 Hello everyone, I once again welcome you on to MSB lecture series on transition elements. Today, let me discuss about the 18 electron rule and also effective atomic number. In my previous lecture, I informed about ligand field theory is important and how ligand field theory effectively explains bonding in coordination chemistry and also it is capable of explaining almost all properties related to transformational compounds and also organometallic compounds. Today, let me introduce another important topic that is the 18 electron rule that means electron counting. And what are its significances? What we should follow while counting electron rule? What does it mean to coordination compounds or organometallic compounds stability and their application? All those things let us study in detail. I shall give you numerous examples so that you will become familiar in electron counting. First of all, let us try to look into what is 18 electron rule. It is a way of expressing the tendency of a transition metal autumn to use all its valence shell. When we talk about transition elements, we have three valence orbitals that is Nd orbitals and N plus 1 s orbital and N plus 1 p orbitals. That means we have a total of 5 plus 1 plus 3 9 orbitals with a capacity of 2 electron each means 18 electrons. So these metals have a tendency to have as many electrons as possible into their valence shell so that it will be having 18 electron at the end and that is called 18 electron rule. Just recall about the octet rule we use in main group compounds especially for p block elements introduced by Gilbert and Lewis. What it says is we have p block elements having s2 p1 to s2 p6 electronic configuration. That means we start with one electron in the p orbital starting from boron group to end with 8 electrons in the valence shell along with s2 electrons that is inert gases. That means most of these p block elements have a tendency to have as many as 8 electrons to have stability and attain next inert gas configuration. And if you just look into any ion whether it is a cation or anion they always make an attempt to have next inert gas configuration. If it is not possible they lose some electrons to become positively charged one to attain previous elements or previous gas inert gas configuration. For example if you take N3 minus it has s2 p3 and another 3 will become s2 p6 so that it becomes a next inert gas configuration. Similarly oxygen with s2 p4 it will take 2 more electrons to become O2 minus and similarly if you go for chlorine bromine iodine or fluorine they have 7 electrons they need one electron that is the reason they make mono anions. That means whenever you look into a stable cation or anion it must satisfy inert gas configuration whether the next one or the previous one. In the same way if you extend the same octet rule to transition elements here instead of having 4 orbitals with capacity of 8 electrons we have 9 orbitals with the capacity of 18 electrons as a result we call it as 18 electron rule. The sum of the number of valence electrons in the gaseous atom plus the number of electrons from neutral or ionic ligands may attain a maximum value of 18. That means is it really true that a 80 electron rule must be obeyed by all transfer elements? Before I answer that question let us look into simple examples here. So I have given here chromium hexa carbonyl you know chromium electronic configuration it is 3D4 4S2 or 3D5 4S1 does not matter. So it has 6 electrons and plus 6 carbonyl groups are giving 12 electrons here so it becomes 18. So it obeys 18 electron rule. Now let us look into this one here and in this one let us assume H is giving 1 electron there are 2 methods to count electrons one is called ionic method other one is called neutral method or covalent method. In ionic method we are defining the axis state of the metal depending upon how many anionic ligands are present. In the neutral or covalent method we are ignoring and we are just counting all electrons from the metal and before its oxidation and then we are giving 1 electron for chloride or 1 electron for hydrogen. So it goes that way in the next slide I will make you familiar with counting electrons and to make you familiar with different ligands can give how many electrons in both ionic method as well as in covalent method. So now if you consider let us consider here using neutral method so here we have 4D6 5S2 so 8 electrons are there and then this hydrogen is giving 1 electron and then this acetate is giving 1 electron and then this they are giving 6 electrons. It becomes 16 electrons but in this acetate what happens this also giving 2 electrons as a result to add another 2 it becomes 18 electron. So now let us go for other method so now we know that axis state of ruthenium is plus 2 here because 1 is H minus 1 is acetate 3COO minus therefore here we have 6 electrons and now H2 is a 2 electron donor and acetate is a 4 electron donor now and plus 6 so this is 18 electron. So this is how you can count and in this case it is very straight forward iron has 8 electrons 3D6 4S2 so we have 8 electrons and then in this case carbon monoxide 5 into 210 it is 18 electron. So whatever the complex have shown here all of them obey 8 electron role so same thing I have shown here. Now I will introduce another term called effective atomic number this was introduced by Sidwick so electron phase from the ligands are added until the central atom is surrounded by the same number