INORGANIC CHEM TEST-03 VIDEO SOLUTION FOR RE NEET -2026

Added: 2026-05-21

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

[00:00:10]Hello Students, We are going to discuss the inorganic question of test number third of test series for the REET 2026 which was held on 19th May 2026. So, let us discuss the organic questions of this examination. The question number one is of these organics, that is question number three is the question of assertion region. He is saying that an element is on the extreme left. Extreme left here means it is not an alkali metal, it is an alkali metal like start with sodium, sodium, magnesium, aluminium, silicon, phosphorus, chlorine, if you go further then it will become argon, so extreme left means it is an alkali metal, extreme left is an alkali metal of the periodic table form acidic oxide, it is not wrong, if you want to make an oxide of sodium, if you put sodium in water then it will form sodium hydroxide and sodium hydroxide is a base. There is no acid. It is of basic nature. So it is wrong.

[00:01:07]Acid is formed during the reaction between water and oxide of a reactive element present in the extreme right of the reactive element.

[00:01:17]Reactive elements i.e. except noble gases, noble gases are called inert.

[00:01:21]So which one would be on the extreme right, except for the noble gas? Halogen. Like there will be chlorine, bromine, iodine. So if you put these in water, they will form the acid. If you put it in water, it will form acid. It may form HCL, it may form HOCL.

[00:01:40]Ok? Acid will form HCL, will form HOCl.

[00:01:42]If you heat it a little, there will be more changes.

[00:01:44]This is part of P block. But one thing is known that its extreme right ones are acidic and the extreme left ones are basic.

[00:01:51]So this is correct. Acid is formed during the reaction between water and the oxide of a reactive element present at the extreme right of the periodic table.

[00:01:59]Perfect acid form. So assertion is our false and reason is true. Assertion Falls. The region is true. So the correct option for the third will be option number. Option number four. Let's move on to the next question.

[00:02:16]Question number five, match the following, match the column one with column two, okay, in column one some of our elements are given, some of its properties are given, which property is this, aluminium +3 magnesium +2 fluoride and so ionization will definitely happen, otherwise this sequence is wrong, it does not even have metallic character, aluminium will have the highest metallicity, fluorine also has the highest electronegativity, but among magnesium and sodium, magnesium has the highest, this is also not correct, ionic radii are the highest, count all the isoelectronic species, all of them are isoelectronic and in isoelectronic species, if you increase the atomic number or the number of protons atomic number, then its Z effective will increase and if Z effective increases, what happens to the size? It keeps getting smaller.

[00:03:06]Ok? So that's why in this the top two will move from aluminum +3 to fluoride.

[00:03:13]So all of them are isoelectronic. The number of electrons in all is 10. The number of protons in this is 13, in this is 12, in this is 11, in this is nine. So the proton which has more z will have more effective. Because z effective is equal to what? The Z effect is directly proportional to the number of protons. Inversely proportional to the number of electrons. The electron has become a bean. The more protons there are, the more Z effective they will be.

[00:03:34]What will be the size if Z effective increases? It will keep getting smaller. This is the concept.

[00:03:37]Well, then this will become the fourth option.

[00:03:40]This order that we have given is the order of ionic radii. Boron, carbon, nitrogen, oxygen. This seems to me to be the order of ionization energy. Look, it will have the highest amount of nitrogen. Boron, carbon, nitrogen, oxygen. So, nitrogen oxygen will have more nitrogen.

[00:03:55]Because it has a 2p3 half field. It has 2p4 incompletely filled. So, nitrogen will be the most. Then there will be oxygen.

[00:04:01]Boron is Z effective at carbon.

[00:04:03]Excess of carbon will result in less boron. So after that will be carbon and the lowest will be boron.

[00:04:07]So this order of ionization energy seems correct. Its first option will be correct.

[00:04:10]Boron, aluminum, magnesium, potassium. This seems to me to be an order of metallic character.

[00:04:17]Potassium is the most abundant and among potassium and magnesium, potassium has more metallic character.

[00:04:21]Aluminium, magnesium in excess. They go here and there. The metallic character will decrease as you move from left to right. The metallic character will increase as you go from top to bottom.

[00:04:28]So among magnesium and aluminium, magnesium will have more metallic character.

