So, very good evening to all of you. We will discuss together some questions on D and F block elements. Now, I have written, you have already got the questions. So, first question, 5 moles of H2S in acid medium fully reacts with KMNF4, A, B, potassium dichromate.
In each of this reaction, the number of moles of water produced is X and number of moles of electrons involved is Y. What is X plus Y in each case? I will do the first one.
You can give me the answer for the second one. Basically, you know very well H2S is a reducing agent. And KMNF4 or K2CR2O7, both are very good oxidizing agents. Medium is acidic.
It is given. So H2S, when it acts as a reducing agent, normally it forms sulfur plus H plus plus two moles of electron. This is H2S as a reducing agent. So the reaction is oxidation. The one which undergoes oxidation is a reducing agent.
Permanganate ion. In acidic medium, forms Mn2 plus plus water. So 8H plus plus 5 moles of electron. So, you multiply this reaction by 5 and the second reaction by 2 and write the net reaction.
So, the balance redox reaction is 5 moles of H2S plus 2 moles of MnO4 minus plus this side you have in the product side you have 10 moles of H plus in the reactant side 16 moles of H plus. So, 6 moles of H plus. Giving 5 moles of sulfur plus 2 moles of Mn2 plus 4 moles of water. I am sorry, 8 moles of water.
So, it is given 5 moles of H2S in each case. So, for 5 moles of H2S, the number of moles of water produced X in this case is 8. And the number of moles of electron transferred is 10. So, X plus Y equal to 18. So you should know the redox reaction and answer it. Now you try the second part B.
Take few minutes and then give me the answer. The final answer must be expressed. If it is a fraction, you express it in the nearest integer. Come on.
Anybody else? So let us see which answer is right. H2S same half reaction for H2S giving sulfur plus 2 H plus plus 2 moles of electrons.
Dichromate ion as an oxidizing agent giving plus 7 H2 so 14 H plus. plus six moles of electron. Okay. Now multiply the first reaction by three and rewrite it and then adjust to what is asked. So three moles of H2S plus dichromate ion.
This side we have 14 H plus, this side we have 6 H plus. So 8 H plus giving 3 moles of sulfur plus 2 moles of CR3 plus plus 7 moles of water. So for 3 moles of H2S reacting. 7 moles of water formed and 6 moles of electron transfer.
6 moles of electron involved. What is asked is for 5 moles of H2S. So for 5 moles of H2S, it will be 35 by 3 moles of water formed. And you have 10 moles of electron transfer.
So, x plus y equal to 35 by 3 plus 10, which will be 65 by 3, which is 22 approximately. I hope it is clear. So 35 by 3 you are asked to write it as an express it as a nearest integer.
So all other things are wrong. So you must be you must try to balance the equation and then correspondingly adjust it for 5 moles. I hope you understand.
Okay. How do you got 18, 13? So you should correct. Okay, the second question.
The magnetic movement of the compound with the strongest oxidizing ability mu of the compound with the strongest oxidizing action or oxidizing ability among MnF4, MnF3, MnF2. So MnF4 manganese is 3D5 4S2. So Mn4 plus will be MnF4 will be manganese in the 4 plus state which will have 3D3. Mn3 plus will have that is MnF3. Mn3 plus will be 3D4.
And Mn2 plus will have the other one 3D5. So the strongest oxidizing agent is Mn3 plus. So Mn3 plus plus electron giving Mn2 plus. So the MnD4 system 3D4 will become 3D5.
Okay, so the number of N-pad electrons here will be 3, N will be 3, N will be 4, N will be 5. So mu equal to root of N into N plus 2. So, root of 24, Bohr banknotes. Okay. The third question. Among acidified K-minofor, alkaline K-minofor, Then cupric sulfate, H2O2, chlorine, ozone, ferric chloride, HNO3, Na2S2O3, sodium thiosulfate, D.
total number of reagents or species whatever you call it that can convert I minus to I2. and what is the redox reaction involved mn4 plus to mn2 plus is also possible but we have to consider m see common mn4 plus becoming mn2 plus even if you include that the redox potential will be different that is also possible in fact mn4 plus giving mn2 plus mn2 plus giving m1 is also possible okay both are possible in this acidified km and a4 mno4 minus we already returned the reaction so i write the net half reaction plus 5 moles of electron giving m and 2 plus plus 4 h2o so 2 i minus giving i 2 plus 2 electron so multiply this by 5 this by 2 so you get the net redox reaction alkaline came in the four MnO4 minus in strongly basic medium it will form MnO4 2 minus and I minus will form Io3 minus. This you can rewrite it by Adding 6 OH minus both sides because it is given in basic medium. This is easier to balance. So you write it as right.
Add 6 OH minus and rewrite it. So I minus plus 6 OH minus. Giving I O 3 minus this side will become 6 H 2 O, this side 3 H 2 O plus 3 H 2 O plus 6 electrons.
Combine these two, you will get that net redox reaction. Multiply this by 6 and add to this. So this will not be the answer because I minus giving I O 3 minus is the product.
Okay, in basic meaning. Then. copper sulphate the net reaction is copper 2 plus the very important reaction plus 4 i minus giving cu i plus i 2 this can also be written as 2 cu 2 plus plus 5 i minus Giving 2 Cu I plus I2 can be written as I3 minus also.
