JEE Advanced 2026 Easiest Questions in Physics, Chemistry & Maths

JEE Advanced 2026 Easy Questions with Solutions

The easiest questions in JEE Advanced 2026 Physics Chemistry Maths were mostly concept-based and did not involve long calculations. Some questions were straight from basic ideas of VSEPR theory, dipole moment trends, optics, and standard integration. Below, we have covered the subject-wise easiest questions from JEE Advanced 2026. The set includes two Chemistry questions, one Physics question, and one Mathematics question.

JEE Advanced 2026 Easiest Chemistry Questions

Chemistry had some of the most direct questions in the paper. That is why many students felt that Chemistry had a better scoring chance compared to the other two subjects. The two questions below can be counted among the easiest problems in JEE Advanced 2026 PCM.

Question 1: Molecular Shapes

Consider the following species:

$\mathrm{SOCl}_2, \mathrm{XeOF}_4, \mathrm{ClF}_3, \mathrm{ClF}_5, \mathrm{XeF}_5^{+}, \mathrm{SO}_3^{2-}, \mathrm{XeF}_3^{+}, \mathrm{SF}_4$

List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

Options

(A) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 3$

(B) $\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 3$

(C) $\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4$

(D) $\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 4$

Detailed Solutions:

This question is based on the VSEPR theory. So the first step is to identify the shap of each species one by one.

Step 1: Find the shape of each molecule

• $\mathrm{SOCl}_2$ Sulfur is the central atom. It has one double bond with oxygen, two single bonds with chlorine, and one lone pair. That gives four electron domains. The molecular shape is trigonal pyramidal.

• $\mathrm{XeOF}_4$ Xenon is the central atom. It forms one bond with oxygen and four bonds with fluorine, along with one lone pair. The electron geometry is octahedral, and the molecular shape is square pyramidal.

• $\mathrm{ClF}_3$ Chlorine has three bond pairs and two lone pairs. Its shape is T-shaped

• $\mathrm{ClF}_5$ Chlorine has five bond pairs and one lone pair. Its shape is square pyramidal

• $\mathrm{XeF}_5^{+}$ Xenon forms five bonds and has one lone pair in this cation. The shape is square pyramidal.

• $\mathrm{SO}_3^{2-}$ Sulfite ion has three bond pairs and one lone pair on sulfur. Its shape is trigonal pyramidal.

• $\mathrm{XeF}_3^{+}$ Xenon forms three bonds and has two lone pairs. The shape is T-shaped.

• $\mathbf{S F}_4$ Sulfur has four bond pairs and one lone pair. The shape is a see-saw.

Step 2: Count the required shapes

See-saw: $\mathrm{SF}_4$ only → one

T-shaped: $\mathrm{ClF}_3, \mathrm{XeF}_3^{+} \rightarrow$ two

Trigonal planar: none → zero

Square pyramidal: $\mathrm{XeOF}_4, \mathrm{ClF}_5, \mathrm{XeF}_5^{+} \rightarrow$ three

Step 3: Match with List-II

$\mathrm{P} \rightarrow 1$

$\mathrm{Q} \rightarrow 2$

$\mathrm{R} \rightarrow 5$

$\mathrm{S} \rightarrow 3$

Final Answer: Option (A)

This was one of the JEE Advanced 2026 easiest questions because it directly tested standard molecular shapes. A student with basic VSEPR revision could solve it without lengthy calculation.

Question 2: Dipole Moment Order

The correct order of dipole moments for the given species is

(A) $\mathrm{BF}_3=\mathrm{NH}_4^{+}<\mathrm{NF}_3<\mathrm{NH}_3$

(B) $\mathrm{BF}_3<\mathrm{NH}_4^{+}<\mathrm{NF}_3<\mathrm{NH}_3$

(C) $\mathrm{NH}_4^{+}<\mathrm{BF}_3<\mathrm{NH}_3<\mathrm{NF}_3$

(D) $\mathrm{BF}_3<\mathrm{NH}_4^{+}<\mathrm{NH}_3<\mathrm{NF}_3$

This question depends on shape and cancellation of bond dipoles

Detailed Solutions:

This question depends on shape and cancellation of bond dipoles.

