 So, suppose I am taking the example of you know the first case, case one we are assuming where for the coordination number 4, case one we are taking for all those complexes whose coordination number number is 4. Coordination number 4 for example, the first question we are discussing here is for this, NiCl4 2 minus ok, NiCl4 2 minus is found to be found to be paramagnetic, paramagnetic with two unpaired electron, unpaired electron explain it is hybridization, hybridization and geometry ok. Now you see how to do this question, how to solve this question, solution. First of all the information we should have to solve this question is the nature of the ligand, nature of ligand and ligand here is chlorine which is nothing but chlorine here Cl minus right and Cl minus is a weak ligand. This is the information I am giving now which ligand is weak or is strong that we will understand later on when we do the crystal field splitting right. For the time being I will give you the nature of the ligand whether it is weak or strong ok. We know from the postulates of VBT that in case of strong ligand there will be a pairing of electrons in the atomic orbital against the Hunt's rule right. If the ligand is weak such kind of pairing against the Hunt's rule is not possible correct. So the first thing is this you should know whether the ligand is weak or strong. Second thing is the oxidation state of metal. So oxidation state of metal you see here chlorine is minus 1, minus 4, minus 2 we have already this nickel has plus 2 oxidation state ok. So oxidation state of metal here is plus 2. So what I will do I will just first drop the electronic configuration of nickel. Nickel we know the atomic number is 28 and its configuration is 1S2, 2S2, 2P6, 3S2, 3P6, 4S2, 3D8. Since electronic configuration since the oxidation state is plus 2. So the electronic configuration of Ni plus 2 will be we have to take 2 electron out and that comes out from the outermost shell which is the 4S shell correct. And that is why the electronic configuration of Ni2 plus is 1S2, 2S2, 2P6, 3S2, 3P6 and 3D8 ok. Now when I draw the orbital diagram here ok. So you see the orbital diagram here it is 3D this is 3D 1, 2, 3, 4 this one is 4S and next we have 4P ok. And the number of electron these are 3D shell and this is 4S and this is 4P correct. So now the number of electrons in 3D sub shell is 8. So we will have 1, 2, 3, 4, 5, 6, 7, 8, 4S and 4P orbitals are vacant empty correct. Now how many orbitals these metal require this metal required to bond with 4 chlorine ligands 4 chlorine chloride ion. The number of bonds number of vacant orbital required is nothing but the coordination number of the metal ok. So coordination number of metal is nothing but 4 because it has with 4 monodentate ligand correct. So 4 chlorine atom you can simply understand this one like that we have 4 chlorine atoms so nickel is here in the center and the 4 chlorine atom is trying to donate its electron right to the metal to form a coordinate bond like this ok. So to accept this pair of electron we must need 4 orbitals because 1 chlorine atom donates 1 pair of electron. So 4 pair of electron we nickel has to accept and for that we require 4 vacant orbitals. Now to form 4 vacant orbitals 4 atomic orbitals must go under hybridization right. 4 atomic orbitals must go under hybridization and it forms 4 hybridized orbital 1, 2, 3 and 4 and what kind of hybridization we have here since 1S and 3P orbital goes into hybridization so we call it as SP3 hybridization correct. Now in this SP3 hybridized orbitals the 4 chlorine atom donates its pair of electron and forms coordinate bond with the metal ok. Hence the hybridization of NICL4 the hybridization is hybridization is SP3 and with SP3 hybridization the geometry of the complex is tetrahedral tetrahedral and magnetic property if you think if you think about the magnetic property of this complex since there are 2 unpaired electron so it is paramagnetic in nature ok. Suppose if the electron suppose the ligand is strong then the pairing of electron takes place against the Hans rule ok. So for that also we will do one you know one example for a strong ligand we will try to understand ok. So I hope this example is clear ok. So by with the help of VBT valence bond theory we can actually find out hybridization geometry and magnetic property ok. So the next one you see we are considering the another molecule which is this the question is NICO4 NICO4 NICO4 is diamagnetic it is given is diamagnetic explain its hybridization geometry explain its hybridization and geometry now you see the solution of this molecule of this question ok. The first thing we should know is what the nature of ligand what is the nature of ligand what is the ligand we have here ligand is carbonyl group that is CO CO it is not cobalt take care of this and this is a strong ligand right. Again I am telling you this information I am giving you cobalt CO is a strong ligand generally carbon when the ligand has carbon atom donor or nitrogen atom donor then those ligands are considered as strong ligands in general ok. But how do we know which one is strong or weak that we will discuss in crystal field theory after this ok after this BBT correct. So nature of ligand is strong then the oxidation state of the metal oxidation state is 0 because this ligand is neutral also neutral ligand and to find out this oxidation state of the metal you should know the charge on the ligand that is also important ok last example if you see we have NiCl4 so chloride ion we have Cl minus that is a negative charge ligand ok. So that is why that is