that we have no3 negative we have ozone we have carbon monoxide and then we have

h3o positive we have ammonium see all of these molecules are having the coordinate bond and one more important thing i need to tell you is one more important thing i need to tell you is that if asked i should write if molecule is having any atom any atom rom third period if molecule is having any atom from third period always always dative bond is present you don't have to see beyond third period onwards beta third period

onwards see in the second period if the middle atom the central atom is from the second period you have to follow these examples okay but if the atom is not from the second period beyond second period that is the third period onwards always there will be the coordinate bond always maybe h2so4 s3po for hclo4 always there will be the presence of the coordinate bond so you have to remember this thing this is something very very important let's say for example if i say let's say for example if i say i have o3 i have ammonium i have h2so4 i have h3po4 i have let's say let's say i have let's say h2o so just tell me which of the following do not have which of the following do not have dative bond just tell me this thing which of the following do not have

dative bond this you need to answer so you don't have to think anything ozone is there ammonium is there now you will have a confusion that h2so4 h3po4 mam has not written in the examples but mam has mentioned over here if the central atom is from third period onwards always there will be the dative bond so the answer will be water since water is not having any dative bond hydronium ion will be half will be having the dative bond so remember this thing okay one more thing i need to tell you is dative bond can be present between the molecules also and in that case there will be the formation of an adduct now you have to understand what is adduct now you have to understand what is adduct let's say i have a molecule over here ammonia and i react this with the bf3 since you know that ammonia is having eight electrons surrounding bf3 is having six electrons surrounding so this bf3 will

be acting as a lewis acid since lewis acid are the ones which are electron deficient it should have eight electrons but it is having six electrons present over here and this is considered as lewis base over here because lewis bases are the ones which are having the lone pair of electrons or the negative charge present so what will be the case this will donate its electron to the bf3 molecule this lone pair will be donated to the bf3 molecule and you will have your molecule like this ammonia with the positive charge and bf3 with the negative charge ammonia positive and the bf3 negative since nitrogen is donating the electron pair to the boron trifluoride you will be having a donor atom as positive and acceptor atom as the negative ones so this is formed adduct this will be called as the adduct now always remember always remember that whenever there is a formation of an

adduct the the hybridization the hybridization the hybridization of acceptor atom the hybridization of acceptor atom increase by one unit hybridization of the acceptor atom increases by one unit since it was sp2 over here now it will be sp3 and i am not talking about the donor atom i'm only talking about the acceptor atom so remember for the acceptor atom only just do not make yourself that donor atom will also have any change in the hybridization no i have just written for the acceptor atom so you have to follow that only just do not use your own brain just follow what ma'am is saying okay okay after that okay so we have understood the formal

charge just follow that okay even i can write the one more formula also also for the formal charts you can use this formula tb minus nv minus half p tv means total valence electrons nv means non-bonding electrons b means bonding electrons so you can use this formula also and you can simply use this trick also whatever is comfortable for you after that we have the fajang's rule now what do you understand by the phajan so since we have understood what are the covalent compounds what are the ionic compounds formal charges and after that we have to understand the fajam's rule now what is

fajan's rule sometimes there is a bond which is not purely ionic or purely covalent we cannot say that the bond is purely ionic or purely covalent it might have a more ionic character it might have a more covalent character so for john's rule actually tells that how much covalent character is present in an ionic bond that we will understand in the fajan's rule okay see there is a ionic bond a positive and b negative present over here

now this is the cation and this is the anion and i am just showing you the electrons of the anions present i am not showing you the electrons of the cations just for my convenience okay just for my explanation i am not writing the electrons of the cations okay so what happens is what catime do it attracts the electrons towards itself okay it is positively charged iron what it will do it will attract all of the electron density of this anion towards itself major of the electron density will be present over here major electron density will be present over here okay okay it was first spherical symmetrical but now it has distorted

its shape is distorted its electron cloud is distorted it is distorted in a way that now some of the electrons are being shared some of the electrons are being shed since these are the nucleus of the two ions henna so electron density is now present between these two nucleus okay now electron density is present between these two nucleus why because the cation has attracted the electron density of the negative ion anion towards itself okay okay fine what is happening over here we can write over here that electron we can write over here that electron cloud electron cloud is distorted we can write over here that electron cloud is distorted electron density is coming electron density is being shared electron density is being shared electron density is being shared

between two nucleus electron density is being shared between two nucleus and just tell me one thing that if i have two nucleus present over here and electron density is shared between these two so what bond it is present we are moving towards the iron we are moving towards the covalent bond right first there was no electron present between the positive and negative ions nucleus but since cation is attracting the electron density of the anion towards itself there is some electron density present between the two nucleus and you are moving you are shifting towards the nature of the covalent bonding okay greater the electron density in between these two nucleus greater is the covalent okay these two these things are if i'll say there are so many things i can say i have to just now write the what is important for you in terms of the examination point of view so just remember this in terms of a cation you have to remember that you have to remember that in the case of

the cation we we have to know the word which is called which is called as polarizing power we have to call it as polarizing power since since cation is distorting the electron cloud of the anion it is polarizing it so the cation will have the polarizing power and the polarizing power which is pp polarizing power is inversely proportional to the size of kit and greater the size of cation less is the polarizing power okay and for the anion for the anion we have to understand the polarizability for the anion we have to understand the polarizability that how easily it can polarize itself polarizability how easily it can polarize itself it's polarizing means that electron density

is being shifted in between the two nucleus it is getting polarized okay this much part is being present between the two nucleus it is getting pulled its electron cloud is getting distorted so for the case of the anion we have to answer the polarizability and the polarizability we are writing polarizability as p star so i will write over here the polarizability is directly proportional to the size of anion greater the size of anion greater it is having the power to get polarized why because if you have a greater size let's say i have a cation over here and i have a anion over here and electrons are present over here and this is the nucleus so they will be easily attracted towards the cation because the distance between these this nucleus which is having the positively charged protons and these electrons is very large so these electrons will be in touch of this cation also that can be easily attracted but in the case of this if you have a smaller size of anion this

nucleus will have a more hold on the electrons of its respective parts so that will not be easily attracted towards the cation and it will not be easily distorted the electron cloud of this small anion will not be easily distorted so you need to remember that greater the size of the anion greater is the tendency of its being polarized or you can say it is having greater polarizability the ability to get polarized polarizability okay perfect now now see written over here polarizing power i have told you pp and polarizability i have told you p star now remember one thing remember one thing greater is the polarizing power i should not write polarizing power i will simply write smaller size of

cation is directly proportional to the greater covalent character inversely proportional to the ionic character and directly proportional to the size of anion so you have to remember every of the thing present over here smaller the size of cation greater is the covalent character less is the greater the covalent character less is the ionic character and greater should be the size so all these things move parallel with each other you have to just remember this much nothing else you need to remember now we'll have some of the uh questions over here let's have some questions over here if i ask you

if i ask you i have lithium chloride i have sodium chloride i have i have cesium chloride all these three are present and i have to tell the covalent character of all these three you know that smaller the size of cation greater will be the covalent character so lithium positive is having the least size over here so one two three this will be the answer and if i take some other question over here i say i have lithium fluoride i have lithium chloride i have lithium bromide and let's say lithium iodide so now cation is same and ion is different just follow the one thing greater the size of the anion greater will be the covalent character so covalent character can be asked over here do not forget to write that what are the factors depending on that greater size of an angle is the covalent character smaller the size of cation greater is a covalent character so do not forget this thing okay

very easy even you can have with the respect of the oxidation state also pbcl2 pbcl4 pbcl2 and pbcl4 this is plus two oxidation state this is plus four oxidation state the covalent character will be greater for the pbcl4 and less for the pbcl2 because the reason is greater the oxidation state even you can even you can have this also factor even i can add this factor also even you can have this factor also i will write over here directly directly proportional to oxidation state and directly proportional to the size of the respective anion because if you have a greater oxidation state if you have a greater positive oxidation state so your size will be less and your size will be lesser covalent character will be more so even you can write over here it is having plus four oxidation strain it is having plus two oxidation state greater the oxidation state less is the size more will be the covalent character and

you will get your answer okay i hope this is clear to you i hope this is clear to you i hope this is clear to you so this is all you need to know about the phajans rule even pseudo-electronic configuration is also having a role in the phajan sul what is that cations having 18 electrons in the outermost shell are considered as pseudo i have already told you polarizing power of cation is in the outermost shell if it is having 18 electrons is more you can simply write as polarizing power of the 18 electrons cation is more let's say for example if i explain you nacl and q plus chloride okay so we will not answer in terms of the anion nacl we are having cucl we are having here we have n a positive here we have copper plus one so any positive is having electronic complication as two eight and if you

write for the copper plus one it will be having argon 3d 10 as the electronic configuration and argon is having if i write the whole one 1s2 2s2 2p6 3s2 3p6 and here i have 3d 10. so this 3s2 3p6 3d 10 is the electron which is 18 electrons cations so the polarizing power is more so if the polarizing power is more greater is the covalent character greater the polarizing power greater is the covalent character greater is the covalent character so i can simply write over here i can simply write over here that my answer will be like this because 18 electron cation is having the greater polarizing power so this can also be asked to not forget you know all the cations which are 18 electron ring plus two cadmium plus two mercury plus two gallium plus three i have already told you all of these and even the silver plus is also a

same example you can understand this with the help of the electronic configuration i have explained you one example okay so let's move forward now we are having the applications of the polarization so the acidic nature of oxides is the major application of the polarization rest any other factor or any other application can also be present but this is the most important one so you need to focus on that see if i ask you a question let's say i have an oxide like i have n2o5 i have p2o5 i have as205 anti-money bismuth and all these things so if i ask you just tell me the order of the acidic strain so what you will say how it will change from top to bottom and similarly if i ask you a question which is n2o and no and no2

so what you will answer me if i ask you the acidic nature from the top to bottom and from the left to right see the size of the cation is increasing over here so i can say the size of the cation i i will simply write over here the top to bottom behavior the size of the cation is getting increased so you can simply say simply say over here with the help of this formula that greater the oxidation state the positive oxidation state or smaller the size of cation smaller the size of cation greater will be the acidic nature just remember this formula and solve all of your questions okay just remember this formula and solve all of your questions so i can simply say size of the cation since increasing so its acidic nature is decreasing so its acidic nature will be decreasing

and what in terms of the oxidation state you know that when you're moving from left to right you're moving from plus one to plus two to plus four so your oxidation state positive oxidation state is increasing when you're moving from left to right okay when you're moving from left to right its positive oxidation state is increasing and you can simply say the acidic nature is also increasing oxidation state positive increasing acidic nature is also increasing right so let's move forward now now we have to move towards the new concept which is the valence bond theory since we have discussed lewis dot theory which has given us so many uh things about how the bond formation takes place how we have we are having the compounds what are the limitations covalent ionic we have discussed so many things now the other theory which was proposed for the formation of the bond versa valence bond theory we will quickly discuss the postulates which are important with

respect to the examination point of view what they are saying as whenever there is a bond formation taking place so first of all what was considered was whenever two atoms combined there are some attractive forces present there are some repulsive forces present if the magnitude of the repulsive forces is more then there will be no bond formation but if the magnitude of the attractive forces is more or even equal to repulsive forces the bond formation will take place okay for example if i write over here as h2 so i have hydrogen over here hydrogen over here okay this is the respective nucleus in which there is the presence of the protons over here okay this is the electron of this hydrogen this is the electron of this hydrogen now i'm going to count the first of all this is the repulsive forces this is the repulsive forces if i ask you about the attractive

forces these are the number of attractive forces so i can simply say that the attractive forces are equal to the repulsive forces so the bond formation will takes place in this case i can say forces of attraction is equal to forces of repulsion so the bond will be formed bond will be formed okay i'm writing over here this is the forces of interaction and these are the forces of repulsions present over here but if i talk about but if i talk about helium 2 you will see in the case of the helium 2 this is the two nucleuses this is the two electrons present over here these are the two electrons present in this now let's calculate the first the repulsive forces this is the repulsive force this is the repulsive force okay these are the two repulsive forces

between the electrons and this is the repulsive force between the protons now if i talk about the attractive forces this is one attractive force this is other attractive force this is one attractive force this is other attractive force so if you talk about the force of the attraction there are only four forces of attraction present and if you talk about the forces of repulsion how many forces of repulsions are present they are present as one two one this two this three this okay so these are the forces of repulsions present over here and even we have to calculate the other repulsions also that we have not considered one this also and other this also so we have over here one two three okay then four and then five so five forces of represent repulsions are present so you can simply say if the forces of repulsion is greater than

forces of attraction you have to you have to consider that bond formation will not be present so i have taken over here crossed over here you are simply seeing over here that forces of repulsions are more so there will be no bond formation takes place so you have to just remember that this kind of a theory was proposed although it was not not considered because all of the molecules cannot be explained by this theory so the next concept came first of all we have to discuss this graph also now what is happening in this graph this graph is considered as variation of potential energy with the nuclear distance here we have on the y-axis potential energy on the x-axis it is internuclear distance in this respective case when i am i want to see the bond length of hydrogen atom i want to see what is happening when the two hydrogen atom is coming closer to each other first they are isolated in this case i will simply write over here that they they are

isolated atoms they are approaching towards each other now they are approaching towards each other i will write over here approaching atoms i will write over here approaching atoms they are approaching each other i will write over here approaching atoms now they have approached they are approached towards the maximum they can be approached towards each other this respective point this respective point is the point at which the bond is forming between the two hydrogen atoms this is the nucleus of one this is the nucleus of two and they have this much of the distance between each other this will be calculated as this distance will be calculated as this distance will be calculated as bond length and if you want to know what this distance is called this distance is called as bond enthalpy if you will break this bond how much energy will be required to break

this point will be bond enthalpy so earlier they are isolated they are moving towards each other each other each other at this respective point the bond formation is taking place the bond formation is taking place and since when they are far they have a greater energy when they are moving towards each other they are having their octet in the case of the hydrogen we call it as due plate the duplet is completed they are stable so the energy is getting decreased when you're moving from this pathway you are moving on this pathway you are moving on this pathway at this respective point at this respective point the energy is minimum energy is minimum why energy is minimum because when h2 is formed hydrogen has one electron hydrogen as other so they have a due plate of electron even octate is also considered as stable and even duplet is also considered as stable the octet and due

