 So, so far today I think magnetism wise we did not discuss too much, we have introduced the topic magnetism where the origin of the magnetism coming or magnetic moment coming into the molecule. It is the molecule unpaired electron that is responsible, number of unpaired electron should technically give you the clear cut idea what is the magnetic moment going to be, there is some more to it. Of course, in reality we know that it is orbital angular momentum and the spin only values that should give you that should combine, but you do not need to combine because it is somewhat restricted, still somewhat allowed that is where when it is allowed then you have a better value or the higher value for magnetic moment right. And when that spin only plus orbital angular momentum is coming into picture that is what we have introduced, but we will be discussing from here on in the next class ok. Now, we will we have two more topic, one is this magnetism the next one will be bioenorganic, two classes we should be able to finish yeah. Oh yes that is a good question sure it I think if I am getting I think in the beginning I did not get your point at the end what I get is. So, you have to think the total ensemble like if of course, if a molecule one molecule has some magnetic moment another molecule associated with it has another magnetic moment you have to basically assemble yes at the end at the very last at the end we will see it is not the individual magnetic behavior will matter it is the total collective molecule whatever molecule you are associated with. If it is only one type of molecule one center then that is of no problem, but the moment you have let us say cluster ok. One thing is attached via another thing through another thing then things are more complicated. So, that is what that is where I would like to take you to at the end of the magnetism. So, they can two magnet can communicate through a mediator one spin up another spin up is it going to be magnetic moment going to be addition of these two can they talk with each other and reverse inverse. Let us say this is up in between these two magnet there is something which is communicating between the two can that communicator influence the total magnetic behavior. So, there is magnetic communication that is actually the basis for you know more fun in this area. We will not discuss a lot of case maybe one or two case we will discuss and then leave ok. So, I think is that the type of queries you are having or maybe I have taken you to a different reaction. Some something related initially I could not hear you, but at the end I think I heard you correctly ok. Now, hopefully the next part will not be that long another 20 minutes or so, or 30 minutes maximum depends on how you how you see the third tutorial which is on coordination compounds you have the print out the question no problem I have that question here ok. Alright, tutorial if you want to go I have no problem, but I think it is important little I will try to get it done very quickly. Tutorial questions were uploaded in the model and nowadays all of you are having smartphone. So, if you have Wi-Fi connection you could have downloaded earlier ok do not come to my office to download it no ok. Just take a one of you should have been able to download it. If you are really dying I you could have asked me I could have send you or give you a print out ok. Anyway that does not matter too much it is in there if you like I can give you the print out the next time which going to pay for it ok we can we can give it IIT can pay for it of course, you are paying IIT right no problem. So, first question crystalline silver oxide is diamagnetic explain question number 1 ok. So, sometime you have to be little bit it causes some some not all of the question either in the exam or in tutorial are going to be straight forward there is something more into it. So, silver oxide you just calculate you figure out what is the silver oxidation state do it plus 2 what is the electronic configuration ok time up silver oxidation state is plus 2. So, it is a d 9 electronic configuration d 9 on electronic configuration means 1 unpaired electron ok. So, it should be paramagnetic diamagnetic means to spin right or unless paired spin how it is this is where I was saying it is a trick question. So, it is a it is a mixture of sorry yeah that is what it is. So, that is where it is a trick question. So, it is a mixture of ag 2 O and ag 2 O 3. Now, you have to look back at each of them let me let me try to discuss you have to look back each of them silver plus silver plus 1 plus and silver 3 plus 1 minus is usually not possible. You want to do not worry I will I will give you I will upload since I have made the slide I will upload ok no need to I mean if you bothered if you are bothered to take the picture and take it it is a public domain thing why why should I not upload it ok. No need to just think little bit silver 1 please if you want to get it done quickly either you answer quickly or I give you the answer and go I will upload the slide ok. Now, silver 1 is going to be detain configuration that is diamagnetic this is silver 1 2 of the silver 1 1 is oxide another over here this is silver 3 plus oxide is always minus 2 6 minus. So, it has to be silver 3 plus silver 3 plus means what 1 d 8 configuration right. So, d 8 configuration can be can