INORGANIC CHEM TEST-17 VIDEO SOLUTION FOR RE NEET-2026

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

[00:00:10]Hello Students, We are going to discuss the Inorganic Question of Test No. 17 of Test Series for RENET 2026 which was held on 12th June 2026. So, let us into the organic question of this examination. The question number first of inorganic, that is question number 136 the first ionization enthalpy of molecular oxygen was almost identical with, okay see, there is one in it, what is molecular oxygen, there was a scientist, you have to study in noble gas, in group 18 in p block, there was a scientist named Neil Bartlett, he did an experiment, he reacted O2 with PTF6, so what he saw was that O2+ became PTF6 minus.

[00:00:53]This means that O2 must have come here from the gaseous state.

[00:00:57]Give it ionization energy. And what they observed is that when O2 is reacting to form O2 plus and in terms of ionization energy, O2 is higher and xenon is lower. There was not much difference between the two. There was a difference of 5 kilojoules. Okay, right? Then he observed that xenon can also react with O2 to form O2+. So xenon can also react. So he also reacted xenon with PTF6. So xenon is also made up of P xc+ and Ptf6 minus and from here he gave a conclusion that xenon can also react with oxygen.

[00:01:27]That is why the ionization energy of molecular oxygen and xenon is almost comparable. Therefore the correct option for 136 will be option number four. Let's move on to the next question. Moving to the next question.

[00:01:42]Question number 141.

[00:01:44]The pair of metal ions that can have three unpaired electrons in the complex MH2O 6Cl2 means what will be the charge on the metal? +2 will come. Ok? The number of unpaired electrons required is three. So Cobalt +2 is in Cobalt +2 4s0 3d7 and the Weak Field Legend for Cobalt +2 will be the same.

[00:02:04]We will not talk about any pairing etc. So the number of unpaired electrons in this will be three. There is no need to make this. Time is passing. The number of unpaired electrons from the number D will be three. So the value of n will be three here. So this work is finished. Yes. So the value of n is three. Correct. Iron +2 Iron +2 has 4s0 3d6. Now there will be a weak field league again. There will be no re-pairing in the Weak Field for Iron +2 either. And the number of unpaired electrons in d6 is four. So the number of unpaired electrons in this will be four. There are no three in both. Therefore, the option we have will be wrong. Chromium is +2. In this it will be 4s0 3d3. Whether it is a weak field or a strong field, it does not matter. The number of unpaired electrons will always be three.

[00:02:46]4s0, 3d5 and 3d5 in mn +2 is a week field league for you mn +2.

[00:02:52]Pairing cannot happen.

[00:02:53]When there is no pairing, the number of unpaired electrons will become five. There are three in one. There are five in one. So this will also not be an option. Vanadium +2 Scandium, Titanium, Vanadium. Vanadium +2 has 3d3. Be it a weak field or a strong field, it may be any.

[00:03:07]In octahedral complexes, the value of n will always be three. As it is, it will remain. There is no possibility of any pairing etc. in this. We just talked about Cobalt +2, in Cobalt +2 n = 3 will be there because it will act like a weak field.

[00:03:19]Pairing etc. is not possible. So here also the value of n will be three. So both of these, vanadium +2 and cobalt +2, with water ligated, both will have the number of unpaired electrons three. The value of magnetic moment will be 15 Bohr magnetons.

[00:03:32]So that's why the third option seems right to us. Let us also see this. In Iron Panadium +2, which was just discovered, n = 3 will come in it and Iron +2 was also discovered in which n = 4 was found. So that's why there wo n't be an option. So he is asking that I want the number of unpaired electrons to be three in both. So vanadium will be +2 and cobalt will be +2. Therefore the correct option for 141 will be option number three. Let us move on to the next question.

[00:03:59]Question no. 142 for bx3 boron halide which is incorrect? Pi Pie Pi Pie Back Donation is possible. Absolutely correct. It is BF3. BF F In this, lone pair of electron is in P.