But for this question it is enough if you write this copper 2 plus, copper 2 plus plus I minus, copper 2 plus plus I minus forming. Cu I that is one half reaction 2 I minus giving I 2 plus 2 electron is other half reaction. So, here plus electron if you add and then multiply the first reaction by 2, you get the net redox reaction.
So, this is a very important redox reaction. This is also used for the estimation of cupric ion. Copper 2 plus can be estimated using iodine evolved.
So, what you find is, so this reaction is used in the estimation of copper 2 plus. in an alloy copper copper 2 plus now number of milliequivalents of copper 2 plus is identical to number of milliequivalents of iodine liberated that is equal to number of milliequivalents of thiosulfate ion consumed using starch indicator the iodine reacts with thiosulfate that is a common reaction s2o3 2 minus forming s4 o6 2 minus plus 2 moles of electron i2 plus 2 electron giving 2 i minus so starch is the indicator starch solution is the indicator iodine plus starch gives a blue color and when blue colored solution is treated with thiosulfate from the burette there is a disappearance of blue color that is the end point okay so this reaction is made use of the estimation of copper or copper 2 plus in an alloy number of milliequivalents of copper 2 plus equal to number of milliequivalents of iodine equal to number of milliequivalents of biosurfing and the equivalent weight of copper 2 plus is same as formula weight because of one mole of electron transfer so this reaction also will will liberate iodine next one is hydrogen peroxide hydrogen peroxide In acidic medium, forms water as oxidizing agent. H2O2 can serve both as an oxidizing agent as well as a reducing agent.
I minus giving I2. This is also correct. Then ozone, ozone forming oxygen. 2i minus giving i2 this is also correct this was asked in one of the advanced questions then ferric ferric chloride fe3 e plus plus electron giving ferrous 2i minus giving i2 plus 2 electron Then nitric acid NO3 minus plus H plus giving NO2 to I minus giving I2. See, even if you do not know, you have to guess, but it is better to know the concept and answer.
So which are the oxidizing agents? Just for information, I am giving all the reaction. See, normal product is, see, you can rule out two things straight away.
One is alkaline K-minofoam. because alkaline came in a four the product is mno42 minus and i minus is converted to io3 minus yes if you know that that is good enough so this will not be the thing okay for all the things and just since redox reaction is asked i have explained to you then then thiosulfate ion s2o3 2 minus forming S4O6 2-. So this is a reducing agent. So two reducing agents cannot react. So this is also not the case.
So straight away you can rule out alkaline K-minofor in this list. and s2o3 2 minus or sodium thiazide both are out the rest of it is not so that will work out to 7 you need not you need not i mean for answering this question we need not know all the reactions but all other things are oxidizing agent that you should know so i minus giving i2 is common so you should know ferric becoming ferrous h2o2 becoming water etc if if it is needed that's okay for answering this question you can rule out alkaline k-amino4 and thiosulfate straight away because thiosulfate is not an oxidizing agent at all it is a reducing agent okay two moles of thiosulfate Then the fourth question peroxidisulfate or it is also called perdisulfate. Ion S2OA2-reacts with Mn2+, BCr3+.
Now S2O3-S2OA2-forming sulfate ion that is a normal half reaction. So, Mn, what is asked is Mn2 plus giving MnO4 minus. Mn2 plus forming MnO4 minus.
It is just the reverse of MnO4 minus reaction. So multiply this by 5, this by 2 and write the net reaction. So that I leave it to you.
Okay. So per sulfate ion giving sulfate ion that you should know half reaction. Rest you can balance it. Okay.
Now same per sulfate ion. CR3 plus forming chromate ion. So, this you combine. So, this you multiply by 3, this you multiply by 2. What you should know is S2OA2 minus forming sulphate ion and AMN2 plus forming MNO4 minus BCR3 plus forming chromate ion. Depending upon the medium, chromate may form dichromate ion.
See, the normal reaction is Mn2 plus forming MnO4 minus. It can form MnO2 also, but in solution, the product, because it is a reversal of MnO4 minus reaction, so it is more common product, unless otherwise given. Now, in connection with this, I will give you a question asked in one of the GE advance. which is based on one of this reaction to measure the amount of that is not in the sheet mncl2 in aqua solution dissolved in an aqua solution few drops of concentrated hcl added to this solution heated further 225 milligram of H2C2O4 added in portions till pink color disappears. That shows KMNFOR is formed, Anushri Mukul.
So till pink color disappears. dash milligram of MnCl2 is present. Initially H2C2O4 molecular weight 90 gram per mole and MnCl2 126 gram per hour. Now this is based on this reaction.
So Mn2 plus so this reaction Mn2 plus the reaction is given unbalanced equation MnCl2 plus K2 is to O8. plus water giving KMnO4 plus K2SO4 plus HCl plus sulfuric acid. This equation is also given in this question.
So, that K2SO2 is involved. So, this reaction unbalanced it is given. added to this solution. So, this solution HCl is added.