[00:04:32]Aluminum will be less. So that's right. The order of metallic characters seems to be correct.

[00:04:36]Second and A will be the electronegativity. Look, chlorine is more electronegative than sulfur than phosphorus than silicon. So the order of electronegativity is correct.

[00:04:46]Third one, this order of electronegativity seems correct to me. So this will be the third one.

[00:04:52]So see. A's fourth, A's fourth, B's first, C's second, D's third. The first option seems correct. So the correct option for fifth will be option number first. Let's move on to the next question.

[00:05:10]Question number seven in which of the following complexes the CFSC will be equal to zero. See when do CFSCs take out zero?

[00:05:17]Octahedral complex is a good complex so we will look into it.

[00:05:20]When will CFSC be zero? This is an octahedral complex. So if or tetrahedral complex will be. So, in the case of d5, this CFSC value becomes zero.

[00:05:30]Or in the case of D10, the value of CFS comes out to be zero. There is no chance of anyone else remaining zero. So, D5 will have zero. But, remember if it is octahedral complex and D5's in weak field not in strong field. And whether it is a weak field or a strong field of D10, the nature of all will be the same. It's not going to make any difference.

[00:05:48]It has zeros in all of the D10.

[00:05:51]But the week field in D5 contains zero. The D5 strong field will not contain zero.

[00:05:54]Then he will get the pairing done. You must have read this, if splitting of octahedral is done and if there is a strong field of D5 then it will get the pairing done. So its -2.0 will become delta0. But if we talk about weak field, then splitting will happen in weak field 1 2 3 4 5 when we calculate delta0 of both of these -0.4 delta0 and + 0.6 delta0 then if we add this into 2 * 3 this will also be -1.2 and this will also be -1.2 + 1.2 delta0 this -1.2 delta0 + 1.2 delta will not become zero.

[00:06:28]So its conclusion means that if we start extracting it from all then it will take time, if we extract it by splitting it in all then it will take a lot of time. So that's why we have to use some concept in this. What to use is that the week field of d5 will always have a value of CFSC of zero and that of d10 will always have a value of zero. In d10, the concept of weak field strong field does not even apply. All the complexes become the same. Okay, fine. Let's see.

[00:06:50]So Fe+2 is in it. Fe+2 cannot have D6. Whether D6 is a strong field or a weak field, it doesn't matter. We will not even think about it. Its Delta Not Zero CF value cannot be zero. Such not equal to 0 is Fe+3. Fe+3 has D5.

[00:07:07]This is true. But in Fe+3, nitrogen is the donor. Nitrogen donors for Fe+3 become strong field. When we have strong fields, we will get the pairing done. If pairing is done then its CFSC value will not be equal to zero. If there is pairing then this one will happen, right? How can there be zero in this? It will not be possible. This will have a delta0 of -2. So it won't happen. Remember.

[00:07:26]For Fe+3, nitrogen is both string field. But for Fe+2 both nitrogen become weak field.

[00:07:32]Ok? Due to the Z effective lower Z effective of Iron +2 this won't happen either.

[00:07:37]CF0 Look at this, charge +2 on Fe [sound of clearing throat] D6 This will not happen. In D6, whether it is a strong field or a weak field does not matter. There is no need to think about this. Let me tell you two conditions.

[00:07:49]We just have to think about this. Look into Fe+3 and sulfur donors for Fe+3. For Fe+3, the oxygen donor is weak field. The sulfur donor will also be a weak field. So in this HCN will work as a weak field. WFL and has a D5 electronic configuration. Tell me, it will become zero, right? There will be D5 splitting in this. 1 2 3 4 5 Its delta not zero. So the correct option for seven would be fourth.

[00:08:16]Hole6 of K3FCN is a complex whose CFS crystal field stabilization energy value will turn out to be zero. So the correct option for seventh will be option number four. Let's move on to the next question.

[00:08:33]10th question. Question number 10th. Which of the following linear combinations of atomic orbitals will lead to the formation of molecular orbitals in homonuclear diatomic molecules? Internal nuclear axis, we have to take the Z axis. Ok?

