 Hello everyone once again I welcome you all to MSB lecture series on transfer metal chemistry. In my previous lecture I was discussing about metal-metal multiple bonding while describing the reason for metal-metal multiple bonding I also showed you how one can use effectively molecular orbital concept to understand the significance of metal-metal bond and also to understand the magnitude of the metal-metal bond. Let us continue from where I had stopped. Today we shall look into more quadruple bond and quintuple bonding we come across among coordination compounds. I have given here two examples look into it carefully one is a chromium complex and another one is also a chromium complex but ligands are very different and you can see in the first one you can see orthometallation is there on opposite side and you see weak interaction with this phenyl group carbon atom. So you are having some sort of interaction here and same thing is true here so that means essentially chromium has one ligand that is a very bulky ligand and one should remember if you want to stabilize any metal in its low valent state which is coordinatively saturated only ways you have to do it kinetically how to do it you have to go for a very bulky ligand so that it will give umbrella protection surrounding the metal atom so that it does not give scope for other bigger ligands to enter. So in this context initially this mono-ligated chromium compound was made and then this interacts to such mono-coordinated chromium species they interact with this ortho substituted carbon here you can see and as a result what happens it appears to be having two coordination with almost linear geometry and in this case what if you carefully count here we have five bonds are there here and if five bonds are there how to explain that one same thing is true in this case also whereas here nitrogen is there central pyridine nitrogen is coordinating and also the nitrogen lone pair is coming here see this is the ligand used in case of the first one and in the second case we have used this ligand here and you can see here first it binds here and you will be having almost one is to one metal to ligand ratio and then since this side is open what happens two such molecules will arrange in such a way that something like this they come and they establish a dive metallic core. So now we have to explain the origin of five bonds in this one how to do that one again we have to go to the same concept that we used earlier to explain multiple bonding like single bond, double bond, triple bond or up to four bonds or quadruple bond same analogy we should use here one interesting thing is here in previous case we were just confined to four orbitals that is dz square something like for me sigma bond and then we had dx z and dy z forming two pi bonds and then we had dx y forming delta bond four orbits we used. So four orbital means if the capacity is 8 then you can have bond order of 4 but if we want to have bond order of 5 in order to see the formation of quintuple bond we need 5 electron on each metal d orbitals that means d5 configuration we had to take in but in earlier case when we took d5 what happened was 2 electrons were going to delta star delta star as a result what happens bond order was reducing from 4 to 3 but in this case what happens since it is using sp hybridization we are left with all the 5 d orbitals for bonding purpose. In square planar complexes we are using dx square minus y square for making method to ligand bond you just recall VLS bond theory where we are using dsp to hybridization that d orbital is dx square minus y square whereas in this case it undergoes sp hybridization to accommodate to 2 ligands surrounding chromium autumn as a result what happens all d orbitals are left unutilized and chromium electronic configuration if you see d4 s2 or d5 s1 and this ligand if you see the top one if it is mono anionic here through CH activation it forms so then it is a mono anionic means at least it should be plus 1 and as a result what happens chromium will be left with 5 electrons so 5 electrons are there and 5 d orbitals are there. Now that means along with dxy dx square minus y square also participates in delta bond formation so that means now we have 1 sigma bond through dz square and 2 pi bonds are there through dx and dyz and then we have overlapping of dxy and dx minus y square that leads to 2 delta bonds that means 10 electrons are there in the bonding 10 divided by 2 equals 5 so bond order is 5 this is how you can visualize printable bonding in this kind of compound so that will become clear when we go to the next one. So again just I have shown here how it happens so next you can see here yeah this is a compound another compound here of course here you can see just go back to this one and here if the delocalization happens here of course this is anionic you should recall here n 2 are coordinated so n minus this is so here almost is similar to n h 2 minus I would say n r 2 minus so now what happens this is delocalized as a result becomes mono anionic very similar to couple of ligates I showed you in my earlier class in couple of examples showed you in my earlier lecture you can see now anionic it is so chromium or molybdenum they are in plus 1 state so d 5 electronic configuration and how this compound is made I have given the method here first you take this compound where chromium is in plus 2 state CrCl2 coordinated to 2 neutral ligates such as THF you take this one and treat this one with lithiated this aromatic group and once you lithiated this one is there LICL comes out and you left with one CL here as a result what happens it forms a dimeric structure having 2 chlorobidges here and now this one needs reduction for that one Kc8 potassium over graphite is taken slight excess in THF then it will abstract to chlorides to generate chromium one species from chromium 2 and then now chromium one species is co-ordinatively unsaturated it interacts with this carbon here and then it establishes a like a bidented made of coordination having this kind of structure and you should remember that here we have 5 electrons are there in the d orbit so this was reported in 2014 in Dalton transaction so another compound is there I will give you the preparation of this compound also before I proceed to show the amort diagram to confirm the presence of 5 bonds so in this one to take it is already