JEE Main 2023 Chemistry April 6 Shift 2 Question Paper with Solution PDF- Download Here

Quick Tip: Amino acids can be represented by their one-letter codes, which are used in protein sequences. This matching question tests your knowledge of such representations.

\( Pb(NO_3)_2 \) is a soluble colourless compound so it cannot be used in confirmatory test of \( Pb^{2+} \) ion.
Correct Answer: (2) lead nitrate
Quick Tip: For confirmatory tests of ions, it is crucial to recognize soluble and insoluble compounds. Lead nitrate is soluble and does not form a visible precipitate with \( Pb^{2+} \) ions.

\[ m.e.q of HBr = m.e.q of Ba(OH)_2 \] \[ M_1 \times n_1 \times V_1 = M_2 \times n_2 \times V_2 \] \[ 0.02 \times 1 \times V_1 = 0.01 \times 2 \times 10 \] \[ V_1 = \frac{0.01 \times 10}{0.02} = 10 \, mL \]
Correct Answer: (2) 10.0 mL
Quick Tip: In titration problems, always apply the principle of equivalence where moles of acid and base are equal. Use the relation \( M_1 \times V_1 = M_2 \times V_2 \) for calculations.

As electropositive character increases, basic character of oxide increases: \[ B_2O_3 < Al_2O_3 < Ga_2O_3 < In_2O_3 < Tl_2O_3 \]
Correct Answer: (4) B\(_2\)O\(_3\)
Quick Tip: When comparing the basicity of oxides, the trend follows the electropositivity of the elements. Oxides of more electropositive elements tend to be more basic.

The IUPAC name of \( K_3[Co(C_2O_4)_3] \) is Potassium trioxalatocobaltate(III).
Correct Answer: (4) Potassium trioxalatocobaltate(III)
Quick Tip: When naming coordination compounds, the ligands are named first followed by the metal center. The oxidation state of the metal is included in parentheses in Roman numerals.

N/A Quick Tip: Use the De-Broglie relation \(2\pi r = n\lambda\) to find the wavelength of an electron in any orbit.

N/A Quick Tip: Remember that DDT and Aldrin are both chlorinated organic compounds used in pest control.

N/A Quick Tip: In chromatography, polar compounds interact more strongly with the stationary phase and take longer to elute.

Hydration enthalpy is inversely proportional to the size of the ion, i.e., \[ Hydration enthalpy \propto \frac{1}{size}. \]
As the size of the ion increases down the group, hydration enthalpy decreases. Therefore, the order of hydration enthalpy is: \[ Be^{2+} > Mg^{2+} > Ca^{2+} > Sr^{2+} > Ba^{2+}. \]
Thus, Be\(^{2+}\) has the highest hydration enthalpy.
Quick Tip: Hydration enthalpy is a key concept for understanding the solvation process. Smaller ions have higher hydration enthalpies due to their higher charge density.

The acid strength of a phenolic compound depends on the electron-withdrawing or electron-donating groups attached to the benzene ring. The electron-withdrawing group increases the acidity of the compound. Among the options, the nitro group (-NO\(_2\)) is a strong electron-withdrawing group through the inductive effect (-I), making the compound with -NO\(_2\) attached to the ring the most acidic.
Quick Tip: When determining the acidity of substituted phenols, remember that electron-withdrawing groups (-I) increase acidity by stabilizing the negative charge on the conjugate base.

The reaction shows a typical mechanism where the protonation of the alcohol leads to a carbocation formation, followed by a shift of the CH\(_3\) group. The major product corresponds to the compound shown in option (4).
Quick Tip: In elimination and rearrangement reactions, the carbocation intermediate plays a crucial role in determining the product.

In the solid state, BeCl\(_2\) exists as a polymer. In the vapour phase, it forms a chloro-bridged dimer. At very high temperatures (above 1200K), it exists as a monomer.
Quick Tip: BeCl\(_2\) is an example of a molecule whose structure changes depending on temperature, showcasing the effect of temperature on molecular structure.

From the diagram, we see that there are two intermediates and three activated complexes. The rate determining step is the second step (II), where the highest energy barrier occurs.
Quick Tip: In reaction mechanisms, the rate determining step is the one with the highest activation energy, which slows down the overall reaction.

During electrolysis of brine solution (NaCl + H\(_2\)O), at the cathode, hydrogen gas (H\(_2\)) is released, and at the anode, chlorine gas (Cl\(_2\)) is produced. Na\(^+\) and OH\(^-\) combine to form NaOH. HCl is not formed during this process.
Quick Tip: The electrolysis of brine is used to produce chlorine gas, hydrogen gas, and sodium hydroxide.

Li and Cs react vigorously with water. Br is a non-metal that does not react in this manner. Ga, however, remains liquid at the boiling point of water, which is 100°C (boiling point of Ga = 2400°C).
Quick Tip: Remember that elements with high melting points, such as Gallium, can remain liquid even at temperatures above the boiling point of water.

The oxidation states of Ni and Fe in Ni(CO)\(_4\) and Fe(CO)\(_5\) are indeed zero. However, while low oxidation states can be stabilized by \(\pi\)-acceptor ligands, this is not the reason why Ni and Fe exhibit zero oxidation states in these complexes.
Quick Tip: Zero oxidation states in metal carbonyls are due to the ligand’s ability to donate electron density through \(\sigma\)-donation and accept electron density through \(\pi\)-back donation.