[00:08:47]This is a good thing. This is a question of the concept of overlapping. Of chemical bonding. This is a great concept. There is something good in it. Who will make bonding, anti-bonding, zero overlapping? This has to be seen. I need to understand this a little. I will discuss all of them one by one. Ok? Listen a little carefully. 2pz, 2px this is 2pz. This is the Z axis. So our X-axis will be perpendicular to this. This one. The internuclear axis is our Z axis. The inner axis is our Z axis. The Z axis has to be considered. So this was our Z axis.

[00:09:13]So this is the internuclear axis. Now bring the molecules along this axis and combine both of them. Combine both atoms and form a molecule.

[00:09:20]How will it be? Zero overlapping. This will not give molecular orbital.

[00:09:24]will give zero overlapping. So this work is finished.

[00:09:25]2s and 2ps look 2s this one.

[00:09:28]is the s orbital. And 2px is this one. Consider this as the x axis. This is px. This x axis is this one. So z will be perpendicular to the x axis.

[00:09:36]So take the z axis.

[00:09:38]Take this one on the z-axis. It became outward or perpendicular, so take it outward. Now you pick it up.

[00:09:45]pick it up from here here and bring it and put it here.

[00:09:49]This one will never overlap.

[00:09:51]When will it do the overlapping? S and P overlapping and will form sigma bond. S and P overlapping. will form a sigma bond.

[00:09:59]When P you take equal to the internuclear axis.

[00:10:01]Which is the internuclear axis. If by the way s and px overlap you will form a sigma bond. When your internuclear axis is the x axis. Overlapping s and py will form a sigma bond. When the internuclear axis is your y axis.

[00:10:15]s and pz will form your sigma bond when the internuclear axis is the z axis. Our internuclear axis here is the z axis. 2px is taken. Never formed any bond.

[00:10:26]This one will be zero overlapping. Look, dxy and dx², I'm erasing this as well.

[00:10:37]Look at this dxy and dx² y² this is x this is y this is x and this is y along the z axis overlapping the z axis this one will be on the outside. Now you take this entire orbital and bring it here and move it along this internal axis to overlap it. So you'll find one orbital with the electron in between the axis and one on the axis. Will they ever be able to overlap each other? Will never be able to do it. So how will this be? No overlapping. No overlapping.

[00:11:08]This too will not be able to form a molecular beta. It is wrong. 2s and 2p, as I was just telling you, if you overlap s and pz along the z axis, it will always result in the formation of a sigma bond. So this will give you sigma bond. He will give it.

[00:11:22]Will form molecular orbital.

[00:11:24]2pz and dx²y² Just look at this. Like this one is 2pz. Let's assume this is the z axis and dx²y², okay? This is dx²y². So our z axis will be this one. Can you see this one? And along with this will be our molecular orbital.

[00:11:45]This one will be the orbital of the atom. This one will be on this.

[00:11:47]So he would like to go along with this and do overlapping like this. So the electron density here is on this axis, this one which is not on the outside, and all the other electron density is in this plane.

[00:11:57]Sometimes it will be able to overlap and understand it in simple language. So it has electron density along the z axis and it has electron density along the x and y axis. The planes of both are different.

[00:12:06]Zeros can never overlap to form molecular orbitals. So he wo n't do this either. Asked what is it? Which molecular orbital can be formed? So my only option coming is D. Rest A, B, C and E none can do it.

[00:12:18]Therefore the option will be third. So the correct option for 10th will be option number three. Let's move on to the next question.

[00:12:30]Question No. 12 Which of the following statements is true with respect to H2O, NH3 and CH4? Ok? The Central Atom of All the Molecules R sp3? Isn't it right? This is a very simple question.

[00:12:42]It also has sp3. It also has sp3.

[00:12:44]It also has sp3. Correct? The h hn bond angle in H2O NH3 and CH4 molecular is also correct. The whole world knows this. Its effect is due to the loan payer.

[00:13:07]All of these also have sp3 reaction. So because of being sp3 the bond angle should have been 109.

[00:13:13]But with the introduction of lone pair the bond angle decreases.

[00:13:14]In this, the bond angle decreases due to the introduction of lone pair.

[00:13:16]So it has no lone pair so the bond angle will be exactly 109.28. But there is a lone pair the bond angle will decrease a little less and comes around 107.5. There are two loan pairs in this. The bond angle will decrease further.