So, I hope you understand the question. So, what is asked is, so you write Mn2 plus, you can write, instead of dealing with this equation, Mn2 plus giving MnO4 minus, that is what is given. So, MnO4 minus plus 8 H plus, plus 5 moles of electron. This will be 4 H2O.
okay and s2oi2 minus giving so4 2 minus plus two moles of electron this is good enough for us right to measure the amount of m1 cl2 few drops of concentrated cl added to this solution and heated so to this solution that is the solution containing m and cl2 and k2s2o8 i hope you get the point okay so that is involved in this it is given here now So this MnO4-this is one redox reaction. The second redox reaction is MnO4-formed in acidic medium forming Mn2 plus 4H2O oxalic acid or oxalate ion forming 2CO2. plus two moles of electron so this is the reaction okay so oxalic acid is a reducing agent permanganate formed is an oxidizing agent now 225 milligram now number of So this is on heating. So the reaction occurs on heating because the reaction, you know very well, permanganate oxalate titration is done under hot conditions.
So number of millimoles of H2C2O4 is 225 milligram divided by molecular weight is 90, which is 2.5. So as per the reaction, 5 millimoles of oxalic acid. requires 2 millimoles of KMnO4.
You have to connect it that's all. Therefore 2.5 millimoles will correspond to 1 millimole of KMnO4. Now from the first reaction which is given as unbalanced equation you find this will be Multiply by 2, this is multiplied by 5. Okay. So, 2 millimoles of MnO4 minus correspond to 2 millimoles of Mn2 plus reacted with Yes to OI2-. I hope you understand the connection.
So 1 millimole of KMnO4 ultimately 1 millimole of KMnO4 correspond to 1 millimole of MnCl2. That is all. 1 millimole of KMnO4 correspond to 1 millimole of MnCl2. So, number of millimoles is weight in milligram divided by molecular weight. So, weight of MnCl2 in milligram is number of millimole 1 into molecular weight 126. So, it will be 126 milligram.
That is the answer. I hope you get the point. So this is the question asked in a GE advance.
The answer is 126 milligram. It is connected to the question I had given. So to measure the amount of MNCL to NACO solution, few drops of H. amcelated Cl added to this solution and heated.
Further 225 milligram of oxalic acid added in portions till the pink color disappears. Dash milligram of MnCl2 is present in the original initial solution. So it is connected to one of the redox reaction involved in this question.
See in this case, in this problem, suppose you have copper 2 plus I minus I2. So this is one reaction. The iodine involved reacts with another reaction.
So according to the law of equivalence, number of milliequivalents of copper 2 plus is same as number of milliequivalents of iodine. equal to number of milli equivalents of i2 liberated equal to number of milli equivalents of thiol. So, we have already seen the reaction 2 Cu2+.
plus four i minus giving two cu i plus i two this is one redox reaction the other redox reaction is s2 o3 2 minus plus i2 giving s4 o6 2 minus plus i minus So this is the reaction. So the number of milliequivalents of copper 2 plus is identical to the number of milliequivalents of I minus consumed equal to number of milliequivalents of iodine iodine liberated milliequivalents of thioconsume. So if you use milliequivalents they all the milli equivalents are same if you want to use it in terms of milli moles only the equations are even okay i have i have tried to explain the problem using milli moles you can also do it in terms of milli equivalents you will get the same answer okay Here also if you use milliequivalents the number of milliequivalents of Mn2 plus reacted equal to number of milliequivalents of MnO4 minus formed equal to number of milliequivalents of S2 O8 2 minus consumed equal to number of milliequivalents of oxalate ion consumed.
So you can decide using I hope you get the point. Next one, MN lanthanoid ions. Fifth question. All these things, it is comfortable if you write ions.
equation instead of full balanced equation. Ionic equation is good enough. That will help you to get the questions quickly. How do you get the mass of MnCl2 is 0.0025?
Because here you find 1 millimole of K-MnO4, 1 millimole of K-MnO4 correspond to 1 millimole of MnCl2. So the number of millimoles of MnCl2 present is 1. So 1 millimole into, so number of millimoles is weight in milligram by molecular weight. Molecular rate is given 126. So, you get 126 milligram or 0.126 gram.
I hope you get the point. Kirtana, number of milliequivalents of Mn2 plus. is same as number of milli equivalents of MnO4 minus consumed equal to number of milli equivalents of oxalate ion. I have tried to explain the problem using millimoles. Both will give the same answer, should give the same answer.
Okay. Fifth one. Mn, the lanthanol ions.
Samarium 3+, so lanthanum 3+, will have F0 configuration, 4F0. Lutetium 3+, as F14. And gadolinium 3+, 4F7. So, no F electron, all the 4F levels are filled here, half filled, so very stable.
So, samarium 3 plus, you need not know, but it is in between. So, samarium 3 plus actually will have 4F5. So you need not know the electronic configuration, but in these things you should know, but samarium is coming in between. So it will not have half filled, it will not have F0, it will not have F14.
So all these are colorless, this will be colored. Then sixth question, magnetic moment of titanium 2 plus. Atomic number will be given to you. Vanadium 2 plus, titanium 3 plus, scandium 3 plus. Follow the order.
Scandium 3 plus, less than titanium 3 plus, less than titanium 2 plus, less than vanadium 2 plus. It is not vanadium 2 minus, it is vanadium 2 plus. I think in my writing it looks like 2 minus. Please correct. So, scandium, EZ is for scandium it is 21, titanium 22, vanadium 23 atomic numbers.
This will be given to you. Okay, so you quickly try. Come on. anybody else good the order is true so scandium 3 plus will have 3d 0 4 0 so n is 0 Then titanium 2 plus will have 3d2. So n is 2. Titanium 3 plus will have 3d1.
n is 1. Then vanadium 2 plus will have 3d3. So the order is right. no no you need not have to remember atomic number will be given to you it is half filled so the x very much stable so only if if the if the electron gets excited it will be colored so gd3 plus is already f7 which is which is of exceptional stability f0 f14 and f7 are very stable lanthanoid ions Okay, that is why it is colorless.
