IR Spectra in metal carbonyl complexes Part 1

追加: 2026-06-06

[00:00:02]So in this class we shall uh discuss about IR spectra in metal carbonal complexes.

[00:00:11]Again it is known to all of you that uh the frequency this bond frequency suppose this is the bond. Now you know that this frequency uh this is known to you. It is given by like this where K is related to the bond strength isn't it? So greater the bond strength higher will be the stretching frequency of this bond.

[00:00:41]So from this idea uh if we look at carbon monoxide now we know that for carbon monoxide there is a triple bond isn't it? For carbon monoxide there is a triple bond. Let us not go into the charges over carbon and oxygen but there is a triple bond it is known to all of us. The a triple bonded you'll find that the stretching frequency rather new bar value this is in the region of 2143 cm inverse 2143 cm inverse for free carbon monoxide. So for free carbon monoxide for free carbon monoxide you know that this is the stretching frequency. Now if we consider a metal carbonial complex for example CR CO6.

[00:01:37]So if you look at this metal carbonial complex. So you know there is uh sigma bond. There is a sigma bond and there is sigma donation rather sigma donation and this sigma donation occurs from homo of carbon monoxide which is of non-bonding character sigma non-bonding to sigma donation and there is pi acceptance by carbon monoxide in the pi star lumo.

[00:02:13]So carbon monoxide will accept back this pi electron density. So pi electron density or what we say is pi acceptance. This pi acceptance occurs in the pi star lumo.

[00:02:28]So sigma donation occurs from sigma star sorry sigma non-bonding homo and pi acceptance in pi star lumo. So if you look carefully see we can see that when this bond is formed between the metal and carbon monoxide electron density is lost from a non-bonding molecular orbital NBMO but it is accepted in an anti-bonding molecular orbital of carbon monoxide.

[00:03:04]So what will be the fate of the bond order of carbon monoxide? So due to due to electron acceptance due to electron acceptance in pi star lumo or theft which is a anti-bonding molecular orbital what will happen the bond order of co will decrease in the complexes this will decrease see electron density from non-bonding when this is lost from the non-bonding MMO there will be no change in bond order of carbon monoxide but when it is accepted in palar lumo the bond order of CO will decrease if the bond order decreases then one should expect that newar CO will be less lesser than 21 43 cm inverse.

[00:04:15]So the major reason for this is the pi acceptance in the pi star numo. So pi acceptance in pi star lumo plays a vital role in lowering the new co in metal carbonial complexes.

[00:04:34]Okay.

[00:04:36]Now let us move to some of the factors some of the different examples and factors that uh are responsible for change in new bario. So let us consider say this complex say titanium this is a this is a very well-known example bhubar exams set ti h26 2 minus if new bar if you look at new bar co carbonal stretch this is 1748 cm inverse P CO6 minus 1858 cm inverse CR CO6 chromium hexacarbonil chromium hexacon 984 cm inverse for manganesees MN H2O O 6 plus value 2094 cm inverse. And for free carbon monoxide you know this is 21 sorry this is 2143 213 cm inverse.

[00:06:11]system.

[00:06:15]We know very well that with respect to free carbon monoxide one should expect that the stretching frequency this is carbon monoxide very sorry the stretching frequency of water leakability once again I'm extremely sorry this will be carbon monoxide very obvious so so the stretching in frequency in of carbon monoxide in all these cases see they are lesser than that of free carbon monoxide.

[00:06:56]Now if you look at all these systems titanium minus2 charge this is oxidation state minus24 + 2 so it will behave as a d6 system and it should be low speed why carbon monoxide is a strong field Venardium minus one venardium 23US 24 so once again 6 so this is also D6 low spin chromium zero valent chromium is already 24 so this is also D6 low spin and MN manganese manganesees is in plus one oxidation state manganese is in plus one oxidation state. So this is what also d6.

[00:08:01]So all these are d6 low spin complexes.

[00:08:05]Title number of electrons over the metal center in all these cases they are same.