plate gives you the stability outermost shell must have either the two electrons or the eight electrons so this is the graph you need to understand okay after that now another concept see first it was proposed that if a forces of attraction is more than forces of repulsion so the bond will be formed but that was not very much applicable to many of the molecules so the next concept which is based on the valence bond theory itself is the orbital overlap concept we have to understand this but before that we need to understand the basic terminologies of this that is covalent bond is formed by the overlapping of the atomic orbitals having the unpaired electrons two atoms their respective orbitals overlap with each other the condition for the overlap will be that is like not applicable every everywhere but since you are understanding the concept so first you should know that the overlapping will take place in such a way that the two atoms the two atomic orbitals must have

one electron each for the sharing okay orbital must be approaching in the proper direction that we will understand and strength of the bond depends on the extent of the overlapping okay all these points are very very important very very important first of all okay the bond strength is depending on the extent of overlapping so how do you define the extent on what factors does the extent depends okay extent of overlapping is actually depending extent of overlapping means at how much you are getting overlapped how much you are getting overlapped what is the extent of the overlapping okay so overlapping depends upon the what type of overlapping exactly you are having okay one is called as the coaxial in which you are having the sigma bond in the coaxial overlapping you are going to form a sigma bond coaxial that means head to head overlapping this kind of overlapping takes place on the internal axis that we will discuss now

after that you have the collateral overlapping which is in the case of the pi bonding that you have the sideways overlap that is not present on the internuclear axis after that another type of the factor on which the overlapping extent of overlapping depends in the types of the orbital see you are forming a sigma bond you are forming a pi bond okay we you people know that the sigma bond is having more extent of the overlapping more stronger bond it is but if we ask in terms of the orbital so even orbital also tells us that what will be the extent of the overlapping see s orbital is non-directional that means that s orbital is having electron density on x-axis also y-axis also and z axis also so its electron density is divided into three axis so every axis is having electron density as one third okay but if i talk

about the p orbital you know that the shape of the p orbital is dumbbell and this orbital is i let's i say this is p x so whole of the electron density of the p x is on the x axis so if i overlap with the any of the other orbital i will have the maximum electron density whole of the electron density of the p orbital present on the x axis but if i overlap any of the orbital with the s orbital let's say if i overlap on the z axis so s orbital is present over here and i overlap this with the p orbital okay this is p orbital this is p z orbital and this is s orbital so the electron density of the s orbital which is present on the z on the z axis is this much only which is one third so not all of the electron density present in the s orbital is participating in the bond formation only the one third is participating while in the case of the p

z whole of the electron density of the pz will be participating in the in the overlapping so extent of overlapping also depends upon what type of orbital you are making s orbital is not using its whole of the electron density in the bond formation because it is diffused it is split in the other axis also so that we have to also consider in our overlapping in the extent of overlapping so remember s orbital is non-directional non-directional means that your electron density is not present only on the one axis it is diffused in all of the axis so during the bond formation only one axis is involved so only one third of the electron density will be involved and directional nature which is in the case of the pdf directional nature which is in the case of the pdf have electron density on the particular axis and that will total will be involved in the bond formation right perfect so we need to remember this also this is something very very important to

understand non-directional s orbital already is having a very less extent of the overlapping next we are having types of overlapping first of all we have head to head this type of overlapping occurs along the internuclear axis okay let's say for example i have s orbital forming a bond with the s orbital this is also s this is also s and they are coming to form bond with each other and this bond is forming on the internuclear axis okay let's say this is z axis i can take any of the axis x y z that is not having any compulsion that you have to take z x or y that is only in the molecular orbital theory right now we can take any axis as the x y or z okay so if the bond is forming on the internuclear axis that respective bond is always called as the sigma bond and that is always coaxial overlapping which is the head to head

overlapping okay sigma 1 formation electron density is present the shared electron density is always present on the internuclear axis sigma bond is stronger than the pi bond that we will understand first the pi bond then we will have the comparison free rotation is possible that also i will explain you after the pi bond okay first of all these are the two major points that you need to understand over here overlapping is occurring along the internuclear axis we will have the example some other examples also but right now i'm explaining you the the basic okay now the collateral overlapping now understand first collateral overlapping which is the sidewise overlapping when two atomic orbitals overlap sidewise okay pi bond formation takes place what is happening over here now see let's say i have two orbitals let's say i have two orbitals one of them is called as px orbital other is also

called as px orbital and i'm making them overlap on the z axis so this is p x this is p x okay so the overlapping region is having electron density present over here this is the overlapping electron density above and below the internuclear axis internuclear axis above shared electron density internal axis below shared electron density okay so this bond is called as a pi bond okay the bond is not present on the internuclear axis this is not head to head overlap head to head overlap always occur on the internal axis since the bond is not formed on the internuclear axis it is not considered as the strong bond this will be the weak bond okay so uh overlapping is perpendicular to the internal axis you have seen that electron density above and below the internal axis you know that no free rotation is possible since

let's say i have this type of a bonding in my pi bond above and below so i cannot rotate this if i rotate this this bond will be broken this bond will be broken and we are not talking about this type of rotation they are not talking about this type of rotation they are talking about this rotation can you break the can you rotate this wand yes you can rotate but the bond will break okay so there can be no rotation possible for the pi bond okay understand this this is your pi bond formation above and below the internuclear axis this is a weaker bond since sigma bond is on the intelligence stronger free rotation is possible because it is a simple bond so there will be free rotation you can revolve it in the either way you can there will be no uh breakage of the bond okay perfect next now the sigma bond can be formed by these all types of the overlapping let's

say i have ss overlapping you can have ss overlapping either on the x-axis y-axis or the z-axis there is no issue in that whatever axis you take you can have the sigma bond okay if you take the x axis you will have the sigma bond formation like this if you have the y axis you have sigma bond formation like this if you have the z axis you have sigma bond formation like this this is x axis this is y axis this is z axis you always have a sigma bond formation like this sp overlapping now understand what do you understand by the sp overlapping you have s p overlapping and you can have this overlapping like this this is s overlapping with the p now you have to remember that if you are taking the x axis this orbital must be of the p x only okay electron density must be present on the x-axis itself let's say if i take

with this on the x-axis if i take the p-y orbital if i take the p-y orbital this is zero overlapping this type of overlapping does not take takes place and if i consider this type of overlapping which is on the z axis so this will be the case even this is considered as the zero overlapping which is also incorrect okay so you have to remember that when you're having the sp overlapping if you have x axis y axis z axis so you have to take the p x you have to take the p y you have to take the p pz if you're taking any other axis and any other p orbital there will be no bond formation this is called as this is called as this is called as zero overlapping that is no overlapping this is called as zero overlapping these two are considered as 0 overlapping this is not any kind of an overlapping now we have

the pp overlapping okay now when you have the pp overlapping so what happens since when you have the pp overlapping there are some of the cases which you need to remember that is you can have p x plus p x you can have p y plus p y and you can have p z plus p z first of all you need to remember that always p orbitals overlap and the two p orbitals will be of the same axis they can never be different of the axis okay so if you have the same axis present x y z always there will be the formation of the sigma bond always there will be the formation of the sigma bond let's say for example if i have this is p x overlapping with the p x this px is overlapping with the px on the x axis always there will be the formation of sigma bond right similarly if i have on the y axis

if i have on the y axis this is p y this is p y and my axis is y similarly you can have on the z axis okay this is p z this is p z overlapping on the z axis overlapping on the z axis always and always there will be the formation of the sigma bond but if you take but if you take this let's say you overlap p x plus p x okay p y plus p y or p z plus p z so always remember that p x plus p x on either of the y or z axis either on the x or z axis or either on the x or y axis always this will be having the pi bond formation always if you have a different axis on which you are having the overlapping so you will have the pi bond let's say i will give you one of the

examples like you have a like you have a y axis over here and you are having the two px orbital overlapping with each other so this will be the overlapping and this is the internal axis so this is a pi bond this is not present on the internal axis the overlapping is not present on the internuclear axis okay okay so this is pi bond and these two are your p x orbital let's say i have p y p y okay this is your p y p y and you are getting overlap on the x axis so you will have the pi bond this is p y orbital this is p y orbital there will be the formation of the pi bond so this table is required to be understood that what should be your overlapping access what type of a bond will be formed how you will understand that this is something very very important very very

important and the question can be asked from this this can be asked okay just have a note down just have a note down access of overlapping okay now let's move forward pi bond now let's discuss the pi bond

pi bond can be formed by the pp overlapping pd overlapping and even the dd overlapping can also form the pi bond okay let's have a pp overlapping you know all these things i have told that how this pp overlapping takes place i don't need to write anything else over here we have already discussed all of these things these things that how the pi bond formation will takes place now we will have the pd overlapping let's say i have an [snorts] orbital over here this is my z axis and this is my p orbital and this is my d orbital let's say i'm making this orbital as dxy and you are overlapping with the p orbital okay so this type of a bond

which is forming over here first of all present above and below the internuclear axis will be considered as the pi bond okay if i make the axis over here let's change the color this is x-axis this is y-axis and here it is present as the z-axis so this respective d orbital will be considered as this is your p y orbital and this respective will be this is x-axis this is y-axis so this will be considered as d x y so this overlapping will be d x y p y overlapping okay d x y p y because this is your respective y axis you have taken the p y orbital you know that how the shapes of the d will be defined you know all these things if you have the x and y axis present this orbital will be called

as d x y if you have y that you it will be called as divisor if you have x it it will be called as the exit orbital okay so this type of overlapping is sideways overlapping and it is also considered as the pi bond it is also considered as the pi bond d x y and the p by okay even you can have the dd overlapping also you can have the dd overlapping also i can write dd since this these are not very much asked in the examination but at least you should have an idea that how the dd type of overlapping takes place like let's have a look over here this is your one of the d orbital this is your other d orbital okay and we have the axis present like this okay this is your y-axis this is your x-axis and this is your overlapping region so we can simply say we can simply they

say that d x y dxy they are forming bond with each other and the bond formed over here is called as the pi bond the bond formed over here is called as the pi bond okay just remember this thing nicely now what we are having is extent of the overlapping whenever you are talking about the whenever you are talking about the extent of overlapping you need to remember that in the case of the pi bond always the pd overlapping extent is more than pp overlapping and and if you have the greater size of the orbital the overlapping extent will be less i am i am explaining you this thing okay let me see if i have written afterwards

let me see if it is not written afterwards i will explain you now itself just a second let's uh first understand this okay now what is happening over here is whenever there is a formation of a pi bond when you are moving from top to bottom whenever there is a formation of a pi bond and you are moving from top to bottom strength of the pi bond decreases for an example if i write sometimes there is a question that is coming that y n2 is having the pi bond but p2 is not present instead of that we have p4 present similarly question is asked o2 is present why s2 is not present why s8 is present and these kinds of questions are coming in your p block chapter which is related to this respective topic now what is happening over here is let's say if i move from

nitrogen to phosphorus to arsenic to antimony to bismuth you know that the size of the orbital increases you know that size of the orbital increases you know that size of the orbital increases so the size of the p orbital will also increase so what is happening over here if i form a pi bond with a smaller size of the orbital and if i make the pi bond with the bigger size of the orbitals from the smaller size if you are moving to the bigger size you have to always remember that your extent of overlapping your extent of overlapping decreases your extent of overlapping decreases so this respective bond will be considered as weak and this respective bond will be considered as strong so if you are trying to make sim the same if you are you are making n2 you are having a triple bond but if you if you are trying

to make p2 with the same triple bonds the bonding will be like this the p orbital will be having the larger size the extent of overlapping will be less what i'm saying is the extent of overlapping will be less the electron density coming in the internuclear region where the sharing is taking place is less because because the size of the orbital is very large the electron density is diffused over the space the greater space greater the electron density will be diffused and less it will be present in the internuclear sharing region right so the bond will be weak because greater the electron density is present in this shearing region more stronger is the bond but in this case because the size is more the extent of overlapping will be less bond is weak you cannot form you cannot form this p2 you cannot form this pi bond effectively instead for that you are having this p4 you have this polymeric structure you have white phosphorus red phosphorus you know the structures of these two right p4 exists like this p4 exists like this if i talk about the

white phosphorus it exists like this s8 exists like this it is having the crown shape same as the reason it can also not form the o2 type structure because of the greater size of the orbitals and the extent of overlapping will be less so the extent of overlapping i will write in the form of formula over here simply that extent of overlapping is inversely proportional to the size of orbitals is inversely proportional to the size of orbitals and you don't you know you are not required to forget this okay size of the orbitals just remember this thing and you can simply say size is defined in the form of the principal quantum number if size is more principle quantum number will be more if size is less principle quantum number will be less for that

itself we have a order over here and now the pd overlapping will be greater than pp overlapping because greater is the electron density present in the overlapping region greater is the strength of the bond overlapping it defines us the strength henna extent of overlapping defined strength stronger will be the bond and greater the n extent of overlapping strength extent of overlapping is inversely proportional to n also henna it is inversely proportional to n also here you have see pd will be having more pd will be having more as compared to pp but 2p 3d 2 is having less number of principal quantum number so its size will be less so its extent of overlapping will be more size is less extent of overlapping is more sizes less extent of overlapping is more so you have to remember that it depends upon the two factors over here okay one is

the n and other is pd and pp and even you have to remember one more thing even you have to remember one more thing that in the case of the that in the case of the pi bond we always consider that pd overlapping is greater than pp we always consider pt is greater than pp and in the case of the normal any kind of a bond let's say sigma always pp overlapping is greater than ps overlapping is greater than ss overlapping same same reason because you know that p orbital is directional s orbital is not direction non-directional non-directional it is so here you have greater extent of the overlapping because all of the electron density is present on the same axis whereas if you have s and p overlapping only the one third electron density of the s will be involved in the bonding and in the case of the s s if they are getting overlapped one third of this and

only the one third of that so the bond will be the weak one okay so you need to remember this very nicely you need to remember this very nicely that these are the important things you need to remember in the case of the pi bond and in the case of the sigma bond and obviously the greater the size less is the strength of the bond so you can relate that with the principal quantum number n okay so i hope you people have understood what i want to tell okay let's move forward now uh there is a new type of bond you need to understand that that is the delta bond that is the delta bond and understand that now what happens when you have a delta bond present with you now let's see first up first of all this is the four lobe interaction ma'am this is four lob interaction so what is this slope

interaction please tell us that also this is this is respectively two lobe interaction okay one interaction this second interaction this okay what about this four lobe interaction now understand let's say i have an orbital present with me which is dxy this is y-axis this is x-axis okay and this is present below and this is present above okay this is y-axis this is x-axis okay one is like this other is like this okay behind each other x and

uh the dx y and the dx y okay just remember so if this is x axis this is y axis there will be an axis present over here there will be an axis present over here okay this axis will be called as the z axis okay this is x this is y this is x this is y and this is the z axis present x axis this is x axis this is y axis x axis and the y axis and the z-axis will be present over here that is pointing towards you okay remember this thing very nicely so what is happening is they are interacting with each other through this z-axis one is above one is below and they are interacting with each other with this respective axis present in between these two which axis will be called as the zx that axis will be called as the z axis

