JEE Main 2023 Chemistry Question Paper Jan 29 Shift 2- Download PDF

Statement I is correct because the decrease in ionization enthalpy from B to Al is influenced by the addition of a new electron shell, which increases shielding, causing a significant drop in ionization energy. However, from Al to Ga, the decrease is smaller due to the poor shielding effect of d-electrons, leading to only a slight decrease in ionization enthalpy.

Statement II is also correct because in Ga, the \textit{d orbitals (3d) are completely filled, contributing to the slight decrease in ionization energy. Quick Tip: Ionization energy generally decreases down a group, but anomalies may arise due to poor shielding by \textit{d and f-orbitals.

The spin-only magnetic moment depends on the number of unpaired electrons. For [FeF\)_6\(]\)^{3-\(, Fe\)^{3+\( has 5 unpaired electrons, resulting in the highest magnetic moment. For [CoF\)_6\(]\)^{3-\(, Co\)^{3+\( in a weak field ligand (fluoride) has 4 unpaired electrons. For [Co(C\)_2\(O\)_4\()\)_3\(]\)^{3-\(, Co\)^{3+\( in a strong field ligand (oxalate) has 0 unpaired electrons. Thus, the order of magnetic moment is [FeF\)_6\(]\)^{3-\( \)>\( [CoF\)_6\(]\)^{3-\( \)>\( [Co(C\)_2\(O\)_4\()\)_3\(]\)^{3-\(. Quick Tip: Magnetic moment (\(\mu\)) is calculated using the formula \(\mu = \sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons.

- Osmosis: The movement of solvent molecules through a semi-permeable membrane towards a solution side (III).

- Reverse Osmosis: Solvent molecules are forced in the reverse direction, from the solution side to the solvent side, under applied pressure (I).

- Electro osmosis: The dispersion medium moves under the effect of an electric field (IV)

- Electrophoresis: Charged colloidal particles move under the influence of an electric potential to oppositely charged electrodes (II).

Hence, the correct match is A-III, B-I, C-IV, D-II. Quick Tip: In electrochemical processes, remember that particle movement depends on the charge and the electric potential applied.

- (i) Manganese exhibits a +7 oxidation state in its oxide (\(Mn_2O_7\)), which is correct.

- (ii) Ruthenium (\(Ru\)) and Osmium (\(Os\)) exhibit a +8 oxidation state in their respective oxides (\(RuO_4\), \(OsO_4\)), making this statement correct.

- (iii) Scandium (\(Sc\)) does not show a +4 oxidation state; it only shows a +3 oxidation state in most of its compounds. Hence, this is incorrect.

- (iv) Chromium (\(Cr\)) in the +6 oxidation state, such as in \(CrO_3\), exhibits strong oxidizing behavior. This makes the statement correct.

Thus, the correct set of statements is (i), (ii), and (iv). Quick Tip: Transition elements often exhibit multiple oxidation states. Pay attention to group trends and stability of oxidation states.

- Neoprene is a synthetic rubber, making it an elastomeric polymer. (A-IV)

- Polyester is a strong and durable material often used as a fibre polymer. (B-III)

- Urea formaldehyde resin is a thermosetting polymer that hardens irreversibly. (C-I)

- Polystyrene is a thermoplastic polymer, which can be reshaped with heat. (D-II)

Hence, the correct matching is A-IV, B-III, C-I, D-II. Quick Tip: Understand the properties and applications of polymer types to differentiate between elastomers, fibres, thermoplastics, and thermosetting polymers.

In this reaction, iodine (\(I_2\)) is produced as a product when \(I^-\) reacts with \(H_2O_2\). Starch is commonly used as an indicator because it forms a blue-colored complex with iodine. The reaction steps are as follows:
\[ I^- + H_2O_2 \rightarrow I_2 + H_2O \] \[ I_2 + Starch \rightarrow Blue complex \]

Thus, the indicator is starch, and the compound forming the blue complex is iodine. Quick Tip: Starch is a specific indicator for iodine, producing a characteristic blue-black complex.

Equanil is a drug that is used as a tranquilizer. It helps in the treatment of anxiety, depression, and related disorders such as hypertension caused by stress. It acts by calming the central nervous system and stabilizing mood swings. Quick Tip: Remember that tranquilizers like Equanil are prescribed for mental health conditions involving anxiety, stress, or depression.

