52 Marks in JUST 12 Minutes🚀 NEET 2026 | Wassim Bhat

Added: 2026-05-05

[00:00:00]My dear students, again this particular video is going to give you some 50 marks in just 8 to 10 minutes. So, I would want you guys to watch this particular video again till the end.

[00:00:10]First concept which I'll be discussing, what is the condition for the photoelectric effect to happen? My dear students, remember energy of incident radiation should be greater or equal to work function or you can say frequency of the incident radiation should be greater or equal to threshold frequency or you can say wavelength of the incident light should be less or equal threshold wavelength, right? For example, you have got a question over here.

[00:00:33]Incident light of wavelength 400 nanometer falls on these metals over here.

[00:00:39]Right? I have to check how many metals will show photoelectric effect. My dear students, incident light have got the wavelength of has got the wavelength of 400 nanometers. That means the incident light is going to carry the energy of 3.1 electron volts. So, this is my incident energy. Now, work function of metal A is 2.1 electron volts. So, incident energy is greater greater than work function, again greater than work function, again greater than work function. So, I would say these are the three metals which are going to show photoelectric effect over here. So, the answer of this question is going to be three. This particular metal, it's not going to show photoelectric effect. Why is that? Because incident energy is not greater or equal to the work function here, neither here. So, these two are not going to show any sort of photoelectric effect over here. This sort of a question is frequently asked in different examinations nowadays.

[00:01:26]My dear students, concept number two.

[00:01:28]How do we determine order from the rate law? For example, you have got a reaction, okay? The question says rate doubles when the concentration of A is increased four times. What is the order of the reaction? How exactly do you approach towards this question? You first of all try to write the rate law.

[00:01:43]R is equal to K, concentration of reactant raised by order. Let me call this as equation number one. Now, the question says the rate is doubling. Rate is doubling when the concentration of A is becoming four times. So, now it becomes four times A raised by N. This is equation number two. My dear students, if you divide equation two with equation one, so R R gets cancelled on left side you get two. K K gets cancelled, A raised by N A raised by N gets cancelled. It becomes four raised by N. So, from this particular equation I can say N value has to be 1 by 2. So, I would say order of the reaction is nothing but 1 by 2. So, like this you are going to solve these sort of questions which are again frequently asked when it comes to the NEET examination.

[00:02:21]My dear students, you must have heard about enthalpy change. Enthalpy change is always equal to activation energy of forward reaction minus activation energy of backward reaction. Perfect. This is one important expression which you need to know. You are given with the reaction which is the exothermic reaction.

[00:02:36]Enthalpy change is given to me as 135.

[00:02:38]That means delta H will be equal 135 kilojoules, right? Since the reaction is exothermic, so the sign has to be negative. Activation energy for the conversion of A into B. A into B means activation energy of forward reaction is given to me. How much is that? That's equal 100 kilojoules. And my dear students, I need to calculate EAB. Now, you already know the equation. Your delta H is nothing but EAF minus EAB.

[00:03:03]Now people, delta H is given to me as minus 135 is equal EAF is equal 100 minus EAB. So, I would say take this 100 on this side it becomes minus 235 is equal to minus EAB or I would say EAB is equal 235 kilojoules per mole. So, activation energy of backward reaction is 235 kilojoules which I was supposed calculate. Have an eye on this particular equation, but remember you have to check in the examination whether the reaction is endothermic or exothermic. If it is exothermic, delta H sign has to be negative. If it is endothermic, delta H sign you have to take positive in this particular equation. Concept number four, my dear students. Sometimes you'll be given you'll be asked what is the rate constant of zero order reaction? What is the rate constant of the first order reaction? What is the rate constant of the second order reaction? My dear students, in general remember rate constant expression, rate constant unit is going to be mole per liter raised by 1 minus N where N is the order of the reaction and second inverse if time is in seconds. Or if pressure is in ATM, time is in seconds, then it's ATM raised by 1 minus N second inverse and N here again represents your order. For example, you have got the zero order reaction whose N is zero. If N is zero, rate constant is going to be this or this. If you have got the first order reaction, N is one. 1 minus 1 is zero.

