Question 1
Which of the following correctly represents the atomic radius order of Group 13 elements?
✅ Correct Answer: $B < Ga < Al < In < Tl$
Explanation: Normally, atomic radius increases down a group. However, Gallium ($Ga$) has a smaller atomic radius ($135\text{ pm}$) than Aluminum ($Al$) ($143\text{ pm}$). This exception occurs due to the poor shielding effect of the fully filled 3d-orbitals in Gallium, which increases the effective nuclear charge ($Z_{\text{eff}}$), pulling the valence electrons closer to the nucleus.
Question 2
In the Etard reaction, which oxidizing agent is used to convert toluene to benzaldehyde, and what is the intermediate complex formed?
✅ Correct Answer: $CrO_2Cl_2$ in $CS_2$; chromium complex
Explanation: According to NCERT, the Etard reaction utilizes Chromyl chloride ($CrO_2Cl_2$) in a non-polar solvent like $CS_2$ to selectively oxidize the methyl group of toluene. It first forms a brown chromium complex, which on hydrolysis yields benzaldehyde. Option A represents the alternative $CrO_3$ + acetic anhydride method.
Question 3
According to NCERT, what is the effect of an increase in temperature on the solubility of gases in liquids, and how does it relate to the Henry's law constant ($K_H$)?
✅ Correct Answer: Solubility decreases, $K_H$ increases.
Explanation: Dissolution of a gas in a liquid is an exothermic process. By Le Chatelier's principle, increasing the temperature decreases solubility. According to Henry's Law ($p = K_H \cdot x$), for a given constant pressure, as the mole fraction or solubility ($x$) decreases, the value of the Henry's law constant ($K_H$) must proportionally increase.
Question 4
The catalytic activity of transition metals and their compounds is ascribed mainly to their ability to:
✅ Correct Answer: Adopt multiple oxidation states and form complexes.
Explanation: NCERT clearly highlights that transition metals act as excellent catalysts (e.g., $V_2O_5$ in the Contact process, $Fe$ in the Haber process) because they can exhibit variable oxidation states and form unstable intermediate complexes with reactants, creating a new reaction pathway that lowers the activation energy.
Question 5
Gabriel phthalimide synthesis is heavily preferred for the preparation of which specific type of amines?
✅ Correct Answer: Pure primary aliphatic amines
Explanation: Gabriel phthalimide synthesis produces pure primary aliphatic amines without the risk of over-alkylation (unlike Hoffmann's ammonolysis). It cannot be used to prepare primary aromatic amines (like aniline) because aryl halides do not easily undergo nucleophilic substitution with the phthalimide anion under normal conditions.
Question 6
For an operating galvanic cell, if the standard cell potential ($E^\circ_{cell}$) is positive, which of the following is true for the standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant ($K_c$)?
✅ Correct Answer: $\Delta G^\circ < 0$, $K_c > 1$
Explanation: Using the thermodynamic equation $\Delta G^\circ = -nFE^\circ_{cell}$, a positive $E^\circ_{cell}$ immediately results in a negative $\Delta G^\circ$ (indicating a spontaneous reaction). Furthermore, since $\Delta G^\circ = -2.303RT \log K_c$, a negative $\Delta G^\circ$ implies that $\log K_c$ must be positive, which means $K_c > 1$.
Question 7
During the $S_N1$ hydrolysis of an optically active tertiary alkyl halide, the stereochemical outcome of the product is predominantly:
✅ Correct Answer: Partial racemization with a slight excess of the inverted product
Explanation: While basic textbook theory states that $S_N1$ leads to 100% racemization via a planar carbocation, NCERT specifically notes that in reality, complete racemization rarely occurs. The departing halide ion momentarily blocks the front face (creating an intimate ion pair), making nucleophilic attack from the back side slightly more favorable. This leads to more inversion than retention.