[00:08:16]So all these are d6 low spin system. So the major differences is the oxidation state oxidation states with increase look carefully with increase in oxidation states from minus2 to minus1 to 0 to + one stretching frequency increases isn't Although they have ex they have all are d6 that means iso electronic all these they are iso electronic iso electronic negative charge monoxide Right.

[00:09:19]The stretching frequency of CO that will depend majorly upon the extent of PI acceptance acceptance acceptance. Greater the pi acceptance lower will be the stretching frequency.

[00:09:51]The pi acceptance in all cases. Going to next in all cases. In all cases, D6 low spin opted. In all cases with increase with increase in negative charge density with increase in negative charge density over metal center with increase in negative charge density over the metal center from negative charge density part from MN N + 1 to chromium 0 to benadium -1 to titanium minus2 with increase in negative charge density the extent the ability the ability of the ability of CO to accept to accept Pi electron density to accept pi electron density in their pi star lumo in their pi star lumo will increase bond order of CO decreases bond order CO decrease. So naturally new CO will decrease. New will decrease.

[00:12:07]We can also think about uh new bar MC metal carbon bond acceptance more with increase with Increase in pi acceptance with increase in pi acceptance from with increase in pi acceptance from metal center.

[00:12:47]Right? So the MC bond will become more stable will become more stable. more stable bond more stable. New bar MC will increase MC will increase MC will increase from new MC.

[00:13:21]It will increase from ML CO6 plus 2 Cr 6 to V CO6 minus to TI CO6 2 minus decrease in this direction. In this direction, it is decreasing.

[00:14:14]New barc this will increase in this order increases pi extent of pi bonding increases with increase in number of electrons by with increase in electron density over the metal center. So if the bond strength increases, if the pi bond strength increases, so naturally metal carbon bond will become most stable and bond order of metal carbon will be highest.

[00:15:14]In this case, stretching frequency will be highest over here. Metal carbon bonded.

[00:15:23]So this is a very common question often asked in the exams.

[00:15:30]Now similar type of question.

[00:15:36]If you look at Fe 2 minus CO4 minus R Ni CO4 nickel tetracarbonate they new stretching frequency 790 cm inverse 1 8 9 0 2 060.

[00:16:40]The reason is same.

[00:16:45]New bar C increases are new bar MC a c inverse order.

[00:17:01]It will decrease in this order. It will decrease in this order.

[00:17:09]complexes are reason 28 art + 10 so this is a d10 system cobalt minus one cobalt is 27 cobalt minus 1 is 28 so this is also a d10 system iron is 26 iron minus 2 28 so this is also a d10 system so once Once again they are iso electronic and since they are D10 systems D10 systems they are always tetradral due to uh because you see for D10 systems CFSE is zero for D10 systems why they are tal CFS is zero CFS zero means steric factor steric factor will decide the geome geometry will decide the geometry cf.

[00:18:14]So the stic factor will decide the geometry stic factor geometry decide obviously tetraal geometry will be favored.

[00:18:25]Now in this case also with in with increase in negative charge density with increase with increase in negative charge density with increase in negative charge density over metal center over metal center from Ira sorry from nickel 0 to cobalt minus1 to iron minus2 the pi acceptance pi acceptance in pi star lumo pi acceptance in pi star lumo of co carbon monoxide will increase will increase.

[00:19:26]So that will lead to lowering of CO bond order and hence lowering of new And it is for the same reason metal carbon bond will become stronger they can ask you multiple choice questions also multicq they arrange in the order of new bar entrance exam they can ask Next, there are some examples.

[00:20:23]If you look at these complexes, Cu Cu4 Cu plus timber PD CO4 2+ plus imma pt CO4 2+ F CO6 2+. So these are some examples where it has been observed that in all these cases in all these cases new CO this is greater than 2143 cm inverse or third that is that is higher than higher than that of free carbon monox oxide. So this is higher than that of free carbon monoxide.

[00:21:44]This is 2184 new bar CO New CO in cm inverse 2259.

[00:21:56]This is 22 261 and this is 2204.

[00:22:07]So in all these cases new CO is higher should have been lower. As per the theory the new CO should be lower. So in all these cases it has been found. These are exceptions to our theory. the theory which we have proposed. So all these are exceptions.

