JEE Main Chemistry Mock Test- 2 - JEE MCQ

Salicylic acid reacts with methanol in the presence of H₂SO₄ to form methyl salicylate, which has a wintergreen smell.

"Methyl salicylate is formed by the esterification of salicylic acid (C₆H₅(OH)COOH) with methanol (CH₃OH), yielding C₆H₅(OH)COOCH₃."

The iodoform test detects compounds containing a methyl ketone group (CH₃CO) or compounds that can be oxidized to such groups, like alcohols that can be oxidized to acetaldehyde or acetone.

Now, let's analyze the options:

1. Ethanol: Ethanol (CH₃CH₂OH) is a primary alcohol, which can be oxidized to acetaldehyde (CH₃CHO), which contains a methyl group attached to a carbonyl group. Hence, ethanol will give the iodoform test.

2. Ethanal: Ethanal (CH₃CHO) is acetaldehyde and directly contains a methyl group attached to a carbonyl group. It will also give the iodoform test.

3. Isopropyl alcohol: Isopropyl alcohol (CH₃CH(OH)CH₃) is a secondary alcohol, and it can be oxidized to acetone (CH₃COCH₃), which contains a methyl group attached to a carbonyl group. Hence, it will give the iodoform test.

4. Benzyl alcohol: Benzyl alcohol (C₆H₅CH₂OH) is a primary alcohol and cannot be oxidized to form a methyl ketone or acetone. It does not contain the necessary group to react with iodine and will not give the iodoform test.

Answer: D: Benzyl alcohol

Phospholipids are esters of glycerol with two carboxylic acid residues and one phosphate group. They are derivatives of glycerol in which two hydroxyl groups are esterified with fatty acids, and the third is esterified with phosphoric acid.

The splitting of spectral lines under a magnetic field is called the Zeeman Effect. The Stark Effect involves splitting due to an electric field. The Photoelectric Effect refers to the emission of electrons when light strikes a metal surface and is unrelated to spectral line splitting.

Iron (Fe) has the electron configuration [Ar] 3d⁶ 4s². For Fe²⁺ ([Ar] 3d⁶), there are 4 unpaired d-electrons in a high-spin state. For Fe³⁺ ([Ar] 3d⁵), there are 5 unpaired d-electrons. Fe (neutral, [Ar] 3d⁶ 4s²) has 4 unpaired d-electrons. Fe⁻ is not a standard ion; assuming Fe⁺ ([Ar] 3d⁶ 4s¹), it has fewer unpaired electrons. Thus, Fe³⁺ has the maximum (5 unpaired d-electrons).

Hyperconjugation effect in Alkyl group:

Alkyl groups are ortho- and para-directing in electrophilic aromatic substitution due to the hyperconjugation effect. Hyperconjugation involves the donation of electron density from the C–H σ-bonds of the alkyl group to the π-system of the aromatic ring, stabilizing the transition state for ortho and para substitution. The inductive effect (option A) also contributes but is secondary to hyperconjugation.

In graphite, each carbon atom uses three of its four valence electrons for covalent bonding with three other carbon atoms in a planar structure, forming a hexagonal lattice. The fourth electron is delocalized in the π-system, spreading out across the structure, which accounts for graphite’s electrical conductivity.

BF₃ has sp² hybridization, with boron forming three equivalent bonds to fluorine atoms, resulting in a trigonal planar structure where all atoms lie in one plane. In contrast, CH₄ (tetrahedral, sp³), PF₅ (trigonal bipyramidal, sp³d), and NH₃ (trigonal pyramidal, sp³) are non-planar.

(A) CF₄: Tetrahedral (sp³, 4 bonding pairs).
SF₄: Seesaw (sp³d, 4 bonding pairs, 1 lone pair).
Different shapes.

(B) BF₃: Trigonal planar (sp², 3 bonding pairs).
PCl₃: Trigonal pyramidal (sp³, 3 bonding pairs, 1 lone pair).
Different shapes.

(C) XeF₂: Xe is sp³d hybridized, 5 electron pairs (2 bonding, 3 lone pairs). Lone pairs in the equatorial positions of the trigonal bipyramidal structure, giving a linear shape.
CO₂: C is sp hybridized, 2 double bonds, linear shape.
Both are linear.

(D) PF₅: Trigonal bipyramidal (sp³d, 5 bonding pairs).
IF₅: Square pyramidal (sp³d², 5 bonding pairs, 1 lone pair).
Different shapes.

Final Answer: Only (C) has identical shapes (linear).
Correct Answer: (C) XeF₂, CO₂.

(A) cis−[Pt(NH₃)₂Cl₂]: Square planar, has a plane of symmetry. : No optical isomerism.

(B) trans−[Pt(NH₃)₂Cl₂]: Square planar, highly symmetric (multiple planes). : No optical isomerism.

(C) cis−[Co(en)₂Cl₂]: Octahedral, Co(III) with two bidentate ethylenediamine (en) and two Cl ligands in cis positions.
Lacks a plane of symmetry, forming non-superimposable enantiomers. : Exhibits optical isomerism.

Optical isomers of cis-[Co(en)₂Cl₂] :

(D) trans−[Co(en)₂Cl₂]: Octahedral, symmetric with planes of symmetry. : No optical isomerism.

Final Answer:
Only (C) exhibits optical isomerism.
Correct Answer: (C) cis−[Co(en)₂Cl₂].

To find the oxidation state of Cr in [Cr(NH₃)₄Cl₂]⁺, sum the charges of all species to equal the complex's charge (+1):
Let Cr oxidation state = x. NH₃ is neutral (charge 0), Cl⁻ has charge -1.
x + (0 × 4) + (−1 × 2) = +1
x − 2 = +1
x = +3
Oxidation state of Cr is +3.
Answer: Correct (C: +3).