Question 8
In an octahedral complex, if $\Delta_o < P$ (where $\Delta_o$ is the crystal field splitting energy and $P$ is the pairing energy), the correct d-electron configuration for a $d^4$ metal ion is:
✅ Correct Answer: $t_{2g}^3 e_g^1$
Explanation: If $\Delta_o < P$, it means the energy required to jump to the $e_g$ level is less than the energy required to force electrons to pair up in the $t_{2g}$ level. This occurs with weak field ligands. Thus, the fourth electron will enter the higher energy $e_g$ orbital rather than pairing up, giving the high-spin configuration $t_{2g}^3 e_g^1$.
Question 9
Arrange the following molecules in the decreasing order of their acidic strength: Phenol (I), p-Nitrophenol (II), p-Cresol (III), m-Nitrophenol (IV).
✅ Correct Answer: II > IV > I > III
Explanation: Electron-withdrawing groups ($-NO_2$) increase acidity, while electron-donating groups ($-CH_3$) decrease it. The $-NO_2$ group exerts both $-I$ and $-R$ effects at the para position, making p-Nitrophenol the strongest acid. At the meta position, it only exerts a $-I$ effect, making m-Nitrophenol weaker than para but stronger than Phenol. p-Cresol is the weakest due to the $+I$ and hyperconjugation of the methyl group.
Question 10
Assertion (A): The rate of a chemical reaction generally doubles for every $10^\circ\text{C}$ rise in temperature.
Reason (R): The number of molecules with kinetic energy equal to or greater than the activation energy nearly doubles with a $10^\circ\text{C}$ rise in temperature.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: According to collision theory and the Maxwell-Boltzmann distribution curve, a small $10^\circ\text{C}$ increase in temperature causes only a minor 1-2% increase in collision frequency, but it causes a massive ~100% increase in the fraction of molecules possessing kinetic energy $\ge E_a$. This doubling of effective collisions doubles the reaction rate.
Question 11
Assertion (A): Halogens are ortho-para directing but deactivate the benzene ring towards electrophilic aromatic substitution.
Reason (R): Halogens have a strong $-I$ effect which decreases overall electron density on the ring, but their $+R$ effect stabilizes the carbocation intermediate more at ortho and para positions.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: Halogens are a unique exception in organic chemistry. Their $-I$ inductive effect is stronger than their $+R$ resonance effect, making them net electron withdrawers (ring deactivators). However, when electrophilic attack occurs, the $+R$ effect (lone pair donation) can stabilize the intermediate arenium ion (sigma complex) specifically at ortho and para positions, dictating the orientation.
Question 12
Statement I: For the combustion of methane in a sealed bomb calorimeter, the heat evolved is equivalent to the enthalpy change ($\Delta H$).
Statement II: A bomb calorimeter measures heat transfer at constant volume, which corresponds to the change in internal energy ($\Delta U$).
✅ Correct Answer: Statement I is incorrect, but Statement II is correct.
Explanation: A bomb calorimeter inherently operates at constant volume. Thermodynamics dictates that heat transfer at constant volume is rigorously equal to the change in internal energy ($\Delta U$), not enthalpy ($\Delta H$). Enthalpy change is defined as the heat transferred at constant pressure ($q_p = \Delta H$). Therefore, Statement I is false, and Statement II is true.
Question 13
Statement I: The first ionization enthalpy of Nitrogen is higher than that of Oxygen.
Statement II: Nitrogen has a stable, exactly half-filled $2p^3$ electronic configuration.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: Nitrogen ($1s^2 2s^2 2p^3$) possesses a half-filled p-subshell, which grants it extra exchange energy and immense stability. Oxygen ($1s^2 2s^2 2p^4$) has one paired electron in the p-orbital, causing repulsion and making it easier to lose an electron to achieve the highly stable half-filled state. Thus, N > O for first ionization energy.
Question 14
What is the molar solubility of $Mg(OH)_2$ ($K_{sp}=1.0 \times 10^{-11}$) in an aqueous buffer solution maintained at a pH of 12?