[00:13:27]So it comes around 104.5. Well then. This is also correct, we have given the order. The increasing order of dipole moment in is CH4NH32 is absolutely correct. H2O has a higher dipole moment.

[00:13:38]It will pull here and pull here. It will pull towards this side. It will pull here, it will pull here. Here also the direction of both is the same.

[00:13:43]Its orbital dipole moment and bond dipole moment both reinforce each other. Because the bond angle between the two is zero.

[00:13:51]The angle is zero. When theta is zero, the value of cos theta is cos 0. It is maximum. So both support each other.

[00:13:59]Will support this also. But in this, the delta negative delta positive between O and H is more than the delta negative delta positive between N and H. So if the charge in it is more then μ = q.d.

[00:14:12]q is large so μ will be large.

[00:14:14]So h2 will be more. It will be less and its μ = 0. It is a symmetrical molecule.

[00:14:19]Therefore, it will have μ0. When μ is 0 it will become non polar. So this will be the least.

[00:14:23]Then NH3, this has also been given to us correctly. Both H2O and NH3 and Lewis base. Correct.

[00:14:28]Both can donate electron pairs.

[00:14:29]So both are Lewis bases so this is also correct. All are correct. A B & A, B, C & D all are correct. The third option will be correct. So the correct option for 12th will be option number three. Let's move on to the next question.

[00:14:45]Question No. 14 Match the column one with column two. In column one, the transition metal cation is given and its spin-only magnetic moment has a value given. This is nothing, right?

[00:14:56]μ is the spin = n * n + 2 Bohr magnetons. Where the value of n is the number of unpaired electrons. So keep counting the unpaired electrons. And whatever its value will come, the number which comes before the decimal, it gives its number of unpaired electrons which will be the first number of decimal, so what will be titanium + 3, d1 number of unpaired electrons in titanium + 3 is one, so its value will be around 1.732, which means first one of decimal is one, so see which one is third, its third will be vanadium scandium titanium vanadium two electrons are removed d3, so the number of electrons will be three, so in this it will be 3 point something, 3.87 given, it will be first.

[00:15:36]In Nick + 2 d8 d8 the number of unpaired electrons is two. Its going to be 2 something. 2.84 will become fourth.

[00:15:46]Scandium +3 has d0. The value of d0 main will become zero. Its dipole moment will be zero. It will be seconds. So look, A's third, A's third, B's first, C's fourth, D's second. The fourth option is matching ours. So the correct option for 14 will be option number four. Let's move on to the next question.

[00:16:08]Question No. 15 When ethane or diamine is added progressively to an aqueous solution of nickel or chloride, the sequence of colour change observed will be. Ok? Look, this is also a good question. There are two ways in this. Either you can remember it or if you don't remember it then you can still manage by applying the concept.

[00:16:29]How can we make things work by applying the concept?

[00:16:30]Please understand what I am saying. If you remember, it is a very good thing. So you will write it down.

[00:16:35]Said NiCl 6 is Nick +2.

[00:16:44]Nickel to chloride. There is two charges on this so it will become 4-. Nick + 2 arrived. Now add one ethane one diamine to it. NI EN and Cl4 will come.

[00:17:04]Ok? Now what happens in this? Just understand.

[00:17:07]As soon as you add EN to it, as soon as you add EN to it, its stability increases due to the addition of EN.

[00:17:22]Its stability will increase. When the stability increases, then its splitting, in this case the delta0 value, will increase.

[00:17:30]Stability will increase if the stability is directly proportional to the crystal field splitting energy. The higher the crystal field splitting energy, the greater the stability. Ok?

[00:17:44]So stability will increase as delta knots increase. Now when the delta knot increases, meaning the splitting will increase, then the splitting will increase, so at this time, to take this electron from here to here, what you call DD transition, which is responsible for the color.

[00:18:01]Mainly there is DD transition, some also have charge transfer spectra.

[00:18:04]So for DD transition, when you give energy from here to here as a form of photon, then it is natural that if you are increasing the delta note value, then the energy required for it will be more. When the energy required is more, the new knot will be more. This means that its wavelength will be less. What will be the frequency? It will be more.

[00:18:23]This means that its delta0 value will be lower. Its delta0 value will be higher.