be diamagnetic ok d 8 going to be square planar because silver is in high oxidation state it is a. So, this is where sometime the problem comes either the high oxidation state. So, it is the combination of both the ligand should be strong field or metal should be high oxidation state or the combination of higher oxidation state and strong field ligand then you are going to get that ok sorry oxidation state ag 2 O plus 1 silver is in plus 1 oxide is oxide all oxide water all oxide metal any oxide you see is minus 2 water if you split H plus H plus oxide 2 minus right. Now, that is fine. So, there the geometry as you can see over here silver is linear and silver 3 which are in grain here it is going to be square planar square planar oxide oxide silver is in the middle ok. Now, that is you know that is some question which is directly I think these questions are something we have given earlier also same questions for the tutorial. Now, work out the hybridization and geometry for the following complexes using the balance bond approach. First one is ok the second one is why is that fantastic. So, nickel is in 0 oxidation state and then therefore, you I think this is may not be feasible 4 of them sp 3 hybridization 4 of the ligand sp 3 hybridization it is going to be tetrahedral nickel tetra cyanide it is going to be strong field ligand and thereby the pairing will occur by balance bond approach. We are going to discuss it by balance bond approach and you know that pairing will occur the elect the hybridization will be then d sp 2 that is going to be the square planar. Of course, you can explain it better by crystal field theory, but that is ok. Now, I think I will skip this this is the summary what is given here. Yes that that that is where over here O S O 4 electronic configuration is going to be 4 F 14 5 D 6 6 S 2 this is what it is little bit difficult over here to do it by V V T. Anyway if you do it by ionic approach osmium 7 plus it is going to be 4 F 14 osmium sorry oxygen osmium is going to be 8 plus 6 plus 2 8. So, 4 F 14 total it is going to be D 3 S. Hybridization and it is going to be tetrahedral we do not call it D 4 or anything it is going to be D 3 S 3 of D and 1 of S D 3 S that is going to be tetrahedral by covalent approach similarly you can look at and you can tell that it is going to be D 3 S as well. It is little bit on a borderline explanation it is it is see I mean these are these are little bit older approach V V T approach. So, the explanation is going to be little bit screwed at some point, but this is ok you can kind of make sense for both of them ok. Now this is most of them are clear cut tetrahedral square planer tetrahedral square planer linear tetrahedral tetrahedral linear is that silver ammonium where another molecule will be coordinating. Anyway this is you should be able to find out by whatever you have done earlier. Tetrahedral place should not be a problem square planer are the one where you have the stronger ligand ok if it is a D 8 configuration that is where the square planer comes into picture, but the real reason that balance bond theory wise explanation was simply given it is a strong field ligand and thereby it should be paired up and shown that is what you have learned in the balance bond approach, but by crystal field approach we have shown how things are going in terms of electron distribution, how orbitals are splitted and thereby why we are saying that D 8 is going to be the square planer one. Now while the most able chloride of zirconium is zirconium tetrachloride that of palladium is palladium chloride PDCL 2 yes it is also called relativistic effect. What is that anyone wants to answer? No it is it is more of a D orbital see what we have seen so far it is there are let me tell you there are these are palladium is going to be down the periodic table. So, you have already seen this z effective because those orbitals the higher orbitals which ever getting involved they are not neutralizing the positive charge effectively. So, as we were discussing in the very early class first class or so, z effective is going to be very very strong. The moment z effective is going to be strong they will be pulling out pulling in those D orbitals which are going to participate into the plus 4 oxidation state up to plus 2 is ok, plus 4 another 2 electron release although technically possible since z effective is very high it is going to not allow those electron last 2 electrons to get oxidized to palladium something like plus 4 relatively speaking. So, those due to the high z effective you are not going to participate very strongly or those plus 4 oxidation state achieving becomes difficult. This is what it is called relativistic theory or relativistic effect or so, called inert fair effect. As you go down in the periodic table the participation of the electrons becomes less and less. I mean if from first row to second row if you go there you do not see much effect as you go down below further it becomes more prominent right. So, therefore, although higher oxidation state is technically feasible, but practically it becomes difficult to access those removing those electrons last third electron, fourth electron becomes extremely difficult because z effective is higher this is going to pull in very tightly. These are becoming more of a core like the electrons become more of a core I mean