[00:04:16]Anything raised by zero is one. So, the unit will become second inverse from this expression as well as this expression. Similarly, if you have got the second order reaction, put the N value as two, you'll be getting the unit of rate constant for this particular reaction as well, right? Now, concept number five. My dear students, one theoretical concept, right? It has got the mathematical background as well. I'm not going into the details. I'm just telling you the crux which you have to remember. Sometimes you'll be given with the cell, you have to check whether the cell reactions will be spontaneous or not. Remember, cell reactions are only spontaneous, cell will only work if delta G for the cell is negative or you can say E cell is positive, okay? For any cell whose delta G is positive or E cell is negative, that means the cell reactions will be non-spontaneous and the cell won't be working, right? And similarly, if delta G for the cell is zero, E cell is zero, remember the cell is at equilibrium at that point of time.

[00:05:07]So, anything can be asked out of these three things which are mentioned over here on the screen.

[00:05:11]Concept number six.

[00:05:13]How do we calculate mole fraction of a component in a vapor phase? For example, you have got a solution of A and B over here. Both A and B I'm considering as volatile. So, you'll have vapors of A as well as B over here. This chi dash represents mole fraction of A in the vapor phase, mole fraction of B in the vapor phase, mole fraction of A in the solution phase, mole fraction of B in the solution phase. Remember this is the expression. Mole fraction of A in the vapor phase is nothing but P not A chi A divided by PS. Similarly, mole fraction of B in the vapor phase is nothing but P not B chi B divided by PS.

[00:05:43]For example, you have got a question. P not A is given, P not B is given, chi A is given, chi B is given. I'm supposed calculate mole fraction of A in the vapor phase. My dear students, all the parameters are given to us except this PS and PS is the total vapor pressure of solution which as per Raoult's Raoult's law is P not A chi A plus P not B chi B.

[00:06:01]So, first of all put the values over here in this expression, get the PS.

[00:06:05]Once you get the PS, put it here and get the mole fraction of A in the vapor phase. Similarly, you can calculate mole fraction of B in the vapor phase as well. This is one more short short question which is frequently asked when it comes to the different competitive examinations. Now, Daniel cell parameters. There might be some theoretical questions which would be asked about the Daniel cell. You have to remember all of these things which I'm telling you over here. In case of Daniel cell, your zinc electrode behaves like the anode, copper behaves like the cathode. Here electron flows from anode to cathode, current flows from cathode to anode in the external circuit. Zinc rod dissolves with time, copper rod's thickness increase with time. This is the net reaction which takes place in the Daniel cell. Salt bridge maintains electrical neutrality in both the solutions or you can say it avoids the liquid junction potential. And how many moles of electrons are exchanged between anode and cathode? Two moles of electrons are exchanged between anode and cathode when it comes to your Daniel cell. These are some of the short short things. Out of these, anything can be asked in your upcoming NEET examination.

[00:07:02]My dear students, calculation of van't Hoff factor. How do we calculate van't Hoff factor? For any solute that undergoes dissociation, van't Hoff factor is equal 1 plus N minus 1 into alpha. For any solute that undergoes association, I is equal to 1 plus 1 by N minus 1 into alpha. For example, let's say you have got the solute this, K3Fe(CN)6.

[00:07:21]How will it undergo dissociation first of all? It will undergo dissociation as three times K positive plus Fe(CN)6 Fe(CN)6 negative. Perfect. So, one particle of solute is giving us three plus one, four particles. So, N value is four.

[00:07:36]N means number of particles produced from one particle of solute. So, four particles are four ions basically are produced from one particle of solute.

[00:07:43]So, N value is four. If N is four, alpha is already given to me. So, I can say I is equal to 1 plus N minus 1 into alpha.

[00:07:51]Alpha here is 0.2. Solve this, get the value of van't Hoff factor. Perfect.

[00:07:55]This is one more important thing, my dear students, which you need to know.

[00:07:59]Then our next concept that is identification of buffer. How there will be some solutions which will be given to you, right? And you have to check which of the following forms the acidic buffer, which of the following solution is the basic buffer. My dear students, there are two things which you need to know. Buffers, we have got acidic buffer and the basic buffer. I believe you would have studied that. Whenever in the container you have got weak acid and the salt of same weak acid with a strong base. For example, this is the weak acid and this is the salt of same weak acid which it has formed with a strong base NaOH. Perfect. So, this particular mixture over here I'll be calling as the acidic buffer. Similarly, whenever you see a weak base and the salt of same weak base which it has formed with a strong acid, right? This entire thing over here I'll be calling as the basic buffer. For example, this is your weak base and this is the salt of same weak base which it has formed with a strong acid HCl. Right? So, this particular stuff over here I'll be calling as basic buffer. So, like this you differentiate your acidic buffers and basic buffers from this.

