JEE Main: Did These Questions Have Wrong Answers In JEE Main Answer Key?

JEE Main: Did These Questions Have Wrong Answers In JEE Main Answer Key?

Updated on May 05, 2022 08:57 AM IST | #JEE Main

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.

JEE Main: Did These Questions Have Wrong Answers In JEE Main Answer Key?
JEE Main: Did These Questions Have Wrong Answers In JEE Main Answer Key?
LiveJEE Mains 2025 Paper 2 Answer Key LIVE: BArch final answer key PDF expected soon at jeemain.nta.nic.inMay 11, 2025 | 4:01 PM IST

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.

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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:

JEE Main 2021: Attempt 3

July 20, Shift 1

Question ID : 86435117255

JEE Main Answer Key: 172

1651642594093


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

1651642593666

____________________________________________________________________________

July 20, Shift 2

Question ID : 86435117528

JEE Main Answer Key: 1

1651642594816


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?

____________________________________________________________________________

July 22, Shift 2

Question ID : 86435117785

JEE Main Answer Key: 86435159952 (Option 2)

1651642595012


Question and Option IDs

1651642592072

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.

____________________________________________________________________________

July 25, Shift 1

Question ID : 86435118234

JEE Main Answer Key: 86435161299 (Option 1)


1651642595826

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)]

____________________________________________________________________________

July 25, Shift 1

Question ID : 86435118225

JEE Main Answer Key: 86435161264 (Option 3)

1651642595404


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.

1651642595160

Hence, Statement I is also correct in the given question.

Correct answer should be 86435161261 [Option (2)]

____________________________________________________________________________

July 27, Shift 1

Question ID : 86435118870

JEE Main Answer Key: 5

1651642594267


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.

____________________________________________________________________________

July 27, Shift 2

Question ID : 86435119236

JEE Main Answer Key: 300

1651642596272

The issue :

The answer in the JEE Main final answer key was wrong. The answer should have been 600.

1651642592994

____________________________________________________________________________

JEE Main 2021: Attempt 4

August 26, Shift 2

Question ID : 86435120226

JEE Main Answer Key: 50

1651642596053


The issue :

The answer given in the answer key is wrong. The answer should be 0.

1651642592640

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.

____________________________________________________________________________

August 27, Shift 1

Question ID : 86435120586

JEE Main Answer Key: 4

1651642593388

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

____________________________________________________________________________

August 31, Shift 2

Question ID : 86435121290

JEE Main Answer Key: 86435170463 (Option 4)

1651642595606

1651642591877

The issue :

Alcoholic NaOH causes dehydrohalogenation (elimination) and not substitution.

The answer should be 86435170462 (Option 2)

____________________________________________________________________________

August 31, Shift 2

Question ID : 86435121287

JEE Main Answer Key: 86435170452 (Option 2)

1651642594461

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

____________________________________________________________________________

September 1, Shift 2

Question ID : 86435121562

JEE Main Answer Key: 86435171281 (Option 4)

1651642594652

1651642591663

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.

____________________________________________________________________________

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.

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Questions related to JEE Main

Have a question related to JEE Main ?

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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