JEE Main Chemistry Mock Test- 3 - JEE MCQ

Salicylic acid (C₆H₄(OH)COOH) reacts with methanol (CH₃OH) in the presence of H₂SO₄ as a catalyst to form methyl salicylate (C₆H₄(OH)COOCH₃), an ester with a wintergreen smell. Other options (succinic, tartaric, oxalic acids) do not produce this ester under these conditions.

1-Phenylethanol (C₆H₅CH(OH)CH₃) is synthesized by reacting benzaldehyde (C₆H₅CHO) with a Grignard reagent. Methyl iodide (CH₃I) reacts with magnesium to form methylmagnesium iodide (CH₃MgI), which adds to the carbonyl group of benzaldehyde. Hydrolysis yields 1-phenylethanol. Other options are incorrect: ethyl iodide forms a different alcohol, and methyl bromide with or without AlBr₃ does not produce the desired product.

When an electron transitions from a higher-energy shell (M, n=3) to a lower-energy shell (L, n=2), it releases energy in the form of a photon, producing an emission spectrum. This is characteristic of emission spectra, not absorption (which involves energy absorption to higher shells), continuous (broad spectra from non-quantized sources), or X-rays (a specific type of emission in heavy atoms). Thus, the answer is emission.

The energy of atomic orbitals is determined by the (n + l) rule, where n is the principal quantum number and l is the azimuthal quantum number (s = 0, p = 1, d = 2, f = 3). Orbitals with lower n+l have lower energy; for equal n + l, lower n is lower in energy. Calculating:

• 5p: n = 5, l = 1, n + l = 6
• 6s: n = 6, l = 0, n + l = 6
• 4f: n = 4, l = 3, n + l = 7
• 5d: n = 5, l = 2, n + l = 7

Order: 5p (n+ l = 6, n = 5) < 6s (n + l = 6, n = 6) < 4f (n+l = 7, n = 4) < 5d (n + l = 7, n = 5).
Thus, the correct order is 5p < 6s < 4f < 5d

According to the Linear Combination of Atomic Orbitals (LCAO) theory, combining two atomic orbitals (AOs) produces two molecular orbitals (MOs): one bonding MO (lower energy) and one antibonding MO (higher energy). Thus, the answer is two MOs (option A).

Elimination of HBr from 2-bromobutane (CH₃CHBrCH₂CH₃) via an E2 reaction with a strong base forms an alkene.
According to Zaitsev’s rule, the more substituted (stable) alkene is favored.
2-Butene (CH₃CH=CHCH₃) is the major product, while 1-butene (CH₂=CHCH₂CH₃) is minor. 2-Butyne requires different conditions (e.g., strong base for alkyne formation), and an equimolar mixture is unlikely.
Thus, the answer is predominantly 2-butene (option B).

The compound

is a butene with a double bond between C2 and C3. According to IUPAC rules, the parent chain is but-2-ene (double bond gets the lowest number). Substituents are:

• C₁: chloro
• C₂: bromo
• C₃: bromo
• C₄: chloro

Alphabetical order prioritizes bromo (2,3-dibromo) over chloro (1,4-dichloro).
Thus, the name is 2,3-dibromo-1,4-dichloro-but-2-ene (option A).
Option B reverses the priority,
C is non-specific, and D is incorrect (no butane)."

The compound CH₃–O–C₂H₅ is an ether. In IUPAC nomenclature, the smaller alkyl group is named as an alkoxy substituent on the larger alkane. Here, the methoxy group (CH₃O–) is attached to ethane (C₂H₅), giving methoxyethane. Other options are incorrect: dimethyl ether is CH₃–O–CH₃, ethyl methyl ether is a common name, and methyl ethyne ether implies a triple bond, which is absent.

Thus, the answer is methoxyethane (option A).

Bond order is proportional to bond strength and inversely proportional to bond length. Higher bond order means shorter bond length.

Given: Bond order of NO = 2.5, NO⁺ = 3. Thus, NO⁺ has a higher bond order and shorter bond length than NO.

Therefore, bond length in NO is greater than in NO⁺.

Answer: Correct (D: Bond length in NO is greater than in NO⁺).

A gene is a segment of DNA that contains the instructions for synthesizing a protein or functional RNA. Nucleosides and nucleotides are DNA components, and ribose is a sugar in RNA, not directly responsible for coding.

