INORGANIC CHEM TEST-02 VIDEO SOLUTION FOR RE NEET-2026

Added: 2026-05-18

[00:00:00]Goal Nations [Music] Leading Institute.

[00:00:09]Hello Students, We are going to discuss the Inorganic Question of Test Series for the REET 2026 which was held on 17th May 2026. So, let us discuss the inorganic question of this examination. Which is question number first and which is inorganic?

[00:00:24]Question number 46 Which of the following set has the same bond order? This is MOT topic of chemical bonding in which bond order is asking the same for everyone. Which has the same bond order? So there is nothing in it. There is a very short trick for this. The formula for finding such bond orders is number of bonding - central bonding by two. We do n't have enough time to do that much. So for this a short trick came up that you can find out the number of electrons of everyone. 14 If so, three.

[00:00:50]Go up or down from 14. Go from 14 to 15, 16, 17, 18, 19, 20. Keep removing.5.

[00:00:57]Come down from 14. 13 12 11 10 9 Let's go up to eight and keep subtracting 0.5 all the way up to here. Ok? This is the short trick we have, we will manage with it.

[00:01:07]Let's count the number of electrons.

[00:01:09]2 out of 14 16 is 14, this also gives 14 73, so the bond order of all will be three. It has beans.

[00:01:16]Absolutely the right option. N2 16 14 2 16 16 8 and 15 16 all of this is also the same. And its bond order will be two. So this will also be our option. 15 * 14 and one 15 16, one 15, all of this is the same. Tell me yes or no? When all of them are same then what will be its bond order?

[00:01:39]2.5 is ok? How much will it be? 2.5 is the same for everyone.

[00:01:46]This one is also the same as the first one.

[00:01:48]This also has the same meaning as everyone else. This also has the same meaning as everyone else.

[00:01:50]So what will be the answer to this? All are correct. All are correct. So the correct option for 46 will be option number four. Let's move on to the next question.

[00:02:06]Question No. 48 Which one of the following orders is not in accord with the property stated against it. Ok?

[00:02:15]We know that electronegativity goes from top to bottom in fluorine, chlorine, bromine, iodine.

[00:02:18]Electro negativity decreases as size increases.

[00:02:21]Fluorine has the best electronegativity.

[00:02:23]The worst is iodine. The order is written correctly. Bond dissociation F2Cl2, Br2 and I2, this is given wrong, right?

[00:02:30]Cl2 has the highest bond dissociation energy.

[00:02:33]Then it is of Br2. Then comes F2.

[00:02:35]Then I2 happens. The reason for the low bond dissociation energy of F2 is that due to the small size of fluorine, there is more lone pair repulsion in it.

[00:02:46]Due to higher lone pair repulsion, its bond dissociation energy decreases slightly.

[00:02:50]Remember one thing, many people say that Sir, loan payer child, there is some confusion that its bond length will not increase due to loan payer loan payer repulsion.

[00:02:55]If the bond length has increased then where is the loan payer repulsion? If the bond length increases then lone pair repulsion ends. Lone pair lone pair repulsion is experienced only because its bond length is small. Its bond length between ClCl and lone pair becomes larger so the lone pair repulsion decreases. So do n't think that if the loan payer ripples then the bond length will increase. If the bond length increases then where will the repulsion remain? The repulsion will end. So that's why its bond dissociation energy of F2 decreases a little. is greater than I2. A little bit is more. So that's why wrong option has been given.

[00:03:24]Oxidizing Power: In the last lecture also I had told you people about extracting oxidizing power.

[00:03:29]In F2 Cl2, Br2, II2 everything you make X- aqueous from X2. Make aqueous from this gas.

[00:03:37]The more negative its DeltaHR, the more spontaneous the reaction will be.

[00:03:42]The better the reduction will be. The better it will behave as an oxidizing agent.

[00:03:45]If we calculate its bond Haber cycle then bond dissociation energy by two x gas is formed. From here, make x minus the electron gain enthalpy in the gas form and from here, take the hydration energy. All this was explained in the last lecture.

[00:03:59]We told him this in the last test.

[00:04:02]This form comes by combining all these. If you combine all of them, then the order you get will be the same order. I had told you another trick for this, a short one, that for oxidizing power, you should generally take the help of electronegativity. The one which has higher electronegativity has higher oxidizing power. But before applying this concept, please check once whether there is any element in the given set of elements whose electron gain enthalpy is positive. If its electron gain enthalpy is positive for any element, then its oxidizing power should be placed last.

[00:04:28]Ok? You people have to remember this.

[00:04:30]Well then. Will have oxidizing power.

[00:04:32]We will again use that concept instead of this. The one with electronegativity. In this, the enthalpy of all electrons is negative. Then we will not be confused. The one with higher electronegativity has higher oxidizing power. If there was an electronegative electron with positive enthalpy, it would have been placed at the last.

[00:04:49]Like in FONCl. Ok?

[00:04:52]This is the correct order. HI HBr HCl HF has perfect acidic property. Go from top to bottom. HF HCl HBr and HI then what is its bond length? Will increase.

[00:05:06]What will happen to the bond energy when the bond length increases? Will decrease.