Step 1: Check $\mathbf{B F}_{\mathbf{3}}$

$\mathrm{BF}_3$ is trigonal planar and perfectly symmetrical. The three B-F bond dipoles cancel each other.

So, dipole moment of $\mathbf{B F}_3=\mathbf{0}$.

Step 2: Check $\mathbf{N H}_4^{+}$

$\mathrm{NH}_4^{+}$is tetrahedral and symmetrical. All four $\mathrm{N}-\mathrm{H}$ bond dipoles cancel.

So, dipole moment of $\mathbf{N H}_4^{+}=\mathbf{0}$.

Step 3: Compare $\mathbf{N H}_3$ and $\mathbf{N F}_3$

Both $\mathrm{NH}_3$ and $\mathrm{NF}_3$ are trigonal pyramidal. The difference comes from the direction of bond dipoles.

- In $\mathbf{N H}_3$, the N-H bond dipoles and the lone pair effect act in the same general direction. So the net dipole moment is higher.

- In $\mathbf{N F}_3$, the N-F bond dipoles act opposite to the lone pair direction to a large extent. So the net dipole moment becomes smaller.

Therefore,

$\mathrm{NH}_3$ has a greater dipole moment than $\mathrm{NF}_3$.

Step 4: Arrange all species

$\mathrm{BF}_3=0$

$\mathrm{NH}_4^{+}=0$

$\mathrm{NF}_3$ is small but non-zero

$\mathrm{NH}_3$ is greater than $\mathrm{NF}_3$

So the correct order is:

$\mathrm{BF}_3=\mathrm{NH}_4^{+}<\mathrm{NF}_3<\mathrm{NH}_3$

Final Answer: Option (A)

This was one of the most direct chemistry questions in the paper because it only required knowledge of geometry and bond dipole cancellation

JEE Advanced 2026 Easiest Physics Questions

Physics usually becomes difficult when a question mixes many concepts. But sometimes the paper contains one very direct conceptual question. The following optics question is a good example of that and clearly fits the JEE Advanced 2026 paper analysis easiest questions category.

Topic: Optical Effects

Question 1:

In List-I, four optical effects are mentioned. The physical phenomena of light which are essential to describe these optical effects are given in List-II. Choose the option which describes the correct match between the entries in List-I to those in List-II

Options

- (A) $\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 3$

- (B) $P \rightarrow 4, Q \rightarrow 2, R \rightarrow 1, S \rightarrow 3$

- (C) $\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 3$

- (D) $\mathrm{P} \rightarrow 5, \mathrm{Q} \rightarrow 4, \mathrm{R} \rightarrow 1, \mathrm{~S} \rightarrow 2$

Detailed Solutions:

This is a pure theory-based question from optics. The easiest way to solve it is to match each phenomenon separately.

Step 1: Match Aurora Borealis

Aurora Borealis occurs when charged particles from space excite oxygen and nitrogen atoms in the upper atmosphere. These excited atoms emit light.

So,

$\mathbf{P} \rightarrow \mathbf{5}$

Step 2: Match partially polarized sunlight

Partially polarized sunlight is mainly produced due to scattering by molecules in the atmosphere.

So,
$\mathbf{Q} \rightarrow \mathbf{4}$

Step 3: Match rainbow

A rainbow is formed due to refraction, dispersion, and internal reflection inside water droplets. Among the given options, the closest correct description is dispersion and reflection.

So,

$\mathbf{R} \rightarrow \mathbf{1}$

Step 4: Match dark and bright fringes

Dark and bright fringes are associated with wave optics. From the given list, the relevant phenomenon is diffraction.

So,
$\mathbf{S} \rightarrow \mathbf{3}$

Step 5: Write the full match

- $\mathrm{P} \rightarrow 5$

- $\mathrm{Q} \rightarrow 4$

- $\mathrm{R} \rightarrow 1$

- $\mathrm{S} \rightarrow 3$

Final Answer: Option (A)

This was among the easiest questions in JEE Advanced 2026 Physics Chemistry Maths because it was fully concept-based and did not involve any numerical work.

JEE Advanced 2026 Easiest Mathematics Questions

Mathematics in JEE Advanced is often time-consuming, so any short and direct integral question becomes important. This integral question from Paper 2 can easily be placed among the JEE Advanced 2026 exam easiest shift questions being discussed by students after the exam.