Morphine is indeed a narcotic analgesic and helps in relieving pain. However, it also causes drowsiness, so Statement I is false. Statement II is true as morphine and its derivatives are obtained from the opium poppy.
Quick Tip: Morphine and its derivatives have medicinal properties but also cause drowsiness, which makes Statement I incorrect.

The reaction is a Grignard reagent reaction with the carbonyl group, followed by reaction with n-PrI. The major product is the one shown in option (3), as a new carbon-carbon bond is formed.
Quick Tip: In Grignard reactions, a nucleophilic attack on the carbonyl group leads to the formation of an alcohol after the addition of a halide.

In neutral or weakly alkaline solution, the oxidation state of Mn changes by 3 units: MnO\(_4^{-}\) \(\rightarrow\) MnO\(_2\). Therefore, the reaction favours the neutral medium.
Quick Tip: The oxidation state change of Mn in reactions involving permanganate depends on the solution's pH. Neutral or slightly alkaline solutions are best for this reaction.

Nessler reagent is \( K_2[HgI_4] \), which contains K, Hg, and I. Nitrogen (N) is not present in the reagent.
Quick Tip: Nessler's reagent is used for the detection of ammonia and is made of mercury and iodine, but does not contain nitrogen.

In the given half reactions, metals V, Fe, and Ag will be oxidized by NO\(_3^-\) because their reduction potentials are less than 0.97 V. Thus, three metals will be oxidized by NO\(_3^-\).
Quick Tip: When comparing reduction potentials, metals with lower reduction potentials are more easily oxidized.

Body-centered cubic (BCC) crystal structure is found in cubic, tetragonal, and orthorhombic systems. Thus, the number of crystal systems where BCC can be found is 3.
Quick Tip: BCC structure is commonly found in metals like iron and chromium.

The iodoform test is positive for compounds that contain a methyl group (\(-CH_3\)) adjacent to a carbonyl group (\(C=O\)) or for alcohols that can be oxidized to acetone. Based on the structure and functional groups, compounds (c), (d), (e), and (f) will give a positive iodoform test. Thus, the answer is 4 compounds.
Quick Tip: For iodoform test to be positive, the structure must have either a \(-COCH_3\) group or a secondary alcohol adjacent to a methyl group.

By Gabriel phthalimide synthesis, i-amine is prepared, which should be aromatic and a-amine. The degree of unsaturation (Du) helps in determining the number of possible isomers.

Du = C + 1 - H - N/2, which equals 4, indicating the presence of a benzene ring. Thus, the number of isomeric aromatic amines that can be synthesized is 5.
Quick Tip: Gabriel Phthalimide synthesis is a key reaction for preparing aromatic amines from halides, especially when the structure is aromatic.

For two solutions to be isotonic, the total number of ions should be the same. Checking the pairs for ion concentration:

- A: 1 M NaCl gives 2 ions, and 2 M Urea gives 2 ions — isotonic.

- B: 1 M CaCl\(_2\) gives 3 ions, and 1.5 M KCl gives 3 ions — isotonic.

- C: 1.5 M AlCl\(_3\) gives 6 ions, and 2 M Na\(_2\)SO\(_4\) gives 6 ions — isotonic.

- D: 2.5 M KCl gives 5 ions, and 1 M Al\(_2\)(SO\(_4\))\(_3\) gives 5 ions — isotonic.

Thus, all 4 pairs are isotonic.
Quick Tip: Isotonic solutions have the same osmotic pressure, which is determined by the concentration of dissolved particles.

Liquid dispersion medium colloidal systems include paints, milk, hair cream, froth, soap lather. Thus, there are 5 such systems.
Quick Tip: Liquid colloidal systems are common in everyday life, especially in products like paints and creams where the dispersion medium is liquid.

In an ice crystal, each water molecule is hydrogen bonded to four neighbouring molecules, forming a tetrahedral arrangement.
Quick Tip: Hydrogen bonding in ice leads to the characteristic structure that makes ice less dense than liquid water.

We can use Hess's Law and combine the equations to calculate the heat of formation of C\(_2\)H\(_5\)OH(l):
\( \Delta H_{f} \) = \( -393.5 \times 2 \) - \( 241.5 \times 8 \times 3 \) + 1234.7 = -277.7 kJ mol\(^{-1}\)
Quick Tip: Hess's Law states that the heat of a reaction is the sum of the heats of the reactions into which it can be divided.

We are given the equilibrium constant Kc for the reaction at 298 K as 20.

K\(_c = \frac{[PCl_5]}{[PCl_3][Cl_2]} = \frac{0.40}{0.20 \times 0.10} = 20
After adding 0.2 mol of Cl\(_2\), the new concentrations become:

PCl\(_3\) = 0.2 - x, Cl\(_2\) = 0.2 + x, PCl\(_5\) = 0.4 + x.

Solving for x, we find that x = 0.084. Thus, the equilibrium concentration of PCl\(_5\) is 0.484 mol L\(^{-1}\).
Quick Tip: For equilibrium reactions, the change in concentrations due to shifting of equilibrium can be calculated using the reaction's equilibrium constant.

The species with square planar geometry include XeF\(_4\), [Cu(NH\(_3\))\(_4\)]\(^{2+}\), [PtCl\(_4\)]\(^{2-}\). Thus, there are 4 species with a square planar shape.
Quick Tip: Square planar geometry is common in transition metal complexes, especially for \(d^8\) configuration metals.