 Hello everyone, once again, welcome you all to MSP lecture series on Advanced Transmittal Chemistry. In my previous lecture, I was discussing about the Redux Potential and also activity series. Let me continue from where I had stopped. So yesterday I showed you this table. I have given Redux Potential for both S&P block elements as well as selected D block elements and here I have given the reactivity series. The reactivity series is a series of metals in order of reactivity from highest to lowest. It is used to determine the products of single displacement reactions. Let us say a metal A will replace another metal B in a solution if A is higher in the series. So only a metal higher in the reactivity series will displace another. A metal can displace metal ions listed below in the series but not above. For example zinc is more active than copper and is able to displace copper ions from solution. If you consider this reaction, why this reaction happens can be clearly seen from the Redux Potential. Here copper 2 plus is reduced to copper but silver cannot displace copper ion from solution. So it is important to distinguish between the displacement of hydrogen from an acid and hydrogen from water. Sodium is highly reactive and is able to displace hydrogen from water. For example if you take sodium solid and add to water I am sure you are all familiar with this equation. So H2 is liberated here. Sodium is highly active and is able to displace hydrogen from water. So less active metals such as iron or zinc cannot displace hydrogen from water but to do readily react with acids and liberate hydrogen. For example if you take zinc and treat with sulfuric acid it forms zinc sulfate plus H2 will be liberated. So let us look into some problems related to these topics. I have given a question here. A polished copper rod is placed in an aqueous solution of zinc nitrate. So in a second experiment a polished zinc rod is placed in an aqueous solution of copper sulfate. Does anything happen to the copper rod? Quantify your answers by calculating appropriate values of delta G0 at 298 Kelvin. So what we have to do is from Redux Potential table you can get the information about Redux Potential for copper and zinc couple. The values are here similarly. So we have to calculate now delta G and I am sure you are all familiar with this equation. F you know Faraday's constant that is equals 96485 coulombs per mole. So just if you use this one in both the cases we should be able to determine delta G in each case. So delta G here if you use in case of zinc it is a 2 into E0 is 0.7696485. So this gives a value of minus 146.6 kilo joules per mole and then if you do the same this is for zinc couple. If we do the same for copper we can get here 2. So what value we get is here 65.6 kilo joules per mole. So this is positive and this is negative you know that zinc cannot replace copper whereas copper can readily replace from copper sulfate and it can form Cu. Let us look into one more question here. Calculate a value of delta G for the reaction. Of course here again the same answer you can anticipate that I showed you in my previous solution. Here what we have to do is of course when once again this we have the value of minus 0.76 volt okay and F is known 96485 coulombs per mole. What we have to do is we have to calculate delta G minus if you put here values. So this comes around minus 146.6 kilo joules per mole. It can be rounded off to minus 147 kilo joules per mole. This is the delta G value for this reaction. Similarly by knowing the redox potential delta G for any reaction can be calculated. So let us look into one more question here. If the standard reduction potential of copper 2 plus to copper and copper 2 plus to copper plus are given as 0.337 and 0.153 volts respectively. What is the standard electrode potential of copper plus to copper half cell? So what is given here? The values for this is given here. This is 0.337 and for this one it is 0.153 volts. So now let us calculate for this one as this one delta G. So for the first one equals minus 2 into 0.33 7. Let us keep F as it is okay. Similarly let us look into this couple here 0.153 because only one electron is involved in this one. Here 2 electrons so here it is 1. So let me write this way and now Cu plus electron gives Cu. So minus delta G equals minus 0.521. We have to sum up so you get the value. Now we can use for this reaction now. For this reaction we know the delta G known and F is known and Z is known. We have to find out E naught. E naught equals what? So let us put this equation minus delta G. So this is for Cu S Cu couple equals minus Z E naught F. So this value is already there. This is F. So I shall put minus 521 F equals here Z is 1. Simply this is E naught F. So it goes therefore and this is also minus therefore E naught. So standard electron potential of copper plus to copper equals 0.521 volts. So this is the answer. So this is how for unknown quantity so can be found out. So let us look into the coordination numbers now. So always consider mononuclear complexes