of electrons as the next noble gas it is essentially same what in 18 electron rule we did was we counted only valence electrons if you count all the electrons including those coming from ligands in a metal complex that is essentially called as effective atomic number EAN metals with odd number of electrons can never attain noble gas configuration that means there are some exceptions. If the electrons are added in phase metals with odd number of electrons can never attain noble gas configuration it is necessary to produce a symmetric structure irrespective of number of electrons involved that means always when we do not know the structure when we make an attempt we should try to write a symmetric structure to a given complex. Now I have given more examples here let us look into these examples to calculate the effective atomic number in each case and you consider chromium here atomic number is 24 chromium as I mentioned electron last during the formation of this one is 0 because 0 oxygen state and electron gained is 12 so what we have is here 24 plus 12 it is 36 so it attains this is the effective atomic number for chromium hexa carbon similarly if you take iron the first one here in this one so here iron is in plus 2 state iron is plus 2 state 2 electrons are lost here so we will be left with 24 electrons and then 6 sinets are giving 12 electrons this is also 36. Now if you take this one so electron last are 3 here so we will be left with 25 and now 12 electrons are coming here then it would be 35 so one electron less than next inert gas atomic number let us look into iron pentacarbonyl here electrons last are 0 and then 10 electrons are coming 26 plus 36 yes this also attains effective atomic number. Now if you look into copper so one electron is lost here copper is in plus 1 state and then ligands are donating 8 electrons so it becomes 36 so it follows nickel if you consider nickel has 8 electrons 2 electrons are lost and 12 electrons are coming so that means 26 plus 38 it does not obey we have 2 electrons excess okay so next same thing is true in this case also instead of 6 ammonia we have 2 ammonia and 2 bidentate ligands ethylene diamine the total number of electrons coming are again 12 so here again 38 it does not obey. If you consider this compound here electron last are because Cp cyclopentadienyl anion one electron is going from ruthenium one electron is going from clore towards chloride so we have ruthenium is having 6 electrons and now what we have is this is 6, 6 electrons are there 6 plus 8, 10 and 12, 54 we have so this satisfies effective atomic this is how you can calculate the I have shown here we can verify whether it is correct or not yes so this is how you can count the electrons and also you can write the effective atomic number and also you will be knowing whether a given complex satisfies 18 electron rule or not. Now the question is why some metal complexes can never obey 18 electron rule is it true If you look into 3D metals 3D metals can have a maximum coordination number of 6 with an octahedral geometry so that means 12 electrons are guaranteed when it forms an octahedral complex with coordination number 6 then if 12 electrons are coming from ligands metal should possess at least 6 electrons in order to satisfy 18 electron rule but by virtue of being a early metal does not have more than 6 electrons in their d orbital then it can never obey 18 electron rule despite having coordination number 6 and 6 ligands surrounding it that means the metals having an electronic configuration of D1, D2, D3, D4 and D5 with monometallic complex formation even with coordination number 6 they can never obey 18 electron rule because we will be having less electrons within the metal itself prior to the complex formation and again metal with all atomic number can never obey 18 electron rule on the other hand if metals have more than 6 electrons with coordination number 6 they also do not obey 18 electron rule because they will be having excess electrons that we saw in case of hexamine nickel 2 plus complex that means outer orbital complexes always have slightly excess electrons then the next inert gas as a result they are unstable and they do not obey 18 electron rule. So now again to make you familiar with the electron counting I am just showing you here 3 different type of ligands we come across one is a 2 electron donors and here in the neutral method they are all 1 electron donors and here of course this does not come into picture here we have Lewis acids and of course in case we come across then you have to subtract 2 electrons for Lewis acids if they are present on a metal center while electron counting. So now I shall make you familiar with the different ligands we come across and how many electrons they will contribute when in ionic method and covalent method H what happens it has 1 electron. So covalent method it is a 1 electron contributor but when you take 1 electron for the metal metal oxygen state increases by plus 1 and hence that electron will come and it becomes H minus now we have 2 electrons so it becomes 2 electron donor in the same way if you take halides they are all 1 electron donor in covalent method but they are 2 electron donor in ionic method because they have this pair of electron that can go to the metal same thing is to in case of hydroxide cyanide and also methyl. So in all these cases covalent method always you consider 1 electron when you are using ionic method you consider 