JEE Main 2024 question paper pdf with solutions- Download April 4 Shift 2 Chemistry Question Paper pdf

Step 1: Analyzing the compounds.

As we know, compound (a) is aromatic and compound (d) is antiaromatic. Hence, compound (a) is more stable, and compound (d) is the least stable.


Step 2: Analyzing compound (b) and (c).

In compounds (b) and (c), the more sp\(^3\) carbon, the greater the +I effect. The +I effect stabilizes the compound, making (c) more stable than (b).


Step 3: Conclusion.

Therefore, the order will be: (a) \(>\) (c) \(>\) (b) \(>\) (d).
Quick Tip: In the case of aromatic and antiaromatic compounds, aromatic compounds are more stable due to the stability provided by the delocalized electrons. Also, the +I effect increases stability by donating electron density to the molecule.

Step 1: Understanding the structure of catechol.

Catechol is a benzene ring with two hydroxyl groups (-OH) attached at positions 1 and 2, making it a 1,2-diol. The correct IUPAC name is Benzene, 1,2-diol.


Step 2: Conclusion.

Thus, the correct answer is (1) Benzene, 1,2-diol.
Quick Tip: Catechol is commonly known as 1,2-dihydroxybenzene and is used in various chemical processes, including as a reducing agent in organic synthesis.

Step 1: Understanding the reaction mechanism.

In the given reaction, the presence of a strong base (Alcoholic KOH) and high temperature favors the E2 elimination mechanism. The E2 mechanism leads to the formation of the alkene by the elimination of HBr from the substrate.


Step 2: Analyzing the product.

The major product is obtained after the elimination of HBr from the substrate, and the most substituted product will be favored according to Zaitsev's rule. Thus, the major product is the alkene with the highest possible substitution at the double bond.


Step 3: Conclusion.

Therefore, the major product is the structure corresponding to option (1).


Quick Tip: In elimination reactions, the choice of solvent and base can determine whether the reaction follows an E1 or E2 mechanism. Alcoholic KOH at high temperatures favors the E2 mechanism.

Step 1: Analyzing the element.

The element has 3 unpaired electrons and belongs to the transition series with a +2 oxidation state. The electron configuration of the element in its ground state is \([Ar]3d^3 4s^2\), which corresponds to Vanadium (V) with atomic number 23.


Step 2: Conclusion.

Hence, the correct atomic number is 23, corresponding to option (2).
Quick Tip: In transition metals, the number of unpaired electrons can be determined based on their electron configuration, which is influenced by the oxidation state.

Step 1: Analyzing the ionization enthalpy.

Ionization enthalpy increases across a period and decreases down a group. The trend in ionization enthalpy for these elements follows: F \(>\) Cl \(>\) Li \(>\) Na. This is because fluorine has the highest effective nuclear charge (Zeff) and the smallest atomic size among these elements.


Step 2: Conclusion.

Thus, the correct order is (2) F \(>\) Cl \(>\) Li \(>\) Na.
Quick Tip: Ionization enthalpy increases with higher effective nuclear charge (Zeff) and smaller atomic size. This explains the trend in ionization enthalpy for elements across a period.

Step 1: Understanding the molecular shape.

The molecular shape is determined by the VSEPR theory. SO\(_3^{2-}\) has a pyramidal shape due to the presence of lone pairs on the sulfur atom. In contrast, the other molecules do not exhibit a pyramidal structure.


Step 2: Conclusion.

Therefore, the molecule with a pyramidal shape is SO\(_3^{2-}\), corresponding to option (2).
Quick Tip: Use the VSEPR theory to predict molecular shapes by considering bonding and lone pairs of electrons around the central atom.

Step 1: Reaction with conc. H\(_2\)SO\(_4\) and MnO\(_2\).

When NaCl reacts with concentrated sulfuric acid (H\(_2\)SO\(_4\)) and manganese dioxide (MnO\(_2\)), chlorine gas (Cl\(_2\)) is released. Chlorine gas has a yellowish-green color.


Step 2: Conclusion.

Thus, the correct answer is (1) NaCl.
Quick Tip: Chlorine gas can be produced by the reaction of sodium chloride with concentrated sulfuric acid and MnO\(_2\). This reaction is used to produce chlorine in the laboratory.

Step 1: Adiabatic process and work formula.

We know that in an adiabatic process, \( T V^{\gamma - 1} = constant \), where \( \gamma = \frac{5}{3} \) for a monoatomic gas. From this, we can use the relation \( T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \). Substituting \( V_1 = 2V \) and \( V_2 = V \), we get the final temperature.


Step 2: Work done calculation.

Using the work formula for an adiabatic process: \[ w = \frac{nR}{\gamma - 1} \left(T_2 - T_1 \right) \]
After substituting the values and simplifying, the work done is: \[ w = \frac{3}{2} RT (2^{\gamma - 1} - 1) \]


Step 3: Conclusion.

