NEET 2026 Question Analysis || Chemistry || Zawlbuk Zirna Run

Ajouté : 2026-05-15

[00:00:00]Foreistry subject.

[00:00:36]Identify the correct statement about CLF3 from the following options.

[00:00:57]Chemical bonding and molecular structure.

[00:01:25]CLF VCP, VBT.

[00:01:45]electron pair.

[00:01:51]Electron pair 10 by 2= 5 electron valence electron 7 minus S bonding divided by true equal to 4 by 2= true true equal half into V + M - C + A/ into C LF3 chlorine valence electron into five SP B3 Doneus hybrid 5 - 3= True.

[00:03:55]Electron 90° 1 2 3 4 5. This one 90° VPR2 reposulate. Last point number seven repuls 90° 180°.

[00:05:07]ABC.

[00:05:14]It has DC with two lone pair on central atom.

[00:05:25]Sorry question.

[00:05:42]Fore formula formal charge FC formula directly. oxygen.

[00:06:23]Number one means oxygen Number two oxygen numberal Number one 0 + one.

[00:07:44]Question number classification of ele elements and periodicity in properties.

[00:08:04]down to phosphorus.

[00:08:34]Decrease silicone phosphorus.

[00:09:03]Sodium, magnesium, berilium, silicone was for us.

[00:09:22]Question incorrect statement.

[00:09:34]Oh 1072 scientist relationship.

[00:10:11]Alenc 5.

[00:10:32]oxidation state.

[00:10:53]X = + 2 + 1 X = + 3 oxidation state one 2 3 4 5 The largest and smallest city are Mg2 Mg2+ Number two group Number 13.

[00:12:55]Incorrect statement.

[00:13:02]Oh.

[00:13:12]A bulb is rated at 150 watt converting 8% energy into light if energy of one photon is 4.42 - 19. How many photons are emitted by the bulb per second?

[00:13:26]Structure of structure oftal energy.

[00:13:53]Single energy one 4 true 10 - 19 8 500 into 150. What an uh true uh 3 to 4 means 4 into 3 to 12 / 4.4 for two 71 into 19 answer to 2.71 into 10^ 19 structure of question to L= 0 1 2 3 quantum number S P D F S S F D 4 S 4 S 5 F 5 F option A2 B3 A2 B3 C4 C4 He can answer to Renee number one now. Oh, next.

[00:16:09]Next.

[00:16:41]Rotation of reactus minus Variation of concentration of reactant with time for zero order reaction.

[00:17:33]Activation.

[00:17:46]First order. Activation activation energy equation natural loal minus e a by rt expression.

[00:18:26]Okay.

[00:18:29]14.34US 1.25 into 10^ 4 by E A R= 1 25 into 10^ 4us this one activation energy equal to 1.25 into 10^ 4 1.987 1.987,000 means 1000 Oh 1.25 1.987 4.84 2.84 84.87in zero order. Uh per liter per second order second inverse second order inverse second inverse order A4 B3 A4 B3 Three. C1. C1 chapter.

[00:21:21]Electrochemistry.

[00:21:28]table.

[00:21:42]SH vol H through H E not H or H+ H true in electron one H+ plus one in true E cell equal to E not cell minus 0.059 059 by N oxidation by reduction 0.059 059 true atmospheric pressure 0.02.02 2 H+ square by hydrogen pressure.

[00:23:44]Next step 59 by 2 H+ H 0.0 square pressure of Halal 0.059 0592 minus 0.03 oh 03 0.02 04 0.03 002 oneus 0.03 03 log 2 into 10 power - 4 - 0.03 logus log 10 - 4us 0.03 03 look 0 3 0 1us 4 - 0.03 4 0.03US 310us 699.

[00:25:32]This one 0109 vol 0.109 volt can answer. Oh, next.

[00:25:52]Oh, ready.

[00:26:13]Forefus.

[00:26:43]deposit.

[00:27:03]Qal I2 1.5 Time to I into by NF3 into 900.

[00:27:45]electron.

[00:27:49]Oh 96 487 5 6 7 0 0 by four.

[00:28:06]>> Yeah. Uh 16 uh 16 17 8 9 12 18 19 0 9 3 09 38 Eight.

[00:28:40]Last question.