2 electrons and this extra electron you should remember there is coming from the metal accordingly metal oxygen state increases and number of D electrons decreases. In case of NO if it is bent it is a 1 electron donor but in ionic method it is also 2 electron donor and carbon monoxide and tertiary phosphine they are whether you consider ionic method or covalent method they are 2 electron donors and then this is ethylene olefins they are 2 electron donors both in ionic method and covalent method and with all neutral ligands if you consider their irrespective of method A or method B they are always 2 electron donors. So if you consider this pi bonding then they are 2 electron donor if you consider oxide O2 minus then they are 2 electron donors or they are 4 electron donors. Similarly if you consider NO with linear they are 2 electron donor in method A and 3 electron donor in method B and ethylene again 2 electron donor in ionic method and covalent method they are 3 electron donor and in ionic method they are actually 4 electron donors. In case of acetylene 3 minus they are 3 electron donor here and here it will be 6 electron donor. In ionic method they are 6 electron donor whether here they are 3 electron donor and butadiene both 4 electrons or 4 electron donor. CP minus it is 6 electron donor whether consider just CP it is a 5 electron donor, eta 6 a benzene for example it is a 6 electron donor here also 6 electron donor and if you consider tropolinium ion it is a 6 electron donor cation otherwise it is 7 electron donor. So now you are familiar with assigning the electrons that are coming towards the metal from these ligands in both ionic method and covalent method. Now let us look into more examples. So now I have given several examples here. Let us try to use both ionic method and covalent method in case if you do not know the oxygen state there is no need to worry. First let us make an attempt to write electron count using covalent method and then if any anion is there we will add one electron depending upon whether mono ion, di anion, tri anion accordingly and then count how many electrons are left on metal you will end up with knowing the precise oxygen state of the metal center in that particular complex. And if you consider here let us start with covalent method. Covalent method this is 3D7 4S2 so we have 9 electrons are there here 9 electrons are there and 2 CPs they are giving 10 electrons. So this is becomes 19 electron species. Let us check now in ionic method. So ionic method it is 7 electrons are coming and CP they are giving 12 electron because CP minus will be 6 electrons so this is 17 electrons. Later we can verify and zirconium this is 4D2 5S2 so this is in plus 2 state here whether it is plus 2 is 4 or there so it cannot be in plus 2 state it has to be in plus 4 state that is zirconium 4 state. So in this case if you go for ionic method sorry covalent method we have 4 electrons here and 10 electrons are coming here and 2 electrons are coming here. So this is 16 electron species but if you go with ionic method here 0 electrons are there and they are giving 12 electrons and they are giving 4 electron 2 each so it becomes 16 electron so it is matching. You see it is matching now. Now if I add 2 electrons here and 2 electrons here obviously you will be knowing the axis state of zirconium is plus 4 and now let us go to iridium. Iridium is cobalt rhodium iridium 5D7 6S2 and 1 chloride is there. So let us not worry about that one first let us write here 9 electrons and 1 electron is coming here and then 6 electrons are coming from 3 triphenyl phosphine so 16 electrons are there and similarly now we know the axis state it is giving 8 electrons and chloride is giving 2 electrons and 6 electrons this is a 16 electron species though 16 electron species having D8 electronic configuration plays a prominent role in homogeneous catalysis for a variety of organic transformation. I shall tell you more details about the catalytic properties and how to use them in catalysis when I go to oxidative addition and ethyl elimination reactions at some point of this course. So now let us look into molybdenum hexacarbonyl 0 that means 6 electrons are there and 12 electrons are coming here 18 electrons and same thing is true here 6 plus 12 18 electrons. So now I believe you are familiar with the electron counting using both ionic method as well as neutral or covalent method let us look into it so 19 yes prediction is correct this is correct and this is 60 electron species and this is also 16 electron species so correct so tally works well so that now you know the axis state of zirconium is 4 here it is not titanium you correct it as zirconium of course titanium zirconium have similar properties when it comes to electron counting chemistry may be little different because one is 3D series other one is 4D series 16 electrons 16 electrons 18 18. So electron count made simple so one should be able to use both ionic method and covalent method what you should do is whenever you come across any metal complex try to do electron counting to make yourself familiar with this process of electron counting. Now CP binding CP binding why we call it as eta 5 hapticity 5 I shall show you so we always make it like this C5H5 then where this 5 hapticity is coming for this one let me write something like this so we are removing one electron so we have a negative charge here and we have 2 bonds here now first this one makes a bond to