INORGANIC CHEM TEST-12 VIDEO SOLUTION FOR RE NEET-2026

Hinzugefügt: 2026-06-06

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

[00:00:09]Hello Students, We are going to discuss the An Organic Question of Test No. 12 of Test Series for Renee 2026 which was held on 5th June 2026. So, let us discuss the organic questions of this examination.

[00:00:21]Which is the first question of inorganic that is question number 92 which of the following molecule has the minimum dipole moment? Ok? HF hf will have the highest dipole moment because it has dipole moment = μ μ = q.d. q will be higher yet the dipole moment will be higher. d will be greater yet the dipole moment will be greater. And if we elaborate further, then μ μ = qd cos theta. Let's take cos theta in the dot product.

[00:00:51]Now the value of theta will increase. The value of cos theta will increase the value of theta. The value of cos theta will decrease. μ net will also decrease.

[00:00:58]When does this happen? When two or three types of dipole moments are occurring in the molecule.

[00:01:02]Then the bond angle is seen in it.

[00:01:04]Ok? In HF, the difference in charge between these delta negative and delta positive is very high. Therefore its mu is higher. μ is very high. There is a problem with NF3. What's the problem with NF3 is that A this is called the dipole moment by lone pair is called as orbital dipole moment. and the bond dipole moment is A. The net Dpol moment will come down through this three. Will come up through this. Now these two are in between the two. Should be cancelled. The angle between these two is 180 and cos 180 = -1. Will be in the mood to cancel. But its magnitude is μ1. Its magnitude is μ2. The magnitude of both is not equal. This is through orbitals and this is through bonds. The magnitude of both will be different. Therefore both will not cancel the net. But the dipole moment will definitely be reduced. Therefore, the dipole moment of NF3 becomes very small. The dipole moment of CHCl3 is not H but H and Cl will pull it towards this side. Carbon chlorine bonds are almost non-polar.

[00:02:05]Ok? Almost non polar.

[00:02:08]Ok? C So you guys must be reading CH4 molecule.

[00:02:10]Methane molecule used to give the free radical reaction. He wants to carry more.

[00:02:14]Free radical reaction is what one wants to carry out. But the bond between carbon and chlorine is polar. And there will be delta negative delta positive charge on it. So there is a good dipole moment in this also.

[00:02:24]Comes around 1.04.

[00:02:27]In H2S the S H lone pair lone pair will pull this and this will also pull that. Net dipole moment above by this and also net dipole moment above by this both of these will support each other. Therefore the dipole moment of H2S will be greater than that of NF3.

[00:02:41]So the order of highest dipole moment will be of HF in this. Then CHCl3 will have. The electronegative difference between carbon and chlorine is greater than the difference between hydrogen and sulfur.

[00:02:51]So therefore q will be greater in this. In this q will be less.

[00:02:53]Hence its dipole moment is more.

[00:02:55]After that it will be H2S and the least amount will be NF3.

[00:02:58]This is the correct order of the dipole moment. And I tell the children again and again to look at the chart given in NCERT once or twice.

[00:03:05]If some data comes to mind then the question will be formed quickly. Ok? And looking at NH3, the dipole moment of N3 is slightly less due to the cancellation of orbital dipole moment and bond dipole moment. Therefore the correct option will be minimum asked. The minimum will be NF3. So the correct option for 92 will be option number second. Let's move on to the next question.

[00:03:28]Question No. 93 Which of the following reactions are carried out by acidified potassium dichromate? This is very simple.

[00:03:38]When potassium dichromate K2Cr2O7 is reacted in an acidic medium, it behaves as an oxidizing agent. It is a strong oxidizing agent. If Sn+4 is not converted to Sn+2 then it will not behave as a reducing agent. It will behave as an oxidizing agent.

[00:03:49]So this will not happen. Will convert Fe+2 to Fe+3.

[00:03:52]Absolutely correct.