It is true in this case. Next question, dash pair of elements will not have similar atomic radii. This is based on lanthanoid contraction. So, scandium and nickel are somewhere else.
Okay, that is not involved in lanthanide contraction. So, A is out of question. B, titanium and hafnium.
Titanium, then zirconium, then hafnium. So, zirconium and hafnium will have similar radii because of the intervening lanthanides. Okay. Then third one is molybdenum tungsten. Molybdenum and tungsten are due to the lanthanide contraction C is right.
Chromium, molybdenum, tungsten. So this will have similar radii because of the intervening lanthanides. And the last one is manganese and uranium.
and rhenium manganese technetium rhenium so this and this will not have similarly so due to intervening lanthanate contraction what are the important lanthanides to remember so the electronic configuration wise you should remember lanthanum then you remember cerium then gadolinium then lutetium. Correspondingly you can get the lanthanide 3 plus cerium 3 plus etc. Okay these things which electronic configuration it is better to remember so that you will not waste time. Okay.
Eighth one, X with excess aqueous OH minus and NaBO3 4H2O. This is NaBO3 4H2O is called hydrated sodium perborate. Anyway, the equation is given for that. Bo3 minus forms what? That is given.
So, forms an yellow solution y. x is likely to be what? So, Explain the reactions involved.
BO3 minus reacts with OH minus to form BO3 3 minus and H2O2. That is given. So, obviously the clue is it must be CR OH trace because it forms a blue solution. It forms a yellow solution. So, the yellow solution is likely to be CRO for 2 minus.
If it is an objective, you need not go further. Now, this BO3 minus reacts in basic medium forming BO3 3 minus. So, one half reaction is. Bo3 minus forming Bo3 3 minus plus 2 moles of electron.
Okay, this is an oxidizing agent. Ferborate is a very good oxidizing agent. The reducing agent is OH minus forming H2O2. So, see H2O2 forming 2 OH minus. plus 2 moles of electron is an oxidizing agent.
Okay. It is a reverse of it. OH minus forming H2O2. So 2 OH minus forming H2O2. So balance plus 2 moles of electron.
So the net reaction is BO3 minus plus OH minus forming. This is given to you. BO3 3 minus plus hydrogen peroxide.
Here this acts as this oxidizing agent, this acts as the reducing agent. Okay. Now, CrOH3 is a blue or green gelatinous precipitate. This with the excess OH-the precipitate dissolves forming CrOH4-. Okay, tetrahydroxidochromate 3 ion, it is a complex ion.
The CrOH4 minus, so X is likely to be X is CrOH3. So this forms a green solution. The green solution CrOH4-reacts with H2O2 to form chromate ion. Okay, now what you do is CrOH4-is also an important reaction.
CrOH4-giving CrO4-. Cr O 4 2 minus plus 4 H plus plus 3 electron. But the medium is basic. So what you do? Add 4 OH minus and rewrite it.
Cr OH 4 minus plus 4 OH minus giving Cr O 4 2 minus plus 4 H 2 O plus 3 electron. Okay, the other reaction is H2O2 giving OH-as an oxidizing agent in basic medium. So combine these two and re-initiate. Yellow solution is chromate ion.
So the Y is chromate ion. So the chromate ion reacts with barium chloride to form a yellow precipitate of barium chromate. It also reacts with the lead acetate. These are the reactions forming lead chromate.
Both are yellow precipitates. With BAM chloride, this is given. Okay.
Then why in acidic solution forms orange solution is that you know very well chromate ion is converted to dichromate ion in acidic medium. This conversion is favorable in acidic medium. The reverse reaction occurs in, I am sorry, this conversion is favorable in acidic medium.
The reverse reaction is favorable in basic medium. This is also very important. Okay, plus water. So, Y forms a yellow precipitate with barium chloride.
Y is an acidic solution forms orange solution. Then potassium salt of Z in the solid state that is potassium dichromate. Okay, so dichromate with concentrated, that is chromyl chloride test. This is very frequently asked. So potassium dichromate plus sodium chloride plus concentrated sulfuric acid on heating forms CrO2Cl2.
So, the reaction is Cr2O7 2 minus plus Cl minus plus H plus forming CrO2Cl2 chromyl chloride. This is labeled as L volatile, red volatile liquid, red liquid which may vaporize, red liquid which may vaporize plus water. So X is CrOH3, Y is chromate ion, then Z is dichromate ion, volatile liquid L is chromyl chromate.
These are the things. And chromyl chloride reacts with NaOH to form sodium chromate plus NaCl plus water. Okay, this reaction is also important.
The ninth question, since we have done enough of redox reactions, I am just skipping it, but I will give you the answer alone. KMnO4 can act only as an oxidizing agent. Dichromate can act only as an oxidizing agent. Na2SO3 can act only as a reducing agent.
Okay. All the others, SO3 2-, SO2 or SO3 2-. Then H2O2.
