 Hello students, today in this session we are going to understand again you know the application of VBT that is Welland's Born theory in the compound or complex whose coordination number is 5. Last session if you remember we have discussed the application of VBT in coordination number 4 such complexes. Now the next thing here we have to understand is the complexes whose coordination number, coordination number is 5. Compound with coordination number 5. For example you see we have a molecule like FeCO5 this is the complex we have FeCO5. We need to find out the dipole moment of this molecule dipole moment then we need to find out hybridization of this molecule central atom hybridization and we also need to find out the geometry of this molecule right. So for all these geometry or hybridization we have to apply Welland's Born theory into it correct Welland's Born theory. So first of all you see we need to understand here we need to understand the bonding of this ligand and metal and for that first of all we calculate or we will find out the oxidation number of ion here in this complex. Since this CO is a neutral molecule so oxidation number of ion is 0 okay. So ion has 26 electron the atomic number is 26 26 electron and its electronic configuration is argon which is 18 4S2 3D 6 4S2 3D 6 oxidation state is 0. So if I write down the orbital diagram okay orbital diagram of this this is 3D 1 2 3 4 this is 4S and this one is 4P okay. In this 3D we have 6 electron so 1 2 3 4 5 and 1 pairing 6 electron 4S has 2 electron in this right. Another important thing we should know that this CO is a strong ligand we know this thing that CO is a strong ligand correct and we know in case of strong ligand the pairing of electron takes place against the Hans rule okay. So what happens here you see this electron pair this 2 electron jumps into this inner d orbital this electron also comes over here and pairing of electron takes place against the Hans rule okay. So in pairing of electron takes place what happens will have this kind of orbital diagram and in this the distribution of electron we have like this pair of electron here pair of electron here and pair of electron here this orbital is empty this is 3D this is 4S and this is 4P okay. Now coordination number is 5 so we need 5 atomic orbital which forms hybrid orbital correct. So this is 1D 1S and 3P this 5 orbital combines atomic orbital combines and it forms 5 hybrid orbitals okay these are 5 DSP3 hybridized hybrid orbitals okay. In this orbital only electron or sorry ligand donates its bond pair of electron and forms bond okay. So first of all the answer for this question the answer for this question which is the first one we have is you know the dipole moment dipole moment will be zero because all the electrons are paired okay all the electrons are paired hybridization we have already done and that is DSP3 hybridization we have okay and its geometry will be with DSP2 hybridization geometry will be trigonal by pyramidal which is TBP trigonal by pyramidal correct. So these are the three information we wanted to find out which we have done according to VVT valence bond theory okay. Another question you see another question we have here is the complex is this NiCn5 and here we have 3 minus this complex the information given is what this complex is diamagnetic diamagnetic with two different different NiCn bond length with two different NiCn bond length out of which out of which four bond lengths four bond lengths are same and one is different four bond lengths same and one is different we need to find out the hybridization and geometry hybridization and geometry of this. So you see how do we do this again you see the coordination number is 5 and the information we have here the first information is what that this Cn minus is a strong ligand strong ligand okay and the oxidation is state of NiCl in this complex you can easily find out is plus 2 because it is negatively charged ligand so this is plus 2 on this NiCl okay we know the atomic number of NiCl is 28 correct. So 28 this becomes argon 4s2 3d8 okay argon 4s2 3d8 but the ions the but the metal has we have Ni2 plus and so configuration of Ni2 plus is what two electrons comes out from the outer motion which is nothing but 4s so it becomes 4s0 3d8. Now when we draw the orbital diagram of this it will be 3s4s and this one is 4p 3s has 8 electron 8 electron so 1 2 3 4 5 6 7 and 8 4s is 0 4p is 0 again this Cn minus is strong ligand so this electron jumps over here okay and it forms and it forms orbital diagram which has this number of electron and the arrangement of electron we can show like this we have 1 2 3 4 5 6 7 8 okay and here you see this orbital goes under hybridization same thing we have and we get 5 dsp2 hybridized hybrid orbital 1 2 3 4 5 dsp2 hybridized hybrid orbital correct in which the ligand donates electron and forms bond okay so first of all you see dsp2 hybridization there are two kind of geometry we have one is trigonal bipyramidal so if you look at the structure here trigonal bipyramidal is and other one is square pyramidal so trigonal bipyramidal means what we'll have metal here right first of all I'll draw this one metal here ligand ligand ligand ligand and one ligand will be on the top like this so all these are bonded with the metal this is square pyramidal okay trigonal bipyramidal you see has this kind of structure we'll have a triangle okay we'll have a triangle and on this from the top we'll have one ligand from the bottom we'll have another ligand all these corners we have ligand present and metal we have here at the center of this triangle okay so this is structure is this structure is trigonal bipyramidal and this structure is square pyramidal square pyramidal correct so you see here the information given here is with different niac and bond length out of which four bond lengths are same and one is different see you see in this trigonal bipyramidal structure there are you know three bond lengths which are same these are these bonds these bonds are same three bond lengths are same and