[00:22:30]In all these cases, in all cases, the sigma non-bonding homo develops develops some amount some amount of anti-bonding character. Anti-bonding character.

[00:23:02]Also, metal carbon bond formation involves involves only sigma donation from the homo and no pi acceptance and no pi acceptance it has been observed. So these are exceptions and no pi acceptance exceptions to our theory of course does upon complexation does upon complexation does upon complexation electron density electron density in anti-bonding AB homo decreases bond order increase decreases. Therefore, bond order of CO increases which implies new bar CO increases.

[00:24:32]New bar CO increases with respect to free carbon monoxide. So new bar CO will also increase with respect to free carbon monoxide. So these are exceptions with respect to the theory which we have proposed.

[00:24:52]Next question regarding the nature of next.

[00:25:03]If you look at say as we move from say chromium hexacarbonil to molydenum hexacarbonel to tungsten hexacarbon tungsten.

[00:25:39]So, malibdenum chromium to malibdenum tungsten that means as we move from 3D to 4D to 5D carbons.

[00:25:52]And if you look at this new CO new chromium, this was around 984 984.

[00:26:12]This is 984.

[00:26:16]C inverse values decreases decreases.

[00:26:32]And these two values, these two values, they are almost similar. Almost similar.

[00:26:42]Almost similar.

[00:26:50]So how to explain this?

[00:26:53]If you look at if you go down from charge the effective nuclear charge further increases.

[00:27:18]If you look at this bonding process a metal to sigma if the effective nuclear charge increases sigma donation is expected to increase.

[00:27:35]Sigma effective nuclear charge.

[00:27:41]So metal effective nuclear charge sigma donation should increase and sigma donation increase acceptance positive charge density acceptance a non-bonding Pi star antibonding sigma donation pi acceptance pi acceptance weaker weak pi acceptance in that case the bond order should increase and stretching frequency should increase but d or vital chromium as you move from chromium to tungsten the d orbital becomes much more diffused.

[00:28:48]So the pi acceptance by co is expected to increase. So there are two factors deciding the stretching frequencies.

[00:29:11]Pi acceptance become weaker but diffusen pi acceptance is expected to become stronger acceptance because density will decrease. New bar CO will decrease.

[00:29:43]But in these two cases, they remain same because the sigma donation and pi acceptance in both cases uh the act two factors factor from chromium 0 to malibdinum 0.

[00:30:29]Chromium 0 to malibum zero.

[00:30:35]Effective nuclear charge increases.

[00:30:41]Effective nuclear charge increase.

[00:30:46]Therefore, sigma donation will increase.

[00:30:55]Sigma donation will increase.

[00:30:58]But due to But due to uh more diffused due to more diffused pi molecular or vital of sorry pi bi molecular more diffused food of malibdum by acceptance by acceptance will also increase. By acceptance will also increase.

[00:31:42]Therefore, therefore CO bond order in MN malibdenamine MO6 will be lower.

[00:31:58]So accounting for lower lower new bar C ever from malibdenum to tungsten from malibdenum to tungsten once again effective nuclear charge increases effective nuclear charge increases.

[00:32:32]effective nuclear charge.

[00:32:34]So one should expect that sigma donation and pi acceptance and pi acceptance sigma donation and pi acceptance for tan 10 will be more favored will be more favored.

[00:33:06]But it is due to orbital mismatch.

[00:33:18]Orbital mismatch by acceptance by W P acceptance in in WCO6 is somewhat reduced is somewhat reduced reduced follow new bar CO in both are close are close to each other.

[00:33:57]New bario in both cases they are close to each other.

[00:34:03]Now let us consider another example that is the second part second aspect of this new bar CO um that is the nature of nature of gas.

[00:34:24]New bar CO also depends upon the nature of a nature of dependent which is also which is also acidic in nature.

[00:34:51]We know that this carbon monoxide electron density donation because that will not lead to any change in bond order of C acceptance regarding new CO if this metal is also pi acidic in nature I'm sorry this is also acidic in nature So there will be competition for pi acceptance.

[00:35:34]There will be competition for pi acceptance.

[00:35:44]So competition acceptance factors consider