H2PO2-hypoposphite ion. HPO3-phosphite ion. Hydrazine.
you can even include n o 2 minus these are some examples of species which can act both as an oxidizing agent as well as a reducing agent so i just skip this so if there is any problem we will sort it out you should remember you should know for objective which of the species which can act both as an oxidizing agent as well as a reducing agent okay but commonly sulfur dioxide NO2 minus nitrate ion, hydrazine, even H2O2, these are common reducing agents, particularly with a stronger oxidizing agent. Not the entire equation. The entire equation is not needed except that is involved in that question. So, I skip this if there is any problem regarding the reactions involved we will try this. Tenth one, a metal ion shows magnetic moment of 5.9 Bohr magneton.
It is likely to be dashed. And the number of hand pad electron is dash. Try this.
Ion is 26 and manganese is 25. So the species is likely to be what among the choices given and what is the number of unpaired electives? Good, B is correct. Mn2 plus will have 3D5 electronic configuration, N is 5. So, mu is root of 35 Bohr magnetons. So, Mn2 plus number of hand-picked electron is 5. Good.
Next one, 11th one, solutions of both KM and A4, K2CR, K2O7 can be used as primary standard in redox titrations. in volumetric analysis. See what is the primary standard?
A primary standard is one where the solid is very much pure, purity. The solid should be stable, pure and it should be stable. In aqua solution, it should not decompose or react with atmospheric species.
or the solvent and reasonably high equivalent weight. Purity, the solid should be stable, then in solution it should not decompose or react with atmospheric species or with the solvent. That means it should not decompose in water. it should have reasonably high equivalent you should you should not try the tenth one it may be asked in a different way so bar go it may be asked in a different way not in the same way but it may be asked in a different way okay as a combined one it may be asked not like this These are the requirements of a primary standard for acid-based titration. You have primary standard for redox titration.
In the case of permanganate, what happens is the permanganate ion, potassium permanganate in aqua solution If you keep it for a long time, it precipitates MnO2. So it is not used as a primary standard. So MnO4-, it reacts with water. forming MnO2 plus oxygen plus OH minus.
So MnO4 minus on standing reacts with water forming MnO2. So this formation of MnO2 will obscure. the endpoint etc that is why it is not con that is directly you cannot weigh and use it as a as a standard solution so that is the so kmn4 is not a primary standard whereas potassium diagram can be used just for information you should know this okay you can take a short break and we'll get back five minutes break and come back okay Thank you. So we will resume. So dichromate ion can be used as a primary standard.
What is meant by primary standard means that which can be used by dissolving it in water and preparing a standard solution. can be prepared by direct weighing and dissolving it. And that solution meets the requirement, namely it will not decompose, it will not react with the solvent etc. That is the thing. okay whereas potassium permanganate solution suppose you prepare 0.1 molar and keep it and this solution will not keep its molarity because the reaction will occur slowly on keeping you will find the deposition of m1 o2 at the bottom due to the attack of solvent that is why permanganate solution is not considered as a primary standard it has to be standardized that is what is important about primary standard which you should know then potassium dichromate see dichromate ion as a oxidizing agent forming so for acidic medium You can use dilute aqueous HCL with this because even though HCL or CL-can act as a reducing agent, this reaction will not occur with dichromate.
That is why for acidification you can use dilute HCL or dilute sulfuric acid. either one can be used for dichromate titration suppose you are titrating dichromate ion with ferrous for acetic medium you can use either one of them because this even though it is a reducing agent this will not occur with this because of the potentials involved if i combine this and calculate the potential it will be negative that means this redox combined redox reaction will not occur in the case of permanganate ion of course if the potentials are given you can even prove it mno4 minus This reaction occurs that is why dilute HCL is never used for acidification. Always permanganate titration we use dilute sulfuric acid. Because HCL not only acts as an acid in this case but acts as a reducing agent.
This reaction is feasible. So the potential is positive. If you are given the potential you can prove it but you should note that dilute HCL why it is not used here.
and why HCl can be used here because the combined redox reaction potential is negative here. The combined E0 cell is positive here. So, that is why you cannot use HCl because HCl acts not only as an acid but also as a reducing agent in this case.
That is the thing which you should know. Then third one. Permanganate ion is unstable in the presence of MnO2+. MnO4-plus MnO2+, in solution forming MnO2.
That is what happens is MnO4- MnO2 plus forming MnO2. So multiply this by 3, this by 2. You will get the net redox. What is important is MnO4 minus and MnO2 plus. in aqua solution forms MnO2.
So because of which it is not stable. Even if you calculate the potential, of course, the potential will be positive for the net reaction. Can you tell me what is the specific redox reaction called?
Quickly, can you tell me? I cannot balance it, you can balance it. So what is the net redox reaction involving permanganate ion and Mn2 plus in aqua solution forming MnO2?
Good. Almost all of you answered correctly. Come proportionation. See, mostly at the advanced level, in this chapter in particular, they ask problems involving redox reactions. involving problems as we have done in the first case.
Mostly it is asked or they may give you ABCD type of problem involving redox reaction using various species. This is what is mostly concentrated in JE advanced in this particular topic. That is why I have given a few questions on that. Next, the oxidation state of chromium D. E0 reduction values for Mn2 plus bar Mn, zinc 2 plus bar zinc are more negative than what is expected because of half-filled D5 and D10 configuration.