[00:05:08]So the bond energy of HI will be lower. So HI will break quickly. It will break quickly. Will give H+ quickly. If it gives H+ quickly then its acidic character will be more. First of all, this is also correct. The second one is another concept of it.

[00:05:19]After removing H+, the conjugate anion formed will be F- Cl- Br- and I-, so if we talk about the highest stability then I- will be the most stable.

[00:05:32]We look at the size in different periods. The anion with larger size will be more stable.

[00:05:36]If there is same period then we look at electronegativity. There is every element of different periods. It is the same group.

[00:05:42]Some things make the child confused because sometimes he learns that same period, same group. Same period is correct. It does n't matter whether it is the same group or not. If there is different group then size is different period then size is different. If it is same period then electronegativity. Ok?

[00:05:58]Same group in same different period. No problem. There is no problem. The period is different. If someone asks the child which is more stable between n- and sulphur minus? Neither is it from the same period nor from the same group. Then the child will get confused. So yes, the thing in this is that sir, this is from a different period. It is not from the same group but from a different period. It is not from the same period. It is from a different period. So we can apply this concept.

[00:06:18]We look at the size in different periods. The anion whose size increases has greater stability.

[00:06:22]The more stable the conjugate anion, the greater is the acidic character of the compound. Therefore, it will have more acidic character. This is also the concept. The second concept is that if we go from top to bottom, the bond length will increase. If the bond length increases, it will be easier for HI to break. When it is easy to break, it will give H+ quickly. If it gives H+ quickly then the acidic character will be more.

[00:06:40]This is applicable to all hydrides. This is n't just about HF. Be it carbon, boron, nitrogen, oxygen or halogen, this one is applicable in all. As we move from top to bottom, bond dissociation energy will decrease. It will give H+ quickly or the acidity of the conjugate anion will increase, increasing the acidic character. It is applicable to all. We have been asked, not in accord is not. So, the second option will be correct.

[00:07:01]Which is given by us wrong, there will be maximum Cl2. F2 will come here. Ok? So the correct option for 48 will be option number second. Let's move on to the next question.

[00:07:15]Question No. 50 Which one of the following has regular tetrahedral shape. Ok? I3- This is very runaround. You guys can also take it out. There are three loan pairs. There is a minus charge I and I.

[00:07:31]Ok? I one it has iodine so the side atom is our monovalent so the central atom will also bear a minus charge. All these anion shapes etc. Where will the minus charge go and where will the plus charge go? I have already told this.

[00:07:43]Plus charge on any species always goes to the central atom. And remove as many electrons as the plus charge.

[00:07:52]Minus that many electrons from the valence electrons. The minus charge will go to the central atom if the side atom is monovalent.

[00:07:57]Like fluorine, chlorine, bromine, iodine, hydrogen. These are all monovalent species. So in this case, if there is negative charge, it will go to the central atom.

[00:08:06]If our side atom is polyvalent, divalent, trivalent like oxygen, oxygen forms two bonds. If it is divalent, whether it is nitrogen, sulphur or phosphorus, then in that case the negatively charged side atom will bear it. Center Atom will not do beer.

[00:08:20]Here the side atom is monovalent, hence negative charge is given to minus. And the more negative charge you give to the centre, the more electrons you will increase. Iodine initially had seven valence electrons. Paid minus charges. The valence electrons became eight. Look, 2 2 4 and one bond has been formed. There are eight electrons. Its hybridisation will be sp3d. The geometry will be triangular by pyramidal and the shape will be linear. So it is not regular tetrahedral. So this is our mistake. SF4 S One fluorine, two fluorine, three fluorine, four fluorine lone pairs. Lone pair on equatorial position In trigonal by pyramidal, the lone pair is on equatorial position.

[00:08:54]Bends tell us what shape it will have? Which is called Seeso.

[00:09:02]This is called Seeso shape.

[00:09:05]It is also called folded square or pie shape.

[00:09:07]So this is also not regular tetrahedral. So this also went wrong.

[00:09:10]BF4- Look and there's fluorine again. There should never be any confusion. Fluorine is our side atom is monovalent. We will give minus charge on central atom. A minus charge is to add one electron. Boron will have three electrons, four electrons.

[00:09:22]Boron was given one electron, two electrons, three electrons, four electrons minus. Fluorine will form a single bond. No loan payer will be left.

[00:09:30]Its hybridisation is sp3 and there is no lone pair.

[00:09:32]Whatever the geometry is, that is the shape.

[00:09:34]Tetrahedral Geometry Tetrahedral Shape. So regular tetrahedral will be of BF4- only. XCF4 XC lone pair lone pair fluorine fluorine fluorine fluorine its shape will become square planar. The geometry will be octahedral.

[00:09:51]Ok? In octahedral, the two lone pairs are always on diagonal positions at 180° to reduce repulsion. There should be more stability.

[00:09:57]Ok? So regular tetrahedral BF4 would be of minus. The correct option for 50 will be option number three. Let's move on to the next question.

