NEET 2026 Chemistry 🔥 | 20 Most Repeated Questions | Must Practice PYQs

追加: 2026-05-05

[00:00:00]Invest the upcoming 25 to 30 minutes with me and learn, attempt, and revise all the most probable and repeated, not 20, but more questions or logic from inorganic side of chemistry, especially chemical bonding, the backbone of inorganic chemistry. And once you see such questions in your NEET examination, so NEET warriors, such questions will be a crime for you to not attempt if you're having a idea or a feeling to land into your dream government medical college.

[00:00:36]Now, why I'm specially doing this for inorganic side of chemical bonding is because of the fact that I am coming live with the physical chemistry, the top part one and part two. Over there, I'll be dealing with the conceptual foundation of physical chemistry, and with these short videos, I'm trying to make you revise all those inorganic basics as well. So, without further ado, let us quickly begin with the process. But yes, I will be really very quick with the discussions. So, for a better understanding, I will be moving aside for the questions. You just read the question. I'm not saying they are just your PYQs. They are those questions which will be helping you to attempt all the logics. So, to begin with, this question says you have to select which of them is incorrect. Now, you can definitely pause. I hope you have attempted the question. Now, definitely, now I'm telling you one thing, more than 90% of the time this topic is repeated in your NEET examination. Like, if you compare ammonia and NF3, then nitrogen being more electronegative than hydrogen makes the dipole moment vector over there. So, dipole moment vector gets added up. Therefore, ammonia has got a higher dipole moment than NF3. Although fluorine is more electronegative than nitrogen, that's why the dipole moment vector is made in this direction.

[00:02:02]Therefore, with the lone pair and the bond pair, the dipole moment vector is partially cancelled, which makes ammonia rather a more polar molecule.

[00:02:11]If you see the resonating structures of carbonate ion, then the double bond can go here, here, and here. So, if you do the resonance hybrid, so it will have a partial bond resonance stabilization, and therefore number of resonating structures are three. Therefore, this is your correct answer because ozone has not three, but two resonating structures, and BF3, because of sp2 hybridization of boron, a symmetrical structure, the net dipole moment vector of BF3 will be zero.

[00:02:43]Moving to the next question, which says, which of the following is least ionic?

[00:02:48]Now, in ionic compound, covalent nature is can be calculated and estimated using the famous Fajans' rule, which says, covalent nature in ionic compound increases when? If you have d-block cation, then all the d-block cation that have pseudo inert gas configuration as in configuration like Ag+, where the they have completely filled d subshell, so they have very high z effective. Why?

[00:03:17]Because shielding of these d and f electrons are very poor. So, d-block cations have very high polarization power of cation, and therefore they are always more covalent salts among all the given ionic salts, mostly.

[00:03:32]Now, what else, sir? Polarization power of like you have to calculate polarization power for s-block elements or p-block elements cations, then how to do it? Calculate their polarization power is calculated using their charge density. What is charge density? Ratio of charge upon size.

[00:03:49]First, we give priority to charge, and if charge is equal, then smaller cation will have high polarization power, and high polarization power that means high covalent nature. And last is for anions, bigger, larger will be the anions, more is their polarizability.

[00:04:08]So, more will be the covalent nature.

[00:04:10]So, among the given options that we have, Ag+ will be the salt which will be having highest covalent nature.

[00:04:18]Moving to next question, which says, among these salt, which is having highest melting point? Now, sir, we already discussed Fajans' rule, and we know if a salt is having more covalent character, it will have less ionic character. So, it will have less ionic character, it will have less of these properties which an ionic compound is supposed to have. Now, if I compare LiCl and NaCl, then as per Fajans' rule, because Li + cation will have more covalent nature, therefore Na+ will have less covalent nature, that is more ionic nature, that means its melting point will be high.

[00:05:01]But sir, how will you compare that for alkali metal chlorides, it is the delta H lattice is decreasing down the group?

[00:05:08]Because that's why we are comparing these two. Because we are thinking that down the group, the delta H lattice as in melting point of metal chloride is decreasing. Why?

