NTPC ACT 2025 Memory-Based Questions | NTPC Assistant Chemist Trainee Recruitment 2026 | Part 1

[00:00:00]Hello everyone. Welcome back to Chemistry with the YouTube channel. I shall NTPC Assistant Chemist Training 2025 memory based and model question for liquid discuss currently. Yeah, video part one.

[00:00:13]I got this video may response actually to make which is coming next part like a and you can leave.

[00:00:20]So, better question here. Consider the following statements regarding the structure and bonding of PCL5 in gaseous state.

[00:00:30]The gaseous statement PCL5 exist as a covalent molecule.

[00:00:39]So, gaseous statement PCL5 exist as covalent molecule and DBP structure.

[00:00:49]Solid statement PCL5 exist as ionic species PCL4 plus and PCL6 minus.

[00:01:08]Better statement here. The molecule belongs to the D3H point group.

[00:01:14]PCL5 structure of the DBP 3 CL equatorial may and those CL axial may.

[00:01:25]So, it's cover point group the key to eat her say X33 chlorine axial P chlorine you guys say X33 point axis of rotation will occur. Yes, CL is my idea. Yes, CL is my idea. Or yes, CL is my idea. 120 degree rotation. So, it's make C3 axis of rotation over.

[00:01:58]Which is about C2 180° Yes, CL Yeah, I could do the CL.

[00:02:15]180° rotation over Are you axial CL?

[00:02:18]Yeah, axial CL go 180° rotate over. I said okay, 3 C2 over.

[00:02:25]It's messy C2.

[00:02:28]Or it's messy C2 total 3 C2 over.

[00:02:33]Next, I got to take a sigma V.

[00:02:36]So is angles over plane pass over so yeah, I could CL is equal to the CL go reflect over or the axial CL is axial CL for reflect over.

[00:02:52]So I said okay.

[00:02:55]It does to be a plane over or it does to be a plane over total 3 sigma V over.

[00:03:02]3 sigma V Next is may Yeah, Joe equal to the plane Yeah, 3 CL one and P explain me.

[00:03:13]Or it's messy Joe C3 XT pass over it is the perpendicular Joe Yeah, Joe plane a equal to the plane over.

[00:03:25]So Yeah, perpendicular principal axis over.

[00:03:29]It's plane to home bullying is sigma V.

[00:03:31]So one sigma H over. So yes, I got symmetry element So it's the point group over. D 3 H I got sigma H absent over.

[00:03:40]So it's the point group over D3 D Okay, so yeah, statement correct here.

[00:03:47]It's may next statement over here. That two axial PCL bonds are longer than the So, PCL2 here PCL2 axial is this up PCL1 gets up 90° repulsion here.

[00:04:04]PCL3 gets up 90° repulsion here and PCL4 gets up 90° repulsion here.

[00:04:10]Or I don't know how equatorial PCL1 could get PCL1 equatorial.

[00:04:16]So, this gets up this PCL2 gets up 90° repulsion and PCL5 gets up 90° repulsion.

[00:04:25]PCL2 and PCL5 90° repulsion. And PCL1 PCL3 gets up 120° repulsion making 90° repulsion are stronger than 120° repulsion.

[00:04:37]So, axial bond experiences greater electron pair repulsion. To reduce this repulsion, the axial bonds becomes slightly longer and therefore slightly weaker.

[00:04:50]So, you see what I'm saying?

[00:04:53]Axial bond covalent data over equatorial bonds. So, it's got a experimental measurement they get axial PCL carbon length is approximately 2.19 angstrom or equatorial PCL carbon length is approximately 2.02 angstrom.

[00:05:17]Hence, the axial bonds are longer than the equatorial bonds. So, the statement two is correct.

[00:05:24]At room temperature, all five chlorine atoms becomes chemically equivalent on the animal time is killed due to rapid Berry pseudorotation.

[00:05:44]Total five chlorine phosphorus as far as the five chlorine environment same as that of >> During this process, actual chlorine atoms becomes equatorial. Equatorial chlorine atoms become actual.

[00:05:58]This exchange occurs continuously and very rapidly at room temperature. Now, consider what happens in animal experiment. Animal spectroscopy observes molecules over a specific time scale. If atoms exchange position must happen than the animal time scale, the instrument cannot distinguish between the different environment.