[00:10:11]Question number 52 Match the column one with column two In column one some properties are given by d block and some of its metals are given which would be the elements with highest second ionization enthalpy Listen all of you listen carefully what I want to tell you, yes, otherwise we will keep on memorizing the data as to whose second ionization energy is more and whose is less, it is not possible to remember so much data, we will do the work by correlating it a little, it will be done quickly.

[00:10:36]Just imagine moving from scandium to zinc. We know that as we move from left to right the Z effective increases and as the Z effective increases the ionization energy will increase.

[00:10:46]Sir, the ionization energy is always left to right, if we go from left to right then the one which is on the right extreme right will have more ionization energy, so second ionization is one more thing for us, whenever higher ionization energy is being asked, then Nick, Copper, Zinc, one of them will have more, let us compare these three, otherwise there is not so much difference that it can be more here also, we will not think that one of these three can be more, the adjacent electronic configuration can be stable or unstable, you must have read this in ionization energy that the effect of stable electronic configuration is only on its adjacent atom.

[00:11:18]In fluorine and nitrogen, nitrogen and oxygen will have more nitrogen and less oxygen. But we cannot compare nitrogen with fluorine.

[00:11:25]Fluorine will be more and nitrogen will be less.

[00:11:27]So the effect of electronic stable electronic configuration is applicable mostly on the adjacent atoms. It is applicable on the adjacent ones.

[00:11:35]So, let us take a look at this. If you have asked for a second, what will you do for a second? We no longer have to remove electrons from nickel.

[00:11:40]We have to remove the electron from nickel plus.

[00:11:41]We don't have to remove electrons from copper, we have to remove electrons from copper plus. Because he is asking us about second ionization energy. So nickel plus copper plus zinc plus sir Z effective, so we know that as we go from left to right the Z effective will increase. But now compare its electronic configuration.

[00:11:56]Whenever successful ionization energy is to be compared to higher ionization energy.

[00:12:00]You must take the help of electronic configuration. Only one topic has to be studied.

[00:12:03]Comparison of successive ionization energies. That topic only has to be studied.

[00:12:08]So nickel plus 4s1 to remove from. It involves rolling from 3d10. And in this we have to remove it from 4s1. There will be more of this. The Z effect is also high and it is also a stable electronic configuration. But copper zinc will have more copper. Because it has to be rolled from 3d10. It has to be removed from the four shells. There is a third shell. It is D10.

[00:12:25]is a stable electronic configuration. Therefore, there will be more copper plus. So the second ionization energy of copper will be the maximum in the entire 3D series.

[00:12:31]Tell me yes or no? Copper will become third.

[00:12:37]Element with highest third ionization. Ok? No problem. Third Ionization Again we will think of the extreme right one. We will not think about the left one because if we go from left to right the Z effect increases. There is a chance of increase in ionization energy. Come again nickel, copper zinc. But now the third one has been asked. Third means remove two electrons.

[00:12:52]You have to remove the electron from Nick + 2. Electron has to be removed not from Nick +1 but from Nick +2, Copper +2, Zinc +2.

[00:12:59]In this you will have to remove electron from 3d 8, in this you will have to remove electron from 3d 9 and in this you will have to remove electron from 3d10. Will move left to right. Z effective will increase. Ionization energy will increase. It does not have any stable electronic configuration. More of this, more of this. Among copper and zinc, zinc has higher Z effective and also has more stable electronic configuration. Therefore, the third ionization energy of zinc will be maximum among scandium and zinc. Zinc will be fourth.

[00:13:27]MNCO6 forms chromium. Cro6 Cro6 It'll be seconds.

[00:13:34]Vanadium can also be made like this. It's not that vanadium can't be made. I am saying chromium, please understand.

[00:13:39]CrO6 of chromium vanadium will be unstable.

[00:13:42]Chromium will be stable. Why? I am going to tell you a very good point, understand.

[00:13:47]When you study effective atomic number (EN), one thing that is taught in it is that whatever carbonyl complexes there are, only those carbonyl complexes are stable which are following EN. The carbonyl complex which does not follow the EN rule is unstable. He has to be made to follow n.

[00:14:04]Now if it is not following n then either it should form a dimer or gain or lose electrons, do whatever you want to do but it will have to be done. Like you guys make this mn co call five. Look, it will contain 35 manganese. 25 + 10 35 35 35 not equal to effective atomic number. So not following N. Now not following n so either gain an electron.

[00:14:26]Gain of electron reduction occurs. If reduction is done then it will be stable and otherwise make this dimer and after making this dimer make hole 10 of MN2CO, now this will become stable sir carbonyl complex, let us leave aside such carbonyl complex, for all others ene has no role in deciding the stability but for a carbonyl complex ene is a very major factor. Carbonyl complex is stable only when it follows the EN rule. So if we take vanadium here, then look at vanadium, the atomic number of vanadium will be 23 23 + 12 35. He is not following this. There is zero charge on the whole. But if we take vanadium, then if we take chromium of vanadium, then in the case of chromium the atomic number is 24 plus 6 * 2 = 12 36, this is following N, right? Its atomic number became equal to that of Zeno Crepan. It became equal to 36. It became equal to the atomic number of the noble gas of its period. So he is following N, he is following I N, so he will become stable. So who would prefer chromium? If not vanadium, the child was right, why wouldn't vanadium be formed?