okay remember this thing very nicely now what is going to happen over here is what is going to happen over here is this dxy combining with this dxy or maybe dx square minus y square so this slope will be interacting with the this slope this this slope will be interacting with this one and this will be interacting with this one so all the four loops will be interacting all the four loops okay so this is called as the four lobe interaction okay so what type of a bond will be formed over here how we can show that this will be the picture present over here okay we can have a color we can have a color

this red color combining with this red color okay this pink color is combining with this pink color this not green this blue is combining with this blue and this let's say green this green is combining with this green so this type of shape will be in terms is present in front of you four lobe interaction formed uh by dxy and dx square minus y square axis of interaction is always red axis of interaction is always set so these types of bonds which are forming as a result is called as the delta bond you need to remember that yes there is a

delta bond also present let's say if i have a a atom and if i have a a atom and i have four bonds present over here how you will define them one of them is sigma other is pi one then another pi two and then there will be delta bond whenever there will be the four bonds you will have this kind of a structure okay so do not forget this do not forget this this is very very important very very important okay so let's move forward after this now we have to understand what all bond parameters are present what is called as bond length bond length is defined as the internuclear distance between the two atoms bond strength how much strong a bond is after that we have one disposition energy how much energy is required to break the bond and bond order that is the number of bonds present between the two atoms okay these all are the bond parameters so what is the what are the factors which affect the bond strength first of all we have

the number of shells greater the number of shells greater the number of shells less is the bond strength less is the bond strength you need to remember this like for an example the one s 1 s bond is stronger than 2 s 2 s bond and that is stronger than 3 s 3 s bond the meaning of this respective factor is this their external nature yes when the principal quantum number n is same the shell is same the size is same so we have to consider in the case of the sigma ss overlapping is the least one then we have the sp and then we have the pp because we are just answering on the basis of the directional nature i have already discussed with you and if you are talking about the pi bond always pd overlapping is greater than pp overlapping always the pd overlapping is greater than pp overlapping let's say for example i have

we are having the examples waiting for us so let's move towards examples which bond is considered as the most stable we have over here 2s2s 2s2p then we have then we have 3s3s and then we have 3p3p so i will tell you the order also and i will tell you the meaning also what will be the answer first of all you have to give the importance to the n first of all n will be given the importance so you have to first answer according to the two is greater than s the bond strength of the two will be greater than the third one so in between these two and these two first you have to answer 2s and 2p will have the the stronger bond as compared to the 2s2s after that you have to look into the s or the p for the third shell and you know that 3p 3p will be greater than 3s3s so always give importance to the

principal quantum number first after that you have to move towards the next answer right let's move forward which bond is having the minimum strength now they're talking about the minimum strain that means that the bond should be the least stable okay 5s 5 is 3p3p 2s2s 2p 3d okay now they are talking about this sideways so sideways is always your pi bond it is always your pi bond and this is your sigma bond so if they are not writing the side wise so you have to understand that that is the sigma bond okay so always and always so always and always sigma is having greater strength than pi so answer is between is in between 3p 3p and 2p 3d both are

sideways so which bond has minimum strength that is the least stable one so obviously this pp is having first the less strength as compared to pd and its principle quantum number is also more so you can simply answer the question 3p3p right its principle quantum number is more and even the overlapping sidewise pp is less than pd i hope you have understood now we have for this example we have 3s3s then we have 4s4s then we have 4p4p and then we have 4s4p 1 2 3 4 we have all of these four examples present minimum strength now what you will answer see this is uh less in size so we have to answer in these three only we know that s p sp and ss so it's very easy to answer

i hope you have understood and these kinds of questions are important these can be asked okay let's move forward even the bond strength depends upon the number of lone pair present on the atoms and this is a very important question even in the p block many questions are coming from this respective part itself okay so number of lone pair of each atom also i should write of each atom or on each atom whatever you want to write henna so of each atom or on each atom whatever you want to write okay so let's see the bond strength will be like this now why if i make the bond of n n if i make the bond of o o and if i make the bond of f what will be the case four electrons present and one two three four five six

one two three four five six in the case of the fluorine one two three four five six one two three four five six now what you need to understand is how many lone pair are present on each item how many lone pair are present on each atom or how many number of electrons are present on each atom greater the number of electrons present on each atom weaker is the bond weaker is the bond this respective bond is having maximum repulsion they are having maximum repulsion between between lone pair of electrons so this bond is very weak you can just supply very less amount of energy and it will break so

maximum strength is of the nn bond greater the number of lone pair on each atom less is the bond strength you can simply write over here you can simply write over here greater the number of electrons on each atom less is the bond strength less is the bond strength greater the number of electrons on each atom less is the bond strength okay so same is the case going with the this respective part this and this so here we have first pp then we have nnn then we have arsenic and then we have anti-money same is the case of the ss then oxygen oxygen then selenium selenium and then tellurium same is the case

and here we have clcl then we have brbr then we have ff and then we have ii all of these results you need to remember very nicely all of these results you need to remember very nicely they are asked in most of the examinations and this is based on that only nitrogen being very smaller in size cannot accommodate that much lone pair of electron so its bond will become very weak it is coming on the second position only remember in the case of the f which is the most exceptional very small size very high number of electrons present on each atom very less amount of energy will be required to break the bond okay so remember this thing these all orders are of the bond strength yet all these orders are of the bond strength we are talking about

bond strength okay bond strength and they can ask you that what is the bond dissociation energy so greater is the bond strength greater is the bond dissociation energy stronger the bond more energy will be required to break the bond okay moving forward now valence bond theory after that we have the hybridization so i'm taking just a one minute break you can have your water i'm also having now the hybridization

valence bond theory we had we were able to explain the molecules many of the molecules were explained with the help of the valence bond theory but there was a problem with the methane okay now let's discuss why we require hybridization let's make a molecule which is methane valence bond theory how does it explains the formation of the bond okay let's say we have the carbon electronic configuration 1s2 2s2 2p2 if we have the excited state of the carbon we have 1 s 2 2 s 1 2 b 3 1 s 2 2 s 1 and 2 p 3 so if i make the orbitals if i make the orbitals it will be like this 2s1 and then we have 2p3

okay so i will bring the hydration towards carbon one hydrogen other hydrogen the third hydrogen and the fourth hydrogen this is forming bond with this this is forming bond with this with this and with this all these are forming bonds with each other okay no problem this overlapping is called as ss overlapping since the electronic configuration is 1s1 this is 1s1 this is 1s1 and this is also 1s1 this is sp overlapping this is also sp overlapping and this is also sp overlapping so i have written all of the overlapping here okay so let's make a diagram this is carbon this is one of the p orbital of carbon other p orbital of carbon and the third

p orbital of carbon okay and i'm also defining the axis over here this axis x-axis and the y-axis and the z-axis this is your y-axis this is your z-axis and hydrogen is coming to overlap on the x on the y on the third let's define this as p x this defined as p y and this defined as p set okay there is a overlapping with the hydrogen no problem now if i make the bonds the bonds will be like this this is one p orbital other and the third one this is carbon combining with all of the hydrogen atoms so by looking at this i am able to see that the bond angle over here is of the 90 degree according to the valence bond theory if i overlap this i am getting the bond angle as 90 degree but this is not correct since you know you know that that this this he is

saying that the bond angle of methane is 90 degree the valence bond theory is explaining the bond angle of methane as 90 degree but we know the ir spectroscopy explains that the bond angle in the methane is not this it is actually 109 degree 28 minutes so how it is possible this bond angle is not correct actual bond angle should be 109 degree 28 minutes but that cannot be explained in the terms of the valence bond theory for this limitation first time we saw that this is not applicable to these types of the molecules like so we can say that valence bond theory is not applicable for all of the molecules so we need something else to explain the structures of the molecules so here it comes the hybridization hybridization is totally hypothetical phenomena there is nothing like hybridization present around us but

it's like uh it's like it's like a it's like a disease someone is having in their brain to explain the structure of the molecules but nothing like that is present around us but just do not listen to all of these things just listen that what mam is trying to explain right now yes hybridization is present and yes it is present now let's discuss what is hybridization i hope you people are enjoying i hope you people are learning new concepts do write in the comment section that how was the lecture okay you might be learning something new something very important some tricks some concepts so do mention in the comments section okay let's discuss now hybridization first of all hybridization is the phenomena what we do there is a mixing of the pr atomic orbitals there is a mixing of the pure atomic orbitals to set the to form

the new set of the equivalent orbitals they are called as the hybrid orbitals what is happening there are some atomic orbitals okay there are some boxes which are having some different energies one is having less energy one is having more energy i will mix all of the boxes with each other and i will do something with them okay i will just mix with the with each other i will just roll and i will do something and i will just give you something else these four orbitals or any number of orbitals which i have mixed i will give them some other shape i will give them some other energy now all will have the same energy okay first of all the meaning of the first postulate is this next is number of hybrid orbitals formed is equal to the number of atomic orbitals combined if you are combining four boxes with each other four atomic

orbitals of each other of the different energies you will after the mixing get hybrid orbitals of the equivalent energies all of the orbitals which you have mixed the hybrid orbitals will also have the same number okay hybrid orbitals are represented as like this they are always represented like this one lobe will be the smaller one and other lobe will be the bigger one why so because you know whenever it combines let's say i combine s with p so i will get this shape now why this is s and this is p now this complete portion will overlap with each other to give you this one will be the bigger and other will be

the smaller the shaded region will come here the shaded region will come over here so this is your hybrid orbital in which electron density is present completely or i should not say completely but 95 percent of the 98 of the electron density is present over the one side okay snb orbitals combines to give you this hybrid orbital okay now hybrid orbital only form the sigma bond do not forget this point i will explain you further hybridization is hypothetical i have already discussed okay now now let's discuss first of all example of the hybridization sp okay

i have an example let's say be f2 now i want to make this example i want to make this respective molecule first of all i will write the electronic configuration of beryllium 1s2 2s2 this is the electronic application of beryllium 1s2 2s2 okay now i will make the excited state of the beryllium this will be 1s2 2s1 2p1 okay so the 2s will have one electron and the 2p will have the other electrons okay this is s let's say this is px this is py this is pz now these two atomic orbitals these two atomic orbitals will combine with each other to give us two hybrid orbitals to give us two hybrid orbitals now these two hybrid orbitals will attach or you know

they will be rearranging themselves in such a way that they have minimum repulsion and maximum stability if i call that these here the presence of electron densities over here now they are present in such a way that their electron density is like present like this so it is not repelling each other this electron density is present in such a way at the angle which is this electron density is present on the angle which is 180 degree which is 180 degree there is no repulsion between this electron density okay so this is the maximum stable position over here i will write over here that hybrid orbital i will write over here that hybrid orbital arrange themselves the hybrid orbital arrange them self in

such a way in such a way that they have maximum stability they have maximum stability and minimum repulsion they have maximum stability and the minimum repulsion okay so this is how this is all we are talking about the beryllium over here we can also write a point point over here that hybridization is a property of the central atom we can write over here that hybridization is the property of the central atom this is the property of central atom it is not property of the whole molecule this is the property of central atom we always talk about the hybridization we always talk about this mixing and all these things with respect to the central atom now the fluorine is coming into the play electronic configuration 1s2 2s22p5

the 5p electrons are present over here how many unpaired electrons it is having there is one unmet electron and that will be involved in the bonding now how the bond formation will takes place the beryllium is coming towards the fluorine this is complete beryllium this is beryllium and the fluorine coming over here is like this the orbital of fluorine which is going to form the bond let's say this is p z p z orbital let's say this is p z orbital this is your z axis and this will be considered as beryllium this is your beryllium this type of overlapping is s p p overlapping this is also sp then why are you calling it as sp please explain this also now listen students what is happening over here one s orbital is combining with the one p orbital so you are having two atomic orbitals combining so you have two hybrid orbitals okay so

this is one sp orbital this is other sp orbitals there will be two hybrid orbitals right there will be two hybrid orbitals there will be two hybrid orbitals present over here sp and sp they will connect with it with each other in such a way that they will have a this type of a bonding so i have explained you bef 2 sp hybridization okay i hope you people have understood now we will discuss that how the sp3 hybridization released the or removed the uh problem which was associated with the methane you know that methane was having the problem of bond angle you know you know that methane was having the problem of bond angle we were not getting the correct value of the bond angle now listen how it was why it was present methane we have the electronic configuration of

carbon as 1s2 2s2 and 2p2 now what is happening you have the carbon in the electronic in the excited state as to s1 2p3 so this is your 2s1 and then we have the 2p3 present like this now the four atomic orbitals combined with each other these all four atomic orbitals combined with each other one s and this all these three p will combine with each other to get us four hybrid orbitals all of them will be called as the sp3 okay you are going to get the four hybrid orbitals you are going to get the four hybrid orbitals and all of them will be called as the sp3 hybrid orbitals now they will arrange themselves now they will arrange themselves in such a way that they have minimum repulsions present in them

and maximum stability present with them minimum repulsions and maximum stability and the bond angle over here will be 109 degree 28 minutes this will be the bond angle 109 degree 28 minutes will be the bond angle of the methane of the methane molecule okay so i don't have to do anything hybrid orbitals are arranging themselves in such a manner we are getting the bond angle as 109 degree 20 minutes so the problem is already solved we don't have to do anything now what is happening now hydrogen is coming now what is happening now hydrogen is coming now hydrogen is coming beta and having the electronic configuration as minus one so no problem present just make the carbon first of all i have told you that the hybrid hybridization is the property

of the central atom so all of the hydrogens are present with the carbon with the bond angle being 109 degree 28 minutes okay the bond angle is 109 degree 28 minutes 109 degree 28 minutes so this is hydration this overlapping is sp3 s overlapping all the four overlapping will be called as sp3s overlapping since we have discussed this over here we have discussed this respective beryllium f2 over here we have discussed this beryllium f2 over here you know that this hybrid orbital which is arranging itself in the angle of 180 degree is having a geometry it is called with the name this geometry is called as linear how we call this as linear because this 180 degree angle the representation the hybrid orbitals

formation was was called as the linear one with the bond angle being 180 degree similarly is the case with the methane okay so it was also associated with the geometry then and that was called as the tetrahedral so you know these names so how it was considered as tetrahedral the one which is having the angle 109 degree 28 minutes with the four bonds and how many number of total tetrahedral angles are there total tetrahedral angle total tetrahedral angles present over here is six how they are six i will explain you this also just note down just a second see