[00:06:23]Instead, it detects the average environment. As a result, all five chlorine atoms appear chemically equivalent and only one chlorine environment is observed. So, partial chlorine one signal I got. First statement to be correct. I got on temperature around minus 79° C got You exchange it not fast may you go to stay may I am uh those two animal signal will be there.

[00:06:54]3 CL equatorial got a lot may you go and those CL actual got a lot animal signal may you go.

[00:07:03]Next statement here, the difference between the actual and equatorial PCL bond length is primarily due to significant participation of phosphorus 3D orbital in bonding.

[00:07:14]So, it's the reason you need it.

[00:07:17]Earlier, chemists believed that phosphorus expanded it is obtained by using vacant 3D orbitals resulting in SP3D hybridization. However, modern theoretical and computational studies have shown that the contribution of phosphorus 3D orbital is extremely small and not the primary reason for bonding.

[00:07:40]Today, hypervalent molecules such PCl5 are better explained using molecular orbital theory three center four electron bonding. Here's your Cl axial phosphorus Cl axial is made three center four electron bonding or type or delocalized bonding models.

[00:08:01]The longer ex- axial bond are mainly a consequence of greater electron pair repulsion not significant participation of phosphorus 3D orbitals. Therefore, statement four is incorrect. Okay?

[00:08:15]So, correct option option A one, two, and three only correct. So, one, two, and three only correct.

[00:08:24]Next, question two. A second order reaction rate constant 0.20 L mol inverse s inverse. If the initial concentration of the reactant is 0.50 mol L inverse, the half-life of the reaction. The half-life of T half equal 1 upon rate constant into initial concentration.

[00:08:49]So, here we have half-life of the formula for second order reaction given. K value given there at 0.20 L mol inverse second inverse.

[00:09:04]And the initial concentration given here A0 = 0.

[00:09:10]50 mol L inverse. So, T half value given here T half = 1 upon 0.20 mol inverse L second inverse into 0.50 mol L inverse. So, L inverse L cancel mol inverse mol cancel which is second inverse which is second over second to calculate about 15 1 upon 0.20 into 0.

[00:09:44]50 second to you calculate about 10 second so correct answer option b 10 second question three which one of the following statement regarding the third law of thermodynamics is not correct not correct which side to get that mirror the entropy of a perfectly crystalline substance the perfect crystalline substance with that was the entropy at zero Kelvin equal zero with that provided that crystal has a unique ground state to your statement correct the third law permits the calculation of absolute entropy by integrating heat capacity from zero Kelvin to the desired temperature your statement be corrected since the entropy at zero Kelvin is known the absolute entropy at any temperature can be calculated using entropy equal zero to t CP by TDT plus summation Delta H transition divided by T transition okay so your statement be corrected to incorrect which side not correct so next statement according to the third law it is possible to reduce the temperature of a system to exactly zero Kelvin by a finite number of thermodynamic process your statement incorrect the un-attainability principle a consequence of the third law states that absolute zero Kelvin cannot be reached by any finite number of thermodynamic process. So, cannot over.

[00:11:36]Okay, so you option you option incorrect over. The answer over C incorrect which I So, incorrect option answer C.

[00:11:47]The degree that they exist that processing residual configuration disorder at zero Kelvin may have a non-zero entropy even at absolute zero Kelvin. So, your statement be correct here.

[00:12:00]You can see I got if a crystal has residual configurational disorder at zero Kelvin just say ice cup it possesses residual entropy. Okay?

[00:12:15]So, it it is entropy is not zero.

[00:12:19]Entropy at zero Kelvin not equal zero what I So, your statement be correct here. So, incorrect statement here option C.

[00:12:31]Answer C.

[00:12:33]Next question here question four the rate constant of a diffusion control reaction is directly proportional to the sum of the diffusion coefficient of the reacting species.

[00:12:45]If the diffusion coefficient of reacting reactants A and B are 2.0 into 10 to the power -9 and 3.0 into 10 to the power -9 m square second inverse respectively and the rate constant is at all. What will be the rate constant if the diffusion coefficient of reactant B is double while all other parameters remain unchanged. So, K2 the value calculate clearly which I So, it's clearly small choice scheme diffusion control rate equation K equal 4 pi R NA DA plus DB into 10 to the power 3 The question make a bullet the rate constant of diffusion coefficient reaction is directly proportional to the sum of the diffusion coefficient of the reacting species.