the metal through covalent bonding sharing electrons now what happen this also a 2 electron donor this can also go to the metal and this can also go to the metal and then this one you can start writing resonance structure this negative charge can keep on coming here coming here coming here accordingly double bonds will be moving as a result what happens it looks like delocalization occurs and when delocalization occurs we write simply something like this a negative charge and we show here so that means here for this is for 2 electron eta 2 2 electron eta 2 eta 4 and then 1 and now this is considered as eta 5 so this is how now one can explain bonding in case of CP having hapticity of 5. Now I have 4 interesting dimeric species by simple electron counting you should be able to tell whether a metal metal bond exists or not if it exists what is its nature whether single bond is there 2 bonds are there or 3 bonds are there and let us begin with simple example of CO 2 CO 9 of course cobalt has 9 electrons in its valence shell 9 means each one so 9 into 2 18 18 is there and another 18 electrons are coming from 9 carbon monoxide 18 so 36 are there divided by 2 so each cobalt has 18 electrons so it obeys 80 electron rule and there is no need to include any metal metal bond between 2 cobalt atoms I will show you at the end its structure so it follows 18 electron rule now let us look into the second one cyclopentadienyl dicharmonyl dimer so here we know now iron is in plus 1 state so let us write directly here per iron I am writing so 6 electrons are coming and then what we have is iron has 3 to 6 4 S 2 1 electron is gone so we have 7 electrons and then we have 4 carbon monoxide are there and that means 2 carbon monoxide will contribute 4 electrons so we have 17 electrons are there that means it is a 34 electron species and if you do it is 17 electron that means each iron cation has 17 electrons in order to satisfy 18 electron rule you have to establish a metal metal bond that means there will be 1 Fe Fe bond will be there in it now let us look into CO 2 CO 8 CO 2 CO 8 it is again 18 plus 16 electrons are there now so that means again 34 are there divided by 2 again you need a cobalt to cobalt single bond then look into the last case here in this case what we have is again 6 electrons are there and 7 are there but 1 carbon monoxide per so 2 are there so that means 15 are there so 15 means you have to you need 3 more electrons so that probably there is a triple bond between 2 metal atoms so this is how you can predict and also you should remember when you make an attempt to write the structure you should try to make it as symmetric as possible you can see here yes it is 18 electron it does not need any so this is how the structure looks like as I mentioned you have to try to write symmetric structure this is the only way you can write symmetric structure each cobalt has 3 terminal and 3 carbon monoxide are bridging it is a very symmetric molecule and does not have any metal metal bond it does not need because both of them have 80 electron each. Now you consider this one as I said one electron is needed again it is very symmetric and also it has centrosymmetry so you have to put a metal metal bond and you should remember by adding a metal metal bond the number of electron does not become 36 you should remember remains 34 only at a given time you can satisfy 18 electron rule for each one when you are satisfying 18 electron rule for this one this one will be having only 16 electrons you should remember and when you are satisfying this one this will be having 16 electron this is very similar to octet rule recall octet rule when I wrote for carbon monoxide I kept 6 electrons between 2 carbon monoxide and a pair of electron and carbon oxygen if you count simultaneously you can never over 18 electron rule but on the other hand when you count for oxygen it has 8 electron and carbon has only 4 electrons but on the other hand when you count for carbon it has 8 electrons and oxygen has 2 electrons so this is how it is so you should remember that simply by adding a metal metal bond you should not make it instead of 34 to 36 number of electrons remains same lot of students they do this kind of mistakes while counting electrons you should not do it electrons are not coming from anywhere within what we do is in order to satisfy its octet rule to make it stable stable we are making a metal metal bond now let us look into this example where we need one cobalt-cobalt bond and that is the reason cobalt octacarbonyl it undergoes dimerations COCO4 if you make it immediately dimerizes in order to satisfy 18 electron rule and the structure looks like this of course it can have both the structures and this is more stable but one can also see this one and of course you can analyze whether we have this structure or this one but just simply look into IR spectroscopy where carbon monoxide stretching frequency can precisely tell you whether you have this one or this one or a mixture of both so here we have 30 electrons are there we need 3 more bonds so this is how it is for example when I am counting electrons for this one you should remember it has 80 electron but this will be having only 12 electrons you should remember that one let me stop here and continue discussing you know 18 electron rule with more interesting molecules in my next lecture until then keep you know calculating this 18 electron rule and try to make yourself familiar with it.