This is true. Okay, compared to the all other things in the series. Then E, the oxidation state of chromium in CrO5, Cr double bond O. So, the chromium is in the plus 6 state.
this is minus 2 each peroxyl linkage is minus 1 so plus 6 is correct then f is cerium shows plus 4 oxidation state because cerium 4 plus will have f 0 configuration which is stable for f 0 and europium 2 plus will have 4F7 configuration which is also stable. That is why for cerium and europium, the all lanthanoid common lanthanoid ions 3 plus oxidation state is common. But cerium shows plus 4 oxidation state, europium shows plus 2 oxidation state, which are uncommon among other lanthanoids because of the exceptional stable configuration. This is important conceptually.
Okay. Then actinoids show more number of oxidation state than lanthanoids. That is because successive, this is true, we have already discussed it, successive ionization energy differ by less than they do for lanthanoids the actinoids the successive ionization energy which in turn is related to the 5f orbitals extend further from the nucleus than 4F of actinide. This is octinoid, this is lanthanides. And hence they are available for bonding to a greater extent than 4F.
Then the difference in energy between 5F, 4D is very much less than 4F, 5D orbitals. This is for actinoid ion, this is for lanthanid ion. For this reason, we may show more number of oxidation state actinoids compared to lanthanids.
this is also related to the actinoy contraction is greater for our element to element that is also true for so h is also related to this okay then i lanthanum 3 plus as f0 configuration gadolinium 3 plus as f7 configuration 4 of 0 4 f7 and lutetium 3 plus f4 f14 so n is 0 here n is 7 here n is 0 here so that is also true then j Lanthanide contraction is due to poor shielding of one 4F electron by the other. This is very much true. Then K, the ionic radii is due to lanthanide contraction. Okay. Don't you get an answer?
So yeah, he said do our first. Thank you. false it is it is false is correct correct because cerium forms cerium 4 plus oxidation state because it has 4 of 0 and europium exists as plus 2 oxidation state because of 4 of 7. So, these are the unusual oxidation state among lanthanides. So, it is false is very much right. The next one chromium 2 plus is a reducing agent see this question chromium 2 plus is a reducing agent it forms chromium 3 plus plus electron in aqua solution if you answer it is 3d4 to 3d3 So, in aqua solution, if it is forming a complex, then if it is an aqua solution, CrH2O6 times 3 plus.
In aqua solution, so if it exists as a complex ion, then T2G level is half filled. so because of which it acts as a cr2 plus acts as a reducing agent in the case of mn2 plus the other one is mn3 plus plus electron giving mn2 plus it is a d5 system which is stable The exact answer should be based on ionization energy, enthalpy of atomization, enthalpy of hydration, everything. So you have to combine all the factors and decide.
But this is one way of explaining it using the electronic configuration assuming it forms a complex ion. half filled in the complex ion t2g is half filled here d5 you can also ask me okay why can't it form a complex ion okay this also can form a complex ion then it is difficult to answer honestly both if they are present in aqua solution then this this question cannot be answered as such but experimentally cr2 plus becoming cr3 plus is the favorable potential and mn3 plus giving m and 2 plus the favorable of cell potential okay i hope you get the point Then third one, all lanthanoids can be called as rareth elements. That means they are not available in nature. That is what it means. The availability is very very little.
Is it true? Actually it is false because except promethium which has no stable isotope, In the case of promethium, we can call it as rareth, but not all other lanthanides. All other lanthanides, they cannot be called as rareths except promethium, which has no stable isotope because it is radiant. I think this is D.
C I have skipped. C you can answer. Quickly you can answer. C is MnO2 plus KOH plus O2 forming KMnO4 and water.
Is it a right reaction? Quickly answer. In a video, I will quickly answer. So it is false because it forms K2MnO4.
Pyrolysiside MnO2 is fused in an oxidizing atmosphere to form K2MnO4. That is the reaction. So this is wrong. So the actual reaction is 2MnO2 plus KOH plus oxygen.
It forms K2MnO4. which is the green solid plus water this is a correct reaction so this as such it is okay then e lanthanum has 4f0 so no f electron and actinium has 5f0 So it is placed along with lanthanoid and actinoid but they have 4f0 configuration 5f0 configuration. So as such the question is false.
What is given is false. Next one you try. 13th question.
Dash cannot react with dichromate H+. 13th question. among sulfite, carbonate, sulfate, nitrate, dash cannot or will not react with potassium dichromate in acidic medium. Is it true or false?
Which among the following cannot or will not react with K2Cr2O7 H+. Now, just to get the idea, I will ask you a question. How will you differentiate between carbon dioxide and sulfur dioxide gas? Tell me, what is a test? which will distinguish between carbon dioxide gas and sulfur dioxide gas.
I will come back answering to this question, if you answer this. So, by what chemical test you will differentiate between carbon dioxide and sulfur dioxide in the lab? Come on. Come on, you answer my question. How will you differentiate between, distinguish between carbon dioxide and sulfur dioxide gas?
Carbon dioxide is related to carbonate, sulfur dioxide is related to sulfite. So this cannot be distinguished using lime water test because both will turn milky. So carbon dioxide will form calcium carbonate.
Sulfur dioxide will form calcium sulfide. And if each one is continuous passing, it will form calcium bicarbonate. and calcium bisulfite both are water soluble so you cannot use lime water test this is also water soluble this is also water solid this is a milky or a white precipitate so using lime water you cannot differentiate so carbon dioxide so sulfur dioxide being a reducing agent it will decolorize acidified gamma in the food decolorize acidified camminophore it turns diachroma acidified potassium dichromate paper green being a reducing agent so sulfite ion is a reducing agent carbonate is not sulf sulfate dioxide or sulfate ion form sulfate ion so sulfate ion is not a reducing agent nitrate ion will form nitrate ion so that is not a reducing so the only thing in this list is sulfite ion all other things are not reducing gauges okay that you should know Next one, 14th question. When KMnO4 acts as an oxidizing agent, it forms MnO4 2-, MnO2, Mn2O3, Mn3+, under different conditions.