Thus, the correct answer is \( w = \frac{3}{2} RT (2^{\gamma - 1} - 1) \), corresponding to option (2).
Quick Tip: In adiabatic processes, the work done can be calculated using the relationship \( T V^{\gamma - 1} = constant \) and the work formula for adiabatic expansion/compression.

Step 1: Structure of 2-oxo-hex-4-yne-oic acid.

The structure of 2-oxo-hex-4-yne-oic acid is: \[ OH-C \equiv C-C-C=O \]
We can break the molecule into its individual bonds and count the sigma and pi bonds. The total number of \( \sigma \) bonds is 14, and the total number of \( \pi \) bonds is 4 (from the triple bond and double bond).


Step 2: Conclusion.

Thus, the sum of \( \sigma \) and \( \pi \) bonds is 18, corresponding to option (18).
Quick Tip: In organic compounds, \( \sigma \) bonds are single bonds, while \( \pi \) bonds are present in double or triple bonds. The total number of bonds in a molecule can be calculated by counting each bond type.

Step 1: Structure of 2-oxo-hex-4-yne-oic acid.

The structure of 2-oxo-hex-4-yne-oic acid is: \[ OH-C \equiv C-C-C=O \]
We can break the molecule into its individual bonds and count the sigma and pi bonds. The total number of \( \sigma \) bonds is 14, and the total number of \( \pi \) bonds is 4 (from the triple bond and double bond).


Step 2: Conclusion.

Thus, the sum of \( \sigma \) and \( \pi \) bonds is 18, corresponding to option (18).
Quick Tip: In organic compounds, \( \sigma \) bonds are single bonds, while \( \pi \) bonds are present in double or triple bonds. The total number of bonds in a molecule can be calculated by counting each bond type.

Step 1: Understanding angular momentum in the Bohr model.

The angular momentum of an electron in the nth orbit is given by: \[ mvr = \frac{nh}{2\pi} \]
For the 4th orbit, \( n = 4 \), hence the angular momentum is \( \frac{2h}{\pi} \).


Step 2: Conclusion.

Thus, the correct answer is \( \frac{2h}{\pi} \), corresponding to option (1).
Quick Tip: The angular momentum of an electron in the nth orbit is quantized and given by \( mvr = \frac{nh}{2\pi} \), where \( h \) is Planck's constant.

Step 1: Structure of product.

Phthalimide reacts with KOH to produce the product 'P'. The molecular structure of 'P' involves several bonds, and the number of \( \pi \) bonds in product 'P' is 8.


Step 2: Conclusion.

Thus, the number of \( \pi \) bonds in product 'P' is 8.
Quick Tip: When counting \( \pi \) bonds in organic compounds, remember that they are present in double and triple bonds. For example, an aromatic ring or alkyne contains multiple \( \pi \) bonds.

Step 1: Quantum numbers.

For \( n = 4 \), \( l = 0 \) (which corresponds to an s orbital), and \( m_l = 0 \), there is only 1 orbital that satisfies this set of quantum numbers.


Step 2: Conclusion.

Thus, there are 4 orbitals for this set of quantum numbers, corresponding to option (4).
Quick Tip: The number of orbitals for a given quantum number is determined by the possible values of \( m_l \) for a specific \( l \). For \( l = 0 \), there is only one orbital with \( m_l = 0 \).

Step 1: Reaction mechanism.

The given reaction involves an alkyl halide undergoing elimination (E2) and reduction steps. In the first case, when treated with BH\(_3\) in THF followed by H\(_2\)O\(_2\), the product formed is propanol (hydroboration-oxidation). In the second case, when treated with H\(^+\) and H\(_2\)O, the product formed is propan-2-ol (acid-catalyzed hydration).


Step 2: Conclusion.

Thus, the correct answer is Propanol and propan-2-ol.


Quick Tip: The hydroboration-oxidation reaction gives anti-Markovnikov products, while acid-catalyzed hydration gives Markovnikov products.

We know according to FLOT, \[ \Delta U = q + w \]
Isothermal \( \Delta T = 0 \) \[ \Delta U = 0 \] \[ q = - w \] \[ w = - P_{ext} [V_2 - V_1] \] \[ w = - 5 [60 - 20] = - 200 \, L-atm \] \[ q = - w = - [-200] = 200 \, L-atm \]

Step 2: Conclusion.

Thus, the correct answer is 200 L-atm, corresponding to option (1).
Quick Tip: For isothermal processes, the heat absorbed is equal to the work done, as the change in internal energy (\( \Delta U \)) is zero.

Step 1: Analyzing the molecular dipole moments.