The Mond process purifies nickel by converting nickel oxides to volatile Ni(CO)4, which is then decomposed to pure nickel. Other processes:

• Bosch process: Produces hydrogen.
• Bessemer process: Refines steel.
• Leblanc process: Produces soda ash.

Answer: Correct (A: Mond’s process).

Polarizability increases with the number of electrons and atomic size, as larger electron clouds are more easily distorted, strengthening van der Waals forces.

Noble gases: He (2 electrons, smallest), Ne (10 electrons), Ar (18 electrons), Xe (54 electrons).

Helium has the fewest electrons and smallest size, making it the least polarizable with the weakest van der Waals forces.

Answer: Correct (A: He).

Longest chain: 7 carbons (hept-). Contains a triple bond (yne) at carbon 1 and a double bond (ene) at carbon 5 (numbering from the left to minimize locants).

Substituents: Chloro at carbon 6, ethyl at carbon 4, methyl at carbon 5.

Name (alphabetical order): 6-chloro-4-ethyl-5-methyl hept-1-yn-5-ene.

Answer: Correct (A: 6-Chloro-4-ethyl-5-methyl hept-1-yn-5-ene).

When HCl gas is passed through a saturated NaCl solution, it ionizes:

HCl → H⁺ + Cl⁻,
NaCl ⇌ Na⁺ + Cl⁻.

The added Cl⁻ increases [Cl⁻], so the ionic product [Na⁺][Cl⁻] exceeds the solubility product Ksp of NaCl, causing precipitation.

Answer: Correct (D: The ionic product exceeds the solubility product).

Stereoisomers have the same molecular formula and connectivity but differ in spatial arrangement.
(A) Chain and rotational isomerism: Chain is structural, rotational is conformational, not stereoisomerism.
(B) Optical and geometric isomerism: Both are types of stereoisomerism (enantiomers and cis-trans, respectively).
(C) Linkage and geometric: Linkage is structural, not stereoisomerism.
(D) Structural and geometric: Structural isomerism involves different connectivity, not stereoisomerism.
Only (B) represents stereoisomerism.
Answer: Correct (B: Optical isomerism and geometric isomerism).

Elements in the same period have the same principal quantum number for their valence electrons.
Calcium (Ca) and zinc (Zn) are in Period 4 ([Ar] 4s² and [Ar] 3d¹⁰ 4s²).
Other pairs are in different periods:
P (Period 3) and Se (Period 4);
Mg (Period 3) and Sb (Period 5);
Ag (Period 5) and Cl (Period 3).

Natural rubber is a polymer of cis-isoprene (cis-2-methyl-1,3-butadiene), formed by addition polymerization. It is obtained as a milky fluid (latex) from rubber trees. Trans-isoprene forms gutta-percha, a different material.

Alkali metals (Group 1) have low first ionization energies due to their ns¹ electron configuration, allowing them to easily lose one electron to form a +1 ion. This makes them strong reducing agents, as they readily donate electrons to other species.

For a non-ideal solution with negative deviation, the volume change on mixing (ΔVₘᵢₓ) is negative due to attractive intermolecular forces between A and B.
The expected volume (100 + 25 = 125 ml) decreases slightly,
so the final volume is close to 125 ml.

The azide ion (N₃⁻) is linear with the central nitrogen having sp hybridization (two bonding domains). We analyze each species:

• N₂O: Central N in N≡N–O is sp hybridized (linear).
• C₂H₂: Each C in H–C≡C–H is sp hybridized (linear).
• CO₂: C in O=C=O is sp hybridized (linear).
• C₃O₂: Central C in O=C=C=C=O is sp hybridized (linear).
• BeF₂: Be in F–Be–F is sp hybridized (linear).
• NO₂: N is sp² hybridized (bent, with resonance).
• PF₃: P is sp³ hybridized (trigonal pyramidal).

Thus, N₂O, C₂H₂, CO₂, C₃O₂, and BeF₂ have at least one atom with sp hybridization, giving a total of 5 species.

A chiral center is a carbon atom bonded to four different groups. Without the specific structure, we assume the compound has 3 carbons, each with four distinct substituents (e.g., a steroid-like molecule with multiple asymmetric carbons).

Upon re-inspection of the structure, there are 3 chiral centers in this compound.
So, the correct number of chiral centers is 3.

For an ideal solution,
Raoult’s Law states: (P₀ – Pₛ)/P₀ = xₛₒₗᵤₜₑ,
where P₀ is the vapor pressure of the pure solvent,
Pₛ is the solution’s vapor pressure (80 torr), and
xₛₒₗᵤₜₑ is the mole fraction of the non-volatile solute (0.2).
Substituting: (P₀ – 80)/P₀ = 0.2 → 0.8P₀
= P₀ – 80 → 80
= 0.2P₀ → P₀
= 80/0.2 = 100^torr.
Thus, the vapor pressure of the pure solvent is 100^torr."

For a monatomic ideal gas,
Cᵥ = (3/2)R = (3/2) × 2 = 3 cal/K-mol.
In an adiabatic process, work done by the gas (W = –75 cal) equals the change in internal energy:
nCᵥ(T₂ – T₁) = –75.
Given n = 0.1 mol, T₁ = 227°C = 500 K:
0.1 × 3 × (T₂ – 500) = –75
T₂ – 500 = –75/0.3
= –250
T₂ = 250 K.
So, the final temperature T₂ is 250 K.