 I once again welcome you all to MSP lecture series on Interpreter Spectroscopy. So in the last few lectures I have been solving problems and then finding out solutions by considering problems having data from more than one type of spectra. So let us continue doing that one in this lecture as well. So there is a problem I will read out while running a new reaction a chemist notices the evolution of a gas a sample of this gas gave a mass spectrum in which the molecular ion m by z equals 44 was the largest ion peak. The only other significant peaks were observed at m by z equals 28 and m by z equals 16. Just for curiosity it is very easy to predict the gas the moment gas means probably we will be thinking of COCO2 N2 or ethylene something like that. So to begin with to examine this rule number 13 and also hydrogen deficiency let us go for again testing this molecule having very low mass to charge ratio 44 here 44. So we divide by 13 C3 H will be 5 now C3 H5 and then evolution of a gas is there. So that means basically there should be carbon monoxide if ethylene is a different thing ethylene does not have but let us see whether we have because this formula gives C3 H5 so it is little more than. So let us add one oxygen and see what would happen for oxygen we have to remove C H4 so C2 H1 so this is not actually it should be more 8 it should be because 36 plus 8 is equal to 46. So remove that one and then we get H4 here C2 H4 and then what would happen if we remove one more oxygen it will become CO2 and then H4 is gone so straight away we get this one this corresponds to 4. So obviously we can tell this is CO2 and then observed peak at m is this is CO and then 16 this is 28 so this information is sufficient I have to tell that the gas that evolved is CO2 you can use to number 13 it will arrive at the structure molecule as well so simply make it C3 H8 and then add one more remove C H4 add one more we are C2 H8 we are taking out then we will be left with only one and CO2 okay easy it works well with your small molecules. Now another example is there here a liquid compound gave a mass spectrum showing a strong molecule ion at m by z decos 156 the only fragment ions are seen at m by z decos 127 and 29 the moment we talk about 127 so one element comes to our mind having atomic weight of 127 that is iodine so probably iodine is there in this molecule let us check first we take 156 divided by 13 and it divides completely so no reminder is there 12 is there so this is C2 and H2 so now let us say if want to add iodine how much I should take out for iodine 127 means C10 H7 so 120 plus 7 so I have to take out C10 H7 here that means C2 H5 iodine well if it is C2 H5 iodine one can write this way CH3 CH2 iodine so it ends here yes possibly this is the C2 iodine it is also very simple but using this rule and then adding iodine here and taking out equivalent to 127 C10 H7 moiety we can arrive at the structure of this iodine here and of course when we look into 1 SNMR it can show very nicely something like this this and also 13 C will give you two signals next look into the lowest wavelength light visible to the human eye is 390 nanometer what is the corresponding frequency it is given in nanometer wave number then we have to convert into frequency here we know that lambda equals C by nu and nu equals C by lambda so for this one we have to consider light velocity of light is 3 into 10 raise to 8 meters and over 390 into 10 raise to 9 meters so now if we divide this one what we get is 0.00769 into 10 raise to 17 here so then if remove decimals here and make it 0.769 then it should be 10 to the power of 14 here this is per second or you can call it as Hertz so this is the corresponding frequency in this case the lowest wavelength light visible to human eye is 390 nanometer the corresponding frequency is 7.69 into 10 raise to 14 Hertz so now let us look into one more in which region of the electromagnetic spectrum is an emission from a neodymium laser corresponding to a transition between electronic energy levels 11502 and then 2111 centimeter minus so we have to find out the energy of this emission between the transition between whatever the energy that corresponds to that means two levels are there so we have to know what we should do is we have to subtract first or take the difference that should give you so one this is the one so now this corresponding wavelength we have to calculate this is the wave number so we know that it is 1 by lambda so here 1 by 9391 so that would be we get something like this so this one if we take out all those things it should come around we can put here 1.065 and or we can put here 1.065 we can make it we can remove 1.065 nanometer we can make it so 9 it becomes 9 10 raise to minus 9 so it becomes nanometer here so it will be 1065 nanometer so here the corresponding transition energy is 1065 nanometer so now what is the color of the solution giving rise to the spectrum shown below so this spectrum is for penta covenanted compound so vanadium is in two means it is plus four states so it is in plus four states due to system it is from this one now let us see whether we should be able to answer this question by looking into this one so here if we focus our attention to the absorption between 500 to 1000 500,000 comes in the red region as well as partly green region you can see here it comes around 500 here and then it is coming up to here so that means basically what it does is it is absorbing from this portion to this portion that means as a result what happens it appears blue here the color of the solution will be blue this is blue in color predict which one of the following complexes will have the more intense dd transitions one is cis compound one is trans compound both of them have two fluorine and two ethylene diamines if we look into the structures here cis one this is the trans compound so one thing we can quickly notice about is inversion center of center of symmetry so in this one we have center of symmetry is there whereas in this case there is no center of symmetry so now the dd of course dd transient itself is leopard forbidden but of course because of mixing of s and p orbitals in a complex d rs and no longer pure d orbitals they have mixing of p as well as s as a result what happens leopard yellow transitions are seen nevertheless their molar absorptive coefficient is very low but here which one will be more intense so here if you look into trans compound inversion center is there so it is forbidden whereas here in this case considering mixing is there in case of cis compound as a result we can say cis compound is more intense in color compared to trans compound because cis compound lacks inversion center of symmetry whereas trans has inversion center so now there is another question what sort of coordination and elements would be best for a commercial pigment based on dd transitions in a transmittal complex I have given some text here the molar absorptive coefficient is greatest for compounds without an inversion center that is the reason if