Question 1
Which of the following $\alpha$-amino acids is NOT optically active?
✅ Correct Answer: Glycine
Explanation: Glycine is the only naturally occurring $\alpha$-amino acid that is optically inactive. This is because its side chain is just a Hydrogen atom ($-H$), meaning its $\alpha$-carbon is attached to two identical Hydrogen atoms, making it completely achiral.
Question 2
A mixture of Nitric acid ($HNO_3$) and water forms a maximum boiling azeotrope. According to NCERT, what is the approximate mass composition of this azeotrope?
✅ Correct Answer: 68% nitric acid and 32% water
Explanation: Azeotropes are binary mixtures having the same composition in liquid and vapor phase. Nitric acid and water form a maximum boiling azeotrope (showing negative deviation from Raoult's law) at a specific composition of exactly 68% $HNO_3$ and 32% $H_2O$ by mass.
Question 3
The strong reducing behavior of Hypophosphorous acid ($H_3PO_2$) is attributed to its structure containing:
✅ Correct Answer: One $P-OH$ bond and two $P-H$ bonds
Explanation: Acids in the oxoacids of phosphorus series that contain $P-H$ bonds act as strong reducing agents. Hypophosphorous acid ($H_3PO_2$) has one $P=O$ bond, one $P-OH$ bond, and two $P-H$ bonds, making it a very potent reducing agent.
Question 4
Maltose is a disaccharide obtained by the partial hydrolysis of starch. Which of the following correctly describes the glycosidic linkage in Maltose?
✅ Correct Answer: Two $\alpha$-D-glucose units linked via C1-C4
Explanation: According to NCERT, Maltose is composed exclusively of two $\alpha$-D-glucose units. The linkage is an $\alpha$-glycosidic bond established between the C1 of one glucose molecule and the C4 of the second glucose molecule.
Question 5
Which of the following is the correct order of increasing ionic radii for the given isoelectronic species?
✅ Correct Answer: $Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$
Explanation: For isoelectronic species (all have 10 electrons here), the ionic radius is inversely proportional to the nuclear charge (number of protons). $Mg^{2+}$ has the highest nuclear charge (+12), pulling electrons closest, making it the smallest. $N^{3-}$ has the lowest nuclear charge (+7), making it the largest.
Question 6
According to Molecular Orbital Theory (MOT), which of the following species is diamagnetic?
✅ Correct Answer: $C_2$
Explanation: In $C_2$ (12 electrons), the valence electrons fill up to the $\pi_{2px}$ and $\pi_{2py}$ bonding molecular orbitals, and all electrons are paired. Therefore, it is strictly diamagnetic. $O_2$ and $S_2$ (vapor) have unpaired electrons in their $\pi^*$ antibonding orbitals. $N_2^+$ has an odd number of electrons (13), so it must be paramagnetic.
Question 7
Which of the following carbocations is the most stable?
✅ Correct Answer: tert-Butyl carbocation ($(CH_3)_3C^+$)
Explanation: While resonance usually provides significant stability, the tert-butyl carbocation has 9 $\alpha$-hydrogens, allowing for 9 hyperconjugative structures along with a strong $+I$ effect from three methyl groups. This extensive hyperconjugation makes the $3^\circ$ alkyl carbocation exceptionally stable, even slightly more stable than the benzyl carbocation in most contexts.
Question 8
An ideal gas undergoes free expansion into a vacuum under perfectly adiabatic conditions. Which of the following sets of thermodynamic parameters is correct for this process?
✅ Correct Answer: $q=0$, $\Delta T=0$, $w=0$
Explanation: For free expansion into a vacuum, the external pressure ($P_{ext}$) is zero. Since $w = -P_{ext} \Delta V$, work done ($w$) is rigorously 0. The process is adiabatic, meaning no heat exchange occurs ($q=0$). According to the first law of thermodynamics ($\Delta U=q+w$), $\Delta U=0$. For an ideal gas, internal energy depends only on temperature, so if $\Delta U=0$, then $\Delta T=0$.