[00:18:29]When delta0 is high, the energy you will have to give for DD transition will be more. Now add another one and progressively another N to it.

[00:18:38]Then what will happen in this also? Nen's Whole Twice cl's Whole Twice. So its stability will increase further because two N's came in it for chelation and it increased, stability increased due to increase in chelation and if you add one more N then the whole three of N will come, its stability will increase further, its stability will increase further, so understand it a little, it means that one, two, three, four complexes are formed in it, the least stable is in it and the most stable will be A, due to chelation the number of chelating will become more in it, so if you talk about VIBGYOR of VIBGOR then the wavelength of red is the least. If the wavelength is low then the energy wavelength is highest. So the energy will be the lowest.

[00:19:17]When the energy is lowest, it means that the red light will be absorbed by the one which has the lowest splitting. And who will have less splitting, who has the lowest stability.

[00:19:25]Who has the least splitting? in which Cl is leagued. In such a situation, here what is given in our NCERT may be that Nick Water may have been left out. The complex of Nick Two is the aqueous solution of Nick. You have written it well.

[00:19:35]Progressively added to an aqueous solution of nic. Aqueous solution of nickel means NiH2O six Cl2 this is bigger.

[00:19:44]This is the complex. Ok? Aquas is not the solution. This is not a complex of nickel Cl2.

[00:19:47]Tell me that only. The hol of NiH2O is 6. Cl2 is. This one is complex. Ok? When aqueous solution of Nick plus aqueous solution of Nick chloride plus aqueous solution of NiCl2 is added, this will form the complex of this type. The bulk of NH2O forms six Cl2a complexes. At its most H2O is a weak field league.

[00:20:05]Add ethyne diamine. Do H2O in everything. There is H2O here. This is also H2O. This will definitely end.

[00:20:11]Now listen, this complex is the least stable. The lace is stable. So its splitting will be less. If there is less splitting then it will require less energy.

[00:20:18]Less energy means more wavelength. So which one has the longest wavelength? Of red. So when the red absorbs then remember the chart in it. Complimentary Chart V I B G Y O R. What does the color look like when it absorbs red? Green. This means the color of A complex will be green. Now come down to this orange. What color will it be when it absorbs orange? The orange one is blue in color. So it will become blue. So NCERT gives it as pale blue and this one gives it as blue. It will turn dark blue.

[00:20:47]Now come down. Come after this. If it comes after blue, it will observe yellow and yellow, then it will observe violet. So what will be its color? Its color gives violet.

[00:20:56]This is the color. Hole six of green NiH2O will be this green. The one with an N will be pale blue. The ones with two N's will turn dark blue.

[00:21:05]And all three NI NI N have a violet color of hole size. Green, Pale Blue, Blue and Violet. The first option seems correct. So the correct option for 15 will be option number first. Let's move on to the next question.

[00:21:18]Look, I will tell you each math.

[00:21:21]Always remember this question about colour.

[00:21:22]So children, there are many variety of questions in this.

[00:21:24]Which of the following complexes will observe the red color in the DD transition? Or who will give the red color in the D? Or whose color will be green?

[00:21:33]What would the color be like when Red observes? Green. The meaning of asking this question is that Sir, the children ask, Sir, now do we have to remember the colour of everyone in this? There is so much complexity. No, like this, which of the following complexes will observe the red color in the transition? So the motive behind asking this question is whether it will absorb red colour. This means the one with the highest wavelength. So the lowest energy means its splitting energy will be the lowest. And the splitting energy will be the lowest. This means which such complex is there in the given set of elements?

[00:22:02]Which complex in the given set of complexes has the lowest stability? Because the one with the lowest stability will have the lowest delta knot value. If the delta knot value is low then the energy will be low.

[00:22:10]Its wavelength will increase.

[00:22:12]Ask this question that which of the following will absorb the violet color or any other color below it. Then we will determine it by the function of stability. The one with higher stability will have higher delta knot and will consume more energy.

[00:22:23]Then its wavelength will decrease.

[00:22:25]Remember this. Let us move on to the next question. The next question is question number 20 the correct order of boiling point of halogens okay F2 this is simple F2 Cl2 Br2 and I2 all of these have van der waal forces and van waal forces are directly proportional to molecular weight so what will be the order of melting and boiling point as we increase from top to bottom so lowest is F2 then Cl2 then Br2 then I2 fourth option will be correct so the correct option of 20 will be option number fourth let's move on to the next question.

