JEE Main 2024 question paper pdf with solutions- Download Jan 30 Shift 2 Chemistry Question Paper pdf

Step 1: Understanding the Color Formation.

KMnO₄ (Potassium Permanganate) shows color due to the transition of a nonbonding 2p oxygen electron to the vacant molecular orbital of a tetrahedral complex. This type of transition involves ligand-to-metal charge transfer, which occurs when an electron moves from the ligand (oxygen) to the metal ion (Mn), leading to a characteristic color.

Step 2: Conclusion.

The transition from ligand to metal causes the color in KMnO₄. Quick Tip: Color in transition metal compounds is often due to electronic transitions between d-orbitals or ligand to metal charge transfer.

Step 1: Mole Fraction Formula.

The mole fraction is given by the formula: \[ Mole fraction of component = \frac{moles of one component}{Total moles of all components}. \]
In this case, for component C: \[ x_C = \frac{n_C}{n_A + n_B + n_C}. \]

Step 2: Conclusion.

The mole fraction of C is \( \frac{n_C}{n_A + n_B + n_C} \), which is the correct answer. Quick Tip: The mole fraction of a component in a solution is the ratio of the moles of that component to the total moles of all components.

Step 1: Understanding the Structure.

The given compound is \( CH_3 - CH = C - CH_3 \), which is an alkyne with a triple bond between the second and third carbon atoms. The numbering starts from the left to give the lowest possible number to the triple bond.

Step 2: Naming the Compound.

The compound has 4 carbon atoms, which makes it a butyne. The position of the triple bond is at the 1st carbon (counting from the left). A methyl group is attached to the third carbon.

Step 3: Conclusion.

The correct IUPAC name is 3-Methylbut-1-yne, as the triple bond is at position 1, and the methyl group is at position 3. Quick Tip: When naming organic compounds, always prioritize giving the lowest number to functional groups like triple or double bonds.

Step 1: Dipole Moment in NH₃ and NF₃.

In \( NH₃ \), the nitrogen atom has a lone pair of electrons and the dipole moment is directed upwards due to the electronegativity difference between nitrogen and hydrogen. In \( NF₃ \), however, the fluorine atoms, being highly electronegative, pull electron density away from nitrogen, but the molecular shape leads to the cancellation of dipole moments. This results in a lower dipole moment in \( NF₃ \) compared to \( NH₃ \), contrary to Statement I.

Step 2: Direction of Electron Flow.

In \( NF₃ \), the electron flow is directed from nitrogen to fluorine because fluorine is more electronegative, not in the same direction as stated in Statement II.

Step 3: Conclusion.

Both statements are false. Statement I is incorrect, and Statement II is also incorrect. Quick Tip: The dipole moment in molecules depends on both the electronegativity difference and the molecular geometry.

Step 1: Understanding the Reaction.

The first reaction involves the formation of an intermediate in a chemical reaction, likely involving a substitution or addition reaction. The second reaction involves the transformation of the molecule into a different structure with the help of reagents like \( B_2H_6 \), which are typical in hydroboration reactions.

Step 2: Identifying the Products.

Both \( A \) and \( B \) involve hydroboration or related reactions, which typically result in alcohols as the products.

Step 3: Conclusion.

The correct identification for both \( A \) and \( B \) is \( OH \). Quick Tip: Reactions involving \( B_2H_6 \) typically lead to the formation of alcohols through hydroboration.

Step 1: Reaction Mechanism.

The reaction between phenol and chloroform in the presence of NaOH at 343 K undergoes a reaction known as the Reimer-Tiemann reaction. This reaction leads to the formation of salicylaldehyde.

Step 2: Hydrolysis.

After the reaction with chloroform and NaOH, the product is subjected to hydrolysis, which results in the formation of salicylaldehyde.

Step 3: Conclusion.

The final product is benzaldehyde, formed after the hydrolysis step. Quick Tip: The Reimer-Tiemann reaction is commonly used to introduce a formyl group (-CHO) into an aromatic compound, particularly phenols.

Step 1: Reaction Mechanism.

In this reaction, the amine group (-NH\(_2\)) of the aniline undergoes diazotization with nitrous acid (\( NaNO_2/HCl \)) after the nitration of phenol. This leads to the formation of the azobenzene compound.

Step 2: Product Identification.

The reaction conditions lead to the formation of p-hydroxy azobenzene due to the position of the substituents in the aromatic ring.

Step 3: Conclusion.

The correct product formed is p-hydroxy azobenzene. Quick Tip: In diazotization reactions, an amine group reacts with nitrous acid to form a diazonium ion, which can then undergo further reactions.

Step 1: Reaction of CrO\(_3\) and NaOH.

When chromium trioxide (\( CrO_3 \)) reacts with chlorine and sodium hydroxide, sodium chromate (\( Na_2CrO_4 \)) and sodium chloride are produced along with water.

Step 2: Reaction of Sodium Chromate.

When sodium chromate reacts with sulfuric acid and hydrogen peroxide, chromium pentoxide (\( CrO_5 \)) is formed.

Step 3: Conclusion.

Thus, the correct answer is Na\(_2\)CrO\(_4\) and CrO\(_5\). Quick Tip: Chromium compounds like \( CrO_3 \) and \( Na_2CrO_4 \) are important in oxidation reactions, especially in organic synthesis.

Step 1: Understanding Chromatography.

Chromatography is a technique used to separate components of a mixture based on their different solubilities in a given solvent. It relies on the differential interaction of the components with the stationary phase and the mobile phase.

Step 2: Conclusion.

Thus, chromatography is the method that separates components based on solubility differences. Quick Tip: Chromatography is used to separate substances in a mixture by passing it through a medium in which components move at different rates.

Step 1: Understanding the Acidity.

The acidity of hydrides in Group 16 increases as we go down the group. Since \( H_2Te \) is in the same group as \( H_2S \), but below it, it is more acidic than \( H_2S \), making Statement I correct.

Step 2: Bond Dissociation Enthalpy (BDE).

The BDE of \( H_2Te \) is lower than that of \( H_2S \), meaning Statement II is incorrect.

Step 3: Conclusion.

Thus, Statement I is correct and Statement II is incorrect. Quick Tip: The acidity of hydrides increases as the atomic size increases because the bond between hydrogen and the heavier element becomes weaker.

Step 1: Identify the Isomers.

The compound in question is \( 2,3 \)-Dichlorobutane. The optical activity of a compound depends on the presence of chiral centers. For \( 2,3 \)-Dichlorobutane, the number of diastereoisomers and enantiomers can be calculated.

Step 2: Conclusion.

The total number of optically active compounds in this case is calculated to be 3, considering the number of stereoisomers and meso forms. Quick Tip: Optically active compounds must have no symmetry elements that lead to internal compensation, such as in meso forms.