[00:05:21]Because delta H lattice is also predominantly decided by the charge density. Charge being same as size of alkali metal cation increases, delta H lattice decreases. Definitely correct.

[00:05:33]So, if this is decreasing, then melting point will also decrease. So, who should have highest melting point? LiCl. But why LiCl is losing battle only from NaCl? Because of high covalent nature.

[00:05:46]And why it is high? That you can only understand using Fajans' rule. So, a very good question. Moving forward now.

[00:05:54]Now, if you want to check ionic nature in a covalent molecule, then you have to check its dipole moment. What is dipole moment? Let's say I have a polar molecule HF, then its electric dipole moment is calculated by the formula charge into distance between charge. What is charge, sir? This is partial positive, delta minus, delta plus. It is partial plus one and minus one. So, what is the magnitude of charge? Electronic charge.

[00:06:24]What is the distance between charges?

[00:06:25]Bond length. Both these values are given. So, we can calculate this dipole moment.

[00:06:32]But this dipole moment we have calculated through theory, so it's a theoretical value of electric dipole moment which we have calculated. Now, if we have gone to mother nature, we have seen what is the actual value. The question is also supplying you with the actual dipole moment. Take the ratio, multiply by 100, and it will give you the percentage ionic nature. And remember one thing, electric dipole moment is a vector quantity with direction from positive to negative. So, in such questions, when you are you are have to compare which is more which is more polar covalent molecule, then we will think about the polarity vectors of electric dipole moment. And obviously, in the given options, you can clearly see C bond F. Sir, it will be most polar. So, over here, it the dipole moment vector is getting added. While over here, the C bond F vector, I mean, not completely, but these vectors some of these components will be canceling each other because it is more or less a trans formation of these geometrical isomers. Whereas this is cis, as in in the same direction. The dipole moment vector you which you can clearly see from the picture which I have made here, the net dipole moment will be getting added up. So, C will be having the most dipole more more magnitude of electric dipole moment vector. Now, the next question says, among the mixture, which of the following is having in dipole dipole attraction? Now, we know all state of matter are attracted to each other through some inter intermolecular forces of attraction. Now, let us check what intermolecular interaction they have.

[00:08:12]Benzene is a nonpolar molecule. Ethanol is a polar molecule. In fact, it is having strong hydrogen bonding attraction. So, these molecules will have what interaction? This dipole will induce a dipole in that nonpolar molecule. So, interaction will be dipole induced dipole, not dipole dipole.

[00:08:32]Option B, which is our answer, this is CH3CN, a polar molecule. This is acetone, another polar molecule. So, it will also have a permanent dipole. It will also have a permanent dipole.

[00:08:43]Interaction will be termed as dipole dipole interaction.

[00:08:47]In option C, this is our answer, but discussing others also. KCl is an ionic compound will furnish K+ and Cl- ion.

[00:08:54]Water is a polar molecule. In fact, hydro it has capacity to do hydrogen bond.

[00:09:00]So, what interaction they will have?

[00:09:01]They will have ion dipole interaction.

[00:09:04]Last option is benzene, we already seen nonpolar. CCl4 is a perfect tetrahedral geometry, so net dipole vector mu net over here will also get cancelled because every symmetrical system, although individual bonds are polar, but the net vector addition of the molecule becomes zero, and therefore the overall molecule having symmetrical structures is always zero. So, this also becomes a nonpolar molecule, and therefore interactions, sir, this is also nonpolar, this is also nonpolar. So, how will they attract? Let's say this one will create an instantaneous dipole. So, and that instantaneous dipole will induce a dipole in that in the another nonpolar molecule. So, interaction between them is termed as instantaneous dipole dipole interaction, commonly known as London dispersion forces. So, our answer is B for this particular case. Which force of attraction is strongest? And what is the relationship between that particular force and the distance between constituents? First of all, let us categorize them.

[00:10:12]Whatever forces which we have studied, that is if I am known nonpolar, how will I attract another nonpolar molecule or any charge?

[00:10:21]Because I can develop some instantaneous dipole. So, these all forces are commonly termed as van der Waals forces of attraction, which is the weakest intermolecular force.