[00:15:29]Vanadium is not being made because the carbonyl complex formed by vanadium will not follow N in coordination number six. If it does not follow N then it will be unstable. If it is unstable then why should we discuss it? If chromium is stable then we will talk about stability.

[00:15:40]Second would be the element with the highest heat of atomization.

[00:15:49]Vanadium has the highest enthalpy of atomization in the third 3D series.

[00:15:51]Vanadium, this has to be remembered. Vanadium will be first. Such enthalpy of atomization depends on the melting point. It remains similar to the melting point.

[00:16:00]But a little vanadium does.

[00:16:02]Chromium has a higher melting point than this. Okay, right? But the enthalpy of atomization of vanadium is higher.

[00:16:06]So it will be first. It should be mixed in it.

[00:16:09]Third of A is A's third, fourth of B is B's fourth is in both. The second of C is in the second of C is in the third. D's first. The second option will be correct. So the correct option for 52 will be option number second. Let's move on to the next question.

[00:16:29]Question number 53. Which of the following is amphoteric oxide? Look, I am telling you.

[00:16:35]Oxides are a very important topic. It is a very important topic. Let me tell you once. I will get it revised.

[00:16:40]Okay, right? What is the concept of oxide? Which oxide is acidic, which oxide is basic, which oxide is amphoteric? All the metals we have whose oxidation states are +1, +2, +3 are basic. Who should be left out? Amphoteric should be omitted.

[00:16:58]I will write amphoteric at last. Metal 's +4 generally want to be amphoteric.

[00:17:02]But they have a lot of basic character. It is very less acidic.

[00:17:07]Remember I am amphoteric but it contains a lot of basic. too much. Okay, right? It is very less acidic. +5 +6 +7 of metal is acidic. Except whom?

[00:17:19]Except amphoteric. Accept amphoteric.

[00:17:22]Secondly, all the oxides of all the metalloids are acidic.

[00:17:29]Except whom? Accept amphoteric.

[00:17:34]And all the non-metals, their oxides are acidic.

[00:17:40]Except whom? Except neutral.

[00:17:42]Accept Neutral.

[00:17:45]Accept Neutral. This thing has to be remembered.

[00:17:47]So which ones are amphoteric? It is amphoteric today, everyone sang the Punjabi song O Janabe Ali, even it went waste. These are all amphoteric. Arsenic +3 Antimony +3 Vanadium +5 Bismuth +3 Chromium +3 The oxides of those on which the charge is mentioned are amphoteric.

[00:18:14]All possible oxides of the element on which charge is not mentioned are amphoteric. You have to remember that they are amphoteric.

[00:18:18]Ok? Neutral oxides N2O, NO and CO are the neutral oxides. Let Mn2 be +7 acidic of manganese +6 acidic of chromium +3 amphoteric of bismuth.

[00:18:33]Bismuth's +5 Acidic.

[00:18:36]Bi2O5 is acidic. Bi2O5 is acidic. It is completely acidic. But Bi2O3 is amphoteric. Let me tell you one more thing. Remember one more thing.

[00:18:47]Listen. The +1 +2 +3 of metal is definitely basic but it is basic. If by the way there is a metal, its +1 +2 +3 amphoteric is formed, then in this amphoteric the basic character will be more and the acidic character will be less. If a metal with +5 +6 +7 forms amphoteric then it will have more acidic character. The basic character will be less. Why? Because the root quality of +1 +2 +3 of metal is of base. So that's why the basic character will dominate in this, sir. In amphoteric the basic character will dominate. But metal has +5 +6 +7. If something is amphoteric then acidic character will dominate in it because its root quality is acidic.

[00:19:23]Please understand what I want to say. Someone ask which has more acidic character Cr2O3 or V2O5? Which acid is better? Both of these are amphoteric, sir. This is also amphoteric. This is also amphoteric. But this is metal. It is +3. Metal is +5. It will have more base in amphoteric. The acid will be less.

[00:19:39]But it will have more acid in it. The base will be less. Because the +5 oxidation state of the metal has the root quality of being acidic. It has become amphoteric, so acidic will dominate in it. Ok? Remember this. Amphoteric: It is acidic. Chromium's +3 is amphoteric. Chromium +2 is a complete base. Vanadium +5 is amphoteric. So which ones are amphoteric? Cr2O3 and V2O5 V2O5 is the first option visible. Mn2O7 is acidic. Amphoteric has asked us. It is basic. This is amphoteric. It is amphoteric. It is acidic. This will also not happen. So the correct option for 53 will be option number first. Let's move on to the next question.

[00:20:22]Question number 55 is giving this assertion.

[00:20:25]COF63- is paramagnetic. Isn't it right?

[00:20:28]Cobalt +3 3d6 electronic configuration F- is a quick field legged.

[00:20:34]Cannot get pairing done. When pairing is not done, four unpaired electrons will remain 1 2 3 4 5 6. So, it will show paramagnetic behavior. This is absolutely the correct statement.