[00:28:59]structure of DNA poses double helix DNA poses double strand helic structure and contain tying as one of the four bases.

[00:29:44]2026.

[00:29:46]Organic chemistry.

[00:29:50]Organic chemistry.

[00:30:02]Fore question number 53. Question number 53.

[00:30:33]53. Select the reagent that reduce nitral through primary amines. Primary amines reduction.

[00:31:01]Feclic hydrogenation nickel platinum platium lithium aluminium hydide sodium borohyden.

[00:31:37]Aluminium 2850 reaction.

[00:32:04]this one true bromine degradation convert.

[00:32:24]This one controversial reducing nitro NCH2 group nitro groupite cyanide snin estas Oh, Stephen reduction.

[00:33:26]This one can answer.

[00:33:29]A C D only.

[00:33:48]The number of chlorine present in the organic product organic compound.

[00:34:05]Last scholarship.

[00:34:26]BHC.

[00:34:28]Hex chloride hexacloride UV lightenzine hexacloride Hydra3itution.

[00:35:29]This one UV light.

[00:35:42]This oneutrient 660 This one. Next question.

[00:36:16]Next question.

[00:36:36]particular formula.

[00:36:38]This one red orange dnping solution. does not produce failing solution. Failing solution.

[00:37:12]Failing solution.

[00:37:40]Option number four solution unreduced. oxidation oxidiz aromatic productidation.

[00:38:43]Sodium bicarbonate.

[00:38:46]Sodium carbon dioxide evolution.

[00:39:04]Oh, Naenic.

[00:39:26]Number four.

[00:39:28]Number four. Number two. Number one, CH3.

[00:39:49]CID acid failing solution oxidation questions.

[00:40:43]This one.

[00:40:46]This one.

[00:41:14]reaction.

[00:41:29]Are you naming Three ethile five methile metropain Hello.

[00:42:11]Alcohol P3.

[00:42:28]This one alcohol primary alcohol PCL5 react to P3 H3 P S2 P3 this one.

[00:43:13]This one C65 antimarov hydrogen Next one. Hey match list one and least trueenhide only.

[00:44:28]Cumin H2O reduction reduction.

[00:44:55]This one alcohol oxidation alcohol primary alcohol secondary alcohol dehydration of alcohols H2S4 of alcohol hydration of dehydration of alcohol.

[00:45:27]This one H2SO4 A3 C4 A2 B3 C4 D1 D1.

[00:46:18]So A2 B3 C4 D1.

[00:46:32]Next, the pair of molecule that are metamars among the following is This one got one example.

[00:47:36]group.

[00:47:49]The functional group that can be identified for resto.

[00:48:18]This one indicator chemistry.

[00:48:30]This one is phenol.

[00:48:52]Question number 78.

[00:48:54]78.

[00:49:10]Organic basone.

[00:49:30]This one in single bone. Single bond.

[00:49:46]This one C2 H4 CH2 CH2.

[00:49:54]This one This one.

[00:50:08]This one.

[00:50:20]A4ig.

[00:50:46]This one.

[00:51:15]The major product Z form in the following sequence of reaction is UV light reaction.

[00:51:41]This one NH2 primary amines. This one to primary amines 0 to 5° This oneable CH3 unstable.

[00:52:37]CH3 CH2Ove 300.

[00:52:59]This one can answer.

[00:53:07]This one direct question. Overall reaction overall reaction.

[00:53:39]This one carbon monoxide hydrogen gas.

[00:53:54]This one carbon carbon monoxide question.

[00:54:33]The following true reaction gives the same false smelling product to false foram.

[00:54:57]Oh, NC Hey group membation membride degradation one ICO Rearrangement C2H5 migration.

[00:55:58]This one, NC25, NCO compound C through H5 C.

[00:56:32]This one through H5.

[00:56:57]product.

[00:57:13]Second product on CH3 substitution reaction electrofilic substitution reaction. Fredal craft alkyation.

[00:57:42]This one to dilute HNO3 HSO4 nitration.

[00:57:46]Huh? Nitration electrofile NO2 plus nitroion.

[00:57:52]This one King Auto products.

[00:58:25]Purification sublimation.

[00:58:42]This one fractional distillation long column tube distillation.

[00:58:53]Sublimation Yes. Oh, so guest explanation.