[00:03:54]Chromium itself will go to +3. This is a correct statement. will form sulfide ion in ppt of sulfur.

[00:03:59]This is also correct.

[00:04:02]I- will convert to I2. This is also correct.

[00:04:05]It will not convert H2O3 into SO4. This is not the case with its acidified dichromate. The reaction takes place in a neutral medium.

[00:04:11]In alkaline medium and KMnO4 generally he wants to do it. So who will do it? Fe+2 to Fe+3, b, c & d. B, C & D. So the correct option for 93 will be option number three. Let's move on to the next question.

[00:04:29]Question No. 95 The preparation of potassium permanganate from MNO2 involves two step process. Absolutely correct. MNO2 which is called paralusite ore. It is called paralysis. It is black in colour.

[00:04:44]This makes it react in basic medium. KNO3 is heated.

[00:04:50]Remember to listen. KNO3 here is basically a source of oxygen. The source of oxygen is oxygen and this oxygen behaves as an oxidizing agent which oxidizes Mn+4 to Mn+6.

[00:05:05]MNO42- forms +6 and the compound formed is K2MnO4 and the colour of K2MnO4 becomes green. This is what creates the green colored compound. Let's read further. Two step process. Two step process. So this is the first step.

[00:05:20]In the second step, either acidify it or neutralize it, then its color changes from green to purple and it also gives MNO2. This reaction is called deprotonation. The deprotonation either oxidizes the acid or neutralizes it and gives the preparation of KMnO4.

[00:05:40]This is a two step process in which the first step is a reaction with KOH and KNO3 to produce.

[00:05:44]Well, what is produced in the first step? So K2MO4 is produced. K MNO2 reacts with KOH + KNO3 and listen again. MNO2 reacts with it.

[00:05:54]KNO3 A little medium reaction needs to be heated. The only purpose of heating is that if potassium nitrate is heated, it gives potassium nitrite plus oxygen. And this oxygen behaves like an oxidizing agent.

[00:06:08]Ok? So it behaves as an oxidizing agent and converts MNO2 into MNO42-.

[00:06:13]Manganese goes from +4 to +6.

[00:06:15]And the medium is kept basic because the green colour of K2MnO4 persists as long as its medium is kept basic. As soon as you acidify or neutralize the medium, its green color immediately converts into what?

[00:06:29]In purple due to the disproportionation reaction.

[00:06:31]Okay, right? This thing has to be remembered. So the question is what will be produced in the first step? So K2MnO4 will give green colour.

[00:06:37]So the correct option for 95 will be option number second. Let's move on to the next question.

[00:06:46]Question No. 96 Consider the following statements. The NF3 molecule has a trigonal planar structure. It is wrong. N F lone pair of electrons. Look, the hybridisation of NF3 is sp3.

[00:07:02]Its shape is pyramidal. It has tetrahedral geometry. It has a pyramidal shape. There is no concept of back bonding in this, sir. Back bonding is not possible in this. Because fluorine does not have a vacant orbital. It also does not have a vacant orbital. So nitrogen cannot give its lone pair of electrons to it.

[00:07:16]So there is no back bonding and so on. Okay, right? If there was back bonding then there could have been sp sp2 hybridisation. If the lone pair of nitrogen gets involved.

[00:07:23]But fluorine has no vacant orbital.

[00:07:25]There for nitrogen's lone pair is perfectly localized. No back bonding concept is applicable here. Therefor hybridization is sp3. Otherwise, sometimes this concept comes to the child's mind, the child who reads a little thinks about it, it seems that the lone pair of nitrogen will not go into back bonding, fluorine does not have any vacant orbital, hence it will not go into back bonding, this is correct, therefore sp3 of n3 is not pyramidal trigonal planar, this is wrong, this is correct, bond order of CO and NO+ is same, say CO has number of electrons 8 6 14, NO+ has 14, bond order of both is three, bond order of both is three. Absolutely correct. Sam has said it. It is written correctly. Dipole moment of CHCl3 is lower than HF. We just talked.