[00:20:44]CO+3 has 3d 3d6 outer electronic configuration. The unpaired electrons do not pair up because of the weak field provided by F- is absolutely correct. That's what I was telling you. f - Cannot get pairing done due to weak field.

[00:20:57]And the region is also correct. Both the assertion reason is correct and the explanation is correct. Therefore the third option will be correct. Both A and R are correct and R is the correct explanation of A. So the correct option for 55 will be option number three. Let's move on to the next question.

[00:21:17]Question No. 58 The ion which contains all equivalent MO bonds. Ok? CrO42- Cr double bond O double bond OO- O- all these are in resonance. You know, right? It's all in resonance. All bonds are equivalent. Partial double bond, partial double bond, partial double bond OO, OO, partial negative, partial negative, partial negative, partial negative.

[00:21:40]Absolutely correct. MNO4- MN double bond O double bond O double bond O- These are also in resonance.

[00:21:47]Everyone will go into resonance. Tell me yes or no? All are equivalent. Do you guys know this? Cr2O7 will not be present in it. Cr double bond O double bond OO- O CR double bond O double bond OO- In this, these three are equivalent, these three six are equivalent but these CRO CRO bonds are different and these six are different. So that is why not all the bonds are equivalent in this. This went wrong.

[00:22:12]Which ones will be correct? A & B. A & B. The third option is the correct one. So the correct option for 58 will be option number three. Let's move on to the next question.

[00:22:27]Question number 59 is a good salt analysis question, it is about cation analysis.

[00:22:34]Group analysis of cations. Now we have to find out which element of which group the cation belongs to.

[00:22:38]Aluminium +3 is a third group element. You will have to remember. Throw crispy potatoes. Throw crispy potatoes. All these are elements of the third group. Zinc +2 Do n't go on Monday. This is the fourth group. Do n't go on Monday. All of them belong to the fourth group. Barium is the Government of India Fifth Group.

[00:23:04]Government of India is the fifth group. Lead plus two also occurs in the first group. It also happens in the second group. I went crazy today. She also became a Punjabi girl. Have you gone mad today? Belongs to the first group. She also became a Punjabi girl. What? He belongs to the second group. The lead ppt formed in the first group PbCl2 and PBCL2 is hot water soluble. Hot water in the lab does not mean that the water is not so hot that it burns the head or body.

[00:23:28]In the lab, hot water means liquid water. Tap water sometimes gets hot. The temperature right now is 40-45°.

[00:23:37]In this, the tap water that will come from the tank will also be hot. In that too, PBCL2 has a chance of being soluble. The distilled water that we have kept in the bottle can also become hot. There may be something soluble in that too. Hot water means leaking warm water.

[00:23:49]Slightly warmer than cold water. Okay, right?

[00:23:51]That is called hot water. Leak is called warm water.

[00:23:53]So to confirm this, there may be a chance that PbCl2 becomes soluble in hot water. So that's why lead plus two to confirm in the first group as well as the second group because PBS is formed in the second group.

[00:24:04]PBS is not hot water soluble. So it is analyzed in the second group also.

[00:24:08]Analysis is done. It happens in both groups. If there is a lead plus two first group, then we will talk in the first group as well as in the second group. A's third A's third fourth option will be correct.

[00:24:17]Fourth group third option of B will be correct.

[00:24:20]Fifth group second option of C will be correct and first option of D. See which one? Fourth of A.

[00:24:25]Fourth of A. Ok? It is in two. B's third, C's second, D's first. The correct option for 59 will be option number four. Let's move on to the next question.

[00:24:40]Question No. 61 Which of the following compounds shows optical isomerism. Ok? Look, this is the best topic of coordination compounds. The best topic.

[00:24:50]If someone wants to ask the best question then he will ask you about optical i.e. stereo isomerism of octahedrol complex. It is very important for you to learn this.

[00:24:57]Learn and practice, only then you will be able to succeed in this. Ok? Let's see the first one.

[00:25:02]Make the cis isomer of it. What we do first is select a pair of legends.

[00:25:06]Select chlorine in this. Two is our bean. We'll put chlorine on lead once because the lead isomer is asking. NH3 NH3 NH3 NH3 This will be it.

[00:25:20]Tell me yes or no? It is optically inactive because it has the same ligand at the trans position.

[00:25:25]It will have POS.

[00:25:29]It will have POS. And when P is the plane of symmetry and POS which will be it? Look, if you want to cut it like this, I will tell you here.

[00:25:36]Understand this. Otherwise the child will not look like a POA. NH3 NH3Cl NNH3 NH3 Look, look once and understand a little.

[00:25:54]Look, this is your octahedral complex. Can you see this one? This is like this and it became octahedral. Ok? This is your ammonia on this one. There is ammonia. There is chlorine here. There is chlorine here. There is ammonia here. There is ammonia here. Ok? Now you make cuts in it like this. Does it look like this? Look at this hand of mine. Now you are making cuts like this. I can see this one. Like this, like this, even if these two groups were different, if there was ammonia here and Cl here, still it would have become PO like this. Okay, right? If these two were different because when we make cuts like this, then there is ammonia here and there is another group here. It is the same group here. Here we have ammonia, here we have another group. So we will cut, this is a plane, these two groups would have cut in the plane.