K₄[Fe(CN)₆] is a complex compound known as potassium ferrocyanide. It consists of a central metal ion (Fe²⁺) bonded to six cyanide ions (CN⁻) through coordinate bonds, forming a coordination complex. The potassium ions (K⁺) are present as counter ions. This compound is not a simple salt but a complex due to the coordination of the metal ion with the cyanide ligands.

XeF₆ reacts with SiO₂ to form XeOF₄ and SiF4: XeF₆ + SiO₂ → XeOF₄ + SiF₄
XeF₆ is a strong fluorinating agent, oxidizing SiO₂ and forming volatile XeOF₄ and SiF₄, a xenon oxyfluoride.

Explanation of the options:

• Option A: XeSiO₄ is not a known compound.
• Option B: XeF₆ is not a reduction product.
• Option D: SiF₂ is unstable.

Answer: C: XeOF4 + SiF4.

Transition metals are hard due to strong metallic bonding, where delocalized valence electrons (including d-electrons) form a cohesive lattice.(A) Large atomic size: Incorrect, as transition metals have relatively small sizes. 
(B) Metallic bonding: Correct, due to strong electron delocalization. 
(C) Covalent bonding: Incorrect, as covalent bonds are not primary in metals. 
(D) High ionization energy: Contributes to strength but not directly to hardness.

Answer: Incorrect (C: covalent bonding). Correct answer: B: metallic bonding.

Sodium nitroprusside (Na₂[Fe(CN)₅NO]) reacts with sulfide ions (S²⁻) to form a purple-colored complex:

• The reaction is: Fe(CN)₅NO²⁻ + S²⁻ → [Fe(CN)₅NOS]⁴⁻
• The purple color is due to the formation of the complex [Fe(CN)₅NOS]⁴⁻.
• Other options are incorrect due to wrong ligand arrangements or charges.

Answer: C: [Fe(CN)₅NOS]⁴⁻.

When more sodium acetate is added to a solution containing an equimolar mixture of sodium acetate and acetic acid, the solution acts as a buffer.

Sodium acetate (the conjugate base of acetic acid) reacts with the added acetic acid (weak acid) to maintain a stable pH. Adding more sodium acetate increases the concentration of the acetate ion (the conjugate base), which will shift the equilibrium to consume some of the added acetate ions by converting them into acetic acid, thus increasing the pH.

Answer: A: increases.

2-Butene (CH₃-CH=CH-CH₃) exhibits geometrical isomerism due to restricted rotation about the carbon-carbon double bond.
The double bond’s pi bond prevents free rotation, leading to cis (same side) and trans (opposite side) isomers.
The double bond’s pi bond prevents free rotation, leading to cis (same side) and trans (opposite side) isomers.
(A) Chiral carbon: Incorrect, as 2- butene lacks chiral centers. 
(B) Single bond rotation: Incorrect, as single bonds rotate freely. 
(C) Triple bond rotation: Incorrect, as 2-butene has a double bond. 
(D) Double bond rotation: Correct, as restricted rotation causes isomerism.

"Hydrogen’s ionization energy (~1312 kJ/mol) is slightly higher than chlorine’s (~1251 kJ/mol). Hydrogen’s small size and lack of electron shielding result in strong nuclear attraction for its 1s¹ electron. In chlorine (Cl, 3p⁵), larger atomic size and electron shielding reduce the effective nuclear charge, lowering the ionization energy.

Chloroform (CHCl₃) undergoes photooxidation in air under sunlight, reacting with oxygen to form phosgene (COCl₂), a poisonous gas, and hydrochloric acid (HCl).
The reaction is: 2CHCl₃ + O₂ → 2COCl₂ + 2HCl.
No explosion, polymerization, or inertness occurs.
Thus, the answer is option B.

In BF₃, boron is sp² hybridized with an empty p-orbital, enabling pπ-pπ back-bonding with fluorine’s lone pairs. This gives B-F bonds partial double-bond character, increasing their dissociation energy (646 kJ/mol). In CF₄, carbon is sp³ hybridized with no empty p-orbital for π-bonding, resulting in a lower C-F bond energy (515 kJ/mol).
Option A is correct;
options B, C, and D are incorrect as they do not explain the back-bonding effect."