[00:08:05]In HF the difference in Q between the two is so high. The difference in charge is so high that its dipole moment is high. Its dipole moment is low.

[00:08:13]Its writes around 1.78. Its writes around 1.04. The dipole moment of CHCl3 is lower than HF. It is written correctly now. The CN bond length is smaller than the CO bond.

[00:08:27]Just listen and understand this also. C double bond N and C double bond O For bond length comparison, we first see whether the comparing atoms are from the same period or different periods. Is it the same period or a different period? If there is a different period then the shell will grow and its size will increase. There is no problem in this. I tell the kids over and over again.

[00:08:48]Like C triple bond N and CH CH, whose bond length will be bigger? So the child thinks it has a triple bond.

[00:08:57]It has a single bond. Triple bond large, triple bond small, single bond large.

[00:09:00]But it is wrong. In this CN will be bigger and CH will be smaller. Why? Because in this both the comparing atoms are from the second period.

[00:09:05]This is the first period. Its bond length will be smaller. If it is of different period and the shell is small then its bond length will be small. Whatever the bond order is, it does n't matter. Ok? This thing has to be remembered. If it is the same period then first of all we look at the bond order. If the bond order is also the same then we look at the percentage character. Did you understand?

[00:09:22]Both of them are of the same period. The bond order is the same.

[00:09:24]You must have read one thing that while finding the radius, the show maker station formula, if the difference in electronegativity becomes high. If the difference in electronegativity is high then the bond length starts getting shorter. If the difference in electronegativity is higher then the bond length will be shorter.

[00:09:39]- 0.09 * Remember the difference of electronegativity? If its difference is more then the bond length will be smaller. If its difference is less then the bond length will be larger.

[00:09:46]Hence the bond length of CN will be larger. The bond length of CO will be smaller. The CN bond length is a smaller than CO is written incorrectly. The bond length of CO will be smaller. The bond length of CN will be larger.

[00:09:56]Ok? It is written wrong. So I asked what is it? Select the correct answer from the options given below. Which will be the correct answer? B &C. B&C. Third option. A and D are wrong. So the correct option for 96 will be option number three. Let's move on to the next question.

[00:10:16]Question No. 98 The reagents NH4Cl aqueous ammonia will precipitate. Look, a little calcium chloride is soluble, sir. There is no problem. How did you know? Get it reacted. Calcium chloride will be formed which is soluble. And when you add ammonia to calcium chloride, it will also react solublely, meaning it will not.

[00:10:35]This will also remain soluble. So the work is finished.

[00:10:42]Add aluminium +3 NH4Cl. Aluminum chloride will be formed.

[00:10:45]Add aqueous ammonia to it. This is a PPT. No change. There will be no change. There is no change in this also.

[00:10:53]Meaning if you get a reaction then there will be no change. There will be no change. Please understand what I am saying.

[00:10:56]Which salts are soluble in aqueous ammonia? Which are from D block? All those in D Block will become soluble.

[00:11:04]Except whom? Except iron and manganese. Except manganese and iron, all others become soluble.

[00:11:12]Aluminium will not be soluble in p block.

[00:11:14]In which will aluminium be soluble? In NaOH.

[00:11:17]Because the compounds of amphoteric metals which are ppt become soluble in excess of NaOH. It will not be soluble in it.

[00:11:23]PPT Will Remain As It Is. So ppt AlCl3 will not have any change in it.

[00:11:28]This will remain. MgCl2 is. This is soluble. Adding aqueous ammonia will make no difference. ZnCl2 ppt. But you will add aqueous ammonia as in the second step. If you add aqueous ammonia, what will be the metal of D block? The soluble forms a soluble complex.

[00:11:42]So this will become Znh3 times 4 + 2 This will be the soluble complex. So the question is who will be left with the PPT? So aluminum will remain +3 ppt. All the rest will become soluble.