The reaction involves the dehydration of a secondary alcohol to form an alkene. Under acidic conditions (\(H_2O^+\)), the \(-OH\) group is protonated and leaves as water, forming a carbocation intermediate. The major product is determined by the stability of the alkene. In this case, the more substituted alkene (Zaitsev's rule) is the major product. The reaction mechanism is as follows:

1. Protonation of the alcohol group.

2. Loss of water to form a carbocation.

3. Elimination of a proton to form the alkene.

Thus, the major product is the one with the double bond in the more substituted position. Quick Tip: Follow Zaitsev’s rule: the major product of elimination is the more substituted, stable alkene.

Dehydrohalogenation involves the elimination of HBr to form alkenes. The number of isomeric alkenes depends on the number of different \(\beta\)-hydrogens that can be removed.

For 2-Bromopentane, two different \(\beta\)-carbons are available, leading to the formation of multiple alkenes (e.g., pent-1-ene and pent-2-ene). Additionally, pent-2-ene can exist as cis and trans isomers, giving a total of three isomeric alkenes.

For the other options:

- 1-Bromo-2-methylbutane and 2-Bromo-3,3-dimethylpentane have only one possible elimination product.

- 2-Bromopropane gives only one product (propene).
Quick Tip: Identify \(\beta\)-hydrogens in the molecule to determine the number of possible elimination products.

The combustion reaction can be represented as:
\[ C_xH_y + O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O \]
Given that 9.5 moles of \(O_2\) are required and 3 moles of water are produced, we can set up the following equations:

- \(\frac{y}{2} = 3 \implies y = 6\)

- \(x + \frac{y}{4} = 9.5 \implies x + \frac{6}{4} = 9.5 \implies x = 8\)

Thus, the molecular formula of the hydrocarbon is \(C_8H_6\). This corresponds to an alkyne or aromatic compound. Quick Tip: Balance the combustion reaction by relating oxygen consumption and water/CO\(_2\) production to deduce the molecular formula.

The reaction proceeds as follows:

1. \(NaCN\) reacts with the carbonyl group to form a cyanohydrin (\(-CN\) and \(-OH\) groups on the same carbon).

2. Ethanol in the presence of \(H_2O^+\) hydrolyzes the cyanohydrin to form a carboxylic acid group (\(-COOH\)) and retains the hydroxyl group.

Thus, the major products are as shown in option (2), where the carboxylic acid (\(-COOH\)) and hydroxyl (\(-OH\)) groups are appropriately positioned. Quick Tip: Understand the reactivity of cyanohydrins and their hydrolysis products to predict major reaction outcomes.

The bond order (BO) is calculated using the molecular orbital (MO) theory formula:
\[ BO = \frac{(Number of bonding electrons - Number of antibonding electrons)}{2} \]
- For \(O_2^-\): Adding one electron to \(O_2\) decreases the bond order from 2 to 1.

- For CO: The bond order remains 3 because of strong triple bonding.

- For NO\(^+\): Removal of one electron from NO increases the bond order from 2.5 to 3.

Thus, the bond orders are 1, 3, and 3, respectively. Quick Tip: For molecular ions, the addition of electrons decreases the bond order, while electron removal increases it.

\(CrO_3\) (chromium trioxide) in amyl alcohol forms a blue complex. This is characteristic of certain chromium compounds when dissolved in organic solvents. The blue colour indicates the formation of a coordination compound involving chromium. Quick Tip: The colour of chromium compounds is an important qualitative test for its oxidation states and complex formation.

The dissolved oxygen (DO) in water is an essential parameter for the survival of aquatic organisms. Fish and other aquatic species depend on adequate oxygen levels to carry out cellular respiration and maintain metabolic processes. In general:

- For healthy growth and reproduction of fish, the dissolved oxygen concentration must be above 6 ppm.

- Levels below 4 ppm can cause stress, and prolonged exposure to such conditions may be lethal to most fish species.


On the other hand, biochemical oxygen demand (BOD) measures the oxygen consumed by microorganisms while decomposing organic matter in the water. It serves as an indirect indicator of the level of organic pollution:

- For clean water, BOD values should remain below 5 ppm.

- Higher BOD levels indicate the presence of excess organic matter, leading to oxygen depletion, which adversely affects aquatic life.