[00:59:29]I don't explain.

[00:59:50]explanation WhatsApp.

[01:00:19]one question number 48 in start 298 kelvin assertian buffer solution contains equal concentration of X minus and HX KB of KB4 X minus is 10 to Buffer solution.

[01:00:44]Buffer solution. Equal concentration.

[01:00:47]Equal concentration.

[01:00:56]In case of buffer solution having equal acid and equal base equal to P K A in true KB is equal to 10 the power may + 14 a into 10 ^ - 10 is equal to 10 ^ -4 k a= 10 ^ -4 by 10 ^ - 10 the power may pa PH is equal PH + P is equal to 14.

[01:02:44]K A BK Ka concentration.

[01:03:00]PH is equal to pKa.

[01:03:05]PH is equal to PH= + 4 option answer.

[01:03:22]Next the ion given below are certain reaction. Identify the reaction for which KP is not is not equal to KC.

[01:03:33]equilibrium.

[01:03:36]Kal Ctal Kal.

[01:04:01]Oh. Number one, one Next one. During las the element present in an organic compound are converted from class uh practical chemistry class organic compound.

[01:04:58]Unknown organic compound.

[01:05:25]To convert to convert coalent from answer to Next mixture from chloroform and acetone form a solution with negative deviation from law due to law positive stronger intermolecular force Strongce hydrogen bonding between hydrogen bonding.

[01:06:38]In a test tube containing a salt, a few drops of dilute H2SO4 was added.

[01:06:58]H2SO4 H2SO4 vapor having this smell of vinegar.

[01:07:10]Vinegar vinegar acetone diluted acetone. So more 2% 3% of acetone in about 100 ml of water.

[01:07:37]formula.

[01:07:42]Few drops of H2SO4 CH3 CH3 E O Hal S4 2 minus SO4US number Identify the incorrect glow number Number one linkage decreases number one. Number two, phosphorus form dy bond with transition metal.

[01:09:16]element overlapping nitrogen can form deep pi bond with oxygenoc location.

[01:09:49]Element second.

[01:10:05]Second.

[01:10:10]Third location.

[01:10:27]Number four. Nitrogen can form multiple bone. Yes, it can form.

[01:10:40]Next qualitative analysis.

[01:10:56]In qualitative analysis, B3+ is detected by appearance of precipitate of B.

[01:11:02]the pH when the following exists.

[01:11:11]B I O gives B IO.

[01:11:32]K2 power is equal to concentration of B IO plus and concentration of O may Our minus 10 is equal to s=<unk> / 4 into 10 ^ - 10 power of minus PH concentration of O PH. H connect to PH is equal PH plus P is equal to 14 P oh P is equal to minus Lo 2 into 10 the power - 5 Hal for 2 plus 10us 2us Five.

[01:13:58]Lookus five. more. Look 2 + look 2 + 10 to the power - 5 - 5 - 0.3 0.310 5 - 0 5 - 0.30 5.00 0 - 0 - 030 4.7 pH is equal to pH is equal to 4.7 pH plus 4.7 is equal to 14 pH is equal to 14 - 4.7 14 - 4 72 10 - 7 3 - 14 9.3 9.3 9.3 2 S2 concentration of O concentration of minus O minus 2 P= 14 Mayority is equal to number of moles of solute by mass of solvent kg.

[01:17:21]Numbers of moles.

[01:17:23]Numbers of moles of solute.

[01:17:31]Numbers of moles of solute.

[01:17:34]Numbers of moles of solute.

[01:17:36]Oh, numbers of moles of solute formula.

[01:17:43]mass by mass. Oh, numbers of moles of solute mass of 35.

[01:18:21]How will we do that? by multiplied by 1,00 2.5 into by 60 into 75 into 1,000 520 0.5 number oneused one Number of moles of solute by volume of solution.

[01:19:49]Volume of solution.

[01:19:57]Mass by volume of solution.

[01:20:17]Volume of solution.

[01:20:20]Volume of solution 450,000 40 uh 450 1,000 true 5 by 2 9 are 18 0.27 True.

[01:21:31]Number three, aquatic species are more comfortable in cold water.

[01:21:39]Aquatic animals. They are more they are more comfortable in in cold water.

