 Once again, it's my pleasure to welcome you all to MSP lecture series on interpretive spectroscopy. In my last lecture, I was discussing about the importance of IR spectroscopy in metallocarbonyls. And also when we have different type of mixidigand complexes having one or more carbonyl groups, it also gives, especially the stretching frequency of carbon monoxide, gives vital information about the electron density surrounding the metal ion and also the donor and acceptor properties of other ligands as well. The donor and acceptor properties of other ligands present along with CO in a metallocomplex have remarkable influence on the stretching frequency of CO. So let us continue some more examples about IR spectra of metallocomplexes and also let's try to learn more about force constant and its relationship with the frequency. So here I have given a list of carbonyls, homolyptic carbonyl complexes, neutral complexes, anionic complexes and also cationic complexes and also I have given the stretching frequency for CO in centimeter minus 1. And some of those things I already discussed in my previous lecture. If you see here nickel tetra carbonyl, it shows around 20, 60 centimeter minus 1. Whereas cobalt tetra carbonyl anion, of course when it is dimeric, it is neutral because it is a 70 electron species and now with COCO4 minus, it is an 80 electron species. Here stretching frequency is 1890 drops considerably because of excess of negative charge on cobalt. So if more and more electron density is there on metal, they will become better by donors and as a result what happens, the stretching frequency of CO drops considerably because of population of electrons into pi star or anti-burning orbitals of carbon monoxide. And similarly if you see here, we have di-anionic compound here, iron compound. For example, Na2, FeCO4 if you consider stretching frequency even further drops to 1790 here because two negative charges are there. And then when we look into positively charged metal ions such as MnCO6 plus, here metal is say reluctant pi donor because of negative charge as a result and not much significant change is there in the stretching frequency of carbon monoxide and it is towards higher set that means CO bond is more stable and Mc bond is relatively longer. And of course CRCO6 we have substantial amount of back bonding because of D6 electronic configuration and also metal in zero valence state. And again, vanadium hexacerminal anion is more or less comparable to that of COCO4 minus and free carbon monoxide is around 2143 and some books say 2130 and all those things but to be precise it should be 2147 centimeter minus 1. And then the effect of binding can be clearly seen here. In terminal binding of a carbon monoxide to metal, the range is around 1850 to 2120. When it acts as a bridging ligand holding two metal centers it drops considerably to 1750 to 1850 and when it is bridging three or more metal centers such stretching frequency even further drops and here when it is bridging three metal centers it comes around 1620 to 1730 centimeter minus 1. That means more and more bridging would leads to almost carbon monoxide behaving like ketonic carbon monoxide group we come across in organic chemistry. Influence of the trans ligand can be clearly seen in this series of complexes I have shown of MOCO3. MOCO3 might remains intact throughout and other three carbon monoxides are replaced by a different type of phosphorus compounds and also nitrogen donors and also diene is there and also pyridine is there. When you see again the same trends are observed here. Here it is not about anionic or ketonic charge present on the metal center. It is about the pi acceptor ability of ligands and if you see here the pi acceptor ability of ligands are considerably dropping when you go from PF3 to pyridine and of course PCL3 and we have trimethylphosphide, triphenylphosphine about these things already we analyzed in my previous lecture and astronitrile it is only sigma donor, diene also partly it can take into pi star but it is a poor pi acceptor and pyridine and in all those things you can see how consistently the CO staging frequency is dropping because when few or more carbon monoxides are left and others are only sigma donor the maintenance of electron are minimizing the electron repulsion by taking back bonding falls only on three carbon monoxide groups as a result what happens? Static frequency drops considerably here and also how immensely this data helps in arriving at different positions of these ligands in spectrochemical series can be seen here and of course when we talk about the position of ligands in a spectrochemical series we are considering both sigma donor ability, pi donor ability as well as pi acceptor ability considering all those things we have made spectrochemical series similarly for back bonding ligands also one can have spectrochemical series in this order and this follows this order here. So now let us look into some problems here and this is related to stretching frequencies of different groups for example here two petro complexes are given of thiocyanate complexes of mercury as well as iron with mercury is in plus two state and also iron is in plus three state and also corresponding stretching frequencies are given here 21, 20 and 710 centimeter minus one in case of mercury complex and in case of iron complex the values are 20, 55 and 18, 28 for assigning the linkage here whether SC is connected to mercury or N is connected to mercury and also