[00:11:54]Ok? So the correct option for 98 will be option number second. Let's move on to the next question.

[00:12:04]Question No. 100 The bond dissociation enthalpy is maximum four.

[00:12:10]You all know that there is Cl2 Br2 F2 and I2. F2 is slightly lower due to lone pair repulsion. If maximum is asked then maximum will be of Cl2.

[00:12:21]Therefore the correct option for 100 will be option number second. Let's move on to the next question.

[00:12:30]Question No. 107 This is an easy question in The Correct Geometry and Hybridisation of Brf5 Brf5. There is a loan pair.

[00:12:39]Its steric number will become six. The hybridisation will be sp3d2 and the geometry will be square pyramidal. The geometry will be a square pyramidal.

[00:12:52]Square sp3d2 and square pyramidal. So the correct option for 107 will be option number three.

[00:13:00][nasal sound] Look, let me tell you one thing, one thing, and a little bit here. will rise again.

[00:13:03]Like in this it is written octahedral geometry, sometimes geometry is octahedral but here the hybridization is d2sp3. d2sp3 does not form these orbital complexes. The outer orbital remains. Therefore, we will not consider it. We will consider this only. So the correct option for 107 will be option number three. Because of this. Otherwise, if this was ours, octahedral at both the places, octahedral here also, sp3d2 here also, sp3d2 here also, this octahedral square pyramidal, then we could have gone to this option also, I could not have crossed it out wrong, yes, it is true that NCERT writes geometry and shape in the same way, but now look at the JEE Mains question once, in JEE Mains at least we can touch the NEET question, in that geometry is our geometry, electronic arrangement is the geometry, the answer has been given on this only, but anyway, we do not have any such dispute in this, d2sp3, we have ended any dispute.

[00:13:51]So we will move on to sp3d2 and square pyramid. Ok? So the correct option for 107 will be option number three. Let's move on to the next question.

[00:14:03]Question No. 108 Which of the following is not correctly matched? Look at acidic oxide.

[00:14:08]Cl2O7 and PBO, you have given it wrongly.

[00:14:12]PbO is amphoteric. Today everyone sang the Punjabi song Janabe Ali, even that went in vain. I have said it many times. If it is amphoteric then it is wrong.

[00:14:19]Neutral oxide NON2O is correct. Amphoteric aluminum is +3 amphoteric of arsenic arsenic antimonic.

[00:14:25]Correctly given. Basic Oxide N2 NMJ This is also given correctly. If you have asked incorrectly then it will be first. The first will be incorrect. So the correct option for 108 will be option number first. Let's move on to the next question.

[00:14:41]Question No. 109 Which of the following represents the correct order of electron affinity? Listen please. I will tell you in this.

[00:14:50]Carbon Silicon Not Nitrogen Nitrogen Phosphorus Carbon Silicon Oxygen Sulphur Fluorine Chlorine If the child has problem with this then understand that Phosphorus is more than Nitrogen Phosphorus is more than Carbon Silicon is more than Carbon Oxygen is more than Oxygen Oxygen is more than Sulphur Sulphur is more than Fluorine Fluorine is more than Chlorine, see this is what I have been given, which one will have the most Chlorine, then which one will have Fluorine, then which one will have Sulphur, then which one will have it? Oxygen.

[00:15:19]Remember this chart. There is a basic and very important chart. Otherwise the child always gets confused as to whom will be taken in silicon oxygen? Sir more electron affinity. Silicon oxygen contains more oxygen. So chlorine, chlorine, fluorine, sulfur, oxygen. The first option seems correct. Therefore the correct option for 119 will be option number first. Let's move on to the next question.

[00:15:39]Question number 120.

[00:15:42]Which of the following lengthwise ions is diamagnetic? Ok? Europium +2 writes direct. You all must be aware that the electronic configuration is 4s7 and it is paramagnetic.