[01:22:01]In cold water oxygen solubility is increased.

[01:22:19]solubility of gas increase with decrease in Henry's law if increase in pressure thenility will also increase for next for a binary mixture of A and be the number of moles of A and B are in A and N B respectively. The fraction the mole fraction of B will be equal to NB numbers of moles of B by fraction of A.

[01:23:37]D and E A B C only answer.

[01:23:48]So next although plus three oxidation state is most common in lanoids serium will show plus4 oxidation state.

[01:24:01]Uh plus three.

[01:24:19]Atomic number 58.

[01:24:25]Electronic configuration.

[01:24:27]Serium electronic configuration 2. What is it?

[01:24:33]X E uh 4 F1 5 D1 62 C E Plus for stable configuration for F noble guest uncover.

[01:25:30]willing 4.

[01:25:46]After losing one or more electron, it acquires four f4 f14 electronic configuration. After losing one or one more electron, it acquires no way 44 configuration.

[01:26:04]It is nearest it nearest inner gas is no it's after losing one or more electron it acquires four f0 configuration Identify P oxygen exhibits only minus true oxidation state. Hey, he 80% of the time oxidation states oxygen oxidation states peroxide. H2O2 F2 plus number one.

[01:27:43]Number two, very large carbon has a bond to form multiple bond with itself.

[01:28:06]Number four number B3 L3 B A B3 One, two, three.

[01:28:47]1 2 3 A C L C Lillain is used as an indicator for titration.

[01:29:21]is used an indicator for titration of sodium hydroxide. Sodium hydroxide base solution again. Standard solution of solution changes observed pH close to the equivalence point during the titration.

[01:29:50]It is colorless when it is acid and neutral. Oh, neutral solution.

[01:30:26]pink.

[01:30:31]So next I don't calculate the spin only magnetic moments T of titanium plus 2 plus 3d2 unknown sorry unpaired electron 3d2 unpaired electronic momental = root over n + 2 where n is the unpaired electrons. One mu is equal to 2 2 + 2 root over 2 4 2 4 2 sorry 2<unk>2 uh 2<unk>2 into 1.4 uh 2.8 A B number.

[01:32:06]Next one. Question number 89.

[01:32:16]Number one, venadium oxide.

[01:32:35]S O2 S O2 in H2 SO4 acid concentrated next E in preparation of NH3.cess process N2 plus H2 gives NH3 PD2 PDL2 number four wers process waker process oxidation oxidation B14 polymerization of alkine.

[01:34:00]mistakes and shape.

[01:34:29]A P C NH3 X plus C.

[01:34:59]NH3 + 2 oxidation PT configuration 5D9 6S1 PT2 plus one electron 5D961 5D 1 D orital 5D 5D 6S 6P 5D 5D 1 2 3 4 Five.

[01:36:02]Washington Square 5D Low spin field 2 3 4 5 6 7 8 5 D S Dsp.

[01:37:05]Easy. Easy.

[01:37:08]Square.

[01:37:15]Option two.

[01:37:29]Coordination number n3 6 CL total x + 3 oxidation state.

[01:38:07]This one.

[01:38:09]Cobalt 3+.

[01:38:13]Cobalt 3 plus one. Iron 26. Cobalt 27.

[01:38:17]Cobalt 3. 3D 7. 3D 6.

[01:38:33]Low speed. High speed.

[01:38:45]Low speed.

[01:38:47]One, two, three, four.

[01:39:15]metal atom oxidation state 1 2 3 4 5 6 it's one d orbital D2 SPG hybridization lowe number always.

[01:40:05]B2 B1 A3 B1 N4 C4 F5 Oh, nicelium.

[01:40:59]Hey Fea enter oxidation X co neutral 3D 3D 64s trueation FD64s 3D orital 3D 4S 4prochemical Strong oxidation state 1 2 3 4 5 6.

[01:42:34]This one or one, two, three, four, five. D3.

[01:42:53]Okay.

[01:42:55]Option number three. Answer.

[01:43:17]Direct formula.

[01:43:32]Internal energy change entropy change 44 kel 298 298 equation.

[01:44:05]Delta G for delta G equals delh S for Temperature internal energy equation provide one equation.

[01:45:08]That's easy.

[01:45:19]Okay.

