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The Joint Entrance Examination (JEE) Main is conducted by the National Testing Agency (NTA), and every year, lakhs of students appear for the exam. It is the entrance test – or the preliminary selection round – for the top engineering colleges in the country including the Indian Institutes of Technology (IIT), National Institutes of Technology (NIT), Indian Institutes of Information Technology (IIIT) and many more government institutions.
Yes, candidates are allowed to challenge more than one question in the JEE Main answer key. For each challenged question, INR 200 is to be paid by the candidate. Objections can be filed through the JEE Main login page.
Clearly, JEE Main is critical. However, JEE Main 2021 had questions which were conceptually or factually ambiguous and some of them were simply incorrect, going by the JEE Main answer key. Last year, the exam was conducted in four phases – February, March, July and August – across 26 Shifts. The result was declared in September.
Here, Careers360 looks into the questions asked in the July and August shifts, and highlights some of the errors in the JEE Main previous year question paper’s Chemistry section. The necessary changes required for these questions were absent from the NTA JEE Main answer keys, which could have affected the scorecards of many candidates in the JEE Main result.
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These questions, along with their respective problems, are listed out below:
Question ID : 86435117255
JEE Main Answer Key: 172
The issue:
The unit in which Kp is to be reported has not been specified in the question.
So, a student solving for the unit as (Pa)-1 will get an answer of 0.172 which would be rounded off to 0 while the given answer will be in units of (kPa)-1
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Question ID : 86435117528
JEE Main Answer Key: 1
The issue:
There is a conceptual error in the question. A negatively-charged sol cannot be precipitated by a negatively-charged ion.
Also, the coagulating value is defined as the minimum concentration of an electrolyte in millimoles per litre required to cause precipitation of a sol in two hours.
So, there arises another problem: How do you extrapolate data for two hours from the one-hour data?
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Question ID : 86435117785
JEE Main Answer Key: 86435159952 (Option 2)
Question and Option IDs
The issue:
Multiple options given in the question (Options 1,2 and 3) are capable of forming tertiary alcohol on reaction with excess of Grignard reagent (CH3MgBr) followed by hydrolysis.
Option 1: Alpha- Beta unsaturated carbonyl compounds show direct addition to the carbonyl group with Grignard reagent. Hence, tertiary alcohols will be obtained from the carbonyl groups
Option 3: Grignard reagent can add to a ketone, forming a tertiary alcohol.
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Question ID : 86435118234
JEE Main Answer Key: 86435161299 (Option 1)
The issue:
Lewis acids are not used with Phenol because Phenol itself forms a complex with the Lewis acid, due to which the ring becomes deactivated towards Electrophilic Substitution.
In fact, this reaction of Phenol with a Lewis acid FeCl3 is actually used to detect the presence of Phenol in solution. Phenol forms a complex with neutral FeCl3 and produces a violet colour.
Correct answer should be 86435161298 [Option (3)]
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Question ID : 86435118225
JEE Main Answer Key: 86435161264 (Option 3)
The issue:
Be is not an alkaline earth metal as its oxide and hydroxide are Amphoteric in nature.
Here is an excerpt from NCERT textbook’s chapter on the s Block elements confirming it.
Hence, Statement I is also correct in the given question.
Correct answer should be 86435161261 [Option (2)]
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Question ID : 86435118870
JEE Main Answer Key: 5
The issue:
Data given is insufficient to solve the problem.
Molality (m) can be calculated from the density of solution only when the Molarity (M) is also given in the question.
In the question, Molarity has not been provided and thus the question cannot be solved.
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Question ID : 86435119236
JEE Main Answer Key: 300
The issue :
The answer in the JEE Main final answer key was wrong. The answer should have been 600.
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Question ID : 86435120226
JEE Main Answer Key: 50
The issue :
The answer given in the answer key is wrong. The answer should be 0.
The answer is 0.02 which would be rounded off to 0.
For the answer to be 50, the value of (Kc)-1 had to be asked.
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Question ID : 86435120586
JEE Main Answer Key: 4
The issue :
The answer given is 4 because the inner 4f electrons have not been counted.
The question does not say that we need to consider only the 5f electrons
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Question ID : 86435121290
JEE Main Answer Key: 86435170463 (Option 4)
The issue :
Alcoholic NaOH causes dehydrohalogenation (elimination) and not substitution.
The answer should be 86435170462 (Option 2)
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Question ID : 86435121287
JEE Main Answer Key: 86435170452 (Option 2)
The issue :
None of the options is correct.
All the given options contain permanganate ion (MnO4-) which contains Mn in its highest oxidation state (+7) and hence, it cannot show disproportionation reaction
Question should have been awarded as bonus
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Question ID : 86435121562
JEE Main Answer Key: 86435171281 (Option 4)
The issue :
Options 1 and 4 can both be correct but only Option 4 has been given as the correct answer.
Acenaphthene (compound in Option 1) is also Aromatic in nature.
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Apart from the above-mentioned questions, there were also some in the integer section that had very lengthy mathematical calculations and whose answers had to be rounded off to the nearest integers.
These questions are usually solved by approximations and it is very difficult for students to arrive at the correct integer answer. Previously, these questions had a correct range for the answer, which gave students some leeway to make appropriate approximations in calculations.
Now, with negative marking in the numerical response questions, this becomes an even bigger problem. Considering the high-stakes nature of the JEE Mains, even a few such errors can have significant ramifications for applicants and be deeply demoralising.
With 987 marks in IPE 2025 and an 80 percentile in JEE Main, you have a moderate to good chance of securing admission to the Computer Science and Engineering (CSE) program at SASTRA University, particularly through Stream 1 admissions.
With a 90.44 percentile in JEE Main, your chances of getting into Computer Science Engineering (CSE) at Amrita University depend on the campus:
Coimbatore Campus: The cutoff for CSE is usually around 95.5 percentile, so admission may be challenging but still possible.
Amritapuri Campus: The cutoff here tends to be around 96.5 percentile, making it less likely for you to get a seat.
Bengaluru Campus: The cutoff for CSE is generally around 94 percentile, so your chances are higher here.
With a 91.74 percentile and a rank of 122000 in JEE Mains 2025, as a female candidate, you may secure Computer Science in some reputed state-level or women-centric engineering colleges through home state or female quota.
With a score of 986/1000 in AP Board (91%) and 82.74 percentile in JEE Mains, you have a good chance of getting CSE in SASTRA University under Stream 1, which considers board marks primarily. Final admission depends on merit rank and seat availability.
Based on your Class 12th score of 986 and JEE Mains score of 82.74 percentile, admission to Computer Science Engineering (CSE) at SASTRA University may be competitive. It depends on the cutoff for that year, which varies based on overall performance and seat availability.
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