[00:04:13]Its vacancy P is vacant. This will give 2p pi 2pπ back bonding. So is chlorine. It has electron in 3p so 3p pi 2p pi will form back bonding in it.

[00:04:25]Everyone else has back bonding. But back bonding is not as effective.

[00:04:28]But the bank will donate. There may be resonance in it. But poor very poor back donations happen. So P Papa P Papa back donation is possible. Correct. This back donation is a kind of resonance. Delocalization will sometimes give this, sometimes this, sometimes that. It cannot give all three together because boron does not have that many vacant orbitals. So, the back donation part, the back bonding part is a part of resonance. Therefore there will be resonance in it also. Acidic character of BF3, BCl3, BBr3, BI3 BF3 has given a very good thing. Understand if our central atom is second period back bonding is a concept to decide the acidic and basic nature of a compound. Very good, this one is a very good example. Ok? Like CH, now you must have seen its part containing CH F3 and CHCl3.

[00:05:17]In this, the acidic basic character seems like fluorine will strongly pull the electron and will become more acidic or less acidic, people think so, but in this, the back donation acidic strength is opposite to what you think from electronegativity, the situation is similar here also, understand if a child thinks like this in BCl3, then it will seem that electronegativity will pull the electron, delta plus will come in it, plus will come more and its acidic character will increase, it will become less acidic character in it, it will come less in bromine.

[00:05:49]but it's not like that. In this, back donation is the dominating factor.

[00:05:53]What actually happens is that boron has a vacant orbital. Vacant P is. is an incomplete octet. It will behave like acidic. He would like to take an electron pair.

[00:06:00]But by donating fluorine to it, its electron hunger was satisfied. Reduced it. Back donation is good, its back donation is very effective. 2p pi is the back donation of 2pπ. So chlorine will not be able to do as good a back donation as fluorine because this back donation of 3p 2pπ has come. Bromine and 4pi 2pπ back donation came and Poor will come equal to no. There is a further Poor Back donation of 5p pie and 2p pie.

[00:06:28]So it will reduce further. So, the best back donation in this will be in BF. After that in BCL, then in BBR and the worst will be in BI. This means that the electron hunger will be satisfied best by fluorine and worst by iodine. Therefore, the acidic character of Bi3 will be more. BF3 will be the lowest.

[00:06:48]So the order of acidic characters is given in reverse here. If this happens then it is our wrong.

[00:06:52]What is asked in the question? You have asked incorrectly.

[00:06:54]So this is our incorrect one.

[00:06:57]Back donation is best in BF3, BCl3, BBr3, Bi3.

[00:07:00]In BF3, then in BCL3, then we just talked about it, it will be in BI3. So this is our right. So if you have asked wrong then our third question is wrong. The order of acidic character of boron halides is wrong. It has been given in reverse. Therefore the correct option for 142 will be option number three. Let's move on to the next question.

[00:07:18]Question number 152.

[00:07:22]The Structure of XCF2 and XCO2F2 Respectively R. This is a simple question.

[00:07:26]XCF2 one fluorine, two fluorines, one lone pair, two lone pairs, three lone pairs. It has eight electrons. Hybridization will be sp3d² trigonal by pyramidal geometry and the shape will be linear. So it will be linear, its linear. This is also linear. Ok? This is over. It's over. xco2f2 xc double bud o double bud o one fluorine two fluorine one lone pair.

[00:07:51]Remove hybridization. 1 2 3 4 5 Hybridization sp3d z² again trigonal by pyramidal geometry there is a lone pair oxygen will remain here double bond will go again after the lone pair multiple bond will occupy the equatorial position remember this thing then do not place the multiple bond wherever you want it will be wrong your lone pair will first occupy the equatorial position first preference will be given to the lone pair after that give preference to the multiple bond so both the multiple bonds will come here one fluorine here one fluorine here what will be the shape cover it ciso ciso linear and ciso of first one linear ciso of second one so the correct option of 152 will be option number third let's move on to the next question.