two bond lengths are different here but here this four bond length are same but one is different so the structure or geometry for this molecule will be hybridization is dsp2 and geometry will be square pyramidal according to the given data in the question okay the two possible geometry we have with this hybridization dsp3 it's not two it's three here change this over here it is dsp three not two okay five are metal we have no dsp3 so dsp3 hybridization there are two geometry possible tbp and square pyramidal according to the given data square pyramidal is the correct point okay so this is we have two examples we have discussed with coordination number five now next we are moving into coordination number six okay so the next thing we have to discuss is for the coordination number six okay so for the complex having coordination number six how do we do this type of questions okay the question is the question is first of all we'll see a few things into this and then we'll move to the question you see here there are two types of a hybridization possible okay first of all this coordination number is the most common coordination number most common and most important type of coordination number okay generally two types of hybridization possible in this kind of coordination number the one type is sp3 d2 other is d2 sp3 the geometry is always octahedral or square by pyramidal sp3 d2 means the complex is high spin complex complex is high spin complex and d2 sp3 means the complex is a low spin complex high spin and low spin complex high spin complex why we are calling it as because the outer d orbital outer d orbital takes part in it takes part in hybridization okay lower the spin complex because the inner d orbital is taking part okay inner d orbital is taking part okay so whatever the you know complex we have whether it is high or low the geometry is always square by pyramidal square by pyramidal or octahedral okay that's what we should know now the first question in this you see the first question is the complex is given which is co nh3 6 3 minus right we need to find out the magnetic property of this complex magnetic property we need to find out hybridization we need to find out geometry of this molecule okay so first of all what we do we'll try to find out the nature of the ligand so n is 3 is a strong ligand nitrogen atom donor donor strong ligand okay oxidation is state of cobalt is plus 3 okay oxidation is state of cobalt is plus 3 correct so now the electronic configuration of cobalt if we draw cobalt has 27 electrons atomic number is 27 so argon 4s 2 3d 7 right 1 2 3 4 and 5 3d orbital is this this is 4s and this is 4p this one is 3d this one is 4s and this one is 4p 3d as how many electrons to understand that if I write down because the metal is in plus 3 oxidation state so for co plus 3 the electronic configuration will be argon 4s 0 3d 6 okay so if this configuration we need to draw so there are six electrons and since the ligand is a strong here so we can pair up the electrons against the rule directly which is nothing but 2 4 6 2 we have inner d orbital 1s and 3p so this orbital goes under hybridization and it forms six different hybrid hybridized hybridized orbital this is d 2 sp 3 hybridized orbital where the ligands donate electron and forms bond correct so to write down the answer of this question you see first of all all the electrons are paired so it is diamagnetic okay hybridization is d 2 sp 3 geometry is what octahedral odd square by pyramidal since inner d orbital is involved it is low spin complex so any of these questions they may ask you regarding the low spin or high spin complex geometry hybridization or magnetic behavior okay one more example we will see here suppose the molecule is fe f6 molecule is fe f6 and here we have 3 minus we need to find out again the hybridization of this we need to find out hybridization we need to find out geometry we need to find out magnetic behavior now for this molecule you see first of all halide ion all halide ions whether it is chlorine bromine whatever it is all halide ions are weak ligands so first of all we will write the nature of ligand f minus is a weak ligand okay oxidation state of ion which is plus 3 in this molecule here it is plus 3 now we know the atomic number of ion which is 26 and the configuration is argon 4s 2 3d 6 so the electronic configuration of fe plus 3 will be argon 4s 0 3d 5 okay orbital diagram if I take suppose it is 3d 1 2 3 4 and 5 this is 4s and this is 4p so this one 3d has 5 electrons so 1 2 3 4 5 since the ligand is weak there's no rearrangement against the hunts rule okay and we have one more thing that we have 4d orbital also vacant here in this state right we need 6 vacant atomic orbitals so that is 1 4s 3 4d and 2 4d okay so this 6 atomic orbital goes into hybridization at forms sp3 d2 hybridized hybrid orbital okay 6 sp3 d2 hybridized hybrid orbital so the hybridization hybridization of this complex is sp3 d2 magnetic behavior electrons are unpaired so it is paramagnetic geometry is again octahedral since outer d orbital involved so it is high spin complex high spin complex okay so this is how we find out hybridization geometry and magnetic behavior of these complexes okay you need to know the oxidation state of the metal and for that you require the charge on the ligand okay and the nature of the ligand nature means whether it is weak or strong you must keep this in mind h2o is a weak ligand all halide ions are a weak ligand generally in general when carbon and nitrogen atom are the donor atom of ligand then the ligand is a strong ligand in general but there are few exceptions again but in general whenever the carbon and nitrogen atom are the donor atoms present in the ligand the ligands are considered to be strong ligand okay this is how we find out the uh the hybridization and geometry of these complexes so this is one theory that we have discussed that is valence bond theory of the complexes the next theory we have to understand is crystal field theory that we will discuss next class thank you very much