Question 9
Assertion (A): Zirconium ($Zr$) and Hafnium ($Hf$) occur together in minerals and are very difficult to separate chemically.
Reason (R): They have almost identical atomic radii due to the poor shielding effect of 4f electrons.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A
Explanation: Hafnium (Period 6) comes immediately after the lanthanide series. The filling of 4f orbitals before 5d orbitals results in a poor shielding effect, causing the effective nuclear charge to pull the valence electrons closer. This is the lanthanoid contraction. Because of this, $Hf$ (159 pm) has nearly the same radius as $Zr$ (160 pm) from Period 5, leading to almost identical chemical properties.
Question 10
Assertion (A): Chlorobenzene undergoes nucleophilic substitution reactions much faster than cyclohexyl chloride.
Reason (R): The $C-Cl$ bond in chlorobenzene acquires a partial double bond character due to resonance.
✅ Correct Answer: A is false but R is true
Explanation: Assertion (A) is completely false. Aryl halides like chlorobenzene are extremely unreactive towards nucleophilic substitution compared to alkyl/cycloalkyl halides (like cyclohexyl chloride). Reason (R) is true and is actually the cause of this low reactivity; resonance gives the $C-Cl$ bond partial double bond character, making it much harder to break.
Question 11
Statement I: The number of angular nodes for a 3d orbital is 2.
Statement II: The number of radial nodes for a 3p orbital is 1.
✅ Correct Answer: Both Statement I and Statement II are correct
Explanation: The number of angular nodes is given by the azimuthal quantum number ($l$). For a d-orbital, $l=2$. So, Statement I is true. The number of radial nodes is given by the formula $n-l-1$. For a 3p orbital, $n=3$ and $l=1$. Radial nodes = $3-1-1=1$. Statement II is also perfectly true.
Question 12
Statement I: According to Valence Bond Theory (VBT), $[CoF_6]^{3-}$ is an inner orbital complex.
Statement II: Fluoride ion ($F^-$) is a weak field ligand and generally does not cause pairing of electrons in the central metal ion.
✅ Correct Answer: Statement I is incorrect, but Statement II is correct
Explanation: Statement II is correct; $F^-$ is a weak field ligand and doesn't force electron pairing against Hund's rule. Because electrons are not paired, the inner 3d orbitals of $Co^{3+}$ remain occupied. The complex must use outer 4d orbitals for hybridization ($sp^3d^2$), making it an outer orbital (high spin) complex. Therefore, Statement I is completely false.
Question 13
For a first-order reaction, the time required for 99.9% completion of the reaction is approximately how many times its half-life ($t_{1/2}$)?
✅ Correct Answer: 10 times
Explanation: For a first-order reaction: $t = \frac{2.303}{k} \log \left(\frac{a}{a-x}\right)$.
For 99.9% completion: $t_{99.9} = \frac{2.303}{k} \log \left(\frac{100}{100 - 99.9}\right) = \frac{2.303}{k} \log(1000) = \frac{2.303 \times 3}{k}$.
Since $t_{1/2} = \frac{0.693}{k} = \frac{2.303 \times 0.3010}{k}$, the ratio $\frac{t_{99.9}}{t_{1/2}} \approx \frac{3}{0.3010} \approx 10$.
Question 14
Calculate the standard reduction potential of a hydrogen electrode placed in a buffer solution with a pH of 10 at 298 K.
✅ Correct Answer: $-0.591\text{ V}$
Explanation: For a hydrogen electrode: $H^+ + e^- \rightarrow \frac{1}{2}H_2$.
Applying the Nernst equation: $E = E^\circ - \frac{0.0591}{1} \log \frac{1}{[H^+]}$.
Since $E^\circ = 0\text{ V}$ for standard hydrogen electrode, and $\log \frac{1}{[H^+]} = -\log[H^+] = \text{pH}$.
$E = -0.0591 \times \text{pH} = -0.0591 \times 10 = -0.591\text{ V}$.
