Question 1
According to NCERT, a mixture of Chloroform ($CHCl_3$) and Acetone ($CH_3COCH_3$) forms a non-ideal solution. Which of the following statements correctly describes this mixture?
✅ Correct Answer: It shows negative deviation due to intermolecular hydrogen bonding between Chloroform and Acetone.
Explanation: When chloroform and acetone are mixed, a new intermolecular hydrogen bond forms between the hydrogen atom of chloroform and the oxygen atom of acetone. This makes the A-B interactions stronger than pure A-A and B-B interactions, leading to a decrease in vapor pressure (negative deviation from Raoult's law) and a negative enthalpy of mixing ($\Delta H_{\text{mix}} < 0$).
Question 2
During the denaturation of a protein due to change in pH or temperature, which of the following structures remains intact?
✅ Correct Answer: Primary structure only
Explanation: Denaturation involves the uncoiling of the protein helix and the disruption of hydrogen bonds and other secondary forces. According to NCERT, during denaturation, the secondary and tertiary structures are destroyed, but the primary structure (the specific sequence of amino acids linked by robust peptide bonds) remains completely intact.
Question 3
In the preparation and properties of potassium dichromate ($K_2Cr_2O_7$), the chromate and dichromate ions are interconvertible in aqueous solution depending on the pH. Which of the following equations represents the correct transformation in an acidic medium?
✅ Correct Answer: $2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O$
Explanation: In an acidic medium (low pH), the yellow chromate ion ($CrO_4^{2-}$) gets converted into the orange dichromate ion ($Cr_2O_7^{2-}$). The addition of $H^+$ shifts the equilibrium to form dichromate and water. Option B ($Cr_2O_7^{2-} + 2OH^- \rightarrow 2CrO_4^{2-} + H_2O$) represents the reverse transformation that happens in a basic medium.
Question 4
Which of the following complexes is used in the estimation of hardness of water?
✅ Correct Answer: $[Ca(EDTA)]^{2-}$
Explanation: The hardness of water is estimated by simple titration with the $Na_2EDTA$ salt. The $Ca^{2+}$ and $Mg^{2+}$ ions present in hard water form highly stable complexes with EDTA. Thus, the formation of the complex $[Ca(EDTA)]^{2-}$ is directly related to this volumetric analysis.
Question 5
Which of the following represents a set of extensive properties?
✅ Correct Answer: Internal energy, Heat capacity, Volume
Explanation: Extensive properties depend directly on the quantity or size of matter present in the system. Internal energy, total heat capacity, and volume all change proportionally with the amount of substance. Temperature, pressure, density, specific heat, and molarity are intensive properties (independent of quantity).
Question 6
Identify the major product formed when 3-methylbutan-2-ol is treated with $HBr$.
✅ Correct Answer: 2-bromo-2-methylbutane
Explanation: Protonation of the alcohol followed by loss of water generates a secondary ($2^\circ$) carbocation at C-2. This carbocation immediately undergoes a 1,2-hydride shift to form a much more stable tertiary ($3^\circ$) carbocation at C-3. The bromide ion then attacks this $3^\circ$ carbon, yielding the rearranged major product: 2-bromo-2-methylbutane.
Question 7
Which of the following species is NOT aromatic according to Hückel's rule?
✅ Correct Answer: Cyclopentadienyl cation
Explanation: Hückel's rule states that for a planar, cyclic, conjugated system to be aromatic, it must possess $(4n+2) \pi$ electrons. The cyclopentadienyl cation has only $4 \pi$ electrons ($n=1$ in $4n$ anti-aromatic rule), making it strictly anti-aromatic. The cyclopentadienyl anion, however, has $6 \pi$ electrons and is aromatic.
Question 8
In a first-order reaction, if the rate constant is $k$, what is the intercept and slope when $\ln[R]$ is plotted against time ($t$)?
✅ Correct Answer: Intercept = $\ln[R]_0$, Slope = $-k$
Explanation: The integrated rate equation for a first-order reaction using natural logs is $\ln[R] = -kt + \ln[R]_0$. Comparing this with the standard straight-line equation $y = mx + c$, the y-intercept ($c$) is clearly $\ln[R]_0$ and the slope ($m$) is exactly $-k$.