The solution’s vapor pressure (600 mm Hg) exceeds the ideal value predicted by Raoult’s law (e.g., for CS₂ = 0.5, P = 0.5·512 + 0.5·344 = 428 mm Hg), indicating positive deviation due to weaker CS₂-acetone interactions compared to CS₂-CS₂ and acetone-acetone interactions. This implies:
A: True (Raoult’s law is not obeyed).
B: False (positive deviation causes volume expansion, so 100 mL CS₂ + 100 mL acetone > 200 mL).
C: True (weaker CS₂-acetone attractions).
D: True (endothermic mixing absorbs heat).

Thus, B is the false statement."

For the reaction CO₂(g) + C(s) ⇌ 2CO(g), let P be the pressure decrease of CO₂ at equilibrium. Initial P_CO₂ = 0.5 atm. At equilibrium:
P_CO₂ = 0.5 – P
P_CO = 2P
Total pressure: (0.5 – P) + 2P = 0.8 → P = 0.3 atm
Kₚ = (P_CO)² / P_CO₂ = (2·0.3)² / (0.5 – 0.3)
= 0.36 / 0.2
= 1.8 atm
Thus, Kₚ = 1.8 atm (option A)."

For Europium (Z=63), a lanthanide, the group number is 3 (f-block).
For Americium (Z=95), an actinide, the period is 7 (5f series).
For Gadolinium (Z=64), a lanthanide with configuration [Xe] 4f⁷ 5d¹ 6s², the most stable oxidation state is +3 (stable f⁷ configuration).
Thus, x = 3, y = 7, z = 3, and x + y + z = 3 + 7 + 3 = 13.

For the first-order decomposition of hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂, with t₁/₂ = 10 min, the rate constant k = ln(2) / t₁/₂ ≈ 0.0693 min⁻¹.

Given:
The volume of O₂ at t = 20 min is 25 ml.
Use the first-order rate law:

t = (t₁/₂ / log 2) * log(V_∞ / (V_∞ - V_t)),

Substitute the known values into the equation: 20 = (10 / log 2) * log(V_∞ / (V_∞ - 25)).

Now solve for V_∞: log(V_∞ / (V_∞ - 25)) = 2 log 2 → V_∞ / (V_∞ - 25) = 4 → V_∞ = 33.33 ml.

At t >> t₁/₂, the reaction is complete, so the total O₂ volume V_∞ = 33.33 ml.

Hence, the total volume of O₂ at the completion of the reaction is 33.33 ml.

The structure is a cyclohexenone with an α-hydrogen, and you're asking how many hydrogen atoms can be exchanged with deuterium when the compound is placed in D₂O (deuterium oxide) solution for a long time.
In this case, the exchange happens at the α-position, which is the carbon next to the carbonyl group. However, the key part of this reaction is the tautomerization process, where the enol form (which has an OH group) and the keto form (which has a C=O group) are in equilibrium. The enol form is where the hydrogen atoms can be exchanged with deuterium.
Since there are eight possible hydrogen atoms (on the carbons adjacent to the carbonyl group) that can undergo exchange through this process, the correct answer is 8.
Thus, 8 H-atoms can be exchanged with D-atoms.
Answer: 8.

Analyzing each order:
(i) Correct: Para-hydrogen is more stable at low T.
(ii) Incorrect: Glycerol > ethanol (viscosity, ~1,412 cP vs. ~1.1 cP).
(iii) Incorrect: He < D₂ (boiling point, ~4.2 K vs. ~23.7 K).
(iv) Incorrect: H₂O > HF (melting point, 0°C vs. ~–83°C).
(v) Correct: H₃BO₃ > BF₃ (melting point, ~170°C vs. ~–127°C).
(vi) Incorrect: NH₃ > SbH₃ (boiling point, –33°C vs. –17°C).
(vii) Incorrect: H₂O > H₂O₂ (H-bond strength, more stable network).
(viii) Correct: D₂O > H₂O (freezing point, ~3.8°C vs. 0°C).
(ix) Incorrect: HF₂⁻ > HF (H-bond strength, ~155 kJ/mol vs. ~29 kJ/mol).
(x) Incorrect: D₂O > H₂O (bond energy, isotope effect).
Total incorrect: 7 (ii, iii, iv, vi, vii, ix, x)."