50 Marks in JUST 10 Minutes! Re-NEET 2026 Hacks | Wassim bhat

[00:00:00]So, my dear students, this video again is going to give you some 48 marks in just 10 to 15 minutes. So, I would want you guys to be with me till the end.

[00:00:09]In this particular short and precise session, these are the sure short topics which are almost almost asked every year in your NEET NEET examination. And here in this particular session, I'll be giving you the glimpses of these particular topics which are written on your screen.

[00:00:24]The first topic, what is that?

[00:00:28]Frequently asked question. Calculate the number of waves made by the electron in the Bohr's orbit.

[00:00:35]My dear students, one simple thing you have to remember. Number of waves made by an electron in an orbit is always equal to the value of n.

[00:00:42]For example, if the electron is moving in the first orbit in the first shell, it is going to make only one wave one wave. If the electron is moving in the second shell, so the electron is going to make some two waves. If the electron is moving so in the third shell, it's going to make some three waves. So, remember in short, number of waves made by the electron in an orbit is always equal to the value of n, the shell number in which electron is moving. As simple as that. Question number one and sure short four marks.

[00:01:10]My dear students, the second question.

[00:01:13]Calculation of series limit.

[00:01:15]Calculation of series limit. What is series limit first of all? Series limit means minimum wavelength emitted or absorbed. Minimum wavelength emitted or absorbed when the electron makes a transition between two energy states. My dear students, first of all, you must have heard about this particular which is the Rydberg's formula. And with the help of this Rydberg's formula, you can calculate easily the wavelength emitted or absorbed by the electron when the electron makes a transition. Now, here comes a point.

[00:01:42]If they ask you calculate the series limit in case of Lyman series. Series limit means minimum wavelength. When can be wavelength minimum? When energy gap is maximum. In Lyman series, when is the energy gap maximum? When the electron makes It's transition from infinity to one. Perfect.

[00:01:58]Right? Similarly, lambda minimum or you can say series limit in case of Balmer series. In case of Balmer series, when can be lambda minimum? Lambda can be minimum if energy gap is maximum. In case of Balmer, energy gap is maximum when the electron makes a transition from infinity to two.

[00:02:12]That's all, you're done. Put these values here in this expression, you can say one by lambda minimum is equal to R.

[00:02:18]Z value is one because we are talking about hydrogen. N1 square minus [clears throat] N2 square. So, one by lambda is nothing but R or you can say lambda is equal to one by R. And remember, one by R value is nothing but 911.5 angstroms. So, this is the minimum wavelength or you can call this as the series limit in case of Lyman series. This you can try on your own. Put the values in this expression and get the value of lambda minimum.

[00:02:41]This is again a sure shot question, my dear students, which can be asked either in your chemistry or in your physics.

[00:02:46]Okay?

[00:02:48]My dear students, one more thing from the chapter chemical bonding. What is it?

[00:02:52]This is something which you call as overlapping. My dear students, first thing which you must know. Whenever S and S orbitals overlap, if considering X as the internuclear axis, Y is the internuclear axis, or Z as the internuclear axis, it will always lead to the formation of sigma bonds.

[00:03:08]Right? In case of SP overlapping, S and PX, if X is the internuclear axis, sigma bond. If Y is the internuclear axis, no bonding at all. If Z is the internuclear axis, no bonding, no overlapping.

[00:03:21]Similarly, your PX and PX, if X is the internuclear axis, sigma bond. Y internuclear axis, pi bond. Z internuclear axis, pi bond. Similarly, when it comes to your PD overlapping, your DXY and PX, keeping X as the internuclear axis, zero overlapping.