why it is difficult it is it become part of the core. So, much attracted it does not want to leave those or the atom is not going to leave those. Next question question number 4 when high pressure is applied what type of electronic configuration is favored for a d 5 transition? When high pressure is applied what type of electronic configuration is favored for a d 5 transition metal complex? It is you are going to put high pressure you are going to take out the electron from those let us say dz square orbital which is shown in there and thereby you are going to kind of pair up the electron. So, because it leads to low electron density between the metal and the ligand that is along the bond axis you are going to end up pairing. So, you will get the low spin complex not clear. When high pressure is applied I will tell you it is not clear to me as well let me see what type of electronic configuration is favored for a d 5 transition metal complex. So, d 5 you are going to have two configuration right t 2 g t 2 g 3 e g 2 and another configuration is going to be t 2 g 5 e g 0 right. Now, I think it is it is you are going to apply more field means you are going to split between the t 2 g and e g e g level. If you are pulling out you are going to separate out the t 2 g versus e g still not clear I am getting more confused I will I will I will bring it back ok leave it I think pulling. So, base it is clear that you pull out then you decrease the electron density along the axis t h can can you explain low electron density between the metal and the ligand that is along the bond axis what exactly is happening repulsion between what ligand and metal ligand and metal ok ok ok ok ok. So, when you are compressing wait wait wait wait I am not clearing it when you are compressing between these two then repulsion force to be much more yeah. Then repulsion for any because it is only thing less than right ok fine I will I think it is still not clear unless I am 100 percent clear I will not ok fine it is it is interesting yeah. I think you are saying exactly what he is trying to say, but explaining explaining becomes little bit difficult let me let me digest it little bit better I will come back ok. Now, provide reasons for the fact that a number of tetrahedral cobalt 2 complex are stable whereas, the corresponding nickel 2 complexes are not anyone provide reasons for the fact that a number of tetrahedral cobalt 2 plus complexes are stable and whereas, corresponding nickel 2 complexes are not nickel 2 is what d 8 d 8 tetrahedral square planar more or tetrahedral versus octahedral even if you are considering and cobalt 2 plus tetrahedral and octahedral you are considering what you see that is what you have to see. So, d 7 tetrahedral C F S E and d 7 octahedral C F S E. So, what you see overall is d 7 tetrahedral complex is greater C F S E is greater than the d 8 tetrahedral complex. Calculate the d 7 C F S E for tetrahedral field what is the d 7 C F S E minus 12 d T right d 7 C F S E tetrahedral d 7 yeah tetrahedral is high spin always high spin minus 12 right 12 d Q what is for d 8 minus 8 minus 8 12 and 4 over there right. 12 which one is more stable d 7 is more stable. So, d 7 is going to be cobalt 2 plus d 7 is going to be tetrahedral you know I am asking you to compare tetrahedral versus tetrahedral tetrahedral for d 7 tetrahedral for d 8 octahedral for d 7 octahedral for d 8 you see the answer is within that ok. So, C F S E of d 8 octahedral d 8 octahedral will be just do the octahedral simple octahedral ok. So, it is going to be minus 1.2 or 12 delta Q or d Q and what is for d d 7 minus no minus 1.2 or 12 delta Q or d Q and what is for d 0.8 or 8 d Q d 7 calculate that is what I was saying maximum cases calculate the C F S E of both the geometry d 7 what is the geometry d 8 what is the geometry what is the splitting or sorry what is the stabilization energy this you should be able to do it in your dream. Now, the statements here is correct C F S E for d 7 is more for tetrahedral case you have find out and C F S E for d 8 octahedral is more C F S E of d 8 octahedral complex is greater than d 7 octahedral you calculate you will find. So, for this d 8 it is going to be minus 12 d Q or minus 1.2 delta 0 for d 7 it is going to be minus 8 d Q or 0.8 delta 0 or delta octahedral right the fact is here. So, the answer is correct the experimental fact is also given over here which you can corroborate fine provide reasons for the fact that a number of tetrahedral cobalt complex are stable whereas corresponding nickel 2 complexes are not that is the answer of course, this is clearly shows that cobalt 2 plus preferred tetrahedral nickel 2 plus are not preferring tetrahedral I am not saying what it is. So, answer is ok. Now, using the crystal field stabilization energy as criteria indicates whether you expect the following spinels to be normal or inverse. Now, calculate calculate and figure it out another 10 minutes you should be done calculate quickly see this is what I was really trying to tell you that you should be able to calculate the C F S E really quick write down calculate C F S E for F E 3 plus what we have asked you to for normal and inverse spinel you do not have to worry about tetrahedral you just think about octahedral F E 3 plus and F E 2 plus octahedral and it is going to be high spin high spin octahedral case you just calculate