[00:23:04]Question No. 22 Which of the following pair is not isoelectronic species? Not isoelectronic. Ok? Not isoelectronic? Remember.

[00:23:15]See, whenever you make this in the exam, don't make it not incorrect or correct. So mark it first. If you mark it then you will keep in mind that otherwise what will the child do that he has asked not. I have just answered no. I asked for it, I gave it correct or incorrect. If it is asked incorrectly and you give the correct one, then children make mistakes in it, so first mark it so that it always remains in your mind that yes, what we have to do in it, okay, so this is not isoelectronic, okay, see, for your convenience, its atomic number has also been given, so cerium +2 will be 60 and erbium +2, two are gone from 68, so 66, this is correct, this is not isoelectric, both are the correct options, erbium + 2 will be 68 and lutetium +3 will also be 68, so this is isoelectronic, so this will not be our option. It went wrong. This is our option. Europium + 2 Europium + 2 Europium + 2 is 63, so it becomes 61, and Terbium + 4 is 65, so it becomes 61. So this is not there, both of these are there. This is also 61 and this is also 61, so this also will not be ours. This is also isoelectronic. TB + 2 TB + 2 63 and TM + 4 TM + 4 65 four goes so 65 two goes so this is also there. So A & D will be our option.

[00:24:32]and D. Forth.

[00:24:35]which is not isoelectronic. B and C are our isoelectronics. And look, if we had not read this note and given isoelectronic, then our B and D options would have been gone.

[00:24:43]This will also be your option B and D. Well, let's see if B and D are not there, who else is not there? If B and D options are not available then you do not face any problem. A and D are the only options.

[00:24:51]Sorry B and C na. B and C are isoelectronic. So look at B and C, you saw an option B and C. B and C are isoelectronic. A and D are not isoelectronic. If you had not seen the knot, you could have also chosen the option of B and C which are isoelectronic. Both of these have happened.

[00:25:03]But we have been asked not to be isoelectronic. Therefore A and D will be its correct option which is not isoelectronic. So the correct option for 22 will be option number four. Let's move on to the next question.

[00:25:18]Question No. 24 for Which of the following diene configuration of octahedral complexes can exist in both high spin and low spin forms?

[00:25:27]Look at the high spin low spin form in the d octahedral octahedral remains D42 D4 to D7. DA to D10 are all sp3d2. There is no octa in this, there can be no low spin high spin govern.

[00:25:45]All the inner orbital complexes of the outer orbits constitute the outer orbital complex.

[00:25:49]And all d0 to d3 are d2 sp3. d2 are sp3. All of them will be inner orbital complexes and all of them will be outer orbital complexes irrespective of the nature of the complex, whatever the nature of the legged may be.

[00:26:04]So the low spin high spin configuration that is defined in octahedral complexes is D4 to D7.

[00:26:10]Not defined for D3. D5 is in the middle of this. Is in D6. Not in D8.

[00:26:15]How is D10 compared to D8? All of them are the same.

[00:26:20]You cannot say low spin or high spin in this. So the correct option for 24 would be B & C. B & C third option is correct. Let's move on to the next question.

[00:26:33]Question No. 28 The number of POP bonds in H4P2O7 Look at H4 H3 H3PO4 Take two molecules of phosphoric acid and remove one molecule of water, it becomes H4P2O7 which is called pyro phosphoric acid or it is also called di phosphoric acid.

[00:27:01]Diphosphoric acid. When we make its structure then POP double bed O B OH OH OH how many POP linkages one is one in it.

[00:27:16]Hole Threshold of HPO3 is also a good thing.

[00:27:18]Look, I am erasing it a little now.

[00:27:25]Our second option is whole threes of HPO3.

[00:27:29]It is made up of three molecules of H3PO4.

[00:27:32]Remove three molecules of water from this and you will get 3 out of 9 H3 P3 and 3 out of 12, giving 9 H3P3O9 and this is written in the simplest whole number ratio as the whole three of HPO3 and its name is cyclo cyclo tri meta phosphoric acid.