[00:08:33]Question No. 161 In dichromate ion dichromate Cr2O72- The number of identical CRO bonds ok make it. So Cr double bd O double bd O- O CR double bd O double bd OO- see this will resonate. This will also resonate. So these three bonds are identical, these three bonds are different, its bond length is bigger.

[00:08:55]All three have short bond lengths.

[00:08:57]So these two are different and these six are similar. So how many identical bonds are there? Six. So that is why the correct option for 161 will be option number two. These three, these three are same, these two are different. Its bond angle also varies. The case of OCL2 occurs due to steric repulsion.

[00:09:14]Remember, there is a loan payer in this, there is a loan payer. It seems to be sp3 hybridisation.

[00:09:18]It should have a low bond angle. But since both of these are bulky groups, its bond angle increases a lot. It becomes much more than 109. Steric repulsion is very high. What will be the number of identical bonds in Cr2O72- CrO bond identity number? Six. Three near this one and three near that one. Correct option of 161 is option number 2. Let's move on to the next question.

[00:09:43]Question No. 167 The Bond Angle of NH3 NH4 + NH2- Yesterday also it was told that for the bond angle first find out the hybridisation.

[00:09:51]Then after that check the steric repulsion. There will be no steric repulsion in any of this. Then come to Wes Par. NH3 NH4+ NH2- Hybridisation will be written as sp3 for all.

[00:10:13]This one is also sp3, this one is also sp3, this one is also sp3, there is no steric repulsion in any of them. Be sure to check steric repulsion in sp3.

[00:10:20]Steric repulsion is not being felt in any of them.

[00:10:22]So we will come in Vesar. In Vaser first of all the number of lone pairs on the central atom will increase. The bond angle will decrease. So the number of loan pairs in this is one. It has zero. There are two in this. The one which has more lone pairs has less bond angle. So the highest bond angle will be of NH4+ having zero lone pair, then one lone pair of NH3, then two lone pairs of NH2- This is the correct bond angle NH4+ Ammonia ion NH2- Ammonium ion NH3 and NH2- You can see the second option, right? So the correct option for 167 will be option number second. Let's move on to the next question.

[00:10:57]Question number 169 Assan is speaking.

[00:11:01]Equated copper one cation undergoes desorption as copper plus. Ok? Copper +1 disproportionates in aqueous medium.

[00:11:12]Copper gives +2 + copper. Aqueous Aqueous goes into solid state. That's perfect. Disruption occurs. Hydration energy of copper +2 is higher than that of copper plus which compensates second ionization energy of copper. This is also correct.

[00:11:30]Which has higher hydration energy, copper +1 or copper +2? The hydration plus charge of copper +2 is higher, right? Therefore, hydration energy will be higher. And compensates for its second ionization energy.

[00:11:38]What is that? Understand the meaning. Electrons will be removed from copper to copper +1, from copper +1 to copper +2, in this the electron has to be removed from 4s1.

[00:11:45]In this, electron has to be removed from 3d10. This is a stable electronic configuration. The value of second ionization energy becomes very high. But what it says is that its hydration energy and its hydration energy will be more. This will be less. So it is said that the second ionization which is the removal of 3d 10 electrons which is very high, compensates the hydration energy of copper +2 which is the released energy and it is said that copper +2 is more stable in aqueous solution.

[00:12:06]Copper +1 becomes less stable. So this is correct. Both of these are correct and also have correct explanations. So the correct option for 169 will be option number three. Let me tell you one thing in this question. Listen to that.

[00:12:24]To explain this, copper dissociates +1 not due to the hydration energy.

[00:12:34]Ok? If it is explained in the true sense, then if we read it in a little detail, then the reason for copper +1 disproportionation, for example, if we talk about Fe+2 going into Fe+3 and we say that Fe is aqueous. It is aqueous, it should become solid. This dissociation happened. He said no, no, it will desorb. And in this, if we talk about hydration energy, then is Fe+2 more or Fe+3 more, still Fe+3 is more. So this does not disproportionate. This would happen if hydration energy was our governing factor.

[00:13:05]But in our routine we say that hydration energy is okay and correct. The explanation is correct. Ours is given in NCERT. Let's accept it.