Question 9
Assertion (A): Benzaldehyde is less reactive than propanal towards nucleophilic addition reactions.
Reason (R): The polarity of the carbonyl group is reduced in benzaldehyde due to resonance with the benzene ring.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: Nucleophilic addition requires a strong partial positive charge on the carbonyl carbon. In benzaldehyde, the $+R$ resonance effect of the phenyl group donates electron density back to the carbonyl carbon, reducing its electrophilicity compared to aliphatic aldehydes like propanal. Additionally, steric hindrance from the bulky benzene ring further blocks nucleophilic attack.
Question 10
Assertion (A): The basic strength of methylamines in aqueous solution follows the order: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.
Reason (R): Basic strength of amines in aqueous medium is solely determined by the +I effect of alkyl groups.
✅ Correct Answer: A is true but R is false.
Explanation: The assertion is factually correct for methyl-substituted amines in water ($2^\circ > 1^\circ > 3^\circ > NH_3$). However, the reason is completely false. Basicity in an aqueous medium is determined by a complex combination of the $+I$ inductive effect, steric hindrance, and the hydration (solvation) energy of the newly protonated amine—not just the $+I$ effect alone.
Question 11
Statement I: The first ionization enthalpy of Nitrogen is greater than that of Oxygen.
Statement II: Nitrogen has a stable, exactly half-filled $2p^3$ electronic configuration, requiring more energy to remove an electron.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: Nitrogen ($1s^2 2s^2 2p^3$) possesses a symmetrical, half-filled p-orbital structure which is exceptionally stable quantum mechanically. Oxygen ($1s^2 2s^2 2p^4$) has one paired electron in the p-orbital, causing interelectronic repulsion, making it easier to remove one electron to attain the stable $2p^3$ state. Thus, Nitrogen > Oxygen for first ionization enthalpy.
Question 12
Statement I: $XeF_2$ and $I_3^-$ are isostructural species.
Statement II: Both $XeF_2$ and $I_3^-$ exhibit $sp^3d$ hybridization with a linear molecular geometry.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: $XeF_2$ has 8 valence electrons + 2 from F = 10 $e^-$ (5 pairs = 2 bond pairs + 3 lone pairs), resulting in $sp^3d$ hybridization and a linear shape. $I_3^-$ (central I) has 7 valence electrons + 2 from other I atoms + 1 negative charge = 10 $e^-$ (5 pairs = 2 bond pairs + 3 lone pairs), also resulting in $sp^3d$ hybridization and a linear shape. They are perfectly isostructural.
Question 13
$18\text{ g}$ of glucose ($C_6H_{12}O_6$) is added to $178.2\text{ g}$ of water. What is the vapor pressure of water for this aqueous solution at $100^\circ\text{C}$? (Vapor pressure of pure water at $100^\circ\text{C} = 760\text{ torr}$)
✅ Correct Answer: $752.4\text{ torr}$
Explanation: Moles of glucose ($n$) = $18 / 180 = 0.1\text{ mol}$.
Moles of water ($N$) = $178.2 / 18 = 9.9\text{ mol}$.
Total moles = $0.1 + 9.9 = 10\text{ mol}$.
Mole fraction of water ($X_{H_2O}$) = $9.9 / 10 = 0.99$.
Using Raoult's Law: $P_{\text{solution}} = P^\circ \times X_{H_2O} = 760 \times 0.99 = 752.4\text{ torr}$.
Question 14
Calculate the pH of a solution formed by mixing $50\text{ mL}$ of $0.1\text{ M}$ $CH_3COOH$ and $50\text{ mL}$ of $0.05\text{ M}$ $NaOH$. (Given $pK_a$ of $CH_3COOH = 4.74$)
✅ Correct Answer: 4.74
Explanation: Milli-moles of $CH_3COOH$ = $50 \times 0.1 = 5\text{ mmol}$.
Milli-moles of $NaOH$ = $50 \times 0.05 = 2.5\text{ mmol}$.
Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
$2.5\text{ mmol}$ of $NaOH$ completely reacts with $2.5\text{ mmol}$ of $CH_3COOH$ to form $2.5\text{ mmol}$ of salt ($CH_3COONa$).
Remaining weak acid = $5 - 2.5 = 2.5\text{ mmol}$.