[01:45:42]Delta N2 Ng2 Number of moles of gaseous product May Gant two moles of gas reactant.

[01:46:32]Hu ng rtus 10 kilo par -1 into r value hey 8.31 8.31 8.31 uh uh 8.31 jewel per kelvin per 298 kelvin temperature.

[01:47:19]First line.

[01:47:23]First line.

[01:47:39]Jeweler.

[01:48:00]Fore 8.314 8.31 left 298* 1 sorry 8.31 into 10 - 3 kilogj - 10 - 8.31 into 29826US 1us 2.476US 12.47 476 kilo per mole.

[01:49:13]Free energy change.

[01:49:26]So delus 12 476 kiloj 28us 44 Four.

[01:50:19]12.476. 476 into 298 into 44 into 104.

[01:50:43]Oh, 44. 48 are 32. Four nines are heart.

[01:50:48]The sixth heart nine. Fours are 8 9 10 11 2 9 1 1 2 9 10 11 9 10 11 9 10 11 1 2 3 1 1 3 1 1 2 1 - 12.476 476 plus 13.11 positive numbers positive 0.63 63 63 56 approximately 63 5 kilogj 8.31 11 8.31 4 0.635 635 positive number nonspontaneous option number one nonspontaneous Hello.

[01:52:57]Almost always equal.

[01:53:05]Next question.

[01:53:20]The following book.

[01:53:27]NH32n square A DB type DSP square isometric.

[01:54:14]C.

[01:54:18]C.

[01:54:24]Bordation number one example basic image.

[01:55:04]optical A3 B1 B1 N O2 N2 oxygen Isomer Electronage is linkage.

[01:56:11]C4 option number C4 True.

[01:56:26]Example text.

[01:56:55]book.

[01:57:20]internal energy.

[01:57:24]Change in internal energy.

[01:57:30]First law of thermodynamics.

[01:57:37]500 Jew is absorbed by the system.

[01:57:42]500 of 500 absorb 500 Jew and work of 200 Jew is done by the system and by the system 500US 200 to 300 Jewy in the form of Heat 500 system work 5000.

[01:59:28]forion.

[02:00:07]DM here on the big one.

[02:00:12]DM one liter is equal to 1 dmq.

[02:00:25]Yeah, I think question foreign CO2 lit.

[02:01:24]Red hot coke.

[02:01:27]Hot reaction.

[02:01:41]CO2 plus hot.

[02:01:54]ratio is one moles of CO and chain quantitative relationship of CO and ratio 1 is to1 combine I don't want STP 1 by 22.4 law of combining volume of combining volumes.

[02:03:10]Okay.

[02:03:20]One mole is one. One lit is one.

[02:03:25]Given temperature one after complete reaction 1.4 4 L C2 L.4 L 1.4.

[02:04:52]Let volume of CO2 that reacted Xain remaining CO2.

[02:05:23]One is to one gives two.

[02:05:52]H2 AC according to CO2 one lit CE fortune.

[02:06:19]L 2 X L C O one to Total volume of gous mixture 1.4 for question 1.4 simplify 2x - x2 1.4 4 - 1 to 0 0.4 X 0.4 CO2 one L 0.4 L C 0.4.8 H2 1us 0.4 0.6 C2 0.6 CO2 one 0.8 C 0.8 C option number 40 answer complicated headlight.

[02:08:48]CO2 plus C gives 2 CO.

[02:08:57]volume one.42 Lording to C 1.4 x 0.4 0.4 into 0.4 4 0.8 co 1 - x 1 0.42 0.6 DMT ladies. Next question.

[02:11:15]A 5 4 g of hydrogen NH through CO NH2 in one molecule of ura hydrogen NH2 CH2 H2 H2 H atom of four H at hydrogen atoms one one molecular four hydrogen atom therefore or one mole of ura 4 5.4 g of ura 5.4 4 5.4 5.4 4 gass 5.4 by 60.0.09.09 09 4 into 0.09 0.36 moles of hydrogen atoms number of particry 6.02 022 one 0.36 0.36 into 6.022 into 10 ^ 23 2.168 into 10^ 23 2.169tus option number 2.168 mass Example, C242in EDTAC moch.

[02:15:51]Electric moment.

[02:16:11]for WhatsApp.