[00:27:56]What is its name? Cyclo trimeta phosphoric acid. Cyclo tri cyclo means it will form a ring. This is how the ring is made. POO PP O double bottom OOH double bottom OOOH double bottom OOH This is the structure of H3 P3O9 How many POP linkages one two one here one two here and the third one here There are three POP linkages in it. H3PO3 Look at H3PO4 Remove one oxygen from it.

[00:28:31]H3PO3 is formed. Remove one more oxygen from it.

[00:28:34]H3PO2 is formed. Its name is phosphoric acid. Its name is phosphorus acid. Its name is hypophosphorous acid.

[00:28:41]Hypophosphoric acid. Look, POP linkage is asked in this. So, you will not think about it at all. There is only one P in it. Two Ps then you will think of writing POP. So in this zero 130 second option is visible, 130 so the correct option of 28 will be option number second. Let's move on to the next question.

[00:29:02]Question No. 30 The electronic configuration of Li2 molecule is KK sigma. Look, in Li2 the number of electrons in it has become six. The number of electrons is equal to six. Now fill it σ1s σ star 1s σ 2s six electrons 2 4 2 6 if you find the bond order then this is finished this will cancel it sigma sigma star 2 - 0 / 2 one bond order one is correct dimg any unpaired electron is visible you are not seeing it dimg this is also correct exist in vapour fuzz this is also correct LI to exist always in vapour fuzz so if you have asked A B and C correct then all three A B and C are correct. So the correct option for 30 will be option number four. Let's move on to the next question.

[00:29:57]Question No. 40 Which one of the following set correctly represents the increasing in magnetic property. Magnetic property means the number of unpaired electrons which has higher value of spin van magnetic moment. So let's check it. The one with higher number of unpaired electrons will show better magnetic properties. Will do more shows. In copper + 2d9 and d9 the number of unpaired electrons is one.

[00:30:16]Vanadium + 2 d3 so the number of unpaired electrons is three. Chromium +2 d4 Number of unpaired electrons is four. And manganese + 2 D5 number of unpaired electrons is five. So the first magnetic property will show better. So first copper, vanadium, chromium and manganese. So the correct option for 40 will be option number first. Let's move on to the next question.

[00:30:47]Question number 41 is an assent question.

[00:30:50]Assumption speaking europium +2 acts as a strong oxidizing agent. Just understand, you know that the most stable and common oxidation state of the entire block of the lanthanoid series, be it 4F series or 5F series, is the most stable and common oxidation state, that is +3, it would like to remain in +3 only.

[00:31:09]So europium is +2 because europium +2 has the electronic configuration 4f7. Half field comes. But it will immediately convert from Europium +2 to Europium +3. When it goes from +2 to +3, it will undergo oxidation. And it will act on which species go into the oxidation process.

[00:31:27]How does it work?

[00:31:29]Acts as a reducing agent.

[00:31:30]Therefore europium +2 would be a very good reducing agent. Will not contain strong oxidizing agents. There is something wrong in this. The most stable oxidation state of lanthanides is +3. Correct? Correct. So the assertion is false, the reason is true. Assertion is false, reason is true. The correct option for 41 will be option number second. Let's move on to the next question.

[00:31:54]Question number 45. The zero probability of finding the electrons in the px orbital.

[00:32:00]Take the px orbital one. See.

[00:32:03]This right here is the x axis.

[00:32:07]This is the y axis and the z axis. So this is the electron density that will be found in it.

[00:32:12]So where will there be zero probability? Two opposite sides of the nucleus along the x axis is the x axis. Look up at the x axis. Its two opposite sides, one opposite side and one will be zero in this that maximum electron density will be found. It will not be zero.

[00:32:26]So you have given it wrongly, right? In the at the nucleus. The electron probability density at the nucleus is always zero. The node is found there. Correct? Same on all sides around the nucleus. It is wrong?

[00:32:38]How will the beans be? Here is the electron density.

[00:32:39]No here, zero on z, zero on y.

[00:32:41]Where is the bean from? This is also wrong. None of this will happen.

[00:32:43]So, the correct option for 45 would be option number second. At the nucleus, the probability of finding the electron is zero. There is a node there. This is all about your inorganic question. Thank you everyone. Thank you so much.