now this is the molecule of methane present over here okay this is a molecule of methane present over here okay one bond is above and other three are present down you can see this molecule so one is the bond angle over here one two three three bond angles are between this green one and all these three white ones three and other three are present in between the below one one over here one between these two other between these two and other between these two so three bond angles from the green and white and three bond angles between the white ones the total number of tetrahedral angles are coming out to be six so do not forget that okay so we have uh completed the problem of the sp3 hybridization which was present in the valence bond theory okay now let's move forward let's move forward with the

sp3 dehydration because this is also something important and we need to discuss this so we have an example of a molecule which is pcl5 the electronic configuration of phosphorus being 1s2 2s2 2p6 3s2 3p5 outer most orbital is 3s and 3p that is the three so electrons which will be involved in the bonding is from the outermost shell now since unmet electron that is having is only one two three four five not three p five beta this is not three p five this is three p three okay so the electronic configuration we are having over here is 3p3 so the number of orbitals the number of unbeared electrons present over here is only three so if you are forming the five bonds you must have the five unpaired electrons so you have to

make the phosphorus in the excited state so this is s orbital one of the electron will be excited to the d orbital okay so now you have how many unpaired electrons now you have this much number of unpaired electrons you have this much number of unpaired electron one s this is p x this is p y this is p z and this is d maybe d x y d y z it can be anything but you have to remember in the case of this hybridization it will be dz square okay so all of these orbitals combined with each other how many atomic orbitals are combining five so there will be five hybrid orbitals and all of them will be given the name as sp3d hybrid orbitals okay and that will combine all of them will combine in such a manner that will be that will be looking like this that will be

looking like this okay this is how they will look so if i uh so if i uh like uh convert them into two parts it will be looking like this okay it will be looking like this okay this is the shape of the sp3d5 hybrid orbitals they will be attached attaching themselves in such a manner they will be attaching themselves in such a manner okay fine now what is happening over here this respective above and below one is having p z combining with the dz square this is having s plus p x plus p y okay so you need to understand that this d is not any of the d one it is d z square okay understand this these bonds these bonds i should make a different color these bonds which are present over here

which is with the green one okay these bonds are called as the axial bond whereas these pink bonds are called as equatorial bonds you know all these things these are known as axial and these are known as equatorial bonds always remember always remember that your axial bonds are longer than equatorial bonds your axial bonds are longer than equatorial bonds yes there are some exceptions but i am just telling you the usual case like 99.9 percent of the cases follow this trend okay why excel is larger than equatorial because that is having the overlapping of pz with a dz square okay which gives us which gives it the larger bond length since you can simply see that in the case of this this is your p z and this is your d z square combining with each other on the same axis so the greater electron density is present on this respective z

axis of the bond and so the bond length will be larger in this case okay the axial bonds will be larger than the equatorial bonds okay remember this thing very nicely after that this arrangement of the hybrid orbitals is giving it the geometry as tbp which is called as trigonal bipyramidal it is called it is called as trigonal by pyramidal it is called as trigonal bipyramidal how many types of a bond angles are present there are 120 degree bond angles 90 degree and 180 123 96 and 181 how now understand now this is your respective trigonal bipyramidal now why do we call it as trigonal bipyramidal three bonds are present on the same

axis one is above and one is below okay so trigonal pyramidal trigonal bipyramidal okay now these all white are forming how many bond angle with each other they are forming the bond angle as 120 degree so how many bond angles of 120 degree are present over here one two and three if i ask you how many 90 degree bond angles are present one two three the green above one is making three with the white ones and the below this green is making 90 degree angles three with the white ones so this much 90 degree angles three this much 90 degree angles three so three and three we are having how much 6 and 180 is formed between these two green ones 180 is formed between these two green ones okay i hope you will understand all

these things and this is important okay let's move forward after that associated with it is a rule which we call it as the bench rule so see this rule is very deep to understand but just remember the result of this okay i'm just explaining you just right okay ben's rule now what is happening with the ben's rule remember i'm just telling you the result apply the result okay because we are having

the one short lecture i cannot explain you the the whole one just go through the result more electronegative atom is attached to the hybrid orbitals having the more percentage p character and less percentage s character okay more electronegative atom okay no problem we have these type of the bonds which is having the which is having the pd overlapping and we have these type of a bonds which have sp3 overlapping i have already told you so what they're telling is more electronegative atom is attached to the hybrid orbital having the more percentage p character so here it is having 50 percentage p character here it is having how many 66.6 percentage of the p character so the more electronegative atom is attached to the hybrid orbitals having the more percentage of p character and less percentage of the s character so

where here we are having at least 33.3 percent of the s character but here we are having no percentage of s character i think you people have understood that if i'm having the shape as like this this one and two bonds are i will be attached with the electronegative atom why because the percentage is correct over here is zero then lone pair is attached to the one which is having the more percentage s character so you know this also so you know this also sat this is like lone pair you have to attach over here because here you are done with the sp3 at least you have percentage s character as 33.3 at least you have percentage s character as 33.3 at least 33.3 percent will be the percentage s character okay

okay now after this what we have over here is let's say if i want to have a molecule as pcl3 f2 okay how you will make this molecule let's say i have to make this molecule so always remember that electronegative atom is always is always coming on the axial position on the axial position always and always remember that lone pair has to come on the equatorial position you don't have to forget this always okay so you have to make this bond pcl3 f2 like this like let's say i have a molecule over here i o2 f2 negative so how you will make fluorine will always come on the on the respective axial bonds and rest of the molecule if it is having lone pair it will be on the

equatorial position always io2 f2 negative pcl3 f2 whatever type of molecule you make you have to follow these two rules and please don't forget this okay let's move forward okay important points that is in sp3 dehydration lone pair always occupy equatorial position i have already discussed electronegative atom always occupy axial position axial bond is longer than equatorial this hybridization is divided into sp2 and pd part which is all the things i have explained these all are important okay let's move forward now there are some hybridizations which are planar what is the planar hybridization that means all of the atoms are present in the same plane okay so we have sp and sp2 as always planar see sp is having the shape like this though so they are

always planar all of the atoms will be in the same plane always and if you talk about the sp2 they are like this so they are all also the planar ones if you want to see the structure i can show you the structure also see this is your respective sp2 so all of the molecules all of the molecule is planar all of the atoms are coming in the same plane all of the atoms are coming in the same plane if you'll cut this all of the atoms will be the part of that knife okay so this is also plainer linear will be planar because always all of the atoms will be in the same plane and if you want to have other hybridizations which are planar so sp3 sp3 dsp3 d2 all of them are also planar but only there is a condition when there are two lone pair present okay then only they will be okay

so if you have sp3 if there is two lone pair present it will become bent so the bend form will have the planar hybridization sp 3d if you have the two lone pair it will become t shape sp3 d2 if you have the two lone pair it will become no sp3 d2 if you have the two lane lone pair it will become square planar okay and if you have sp3 d3 and you have two lone pair it will become pentagonal it will not be pentagonal but with the two lone pair it will be you can simply write as pentagonal because all of the above and below positions will be replaced by the lone pair okay so all of these hybridizations which is accommodated with the two lone

pair of electrons will be called as the planar for example if i say if i say in the case of the tetrahedral if i say in the case of the tetrahedral okay in the case of the tetrahedral if i have this is tetrahedral okay this is tetrahedral this is the molecule of the water these two are the lone pairs lone pairs so this whole molecule with the blue and the red one is considered as a same as triangular planar i'm just removing one of the bond so this molecule is now the planar molecule this molecule is now the planar molecule okay now if i talk about the octahedral if i talk about the octahedral so it will be this is octahedral okay there was one lone pair above there was one lone pair below now i cannot see the lone pair spectrometer cannot see the lone pair so it will be square planar there are lone pair present over here

but spectrometer cannot see that there is lone pair present over here there is lone pair present over here but this is square planar so you have to understand that all these hybridizations with the two lone pair of electron is considered as the planar ones okay this question can be asked hybridization in the case of the odd electrons pc now what is the answer for this question how you will identify that what will be the hybridization in the case of the odd electron species you see there are two cases if electronegativity of the surrounding atom is less than central atom okay so the uh but before that but before that i need to explain you one more thing how you calculate the hybridization before that you need to understand this okay see the formula for the calculation of the hybridization is number of sigma bonds plus number of lone pair which are localized localized means that

do not participate do not participate in resonance do not participate in the resonance that is the meaning of the localized electrons so what is happening over here let's say i have a molecule bef2 if you want to simply see the what is the hybridization just make the structure of beryllium f2 just make the structure of beryllium f2 how many sigma bonds are there two so the hybridization will be sp let's say i have an example of the bf3 okay i'm giving you the very easy easy examples bf3 how many sigma bonds are there three sigma bonds are there so hybridization will be sp2 let's say i have another example as so2 so if i make the structure of so2 there will be one lone pair present over here

and this is your structure so there will be only sigma 2 and there will be lone pair as one so the hybridization will be sp2 the hybridization will be sp2 this lone pair is not taking any participation in the resonance so it will be counted it will be considered as a localized lone pair okay after that let's say i have so3 so the structure of so3 will be like this how many sigma three sigma so the hybridization will be sp2 how many sigma there are three sigma so the hybridization will be sp2 now five i have let's say pcl5 so make the structure of pcl5 one chlorine above one down and you have over here five sigma so hybridization will be sp3d hybridization will be sp3t okay after that i have let's say i have let's say this example nitrogen and hydrogen over here so this

nitrogen is having three sigma and one lone pair so four is coming so the hybridization will be sp3 four is coming so hybridization will be sp3 but let's say i have a case over here like this this nitrogen is present lone pair is present and there is a pi bond also present so this respective lone pair of electrons can be used in the delocalization so these cannot be counted in the hybridization okay so there are only three sigma so the hybridization will be only sp2 so you need to understand what is happening you need to understand what is happening you need to understand what is being asked in the question what is the formula you need to remember all of these things okay so this is the way in which you have to calculate first of all the hybridization and now i will tell you

that how we have to answer the hybridization of the odd electrons pc now you make the ch3 radical okay now ch3 radical if i make normal the structure of the ch3 radical it will be like this how many sigma there will be three sigma but if the electronegativity of surrounding atom is less than the electronegativity of the central atom central atom is carbon surrounding atom is hydrogen if the electronegativity of central atom is more and the surrounding atom is less so this lone electron will not be counted so only three sigma is present and the hybridization sp2 is present only this much but if you have clf three right and the electronegativity of surrounding atom is more than the electronegativity of the central atom you have the structure like this this is not clf3

this is cf3 this is not clf3 this is cf3 this is cf3 radical cf3 radical fluorine over here carbon over here fluorine over here carbon overhead this lone electron count three sigma and one electron number is coming out to be four and the hybridization will be now sp3 because this electron will be counted as in the hybridization same you can answer for the case of the no2 also see the structure of no2 is like this the structure of no2 is like this and the electronegativity of the surrounding atom is more than the electronegativity of the central atom so the hybridization will be you have two sigma plus one electron so you have hybridization as sp2 do not write the hybridization of no2 as sp you have odd electrons pc so the electron will be counted present on the nitrogen because the electronegativity of the surrounding

atom is more than the central atom okay let's move forward now one more thing now this table is very very important this table is very very important go through this table types of hybridization how many types of hybridizations are present how many types of hybridizations are present you have sp you have sp2 you have sp3 you have sp3d you have sp3 d2

and you have sp3 d3 all these types of hybridizations are present how many orbitals s plus p shape is linear how many percentage s character how many percentage p character you can simply calculate sp divided by 100 there are two numbers s and p you will get 50 50 sp2 you have three number 100 divided by three you have how much s and three p total four hundred divided by four you can calculate the percentage s character p character d character all of this whole table is very very important please remember in the case of sp3d which d orbital is involved sp3 d2 which d orbital is involved sp3 d3 which d orbital is involved you have to remember everything you have to remember everything every of the single thing you have to remember in this okay now moving the moving towards the next point now you have a question hybridization of the

solid molecules hybridization of the solid molecules pcl5 pbr5 and if5 see pcl5 exists in the form of pcl4 positive pcl6 negative this is solid pcl5 i am not talking about the gaseous pcl5 that is totally trigonal bipyramidal but in the solid state it exists like this pbr5 exists like p b r 4 positive and b are negative p b r 4 positive and b are negative now what happens is i cannot write over here pbr6 negative because phosphorus is having very less size it cannot accommodate six bromine atoms

around it so the pbr6 negative is not possible so only the br negative will be [clears throat] present now moving towards the next point which is the number of p pi p pi and the p by d pi bond sometimes this question can also be asked so always remember s p can have only two pi bonds that will be p by p pi sp2 will have one p pi p pi bond and rest all d let's say for example if i talk about this so you have this kind of a structure so there is two pi bonds so both of them are p by p by bonds and now if you have this structure sp2 let's say if i have so2 molecule so it will be uh drawn like this so if you have two pi bonds first of all one will be the p pi p pi rest all will be the d pi p pi so this question is always asked in the p blocks that what type of

pi bonds are present over here so just follow this if you have for the so3 if you have for the so3 all of the bonds are the pi bonds so one [clears throat] only will be the p pi in the case of this rest 2 r d by p pi with any confusion when with any other confusion you can write this as like this okay sp3 all d pi p pi all d pi p pi all d pi p pi so this you can have a look next we have over here in the next [clears throat] case we have to discuss isostructural isoelectronic isostructural and isoelectronic now what do you understand by that i've written over here same number same shape or same number of electrons now let's have an example let's say

i have over here no3 negative and co3 two negative let's first calculate the total number of electrons so nitrogen is having electrons as seven oxygen is having eight with the three and plus one eight threes are 24 and this is 25 30 to 32 electrons are present in no3 negative whereas in the co 3 2 negative carbon is having 6 and 8 threes are 24 plus two so 24 plus two is 26 and six is 32. so yes these two are having same number of electrons so yes they are isoelectronic i can call them as isoelectronic what about the shape what about the shape so no3 negative is having the structure like this so hybridization is sp2 this will be considered as planar triangular planar if i write the exact name this will be considered as this

will be called as triangular planar what about co32 negative what about co32 negative so it is sp2 with three sigma and the same shape will be for this also the shame the same shape will be for this also isostructural and isoelectronic same shape and the same number of electrons is present same shape and the same number of electrons is present in the isostructural and the isoelectronic okay let's move forward now we have to discuss the another topic which is called as bond angle but before bond angle i want to explain you va cpr theory because we need to understand now that what types of shapes and all these things are present we need to discuss that first and for that you have to

