 once again welcome you all to MSP lecture series on interpretive spectroscopy. This is going to be the last lecture on IR spectroscopy. At the end as I mentioned I would come up with more examples in each case, but in this lecture let me conclude with few more problems and what you should remember when we are looking into stretching force constant and its relationship with stretching frequency. The stretching frequency is inversely proportional to the reduced mass and it is directly proportional stretching force constant that you should remember and also when we are replacing one with other isotope say H with deuterium or C 12 C with 13 C or O 17 with O 18 what you should remember is we should assume that the stretching force constant is more or less same in case of both the bonds. With that one new bar is inversely proportional to the square root of mu of that bond. So, that we should reduce it mass. So, that we should remember with that one let us try to make you familiar with this kind of problems by giving another 2 or 3 problems. Let me continue with such problems again. So, now I have another problem here and absorption at 31 61 centimeter minus 1 in an IR spectrum is assigned to CH stretching mode at what wave number will this band appear upon deuteration. So, that means if I say if I replace CH with CD what would happen to the stretching frequency presently it is 31 61. So, again we have to go for the same relationship as I had mentioned we know that this one equals 2 pi C square root of F over mu. So, F is force constant mu is reduced mass and then if we are considering now CH. So, this is inversely proportional to CH. So, now let us calculate the value in previous problems I took like this one or we can directly calculate also by considering M 1 M 2 over M 1 plus M 2. So, this is for CH. So, for this one what we will do is we can calculate here this is 12 12 carbon 1 H and 2 D. So, here it is 12 into 1 over 12 plus 1 similarly CD equals 12. So, now we know that the relationship value is given here for mu CH. So, this is the stretching frequency value for CD in place of CH that means the band after deuterating the H would appear at 23 20 centimeter minus 1. So, this is how you should I think you have understood now you should be able to do any such problems without any problem. So, now let us look into one more example here it is quite interesting an absorption in the IR spectrum of a compound containing an X H bond shifts from 36 57 to 26 61 upon deuteration find out whether X is C or O. So, that means we have a X H bond is there for which shift is 57 to 26 61 then we have to find out whether X is C or O. So, what we should do for this one we have to consider in both the cases we have to calculate. For example, let us take mu OH equals this is the one and then if you consider mu OH this will be 16 by 17 this will be equal to 0.94 4 1 1 7 well I am giving the whatever the value I get when I do calculation from the calculator. So, do not worry about it up to 2 decimals is good enough similarly I can calculate mu O D this is 32 by 18. So, this will give you 1.7777. Similarly, I would also like to calculate and keep it CD for example, here it is 12 by 13 this is 9.92 370 and here it is 24 by 14. So, this will be 1.71. So, now we have to calculate and we use the same analogy here mu CD over mu CH equals square root of CH over CD. So, if you put here mu CH into this value put here 3657 and if you this value if you put it here CH 0.92 307 or 1.71 this will be 3657 into 0.7347. So, this comes around 2687 centimeter minus 1. So, now let us do the same calculation for mu O D. Similarly, if I do for mu O D which will be mu OH into and here the value is again same because we do not know which is the value into if you put here the value I am just taking up to 3 here and then 1.777. So, this equal 3657 into 0.7277. So, this will come around 2661 centimeter minus 1. So, this is the value given here. So, this value is given here. So, we got this one. So, we can say now X equals O and also simply by looking to the data also you can see if the stretching frequency comes around 3657 centimeter minus 1 certainly it is not for CH it is for OH. OH usually shows around 3600, but on the other hand CH shows around 3000. So, that information also you can see. However, to rationalize our understanding we have to go through this process even let us say we get it in the first place itself it is always advisable to check for other option to conclude without any ambiguity that this bond we are referring to is OH. OH initially shows 3657 centimeter minus 1 after deuteration we are getting O D that comes to 2661 centimeter minus 1. So, this is how of course, we can take the hint from the chemistry knowledge we have and also spectral bands we have we know, but on the other hand when such problems are there always to learn more and get some confidence over solving the problems I think we should do all this exercise and we have to conclude. Now, it matches perfectly to this value here. So, you can say without any hesitation it is the X is O and we are talking about OH bond and its shift to lower frequency upon deuteration. So, enough let me come to some understanding of CO stretching frequencies here. So, here which of the two isoelectronic compound C R CO 6 and V CO 6 minus both are 18 electron complexes it is a D 