[00:10:32]Followed by dipole-dipole attraction, because permanent dipole plus-minus will always attract permanent dipole plus-minus. But what is the extreme case of dipole-dipole attraction? Sir, hydrogen bonding, strongest bond, strongest intermolecular force, even also can be regarded as a chemical bond because strength is so high.

[00:10:51]Sir, how will we know this?

[00:10:53]Statistically, we have seen the energy for hydrogen bond has a ratio of has a relationship with the distance between constituents as 1 upon R.

[00:11:04]For dipole-dipole, it is 1 upon R cubed.

[00:11:07]Although this is for stationary dipole, but for mains, you just learn this that it is inverse it is inversely proportional to 1 by R cubed.

[00:11:14]And van der Waals forces are weakest because energy is proportional to 1 by R raised to the power 6. So, these interaction forces are weakest. Again, talking about hydrogen bonding, who has the capacity to bond to have strong hydrogen bond? O bond H, N bond H, and F bond H. So, let us check which of them which of these molecules have hydrogen bonding between them. Sir, O bond H, so this will have hydrogen bonding, option one.

[00:11:44]Uh water, definitely. Second, no. This is hydrocarbon, so nonpolar.

[00:11:50]Benzene, nonpolar. This will also have hydrogen bonding, although intramolecular hydrogen bonding, because intermolecular hydrogen bonding is always operational when a cyclic chain can be formed. So, ortho isomers can form cyclic chain. So, this will also be this also can do hydrogen bonding.

[00:12:05]HF also can do hydrogen bonding. NH3, so our answer will be five. Moving to the next question.

[00:12:12]You attempt, I'm moving aside. So, I hope you attempted it, and such questions are very very very probable for JEE Mains. So, out of these followings, the one with nonzero dipole moment and the central atom having maximum number of lone pair. So, let us decode each of them through VSEPR theory.

[00:12:31]So, for SO2, in order to calculate hybridization, what we calculate steric number. Steric number is summation of bond pair, which is only sigma bond and lone pair. So, steric number for SO2 is three, hybridization is sp2.

[00:12:45]Geometry is trigonal planar, but shape because of lone pair, we do not see that lone pair. We only see nucleus with respect to other nucleus, so shape is just V shape or bent shape. So, yes, because it is not a symmetrical structure, so it will have a net dipole moment, so its mu net is not zero, and therefore it's a polar molecule. So, lone pair it has is one. But we have to give hybridization of that molecule which has highest number of lone pair.

[00:13:14]So, let us check all the options what lone pair they have. Whatever is having highest lone pair and a polar molecule with mu net not equal to zero, that will be our answer. So, sp2 is for this, but let us check for NF3 and NH3, both hybridization is steric number is four, so hybridization is sp3, and both are no both are polar molecule with shape of trigonal pyramidal. But one more important thing I'm telling you, the net vector addition of NF3 is getting cancelled, while for NH3 it is getting added up. So, if you somebody is asking you this is among ammonia and NF3 which is more polar, so it is also a very famous question. NH3 is more most polar among these two.

[00:14:01]So, but not asked, but a very important point. Now, XEF2, now it is sp3d hybridized because it is having three lone pair, so these lone pair will go in the equatorial position because lone pair has to be there has to be kept at maximum distance, so they always occupy equatorial position in a trigonal bipyramidal geometry. But what is shape?

[00:14:23]Shape with respect to nucleus we see, so shape is a linear molecule. So, because it's a linear molecule, therefore net dipole moment will get cancelled, so it becomes a nonpolar molecule. And we are asked polar. So, this is our answer, ClF3 with hybridization sp3d and maximum number of lone pair it has is two.

[00:14:42]That's why because maximum number of lone pair this guy is having, therefore this will be our answer, so our answer will be sp3d. Why?

[00:14:50]Because SF4 is also given, although it is also a polar molecule, but number of lone pair SF4 has is one. Its shape is its geometry is trigonal bipyramidal, but its shape is seesaw or pi shape molecule. Let us check and attempt this another question. I hope you got your answer. So, number of species having nonzero dipole moment are five, and those species are nonpolar. Moving to next question. Pause, attempt.

