IIT JEE Prelims Chemistry Test

Solution

Correct: D

NO (Nitric Oxide) has an odd number of electrons (15). According to Molecular Orbital Theory, this results in one unpaired electron in the π*2p molecular orbital, making it paramagnetic. The other options have all their electrons paired.

Solution

Correct: B

Using the Arrhenius equation: ln(k2/k1) = (Ea/R) * (1/T1 - 1/T2). ln(2.57 x 10^-2 / 1.78 x 10^-3) = (Ea/8.314) * (1/300 - 1/310). Solving for Ea, we get approximately 63969 J/mol, which is about 64.0 kJ/mol.

Solution

Correct: C

Since the concentration of HCl is very low, we must consider the auto-ionization of water. [H+] = 10^-8 (from HCl) + 10^-7 (from water) = 1.1 x 10^-7 M. pH = -log[H+] = -log(1.1 x 10^-7) ≈ 6.96.

Solution

Correct: D

Acetaldehyde, being an aldehyde, will reduce Tollen's reagent and Fehling's solution, forming a silver mirror or a red precipitate of Cu2O, respectively. Acetone, being a ketone, does not reduce these reagents. I2/NaOH (iodoform test) will also distinguish, with acetaldehyde giving a positive test (yellow precipitate of iodoform). Therefore, all are valid

Solution

Correct: A

De Broglie wavelength (λ) = h/p, where p is momentum. Kinetic energy (E) = p^2 / 2m, so p = √(2mE). Therefore, λ = h/√(2mE).

Solution

Correct: A

Alkyl benzenes are oxidized to benzoic acid by KMnO4/H+ regardless of the length of the alkyl chain (except for tertiary butyl groups).

Solution

Correct: B

The number of radial nodes = n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number. For a 3p orbital, n = 3 and l = 1. Therefore, the number of radial nodes = 3 - 1 - 1 = 1.

Solution

Correct: C

In polar protic solvents, nucleophilicity increases down the group because larger ions are better solvated and less tightly bound to the protons of the solvent, making them more available to attack. Therefore, I- is the strongest nucleophile.

Solution

Correct: B

The IUPAC name is Bis(ethylenediamine)dichlorocobalt(III) chloride. 'en' stands for ethylenediamine. The oxidation state of Cobalt is +3 (x + 0 + (-2) = +1, so x=3). The ligands are named alphabetically before the metal.

Solution

Correct: C

Bakelite is a thermosetting polymer, meaning it undergoes irreversible chemical change upon heating, forming a rigid network. The other options are thermoplastics, which can be repeatedly softened by heating and hardened by cooling.

Solution

Correct: C

K2SO4 dissociates into 2 K+ ions and 1 SO42- ion in water. Therefore, i = 1 + (n-1)α, where n is the number of ions produced (3 in this case) and α is the degree of dissociation. Assuming complete dissociation (α=1), i = 3.

Solution

Correct: A

A positive iodoform test is given by compounds containing the CH3CO- group or compounds that can be oxidized to form this group. Ethanol (CH3CH2OH) can be oxidized to acetaldehyde (CH3CHO), which contains the CH3CO- group.

Solution

Correct: B

ΔG° = -nFE°, where n is the number of moles of electrons transferred, F is the Faraday constant, and E° is the standard electrode potential. -212300 J/mol = -2 * 96500 C/mol * E°. Solving for E°, we get E° ≈ -1.10 V.

Solution

Correct: B

The nitro group (-NO2) is a strong electron-withdrawing group, which stabilizes the phenoxide ion after deprotonation, making o-nitrophenol the most acidic. Methoxy is electron donating so it reduces acidity. Cresol is less acidic than phenol itself.

Solution

Correct: C

Kp = Kc(RT)^Δn, where Δn = (moles of gaseous products) - (moles of gaseous reactants) = 2 - (1+3) = -2. So, Kc = Kp / (RT)^Δn = 1.64 x 10^-4 / (0.0821 * 673)^-2. Solving results in Kc = 1.64 x 10^-4 / ((0.0821 * 673) ^ -2)

Solution

Correct: D

Peptide bonds are not easily hydrolyzed under physiological conditions. Strong acids or bases are required for hydrolysis. The other statements are correct.

Solution

Correct: C

The overall order of the reaction is the sum of the exponents in the rate law. Therefore, the overall order is 1/2 + 2 = 5/2.

Solution

Correct: B

Trimethylsilyl chloride (TMSCl) in the presence of a base like triethylamine (Et3N) is used to protect alcohols as silyl ethers. The silyl ether can later be removed with fluoride ions (e.g., TBAF).

Solution

Correct: B

Bond dissociation energy generally decreases down the halogen group. However, F2 has a lower bond dissociation energy than Cl2 due to the small size and high electron density of fluorine, leading to interelectronic repulsion. Cl2 has the highest bond dissociation energy among the given options.

Solution

Correct: B

In the presence of peroxide, HBr undergoes anti-Markovnikov addition. Thus, the bromine adds to the less substituted carbon. The product is 1-Bromo-2-methylpropane