follow a simple formula and you will be able to calculate that so before bond angle let's start the va cpr theory now let's start the va cpr theory and if i tell you this is the most important part of the chemical bonding all of the examinations ever held for the chemistry have the role of this respective topic you should know the shape the hybridization uh the geometry of all of the molecules including the bond angles also so let's start first of all we have for example first of all we should not start with the example first we should know the formula that is va cpr theory that is valence shell electron pair repulsion theory we have to start now via cpr theory so what formula you are going to use half

number of valence electrons number of valence electron plus number of monovalent atom plus number of monovalent atom plus you will have an ionic charge and minus you will have cationic charge first of all this will be the formula for the calculation of a steric number with the help of this we are going to calculate the steric number and the steric number will help to decide so many things let's discuss first this is called as number of monovalent atom is the one which is having valency either equal to plus one or equal to minus one

this is the number of monovalent atoms an ionic charge means like you have no three negative so you will have minus one as the anionic charge c o three two negative so minus two will be the anionic charge same is going for the cationic charge let's say i have ammonium so plus one will be my cationic charge so you should know the meaning of all of the terms first okay and you are going to calculate with the help of this a number which is known as the steric number now let's discuss with the help of the examples first first in the category we have let's say the geometry the the geometry which is linear first of all we have the geometry which is linear it is shown by let's say bef2 let's say co2 etc there are many molecules which are going to show the geometry linear i have just written these two now how you will

calculate let's say for example i have for the bef2 right so first of all half beryllium has valence electron as 2 so just write 2 over here and fluorine is a monovalent atom so i have to write 2 over here for the fluorine because there are two fluorine atoms number of monovalent atom i have to write two fluorine are present so i have to write the number two okay now if this linear is they are telling you so the steric number will come out to be two this is for sure if you have linear geometry or static number will be 2 your hybridization will be sp your hybridization will be sp all these things are associated with each other all of these things are associated with each other so this is coming out to be linear whenever there is lone pair zero whenever there is lone pair equal to zero i'm writing this over here whenever there is lone pair equal to zero so always your geometry is equal to shape

always your geometry is equal to shape and your steric number decides your hybridization also your steric number also decides your hybridization so you should know all of these things which are related with the steric number okay perfect now you can make the shape very easily for the ef 2 you know all of the things associated with it now let us move towards the second one which is geometry equal to triangular planar which is geometry equal to triangular planar the geometry we are having over here is triangular planar this is triangular planar so if you have in the first case lone pair equal to zero so i have an example over here which is bf3 so you will write over here half for the boron you will write over here 3 and 3 is the monovalent atom so this number is coming out to be 3 so the steric number

is 3 so you will be having the geometry and shape same which is equal to triangular planar if you want to have the bond angle this is coming out to be 120 degree and if you want to have the hybridization that will be equal to sp2 because the number is coming out to be 3 the steric number is coming out to be 3. for this case the steric number will be always 3 the steric number will always be equal to 3 so this will be having the bond angle as 120 degree and it will look like this it will look like this now i have another case of this where lone pair is equal to 1 and bond pair is equal to 2 lone pair is equal to 1 and bond pair is equal to 2 what can be the case example we have over here so2 now apply this formula for the so2 sulfur is having valence electrons as six and oxygen is not a monovalent atom since the steric number is coming out to be three your lone pair is one your bond

pair is two two oxygen are connected with the sulfur so the bond pair is two so just for the calculation of loan pay how do you calculate the loan pair okay if you want to calculate the loan pay just do one thing steric number minus one pair just do one thing steric number minus one pair you will get the lone pair okay now the shape over here will be shape over here will be bent or maybe you can call it as v shape or maybe you can call it as angular all of these are associated with one of the lone pair and two of the bond where all of these are associated with the same thing so it will look like this it will look like this okay now you have to remember one thing the angle over here will not be equal to 120 degree this respective bond angle this respective bond angle will be less than 120 degree because of the reason that these lone pair will shift this

inside because of the lone pair bond pair repulsions present in this respective molecule whenever you have a lone pair you have a different shape geometry will be same geometry will be same triangular planar but shape will change shape is told with respect to the lone pair see whenever we are looking the shape we look that with the help of the spectrometer and spectrometer cannot uh identify where exactly the lone pair is present so for the spectrometer it will be like this it will be like this okay now what is this it is a bent shape it is a bent v shape okay if this is v so we are having this kind of a shape it is an angle form so angular shape the spectrometer is not able to see any lone pair present it only tells us about the bonds present so it will be bent angular or the v shape okay let's move forward the second case is done now we have over here the next case which is the third case

where you have geometry equal to always equal to tetrahedral geometry is always equal to tetrahedral your hybridization is always equal to sp3 as your steric number as your steric number is coming out to be 4 always all these three things are associated with each other now we will be having the first case that our lone pair is equal to 0 so we have an example over here which is methane so half carbon having four valence electron hydrogen is a monovalent atom because it is having valency as plus one so this number is coming out to be four okay now if you have lone pair equal to zero so your geometry is equal to shape so geometry and shape will be same now what will be the structure over here the structure will be like this this is the structure of methane the bond angle over here is we know that 109 degree 28 minutes and total number of bond angles are six

which is i have explained you in the theory itself okay now let's say i have lone pair equal to one so you have the case over here which is of the ammonia so half ammonia is having valence electrons as how much it is having five valence electron attached with the three hydrogen atoms so this number is coming out to be four so if you have three bond pair of electrons your lone pair will be 1 as the steric number is coming out to be 4 right after that we have to make the shape of this now since you have lone pair present your shape and geometry will not be equal so the shape will be the shape will be pyramidal over here the shape will be pyramidal over here and you are going to make ammonia like this since here also we have lone pair bond pair repulsion so bond angle will not be equal to the 109 degree it will be around 107 degrees so you have to

remember the bond angle of ammonia 1 0 7 degree now there can be another case present over here which is having the lone pair is equal to 2 which is the case of the water now for water half valence electron 6 2 monovalent atom coming out to be 4 now this number is having shape as equal to bend same v shape same angular all the three similar okay now how it looks it looks like this how it is going to be it will be like this now it is having angle as 105 degree now if you are able to see first we are having the tetrahedral angle as 109 degrees then we have 107 degree and then we have even the less angle which is one zero five degree what is the reason the reason is we have two types of

repulsions present over here the pink sign represents lone pair bond pair repulsions whereas if i draw the green color it represents lone pair lone pair repulsion and you know that the order of the repulsions is highest for the lone pair lone pair then lone pair bond pair and then bond pair bond pin so these two lone pair will shift away from each other and as a result the bond will be shifted inside so your bond angle will change from one zero nine degree to one zero five degree right so all the things you need to remember all the things you need to remember now since you if you just not see the lone pair of electrons so this is also this kind of a shape so this is bent this is angular this is v shape similarly we have here also now similarly we have here also here also we are having the same thing we are not identifying any kind of a lone pair present so we have to just write bent v shape and angular all of these things right after that we have the steric number

after that we have the steric number as fourth number the steric number as the 5 which is having the geometry which is having the geometry equal to tbp which is triangular which is triangular by pyramidal triangular by pyramidal and hybridization being sp3d since you have the five asymptotic number so the hybridization will be sp3t now in the first case we have no lone pair lone pair is zero so example will be pcl5 now half phosphorus is having five chlorine five so this number comes out to be 5 right if i draw the structure it will be like this i have explained every of the important thing related with this pcl5 and you know everything of this we have

bond angle of 120 degree as 3 90 degree as 6 and 180 degree as one we know all of the bond angles present in this okay now after that we have the second case in which lone pair is one for that we have the example sf4 now apply the formula half sulfur is six four monovalent atom this number is coming out to be five so if you have five steric number and four bonds are associated so one will be the lone pair okay in this case you have a shape present that will be seesaw the shape will be present as seesaw now sulfur one fluorine above one fluorine down and then we have these types of fluorine and a lone pair present over here now if you see this is the stand of the theta and the two people are sitting like this this is looking like a seesaw and just remember this okay shape has changed geometry i have already written geometry

is always same if the steric number is same geometry will always be same only the change comes in the shape when there is a change in the lone pair of electrons right now third number we are having over here is your lone pair is equal to 2 so you can have clf 3 right so half chlorine is having 7 this is 3 this is coming out to be 5. now you have lone pair as 2 and bond pair only as three so you have a different shape now the shape will be called as bent t shape the shape here will be called as the bent t shape bent t shape okay now we have cl and then we have f3 clf3 and these are the two lone pair of electrons cl f3 okay you can have we if you're not having very straight i can tell you one thing that these bonds are not very much straight they are slightly bent they are slightly bent because

these lone pair will repel these bonds and they are slightly bent so we will call it as bent t shape we will not simply call it as t shape okay and the last case over here is the fourth one which is having your lone pair equal to three in that case i have over here in that case i have over here xz f2 i have i f2 negative i have three negative i have let's say i make xcf2 so half the non is eight plus two so i have five over here okay now five i'm having and only two bonds i'm having so i'm having the two as the bond pair and three as the lone pair so my shape will be my shape in this case will be linear how you are having xenon over here and you know that the electronegative atoms are coming on the axial positions and the lone pair are always coming at the equatorial positions so you will see this molecule

like this xz f2 you can simply see the structure this is coming out to be linear since you whenever you have the shape you only see the bonds and they are looking like a linear shape so i don't think so there should be any doubt present over here right after that the next case we have over here is the steric number steric number six in that case you have geometry in this case geometry is octahedral in this case geometry is octahedral and you have square by pyramidal and you have square bipyramidal as your geometry right now you have exactly first case which is lone pair equal to zero now let's say i have an example sf6 so sulfur half six monovalent atom six this number is

coming out to be six after that we have the structure like this sf6 so we have total bond angles as 90 degree being 8 not even 8 we have 12 180 as 3 okay so the 180 angles are 3 and the 90 degree angles are 4 above 4 down and 4 on the plane so it is coming out to be 12 90 degree angles okay after that we have over here the lone pair equal to one like we have in the case br f5 we have the case brf5 now apply bromine for the seven and then five this number is coming out to be six now if this number is coming out to be six just minus the number of bonds so six minus five one lone pair will be

present in that case the shape is coming out to be square pyramidal square pyramidal will be the shape square pyramidal will be the shape so we are having over here f5 br f5 this will be the shape brf5 okay now let's say another one which is having the lone pair equal to two so your shape in this case will be square planar your shape in this case will be considered as square planar so whenever you have a octahedral geometry and you have a two lone pair so you will be having the shape as the square planar okay now let's say i have an example xd f4 so you have half the non as eight plus four as a monovalent atom coming out to be six right so lone pair is two bond pair is four you already know one pair

is four so lone pair will be two and this will look like this this will look like this one lone pair above one lone pair down we are not able to see any lone pair when we are looking through the spectrometer so it will look simply like this okay now let's move forward towards the last case i'm uh one thing more i went to i want to tell you is that never you have to assume yourself exhale equatorial do not exist do not exist in octa hedral geometry many of you have this doubt that ma'am why are you placing this lone pair in the exhale position why not on the equatorial position there is no axial or equatorial present in the octahedral all of the bonds are equal

from wherever you see this is a symmetrical symmetry okay so there is no axial or equatorial present in the case of the present in the case of the octahedral last case we are having is steric number as7 in this case we have shape equal to pentagonal [laughter] pentagonal bipyramidal pentagonal bipyramidal will be the shape present over here and the hybridization will be sp3 d3 sp3 d3 now let's have an example of if7 where lone pair is equal to 0 so iodine is half 7 plus 7 is coming out to be 7 so the structure will be like this now other atoms will be

that's why you're calling it as that's why you're calling it as pentagonal bipyramidal pentagonal bipyramidal now remember one thing over here that that these bonds are your equatorial bonds since i'm writing everything over here so that means that pentagonal bipyramidal is also having axial and equatorial bonds present with them okay and this respective bond is considered the axial and here the bond length remember one thing very nicely here the bond length of xzl is less than equal to i have made in such a way only that you will able to see that which is uh larger in length and which is smaller in length right now we have the lone pair equal to 1 and that will be the case of xz f6 right so half xenon will be having 8 plus 6 equal to 7 so if it is xz f6

lone pair will be equal to 1 okay now let's make the shape first what will be the shape what it will be called as it will be called as distorted distorted octahedral the shape will be called as distorted octahedral after that after that if i make the shape just do one thing just make an octahedral distorted octahedral just make a octahedral which is distorted which is not the proper one and just have a lone pair present anywhere just have a lone pair present anywhere because now it will not be the proper octahedral no bond angle will be of the 90 or 180 so the bond angles all are changed because because of the presence of the lone pair there will be a lone pair bond pair repulsions and that lone wave will shift the bonds towards any of the direction the octahedral will distort itself it will be distorted octahedral right

now i have told you all of the shapes and these things are very very important via cpr theory you don't have to forget you have to go through every of this thing very very nicely okay now let's move forward towards the next point which is the bond angle now since you have discussed all of the things regarding the vsepr theory now we have to move towards the other bond parameter which we were discussing before vscpr theory so now we are moving towards the bond angle now bond angle are having certain rules present with them okay you have to follow those rules you cannot simply just apply any rule anywhere okay you have a synchronization that first this will be applied then this will be applied right if you'll follow this you will have all of your questions correct you will not be having any of the doubt present okay now let's discuss first rule is whenever you have to to tell the bond angle your first rule is

to check the hybridization your first rule is to check the hybridization and that tells us that if you have a greater percentage s character if you have a greater percentage s character your bond angle will be greater if you have a greater percentage s character your bond angle will be greater do not forget this thing so if i talk about the hybridization of all these three so i have sp3 over here i have sp over here and i have sp2 over here if i write the percentage s corrector of all of these so sp3 is having 25 this is having 50 percent and this is having 33.3 percent you are simply having the answers present first you will be having the bond angle highest for be f2 then you [snorts] will be having for be bf3 and then you will be having for me this will be the answer because greater the percentage s character greater will be the bond angle this is the case when you're having no

lone pair present nothing present all of the hybridizations are coming out to be different so you have to first check the hybridization next you have to check is there any regular geometry present or not now what do you understand by this word now what do you understand by this word now what do you understand by this word and what is regular geometry now understand regular geometry is you have hybridization as same you have no lone pair on central atom you have no lone pair on central atom if you have these two things present with you you are following the regular geometry and in that case the bond angle is always equal and in that case your bond angle is always equal now let's have first example of bf3 bcl3 and then we have bbr3 now let's check