[00:26:31]His work is finished. And here's Cl, here's Cl. There's ammonia here. There's ammonia here. Here's Cl, here's Cl, here's ammonia, ammonia. These two will superpose each other. The child cannot see this POA.

[00:26:41]Look at this POS, this one looks like a plane. This one is a plane.

[00:26:44]Look, this plane is coming like this with my mobile. See, this one is the plane and the child can see the other plane and not this one. The child cannot see this kind of plane. It is very important to see this. Even if by the way both of these are same, still PO will be deducted.

[00:26:57]If both of these become the same. Look, from here, these two ClCl ammonia ammonia here, still it will have POS. Look, this is the POS when both of them are same and this is the POS when both of them are same. These two are beans. Okay, right? So this is how it is taught to such children.

[00:27:10]What is taught?

[00:27:12]Understand that if two trans pair same NaCl3 NaCl3 two trans pair same or trans pair same legged previous PO in this there are many PO one PO one PO so you saw it right? Not a P.O. It has a lot of POS. So learn it.

[00:27:30]NH3NH3 This is also a POS in this. There will be a cut in this also.

[00:27:34]Make one such cut here, Clcl NH3NH3, it will give POH. There's another POH in this, look, cut from here to this entire plane.

[00:27:42]Cut it like this and it will become its mirror image.

[00:27:44]No, look at this one, cut it like this, ammonia ammonia like this. The plane is visible, this plane.

[00:27:52]This will be completely cut. Ammonia Ammonia will be superimposed. It has POS.

[00:27:55]So there's a lot of POS in this. So we cannot see it like this. So how will you make it in the exam? So what should I tell you for making it in the exam that if there is same ligand or two trans pairs are same in transposition then it will be POS and if POS is there then it will become optically inactive. So this is optically inactive sir, look at its transform, is it taught like this that two bi dentate and cis orientations are optically active, two bi dentate trans trans orientations are optically inactive, as if it is told, it should be made as transposition of clcl, transposition n, this is what will happen, this is ethyne diamine, trans, what was told, transposition of pe, same ligand, PO and POS, so how is it optically inactive, cut it like this, both of them will become mirror images of each other, it will become optically inactive, this will also not happen. It is wrong. It is also not optically active.

[00:28:44]Look at its cis isomer, I'm erasing it a little bit here.

[00:28:48]I am deleting this. Look at its cis isomer.

[00:28:51]Place the clcl on the mirror. n and n listen to one thing. Do n't even try to see POS etc. here. There is a different concept here. What do you think, two bidentates are on cis orientation. Both of these are more dentate. There is bulkiness.

[00:29:08]Bydentate is not just a small group. But it is a group. Due to high bulkiness, there is steric repulsion between these two. And due to steric repulsion, both of them go out of plane. One two these two go into different planes from each other.

[00:29:22]When you go to a different plane, you will be dead. There will be no chance of getting a PO in this.

[00:29:26]And when there is no chance of getting POA, the compound will always become optically active.

[00:29:30]That's why three bidentate so three bidentate when three in a compound three bidentate. All three are ethyne diamines, which is your second option.

[00:29:38]Look at the next one, it is always said optically that three by dentate always optically active. Why? Because you will always get this orientation in three by dentate.

[00:29:46]When this orientation is always found, it will always be optically active. And when it is optically active, it is said that two bi-dentate cis orientation is always optically active. Three by dentate always optically active. So this is also optically active and this is also optically active. One key point is optical isomerism. Please remember that. Correct? The next question is question number 65.

[00:30:06]The incorrect statement regarding 15th group hydride. Ok? See, whenever you have to talk about hydrides like NH3 PH3 SH3 SBH3 and BH3, whenever you have to talk about hydrides, start with the bond length of any hydride, be it boron family, carbon family, nitrogen or any other family, the bond length will increase, bond energy will decrease, thermal stability will also decrease, reducing power and acidic character will increase. The reducing character will increase from top to bottom.

[00:30:47]Because the bond length is increased the bond length of BH is increased. When the bond length increases, the bond length of BH will be easier to break.

[00:30:55]When the bond is easy to break, hydrogen will release quickly from here. End loss of hydrogen oxidation occurs. Which goes into the oxidation process acts as reducing agent. So the best reducing agent would be bismuth with the best tendency to oxidize. It will be of BH3.

[00:31:10]Therefore the best reducing agent would be BH3. Top to bottom reducing character will increase. Now let's see what property has been asked for.

[00:31:16]Thermal stability will decrease from top to bottom. The correct order has been placed.

[00:31:20]given in the decreasing order. The correct order has been placed. The reducing character will increase from top to bottom.

[00:31:24]It has been given in descending order, right?

[00:31:26]given incorrectly. This will be the most. then this then this will be the least least most... than most most most most most than most. given incorrectly. The bond dissociation energy of NH will be higher.

[00:31:33]Then PA then SA then SBA then BIA. The order is correct. You have given it absolutely correctly.