you know quadruple bonded compound very similar to rhenium dimer Re2Cl82- so just look into this one so this is a potassium salt because tetra anionic so what one should do is add a ligand that ligand I will show you what it is and then you can make a compound of this type here I will show you what is A and then again you use the reduction process using potassium over graphite that leads to the formation of a quadruple bonded compound like this so you may be wondering what is A so A is this one here lithiated this bulky aromatic compound here this is the lithiated one and then if you just see here two lithiums here two lithiums are there that means when you are using two equivalent one lithium is coming out as LICL so you can see the missing of one chlorine here and then another one another one will remain here coordinated to lone pairs of chlorine here and then of course once when you reduce with Kc8 everything is gone and you simply ended up with a malbeden plus one the compound having this mono anionic ligands bridging in this fashion I am sure it is clear about repetitive methods so this came in 2009 in JAGS so now let us see here malbeden means plus one state d5 system to form this complex as I mentioned earlier malbeden has not used dx minus y square otherwise it would have used if it had squared by the geometry so now this is also available along with dx y you know oriented in this fashion something like this one is like this and another one is at 45 degrees both of them can only overlap in this fashion when they stacked in along z axis you can see here 5 are there now just see I have added dx minus y square also for this bonding scheme here we have total 5 electrons whether it a chromium or malbeden first to establish a sigma bond by utilizing 2 electrons and now pi bond we are using these 2 these 2 4 electrons so 2 pi bonds are there and now we are using now 2 electrons here one each from dx y and dx y square from both the metal centers and then we establish a 2 delta bond so then the bond order is 5 so this is how you can explain quintuple bonding among coordination compounds having d5 electronic configuration with dx square minus y square available for metal metal bonding. Now let us go to another interesting compound here if you recall my previous lecture where I classified metal metal bonding compounds into covalent bonding, dative bonding and symmetric metal metal bonding so now let us look into dative bond scheme so when a metal with at least 2 d electrons and a moderately high electron count is adjacent to a metal that is coordinatively unsaturated and electron deficient in that case what happens both of them can have some sort of Lewis acid base interaction the one metal that is electron rich can give a pair of electron to the one which is electron deficient provided it has empty orbit of suitable symmetry to take those electrons in that case what happens you can establish a dative metal metal bond or you can call it as a adduct between 2 metal centers so I have given one ideal example here and it is very interesting to analyze this one and there are several ways one can analyze this compound this is a phosphide ligand here you can see pr 2 minus and it is also pr 2 minus and we have 3 carbon groups all together and if you just look into this one well of course this can be plus 2 state nickel can be plus 2 state provided it anionic establishes a bond with nickel in that case what happens this will be plus 2 state that means d8 system d8 plus 4 12 and plus 14 so this becomes a 14 electron species on the other hand this one is nickel 0 now pr 2 acting as a neutral ligand towards this nickel so d10 plus 8 is 18 electron system that means it is very similar to NiCO4 or NiPPH3 4 times now so this is 18 and this is 14 so this electrons can go here so that it becomes a 16 electron species and to an extent electron deficiency can be removed here so you can see how dative bond is directed from this co-ordinatively saturated and electronically rich nickel to this 14 electron species that means what are the other alternates for example we can have phosphide one each on these two metals in that case what happens both of them will be nickel plus 1 state if they are in nickel plus 1 state what happens the d9 system they become both of them become d9 system and then we have anionic on both sides so two electrons are coming here and bridging one neutral also two electrons are coming whereas in this case one carbon monoxide is there and of course there is one metal metal bond you can think of d9 system so 16 electrons will be there for this one and then here in plus 1 state and then you have 4 electron this will be 80 electron but here if it is 16 many d8 systems are stable with 16 electron species you cannot think of a dative bond so since it is an example that is proved through experiments and also x-ray structure so this kind of analysis is incorrect so nickel plus 1 is not correct instead what I said earlier so this is in 0 state and this is in plus 2 state is the correct analysis and nickel with 18 electron is donating 2 electrons to this one so let us analyze for better understanding so now you can see here 2 options so now I have shown both the terms here so I have shown here nickel this is binding to 4 neutral ligands with 18 electrons with d10 system that is 0 and now this is going this way that means now this is in nickel plus 2 because here we are talking about phosphate anionic so this is a covalent bond and this is a coordinate bond now this is in nickel plus 2 and 60 electron species and now this will be nickel 0 so this also becomes 10 plus 60 electron species both are 60 electron species in that case what happens none of them can give electrons to this on the other hand other alternative is as I mentioned nickel 1 nickel 1 can be there again we overruled from my previous discussion and argument and then what would happen here 17 and 18 and now it is 16 and 18 but when we looked into the x-ray data what it shows is nickel 2 NiCO 170 picometer and Ni2 phosphorus 216 picometer now let us look into the bond parameter for this 1 nickel 0 NiCO 178 picometer that means it is longer why it is longer because it is taking some back bonding is there because this electron is rich when the back bonding is more to CO what happened NiCO bond shrinks and CO bond distance increases so here it is electron deficient it is not readily