The sum of the number of moles of electron transfer per mole of MnO4-. So, try to work it out and get me the answer. MnO4-.
forming MnO4 2-. MnO4-forming MnO2. MnO4-forming Mn2O3.
MnO4-forming Mn2+. The sum of the number of moles of electron transferred per mole of permanganate ion in each case. What is the sum?
Quickly answer this. I hope you understand my question. Per mole of MnO4 minus use, what is the total number of moles of electrons transferred or involved together.
Good. 13 is right. So MnO4 minus plus electrons.
electron giving MnO4 2 minus. One mole of electron involved. Then MnO4 minus giving MnO2. So, three moles of electrons involved. Then MnO4 minus giving Mn2O3.
So 8 oxygens here 3 so triphi 10H+. So 10H+, 2-so 8 moles of electron. So per mole I hope it is balanced. So per mole it is 4 moles of electron. I hope you get the point.
Last one MnO4-. plus 8 H plus plus 5 moles of electron giving Mn2 plus plus 4 H2O per mole of permanganate ion 5 moles of electron. So, 1 plus 3 plus 4 plus 5 13 moles of electron.
Question 13. Correct. Two oxidizing agents will not get. Exactly. So you are right.
Manasabhiji, you are right. So permanganate, dichromate ion. is an oxidizing agent.
Nitrate ion is an oxidizing agent. Sulfate ion is an oxidizing agent. So these will not react. This question is related to, for example, ferric iodide, PbI4.
They are not stable. I can mix ferric chloride. with potassium iodide. It may form ferric iodide but this is not stable.
Why? Because it is a combination of this question we already discussed in the class. So ferric is an oxidizing agent, I-is a reducing agent. So ferric iodide will form ferrous and iodine.
Same with Pb4 plus I minus. So Pb4 plus will form Pb2 plus. I minus will form iodine.
Okay. Same type of concept involved in this. They are not stable because it is a combination of oxidizing and reducing agent together.
Okay. Next one. 15. You try.
Take a few minutes and try this question. So one clue is given which I have missed it. M2 plus colored. That is given. Carbonate is not.
Carbonate ion is not. So M2 plus with H2S acidic medium it forms a black precipitate. So it is most likely to be definitely it is second group cation in cation analysis. Second group of your cation analysis, not second group of periodic table.
So it is most likely to be copper. The black precipitate is CUS. Because if it is cadmium, it will form a yellow precipitate. So in the second group, copper 2 plus is colored.
If it is given as blue color, the conclusion is very obvious. That is why it is simply given as colored. Right?
The colored I missed it. Then copper 2 plus forms CUS. The black precipitate is CUS. So CUS.
plus nitric acid i write the net reaction you can split it and write it also it reacts with concentrated nitric acid forming copper 2 plus then nitrate ion sulfur that is the yellow substance plus no plus 4H2. This is the redox reaction. You can split it and write also. Then copper 2 plus B plus sulfur is given.
So B is the copper 2 plus. that m2 plus okay then m2 plus with ki so that also i have shown it to you in the middle in the beginning 2 cu 2 plus plus 4 i minus giving cu i which is a white precipitate plus cui plus i2 okay so the white precipitate is cuprous iodide this can be written as cu2 i2 also then how does m2 plus reacts with excess cyanide we have already seen this 2 cu2 plus plus 10 cn minus giving copper one cyanide complex cu cn43 minus plus cyanogen so the ion involved is copper 2 plus okay the colored ion with acidic medium in h2s it forms a black precipitate is the clue for copper 2 plus Next one, 16. This was also asked in GE Advance. I have just shortened or modified the question.
5.6 gram of iron dissolved in cold diluted HCL and the solution is made up to 250 cc. 25 cc of this require 12.5 cc of 0.03 molar KMnO4. what is the number of moles of ferrous in 250 ml what is the percentage of ion in the sample by weight now ferrous ion forms ferrous ferrous form spheric and permanganate ion so this is a redox reaction now it's made up to 250 cc 25 cc of a resolution correspond to 12.5 cc of 0.03 molar km in a four one molar km in a four as an oxidizing agent in acidic medium is phi normal because of 5 moles of electron transfer so 0.03 molar is 0.15 normal into 5. so number of milli equivalents of kmn of 4 consumed is 12.5 into 0.03 into 5 that is 0.15. This is same as number of milli equivalents of ferrous reacted with it.
This is for 25 cc. So 25 cc of the solution correspond to 12.5 into 0.15 milli equivalents of ferrous. Therefore, 250 cc will correspond to 125 into 0.15 milli equivalents of ferrous.
Milli equivalents of ferrous is same as milli moles of ferrous because of one mole of electron transfer. So, number of milli moles of ferrous is 12.5 into 0.15. Okay.