Topic: Definite Integral

Question 1:

The value of the definite integral

$\int_0^2 \frac{1}{3^x+3} d x$

is

- (A) $\frac{1}{2}$

- (B) $\frac{1}{3}$

- (C) $\frac{\log _e 3}{3}$

- (D) $\frac{\log _e 3}{2}$

Detailed Solutions:

Step 1: Substitute $t=3^x$

Then,

$t=3^x$

Differentiating both sides,

$d t=3^x \ln 3 d x=t \ln 3 d x$

So,

$d x=\frac{d t}{t \ln 3}$

Step 2: Change the limits

When $x=0$,

$t=3^0=1$

When $x=2$,

$t=3^2=9$

So the integral becomes

$I=\int_1^9 \frac{1}{t+3}\cdot\frac{1}{t\ln 3},dt$

$I=\frac{1}{\ln 3}\int_1^9 \frac{1}{t(t+3)},dt$

Step 3: Use partial fractions

We write

$\frac{1}{t(t+3)}=\frac{A}{t}+\frac{B}{t+3}$

Multiplying by $t(t+3)$,

$1=A(t+3)+Bt$

$1=(A+B)t+3A$

Comparing coefficients,

$A+B=0,\quad 3A=1$

So,

$A=\frac{1}{3},\quad B=-\frac{1}{3}$

Hence,

$\frac{1}{t(t+3)}=\frac{1}{3}\left(\frac{1}{t}-\frac{1}{t+3}\right)$

Step 4: Substitute back into the integral

$I=\frac{1}{\ln 3}\int_1^9 \frac{1}{3}\left(\frac{1}{t}-\frac{1}{t+3}\right),dt$

$I=\frac{1}{3\ln 3}\int_1^9\left(\frac{1}{t}-\frac{1}{t+3}\right),dt$

Step 5: Integrate

$I=\frac{1}{3\ln 3}[\ln t-\ln(t+3)]_1^9$

Now apply the limits:

$I=\frac{1}{3\ln 3}[(\ln 9-\ln 12)-(\ln 1-\ln 4)]$

Since $\ln 1=0$,

$I=\frac{1}{3\ln 3}(\ln 9-\ln 12+\ln 4)$

$I=\frac{1}{3\ln 3}\ln\left(\frac{9\times4}{12}\right)$

$I=\frac{1}{3\ln 3}\ln 3$

$I=\frac{1}{3}$

Final Answer: Option (B)

This was one of the JEE Advanced 2026 easy questions with solutions because the substitution was straightforward and the algebra after that was standard.

How Many Easy Questions Questions are there in JEE Advanced?

Many students ask, 'How many easy questions are there in JEE Advanced?' The exact number changes every year, but every paper usually contains a small set of direct questions that are easier than the rest.

In the present set, Chemistry had two very direct conceptual questions, while Physics and Mathematics had one each. This is why, in the JEE Advanced 2026 paper analysis of easiest questions, students are discussing these questions as clear scoring opportunities. Students searching for JEE Advanced 2026 exam easiest shift questions are mainly looking for such straightforward questions from Paper 1 and Paper 2.

Why were these JEE Advanced 2026 Questions the Easiest?

• These questions were easier because the idea behind them was clear.

• They came from topics that most students study again and again.

• The Chemistry questions were from basic chapters like molecular shape and dipole moment.

• A student who has revised well should be able to solve these questions easily without getting stuck.

• The question on Physics was simple as it involved common concepts of optics.

• The Maths question was also direct because the method was easy to see.

• These questions did not need long calculations.

• They also did not mix too many concepts in one question.

• The language of the questions was simple, so students could understand them quickly.

• That is why these questions became good scoring chances in the paper.

Key Takeaways from JEE Advanced 2026 Easiest Questions

• Even in a tough JEE Advanced paper, a few questions are always easier than the rest.

• Students should try to find such questions early in the exam.

• Solving easy questions first can save time.

• It also helps in building confidence during the paper.

• These questions show that basic concepts are very important.

• Students should not ignore simple and familiar topics during preparation.

• Easy questions are not always numerical.

• Some easy questions are theory-based and can be solved quickly with a clear understanding.

• Previous year papers help a lot in learning this skill.

• The main lesson is simple: good preparation and smart question choice can improve the score.