JEE Advanced Chemistry 2025 Analysis - Top Skipped, Wrongly Attempted and Correctly Solved Questions

JEE Advanced 2025 Chemistry Analysis - Paper 1 and Paper 2 Question-Wise Students Performance

Chemistry of JEE Advanced 2025 in both Paper 1 and Paper 2 is a perfect blend of simple and tricky questions. Given below a closer look at candidate wise performance on which types of questions were solved accurately, skipped frequently, or answered incorrectly from Chemistry.

Most Skipped Chemistry Questions (High Non-Attempt %)

Some JEE Advanced Chemistry questions were left unanswered. Analysing these non attempt questions helps identify the topics that are difficult to understand.

Paper 1

Paper 2

JEE Advanced Chemistry Paper 2 Question 7

Question 7 of Chemistry JEE Advanced is one of the key problems that tested both conceptual clarity and application skills

Solution:

Oxidation

Tetralin on strong oxidation with acidic permanganate forms phthalic acid, which on heating with ammonia forms phthalimide.

$\text{Tetralin} \xrightarrow[ \Delta]{KMnO_4/H^+} C_6H_4(COOH)_2$

$C_6H_4(COOH)_2 + NH_3 \xrightarrow{\Delta} C_6H_4(CO)_2NH + 2H_2O$

Thus,

$X = \text{Phthalimide}$

Gabriel synthesis

Deprotonation and alkylation:

$C_6H_4(CO)_2NH + KOH \rightarrow C_6H_4(CO)_2N^-K^+ + H_2O$

$C_6H_4(CO)_2N^-K^+ + RBr \rightarrow C_6H_4(CO)_2NR$

$Y = \text{N-alkyl phthalimide}$

Hydrolysis

$C_6H_4(CO)_2NR + NaOH \rightarrow C_6H_4(COONa)_2 + RNH_2$

$Z = RNH_2$

(A) $X$ and $Y$ contain carbonyl groups → True

(B) Carbylamine reaction:

$
RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O
$

Occurs only for primary amines, not $Y$ → False

(C) $Z = RNH_2$ reacts with Hinsberg reagent → True

(D) $Z$ is aromatic primary amine → False

$\boxed{A \text{ and } C}$

Hence, the correct answers are option 1,3.

Why most students skip this question?

1. First step to solve this question is Multi step Conversion where the reaction required linking oxidation → phthalimide formation → Gabriel synthesis → hydrolysis.

2. The key intermediate here is phthalimide, which is not a very common compound in everyday practice questions. Since many students had not memorized that naphthalene oxidation → phthalic acid → phthalimide, they struggled to identify “X” confidently.

3. In this question Option (B) tempted students with the carbylamine test, but the intermediate Y is not a primary amine. While option (D) suggested Z is an aromatic amine, but Gabriel synthesis doesn’t yield aryl amines.

4. This questions was placed among organic reaction sequence questions, which are usually time-consuming. Many aspirants skipped it to avoid losing time on an unfamiliar, lengthy pathway, preferring quicker physical or inorganic chemistry questions.

Most Wrongly Attempted Chemistry Questions (High Wrong Response %)

Some Chemistry JEE Advanced 2025 questions have a high percentage of wrong responses, which shows common misconceptions and calculation errors among candidates. Analysing these questions helps students understand the tricky concepts.

Paper 1

Paper 2

JEE Advanced Chemistry Paper 2 Question 14

Question 14: An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $\frac{X}{F} \times 10^3$ volts, where $F$ is the Faraday constant. The value of $X$ is $\_\_\_\_$.

Use : Standard Gibbs energies of formation at 298 K are :

$\begin{aligned}
& \Delta_f G_{\mathrm{CO}_2}^{\mathrm{o}}=-394 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \\
& \Delta_f G_{\text {water }}^{\mathrm{o}}=-237 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \Delta_f G_{\text {butane }}^{\mathrm{o}}=-18 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$

Solution:

Given

1. Combustion reaction of butane

$C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O$

2. Standard Gibbs free energy change

$\Delta G^\circ = \sum \Delta_f G^\circ(\text{products}) - \sum \Delta_f G^\circ(\text{reactants})$

$\Delta G^\circ = [4(-394) + 5(-237)] - [(-18)]$

$\Delta G^\circ = (-1576 - 1185) + 18$

$\Delta G^\circ = -2743 \ \text{kJ mol}^{-1}$

3. Number of electrons transferred

Carbon oxidation state change:

$-2.5 \rightarrow +4$

Increase per carbon:

$4 - (-2.5) = 6.5$

For 4 carbons:

$n = 4 \times 6.5 = 26$

4. Cell potential relation

$\Delta G^\circ = -nFE^\circ$

$E^\circ = \frac{2743 \times 10^3}{26F}$

$E^\circ = \frac{105.5}{F} \times 10^3$

$\boxed{X = 105.5}$

Hence, the correct answer is 105.5

Why many students got it wrong?

1. Combination of Gibbs energy calculations with electrochemistry concepts is asked here. Finding and applying the link between $\Delta \mathrm{G}^0$ of reaction and cell potential can be difficult.

2. Students often do mistakes in Balancing of reactions or Electron Count

3. Errors may arise in the step where $\Delta \mathrm{G}^0$ for the overall reaction is calculated using standard Gibbs free energies of formation.

4. This question involves multiple calculation steps like reaction balancing → $\Delta \mathrm{G}^0$ → electrons transferred → unit conversions, under time constraints students tend to skip detailed re-checks. Any small slip along the way compounds to an incorrect final value.

Most Correctly Solved Chemistry Questions (Higher Accuracy)

Some JEE Advanced 2025 Chemistry questions were answered correctly by the majority of candidates, showing clear understanding and strong command over fundamental concepts.

Paper 1

Paper 2

Distribution of Total Marks in Aggregate (all, qualified, and allotted candidates)

Below is the total marks distribution in Chemistry JEE Advanced 2025 that shows how candidates performed overall, including those who qualified and those who were allotted seats.

For All Candidates

For Qualified Candidates

For Seat Allotted Candidates

Most Difficult Questions of JEE Advanced 2025 Chemistry

Some Chemistry questions in JEE Advanced 2025 were difficult. These questions had high wrong response or non attempt rates.

Paper 1

The most difficult questions in JEE Advanced chemistry Paper 1 are Q13, Q11, Q10, Q9. Questions 9 and 11 have, on average, 75.35% wrong attempts and Question 13 is the least full-mark (0.48%) question.

Paper-2