[00:15:21]And these questions are also very easy as per VSEPR theory. So, BrF5 and XEF4, that is option D, are species having square pyramidal geometry with steric number six and presence of one lone pair.

[00:15:37]Attempt this question. Pause.

[00:15:40]I hope you got this answer also. Option answer is option A. Statement A, B, C are correct because ammonia sorry, ammonia, methane, and water all are sp3 hybridized, and this ammonia, because of presence of lone pair, acts as a Lewis base, which forces water to act as Lewis acid. Moving to the next question. Pause and attempt.

[00:16:08]I hope you got the number of linear as per VSEPR theory. Over here, six molecules are linear. People make mistake in this NO2, but NO2, because of the presence of odd electron, is not linear, it is bent shape molecule. Now, let us attempt this question from molecular orbital theory.

[00:16:30]I hope you got your answer because it will be a crime of not taking marks from such question. Answer is NO to NO+ because NO what follow what electronic configuration of molecular orbital they follow, sigma 2pz is filled first, followed by pi 2px and pi 2py. So, its bond order for NO will be 2.5.

[00:16:51]For NO+, I'll remove this electron because it's an antibonding electron, so bond order will increase to three, so bond order is increasing, and from paramagnetic it has become becoming diamagnetic. Only for nitrogen, you will follow the only thing you have to remember is these two molecular orbitals gets interchanged. Pi 2px is filled first, and then after that sigma 2pz and uh sigma 2pz. And N2 when becomes N2+, it becomes from paramagnetic it becomes from diamagnetic it becomes paramagnetic. A very very religious molecular orbital diagram you should follow. Now, attempting one more question from MOT for heteroatomic molecule, let's try to solve this. Let us summarize MOT again. For homonuclear molecules having number of electrons lesser than 14, or maybe for lighter molecules like N2, we follow MOT of nitrogen for lighter molecules also. And for homonuclear diatomic molecules having electrons more than 14, like for oxygen and F2, we follow MOT of O2. So, actually what we do is you we check the neutral molecule, like in case of NO, NO has number of electrons more than 14, it has number of electrons 15. We check of the neutral molecule, and then make cation ion of that neutral molecule. So, because NO has total electron 14 15 NO, and because it is 15, so we will follow MOT diagram of oxygen. Hence, this we we have already done this, so bond order of NO+ will be three. Now, for CO, actually for NO also, because participating atomic orbitals differ in energy. So, this is a generalization we are making, and similarly when we make generalization for CO, number of electron is 14, so we should follow MOT diagram for N2. But actually what happens for CO specially is because carbon and oxygen have dissimilar energies of combining atomic orbitals.

[00:18:50]So, what changes what change is proposed in this case is actually this sigma all the antibonding molecular orbital becomes of higher energy, and after sigma 2s, pi 2px, pi 2py, and sigma 2pz comes. And post these bonding molecular orbitals uh energy of antibonding molecular orbital come. Although if you do not even know this, if you just follow this rule that for 14 we'll do electronic configuration of molecular orbital following nitrogen. So, then then also you will get your bond order to be three, so summation will be six, but I guess you should know it for some good questions. So, moving to the moving to the last question. Now, let us attempt this one.

[00:19:34]Yes, so I hope you got your answer for this beautiful example of back bonding, which clearly says that trimethylamine is a simple sp3 hybridized nitrogen atom, which is a strong base because of lone pair of on nitrogen atom, but if instead of carbon I have this silicon over here, then this silicon has empty d orbital in which the lone pair of nitrogen can be easily coordinated. So, there this is a clear case of backbonding. And due to this backbonding, sometime bond will be over here, sometime here, sometime here. So, these three resonating structures suggest the shape of nitrogen is trigonal planar because lone pair is not available in the central atom, and therefore it is definitely a weaker base. So, our answer will be beautifully option A. So, this is me, Gaurav Singh Arora, signing off with all the best wishes, and we'll meet in the next chapter.