[00:31:38]Basic top to bottom basic character decays all hydrides.

[00:31:42]Acidic character will increase from top to bottom. Basic character decreases from top to bottom for all hydrides.

[00:31:47]Acidic character increases from top to bottom. You have asked me about an incorrect statement, so which one is incorrect? Second one. So the correct option for 65 will be option number second. Let's move on to the next question.

[00:32:00]Question No. 68 Which of the following species can act as a reducing agent?

[00:32:05]Ok? A carbonyl complex, we just talked about carbonyl complex is stable only when it follows the EN rule.

[00:32:13]If it does not follow the EN then it will either dimerize or undergo oxidation or reduction. Depending on the conditions, it will do whatever it deems appropriate to do to obtain the atomic number of its nearest noble gas configuration.

[00:32:30]Whoever followed EN became stable.

[00:32:32]Neither oxidizing nor reducing and the one which does not follow N will either become oxidizing or reducing or will demagnetize, let us check who is following N and who is not following coco4 - Cobalt has charge -1, atomic number is 27 so 27 + 1 + 2 * 4 28 + 8 36 = crepen ok it is following N 25 25 + 12 6 isn't it? 6 * 12 37 Not Equal to Krypton is not following N.

[00:33:09]This is unstable. To achieve stability, it will lose electrons to get 36 electrons.

[00:33:13]When there is loss of electrons, the oxidation process will take place.

[00:33:18]Acts as reducing agent. Loss of electrons is oxidation. Acts as reducing agent. So who will behave like a reducing agent? Hole six of MnCO.

[00:33:27]But let's see the next one also. The charge over mn is again zero. 25 In this the coordination number is five so it will become 10.

[00:33:33]35 This is also not following. There is no equivalent to Krypton. But this will gain off electrons. Gain of electrons to get 36. Gain of electron reduction occurs.

[00:33:43]Act [sound of clearing throat] as an oxidizing agent.

[00:33:49]Chromium 0 24 plus 24 + 12 36 is following this. If you follow it, it will be stable. We just read that chromium was the same complex in the previous question, why were you making only CrCO6? Because Chromium is following EN in that case.

[00:34:12]This will be neither oxidizing nor reducing. So that's why A will be our reducing agent and A will be our oxidizing agent. Both of them are following EN.

[00:34:20]Hence neither oxidizing nor reducing. So our correct option will be 68 seconds. Let's move on to the next question.

[00:34:30]Question No. 69 The correct set of species with zero dipole moment. Ok? CO2 C double bond O double bond O is a linear species. Linear species are always non-polar. μ is its zero. Look, it is better to call a linear species or a symmetrical molecule. Ok? Ok? Symmetrical molecules are always non-polar.

[00:34:51]Speaking linearly creates a bit of a problem.

[00:34:53]Like By The Way is C2. Look, linear C double bottom C double bottom O, this will also be linear. But this will be its μ net not equal to 0. Because the dipole moment of CS and the dipole moment of CO cannot be the same.

[00:35:03]Cannot cancel. So that's why the linear one can sometimes be wrong.

[00:35:07]But this is a symmetrical molecule.

[00:35:10]All surrounding atoms are the same. It is always zero.

[00:35:13]Its dipole moment μ is always zero.

[00:35:15]But it will have zero. Ok?

[00:35:18]COCL2 C BO CLCL This will pull here and this will pull here will pull here. Its net will come down.

[00:35:24]Look, the angle between both its dipoles and dipole moments is 180°.

[00:35:30]Ok? Both are in opposite directions.

[00:35:32]So it should have been cancelled and made zero.

[00:35:34]But since A belongs to CO. So its magnitude μ1 is different, μ2 is different. Hey, what do you write for μ equal to? Root under μ1² + μ2² + 2 mμ1 mμ2 cos theta this does not happen and cos will become 180° so the value of cos 180 is -1 then both the -1's get cancelled and become zero. When will it become zero?

[00:35:54]When μ1 and mμ2 are equal, then μ1 and mμ2 are not equal, so how will it become zero? So this is CO's and CL's. Both μ1 and μ2 cannot be equal. Therefore its dipole moment cannot be zero. CCL2 CHCl C H Cl ClH This will pull this way and this will pull that way. Its dipole moment will also not be zero.

[00:36:19]BCL3 BCL Clcl is a symmetrical molecule. The dipole moment of a symmetric molecule is always zero. This will pull upwards. Together these two will pull downwards. But both BCLs, these are also BCLs.

[00:36:31]This is also BCL. Its μ will become 0. He will cancel both. μ1 μ2 will be equal. And the dipole moment of a symmetric molecule is always zero. So its dipole moment will also be zero. This will not happen. A and D, its correct option will appear. The correct option for 69 will be option number first. Let's move on to the next question.

[00:36:50]Next question.

[00:36:54]Question No. 72 Which of the following order regarding ionization energy is correct? Ok? Ionization energy is asked. No one gives orders in this. So we will talk only about first ionization energy. There is nitrogen, there is oxygen, there is fluorine. Z effect increases as we move from left to right.