[00:13:55]Okay. directly proportional to DA plus DB lucky stuff constant A DB K A diffusion coefficient of reactant A reactant A and DB K A diffusion coefficient of reactant B So, initially initially DA and DB K value kitna tha DA K value the initially 2.0 into 10 to the power minus 9 meter square second inverse and DB K value tha 3.0 into 10 to the power minus 9 meter square second inverse or rate constant K value kitna tha K1 equal 5.0 into 10 to the power 9 liter mole inverse second inverse When the diffusion coefficient of B is doubled So, DB prime K value kitna hua 6.0 into 10 to the power minus 9 meter square second inverse value 3.0 into 10 to the power minus 9 is called double K value 6.0 into 10 to the power minus 9 and lucky stuff parameters same over to DA kitna over DA K So, divided by K1 DA prime plus DB prime divided by DA plus DB K1 5.0 into 10 to the power 9 2.0 plus DB prime got 6.0 into 10 to the power minus 9 and 2.0 plus DB got 3.0 into 10 to the power minus 9 next I got 6 plus 2 8 divided by 5 8 divided by 5 into 5 into 10 to the power 9 cancel out ho jayega so 8 into 10 to the power 9 option option B 8.0 into 10 to the power 9 liter mole inverse second inverse so correct option answer B question five in the preparation of salicylic acid from phenol by Kolbe Smith Smith reaction the reaction carried out using sodium phenoxide and CO2 under high pressure followed by acidification which of the following statement is correct the correct statement phenol so OH minus is a strong will take it.

[00:17:22]Here O- This O- conjugation will take it.

[00:17:39]O with this sodium plus and C double bond O here minus.

[00:17:48]Okay?

[00:17:49]Here minus will take it either CO2 will attack here.

[00:17:54]Here double bond O hydrogen or either C double bond O O- sodium plus.

[00:18:10]Here H3O+ will take hydrogen either with love acidification will take it.

[00:18:20]Or either see this hydrogen will attack here either double bond will take it aromatization will take it.

[00:18:29]So, aromatized here.

[00:18:36]Double bond O O- this O- hydrogen will take it. So, phenol say salicylic acid preparation will take it.

[00:18:50]This is phenol.

[00:18:53]This is salicylic acid.

[00:18:58]Okay?

[00:19:02]And this is sodium phenoxide.

[00:19:08]Okay?

[00:19:10]Option A, the reaction proceed efficiently with phenol itself.

[00:19:15]Formation of sodium phenoxide is not essential. This statement incorrect you keep keep phenol itself is much less reactive to what carboxylation formation of sodium phenoxide is the essential because the phenoxide ion strongly activate the aromatic ring to what electrophilic attack by by CO2. So, option A incorrect.

[00:19:41]Next option, carboxylation occur predominantly at the ortho position because the sodium ion coordinate with oxygen favoring ortho attack of CO2 under the reaction condition.

[00:19:55]The option correct here. In the Kolbe-Schmitt reaction, sodium phenoxide reacts with CO2 under high pressure. The sodium plus ions remains associated with the phenoxide oxygen which favor ortho for carboxylation.

[00:20:12]Yeah.

[00:20:13]Ortho carboxylation favor carbon.

[00:20:16]After acidification, the major product obtained is salicylic acid.

[00:20:21]Okay, so option B correct here.

[00:20:25]And option B answer here.

[00:20:27]Next option, the major product obtained after acidification is para-hydroxybenzoic acid due to lower steric hindrance at the para position.

[00:20:36]The option incorrect here. Although a small amount of para-hydroxybenzoic acid may be formed, the ortho isomer salicylic acid is the major product under the standard Kolbe-Schmitt reaction condition. So, option C incorrect.

[00:20:56]The reaction involves electrophilic substitution by the carbonate ion. Yeah, incorrect here. Electrophilic substitution carbon or CO2 may not carbonate ion pay. So, option D incorrect. So correct option answer B Okay.

[00:21:17]Question six general English say which statement correctly distinguishes linguistic from philology.

[00:21:26]Okay.

[00:21:27]Let's understand the difference carefully. Linguistic is the scientific study of language.

[00:21:35]So linguistics Okay.