you know negative charge ligand so and if you know the charge on the ligand you can find out the oxidation state of the metal to find out oxidation state of the metal you should know the charge on the ligand also this is neutral so no charge oxidation state of the metal is 0. Now the same thing again NiCl has 28 electrons and the oxidation state is 0. So the configuration of this is again 1s2, 2s2, 2p6, 3s2, 3p6, 4s2 and 3d8 orbital diagram if I draw here you see orbital diagram we have 3d sub shell has 8 electrons 4s has 2 and after this 4p which is vacant 1, 2, 3, 4 and 5 number of electrons present you see 3d has 8, 1, 2, 3, 4, 5, 6, 7, 8 and this has 2 electrons like this if you are filling the electrons 1, 2, 3, 1 by 1 it means we are following the Hans rule okay but that Hans rule is not applicable here because the ligand is strong in case of strong ligand there will be pairing against the Hans rule one more thing what you can do that all these electrons you can fill directly 2, 4, 6, 8 like that also you can do okay and another way is what the 2 the unpaired electron present here gets paired why because the ligand is strong so one electron from 4s jumps over here and again the another electron jumps over here and this orbital becomes empty and this orbital becomes completely filled so this converts into this one and then we have 4s and this is 4p right so 1, 2, 4, 6, 8 and 10 and these three orbitals are empty so this is 3d orbital this is 3d this is 4s and this is 4p now again the same thing since the coordination number is 4 coordination number is 4 it means we require 4 hybrid orbitals and to get 4 hybrid orbitals 4 atomic orbitals combines and these atomic orbitals must be vacant so that it can accept the pair of electron which is coming from the ligand correct so ni co4 these three 4 orbitals combines and forms 4 hybrid orbital 1, 2, 3 and 4 this is the hybrid orbital and what kind of hybridization it is it is again 1sn3p so 4sp3 hybridized orbital which is completely vacant and the ligand donates pair of electron to these orbital and forms a coordinate bond hence since the hybridization is sp3 right to summarize this what we can write that hybridization of this now to summarize this hybridization of this molecule is sp3 sp3 geometry is tetrahedral and the magnetic property magnetic property since all electrons are paired you see it is diamagnetic diamagnetic in nature so like this you can understand the magnetic behavior hybridization and geometry one more question we will see here for coordination number 4 suppose suppose we have a molecule say ni cn4 2 minus we need to find out we need to find out the hybridization hybridization next is geometry next is magnetic property these three things we need to find out again you see what all things we should know here the nature of ligand ligand is what cn minus and it is a strong ligand cn minus is a strong ligand you should memorize this it is a negative charged ligand also negative charged and if it is negative charge the oxidation state of nickel if you calculate it is plus 2 so oxidation state of metallis plus 2 nickel atomic number is again 28 configuration is argon 4 s 2 3d 8 electronic configuration of ni plus 2 is argon and 3d 8 there's no electron in 4s uh shell okay so you see next here um orbital diagram if i draw this is 3d 4s and next is 4p 3d has 8 electrons so 1 2 3 4 5 6 7 8 and 4s has zero electron right but since the ligand is a strong right so this electron jumps over here and pairing takes place against the unstroke why because we have strong ligand so in this you see what happens here we have 2 4 6 8 so 1 d orbital is vacant now right again the coordination number is what the coordination number is 4 it means we required 4 vacant orbital 4 atomic orbital combines and forms the hybrid orbital so we have 1d 1s and 2p so we'll get 4 hybrid orbital here 1 2 3 and 4 and what kind of hybridization ds p2 right because 1d 1s and 2p involved so ds p2 hybridization okay so now to write down the answer for this hybridization is what dsp2 and dsp2 hybridization means the geometry is a square planet and magnetic properties since all the electrons are paired so it is diamagnetic it is diamagnetic one type of question also they ask that what is the d which d orbit because there are five d orbitals no so in this geometry in this hybridization dsp2 what like which d orbital is involved over here okay so for that i'll write down just one table here okay and you see this geometry first of all on one side then other side the d orbital involved okay so geometry if it is the geometry if it is a square planet square planet involves dx square y square orbital if it is trigonal bipyramidal tbp trigonal bipyramidal the d orbital involved is dz square if it is a square pyramidal if it is square pyramidal then it is dx square y square if it is octahedral then the orbital involves are dz square and dx square y square okay these few things you must remember this type of question they have asked many times in the exam especially this hybridization magnetic property geometry and all this kind of question they have asked the most important part in this we have for the coordination number six okay so four we have discussed we'll discuss five next and then we'll discuss for the same thing for coordination number six okay so one type of question related to this hybridization geometry magnetic properties the another type is which d orbital involved in this various geometry okay so we are done with the coordination number four next session we'll start with coordination number five and six thank you very much