JEE Main 2023 Chemistry April 8 Shift 2 Question Paper is available here for download. Candidates can download official JEE Main 2023 Chemistry Question Paper PDF with Solution and Answer Key for April 8 Shift 2 using the link below. JEE Main Chemistry Question Paper is divided into two sections, Section A with 20 MCQs and Section B with 10 numerical type questions. Candidates are required to answer all questions from Section A and any 5 questions from section B.
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JEE Main 2023 Chemistry Question Paper April 8 Shift 2 PDF
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JEE Main 2023 8th April Shift 2 Chemistry Question Paper with Solution
The statement/s which are true about antagonists from the following is/are:
A. They bind to the receptor site.
B. Get transferred inside the cell for their action.
C. Inhibit the natural communication of the body.
D. Mimic the natural messenger.
Choose the correct answer from the options given below:
View Solution
The correct reaction profile diagram for a positive catalyst reaction is:
Choose the correct answer from the options given below:
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Given below are two statements: One is labeled as Assertion A, and the other is labeled as Reason R.
Assertion A: Sodium is about 30 times as abundant as potassium in the oceans.
Reason R: Potassium is bigger in size than sodium.
In the light of the above statements, choose the correct answer from the options given below:
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Which of these reactions is not a part of the breakdown of ozone in the stratosphere?
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The correct IUPAC nomenclature for the following compound is:
Replace with actual compound structure
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Henry Moseley studied characteristic X-ray spectra of elements. The graph which represents his observation correctly is:
Given: \(\nu\) = frequency of X-ray emitted
\(Z\) = atomic number
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Match List I with List II.
List I (Coordination Complex) & List II (Number of Unpaired Electrons)
A. [Cr(CN)\(_6\)]\(^{3-}\) & I. 0
B. [Fe(H\(_2\)O)\(_6\)]\(^{2+}\) & II. 3
C. [Co(NH\(_3\))\(_6\)]\(^{3+}\) & III. 2
D. [Ni(NH\(_3\))\(_6\)]\(^{2+}\) & IV. 4
Choose the correct answer from the options given below:
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The major product ‘P’ formed in the following reaction is:
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For a good quality cement, the ratio of lime to the total of the oxides of Si, Al, and Fe should be as close as to:
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Match List I with List II.
List I (Natural Amino Acid) & List II (One Letter Code)
A. Glutamic acid & I. Q
B. Glutamine & II. W
C. Tyrosine & III. E
D. Tryptophan & IV. Y
Choose the correct answer from the options given below:
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Which of the following have the same number of significant figures?
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Given below are two statements:
Statement I: Methyl orange is a weak acid.
Statement II: The benzenoid form of methyl orange is more intense/deeply coloured than the quinonoid form.
In the light of the above statements, choose the most appropriate answer from the options given below:
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The descending order of acidity for the following carboxylic acids is –
A. CH\textsubscript{3COOH
B. F\textsubscript{3C–COOH
C. ClCH\textsubscript{2–COOH
D. BrCH\textsubscript{2–COOH
Choose the correct answer from the options given below:
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In the Hall-Héroult process, the following is used for reducing Al\textsubscript{2}O\textsubscript{3}:
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Arrange the following gases in increasing order of van der Waals constant \(a\):
A. Ar
B. CH\textsubscript{4
C. H\textsubscript{2O
D. C\textsubscript{6H\textsubscript{6
Choose the correct answer from the options given below:
View Solution
Given below are two statements:
Statement I: In redox titration, the indicators used are sensitive to change in pH of the solution.
Statement II: In acid-base titration, the indicators used are sensitive to change in oxidation potential.
In the light of the above statements, choose the most appropriate answer from the options given below:
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Which of the following can reduce decomposition of H\textsubscript{2}O\textsubscript{2} on exposure to light?
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The correct order of reactivity of the following haloarenes towards nucleophilic substitution with aqueous NaOH is:
Choose the correct answer from the options given below:
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A compound ‘X’ when treated with phthalic anhydride in the presence of concentrated H\textsubscript{2}SO\textsubscript{4}, yields ‘Y’. ‘Y’ is used as an acid-base indicator. ‘X’ and ‘Y’ are respectively:
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The reaction between a compound and phthalic anhydride in the presence of concentrated sulfuric acid (H\textsubscript{2SO\textsubscript{4) leads to the formation of various derivatives depending on the starting compound. Phenolphthalein is formed when carbolic acid (phenol) reacts with phthalic anhydride.