donate electrons to pi star of CO as a result what happens NiC bond is little longer as a result CO bond still has triple bond character so it is slightly shorter so that means this proves that this is in plus 2 state and this is in 0 valent state and again that is also reflected from NiP 224 picometer this is a coordinate bond whereas here we have a covalent bond because it is an anionic P minus so 216 this is shorter always covalent bonds are shorter compared to coordinate bond so this is another proof to say that this has a dative bond with nickel 0 18 electron species and nickel 2 14 electron species so this proves beyond any doubt that there is a dative bond between 2 nickels and this 0 valent nickel is donating a pair of electrons to this one to overcome its electron deficiency this is how one should analyze carefully by looking into all parameters and NINI distance is 241 picometer or 2.41 anion strength units so now let us come to the last one I did mention about symmetry metamethyl bonding so what would happen here is weak metamethyl interactions observed due to the molecular orbital symmetry interactions of field and the empty metamethyl bonding orbitals under anti-bonding orbitals observed in D8 system and is very rare D8 system should not have any metamethyl bonding due to the zero bond energy as both bonding and anti-bonding orbitals are completely filled that means you can see here for example if you recall a more diagram just I am writing here 2 electron sigma 2 pi here and 2 delta and then again delta is filled and pi is filled and then sigma is filled so this is sigma 2 pi and then delta delta star pi star 2 pi star and sigma star so all 10 electrons are there so bond order is 0 so that means you do not expect any metamethyl bonding and in a square panner complex having D8 system but HP gray and others observed in a few bi or polymetallic complexes having D8 electronic configuration and showed the presence of weak metamethyl interactions both in solution and in solid state that means if you observe in solution these molecules retaining weak interactions and stacking along the axial position probably there is bonding that has to be explained so one such complex of iridium with isocyanate showed weak metamethyl interactions and that gave enough proof to go further you can see here this iridium complex here iridium plus compound D8 system having neutral isocyanate ligands are stacked up in this fashion having unusually shorter iridium iridium bond despite war rule we war rule any metamethyl bonding as bond order comes to zero then how to explain this one so gray proposed in 1974 that these weak metamethyl interactions were caused by a molecular orbital symmetry interaction between the field sigma metal metal bonding and sigma star anti-bonding orbitals with the mt pz sigma and sigma star orbitals you should remember when we think of valence bond theory to explain square planar complex geometries we are using dsp2 that means d orbital is dx my square and then we are using s orbital along with px and py orbitals and pz orbital is left unutilizable in a square planar complex along with four other orbitals but in the d8 what happens everything is filled that means now the interaction between dz square where we have sigma sigma star field and unfilled they interact with mt sigma sigma star of pz orbitals that means take pz orbitals from two metal centers and try to establish a molecular orbital sigma pz and sigma star pz and then they are empty and whereas this is filled that means there is some interaction happens between these two pair of molecular orbitals the field dz square bonding and anti-bonding orbitals with mt pz due to this kind of interactions the mt orbitals are pushed up in energy and the field orbitals pushed down in energy by this symmetry interaction so due to this kind of arrangement this generates a weak metamethyl bond strong enough to allow these complexes to form mm bonds and retain them even in solution so this was observed in many complexes of iridium and even palladium and platinum compounds with the d8 electronic configuration probably if you look into literature and work by gray you will come across more examples with this I am completing the discussion on metamethyl multiple bonding so once this is done you should remember one thing when we predict metamethyl bonding in these cases where you can have single bond double bond triple bond quadruple bar or two quintuple bonding the concept usually is very different compared to what we come across in predicting metamethyl bondings while using 18 electron rule sometime they may be in some cases it does not apply do not try to mix up their different entities and different type of components we come across these things maybe there may be some examples where we can see both merging to give a correct explanation about 18 electron rule nevertheless their entities but beautiful concepts and without any problem one can explain using simple molecular orbital concept what one should remember is again in case of square planar complexes we have four d orbitals are there d z square d x y d x z and d y z and d x z and d y z generate two pi bonds d z square generates one sigma bond and because of their orientation d x y will overlap only like this only like this and they generate delta bonding and then we have eight electrons in the bonding and if bond order is nothing but number of electrons present in bonding orbits minus number of electrons present in anti-bond divided by 2 so that gives the bond magnitude of 4 to have four bonds or quadruple bond though a square planar metal complex should have d 4 electronic configuration or they should have four electrons in their d orbitals and if we have unutilized d x y square also with lower symmetry or sp hybridization for example in that case what happens d x square minus y square is also made available for bonding in that case all 5 d orbits are available for bonding you can imagine or you can explain the quintuple bonding there may be more examples of course if you are more interested you can always go to literature and look into more examples in my next lecture I shall discuss about the preparation of some of the important coordination compounds and argumentary compounds keeping their utility in further reactions and then making ideal compounds for homogeneous catalysis until then have an excellent time reading chemistry.