JEE Main 2026 April 2 Shift 1 Chemistry Question Paper is available here with answer key and solutions. NTA conducted the first shift of the day on April 2, 2026, from 9:00 AM to 12:00 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 2 Shift 1 chemistry question paper along with detailed solutions to analyze their performance and understand the exam pattern better.
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JEE Main 2026 April 2 Shift 1 Chemistry Question Paper with Solution PDF |
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The mass of iron converted into \(Fe_3O_4\) by the action of \(18\) g of steam is: (Given: Molar mass of H, O and Fe are \(1, 16\) and \(56\) g mol\(^{-1}\) respectively). Assume iron is present in excess.
What is the energy (in J atom\(^{-1}\)) required for the following process?
\[ Li^{2+}(g) \rightarrow Li^{3+}(g) + e^{-} \]
(Take the ionization energy for the H atom in the ground state as \(2.18 \times 10^{-18}\) J atom\(^{-1}\)).
Given below are two statements:
Statement (I): The correct sequence of bond lengths in the following species is: \[ O_2^+ < O_2 < O_2^- < O_2^{2-} \]
Statement (II): The correct sequence of number of unpaired electrons in the following species is: \[ O_2 > O_2^+ > O_2^- > O_2^{2-} \]
In the light of the above statements, choose the correct answer.
Consider the following equations:
(i) \[ 2Al(s) + 6HCl(aq) \rightarrow Al_2Cl_6(aq) + 3H_2(g) + 1200 \, kJ/mol \]
(ii) \[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) + 164 \, kJ/mol \]
(iii) \[ HCl(g) + aq \rightarrow HCl(aq) + 83 \, kJ/mol \]
(iv) \[ Al_2Cl_6(s) + aq \rightarrow Al_2Cl_6(aq) + 663 \, kJ/mol \]
The enthalpy of formation of anhydrous solid \(Al_2Cl_6\) is:
19.5 g of fluoro acetic acid (molar mass = 78 g mol\(^{-1}\)) is dissolved in 500 g of water at 298 K. The depression in the freezing point of water was \(1^\circ C\). What is \(K_a\) of fluoro acetic acid? (For water, \(K_f = 1.86\, K\,kg\,mol^{-1}\)). Assume molarity and molality to have same values.
The solubility product constants of \(Ag_2CrO_4\) and \(AgBr\) are \(32x\) and \(4y\) respectively at 298 K. The value of
\[ \left(\frac{molarity of Ag_2CrO_4}{molarity of AgBr}\right) \]
can be expressed as:
An electrochemical cell is constructed using half cells in the direction of spontaneous change
\[ Fe(OH)_2(s) + 2e^- \rightarrow Fe(s) + 2OH^- (aq) \qquad E^\circ = -0.88\,V \]
\[ AgBr(s) + e^- \rightarrow Ag(s) + Br^- (aq) \qquad E^\circ = +0.07\,V \]
Which of the following option is correct?
\(t_{100%}\) is the time required for the 100% completion of the reaction while \(t_{1/2}\) is the time required for 50% of the reaction to be completed. Which of the following correctly represents the relation between \(t_{100%}\) and \(t_{1/2}\) for zero and first order reactions respectively?
Given below are two statements:
Statement (I): The first ionisation enthalpy of the elements Na, Mg, Cl and Ar follows the order \[ Na > Mg > Cl > Ar \]
Statement (II): Among Ca, Al, Fe and B, the third ionisation enthalpy is very high for Ca.
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement (I): Oxidising power of halogens decreases in the order \[ F_2 > Cl_2 > Br_2 > I_2 \]
which is the basis of the “Layer test”.
Statement (II): “Layer test” to identify \(Br_2\) and \(I_2\) in aqueous solution involves the oxidation of bromide or iodide into \(Br_2\) or \(I_2\) respectively with \(Cl_2\), which is a type of displacement redox reaction.
In the light of the above statements, choose the correct answer:
Which of the following sets includes all the species that will change the orange colour of \(K_2Cr_2O_7\) in acidic medium?
Match List-I with List-II.
List-I (Chromium(III) complexes, en = ethylenediamine)
A. \([Cr(CN)_6]^{3-}\)
B. \([CrF_6]^{3-}\)
C. \([Cr(H_2O)_6]^{3+}\)
D. \([Cr(en)_3]^{3+}\)
List-II (\(\Delta_0\) in cm\(^{-1}\))
I. 15,060
II. 17,400
III. 22,300
IV. 26,600
View Solution
Concept:
Crystal field splitting depends on ligand strength.
Spectrochemical series:
\[ CN^- > en > H_2O > F^- \]
Thus the splitting energy order:
\[ [Cr(CN)_6]^{3-} > [Cr(en)_3]^{3+} > [Cr(H_2O)_6]^{3+} > [CrF_6]^{3-} \]
Step 1: Arrange \(\Delta_0\)
Largest \(\Delta_0\):
\[ 26,600 \]
Smallest \(\Delta_0\):
\[ 15,060 \]
Step 2: Match complexes
\[ [Cr(CN)_6]^{3-} \rightarrow 26,600 \]
\[ [Cr(en)_3]^{3+} \rightarrow 22,300 \]
\[ [Cr(H_2O)_6]^{3+} \rightarrow 17,400 \]
\[ [CrF_6]^{3-} \rightarrow 15,060 \]
Thus
\[ A-IV,\ B-I,\ C-II,\ D-III \]
\[ \boxed{A-IV,\ B-I,\ C-II,\ D-III} \] Quick Tip: Ligand field strength increases in the order \(F^- < H_2O < en < CN^-\).
Given below are two statements:
Statement (I): 1,2,3–Trihydroxypropane can be separated from water by simple distillation.