what is the hybridization it is sp2 it is sp2 it is sp2 everybody is having the same hybridization yes correct so first point is done hybridization is coming out to be same there is a lone pair present on the central atom check there is no lone pair present on the central atom if i say there is any lone pair there is 0 0 0 that means that they are following the rule of the regular geometry and yes and yes one more thing i have to need to tell you is that surrounding atom must be same surrounding atom should be same surrounding atom must be same you have to follow this also that surrounding atom must be same so just follow this just follow this and get your answer bf3 bcl3 and bbr3 so you have the same bond angle let's uh solve for this also phosphate and sulfate are they having the regular geometry so hybridization this is having sp3 this is

also having sp3 hybridization is there any lone pair present is there any lone pair present so this is zero and zero lone pair so obviously they will also have the equal bond angle if you want to know the hybridization of phosphate i can tell you if you have any confusion o negative and then you have this is the structure of phosphate so there are sigma coming out to be four the hybridization will be sp3 if anybody of you is having any doubt in the phosphate so this is a structure of phosphate you can simply calculate with the help of the sigma bonds to the number of sigma bonds is four so hybridization will be sp3 so check if there is any kind of a regular geometry present or not so they are also coming under the category of regular geometry so they will be having the equal bond angle okay now let's move forward now rule 3rd is coming now what is what we have to follow according to the rule

third now let's say you have the same hybridization but the lone pair present on the central atom is different so what you have to follow bond angle is inversely proportional to the lone pair present on the central atom if you have a greater lone pair present on the central atom you will be having the less bond angle now let's follow this respective rule now hybridization first i will be calculating okay so this is sp3 hybridized this is sp3 hybridized and this is also sp3 hybridized that means that yes hybridization is same okay now i'm calculating the lone pair this is having 0 this is 1 this is 2 it means that this is not a regular geometry you are having lone person you cannot be having the regular geometry so if you have such kind of a case you have to follow that rule so the answer will be greater is the bond angle for the methane then we have for the ammonia and then we have for the h2 the reason we have written over here we have to follow this respective box that if you have the

same hybridization next thing you have to follow after checking the legal geometry that greater the number of lone pair present on the central atom less will be the bond angle okay perfect now the rule 4 hybridization is same lone pair is present is also same on the central atom the lone pair is also same so the first rule cannot be followed because hybridization is same second rule is there any kind of a regular geometry no because there is a presence of lone pair third rule different number of lone pair on central atom no lone pair is also same then what we should do now we should follow this bond angle is directly proportional to the electronegativity you know which is the sign of this this this tells you about this tells you about electronegativity what this sign tells you is about the electro negativity it tells you about the electronegativity so the bond angle is directly proportional to the electronegativity of the central atom so follow this now first of all we need to calculate the

hybridization first of all hybridization this is sp3 this is also sp3 and this is also sp so the first rule is cancelled after that we are having the lone pair this is also having one one one all are having one lone pair so just have to follow the electronegativity of central atom is more so the bond angle will be more nothing else we need to follow nothing else we need to follow we are done with the answer we are done with the answer nothing else we have to follow but we have to move according to the rules if we are not following according to the rules we will be confused like in the organic we are having so many effects present with us but if we are not going through the effects like which we should follow first and which we should follow afterwards we will be confused in all of the you know uh order uh questions acidic strength basic strength carbocations we will be confused because when we are studying those effects those

effects seems to be very easy like if i tell you that i just don't explain you any of this rule i just write this over here you will be very you will say that this is so easy but if you have so many uh different different types of things present with you so different type of compounds present with you so what you will do that you need to understand you have to follow the rules over here right next rule number five hybridization is same lone pair is same and the central atom is different lone pair is same and even you have to write over here the central atom is different because in the previous case [snorts] in the previous case center atom is same hello because in the case that was previous central atom was different now so what was happening so we were able to uh calculate with the help of the electronegativity of the central atom we were able to calculate through the

electronegativity of the central atom now you are having the same atom same central atom so now what you have to see hybridization same lone pair on the central atom same central atom is same now the next rule bond angle is inversely proportional to the electronegativity of the side atom see okay so we should write over here we should write over here hybridization hybridization yes this is sp3 this is sp3 this is also having hybridization as sp3 let's see the lone pair it is having one it is having one this is having one lone pair now what do we should now what we should say about the electronegativity electronegativity is same because central atom is same so bond angle is inversely proportional to the electronegativity of the side atom this is having the highest electronegativity this will be having the smallest bond angle so highest will be for the pi3 and

then we have pbr3 and then we have pcl3 some of you answer this question with the help of the size of the surrounding atom but that is not always applicable so you should must follow the electronegativity that will never give you the wrong answer but the size sometimes we are not knowing because they are very uh very like near to each other so in that case you should must follow the electronegativity now now there are some compounds now there are some compounds they are coming under the rule drago rule they are coming under the rule drago rule now you have there is very this is a very deep topic drago rule but i'm just telling you the result over here that you need to understand you need to learn and you need to follow okay drago rule that tells us that there are some of the compounds which are not undergoing the concept of the hybridization okay if you have a

if you have a element beyond second period that is third period onwards and electronegativity of side atom is nearly equal to 2.5 electronegativity nearly equal to 2.5 of the surrounding atoms don't follow they do not follow hybridization they do not follow hybridization they do not follow any kind of a hybridization for example and they have one more thing they have

bond angle nearly equal to 90 degree they have a bond angle nearly equal to 90 degree for example i have ammonia phosphene ash3 sbh3 bih3 so all these molecules all these molecules follow drago you can see from here electronegativity of hydrogen is nearly equal to 2.5 all these central elements are having the uh they are from third period onwards this is the third period fourth period fifth period and the sixth period okay so these all follow the drago concept okay there should be the presence of lone pair also lone pair must present lone pair should be present so the order of the bond angle of all these three all these uh examples will be this is having a very high bond angle and then bond angle is decreasing this is having the

bond angle as 107 degree rest of all of them are having bond angle exactly equal to 90 degrees let's say this is around 92 degree and this bi h3 is exactly equal to 90 degree they have a very slight difference over here 92 92.1 92.3 then 91 then it is like 90. 90.9 and then it is coming out to be 90. so all of these present are having exactly like very near bond angles but ammonia is not following any kind of a dragon rule because there is a condition which i have written the element should be beyond second period that is the third period onwards so you have to remember all of these things so whatever i have written in the boxes will follow the drago rule they will have the bond angle nearly equal to 90 this will be 92 so this will be 90. i can i'm not going very much deep into this but i am writing the result and if you're following the result you'll get the all of the questions correct okay after that we have water

we have h2s we have h2se we have h2te now in all of these these are following the bond angle of the 90 degree value because they are the drago compounds okay so all these three are having bond angle nearly equal to 90 degree and this is having bond angle as one zero five point four okay so this is the bond angle 105.4 they are having around the 90 bond angle okay so please don't forget in our course we have to remember these molecules which are following the drago rule let's say i give you a question let's say i give a give you a question let's say i tell you there is methane present there is bef2 present there is the of2 present there is the ph3 present okay so let's say these all molecules are present and you have to arrange them

in the order of the bond angle so what you will do so first rule is the hybridization this is the hybridization as sp3 this is having sp this is also having sp3 and this is there is no hybridization because it's a drago compound it is having the bond angle nearly equal to 90 degrees so obviously this will be coming at the last now compare all these three this is hybridization now check the lone pair is there any lone pair zero lone pair zero lone pair two don't pair one huh if you have to write the lone pair lone pair is present on phosphene there is one lone pair present so first follow the rule of the hybridization which is having the highest percentage s character which is bef2 so bef2 is coming first because first we have to follow the hybridization greater the percentages character greater is the bond angle so first sp will come now check the methane and of2 if you have the same hybridization but you have the different

lone pair so greater the number of lone pair less is the bond angle greater the number of lone pair less is the bond angle okay so it is coming out to be methane and then it is coming out to be of 2 and obviously it is having very less bond angle phosphine is a drago compound so you have to uh you have to like apply these rules like this okay so if you are following according to the rules uh order i have written you will have all of the answers correct okay just have a look very nicely very deeply with no other distractions present with you along of that like mobile phones or all of these because you know that uh this is a very crucial time for you we are doing we are giving you so much of the content you have to just go through it just have a look at this and you will be having all of your topics revised if you are not knowing something which is something new to you we are explaining that in a very less time so that you can apply that so just have a uh you know

use of all these things okay rest is up to you now the next topic we are having over here is the dipole moment now what is dipole moment what uh all the important things are associated with that you know all of this all of those things right now i'm going through the topic what what all important things you are requiring okay now let's have a look now what is a dipole moment dipole moment actually tells you about the ionic character in a covalent bond right so dipole moment tells us about the ionic character in the covalent bond it is actually it is actually the product of the it is actually the product of the charge into the distance now how do you define and how do you get to know that there is a dipole present in a bond let's say i have a bond of like i have a water molecule so i have this kind of a bond

so i know that the electronegativity of oxygen is greater than electronegativity of the hydrogen so i have to write del negative over here and del positive over here there is a difference of the electronegativity so we have to write the charges the slight charges according to the electronegativity right now what is happening over here this respective distance is representing d over here and the product of the charges q is represented as del positive and del negative okay now what is happening over here in the next point like it's a vector quantity okay unit is d by represented as this and one more important thing you need to understand is uh i will write over here like i have written water i will explain you with the help of the water only whenever you are having the dipole so you should have a direction from the less electronegative atom to more

electronegative atom with the cut sign on the arrow this is always you have to this is how you always have to represent the dipole okay now the next thing we need to understand for this is calculation of the dipole moment now understand the calculation see first of all we have the two kinds of the units in which we have to calculate one is electrostatic unit of charge which is esu so here if you will calculate the dipole moment you have the charge equal to 4.8 4.8 into 10 to the power minus 10 esu and distance is 10 to the power 8 minus 8 centimeter okay now if you will calculate the dipole moment which will be coming out to be q into d multiply 4.8 into 10 to the power minus 18 esu centimeter sometimes it is written as

4.8 dbi where one d by where one divide will be equal to 10 to the power minus 18 esu centimeter so you need to remember the value of 1 dbi also in the esu centimeter even you can have the unit in the cgs that is here the charge will be 1.6 into 10 to the power minus 19 coulomb and the distance will be 10 to the power minus 10 meter the distance will be 10 to the power minus 10 meter now the dipole moment mu will be equal to q into d whereas mu will be equal to charge equal to 1.6 into 10 to the power minus 19 coulomb and the distance will be 10 to the power minus 10 meter you can write it as 1.6 into 10 to the power minus 29 coulomb this is not centimeter this

coulomb meter okay so always remember this even you can have one more thing you can compare these two dipole moments you can compare these two dipole moments you can actually compare these two dipole moments so you can have a unit of dubai in terms of the coulomb meter also so what you can do is one debye okay that will be equal to that will be equal to 4.8 into 10 to the power minus 18. esu centimeter and 1.6 into 10 to the power minus 29 coulomb meter and change this into 4.8 dbi which is d you can simply write 1.6 into 10 to the power minus 29 coulomb meter so 1 divide can have value

equal to 3.33 into 10 to the power minus 30 coulomb meter so even you can have the value in the coulomb meter also and even you can have the value in the esu centimeter also because you need these values because you have to solve the uh numericals also right now applications of the dipole moment what all things dipole moment can decide this is something it can decide the polar and the non-polar molecules like for the homodiatomic molecules like i have i2 i have h2 in all these molecules the dipole moment is zero so we simply write these molecules as non-polar i've explained you this thing but actually this is explained with the help of the dipole moment i cannot that i cannot draw the direction of the dipole so there is no polarity present okay electronegativity difference is zero the non-polar molecule examples i have written for the heterodiatomic molecules there is a difference in the electronegativity and there is

non-zero that the molecule is having it is a polar molecule okay just simply right this is a polar molecule electronegative difference is non-zero it's not equal to zero i have explained you all these things so all these molecules are having dipole and if you want to have the order of the dipole moment present over here first it's hf then hcl then hpr so here the dipole moment is decided by the charge the charge greater the electronegativity difference greater will be the q i will try it over here greater the difference of electronegativity greater will be the q this is something you need to remember for the answering of the questions okay now the another application is the percentage of the ionic character or the charge fraction so the formula will be mu experimental divided by the mu calculated mu experimental will be given in the question

and mu you have to self-calculate you have to self-calculate the question can even ask you percentage [clears throat] covalent character also that you have to subtract from the 100 and you will get your answer now we have a question present over here in a diatomic molecule the dipole moment is 1.2 dbi bond length is one angstrom what is the charge fraction no problem we are having first the dipole moment calculation mu is equal to q into distance the charge we have to take is 4.8 into 10 to the power minus 10 asu and distance is given as one angstrom so we can take it as 10 to the power minus 8 centimeter so the dipole moment will be 10 to the power the dipole moment will be 4.8 into 10 to the power minus 18 esu centimeter that is coming out to be 4.8 dbi

that is coming out to be 4.8 device so this is the dipole moment which is coming out after calculation now percentage ionic character which is given dipole moment as 1.2 divide you have calculated 4.8 into 100 so this is coming out to be 4 and it is coming out to be 25 so you have the charge fraction of the percentage ionic character as 25 okay let's move forward dipole moment of the regular geometries is zero this is something you need to understand very very nicely see if you have the geometries like if you have the geometries like ab2 you have a b3 you have ab4 ab5 ab6 and ab7

all of these geometries are regular geometries okay this is linear this is triangular planar this is tetrahedral this is triangular bipyramidal this is square by a pyramidal and this is pentagonal bipyramidal all of these have dipole moments equal to zero okay surrounding atom is same there is no lone pair present because i have already written the regular geometries so there will be lone pair equal to zero i have already told you what is the meaning of the regular geometries so there will be no doubt in that so all of these molecules all of these molecules combinedly all of these molecules combinedly have their dipole moment equal to 0 and there should be no exception in this there should be no exception in this after that we have to discuss some other cases also

now let's start now this is about the regular geometries now we will understand the other cases now if you don't have the regular geometry so what will be the dipole moment see if i take the example of co2 right so you will simply say that draw the dipole over here draw the dipole over here you will cancel both of these dipoles you will have a regular geometry or dipole moment will be zero but what if i have this type of a molecule so there will be not the equal dipole obviously the direction will be same but this will be the bond dipole of the carbon oxygen and this is of the carbon sulfur so if the bond dipole is not of constituting the same atoms it cannot cancel because the difference of electronegativity over here will be something different than this so these dipoles cannot be cancelled so the dipole moment of this molecule will be non-zero okay after