while deciding the coordination number and coordination numbers and the corresponding geometries should be looked into it. For example if you have coordination number 2 it is supposed to have linear geometry. If coordination number 3 it should have trigonal planar geometry. So it continues. I shall elaborate more when we go to Weyland's Bond theory and the bonding concepts. Although the coordination environments are often described in terms of regular geometries. For example when coordination number 4 is there a sheer number can only tell you about the geometry whether tetrahedral geometry or square geometry but it does not tell you information about whether it is a regular geometry or distorted geometry. So this information can come from actual bonding and also by looking into the nature of the ligands and all these things are very nicely understood when we go for a crystal field theory where we come across spectrochemical series where we know whether a given ligand is strongly and or weak ligand. So that means the consequences of steric effects and other factors can influence the geometry and hence we can see some distortion in them. So discussion of a particular geometry usually involve bond parameters such as bond lengths and bond angles that can be determined in the solid state by doing single crystal x-ray diffraction. And of course here several other factors also influence these parameters including crystal packing forces. And when the energy difference between possible structures is small. For example when we have coordination number 5 and coordination number 8 if they possess more than one type of geometry and the energy difference between those two geometries are small then they can exhibit flexional behavior in solution. And sometime that can be monitored by going to the lower temperature using various spectroscopic methods. The small energy difference may also lead to the observation of different structures in the solid state. That means if they have a preference for a particular geometry at a particular temperature by controlling the temperature we can crystallize in a particular geometry for a given complex or a compound. If you consider pentacyononucleate anion where nickel is in plus 2 state the shape of the anion depend upon the cation present. So that means here when the coordination number is 5 in this case you can anticipate trigonal bipyramid geometry or it can also have square pyramid geometry. However the geometry taken adopted by this anion depends on the type of cation we have with this. And similarly in case of trisethylene chromium is a counter cation here chromium is in plus 3 state and in this one nickel is in plus 2 state it is a tri anion. So here this one can exhibit both TBP and square pyramidal geometries. So this particular combination of cation and anion can exhibit both TBP as well as square pyramidal geometries. That means it is not appropriate to consider just ionic lattices instead should involve complexes in which the metal center is covalently bonded to the atoms in the coordination sphere. The metal ligand bonding also should include not just sigma donor but also pi donor and also sigma acceptor as well as pi acceptor components. Of course you can understand these things very clearly when we talk about the classification of ligands. I should tell you again whatever the ligands we have at our disposal as coordination commits can be simply classified into three categories. One category is pure sigma donors the complexes having pure sigma donors. Examples are ammonia and water complexes aqua complexes or primary and secondary aliphatic amine complexes. And the second category belongs to those complexes having ligands which are capable of acting as sigma donors as well as pi donors. That means electron rich ligands exhibit this property. For example if you consider halides such as fluoride, chloride, bromide and iodide. So they have low energy field sigma orbitals and low energy field pi orbitals. So they are called as sigma donor and pi donor complexes. And the third one is carbon monoxide and phosphine and other related compounds. We call them as sigma donor and pi acceptor. So that means all ligands should fall into one of these categories. We shall look into more and more examples when we go to the bonding concept. That means these properties, ligating properties have significance influence on the geometry adapted by a particular metal complex. Let us look into the reactivity of the metals. In general the metals are moderately reactive and combined to give binary compounds when heated with oxygen, sulfur or halogens. And product formation depends on stoichiometry and also the oxygen state of the metal we are considering. For example when you treat metals such as transient metals with hydrogen, boron, carbon or nitrogen. The corresponding interstitial species such as hydrides, borides, carbides or nitrates are formed. For example we take osmium here and treat with two molecules of oxygen at high temperature then it can form osmium tetra oxide. And of course here osmium