[00:13:13]But the biggest reason for disproportionation is its note value.

[00:13:17]Just understand what needs to be said. Like copper +1 went into copper +2 and lost one electron.

[00:13:23]Now you must have read about this electron in electrochemistry that if any of its aqueous is in aqueous state.

[00:13:30]In aqueous, there are two types of cations. There is also an H Plus and a Copper Plus. And there are two types of cations and they have to be reduced, then in electrochemistry, the principle of electrolysis says that if there are two types of cations in a solution, then the reduction of that cation will happen first which has higher SRP value. So there is competition between Copper Plus and H Plus, then which one has higher SRP value? You can check the copper plus electrochemically.

[00:13:50]Copper Plus has a higher SRP value.

[00:13:52]H+ has a lower SRP value. So that's why it doesn't get reduced.

[00:13:55]It gets reduced. And it becomes copper +1, it gains this electron and becomes copper. So disparate it does. Because there is copper +1 disproportionate in it, the reduction potential of copper plus is less than hydrogen plus. If it contains H+ and Fe+2. So H+ has higher SRP value.

[00:14:12]Fe+2 has a low SRP value. Therefore H+ will disproportionate there.

[00:14:16]Meaning it will be reduced. It will give H2 gas.

[00:14:18]And here Iron Plus will not allow dissorption to happen. Meaning it will not allow Iron Plus to convert Iron +2 into Iron. So its main reason is anode value. But this becomes a bit of a vast concept. That is why we manage our work with hydration energy because the hydration energy is more and hence it is disproportionating itself. Ok? So now let's get this going. Both assertion reason are true and reason is the correct explanation of assertion. The correct option for 169 will be option number three. Let's move on to the next question.

[00:14:47]Question number 174 is statement one.

[00:14:50]Formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given one.

[00:14:56]Absolutely correct. You guys have read Resonance. The one with higher stability will have lower energy. Due to which his energy will be less. There will be more stability. A lot. Now tell its stability for RS. The one which is neutral will be more stable.

[00:15:09]In charge, then octet rule, number of pa, number of bonds, number of bonds, after electronegativity rule, which one has more charge, for the more electronegative, we will decide from this. So the one which has many RS in many structures and has less energy will have more stability. We decide this.

[00:15:28]Absolutely correct statement.

[00:15:30]The extent of overlap does not decide the strength of the covalent bond. This is written wrong, isn't it? The more overlapping the better. The greater the extent of overlapping, the better the bond will be. The extent of overlapping is greater in sigma bonds. The extent of overlapping of pi bonds decreases. That's why Bond is weak.

[00:15:44]This bond is strong. This is written wrong. Extent of overlapping surely decides the strength of Extent of overlapping surely decides the strength of covalent bond. This is written wrong. Does not have been written. This is wrong. Statement one true. Statement one true. Statement to False. Statement one is correct. Statement to is incorrect. So the correct option for 174 will be option number three. Let us move on to the next question.

[00:16:11]Question No. 175 Which of the following statements are correct? Both LiCl and MgCl2 are soluble in ethanol? Absolutely correct statement. This is the concept of this block.

[00:16:21]You have to study ionic compounds.

[00:16:22]Just understand. LiCl and MgCl2, both are most polar covalent bonds.

[00:16:38]Okay, right? These people are covalent bonds.

[00:16:42]is a polar covalent bond. You have to remember this, sir. It contains covalent. The molecule is not completely ionic. Not 100% Ionic.

[00:16:48]But the ionic character remains in it. The covalent bond is polar. And since there is a polar covalent bond, ethanol is also a polar solvent. It is a polar solvent. And being a polar solvent, it will dissolve the polar solvent in itself. Therefore its solubility is good in ethanol.

[00:17:07]Look, people think that LiCl is completely ionic. Not ionic. Similarly, the compounds of lithium and beryllium are mostly ionic. Mostly covalent. Because their polarizing power is higher due to their smaller size.

[00:17:18]But this is a polar covalent bond. The polarity in this is quite good.