that if i say this is bf3 obviously the dipole moment of bf3 will be 0 but if i talk about bf2cl so this dipole moment will be equal to non-zero now how understand if i have b f 3 c l b f 2 c l so i have a bond dipole like this this is my bond dipole this is my bond dipole and this is my bond dipole now if i make the resultant the resultant of the bf bond will be here this resultant will be here and this resultant will be here this resultant will be of the bf this resultant will be of the b c l f and this resultant will be of the b c l f so if you try to cancel it will not be cancelled why because here the bond dipole is of the b f

whereas the resultant is of the bclf if you see here the bond dipole is is of the bcl here it will of the bf so it will also not be cancelled here the bond dipole is of the bf here the resultant dipole is of the bcl f so they are not same so they cannot be cancelled okay so the dipole moment here also it is coming out to be non-zero so if you have the regular geometries you have dipole moment zero if you have non-regular geometries then your dipole moment is non-zero okay let's say another example if i take i have nh3 over here and i'm comparing this with i'm having nh3 over here and i'm comparing this with nf3 so what will be the case let's say i draw this let's say i draw this and then i will explain you this is h this is h and this is h and similarly if i draw n f 3 this is f

this is f and this is f now first of all we have to make the bond dipoles this will be towards the more electronegative atom towards the more electronegative atom and towards a more electronegative atom first of all i am making the bond dipoles whenever there is a lone pair whenever there is a lone pair always a dipole will be towards the lone pair always a dipole will be towards the lone pair and there is no exception in that now understand that if there are three dipoles pointing towards the similar direction the resultant will also move in the same direction they are all pointing downwards so the resultant will point in the downward direction this resultant is of the nf this resultant is of the nh whereas the uh above red colored dipole is from the nitrogen to the lone pair now what is happening these

arrows are pointing towards the upward and this arrow is also pointing towards the upward direction so these two are adding into each other but if you see here the bond dipole of the nitrogen and lone pair is pointing towards the above direction whereas this is pointing towards the downward direction they are not exactly cancelling each other but somehow there will be less dipole moment as compared to the first one so the dipole moment order will be like this if the resultant and the dipole is uh adding into each other and other is cancelling not exactly cancelling but there will be the difference in the dipole moment in the values itself okay none of them is zero but this is more and this is less this you need to understand okay perfect let's move forward now i have another question let's say i have pcl3 f2 and i have p

f3 cl2 so phosphorus i'm having over here first of all the making of the diagram is very very important if you're not able to make diagram effectively your answer is not coming out to be correct p f3 and then we have cl2 pf3 and then we have the cl2 now make the dipoles first above dipole below dipole okay and now i'm making a different color for this so the two fluorines which are present above and below will cancel each other and these three cl are present as the angle being 120 degree they will also cancel each other so the dipole moment will be coming out to be zero right if i talk about this case in this case these two will cancel each other that i

understand but what about this this is a bond dipole pf this is a bond dipole pcl and this is a bond dipole pcl now what i am going to do is i am going to break this molecule into the path this is pf this is pcl and this is pcl will it come out to be 0 that we have to see dipole this is also dipole this is also dipole now i'm going to make the resultant resultant of these two will be here these two will be here and these two will be here is anyone cancelling each other that you have to see see this is the bond of pf the resultant will be of the pcl not canceling each other this respective resultant is of the pfn pcl whereas the bond is of the pcl okay i have to write it over here this is mu of the pf this is mu of the pcl this is mu of the pcl okay if i write about the resultant

this resultant is of the p clf this resultant is also of the p c l f and this resultant is of the p c l so you have to see if they are cancelling each other or not okay p f the resultant pcl not cancelling resultant pclf bond pcl not cancelling so the reason for this i have explained the dipole moment is not zero whereas here the dipole moment is zero so you nee if you are not able to first identify by seeing the molecule you can uh cut it into the path and then see the dipole moment you will be able to get the answer any other molecule if you want to discuss we can discuss that also let's say if i have something which is associated with the benzene let's say benzene first of all is a planar molecule benzene first of all is a planar molecule do not get confused that it is

having difference in the planes of any atoms so if i make the molecule like this this is cl with the ortho position this is cl at the meta position whereas this is cl at the para position this is ortho this is meta and this is pera okay so whenever you have benzene and you have the similar groups if you have the similar groups and both are the minus r groups i should say both are the minus i groups in that case dipole moment is inversely proportional to the angle in that case dipole moment is inversely proportional to the angle what is the angle is the highest angle present over here this angle is the highest 180 degree so this will be the order of the dipole moment highest the bond angle less is the dipole moment highest the

bond angle less is the dipole moment whereas in this case whereas in this case if you have such kind of a structure let's say you have ch3 you have no2 and you have over here this is ch3 and this is no2 whereas here you have this is ch3 this is no2 in this case what is happening one of the group is plus i and other group is minus i if you have plus i minus i then in that case what is happening dipole moment is directly proportional to the angle in that case dipole moment will be directly proportional to the angle so the dipole moment order will be 3 2 1.

what is the reason for that this is the direction of the dipole this is the direction of the dipole this is the direction of the dipole so in this case it will be all added so if the all the dipoles will be added so it will be having the highest dipole moment okay now how you are able to decide between the ch3 and this respective benzene that why i'm making this as the arrow now understand understand that we have ch3 over here and the molecule is like this now we have to decide the hybridization for this carbon and the hybridization for this carbon okay this carbon is having hybridization as sp3 and this carbon is having hybridization as sp2 and you know that percentage s character is highest for sp then for sp2 and then for sp3 and same is the electronegativity order as the percentage s character so greater the

electronegativity there will be the sign of the arrow so sp2 will have the greater electronegativity so that's why i have to make the arrow like this so understand this thing and remember this thing okay anything else if you want to discuss it another example i want to tell you is that let's say you have methane you have ch 3cl you have ch2cl2 you have chcl3 and you have ccl4 henna we should take all of these molecules so when you're moving from left to right as number of chlorine atoms increases dipole moment decreases it's just a trick i'm telling you if you will make the resultant diagram you will be able to understand how we are getting this i'm just telling you in the form of the trick you need to remember because this

is a one shot lecture and one other thing i need to tell you is such kind of a molecule which is having oh in the para direction so whenever you have this kind of a molecule always you have to make the bond dipoles like this okay this molecule is a trans molecule whereas if you make this molecule this is o h and this is also o h so arrow will be like this and this is a this molecule this is trans molecule this is cis molecules you can simply see with the help of the direction of the arrows these are going to be added and these are going to be subtracted so cis will have the greater dipole moment as compared to the trans so you need to understand this molecule also so i think i have explained every of them do not forget these i have explained the phenomena which you have to follow another example i need to tell you is

this kind of a molecule this is fluorine this is chlorine this is bromine whereas this is iodine whenever you solve the question of the uh dipole moment you always consider the difference of electronegativity as the priority and you will write the answer first second third fourth you will write the answer first second third fourth but the order of the dipole moment over here is not like that it will be two one three four now what is happening that chlorine is coming first and fluorine is coming afterwards because in this case because in this case distance dominate distance dominant you know that the dipole moment depends upon the distance between the charges and the uh product of the charges which is q okay so distance dominates in the case of the cl here the distance between carbon and

chlorine is larger and carbon and fluorine is less okay i will make a a significant difference in the size and distance so you will be able to understand the answer very nicely this is only this case okay this is something exceptional so you need to understand and you need to remember it very very nicely do not forget this do not forget this and same goes for the ch3cl ch3f ch3br and ch3i and same question is going for this one also same reason here also distance dominate for this these two questions you have to understand that distance dominates and the answer will be for the first chlorine then fluorine then bromine and then iodine do not forget this do not forget this at all okay let's move forward towards the topic which is called as the molecular orbital theory see molecular orbital

theory was discovered when it was when we were not able to understand that why o2 is a paramagnetic molecule because if we if we make the diagram of o2 so it is like this if we make the diagram of o2 it is coming out to be like this by looking at this diagram i cannot tell that o2 is paramagnetic because odd electron species are paramagnetic that i understand all the odd electron species are paramagnetic that i understand but this o2 this is not odd electron species since uh it is not paramagnetic but it was the theory which has explained this is the molecular orbital theory right let's discuss see it can be used for any of the molecules present in this world but we are going according to the ncert so it is telling us to write diatomic molecule but this is not something very very like uh stand by that this is only the condition no

diatomic molecules try atomic molecules any molecules in the world can be used to explain the molecular orbital theory or molecular orbital theory can explain those molecules okay vice versa i have explained you applied for diatomic now there are so many points written over here that i will explain you then i will come to these points again just i will read and just you have to listen because i don't like this i'm just reading and i'm underlining i'm reading and underlining i like to explain and then relate so that you have a better understanding i hope you people have enjoyed the lecture i hope you people have understood the lecture i hope you people are studying very much i hope you people are going to get a very good marks in heat examination specifically the chemistry examination i need the marks of you all of you all in my comment section that we have got this much marks in the chemistry i want

all the marks okay now okay two atomic orbitals combine to form two molecular orbitals one is bonding one is anti-bonding ma'am yes i have heard or we have heard all these names ma'am one is bonding mot is explained ma'am on the basis of the wave nature of electrons yes ma'am them that is always taken as ma'am internuclear axis ma'am yes ma'am that is always taken ma'am always take another internuclear axis ma'am this we will never forget ma'am never forget homo ma'am what is homo am highest occupied molecule orbital man what is lumo lowest unoccupied remember what all these things are please explain let's start see how molecular orbitals is formed before that i want one minute break to have a water so that i will have my train non-stop okay you can also have a water break okay

now molecular orbital theory we understand the molecular orbital theory with the help of the wave nature we explain the atomic orbitals with the help of the wave function there are two types of interference of the wave one is called as the constructive and other is called as the destructive interference now what do you understand by the constructive interference now let's see see i have two types of the waves one wave is looking like this above part is called as crest the below part is called as stuff i want them to overlap in such a way that my crest comes on the crest my trough should come on the trough okay so as a result as a result i am going to get a wave i am going to get a wave which is having

which is having very high amplitude okay if i am going to add psi one plus psi two this is called as the wave as psi one this is called as the wave as side two when i'm going to add these two wave when i'm going to add these two wave they will be added in such a way that the amplitude of the wave will increase okay first i will explain you in a very bit bit manner okay so first of all the meaning of the constructive interference is this okay now the destructive interference destructive interference now we have a one wave over here we have other wave over here now what i'm doing here is now the crest is not on the crest crest is uh interfered with the trough okay they are overlapping in such a way that crest is not coming over the crest it is coming over the trough okay so as a result

so as a result the amplitude is getting decreased as a result the amplitude is getting decreased the amplitude is getting decreased okay this is also the overlapping of the wave and this is also the overlapping of wave we will write over here psi 1 minus i2 whenever we are having the destructive interference we write it like this in the constructive interference we write it like this ok ok now waves overlap constructively waves overlap destructively they will have the same face crest plus crest trough plus trough different face crest trough trough crest okay bonding molecular orbital is formed whenever we have a constructive interference we have a bonding molecular orbital it is of the lower energy probability of finding the electron is very high we will understand all of

these things okay now what is happening let's see whenever there is one atom a it is combining with the other atom b and there are some electrons present in a there is some electron present in b so the atomic orbitals are overlapping with each other the atomic orbital of this electron is overlapping with the atomic orbital of this electron now this atomic orbital now this atomic orbital is represented in terms of the wave function of this also and of this also i am writing it as i a i am writing it as sin b they are overlapped in two manners one is psi a and psi b and other is psi a minus sign b this is the way in which they overlap in which they overlap okay now

now now whenever they overlap okay whenever they overlap these two atomic orbitals when they overlap with each other there is a formation of two types of the orbitals molecular orbital okay this is one molecular orbital this is other molecular orbital this is formed by the constructive interference this is formed by the destructive interference you know what is constructive you know what is destructive okay the of the molecular orbital which is formed by the constructive interference is called as which is called as the bonding molecular orbital whereas this is called as the antibonding whereas this is called as the antibonding molecular orbital okay these all are always high in energy it is

always high in energy it is always low in energy one is always high in energy and other is always low in the energy okay these are all represented these are all represented as sigma star pi star these are represented as sigma and the pi okay first of all i have written all of the things associated with these okay why they are high in energy why they are low in energy that i will explain you further but two atomic orbitals combine to form two molecular orbitals two molecular orbitals i have written over here that according to two atomic orbitals combined to give two molecular orbitals do not forget this thing two atomic orbitals now you will be able to understand you will be able to relate with me to get two molecular orbitals to get two molecular orbitals

okay so i have written this over here two atomic orbitals combined to give two molecular orbitals one is the bonding and other is the anti-bonding molecular orbital other is the antibonding molecular orbital higher in energy lower in energy now if i tell you with the help of the overlapping diagram so one is overlap with the 1s so what is the structure you are going to get ok this is the nucleus this is positive this is negative wave function this is positive this is negative whenever these combine the positive electron density is combined with the positive electron density and you are going to get the shape like this the nucleus will be here and the electron density and the electrons density is coming in between and the electron density is coming in between this orbital is called as the bonding molecular orbital the name of this is

sigma 1s now why it is called as bonding why it is having lower energy why it is stable why it is stable why it is stable this is unstable this is unstable because higher in energy so it will be unstable whereas this is considered as stable whereas this is considered as stable okay you have to remember this thing then why it is stable why it is considered as stable because the two nucleus are very far away from each other whereas if you have the destructive interference the crest is operating with the iso overlapping with the trough and it is not overlapping with the crest so in this case the orbital you are going to get in is in such a way that the two nucleus are directly approaching towards each other they are repelling each other there is a place present over here which

is called as a nod what is a node you have studied in the structure of atom chapter where the electron density where the electron density is probability of finding the electron is zero i can never say i can never say that electron density zero i have to just say that the probability of finding electron is very less probability i have to write over here probability of finding probability of finding electron density is almost equal to zero so this is called as a note okay what it will be represented as it will be represented as sigma star 1s it will be represented as sigma 1 sigma 1 sigma star 1 s why it is anti-bonding molecular orbital why it is anti-bonding molecule orbital why it is

higher in energy because the two nuclears are directly approaching towards each other and electron density is here the two nucleus are directly seeing each other what is present in the nucleus protons are present so they are repelling each other when whenever there is something which is repelling each other the energy will be increased tubby that's why only they are higher in energy bonding is lower in energy right okay now moving forward to be said two 2pz always remember always remember the internuclear axis will be z the internuclear axis will be z this is the internuclear axis always there internuclear axis is always there this can never be changed in the molecular orbital theory if you're talking about any other theory then it is fine you can take any axes as x y or z but in the molecular orbital sorry it is a rule that you cannot change you cannot change beta okay now sigma 2pz in the first case when i'm forming