Why X = 6 ppm and Y = 5 ppm?

1. Dissolved oxygen levels above 6 ppm ensure a favorable environment for fish, supporting their growth, activity, and reproduction.

2. A BOD below 5 ppm reflects clean water, indicating minimal pollution and sufficient oxygen for aquatic organisms.


Thus, the correct values for X and Y are 6 ppm and 5 ppm, respectively. Quick Tip: Dissolved oxygen levels above 4 ppm and low BOD values are critical for maintaining healthy aquatic ecosystems.

This reaction is the Hofmann bromamide reaction, where amides are converted to primary amines with one fewer carbon atom. For propanamide (\(CH_3CH_2CONH_2\)):
\[ CH_3CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2NH_2 (Ethylamine) \] Quick Tip: In the Hofmann bromamide reaction, the product has one less carbon than the starting amide.

The given tetrapeptide contains the following amino acid residues:

- Phenylalanine (\(F\))

- Leucine (\(L\))

- Aspartic acid (\(D\))

- Tyrosine (\(Y\))

Thus, the sequence of amino acids is FLDY. Quick Tip: Learn the one-letter codes for amino acids to identify peptide sequences quickly.

- (A) \(\Delta U = q + p\Delta V\) is incorrect because the first law of thermodynamics states \(\Delta U = q + w\), where \(w = -p\Delta V\). Hence, the correct relation is \(\Delta U = q - p\Delta V\).

- (B) \(G = H - TS\) is correct, as it is the definition of Gibbs free energy (\(G\)).

- (C) \(\Delta S = \frac{q_{rev}}{T}\) is correct, as it represents the change in entropy (\(S\)) under reversible conditions.

- (D) \(\Delta H = \Delta U - nRT\) is incorrect because for an ideal gas, \(\Delta H = \Delta U + nRT\). Quick Tip: Remember that \(\Delta U = q + w\), where \(w = -p\Delta V\), and \(\Delta H = \Delta U + nRT\) for ideal gases.

- Calamine (\(ZnCO_3\)) is a carbonate-based mineral.

- Siderite (\(FeCO_3\)) is an iron carbonate mineral.

- Sphalerite (\(ZnS\)) is a sulphide-based mineral, making it the correct answer.

- Malachite (\(Cu_2CO_3(OH)_2\)) is a copper carbonate hydroxide mineral. Quick Tip: Sphalerite (\(ZnS\)) is a sulphide ore commonly associated with zinc extraction.

- Statement I is correct because nickel is widely used as a catalyst in hydrogenation reactions for producing edible fats (like margarine) and in the production of synthesis gas (syn gas).

- Statement II is incorrect as hydrides of silicon (\(SiH_4\)) are electron-precise and neither electron-rich nor electron-deficient. Quick Tip: Silicon hydrides are electron-precise, unlike boron hydrides, which can be electron-deficient.

- \(i\) (van’t Hoff factor) is associated with the abnormal molar mass (\(M_{ab}\)), making A-III correct.

- \(k_f\) (cryoscopic constant) relates to the depression of freezing point, making B-I correct.

- Solutions with the same osmotic pressure are isotonic solutions, making C-II correct.

- Azeotropes are solutions with the same composition in both the liquid and vapour phases, making D-IV correct. Quick Tip: Understand colligative properties and their relation to the van’t Hoff factor and isotonic solutions for better accuracy.

The decomposition of lithium nitrate (\(LiNO_3\)) is as follows: \[ 2LiNO_3 \rightarrow 2Li_2O + 2NO_2 + \frac{1}{2}O_2 \]
The reaction produces three compounds: \(Li_2O\), \(NO_2\), and \(O_2\). Quick Tip: Thermal decomposition of nitrates often produces oxides, nitrogen oxides, and oxygen depending on the metal.

The equilibrium constant for the given reaction is calculated by combining the given equilibria: \[ K_{eq} = \frac{K_2 \times K_3^3}{K_1} \]
Substituting the values: \[ K_{eq} = \frac{(1.6 \times 10^{12}) \times (1.0 \times 10^{13})^3}{4 \times 10^5} \] \[ K_{eq} = \frac{1.6 \times 10^{12} \times 10^{39}}{4 \times 10^5} = 4 \times 10^{33} \] Quick Tip: For combined equilibria, multiply or divide the equilibrium constants based on how the reactions are combined.