[00:37:08]Increasing the Z effective will increase the ionization energy. But before applying Z effect in it, one more thing has to be checked whether anyone has stable electronic configuration or not.

[00:37:18]It has a stable electronic configuration so the ionization energy will increase.

[00:37:22]Nitrogen has more nitrogen in oxygen than oxygen has more fluorine in fluorine. If we want to write all three by combining them, then fluorine will be the most abundant.

[00:37:29]Nitrogen will have more fluorine than oxygen. There will be no more than fluorine. It is applicable only on adjacent atoms.

[00:37:34]Fluorine is the most abundant, then nitrogen and then oxygen. See which order we have placed. It contains more nitrogen.

[00:37:40]given incorrectly. Fluorine will not cause excess nitrogen. It contains more fluorine.

[00:37:44]Ok? But nitrogen has been given wrongly in oxygen. There will be more nitrogen.

[00:37:48]Oxygen will decrease. In this, there is more nitrogen and less oxygen in nitrogen oxygen. The more correct order of fluorine in oxygen fluorine is given. I have ordered this one. We were reading this order. Nitrogen oxygen fluorine has given more wrong of oxygen in nitrogen oxygen. This is also given wrong. Both have been given wrong in this.

[00:38:04]So the correct option will be option number three. The correct option for 72 is option number three. Let's move on to the next question.

[00:38:16]Question No. 81 Match the column one with column two. In column one, right? This is a question from the application of complex compounds.

[00:38:26]Children sometimes take it lightly. Does not read properly. It also gets missed. Therefore, it has to be read carefully. It's a good thing. Photography.

[00:38:33]We will use Ag2O3 in photography.

[00:38:36]Please understand the third option. Let me tell you once.

[00:38:38]The earlier black and white photography was based on the photosensitivity of silver halide. And the most photosensitive among the silver halides was AgBr.

[00:38:50]What about AgBR was that AgBR was so sensitive to light that it would dissociate as soon as light fell on it. So that's why we used to get the reel made like this.

[00:38:58]We got the reel made like this. It had such a coating of AgBr. As soon as the flash light of the camera fell on the AGBR, the AGBR broke as soon as the light fell on it and after breaking the AGBR gave you such a photo of itself. But it was not possible for the flash to hit your face and then go back and strike it on the reel. It was not possible to break the entire A to Z exactly as it was. So what we used to do was that we used to take out the reel and after taking it out, we used to give it back to the studio people. The studio owner used to do the developing of the film.

[00:39:26]So what happened in that was that this reel was put in a solution. That solution was called hyposolution.

[00:39:33]Hyposolution. What was hyposolution? Na2H2O3 was called hypo solution. What did he do? AgBr was used for this reaction and a complex was formed. H2O3 Whole Twice Charge -3 Its +1 Its -4 Whole Pay Charge -3 will be left. If you want to remove -3 then write Na2 here and Na3 and along with it NaBr was formed so to use undissociated AgBr, hypo solution was used. He used to form the complex with Ag and then after cleaning our entire face, he used to take out the printout and give us Isdola. So that was its job. The third option is chelating ligand, chelating EDTA and penicillin, listen here EDTA can also be a chelating ligand but deep penicilla amine, we are seeing another use of EDTA in this option, in hardness of water estimation, when we wanted to remove the hardness of any hard water, then we used to add EDTA solution, Na2 EDTA solution in it, that EDTA was a complex compound of sodium or magnesium or calcium due to which it had hard water. You will read hard water. Hardness of water is due to the presence of chloride and sulphate salts of calcium or magnesium or bicarbonate salts of calcium or magnesium.

[00:40:56]Temporary hardness is caused by bicarbonate. If you boil it, it will go away.

[00:40:58]But permanent hardness is chloride or sulphate which is a little tough to remove. So what did we do for this? Add EDTA. What EDTA used to do was that it used to eliminate the sulphate or chloride of calcium or magnesium and form its own EDTA complex with calcium or magnesium, hence it used to remove the hardness slowly.

[00:41:16]To find out how much hardness is present, we used to add EDTA, so for chelating ligand, P D Anisyl Penicillamine, or more Disferon B and Dipenicillamine, which is a chelating ligand, is used to remove excess of iron, if this one is correct then the fourth option will be correct. Chlorophyll is a major pigment of photosynthesis which is a complex compound of magnesium. And hardness of water will be EDT.

[00:41:43]First will be its second. Look, A's third, B's fourth, A's third, B's fourth, C's second, D's first. So the correct option for 81 will be option number three. Let's move on to the next question.

[00:42:00]Question number 85. Among Group 14 Elements The Least Value of Melting Point Group 14 Carbon, Silicon, Germanium, Tin, Lead. The order of its melting point is top to bottom, melting point decreases. But lead and tin have more lead. Due to lengthwise contraction the force of attraction is slightly increased and hence the melting point is slightly higher than tin. So the tin [throat clearing sound] would have the lowest.

[00:42:22]If we write the same order of boiling point then top to bottom is regular decrease. There is no problem. We have been asked the least value of melting point that will be tin. So the correct option for 85 will be option number second. This is all about your inorganic question. Thank you everyone.

[00:42:37]Thank you so much.