Since $[Salt] = [Acid]$, it forms an ideal acidic buffer. Using the Henderson-Hasselbalch equation: $pH = pK_a + \log([Salt]/[Acid]) = 4.74 + \log(1) = 4.74$.
Question 15
The oxidation state of Chromium in Chromium pentoxide ($CrO_5$) is:
✅ Correct Answer: +6
Explanation: If calculated algebraically ($x - 10 = 0$), the oxidation state falsely appears to be +10, which is physically impossible as Cr only possesses 6 valence electrons ($3d^5 4s^1$). Structurally, $CrO_5$ has a "butterfly" geometry containing two peroxy linkages (4 oxygen atoms with a -1 state) and one standard double-bonded oxygen (-2 state). Thus: $x + 4(-1) + 1(-2) = 0 \implies x = +6$.
Question 16
Identify the final product 'C' in the following sequence of reactions:
Benzene $\xrightarrow{CH_3Cl/AlCl_3}$ A $\xrightarrow{KMnO_4/KOH, \Delta}$ B $\xrightarrow{SOCl_2, \text{ then } NH_3, \text{ then } Br_2/KOH}$ C
✅ Correct Answer: Aniline
Explanation: 1. Friedel-Crafts alkylation of Benzene yields Toluene (A).
2. Strong oxidation of Toluene with alkaline $KMnO_4$ yields Benzoic acid (B).
3. Benzoic acid with $SOCl_2$ yields Benzoyl chloride, which upon treatment with $NH_3$ forms Benzamide.
4. Benzamide finally undergoes Hoffmann bromamide degradation ($Br_2/KOH$) to step down the carbon chain (removing the carbonyl group entirely), ultimately forming Aniline (C).
Question 17
Which of the following amines CANNOT be prepared using Gabriel phthalimide synthesis?
✅ Correct Answer: Aniline
Explanation: Gabriel phthalimide synthesis is exclusively reserved for preparing aliphatic primary amines. Aniline (an aromatic primary amine) cannot be synthesized this way because the required intermediate step involves nucleophilic substitution. Aryl halides (like chlorobenzene) do not undergo simple nucleophilic substitution with the bulky phthalimide anion under normal conditions due to the partial double bond character in the C-X bond.
Question 18
The standard reduction potentials of three metals A, B, and C are $+0.5\text{ V}$, $-3.0\text{ V}$, and $-1.2\text{ V}$ respectively. The correct order of their reducing power is:
✅ Correct Answer: $B > C > A$
Explanation: A lower (more negative) standard reduction potential means the substance has a higher thermodynamic tendency to get oxidized, making it a stronger reducing agent. Therefore, the metal with the most negative value (B: $-3.0\text{ V}$) is the absolute strongest reducing agent, followed by C ($-1.2\text{ V}$), and then finally A ($+0.5\text{ V}$).
Question 19
At what temperature will the root mean square (rms) speed of an $O_2$ molecule be equal to the rms speed of an $H_2$ molecule at $27^\circ\text{C}$?
✅ Correct Answer: $4800\text{ K}$
Explanation: The formula is $v_{\text{rms}} = \sqrt{3RT/M}$. For the speeds to be identically equal: $\sqrt{T_{O_2} / M_{O_2}} = \sqrt{T_{H_2}/M_{H_2}}$.
Given $T_{H_2} = 27^\circ\text{C} = 300\text{ K}$, $M_{H_2} = 2\text{ g/mol}$, and $M_{O_2} = 32\text{ g/mol}$.
$T_{O_2} / 32 = 300 / 2 \implies T_{O_2} = 150 \times 32 = 4800\text{ K}$.
Question 20
One mole of an ideal gas expands freely into a vacuum (free expansion) at a constant temperature of $300\text{ K}$. What is the work done ($w$) and the change in internal energy ($\Delta U$) for this process?
✅ Correct Answer: $w=0, \Delta U=0$
Explanation: In a free expansion, the gas expands against zero external pressure ($P_{\text{ext}} = 0$). Since $w = -P_{\text{ext}} \Delta V$, the work done is rigorously 0. Furthermore, because it's an ideal gas expanding at a constant temperature (isothermal conditions), there is absolutely no change in internal energy, so $\Delta U = 0$ as well.
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