we look into compounds having tetrahedral geometry as well as octahedral geometry again whether it is homolyptic or heterolyptic in case of homolyptic again tetrahedral compounds lack of inversion center they are more intense whereas octahedral compounds homolyptic ones they have center of symmetry and they are less intense in nature or leporta forbidden or they show weak transition therefore among all preferred geometries if we consider adapted by trans metal complexes tetrahedral geometry would be the best suited to think of such complexes as commercial pigments because it doesn't have an inversion center as this would give maximum color per metal center based on the type of dd transition as well however for the utility as pigments the highest color density per metal center is required that is a primary objective of commercializing any compound as a pigment so here in this case the preferred ones are those complexes have charge transfer transitions because most of the transfer transitions are very intense and then the highest color density comes per metal and as a result preferred is tetrahedral and also on and above the most preferred ones will be complexes which show charge transfer transitions there is one more question here the spectrum of hexachloro titanate 3 minus has an absorption at 760 nanometer and that of hexachloro titanate is 518 nanometer and the corresponding gromo is at 877 nanometer so calculate delta Akta in centimeter minus 1 for each of those things and comment on the values so all these complexes are titanium 3 plus d1 configurations are there and then one can simply by taking the reciprocal of this one we should be able to calculate the values here we should take the reciprocal and we have calculated that one this comes around 11400 centimeter minus per titanium gromo compound 13000 for chloro compound and 19300 for chloro compound here and of course when we look into this data we can you know recollect the position of these halides in the spectrochemical series it gives this information here so relative strength of these ligands can also be obtained directly from these values and of course I have also given an extended spectrochemical series here and this is also I have given in the beginning also while discussing about UV spectra of d1 to d10 system so now here one more example is there electronic absorption spectra of hexalquan nickel 2 plus and tris ethylene diamine nickel 2 plus are shown below determine delta Akta for these two complexes and the electronic spectra are given here this one what you should see here is both are octahedral nickel 2 complexes and both are d8 system and if the d8 system is there and of course they show three spin allot transitions I already told many times while discussing UV UV spectroscopy that d1 d4 d6 and d9 show one single dd transition because all of them were put into one group because d1 has one electron and d4 has one less than half field d6 has one more than half field and then d9 has one less than completely filled so all of them were put into one group and they show invariably one dd transition whereas d2 d3 d7 d8 all of them again they can be put into another category like having two electrons less than half two electrons less than half field two electrons more than half field and two electrons less than completely filled d2 d3 d7 d8 they show three transitions so now d8 nickel belongs nickel is a d8 system nickel 2 plus d8 system it belongs to this one so here in this one the from the spectrum what you can notice is we are seeing here if you just see here this is the first one here the first one here we are seeing this one in case of nickel compound here this shows 8000 approximately 500 centimeter minus 1 and then whereas in case of this one it is coming here this for this it is coming around 11,250 centimeter minus 1 so therefore for nickel compound is 8500 centimeter minus 1 and then for this one this is equal to delta of n i of course one can also look into d8 thin base overall diagram also to see because there we are keeping all the ligands are there and one can also see what kind of transitions we come across and how to determine delta O so one more spectrum is given one more problem is given here the spectrum of hexamine rhodium 3 plus has 2 dd transitions 32,800 and 39,200 centimeter minus and in the spectrum of iridium n hen 3 6 3 plus they are at 39,800 and 46,800 centimeter minus 1 estimate delta O for these complexes and compare the values to that of hexamine cobalt we know cobalt rhodium iridium they are all d7 s2 system all of them are d7 s2 system we are going from 3d to 4d to 5d here and here is cobalt rhodium and iridium so 3d 4d 5d system here and then how to calculate that one so to calculate delta what we should do is we should take the difference of these two and then one quarter of that one should be added to the lowest one so that would give you approximately delta O so for example 32,800 subtract from this one we get 6400 so divided by 4 it will be about 1600 plus 32,800 so this would give you 34,400 so this this is nothing but delta O for rhodium 3 here rhodium is in 3 system so d6 system it is d6 system you know that it will show only one transition and it is showing two transitions here same calculation we can do here also 46,800 minus 39,800 it is 7000 4 times will be 1750 1750 plus if we add 39,800 so that would give this one fourth of this one that would give you 41,550 centimeter minus 1 that is equal to octahedral value of iridium 3 plus so that means now if we just compare cobalt cobalt value is given is 23,000 so then if you if you look into rhodium it is 34,400 and then if you go for this is cobalt this is rhodium and this is iridium will be 41,550 minus 1 so this one can very nicely explain when we when we go from 3d to 4d to 5d what basically happens is your crystallization energy increases why that happens as we go down higher orbitals are there the orbits become larger in size when they are larger in size the greater the difference between the extent of interaction of the T2G and EG orbital and then increase in the energy between T2G and EG is possible as a result of this one and hence what happens delta increases down the group. So I think let me stop today and consider more problems in my next lecture so what you should remember is of course we are considering cobalt rhodium iridium 3d, 4d, 5d and also the values also you can see here calculated so it is very simple if two transitions are there and take the difference of the transition and then consider one-fourth of that one and add that one to the lower absorption band that is 30 to 800 in case of rhodium and 39,800 in case of iridium and then that would give you the corresponding octahedral delta value for the corresponding metal complex and then you can compare of course 3d to 4d to 5d we know that the incremental increase there in the critical friction energy and then the reason is what happens there increase in the size of orbitals. So let me stop now and continue in my next lecture. Thank you.