The unit of the rate constant is given as \(L \, mol^{-1} \, s^{-1}\), which corresponds to a second-order reaction. The general formula for the unit of a rate constant is: \[ Unit of k = (concentration)^{1-n} \times time^{-1} \]
where \(n\) is the order of the reaction. For \(n = 2\), the unit becomes \(L \, mol^{-1} \, s^{-1}\). Quick Tip: The units of the rate constant can be used to quickly determine the order of the reaction.

Acidic oxides react with water to form acids. Among the given oxides:
- Acidic oxides: \(N_2O_3\), \(Cl_2O_7\), \(SO_2\), \(NO_2\)
- Neutral oxides: \(NO\), \(N_2O\), \(CO\)
- Basic oxides: \(CaO\), \(Na_2O\)

Thus, there are 4 acidic oxides. Quick Tip: Classify oxides as acidic, basic, or neutral based on their reaction with water and acids/bases.

- Mass of carbon in the compound: \[ 0.01 \times \frac{60}{100} = 0.006 \, g \]
- Moles of CO\(_2\) produced: \[ \frac{4.4}{44} = 0.1 \, mol \]
- Mass of carbon in CO\(_2\): \[ 0.1 \times 12 = 1.2 \, g \]
- The molar mass \(M\) of the compound: \[ M = \frac{mass of compound}{moles of compound} = \frac{0.01}{0.006} \times 12 = 200 \, g/mol \] Quick Tip: Relate the mass and moles of carbon to deduce the molar mass using the percentage composition.

Fehling’s reagent contains \(Cu^{2+}\) ions complexed with tartrate ions. Tartrate is a bidentate ligand as it binds to the metal through two donor atoms. Quick Tip: The denticity of a ligand refers to the number of donor atoms through which it binds to a metal ion.

- One unit cell of hcp contains 18 voids.

- Total number of voids in 0.02 mol of hcp: \[ No. of voids = 18 \times 6.02 \times 10^{23} \times 0.02 \] \[ = 3.6 \times 10^{21} \] Quick Tip: In a hexagonal close-packed structure (hcp), the voids are always proportional to the number of atoms.

The radius of the \(n\)-th Bohr orbit is given by: \[ r_n = r_1 \frac{n^2}{Z} \]
For \(n = 3\) and \(Z = 2\): \[ r_{He^{+}} = 0.6 \times \frac{3^2}{2} = 2.7 \, Å \]
Converting to picometers: \[ 2.7 \, Å = 270 \, pm \] Quick Tip: For hydrogen-like atoms, the radius is inversely proportional to the nuclear charge (\(Z\)).

The equilibrium constant is related to the electrode potentials: \[ \Delta G^\circ = -nF E^\circ = -2.303RT \log K \] \[ E^\circ_{cell} = \frac{0.059}{2} \log K \]
Substituting \(K = 1 \times 10^{20}\): \[ E^\circ_{cell} = \frac{0.059}{2} \log (1 \times 10^{20}) = 0.059 \times 10 = 0.59 \, V \]
Using \(E^\circ_{cell} = E^\circ_{Sn^{2+}/Sn} - E^\circ_{Zn^{2+}/Zn}\): \[ 0.59 = E^\circ_{Sn^{2+}/Sn} - (-0.76) \] \[ E^\circ_{Sn^{2+}/Sn} = 0.59 - 0.76 = 0.17 \, V = 17 \times 10^{-2} \, V \] Quick Tip: For electrochemical cells, use \(\Delta G^\circ\) or \(K_{eq}\) to find electrode potentials.

Moles of Na: \[ Moles of Na = \frac{0.69}{23} = 3 \times 10^{-2} \] \[ Na + H_2O \rightarrow NaOH + \frac{1}{2} H_2 \]
Moles of NaOH produced = \(3 \times 10^{-2}\).

No. of equivalents of NaOH = No. of equivalents of HCl.

Mass of HCl = \(73 \, g/L\). Normality: \[ Normality = \frac{73}{36.5} = 2 \, N \]
Using \(N_1 V_1 = N_2 V_2\): \[ 2 \times V = 3 \times 10^{-2} \quad \Rightarrow \quad V = 15 \, mL \] Quick Tip: Relate the moles of reactants to equivalents and use the concept of normality for neutralization reactions.