Statement (II): An azeotropic mixture cannot be separated by fractional distillation.
In the light of the above statements, choose the correct answer from the options given below:
Given below are two statements:
Statement (I): Benzyl chloride reacts faster in \(S_N1\) mechanism than ethyl chloride.
Statement (II): Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance.
In the light of the above statements, choose the correct answer from the options given below:
In IUPAC names the correct order of decreasing priority of functional groups is:
For the given molecule \(X\), the preferred site for the attack of the electrophile is:
Match List–I with List–II.
List–I (Mixture of Compounds)
A. Diethyl amine + Ethyl amine
B. Acetaldehyde + Acetone
C. Ethanol + Phenol
D. Benzoic acid + Cinnamic acid
List–II (Reagent used to distinguish)
I. Bromine water
II. \(CHCl_3 + KOH,\ \Delta\)
III. Neutral \(FeCl_3\)
IV. Ammoniacal silver nitrate
Choose the correct answer.
View Solution
Concept:
Different functional groups respond differently to specific qualitative reagents.
Step 1: A. Diethyl amine vs Ethyl amine
Ethyl amine is a primary amine and gives the carbylamine test.
\[ CHCl_3 + KOH,\ \Delta \]
Thus
\[ A \rightarrow II \]
Step 2: B. Acetaldehyde vs Acetone
Acetaldehyde gives Tollens' test but acetone does not.
Reagent:
\[ Ammoniacal silver nitrate \]
Thus
\[ B \rightarrow IV \]
Step 3: C. Ethanol vs Phenol
Phenol gives a violet complex with neutral \(FeCl_3\).
Thus
\[ C \rightarrow III \]
Step 4: D. Benzoic acid vs Cinnamic acid
Cinnamic acid has a C=C double bond which decolorizes bromine water.
Thus
\[ D \rightarrow I \]
Hence the correct match is:
\[ A-II,\ B-IV,\ C-III,\ D-I \]
\[ \boxed{A-II,\ B-IV,\ C-III,\ D-I} \] Quick Tip: Carbylamine test identifies primary amines, Tollens' test identifies aldehydes, and \(FeCl_3\) test identifies phenols.
Consider the three aromatic molecules (P, Q and R). The correct order regarding the reactivity of these compounds with \(Ph{-}N=N^+Cl^-\) under optimum but slightly acidic medium is:
Match List–I with List–II.
List–I (Vitamin)
A. Vitamin \(B_1\)
B. Vitamin \(B_2\)
C. Vitamin \(B_6\)
D. Vitamin \(C\)
List–II (Name)
I. Pyridoxine
II. Ascorbic acid
III. Thiamine
IV. Riboflavin
Choose the correct answer from the options given below:
View Solution
Concept:
Different vitamins have specific chemical names.
Step 1: Identify names
\[ Vitamin B_1 = Thiamine \]
\[ Vitamin B_2 = Riboflavin \]
\[ Vitamin B_6 = Pyridoxine \]
\[ Vitamin C = Ascorbic acid \]
Step 2: Match
\[ A \rightarrow III \]
\[ B \rightarrow IV \]
\[ C \rightarrow I \]
\[ D \rightarrow II \]
Thus
\[ \boxed{A-III,\ B-IV,\ C-I,\ D-II} \] Quick Tip: Vitamin \(B_1\) = Thiamine, \(B_2\) = Riboflavin, \(B_6\) = Pyridoxine, and Vitamin C = Ascorbic acid.
A salt with few drops of HCl gives apple green colour on flame test. The group precipitate of the salt if dissolved in acetic acid and treated with \(K_2CrO_4\) gives yellow precipitate. When the sodium carbonate extract of the salt solution is heated with conc. \(HNO_3\) and ammonium molybdate, it results in a canary yellow precipitate. The cation and anion present in the salt are respectively:
5.33 g of \(CrCl_3 \cdot 6H_2O\), which is a 1:3 electrolyte, is dissolved in water and is passed through a cation exchanger. The chloride ions in the eluted solution, on treatment with \(AgNO_3\), results in 8.61 g of \(AgCl\). The ratio of moles of complex reacted and moles of \(AgCl\) formed is ______ \(\times 10^{-2}\). (Nearest integer)
\[ [Molar\ mass: Cr = 52,\ Ag = 108,\ Cl = 35.5,\ H = 1,\ O = 16] \]
Consider the isomers of hydrocarbon with molecular formula \(C_5H_{10}\). These isomers do not decolourise \(KMnO_4\) solution. These isomers are subjected to chlorination with chlorine in presence of light to give monochloro compounds. The total number of monochloro compounds (structural isomers only) formed is ______.
One mole of an alkane (\(x\)) requires 8 mole oxygen for complete combustion. Sum of number of carbon and hydrogen atoms in the alkane (\(x\)) is ______.
For reaction \(A \rightarrow P\), rate constant \(k = 1.5 \times 10^3\ s^{-1}\) at \(27^\circ C\). If activation energy for the above reaction is \(60\ kJ\ mol^{-1}\), then the temperature (in \(^{\circ}C\)) at which rate constant \(k = 4.5 \times 10^3\ s^{-1}\) is ______. (Nearest integer)
\[ Given: \log 2 = 0.30,\ \log 3 = 0.48,\ R = 8.3\ J\ K^{-1}\ mol^{-1},\ \ln 10 = 2.3 \]
At the transition temperature \(T\), \(A \rightleftharpoons B\) and \(\Delta G^\circ = 105 - 35\log T\), where \(A\) and \(B\) are two states of substance \(X\).
The transition temperature in \(^{\circ}C\) when pressure is 1 atm is ______.
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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 |
JEE Main 2026 Chemistry Revision
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