[00:17:22]Therefore it is soluble because it is also a polar solvent. This is also a polar solvent.

[00:17:25]Solubility will be very good. It is written correctly.

[00:17:27]The oxides Li2O and MgO combine with access of oxygen to give super. No, this has been written wrong. Listen. Superoxide Generally, among the alkali metals, lithium, sodium, potassium, rubidium and cesium form superoxide well.

[00:17:45]This forms sodium peroxide and lithium oxide. Therefore their superoxide is not possible. Peroxide of sodium is possible and superoxide of lithium, rubidium, cesium is possible. Lithium will form peroxide. Peroxide forms in the major case and a little bit of superoxide or oxide will form in the minor case.

[00:18:02]But potassium, rubidium, cesium give superoxide as their major product when react with oxygen. So, lithium and magnesium are in a diagonal relationship. The work that our lithium does is the same property magnesium wants to show. You have to read this. Similarities between lithium and magnesium come in similarities. What shows do you do? Both will not form superoxide. It is written wrong. LIF is more soluble in water than others. This is also written wrong. It is also wrongly written about oxides.

[00:18:29]Just listen. Generally speaking, in the first and second groups, solubility decreases from top to bottom. Ok? Lithium will be more soluble than lithium and rubidium will be less soluble than acidium. But which ones come in the first group where solubility increases down the line?

[00:18:51]Fluoride, oxide, hydroxide, carbonate, bicarbonate and sulphate. There are so many anions with which the solubility increases from top to bottom.

[00:19:03]And in which of the secondary groups will it increase? Fluoride, oxide, hydroxide and oxalate. And one more thing is mentioned in this that is the solubility of beryllium salt.

[00:19:15]Beryllium Salts Are Most Soluble in Its Group. Except oxide and hydroxide. Except for the oxide and hydroxide, the solubility of all other beryllium salts is maximum in their group. One thing should be known. This is meant to be made quickly.

[00:19:35]So if we talk about fluoride in the first M, what will be the top to bottom solubility? Will increase. Meaning lithium fluoride sodium fluoride, potassium fluoride, rubidium fluoride, cesium fluoride, what will be the top to bottom solubility? Will increase. So the solubility of lithium fluoride will be low.

[00:19:51]What will be the top to bottom solubility of the oxide in the oxide as well? Will increase.

[00:19:53]What will be the solubility of lithium oxide? Will reduce.

[00:19:54]In this, reason can be said that polarizing power dominates here.

[00:19:58]Ok? Or another one is that fluoride oxide is an even better thing for making a baby early, that fluoride oxide, these are small anions.

[00:20:07]Want to be happier with little Katyan.

[00:20:09]When happy with smaller anions means more stable. If there are more stables, it will be more difficult to break them. When it is difficult to break, its lattice energy increases. If there is more lattice then its stability is more. Dissose solubility lattice more solubility less.

[00:20:22]Higher the hydration, higher is the solubility. So that's why smaller they are the smaller anion and they are more stable in with the smaller cation.

[00:20:30]This is it. But if we talk about it concept wise, it has polarizing power. And even better than that, what needs to be remembered generally is that in the first A and second A groups, the solubility generally decreases. In what does he increase? In the first A, fluoride oxide, hydroxide, carbonate, bicarbonate sulphate and in the second A, fluoride oxide, hydroxide and oxylato. But one more thing to remember in the second A group is that the solubility of beryllium salts is maximum in its group except for oxide and hydroxide. Ok? So lithium fluoride is more soluble in water than the other. This is written wrong. And lithium oxide is more soluble in water than this, this is also written wrong. So we have been asked which one is correct? So correct is our first.

[00:21:07]Lithium chloride and magnesium chloride are soluble in ethanol because it is both polar and non-polar. Hence solubility occurs. Everything else is also written wrong.

[00:21:15]So the correct option for 175 will be option number first. Let's move on to the next question.

[00:21:23]Question No. 176 Consider the following statements. Lanthanum hydroxide of LaOH is the least basic among hydroxides of lanth. It is written wrong. We go left to right no matter what.