oxygen state is plus 8. And then if we take iron and heat with sulfur it can form FeS. And if you take vanadium and treat with halides it can form. So here it depends on the stoichiometry and also the reaction conditions. All these things are on heating happens at high temperature reactions where x can be F N equals 5 and when x equals Cl N equals 4 and when x equals bromide or iodide N equals 3. So depending upon the type of halides we are using and the stoichiometry and reaction condition we can get the corresponding homolyptic oxides, sulfides or halides. So many D block metals should on thermodynamic grounds liberate H2 from acid but in practice many do not liberate. The reason is they are passivated by a thin surface coating of oxide or by having a high dihydrogen ore potential or both. For example if you consider silver, gold and mercury they are least reactive elements. Gold is not oxidized by atmospheric oxygen or even by acids except aqua regia that is 3 is to 1 mixture of hydrochloric acid and nitric acid. And then when we talk about color of metal complexes of course electronic spectroscopy comes very handy in explaining and that we can understand better when we go to crystal field theory. And of course valence bond theory can also tell you about magnetic properties but not in detail especially the temperature dependence of magnetism and all those things can be understood through crystal field theory. And of course complex formation and why a complex is more stable and why a particular complex adopt a particular geometry despite other options are there. So all these things we should discuss once after completing coordination theory are proposed by Werner and then by understanding crystal field theory and molecular orbital theory. As I mentioned about interstitial hydrates. So interstitial hydrates most commonly exist within metals or alloys and their bonding is generally metallic. So transient metals form interstitial binary hydrates when exposed to hydrogen. These systems are usually non-starchometric with variable amounts of hydrogen atoms in the lattice. In material engineering one should be extremely careful about interstitial hydrogen present in the lattice especially when we are using such material for construction. Here if interstitial hydrogen is present in construction material. So the phenomenon of hydrogen embrittlement results and then what happens where the structure can be weak that is the reason usually in material engineering they will ensure that metal do not contain any interstitial lattice. In case if they found it has interstitial lattice some gases one has to heat it to very high temperature and then under vacuum and get rid of all the gases so that this hydrogen embrittlement does not damage the structure. Let us consider a reaction in which manganese is oxidized from the plus 2 state to plus 7 state. So when the manganese atom is oxidized it becomes more electronegative. In the plus 7 oxygen state this atom is electronegative enough to react with water to form a covalent oxide MnO4 minus. So it is useful to have a way of distinguishing between the charge on a transition metal ion and the oxygen state of the transition metal by convention symbols such as Mn2 plus refer to ions that carry plus 2 charge. Of course you are familiar with this kind of notation and one can also use for plus 7 oxygen state one can also write something like this or someone can also write like this but this is more appropriate to write like this rather than writing like this. So when we completely expand then the metal so for example when we expand something like this metal then one should write in Roman italic like this. It is not really appropriate to write like this so when we have to use this arabic numeral it is better to use symbol and write 7 plus. So manganese 7 is powerful enough to decompose water so that means if someone is looking for renewable energy and also to liberate hydrogen through oxidation of water one should know what kind of metal complex one should use usually they should use early transformation in their highest possible oxygen state and as you can see clearly when the metal is oxidized into highest possible oxygen state they become very strongly you know oxidizing in nature because of the increase in the electronegativity. So now in case of manganese 7 the electronegativity can be compared to that of fluorine and hence you can see how effective it is in oxidizing most of the substances. A similar phenomena can be seen in the chemistry of both vanadium and chromium. Vanadium exists in aqueous solution as the V2 plus ion but once it is oxidized to plus 4 or plus 5 oxygen state it reacts with water to form Vo2 plus or Vo2 plus ion. Similarly chromium 3 ion can be formed in aqueous solution but once this ion is oxidized to chromium 6 as in case of potassium dichromate it reacts with water to form CrO4 2 minus and Cr2O7 2 minus ions. Let me stop here and continue discussing about reactivity of metal complexes in my next lecture until then have an excellent time of learning chemistry.