[00:08:56]Again, a theoretical question can be asked. My dear students, concept number 10.

[00:09:00]How do you check whether the given species aromatic, non-aromatic or anti-aromatic? For the aromatic compounds, aromatic compounds are considered to be maximum stable, right?

[00:09:09]Anti-aromatic are considered to be least stable. The stability of non-aromatic is between aromatic and anti-aromatic.

[00:09:15]Perfect. Now, these are the parameters of your aromatic compounds. Molecule has to be cyclic, planar, complete cyclic conjugation. It should follow the Huckel's rule which means 4n plus 2 pi electrons should be present in the complete cyclic conjugation. For example, look at your benzene. There are 2 4 6, six pi electrons involved. So, it is your aromatic compound. Over here, this particular compound, two pi electrons are involved in the complete cyclic conjugation. So, it is aromatic.

[00:09:39]In case of anti-aromatic, only differentiating factor is there are 4n pi electrons involved in the complete cyclic conjugation. That means either four pi, eight pi or 12 pi electrons are involved. For example, over here. Look at this molecule. Over here, four pi electrons are involved in the complete cyclic conjugation. So, this is your anti-aromatic one. Now, my dear students, non-aromatic. The molecule which is neither aromatic nor anti-aromatic makes it non-aromatic. For example, look at this particular carbon.

[00:10:03]This carbon is sp3 hybridized. If it is sp3, the molecule is non-planar. If the molecule is non-planar, non-planar neither comes in this category nor in this category. So, it is the non-aromatic molecule. Perfect. Now my dear students, one table which I want every one of you to remember. Anything can be asked from this particular table.

[00:10:20]Please and please take the screenshot.

[00:10:21]Anyways, I'll be sharing the PDF of this particular session in the Telegram and the name of the Telegram channel is Wasim Butt Chemistry Official.

[00:10:29]Wasim Butt Chemistry Official.

[00:10:32]Chemistry Official is the name of the Telegram channel on which I shall be sharing the PDF. Anything can be asked from this particular table. Products of electrolysis. For example, you are keeping the electrolyte as aqueous NaCl.

[00:10:43]Electrode used is platinum or graphite.

[00:10:45]At anode, Cl2 gas is liberated. At cathode, H2 gas is liberated. For example, you have got fused NaCl. Again using the same electrode. At anode, Cl2 is liberated and Na is deposited. So, like this, this particular table is super important. I would want every one of you to remember this particular table on priority. Perfect.

[00:11:02]Now my dear students, how do we balance the reaction in the basic medium? My dear students, there's a trick for that.

[00:11:08]Let me tell you that. First of all, get the n factors of reactants. If you look at the n factor of this reactant, it's two. n factor of this particular reactant is also two. Cross cross the n factors. So, make this two here. Make this two here. Perfect. After cross crossing the n factors, balance all the atoms except hydrogen oxygen. So, there are four chlorines here. Four chlorine.

[00:11:26]Two iodines here. Two iodines here. Now, in case of basic medium, we are supposed to balance charge first. Now my dear students, on reactant side we have got total minus two charge. On product side, we have got minus four minus two. That means minus six charge. Which side is negative charge deficient? Left side.

[00:11:41]How many negative charge deficiencies are there? Four negative charge deficiencies. So, add four OH negatives on this side. Now, after this, balance oxygen with the help of water. So, you have got four plus six. So, four plus six is 10 oxygen on this side. Eight oxygen on this side. So, this side is oxygen deficient. So, add two water molecules on this side to make the reaction balanced. So, now it's a balanced chemical equation. So, my dear students, these were some short short 12 concepts. 12 concepts means 48 marks which you are getting from this particular video. Perfect. I'll keep on coming with more videos tomorrow as well as today. Perfect. So, stay tuned to the channel and do not forget to join the Telegram channel for more things. So, take care. God bless you all and love you all guys. Bye-bye.