should I calculate no I can calculate F E 3 plus is D 5 system high spin D 5 system is 0 D T 2 G 3 E G 2. So, there is 0 C F S E high spin iron 2 plus that is D 6 D 6 means high spin T 2 G 4 E G 2 F E 2 plus having minus 4 D K 2. As a stabilization energy F E 3 plus having 0 stabilization energy. So, higher oxidation state is having less stabilization lower oxidation state is having higher stabilization you are going to get an inverse spinel I guess I was discussing in the class as well. Now, CO 3 O 4 is a little tricky case because CO 3 is less stable stable stable stable low spin high charge and iron cobalt D 6 iron is low spin because high charge even with wheat ligand this is a tricky case. CO 3 plus has a similar structure with D 7 and D 6 configuration this is going to be a normal spinel D 7 D 7 is what? D 7 is what? CO 2 cobalt D 6 CO 2 plus is D 7 iron cobalt D 7 is CO 2 plus. If you CO 2 plus with oxide it is not going to be the low spin it is going to be the high spin. CO 2 plus lower oxidation state oxide is not that of a great ligand. So, it is going to be the high spin CO 2 plus high spin D 7. So, it is going to be minus 8 D Q ok. How about CO 3 plus? CO 3 plus it is going to be D 6 D 6 it is going to be minus 4 D Q if it is high spin 3 T 2 G 4 E G 2 T 2 G 4 E G 2 if it is high spin D 6. Now, the problem is that is minus 4 D Q technically speaking it should be inverse spinel higher oxidation state is having lower stability compared to the lower oxidation state. But, CO 3 plus this is an exception you have to kind of remember, but we will try in the exam we will try to give not such example ok. CO 3 plus being high oxidation state and even with the lower oxygen lower you know even with the you know weaker filled ligand such as oxide we still are going to get these are normal spinel ok. Next few more question left by showing the details determine the CFSE for the following complex these are the very simple question. CFSE for Fe 2 plus Fe 2 plus is D 6. CL is a weak filled ligand iron is 2 plus low oxidation state. So, it is going to be the high spin D 6 high spin T 2 G 4 E G 2 T 2 G 3 E G 2 cancels out minus 4 D Q all of you got it or should I wait a little bit 30 seconds. The titanium titanium is going to be sorry not tungsten my bad ok tungsten. Tungsten is going to be carbonyl is going to be strong filled ligand almost the strongest you can get out there. So, it is going to give you low spin for sure T 2 G 6 E G 0. Yeah it is usually we when we count we just count for D until these like D 6 that is why we are counting because that was atomic orbital you have to take into the atomic orbital into the more of a complex orbital or molecular orbital. We never so far we never talk about D 2 S you know atomic orbital wise it is correct D 4 S 2 D 4, but in reality when you are doing anything with the hybridization or anything with specially with the complexation you have to say it is a it is a it is a mixing orbital mixing is happening yeah you have to say it is a D 6. So, T 2 G 6 E G 0 ok. Now explain what is meant by the term synergic bonding this is the textbook question synergic bonding I think you have studied for your exam before it is the it is the pi bonding between the between the ligand sorry sigma bonding between the ligand this ligand orbital and the metal orbital and pi back bonding between the metal orbital and the anti-bonding orbital of the carbon monoxide ok. It is written in the way I think you have studied before synergic bonding it is the ligand donates metal gives back. So, it is I am teaching that is what I was trying to say you tomorrow you will give me money that is the synergic bonding right once you become trillion year I am sure some of you billion year or whatever ok. Now the chromium 2 plus ion in CRF 2 is surrounded by 6 fluoride ions chromium 2 plus this is this is this is a clear cut depth I mean statement is given it in CRF 2 you are saying, but it is surrounded by 6 ligand ok. It is surrounded by 6 fluoride of these 4 are at a distance of 2 angstrom 4 are shorter and the other 2 are at a distance of 2.43 angstrom explain this observation ok. So, chromium 2 plus is surrounded by 6 fluoride ions in an octahedral environment it is surrounded by 6 fluoride ion chromium 2 plus is what d 4 scandium titanium vanadium chromium d 4 S 2. So, 2 plus is going to be d 4 now it is going to be T 2 G 3 E G 1 right. So, the unsymmetrical distribution of electrons in E G orbital that is what we were trying to show right. E G orbital whenever you are having unsymmetrical feeling then there is a possibility of further splitting of the E G orbital. E G orbital can be splitted this observation suggest that E G electrons in d G 2 orbital because the 2 of them are longer 4 4 are shorter 2 are longer little bit longer. 2 longer means that direction it is getting stabilized. The ligands are getting longer that means, ligands are not coming close I think now I can explain little bit that question. Anyway, ligands are further and therefore, Z it is going to be Z out right slightly Z out and you are going to get d Z square orbital stabilized ligand are far at the Z direction. Therefore, d Z 2 orbital is going to be stabilized slightly lower in energy and the electron since it is stabilized electron is going to go there because you it wants to achieve the stabilized state ok move on that is it.