Reaction Mechanism:
Phenol reacts with phthalic anhydride in an electrophilic aromatic substitution reaction catalyzed by concentrated sulfuric acid. This results in the formation of phenolphthalein, which is a widely used acid-base indicator.
Analysis of options:
1.Anisole, methyl orange: Incorrect, as methyl orange is formed by a different reaction pathway involving sulfonation of dimethylaniline, not phthalic anhydride.
2. Toluidine, Phenolphthalein: Incorrect, as toluidine does not react with phthalic anhydride to yield phenolphthalein.
3. Carbolic acid, Phenolphthalein: Correct, as carbolic acid (phenol) reacts with phthalic anhydride to produce phenolphthalein.
4. Salicylaldehyde, Phenolphthalein: Incorrect, as salicylaldehyde reacts with phthalic anhydride to form other derivatives, not phenolphthalein.
Thus, the correct answer is option (3). Quick Tip: Phenolphthalein is a common acid-base indicator prepared by reacting phenol with phthalic anhydride in the presence of concentrated H\textsubscript{2}SO\textsubscript{4}.
The product (P) formed from the following multistep reaction is:
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The observed magnetic moment of the complex [Mn(NCS)\textsubscript{6}]\textsuperscript{4−} is 6.06 BM. The numerical value of \(x\) is:
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The magnetic moment (\(\mu\)) is related to the number of unpaired electrons (\(n\)) by the formula: \[ \mu = \sqrt{n(n+2)} \, BM. \]
Given \(\mu = 6.06 \, BM\), solve for \(n\): \[ 6.06 = \sqrt{n(n+2)}. \]
Squaring both sides: \[ 36.72 = n(n+2). \]
This gives \(n \approx 5\) (nearest integer).
Since Mn has 5 unpaired electrons, its oxidation state must be \(+2\), as follows: \[ Mn^{2+} \implies x = 4. \] Quick Tip: Magnetic moment calculations provide insights into the number of unpaired electrons in coordination complexes.
For the complete combustion of ethane, \[ C\textsubscript{2}H\textsubscript{4}(g) + 3O\textsubscript{2}(g) \rightarrow 2CO\textsubscript{2}(g) + 2H\textsubscript{2}O(l), \]
the amount of heat produced as measured in a bomb calorimeter is 1406 kJ mol\textsuperscript{−1 at 300 K. The minimum value of \(T\Delta S\) needed to reach equilibrium is:
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The solubility product of BaSO\textsubscript{4} is \(1 \times 10^{-10}\) at 298 K. The solubility of BaSO\textsubscript{4 in 0.1 M K\textsubscript{2SO\textsubscript{4(aq) is \(x \times 10^{-9}\) g L\textsuperscript{−1. Calculate the value of \(x\).
View Solution
Given: \[ K_{sp} = [Ba^{2+}][SO\textsubscript{4}^{2-}] = 1 \times 10^{-10}. \]
Let the solubility of BaSO\textsubscript{4 in 0.1 M K\textsubscript{2SO\textsubscript{4 be \(x\). Since the concentration of SO\textsubscript{4\textsuperscript{2− is dominated by the dissociation of K\textsubscript{2SO\textsubscript{4, we approximate: \[ [SO\textsubscript{4}^{2-}] \approx 0.1. \]
From the solubility equilibrium: \[ [Ba^{2+}] = \frac{K_{sp}}{[SO\textsubscript{4}^{2-}]} = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9} \, mol L\textsuperscript{−1}. \]
The molar mass of BaSO\textsubscript{4 is \(233 \, g mol\textsuperscript{−1}\). Thus, the solubility in grams per liter is: \[ Solubility = (1 \times 10^{-9}) \times 233 = 233 \times 10^{-9} \, g L\textsuperscript{−1}. \]
Hence, \(x = 233\). Quick Tip: In the presence of a common ion, the solubility of sparingly soluble salts decreases significantly.
The number of atomic orbitals from the following having 5 radial nodes is: 7s, 7p, 6s, 6p, 8d.
View Solution
The number of radial nodes for an atomic orbital is calculated as: \[ Radial nodes = n - \ell - 1, \]
where \(n\) is the principal quantum number, and \(\ell\) is the azimuthal quantum number.
1. For 6s:
\[ n = 6, \, \ell = 0. \]
Radial nodes = \(6 - 0 - 1 = 5\).
2. For 7p:
\[ n = 7, \, \ell = 1. \]
Radial nodes = \(7 - 1 - 1 = 5\).
3. For 8d:
\[ n = 8, \, \ell = 2. \]
Radial nodes = \(8 - 2 - 1 = 5\).