Question 15
$16\text{ g}$ of oxygen gas ($O_2$) and $4\text{ g}$ of hydrogen gas ($H_2$) are ignited in a closed vessel to form water. Identify the limiting reagent and calculate the maximum mass of water formed.
✅ Correct Answer: $O_2$ is limiting, $18\text{ g}$ of water is formed
Explanation: Balanced equation: $2H_2 + O_2 \rightarrow 2H_2O$.
Moles of $O_2 = 16 / 32 = 0.5\text{ mol}$.
Moles of $H_2 = 4 / 2 = 2\text{ mol}$.
According to stoichiometry, 1 mol $O_2$ reacts with 2 mol $H_2$. So, 0.5 mol $O_2$ requires 1 mol $H_2$.
Since we have 2 mol $H_2$, $O_2$ is heavily limiting. 0.5 mol $O_2$ produces 1 mol $H_2O$. Mass of 1 mol $H_2O = 18\text{ g}$.
Question 16
When an aliphatic or aromatic primary amine is heated with chloroform and ethanolic potassium hydroxide, it forms a product with a highly offensive smell. What is this product?
✅ Correct Answer: An isocyanide
Explanation: This is the famous Carbylamine reaction (or isocyanide test), an important chemical test exclusively for primary amines. Primary amines react with $CHCl_3$ and ethanolic $KOH$ to form alkyl/aryl isocyanides (carbylamines), which possess a characteristic foul, highly offensive smell.
Question 17
Arrange the following alkyl halides in the decreasing order of their reactivity towards $S_N2$ mechanism:
I. $CH_3Cl$
II. $CH_3CH_2Cl$
III. $(CH_3)_2CHCl$
IV. $(CH_3)_3CCl$
✅ Correct Answer: I > II > III > IV
Explanation: The $S_N2$ mechanism proceeds via a transition state where the nucleophile strictly attacks from the back. Bulky alkyl groups block this backside attack (steric hindrance). Therefore, the reactivity order for $S_N2$ is Methyl > $1^\circ$ > $2^\circ$ > $3^\circ$. Order: $CH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl$.
Question 18
An organic compound $X$ with the molecular formula $C_9H_{10}O$ forms a 2,4-DNP derivative, reduces Tollens' reagent, and undergoes Cannizzaro reaction. On vigorous oxidation with $KMnO_4$, it gives 1,2-benzenedicarboxylic acid. Compound $X$ is:
✅ Correct Answer: 2-Ethylbenzaldehyde
Explanation: 1. Forms 2,4-DNP and reduces Tollens' -> It is an aldehyde.
2. Undergoes Cannizzaro -> Has NO $\alpha$-hydrogen. Thus, the $-CHO$ group is directly attached to the benzene ring.
3. Oxidation gives 1,2-benzenedicarboxylic acid (phthalic acid) -> The substituents on the benzene ring must be ortho to each other.
Given formula $C_9H_{10}O$: Benzene ring (6C) + $CHO$ group (1C) leaves 2 carbons, an ethyl group. It must be an ethyl group ortho to the aldehyde group. Thus, 2-Ethylbenzaldehyde.
Question 19
Consider the synthesis of Ammonia by Haber's process:
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92.4\text{ kJ/mol}$
If an inert gas (like Helium) is added to the system at constant volume at equilibrium, the reaction will:
✅ Correct Answer: Remain unaffected
Explanation: According to Le Chatelier's principle, the addition of an inert gas strictly at constant volume does not change the partial pressures or the molar concentrations of the reacting gases. Since the active reacting concentrations remain unchanged, the equilibrium is entirely unaffected.
Question 20
Which of the following hydrides of Group 15 elements has the highest boiling point?
✅ Correct Answer: $BiH_3$
Explanation: Boiling points of group 15 hydrides normally increase down the group due to an increase in molecular size and van der Waals dispersion forces. $NH_3$ is an exception and has a much higher boiling point than $PH_3$ and $AsH_3$ due to hydrogen bonding, but it is still fundamentally lower than the massive $SbH_3$ and $BiH_3$. Order: $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$. Hence, $BiH_3$ has the absolute highest boiling point.
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