[00:03:36]Keeping Z as the internuclear axis, zero zero overlapping. Keeping Y as the internuclear axis, it is going to lead to the formation of pi bond. These anything can be asked out of this particular slide as well. Keeping, for example, Y axis as the internuclear axis, what will be the overlapping here or there? Something can be asked from this particular slide only. Okay, so have an eye on this particular slide as well. And my dear students, this particular PDF I'll be sharing on my Telegram, right? The name of the Telegram channel is Wasim Bhat Chemistry Official, right? Let me write it over here. It is Wasim Bhat Wasim Bhat Chemistry Official is the name of the Telegram channel on which I will be sharing this particular handwritten PDF. If you are not the part of that Telegram channel, be the part of the Telegram channel right now. Moving on to the next topic, my dear students.

[00:04:23]A frequently asked question in your JEE Mains, but the level of the question is exactly same as if as of your NEET, right? So, you can expect this sort of a question for sure when it comes to the relationship between delta H and delta U. My dear students, a general result is delta H is equal delta U plus delta of PV. Now, the question says, you have got 1 mole of non-ideal gas undergoing a change in state from this to this with delta U is equal to 40 atm liter, calculate delta H. My dear students, if you look carefully, this is your P1, this is your V1, this is your T1. This is your P2, this is V2, and this is T2. All the parameters are changing. Pressure is changing, volume is changing, temperature is changing.

[00:05:00]And when all the parameters keep on changing, at that time remember, delta H is written as delta U plus it is written as P2V2 minus P1V1. This is the formula used when all the parameters keep on changing in case of these sort of questions. So, my dear students, you are supposed to calculate delta H. Delta H is nothing but delta U. Delta U is 40 atm liter plus P2 multiplied by V2. 5 * 4 = 20 minus P1 multiplied by V1. 3 * 1 = 3. 20 minus 3 is nothing but 17. So, 40 plus 17 makes it 57 atm liter. Right? So, this is the value of delta H which you were supposed to calculate. Perfect.

[00:05:37]Just remember this particular result.

[00:05:38]Accordingly, you can use this result if all the parameters were changing, right?

[00:05:42]This is again a sure shot question which I wanted to give you.

[00:05:45]My dear students, there can be question asked related to the graphs. Which of the following graphs correctly represent equilibrium? My dear students, these are the only graphs These are the only graphs which correctly represents equilibrium. Perfect. Concentration versus time. When you define equilibrium, we say at equilibrium, the concentration of reactants and products they become fixed. Right? This particular graph is showing that concentration of reactant and product is becoming fixed after some time. Right?

[00:06:09]So, it is representing equilibrium. Even this is representing equilibrium. This is representing rate versus time. You know, at equilibrium, rate of forward and backward reaction becomes equal. So, here you can see rate of forward and backward reaction is becoming equal after this particular point of time.

[00:06:23]Perfect. So, anything apart from these will not represent your equilibrium. So, remember, these are going to be your sure shot graphs which correctly represents your equilibrium state. So, have an eye on these graphs as well.

[00:06:34]Okay.

[00:06:35]Question again has been asked from Henry's constant as well. Question will be asked, what will be the solubility of the gas? Arrange the following gases on the base of their solubility. Remember one simple thing, solubility of the gas in the liquid is inversely proportional to Henry's constant. More the Henry's constant of the gas, lesser is going to be the solubility of the gas. Already one question has been asked from this topic. Again, there are high chances that question might be asked asked from this particular topic again. As you are given with three gases, X, Y, and Z, their Henry's constant values are given.

[00:07:05]Henry's constant values are given. More the values of the value of Henry's constant, lesser is going to be the solubility. Right? So, I would say this is the maximum value of your Henry's constant, that's for Z. So, Z will have minimum solubility followed by Y, followed by X. This is the order of the solubility which you have to remember.

[00:07:21]Just base Just considering this particular thing into your mind.

[00:07:24]Solubility is inversely proportional to Henry's constant. That's all. That's more than sufficient.

