JEE Main 2026 April 8 Shift 2 Chemistry Question Paper is available here with answer key and solutions. NTA conducted the second shift of the day on April 8, 2026, from 3:00 PM to 6:00 PM.
- The JEE Main Chemistry Question Paper contains a total of 25 questions.
- Each correct answer gets you 4 marks while incorrect answers gets you a negative mark of 1.
Candidates can download the JEE Main 2026 April 8 Shift 2 chemistry question paper along with detailed solutions to analyze their performance and understand the exam pattern better.
JEE Main 2026 April 8 Shift 2 Chemistry Question Paper with Solution PDF
Also Check:
Calculate \(|\Delta H^{\circ}|\) for the following reaction (in kJ/mol):
\[ 2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(l) \]
Given data:
Statement-I: When temperature is increased from 298 K to 308 K, having \( E_a = 12.72 \, Kcal mol^{-1} \), Rate constant (K) is doubled.
Statement-II: For a first (1\text{st) order reaction \( A \rightarrow B \), the following graph is valid.
Given the standard electrode potentials: \\
\( E^{\circ}_{Fe^{2+}/Fe} = x \, \text{volts}, \, E^{\circ}_{Fe^{3+}/Fe} = y \, \text{volts} \). \\
Calculate \( E^{\circ}_{Fe^{2+}/Fe^{3+}} \) in volts.
Calculate the equilibrium constant for the following equilibrium:
\[ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \rightleftharpoons XYZ \]
Given: \[ 2XY \rightleftharpoons X_2 + Y_2, \quad K_1 = 2.5 \times 10^{-5} \] \[ \frac{1}{2} Z_2 + XY \rightleftharpoons XYZ, \quad K_2 = 5 \times 10^{-3} \]
Wavelengths of two photons are given as \(\lambda_1 = 3000 \, Å\) and \(\lambda_2 = 6000 \, Å\) respectively. Calculate the ratio of their energies \(\left( \frac{E_1}{E_2} \right)\).
Statement-I: 30% (w/w) methanol in CCl₄ have mole fraction of solvent equal to 0.33.
Statement-II: CCl₄ and methanol mixture show positive deviation from Raoult's law.
Statement-I: \( K_b \) is more than \( K_f \) for water.
Statement-II: When we add a non-volatile solute in water, elevation in boiling point is more than depression in freezing point.
CaCO\(_3\)(s) \(\rightleftharpoons\) CaO(s) + CO\(_2\)(g), \( K_{p1} = 8 \times 10^{-2} \)
C(s) + CO\(_2\)(g) \(\rightleftharpoons\) 2CO(g), \( K_{p2} = 2 \)
If the partial pressure of CO at equilibrium is \( x \times 10^{-1} \) atm, then find the value of \( x \).
Match the following:
Statement I: Among the following set of ionic species, [Cr³⁺, Mn²⁺]; [Ti⁴⁺, V²⁺]; [Sc³⁺, V⁴⁺] and [Co³⁺, Cu²⁺] three sets have both ions are coloured.
Statement II: Among the following set of ionic species [Lu³⁺, La³⁺]; [Ln³⁺, Ce⁴⁺]; [Yb²⁺, Eu²⁺] and [Nd³⁺, Sm³⁺] three sets have both ions diamagnetic in nature.
For the given coordination compounds:
The number of paramagnetic species are:
Statement I: d-Block elements utilize their only 3d electrons for bonding on the surface of catalyst.
Statement II: This leads to strengthening of bonds in the reacting molecules.
Consider the statements related to group 15 hydrides.
(A) Stability of hydrides decreases down the group.
(B) Reducing nature of hydrides increases down the group.
(C) Lone pair donating tendency increases down the group.
(D) H–E–H bond angle decreases down the group.
Question 14:
View Solution
Step 1: Explanation of Ionization Energies.
Ionization energy is the energy required to remove an electron from an atom in its gaseous state. The configuration plays a key role in determining the ionization energy. For atoms with the same principal quantum number, the order of ionization energy can be explained by electron shielding and the effective nuclear charge.
Step 2: Understanding configurations and IE.
- (A) ns²np¹: The element with this configuration has an electron in an unfilled p-orbital, which results in a lower ionization energy compared to a fully filled p-orbital (ns²np⁶). Therefore, its IE is lower. This corresponds to value (Q) 801 kJ/mol.