[00:21:36]If we go from left to right during periods. If you move from left to right in a period, the acidic character increases due to increase in electronegativity. Whose basic character decreases? Be it oxide or hydroxide, the acidic character of both increases and the basicity will decrease, hence lanthanum hydroxide will be the most basic and latium hydroxide will be the least basic, so it is written wrong, lanthanum hydroxide is the least basic, it is not the most basic, latium hydroxide is the least basic, it is written wrong, zirconium +4 and hafnium +4 have almost the same ionic, it is absolutely correct, due to the size in 4D and 5D what happens. It becomes almost the same. The charges are same on both. Therefore, due to lanthanoid contraction, the ionic radii of both become the same.

[00:22:22]Cerium +4 can act as an oxy, absolutely correct. Cerium +4 wants to convert immediately to Cerium +3. Cerium exhibits +4 because it has a noble gas configuration.

[00:22:34]Wants to go to Serialium +3 immediately. Some people by the way think that cerium has +4 noble gas configuration. Why do you still want to go? In cerium +3. The Nobel gas configuration is a secondary factor.

[00:22:43]is the primary factor. The note value of this reaction can be seen in NCERT, it is written there. Its E note value was found positive. Found around 1.74 volts. When E note value is positive then delta G note is negative.

[00:22:55]How? Because we have to study an equation in electrochemistry. DeltaG Note = - NF * E Note Cell. n is our number of electrons involved. f is our Faraday constant. Note e is our cell constant.

[00:23:06]is the cell potential. If f is positive then the value of delta G note will be negative and negative delta G note value means reaction is spontaneous and reaction spontaneous means that the product is moving towards stable. Therefore celium +3 is more stable. So this will do the reduction. Calcium +4 acts as an oxidizing agent? It is written correctly.

[00:23:22]Which of the following statements is or are correct? Which one is correct? The first one was wrong, right? Second and third are correct. So the correct option for 176 will be option number second. Let's move on to the next question.

[00:23:36]Question No. 179 The incorrect matching pair is incorrect.

[00:23:43]Gave that molecule its shape. XCF4 Square Pair. Is it written correctly? Make it and show it. XC F Two lone pairs total number of electrons is eight.

[00:23:53]sp3 due D2 hybridisation.

[00:23:55]Cover the loan pair. You will see a square planer.

[00:23:57]BRF5 Square This is also written correctly.

[00:23:59]Brf also has octahedral geometry.

[00:24:02]A loan payer will arrive. The remaining five are fluorine. The shape will be square pyramidal. Ok? This is also written correctly. This is also correct. Clf3 T shape this is also correct. Cl one lone pair second lone pair one fluorine two fluorine three fluorine. SP3D hybridization T shape is bent T. T shape will also work.

[00:24:24]No problem. The geometry of NH3 is tetrahedral. The shape will be triangular pyramidal. It has been written wrong.

[00:24:30]NH3 is also a lone pair, right? So this will become triangular pyramidal. So this is written wrong. So you have asked incorrectly.

[00:24:37]Incorrect matching pair will become four. So the correct option for 179 will be option number four. Let us move on to the next question.

[00:24:47]Question number 180. The magnetic properties of B2 and C2 will be magnetic properties. B2 has number of electrons 12. No, it's 10, right? Boron has 10 and C2 has number of electrons 12. This one has 10, this one has 12. Have you already told me how all even numbers are accepted except 10 and 16?

[00:25:15]Diamagnetic.

[00:25:16]And all odd numbers plus 10 and 16 electrons. How do these happen? Paramagnetic. Let's make it 10. The atomic number of boron in B2 is five. 5 became 10.

[00:25:29]How are the 10 ones? This will be paramagnetic.

[00:25:32]And what will this be like? Die is 12, right?

[00:25:35]See if it is the one with 12. The one with 12 will become diamagnetic. So B2 is paramagnetic and C2 is diamagnetic. The first option is correct. So the correct option for 180 will be option number first. This is all about your inorganic question. Thank you everyone. Thank you so much.