iron we have to have some information about free SCN. So free SCN stretching frequency is given here the C triple bond N stretching frequency is 2053 and also for CS it is 748 centimeter minus one just for your information I have also given isocyanate in this one C triple bond N resonates around 2060 centimeter minus one and also it can have ranges depending upon what kind of R groups we have it can have anywhere between 2060 to 20 to 20 to 60 to 20 to 40 centimeter minus one. Now the question is in this complex which is the donor side and if you go back to linkage isomerism whether SC is coordinated to mercury or N is coordinated to mercury and same thing is to in case of iron whether S or N. Now let us compare the data here with the data we obtained for this mercury and cyanate complexes. You see because the wave number of CN for the complex is higher so this is higher compared to this one for SCN and the wave number for CS is lower in the complex compared to free CS. Because of this one we can conclude that SC is coordinating to mercury. This is analogy and let us look into iron complex in case of iron complex if you compare the stretching frequency of CN with that of that one this is much lower here on the other hand CS stretching frequency is much larger compared to free ligand that indicates probably in case of iron N is coordinating to FV3 center. From coordination chemistry point of view let us say I do not have data for comparison for the free ligands then how to assign simply by looking into it yes. We can see here two things about looking to soft and hard interactions mercury 2 plus is a soft center so if since it is a soft center it would prefer to have a soft donor atom and hence SC is a soft donor atom probably you can think that mercury is corrected to sulfur on the other hand iron 3 plus is hard center it would prefer the hard site that means probably nitrogen is going in this way also you can analyze and you can arrive at the right answer and also if you see mercury war is in our mercury sulfide so that also tells that mercury has more affinity towards yes from that point of but here without any ambiguity simply by comparing the data with the data for free ligand the stretching frequencies we should be able to tell the correct linkage of such ligands which are capable of showing linkage isomerism whether SC is coordinating SC is binding or N is binding. Let us look into another example here this is a octahedral complex of cobalt pentasino thiocyanate or cobalt date it can appear of course because of linkage isomerism it can appear in two isomeric forms which give rise to the following IR absorption bands so here we have 26, 2065 and 810 and another one is 2010 and 700 centimeter minus 1 so by simply comparing to whatever the discussion I had with respect to mercury and complexes I showed in my previous slide you should be able to judge that yes in case of isomer 1 N is binding whereas in case of 2 SC is binding you can say without any problem and of course you can also write the corresponding isomers having octahedral geometry without any problem I am not going to go into write those things probably you can write and examine those things. Let us look into another example here the following infrared absorption bands are observed for the complex containing a bipyridine ligand and also thiocyanate ligand it is NCS not NSC in the region of NCS stretching modes that is we have a total of four stretching frequencies are there 2117, 295, 842 and 700 now that means it is say probably a mixture of two isomers are there and we should be able to identify you know stretching frequencies corresponding to N coordination to palladium and also S coordination to palladium again with our previous experience of interpreting data in case of mercury 2 compound and iron 3 compound we can conclude here that when we have 20, 95 and 842 this can be assigned to palladium having NCS bond and similarly when we look into the data 2117 and 700 this can be assigned to palladium having SCN type of linkages. So this how we can interpret and we can say without an ambiguity how the linkage of NCS takes place and then how we can interpret and elucidate the structures by giving the right kind of stretching frequencies to N as well as S bond complexes. So now another problem is it is not a problem it is just for analysis you are already familiar I have spoken many times about CO stretching frequencies CO stretching vibrations of CO can be formed at 2155 centimeter minus 1 the free karma monoxide shows at 2150 again here it is at a different region that is what I am telling you. So books from books to books this value varies 2130 to 2160 it goes up to that one but nevertheless the stretching frequency of free CO is always greater than what we see in case of metallocarbonyless unless there is no back bonding at all in those metal complexes. But lies in the range of 2100 to 1800 in metallocarbonyl compounds that is very much true. The chemical bond in metallocarbonyl compounds is a super position of a sigma metal to carbon bond and a metal to carbon back donation so that you are already familiar I have showed you examples and also I have showed you bonding modes through orbital interactions what information is obtained by the mean value of CO stretching modes according to the part of both contributions to the chemical bond in the following charged metal carbonyl ions. That means whether we can look into the sigma donor component and pi acceptor component whether we