- (B) ns²np⁶: This configuration is highly stable because the p-orbital is fully filled. It has a higher ionization energy, as it requires more energy to remove an electron from a stable, filled orbital. Therefore, its IE corresponds to value (P) 2080 kJ/mol.
- (C) ns²: This configuration represents a noble gas, which is highly stable with a completely filled s-orbital. The ionization energy here is relatively low compared to other configurations, corresponding to (S) 899 kJ/mol.
- (D) ns²np³: This configuration corresponds to an atom with half-filled p-orbitals, which is relatively stable. Its ionization energy is moderate, and it corresponds to value (R) 1402 kJ/mol.
Step 3: Conclusion.
The correct matching based on the ionization energies is:
- A → Q (801 kJ/mol),
- B → P (2080 kJ/mol),
- C → S (899 kJ/mol),
- D → R (1402 kJ/mol).
Final Answer: (C) A → Q; B → P; C → S; D → R Quick Tip: When comparing ionization energies, note that fully filled or half-filled orbitals (such as ns²np⁶ and ns²np³) are more stable and generally have higher ionization energies compared to configurations with unfilled orbitals (such as ns²np¹).
Bromine trifluoride auto ionizes into BrF\(_2^+\) and BrF\(_4^-\). The geometry of these ions respectively are:
Benzene reacts with CO/HCl/AlCl₃ and give 'P'.
Benzene reacts with CH₃-C-Cl / AlCl₃ to give 'Q'.
P and Q react with each other in presence of NaOH / H₂O give product Z. Find number of π e⁻ in product Z.
Statement 1:
Boiling point order is in decreasing order due to decreasing Vanderwaal's forces.
Statement 2:
P-dichlorobenzene has more melting point value due to its symmetric structure than O-dichlorobenzene while it has less boiling point than O-dichlorobenzene.
How many dibromo products are possible for ‘B’ (including stereoisomer)?
Which of the following compounds give a positive neutral FeCl\(_3\) test?
Statement-1: During fractional distillation, higher boiling point will condense first.
Statement-2: In fractionating column, one which has higher boiling point value will be more in concentration above the fractionating column.
Which of the following statement(s) are incorrect?
Find total number of aromatic compounds from the given organic molecules.
Question 23:
View Solution
The reaction follows the process of reduction and hydrolysis. Let's break it down step by step:
Step 1: Reduction by NaBH_4.
Sodium borohydride (NaBH\(_4\)) is used for reducing carbonyl groups. The reaction involves the reduction of an amide (or possibly a nitrile) to the corresponding amine. Therefore, the first step reduces a carbonyl group to an alcohol.
Step 2: Reaction with aqueous NaOH.
The second step involves treatment with aqueous NaOH (\(\Delta\)), which induces hydrolysis. This step would convert any remaining amide group into a carboxylic acid group.
Step 3: Acidic hydrolysis.
The final step involves acidic hydrolysis (H\(_2\)O\(^+\)), which typically works on ester or carboxylate salts, leading to the formation of an alcohol or an acid.
Thus, the product formed corresponds to the structure in option (B), which contains both an alcohol and a carboxylic acid group.
Final Answer: (B) Quick Tip: NaBH\(_4\) reduces carbonyl groups (like in amides or esters) to alcohols, and NaOH hydrolysis helps in converting amides into carboxylic acids.
Which of the following statement is correct about Hoffmann’s bromamide degradation reaction?
(a) Alkyl amide do not react.
(b) Secondary amide do not form secondary amide.
(c) Ratio of NaOH and Br₂ is 4 : 2.
(d) Na₂CO₃, NaBr and H₂O also formed along with amine.
JEE Main 2026 Chemistry Exam Pattern
| Particulars | Details |
|---|---|
| Exam Mode | Online (Computer-Based Test) |
| Paper | B.E./B.Tech |
| Medium of Exam | 13 languages: English, Hindi, Gujarati, Bengali, Tamil, Telugu, Kannada, Marathi, Malayalam, Odia, Punjabi, Assamese, Urdu |
| Type of Questions | Multiple Choice Questions (MCQs) + Numerical Value Questions |
| Total Marks | 100 marks |
| Marking Scheme | +4 for correct answer & -1 for incorrect MCQ and Numerical Value-based Questions |
| Total Questions | 25 Questions |
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