6 12 electrons from carbon monoxide and 6 electrons from the metal 18 electron and here vanadium 3 D 3 4 S 2 plus 1 electron it becomes 6 and again 12 electrons both are isoelectronic will have the highest CO stretching frequency. So, which one will have the highest stretching frequency which of the two chromium compounds again this first question is among isoelectronic species such as chromium hexacarbonyl and vanadium hexacarbonyl anion which one would be having higher stretching frequency and the next question is if we have two chromium compounds having C R CO 4 mighty and two different phosphines one is triethylphosphine one is triphenylphosphine which one will be having lower stretching frequency and which will have the short ray MC bond. The one which has higher stretching frequency will be having longer MC bond and which one having lower stretch frequency will be having short ray MC bond because as more and more CO bond is weakened metal to carbon bond will be strengthened. This is again about metal to carbon back bonding ability and the negative charge on vanadium complex results in greater metal d pi d pi is here T 2 g d x y y z z x giving the electrons to pi star of CO compared to the chromium complex that means compared to chromium vanadium is a good pi donor as a result stretching frequency of CO drops considerably in case of vanadium complex compared to chromium complex. So, job is done. So, this results in weakening of CO bond with the corresponding decrease in stretching frequency. So, thus chromium complex has higher stretching frequency and its bond chromium to carbon bond length will be little longer ok. In case of two phosphines we have triethylphosphine and triphenylphosphine. So, ethyl groups are electron donating groups whereas phenyl group is electron withdrawing group. So, that means when we have electron withdrawing groups on phosphorous it makes them poor sigma donor, but excellent pi acceptors. On the other hand electron donating groups on phosphorous makes it very good sigma donor, but poor pi acceptor. So, that means here. So, ethylphosphine complex will have greater electron density because triethylphosphine is a good sigma donor and also relatively it is a weak pi acceptor compared to carbon monoxide. As a result what happens more and more electron density goes to carbon monoxide rather than going to phosphine. So, here stretching frequency drops considerably and metal to carbon bond strengthens. Whereas, in case of triphenylphosphine what happens it is also a good pi acceptor as a result back bonding will be little less compared to what we see in case of chromium compound. So, this is how we can analyze which of the two iron compounds FeCO5 and FeCO4PET3 will have the higher new CO which will have the longer MC bond. So, same analogy you can use here. In case of iron pentacerminal we have 5 carbon monoxide groups are there and equally electron is distributed probably less lessening of frequency is less because 5 are there and CO stretching frequency is little bit higher here. Whereas, in this case what happens one carbon monoxide is replaced by another electron pumping group as a result now more electron density is there on iron and also it has to take care by only 4 carbon monoxide. So, here the stretching frequency will be less here and iron to carbon bond is more stronger. So, this is how simply by looking into the nature of the ligands present we can analyze and we can conclude about the stretching frequency trends among these complexes. So, let me look into one more example before I conclude on IR spectroscopy in this lecture. This is again about isotopic labeling or exchanging. The IR spectrum of free carbon monoxide shows an absorption band at 2170 centimeter minus 1 assigned to vibration mode of the molecule. If the sample is enriched in 13C what change do we expect to see in its IR spectrum? So, this is again about exchange. So, here earlier we are talking about H with deuterium here we are talking about 12C with 13C and O remains same. So, that means, basically we have to look into again this relationship is inversely proportional to. So, now let us try to find out a reduced mass. So, that here we have to find out 13C O versus 12C O. This relationship if we write is will be 12C O. So, we have to calculate now mu 12C O this equal to m 1 m 2 by m 1 plus m 2. So, this is 12 into 16 over 12 plus 16 is 28. So, this comes around 6.8571 and similarly if we calculate for 13C O reduced mass this will be 13 into 16 by 29 this comes around 7.1724. Now, if we apply the same here. So, value is given here 2170. So, that means, the answer is this one. What change we expect to see in its IR spectrum? So, IR spectrum stretching frequency due to 13C O drops considerably from 2170 to 2122 centimeter minus 1. Again here the assumption what we made is the stretching force constant in case of both 12C O and 13C O would remain more or less unchanged. Then only we can arrive at this relationship here. With this one let me stop today's lecture and conclude about IR spectroscopy and then probably in my next lecture I am going to start about mass spectrometry. And as I said at the end about 10 to 12 lectures I devote completely to solving all kinds of very interesting problems having one or more spectroscopic components in it. So, until then have an excellent time. Thank you.