[00:07:29]One more question can be asked, my dear students. You can be given with any of these four to five graphs, right? And you will be asked, what is the order of the reaction? Remember all these graphs, rate versus concentration of reactant, concentration of reactant versus time, half-life versus initial concentration, completion time versus initial concentration, degree of dissociation versus initial concentration. All these are basically the graphs for the zero order reactions. Perfect. Similarly, all these are here the graphs of the first order reactions. So, they might give you certain graphs and they will ask you what is going to be the order of the reaction. So, remember these few graphs over here. One is for zero order and the second part is for the first order reactions. One again important part. My dear students, there is again one more important thing which is frequently asked, that is calculation of E naught cell. How do you calculate standard EMF of the cell? And you know, standard EMF of the cell is nothing but it is E naught cathode minus E naught anode.

[00:08:22]What is E naught cathode? That is SRP, standard reduction potential of cathode minus standard reduction potential of anode, right? So, my dear students, if you look at this particular question, I'm given with certain values. E naught of A positive gives A is equal to 1.1, E naught of B gives B negative 2 is equal to 3.1. So, over here, this is the SRP value, right? Again, here, this is also the SRP value. Perfect. So, first of all, let's try to check which one has got more SRP. This one has got more SRP.

[00:08:49]SRP means more the SRP, more tendency to undergo reduction, and reduction takes place at cathode. So, this is your cathode here, and this becomes your anode here. So, you can easily calculate E naught cell, which is E naught cathode, SRP of cathode, that's 3.1, minus SRP of anode, that's 1.1. The value comes out to be 2 volts, right? Now, it's not necessary that they will give you SRPs only. They might give you SOP values also. Perfect.

[00:09:13]You just need to convert those SOPs into SRPs by just changing their signs, and accordingly use this particular formula.

[00:09:19]But take care of this particular thing, E naught cell is E naught cathode minus E naught anode, SRP of cathode minus SRP of anode. Perfect. Now, my dear students, uh again, a frequently asked question, that is hybridization in case of coordination compounds. If my dear students, look at this particular complex. Perfect. Iron here is in dipositive state. I want to calculate the hybridization of the central metal ion over here in this particular complex. First of all, Fe di-positive configuration will be 3d6 4s0. Perfect.

[00:09:48]3d6 4s0, that will be the Fe di-positive outermost configuration. Now, my dear students, look carefully. This cyanide, it's a strong field ligand. Remember, strong field ligand always causes pairing of electrons. So, there will be pairing of electrons here. One will go here, one will go here. Perfect. So, the new configuration after the pairing is done will be 1 2 3 4 5 6. And this is S, this is your P, and this is your D. Now, my dear students, after this, after this, what is the coordination number here? Coordination number here is six.

[00:10:17]So, this central metal ion, it has to arrange some six vacant orbitals for hybridization. Those six vacant orbitals will be 1 2 3 4 5 6. So, the hybridization involved is d2sp3.

[00:10:27]Perfect. Look at this particular one.

[00:10:29]NiCl4 di-negative. Nickel here is in di-positive oxidation state. So, its outermost configuration will be 3d8 4s0.

[00:10:35]Now, this Cl negative, it's a weak field ligand. If it's a weak field ligand, it it is not going to cause any sort of pairing, right? It is not going to cause any sort of pairing. What is the coordination number here? Coordination number is four. So, this nickel di-positive has to arrange some four vacant orbitals for hybridization. Those are going to be 1 2 3 and 4. So, the hybridization involved is simply going to be sp3. Nothing more than this.

[00:10:56]Perfect. [snorts] You just have to check which is the strong field ligand, which is the weak field ligand. Strong field ligand will cause pairing, weak field ligand will not cause any sort of pairing. That's all. And my dear students, at the end, this is something which is which I'm giving you additional. So, remember this particular sequence as well.

[00:11:11]Stability of carbocation. That That carbocation wherein you see a group showing plus M, right? That will stabilize the carbocation the maximum, followed by plus H showing group, followed by plus I showing group, minus I, minus H, minus M. Similarly, your stability of carbon carbanions follows this particular order. Acidic strength order. Acidic strength order will follow the same order as that of your carbanion order and basic strength order will follow the same order as that of your carbocation order.