[00:21:40]Scientific study of language.

[00:21:50]Okay.

[00:21:52]It examines how language are structured, how speech sound are produced, how words and sentences are formed, how meanings are conveyed, and how language are used in communication. You okay?

[00:22:06]Linguistics And on the other hand, philology focuses on philology focuses on the historical historical development of language.

[00:22:21]Development of languages.

[00:22:26]Okay.

[00:22:29]By studying ancient manuscript written text, it help us understand how language evolve over time.

[00:22:39]So Now, let's eliminate the incorrect option.

[00:22:43]Option A incorrect Linguistic is not limited to ancient languages. It studies both modern and ancient languages.

[00:22:56]Option B incorrect. Philology mainly studies written text, not spoken language. Okay, you be incorrect.

[00:23:06]Option D You be incorrect. Linguistic and technology are related field, but they are not the same. So, you be incorrect.

[00:23:17]So, correct option answer C. Linguistic scientifically analyze language, while technology focuses on the historical development of languages through text. Okay. Answer option C.

[00:23:32]Next, question seven. What is the meaning of the word predicament? So, the word predicament You do predicament is the meaning here difficult tricky or unpleasant situation. Okay.

[00:23:57]So, predicament coming here difficult, tricky, or unpleasant situation from which it is hard to find solution.

[00:24:06]For example I got a example here.

[00:24:11]After losing his wallet while traveling he found himself in a serious predicament.

[00:24:43]Okay. Now, let's eliminate the incorrect option.

[00:24:48]Option A incorrect.

[00:24:50]A predicament is not a pleasant sub- surprise.

[00:24:54]Option C incorrect. It does not mean a success or achievement.

[00:25:01]Option D incorrect. It is not related to an agreement or peace.

[00:25:08]Correct answer option B. So, a difficult or unpleasant situation. So, correct answer option B. Next question, question eight. Choose the correctly spelled word. The correct spelling is pre dece ssor.

[00:25:30]Okay?

[00:25:31]Which means a person who held a position or job before someone else.

[00:25:38]It's a meaning here.

[00:25:40]A person who held a position or job before someone else.

[00:26:00]Okay?

[00:26:02]So, it's the correct option here. B.

[00:26:05]Predecessor.

[00:26:07]Okay?

[00:26:08]Next question nine. Find the odd one out.

[00:26:13]Number eight 9 27 15 11 39 42. So, it's me say common here.

[00:26:19]Nine multiple of three. 27 multiple of three. 15 multiple of three. 11 not multiple of three. 39 multiple of three.

[00:26:30]And 42 multiple of three. So, 11 not a multiple of three. Since all the other number are divisible by three, 11 is the odd one out. Therefore, the correct answer is option A.

[00:26:46]11.

[00:26:48]Next question, question 10. The ratio of the ages of A and B 6 years ago was 2:3.

[00:26:55]Their present age ratio is 5:7.

[00:26:59]So, let the present ages of A and B A guy at 5 into X and B guy 7 X.

[00:27:08]So, 6 years ago 6 year ago kitna hoga?

[00:27:14]5 X say minus 6 hoga divided by 7 X say minus 6 ratio kya hai?

[00:27:20]2:3.

[00:27:22]Isko cross multiplying karne ke baad kya aayega? 3 into 5 X minus 6 equal 2 into 7 X minus 6.

[00:27:33]So, 15 X minus 18 equal 14 X minus 12.

[00:27:40]X ka value aayega 15 X minus 14 X equal X equal 6.

[00:27:47]So, the present ages are A guy at 5 into X ka value kitna hai? 6.

[00:27:55]30 years and B guy kitna aayega? 7 into 6 equal 42 years.

[00:28:04]So, after 3 years after 3 years A guy value kitna hoga? A guy age 33 years and B guy kitna hoga?

[00:28:18]42 plus 3 equal 45 years.

[00:28:25]Okay, therefore the correct answer is option A, 33 years and 45 years.

[00:28:33]Thank you so much. Video last tak dekhne ke liye. Yeh video bahut lengthy ho gaya, isliye main is idhar hi end kar raha hu. So, agar part 2 chahiye iske liye comment kar dena aur video ko like kar dena.

[00:28:47]So, is video mein agar theek thaak like aaya na favor like I am I will push it part two or three like I did thank you so much.