So number of moles will be 12.5 into 0.15 divided by 1000. Moles of ferrous. Now weight will be 12.5 into 0.15 divided by 1000 into 56 grams of Fe, atomic weight of iron is 56. So 12.5 into 0.15 by 1000 into 56. This is the iron present in the sample. So 5.6 gram of the sample.
correspond to 12.5 into 0.15 divided by 1000 into 56 gram of iron present as Fe2+. Therefore, 100 gram of the sample correspond to how much? I hope you understand. So, the answer is 18. 75% that is for the later part. The first part is 1.875 into 10 power minus 2 moles of ferrous.
See, as you find, most of the questions asked in advanced level are combination of D-block and redox. And sometimes the D block is combined with complexes and redox reaction is lost. Okay.
For example, suppose I give a question which is based on redox and complex. Just to show you. In dilute, aqua sulfuric acid, Fe, H2O twice.
Oxalate ion with permanganate ion acidic medium forms the corresponding ferric complex. Okay, what is the ratio of rate of change of H plus consumption. So kinetics, redox equation, everything combined divided by rate of change of permanganate consumption is what?
So what you do here, you write the balanced equation Fe H2O twice dioxylator complex. 2 minus giving Fe H2O twice C2O4 twice minus plus electron whereas complex becoming ferric complex MnO4 minus plus 8 H plus plus 5 moles of electron giving Mn2 plus plus 4 H2O. So, multiply this by 5. So, 5 moles of ferrous complex plus MnO4 minus.
plus 8 moles of H plus giving 5 moles of ferric complex plus m and 2 plus plus 4 H2O. So, minus 1 by 8 into d of H plus by dt. just recall your kinetics that is equal to minus d of mn04 minus by dt from the net reaction so what is asked is rate of consumption of h plus divided by rate of consumption of permanganate ion so the required ratio will be 8 so rate of Consumption of H plus divided by rate of consumption of permanganate ion will be 8. So it is a combination of redox reaction and kinetics.
So far in these questions, which question you are asking, Bargo? I am sorry, correct. Correct, you are right. I am very, very sorry. The reaction should be plus electron.
Ferrous complex becoming ferric complex. Correct. You are right. Anish, thank you. Vargo, so both are in the reactant side.
This is also in the reactant side. This is also in the reactant side. Both are in the reactant side.
So, minus 1 by 8 into d of H plus by dt is minus d of Mn over minus by dt. See, if it is a complex, it should be given, that's all. So, otherwise, ferrous becoming ferric is the only reaction possible with permanganate ion, diagrammate ion, etc.
The last question, 1 molar permanganate ion, this is also frequently asked as a problem also, reacts in H plus with dashed modes of ferrous oxalate. So, ferrous oxalate is an interesting case because it can act as reducing agent at room temperature. it corresponds to one mole of electron transfer.
The same ferrous oxalate can act as a reducing agent at 70 to 80 degree centigrade on warming. It corresponds to three moles of electron transfer. It simply acts as a salt that is non-redox reaction. No electron is transferred.
So, the equivalent weight of ferrous oxalate is molecular weight by one in the first case. molecular weight by 3 in the second case. Molecular weight by 2 as a salt because the total charge on cation are anion. So this will be the first one.
This will be the second one and as salt this will be the by 2. Okay now so ferrous oxalate at 70 to 80 degree centigrade ferrous plus oxalate ion. forming ferric plus 2 CO2. Okay.
Ferrous oxalate ion forming carbon dioxide. So, ferrous plus oxalate ion becoming ferric plus then permanganate ion plus 8 h plus plus 5 moles of electron giving mn2 plus plus 4 h2 so what is asked is one mole of permanganate ion so this will be into 5 into 3 into 5 3 moles of potassium permanganate will correspond to 5 moles of ferrous oxalate at 70 degrees. 1 mole will correspond to 5 by 3 moles. That is the answer.
Write the ionic reaction. Ionic half reaction. unbalanced not the whole reaction okay which is the most important j means it is very difficult to answer see kinetics electrochemistry then thermodynamics equilibrium Then complexes, bonding, then organic, the whole thing, particularly isomerism, etc. 15th question again. The colored metal ion M2 plus reacts with hydrogen sulfide in acidic medium.
I suppose both are given in the question. It forms a black precipitate. So it is a metal ion.
It's colored. So it is more likely to be copper 2 plus because second group metal ions will form a precipitate with H2S in acidic medium. So it must be CUS. Then CUS. COS reacting with concentrated nitric acid to form copper 2 plus that is the solution plus sulfur plus NO that those are the main products.
Okay that is what is given. Part of it is present as nitrate that you need not worry plus water. this is unbalanced then copper 2 plus reacts with i minus to form cu i plus i2 this is a white precipitate and then the metal ion m2 plus that is copper 2 plus Reacting with cyanide ion, this we have discussed many times.
It forms copper 1 cyanide complex CuCn4 3-plus cyanogen which is a poisonous gas. These are the things. So black precipitate is A is CUS, B is if it is given as a gas B may be NO.
Okay. It is not in B is water. So B may be copper tubeless or B may be NO.
So sulfur already is given. So if it is given B as a gas then it is NO. Then C is a white precipitate which is cuprousite which can be written as CuI or Cu2I2. So all the best to you. I have concentrated on redox reaction because that is mostly asked in this particular chapter.
Plus I just gave a survey of other things involved. Okay. All the best to you.
So we will see the combined redox reaction questions in complexes later on when we discuss. advanced questions on coordination compound. There we can combine D-block and coordination and redox reaction. All the best to all of you.