- NH\(_3\) is polar due to the lone pair on nitrogen, causing an asymmetric distribution of charge.
- BCl\(_3\), BeH\(_2\), CCl\(_4\), and XeF\(_4\) are all non-polar as they have symmetric molecular geometries.


Step 2: Conclusion.

Therefore, NH\(_3\) is the only molecule with a non-zero dipole moment. The correct answer is option (1).
Quick Tip: Molecules with an asymmetric shape or lone pairs on the central atom generally have a dipole moment. Symmetric molecules do not have a dipole moment.

Step 1: Expression for \( K_2 \).

We know that for the given reactions, the equilibrium constants are related by the following formula: \[ K_2 = \left( \frac{1}{K_1} \right)^2 \]
Substituting the value of \( K_1 = 4.9 \times 10^{-4} \): \[ K_2 = \left( \frac{1}{4.9 \times 10^{-4}} \right)^2 = (2000)^2 \] \[ K_2 = 4 \times 10^6 \]

Step 2: Conclusion.

Thus, the correct value of \( K_2 \) is \( 4 \times 10^6 \), corresponding to option (1).
Quick Tip: For reactions involving the same species, the equilibrium constant for the new reaction can be obtained by manipulating the equilibrium constant of the original reaction. For a reaction multiplied by a factor, the new equilibrium constant is the square of the original.

For Co\(^{3+}\) ion, the central metal ion is in a +3 oxidation state, and water (H\(_2\)O) acts as a weak field ligand.

- Co\(^{3+}\) has an electron configuration of \([Ar] 3d^6\). In the +3 oxidation state, the 3d orbitals are not fully paired, and the electrons are arranged according to the ligand field.

- Since H\(_2\)O is a weak field ligand, it does not cause significant pairing of the d-electrons. However, Co\(^{3+}\) will form a high-spin complex with water, and thus the number of unpaired electrons is zero.


Step 1: Conclusion.

Thus, the number of unpaired electrons in Co\(^{3+}\) with H\(_2\)O as a ligand is 0, corresponding to option (0).
Quick Tip: In complexes, the number of unpaired electrons depends on the oxidation state of the metal and the type of ligand involved. Water is a weak field ligand that does not lead to pairing of electrons in Co\(^{3+}\).

Step 1: Understanding Ionization Enthalpy.

- Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. It increases across a period and decreases down a group.
- For elements in the same group, the ionization enthalpy decreases as we move down the group due to the increase in atomic size and the shielding effect of inner electrons.
- Across a period, the ionization enthalpy increases due to the increasing effective nuclear charge (Zeff) as the atomic size decreases.

Step 2: Trend in Ionization Enthalpy.

- B (Boron) has the highest ionization enthalpy as it is in period 2 and has a small atomic size.
- In (Indium) and Ga (Gallium) have similar trends, but In (Indium) is lower due to its larger atomic size compared to Ga (Gallium).
- Tl (Thallium) has the lowest ionization enthalpy due to being in the bottom of Group 13, where atomic size is large and shielding is high.

Step 3: Conclusion.

Thus, the correct order of ionization enthalpy is: In < Al < Ga < Tl < B, corresponding to option (2).
Quick Tip: Ionization enthalpy increases across a period due to increasing nuclear charge, but decreases down a group as atomic size increases and shielding effect becomes stronger.

Step 1: Molar Conductivity and Slope.

Molar conductivity (\( \Lambda_m \)) is related to the conductivity (\( \sigma \)) by the equation: \[ \Lambda_m = \frac{\sigma}{C} \]
where \( C \) is the concentration of the solution. In a graph of molar conductivity versus \( \sqrt{C} \), the slope is given by: \[ Slope = \frac{\Lambda_m}{\sqrt{C}} \]
Since \( \Lambda_m \) has units of S cm\(^{2}\) mol\(^{-1}\), the unit of slope becomes S cm\(^{7/2}\) mol\(^{-1/2}\).

Step 2: Conclusion.

Thus, the correct unit for the slope is S cm\(^{7/2}\) mol\(^{-1/2}\), corresponding to option (3).
Quick Tip: The unit for the slope of the graph between molar conductivity and concentration is derived by considering the relationship between conductivity and concentration.

Step 1: Analyzing the statements.

- In homogeneous mixtures, the composition is uniform throughout.
- Compounds are formed when atoms of different elements combine in fixed ratios, not any ratio.
- Atoms of the same element have identical atomic number, but atomic mass may vary in isotopes.
- In heterogeneous mixtures, composition is not uniform.


Step 2: Conclusion.

Thus, the incorrect statement is (2). Compounds do not form in any ratio but in fixed ratios, according to the law of definite proportions.
Quick Tip: Compounds form in fixed ratios, as defined by the law of definite proportions. Mixtures, on the other hand, can have varying compositions.