forming the bonding molecule orbital this positive is combining with the positive to have a to have a constructive interference first we are showing the constructive interference and we are going to get over here this kind of a structure we are going to get this kind of a structure okay this positive will come over here and the two nucleus and the two nucleus are far away from each other and the electron density is present between them electron density is present between them this is called as this is called as sigma 2pz this is called as sigma 2p z whereas if you have the destructive interference whereas if you have the destructive interference what you're going to get

this orbital this orbital overlaps this orbital and this orbital overlaps and it is going to give you this kind of a orbital and here we are having the node also the node is also present over here and this is the nucleus and if i draw the electron density it is coming over here so you know what is not the position where the electron density is almost zero we are not able to find electron in that respective area so what will be the name of these orbitals this will be sigma 2pz this will be the sigma star 2pz so you have to remember this you have to remember this just make the axis over here this is the z axis positive is overlapping with the negative next we have 2px or 2px you know that whenever the pi

orbitals overlap this is overlapping let's say this is px orbital this is px orbital they are coming to overlap this is positive this is positive this is negative and this is negative so what we are going to get we are going to get this kind of a orbital diagram this whole electron density as positive this is as negative and we are going to call this area as not here also the node is present here also the node is present and the name of this orbital will be the pi 2 p x or the pi 2 p y pi 2 p x and pi 2 p by you have to call them as like this and if you want to have the destructive interference in that case the overlapping will not be in this case it will be negative positive destructive interference and you will have the

orbitals like this okay the orbitals will be like this with their nucleus present over here and this will be the node there will be not present over here and there are two nodes present over here there are two nodes present over here this orbital will be called as pi star 2px and pi star 2py the orbitals will be called as pi star 2px and the pi star 2py now we have to answer now what is the degenerate orbital see if you're having so your nucleus directly approaching towards each other these are the nucleuses they are directly approaching towards each other that's why it is called as the anti-bonding molecular orbital now what are the degenerate orbitals you know that orbitals which are having same energy the orbitals which are having the same

energy is called as the degenerate orbital so orbitals which are having the same energy is called as the degenerate orbitals okay now what are the orbitals pi star 2px will always have the energy equal to pi star 2py after that pi 2px is also having energy equal to pi 2py and you don't have to forget this and you don't have to forget this that they are having the same energy this is very very important because you know in the molecular orbital theory we have to follow all of the rules probably we have to follow off but we have to follow ones we have to follow and we make a diagram like this we make the diagram like this that we are when we are moving from top to bottom to top our energy is increasing we write over here 1s then we write 2s then we write 2p we are following the we are following the softbow principle this is softball principle now you have

to make the diagram you have to make the diagram 1s is going to combine with the 1s 1s is going to combine with the 1s and you are going to get this respective orbital 1s is combining with 1s to give you sigma 1s and sigma star 1s one is bonding other is antibonding after that you are having 2s this 2s is combining the 2s is combining the 2s is combining now what you're going to get 2s over here 2s over here this is sigma 2s this is sigma star 2s okay so this is the energy order first time writing 1s and i'm writing 2s then i am going to write the 2p now for the 2p i have such kind of a case the above one will be the same the above one will be the same always okay

but the below one can change and how it will change i will tell you okay this can be the one case i'm not designing the right part okay that you can design yourself and one case will be like this and the one case will be like this the above part is always same there is no difference in the above part but this part is changed according to the atomic number okay according to the atomic number this is 2p present over here this is also 2p present over here if i write the names of the orbitals it will be like this this is pi 2px this is pi 2py this is sigma 2pz right after that we have pi star 2px then we have pi star

2py then we have sigma star 2pz after that we have sigma 2pz we have pi 2px pi 2py okay after that we can have pi star 2px pi star 2p by sigma star 2pz where you have to follow which one see this part is always same for any of them this part is always same we have two different type of a diagram we have to use this respective diagram when our total electrons is less than 14 less than or equal to 14 and we have to use this diagram when our electrons total is greater than

14 and we don't have to forget this thing we don't have to forget this thing this is something very very important the 1s2s diagram will be same but the above one will be different the above one will be different one will be for the total electrons that will be less than equal to 14 and other will be uh greater than 14. don't forget this thing okay remember this thing very nicely now the electronic configuration that you need to remember very nicely whenever you have total electrons greater than 14 15 to 20 order will be sigma 1 is sigma star 1s sigma 2s sigma star 2s then we have sigma 2p set then we have pi 2px equal to pi 2 p y

then we have pi star 2 p x equal to pi star 2 p y and then we have sigma star 2 p that this is the order of the energies when you have total electrons greater than 14 and you don't have to forget this less than equal to 14 1 to 14 any electrons is present okay so the order of the energies will be sigma 1 is sigma star 1 is sigma 2 s sigma star 2 s then we have uh pi 2px equal to pi 2py then we have sigma 2pz and after that whatever is present is always same and that is pi star 2 p x is equal to

pi star 2 p y and then we have sigma star 2 p z so only the change is coming in this nothing else bond order always number of bonding electrons minus number of antibonding electrons divided by two if the bond order is zero molecule will not exist because bond order is zero that means the bond does not exist between two atoms bond order is directly proportional to the stability if you have a greater bond order you have a greater stability if you have bond orders same less the number of antibonding molecular orbitals electrons more is the stability is something very important this can be used in any of the question okay let's discuss the next part hydrogen molecule now if i make the

diagram for the h2 molecule you can have like this the very simplest diagram i am going to make is like this this is 1s this is 1s now if i combine the two hydrogens with each other this is 1s and this is 1 so both electrons will be combining to give you the electronic configuration to give you the electronic configuration as as sigma 1 s2 and if i make the bond order of this bonding electrons to antibonding 0 the bond order is coming out to be 1 so this is the calculation for the h2 bond order is one if you want to have the magnetic magnetic behavior paramagnetic behavior everything you can calculate since there is no unbed electron this molecule is diamagnetic since there is no unpaired electron this molecule is considered as diamagnetic now let's say

i have over here h2 positive and then i have over here h2 negative what about these two if i make the diagram for these two it will be like this if i make the diagram for these two it will be like this it will be like this see the diagram because the last point which i have written in the molecular orbitals features is required in this okay so this is sigma 1 s this is sigma star 1 s this is sigma this is sigma 1 is this is sigma star 1s so if you're talking about the h2 positive you have only one electron it will come over here if you're talking about the h2 negative so it will be like this so electrons will be filled like this you have to fill the given number of electrons into the molecular orbital now electronic configuration sigma 1s1 electronic configuration sigma

1s2 sigma star 1s1 okay if i talk about the bond order it will be 1 minus 0 divided by 2 coming out to be 0.5 what it is 2 minus 1 divided by 2 it is also coming out to be 0.5 so how should i decide the order see this is paramagnetic because there is a presence of one in unpaid electron and this is also paramagnetic because here also there is one unpaired electron present in the sigma star 1s now if i want to calculate the stability order if i want to know the stability order so how i will decide first of all the highest bond order is of the h2 so h2 will come the in the front what about h2 positive and h2 negative both are having their dipole the bond order as 0.5 but this is having more number of antibonding electron

there is no anti-bonding electron so h2 positive will be more stable as compared to h2 negative h2 positive is more stable as compared to the h2 negative so this is the required diagram of the h2 molecule and you need to understand every of them very very nicely let's move forward you can have the diagram for the c2 for n2 for oxygen molecules everything i have written on the blackboard you have to practice all of them i will make only one of them any of them which you will like and after that we have for the heterodiatomic molecules let's discuss for the oxygen molecule rest of them you have to discuss and you have to draw i will make for the o2 now let's come let's make

here itself or somewhere else we should make on the new page if you have any doubt you can ask first of all i'm making for the 1s first of all i am making for the 1s then i will make for the 2s and then i will make further 2p now let's draw

this is 2p this is 2s this is 1s 1s 2s and 2p now let's write the electronic configuration 1s2 completely filled then we have for the 2s 2 is completely filled then we have 2p4 present so let's fill now let's write the names of the orbitals sigma one is sigma star one is sigma two s sigma star two s this is sigma two p z this is pi 2px this is pi 2py pi 2px pi 2py then we have

pi star 2px and pi star 2py and then sigma star 2pz so this is how the diagram of o2 is made i hope this is clear to everyone okay now let's first know all of the important things which are required always remember always remember whenever we have the total electrons whenever we have total electrons greater than 14 the molecular orbital diagram is the correct one the molecular orbital diagram is the correct one because you know whenever you have the molecular orbital diagram like this if you see the molecular orbital diagram of this total electrons less than 14 or equal to 14 you can simply see that sigma is coming above higher in energy and pi is coming down

which is lower in energy which is practically not possible because sigma bond is a stronger bond it should have a less energy so this diagram is not considered as correct why because in this diagram there is always 2s 2p mixing what happens what happens exactly is there is 2s 2pz mixing sigma so it goes above in energy so that this sigma star 2s do not mix with the sigma 2p3 this sigma star 2s should not mix with the sigma 2pz so it goes above to minimize the interaction between the two orbitals whereas that respective diagram is considered as the correct one is considered as the correct one okay it's considered as the correct one so first of all you know sometimes question can come

make the correct molecular orbital diagram for the b2 according to the rules but b2 is having boron 2 is having total electrons less than 14 so it should have a diagram according to that what i have explained but they are telling you to draw the correct according to the molecular orbital theory so what is the correct diagram according to the molecular orbital theory this is the correct diagram now sigma should have the less energy pi should have the more energy when a term is written in the question draw the correct diagram according to the molecular orbital theory you have to follow this diagram whatever i have written whatever i have written if they're just simply writing draw the diagram of the b2 draw the diagram of the n2 then follow whatever i've explained you earlier okay remember now if i calculate the bond order see do not inculcate your uh electrons of 1s and 2s when you are

calculating the bond order because you know that bonding is cancelled by the bonding bonding is cancelled by the antibonding how many bonding electrons are there one two three four how many antibonding electrons are there one two three four so they are cancelled so why i should have a bigger calculation using this so just have the look on this only no so one two three four five six six are the bonding electrons too as the antibonding electrons so it is coming out to be two bond order is two is it paramagnetic or diamagnetic it is paramagnetic it is para magnetic because you can see the two unpaired electrons the main point through which i have started this respective theory was o2 being paramagnetic okay now what else uh is required to understand over here i have explained you one more term which is called as the homo lumo see this is considered as the occupied

molecular orbital this molecule orbital is having electrons so it is occupied molecular orbital so it will be called as the homo highest in energy highest occupied highest occupied molecular orbital and this is considered as unoccupied is there any molecular orbital present here which is unoccupied no only one you can see over here that is that only one you can see highest and occupied the lowest unoccupied molecular orbital okay just don't see the energy order you will say mom this is so much high in energy why are you calling it as lumo lowest unoccupied beta there is no unoccupied molecular orbital in between so how i can how i can say it something else i have to call it as low waste unoccupied word is having importance

over here an occupied world is having importance over here lowest unoccupied lowest unoccupied molecular orbital lowest unoccupied molecule orbital lumo lowest unoccupied molecular orbital it's not looking like leo motors looking like luma low waste unoccupied molecular orbital this is homo so please do not forget this do not forget this okay and rest you can calculate the bond order of every one o two is two o two positive you can calculate yourself or to negative or to two negative okay just remember this this is considered as superoxide ion okay this is required in the uh s block element and this is your peroxide ion you sometimes make ko2

you sometimes make na2o2 the oxides of the s block elements okay bao2 there are many i'm just writing some of them after that we have over here after that we have over here one more thing present yes the n2 diagram i think if you have understood the o2 diagram you can make the entro diagram yourself i have just left the space and you have to draw yourself you can refer your or in or any of the books and please draw once you have to remember everything over here okay let's move forward now i'm writing some points so you need to learn them okay i'm writing in the diagram of the c2 you will make the diagram yourself i have explained you the diagram now remember that c2 is having bond under order as to

both bonds are pi okay this is c2 molecule and both of the bonds are considered as the pi bond there is no sigma bond present over here do not forget this same is going for the b2 whereas the b2 is having only one bond between them so it is also considered as the pi one when you will make the diagram of this you will understand why because there is no electron in the sigma 2pz orbital so it is considered as having no sigma bond okay because there is no bond formed on the internuclear axis so how we will call it as sigma bond okay after that remember one thing also co is having your bond order as three whereas co positive is having the bond order as three point five you can make the diagram also carbon is having electrons as six oxygen is having eight just add them in which category they are falling just make the diagram but the molecular orbital diagram of co is slightly

different which is not included or studied in the syllabus of the j mains or neat examination but these two questions can be asked so you need to remember this okay this much is required in the molecular orbital theory now let's move forward okay now we are moving towards the last slide which is the thank you slide now we are moving towards the last last slide which is considered as the thank you slide i hope you people have covered all of the chemical bonding chapter through this lecture i have not included much questions over here but i have included many questions like 50 around questions in the dbp given so you have to solve all the questions i have made solutions for you you will be able to understand

the solutions very effectively maybe some new questions will also be present so there will be better understanding for you through dpp okay because already we are having a very big lecture so questions i have included in the dpp and they are very very important questions just go through them kindly watch the lecture give some comments how did you like the lecture or whatever you like okay we we are also motivated by looking all those comments okay so thank you so much for attending love you all and uh let's hope we have a life class someday so that we are able to uh interact with each other more effectively right thank you so much for attending and just to remember one thing you have to do the hard work plus the smart work for your examination right because time is very less it's the most precious time present right now just have a good time table prioritize your all of the things present in the day

have a good food do not have a uh like so much amount of food in a single time eat in the like periods like you have something at eight o'clock something like 10 o'clock something at one o'clock okay so you should have a timetable that when you have to exactly eat okay and do not over sleep do not less sleep okay you should have at least six hours of sleep in a day for the effectiveness you know that your need examination is between two to five so you should maximum study in that time two to five is the best time for you because that time should be that you are having your best efficiency if you are sleeping every day between 2 to 5 now what will be the problem that you will face at the day of the examination you will feel sleepy at that time so what you have to do just to have your best efficiency at that time okay and uh for the inorganic chemistry you have to multiple times you have to go for the revision for the s block p block your

all the chemical bonding is like something you have to understand coordination compounds is something you have to understand rest the chapters you have like periodicity you have to understand but rest of the chapters you know that are total learning chapters and you have to learn them so multiple revision is required for them physical chemistry you need to have a tabular form of the formulas the concepts you can watch a video of my which is present on the computation vala that how to to study how to make chemistry easy i have explained so many things over here that is in the hindi language i think you people can understand okay so you can go through that so many things i have explained over there and i will explain so many things in the coming lectures also thank you so much for attending