[00:15:53]Ibium +2 becomes 414. It is diamagnetic. Cerium +2 Look at Cerium +2 becomes Cerium +3d0. Cerium is +1 d1. So it will remain paramagnetic.

[00:16:05]Cerium +4 becomes cerium +4.

[00:16:09]Look, they are writing D in this instead of D [sound of clearing throat].

[00:16:11]What the f would be right? Cerium has 58.

[00:16:13]Xenon is 6s2 5d 4f0 4f 56 2 58 but in exception one electron comes here. One electron resides in it. Two electrons have to be removed in this. So it contains both 4f1 and 5d1.

[00:16:33]This happens because Serium is in exception. So this will also show parametric. Ceramium +2 is 62 of ceramium. So xenon 6s2 5d 0 4f so 54 2 56 56 62 it has two zeros, it will have 4f6 so it will also become paramagnetic, it is also paramagnetic, what have you asked us in the question, diamagnetic, diamagnetic, it is also para, it is diamagnetic, the second option will be correct, our cerium +2 has 4f1 5d1 so it will also be paramagnetic, cerium +2 has six, hence it is also para, it also has unpaired electrons, hence it will also be paramagnetic, okay, so the correct option for 120 will be option number second. Let's move on to the next question.

[00:17:20]Question number 123 The basicity order is this is the nitrogen family. You guys know.

[00:17:25]Go from top to bottom. Basic character decreases. So nitrogen, phosphorus, arsenic, antimony, bismuth is written correctly. The first option is correct. Phosphorus is the more common form of phosphine. Phosphine is more wrong.

[00:17:36]NH This also went wrong. So which would be the right option? Option number first. So the correct option for 123 will be option number Fux.

[00:17:42]In the last class it was told that the basic character of hydride increases from top to bottom.

[00:17:46]Acidic character increases top to bottom Basic character decreases top to bottom. The acidic character of all hydrides increases from top to bottom. There is no exception to all hydrides. Be it boron, carbon, nitrogen, oxygen, fluorine, the acidic character of all hydrides will increase from top to bottom. The base basic character will decrease. Ok? This thing has to be remembered. Let's move on to the next question.

[00:18:09]Question No. 129 The complex compound having paramagnetic nature is Fe+2 4s0 3d6 strong field legund. Will get the pairing done. Number of unpaired electrons zero. It will become diamagnetic.

[00:18:24]This went wrong. Titanium +3 4s0 3d1 This is Strong Field Leagued. It doesn't matter. Whether it is parametric, strong field or weak field, it doesn't matter. At least one unpaired electron will remain. There will be no pairing etc. in this. So that's why it will always be paramagnetic. Whatever it may be.

[00:18:41]Cobalt would be +3 d6.

[00:18:45]Again Strong Field is leagued. This will also get the pairing done. This too will become dynamic.

[00:18:48]Nick + 2 is d8. Coordination number is four. If you get the pairing done then this will also become zero.

[00:18:55]n = 0 is also diamagnetic. Asked what is it? Paramagnetic. So titanium complexes will be paramagnetic. Therefore the correct option for 129 will be option number second. Let's move on to the next question.

[00:19:10]Question number 131.

[00:19:12]Question. Mohr's salt gives NH4+ Fe+2 and SO42- in the aqueous solution. Absolutely correct. Mohr salt is the whole twice SO4 and 6H2O double salt of FeSO4 NH4. If you put it in water, you will get Fe+2 + SO42- and it will break down in water to give 2NH4+ + SO42-. Fe+2 test will be available.

[00:19:36]Sulphate will be available and ammonium ion will be available. Absolutely correct. Mohr salt is a coordination compound Na. If it was a coordination compound, we would not have been able to test it. There is a double salt. Complex is not a compound. So the assertion is true. Region is Falls. The correct option for 131 will be option number three.

[00:19:53]This was our last question. 131 This is all about your inorganic question. Thank you everyone. Thank you so much.