4. For 7s:
\[ n = 7, \, \ell = 0. \]
Radial nodes = \(7 - 0 - 1 = 6\).
5. For 8p:
\[ n = 8, \, \ell = 1. \]
Radial nodes = \(8 - 1 - 1 = 6\).
The orbitals with 5 radial nodes are: \[ 6s, \, 7p, \, 8d. \]
Thus, the total number of atomic orbitals with 5 radial nodes is: \[ \textbf{3}. \] Quick Tip: Radial nodes depend on both the principal quantum number (\(n\)) and the azimuthal quantum number (\(\ell\)). Larger \(n\) values increase the number of nodes.
The number of incorrect statements from the following is:
1. The electrical work that a reaction can perform at constant pressure and temperature is equal to the reaction Gibbs energy.
2. \(E^\circ_{cell}\) is dependent on the pressure.
3. \(\frac{dE^\circ_{cell}}{dT} = \frac{\Delta S^\circ}{nF}\).
4. A cell is operating reversibly if the cell potential is exactly balanced by an opposing source of potential difference.
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The coagulating value of the electrolytes AlCl\textsubscript{3} and NaCl for As\textsubscript{2}S\textsubscript{3} are 0.09 and 50.04, respectively. The coagulating power of AlCl\textsubscript{3} is \(x\) times the coagulating power of NaCl. The value of \(x\) is:
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If the boiling points of two solvents X and Y (having the same molecular weights) are in the ratio 2:1, and their enthalpy of vaporizations are in the ratio 1:2, then the boiling point elevation constant of X is \(m\) times the boiling point elevation constant of Y. The value of \(m\) is:
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The number of species from the following carrying a single lone pair on the central atom Xenon is:
Given species:
XeF\textsubscript{5\textsuperscript{+, XeO\textsubscript{3, XeO\textsubscript{2F\textsubscript{2, XeF\textsubscript{5\textsuperscript{−, XeO\textsubscript{3F\textsubscript{2, XeOF\textsubscript{4, XeF\textsubscript{4.
View Solution
The ratio of sigma and pi bonds present in pyrophosphoric acid is:
View Solution
The sum of oxidation states of the metals in Fe(CO)\textsubscript{5}, VO\textsuperscript{2+}, and WO\textsubscript{3} is:
View Solution
Calculate the oxidation state of each metal:
1.Fe(CO)\textsubscript{5:
Carbon monoxide (CO) is a neutral ligand, and the oxidation state of Fe in Fe(CO)\textsubscript{5 is:
\[ Oxidation state of Fe = 0. \]
2.VO\textsuperscript{2+:
Oxygen has an oxidation state of \(-2\). For VO\textsuperscript{2+:
\[ x - 2 = +2 \implies x = +4. \]
The oxidation state of V is \(+4\).
3.WO\textsubscript{3:
Each oxygen contributes \(-2\), and tungsten (W) has:
\[ x + (-2 \times 3) = 0 \implies x = +6. \]
The oxidation state of W is \(+6\).
Now sum the oxidation states: \[ 0 + 4 + 6 = 10. \]
Thus, the total sum of oxidation states is: \[ \textbf{10}. \] Quick Tip: To calculate oxidation states in coordination complexes, consider the charge of ligands and the net charge of the compound.
JEE Main 2023 Chemistry Paper Analysis April 8 Shift 2
JEE Main 2023 Chemistry Paper Analysis for the exam scheduled on April 8 Shift 2 is available here. Candidates can check subject-wise paper analysis for the exam scheduled on April 8 Shift 2 here along with the topics with the highest weightage.
JEE Main 2023 Chemistry Question Paper Pattern
| Feature | Question Paper Pattern |
|---|---|
| Examination Mode | Computer-based Test |
| Exam Language | 13 languages (English, Hindi, Assamese, Bengali, Gujarati, Kannada, Malayalam, Marathi, Odia, Punjabi, Tamil, Telugu, and Urdu) |
| Exam Duration | 3 hours |
| Sectional Time Limit | None |
| Chemistry Marks | 100 marks |
| Total Number of Questions Asked | 20 MCQs + 10 Numerical Type Questions |
| Total Number of Questions to be Answered | 20 MCQs + 5 Numerical Type Questions |
| Marking Scheme | +4 for each correct answer |
| Negative Marking | -1 for each incorrect answer |
Also Check:
JEE Main Previous Year Question Paper
| JEE Main 2022 Question Paper | JEE Main 2021 Question Paper | JEE Main 2020 Question Paper |
| JEE Main 2019 Question Paper | JEE Main 2018 Question Paper | JEE Main 2017 Question Paper |
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