can separate these two entities based on comparing the data for a series of compounds keeping the metal constant and changing the number of carbon monoxide are going for cationic anionic and neutral and also replaces some of the carbon monoxide in a metal complex with sigma donor pi acceptor sigma donor pi donor and also sigma donor and pi acceptor type of ligands and then we can analyze and probably we can but still the separation of sigma donor and pi acceptor abilities are components is very difficult because often we see they complement each other but nevertheless it gives some idea about the exact position of some of these ligands in the spectrochemical series so now let us look into the values given for a series of compounds here NSEO for 2094 CO2 for mass 1946 FCO4 2 minus so here we can conclude very nicely that when you go from early metals to late metals, late metals despite having enormous electron richness they are reluctant pi donors that can be seen in case of whether nickel tetracharbonyl or palladium or even silver complex having carbon monoxide whose stretching frequencies are much higher but on the other hand early metals despite having electron deficiency and having less than 6 electrons they always they act as excellent pi donors and positive charge makes them reluctant pi donors but negative charge on metal makes them better pi donors and as a consequence of this one what happens the influence is on metal to carbon as well as carbon to oxygen bonds as more and more backbone it comes here not in this direction it comes in this way what would happen is this bond become more and more elongated and stretching frequency drops considerably so this information already we have discussed in N number of times let us move on to other problems now a very interesting problem is there let me read it out an absorption at 3650 centimeter minus 1 in the IR spectrum of a compound say X has been assigned to an OH stretching mode to what wave number is this band expected to shift upon deterioration that means the complex is there we have a OH stretching frequency is there and if you replace H with D what would happen to the stretching frequency that is the question and also during this one certain assumption has to be made what assumption one has to make while doing this calculation to find out approximate stretching frequency or absorption band for a compound having body instead of OH group so now from the standard expression that you already know nu bar equals 1 over 2 pi c into square root of f over mu where f is a force constant stretching force constant and mu is the reduced mass and that means basically we can consider here this nu is inversely proportional to square root of mu reduced mass that means this is equal to we can say stretching frequency or the frequency absorption frequency for nu bar is inversely proportional to OH band reduced mass so in that case what happens assuming both the OD and OH have very similar stretching force constant we can arrive at this equation here if you write for this one nu OD over nu OH equals nu OH square root of mu OH over mu OD so now we know this relationship and now we know these entities we can calculate and we know this one now we have to calculate what is this so now what we do is reduced mass we have to find out in this fashion and this is one way of doing it or simply we can do this way also mn m2 by m1 plus m2 this would be rather easier but if you do with the first one what we have to do is here oxygen is there this is an atomic weight is 16 and H is 1 and then D is 2 we should remember this one so 1 by 16 plus 1 by 1 this is equal to 1.0625 and then this is for 1 by mu of OH so mu OH equals what we get is 0.94 1 2 atomic mass units we are getting and similarly we can also calculate now mu OD this is equal to 1 by 16 plus 1 by 2 so for this one we get 0.5 625 if we go for mu OD it comes around 1.7 7 7 8 this you can do calculation now we put here the values so this value is already given here 3650 and if you put here values this comes around 0.7 2 7 5 7 so this gives approximately 26 56 certificate minus 1 so that means if you replace H with D then the stretching frequency of OD drops considerably from 30 650 to 2656 centimeter minus 1 so here the assumption is very important in most of such problems where we come across replacement of one isotope with another one probably we should assume that the stretching force constant would be very similar then only we can arrive at this kind of relationship here and hence we can calculate the values and solve the problem so let's look into one more example here an absorption band at 33 37 centimeter minus 1 in the vibration spectrum of NH3 shift to X value in ND3 find out the value of X so again here we should come up with the same relationship whatever we saw we know now so this is inversely proportional to square root of mu if we use here so now let's calculate mu N H first of course N we are considering 14 H 1 and D 2 per calculating reduced mass and here also we can do directly or something like this also we can do it it's 1 over 14 plus 1 by 1 so this is equal 15 by 14 so this value will be 1.0 7 1 4 and mu N H will be similarly this value is going to be 0. and here the value will be 5 7 1 4 so equals 1.7 5 0 so now we have to apply here this is given here that means the value of X is this one so this how you can solve this very interesting problems let me come up with more examples in my next lecture until then enjoy these problems thank you