✅ Correct Answer: $1.0 \times 10^{-7}\text{ M}$
Explanation: $pH = 12$ means $pOH = 2$, so the concentration of $[OH^-]$ is strictly maintained at $10^{-2}\text{ M}$ by the buffer.
The solubility product equation is: $K_{sp} = [Mg^{2+}][OH^-]^2$.
Let the solubility of $Mg(OH)_2$ be $S$, so $[Mg^{2+}] = S$.
$1.0 \times 10^{-11} = (S) \times (10^{-2})^2 = S \times 10^{-4}$.
$S = 1.0 \times 10^{-7}\text{ M}$.
Question 15
A first-order reaction is 75% complete in 60 minutes. What is the half-life ($t_{1/2}$) of this reaction?
✅ Correct Answer: 30 minutes
Explanation: For a first-order reaction, completing 75% means $25\%$ of the reactant remains.
$100\% \xrightarrow{t_{1/2}} 50\% \xrightarrow{t_{1/2}} 25\%$.
This requires exactly two half-lives. Therefore, $t_{75\%} = 2 \times t_{1/2}$.
$60\text{ mins} = 2 \times t_{1/2} \implies t_{1/2} = 30\text{ minutes}$.
Question 16
A current of $9.65\text{ A}$ is passed through an aqueous solution of $CuSO_4$ for $1000$ seconds. What mass of copper is deposited at the cathode? (Atomic mass of Cu = $63.5\text{ g/mol}$, $1\text{ F} = 96500\text{ C}$)
✅ Correct Answer: $3.175\text{ g}$
Explanation: Total charge $Q = I \times t = 9.65\text{ A} \times 1000\text{ s} = 9650\text{ C}$.
Convert to Faradays: $9650 / 96500 = 0.1\text{ F}$.
Reaction: $Cu^{2+} + 2e^- \rightarrow Cu$. It takes 2 Faradays to deposit 1 mole of Cu ($63.5\text{ g}$).
Therefore, $0.1\text{ F}$ will deposit: $(63.5 / 2) \times 0.1 = 3.175\text{ g}$.
Question 17
Which of the following alcohols will show immediate turbidity when treated with Lucas reagent (anhydrous $ZnCl_2$ and conc. $HCl$) at room temperature?
✅ Correct Answer: 2-Methylpropan-2-ol
Explanation: The Lucas test relies on the stability of carbocations formed via an $S_N1$ mechanism. Tertiary alcohols form highly stable $3^\circ$ carbocations and react immediately to form insoluble alkyl chlorides (turbidity). 2-Methylpropan-2-ol is a tertiary alcohol, hence it reacts instantly. Primary alcohols do not react at room temperature, and secondary alcohols take 5-10 minutes.
Question 18
Which of the following pairs of species have identical molecular shapes (isostructural)?
✅ Correct Answer: $XeF_2$ and $I_3^-$
Explanation: $XeF_2$ has 2 bond pairs and 3 lone pairs on Xenon ($sp^3d$ hybridization), resulting in a linear shape. $I_3^-$ has a central Iodine with 2 bond pairs and 3 lone pairs ($sp^3d$ hybridization), also resulting in a linear shape. The other pairs have different shapes (e.g., $NH_3$ is pyramidal, $BF_3$ is trigonal planar).
Question 19
An unknown organic compound $X$ ($C_5H_{10}O$) forms a solid phenylhydrazone derivative but gives a negative Tollens' test. It also does NOT give a yellow precipitate with $I_2/NaOH$. Compound $X$ is most likely:
✅ Correct Answer: Pentan-3-one
Explanation: 1. Forms phenylhydrazone $\Rightarrow$ It is a carbonyl compound (aldehyde or ketone).
2. Negative Tollens' test $\Rightarrow$ It is a ketone, not an aldehyde (rules out Pentanal and 2-Methylbutanal).
3. Negative Iodoform test ($I_2/NaOH$) $\Rightarrow$ It is NOT a methyl ketone. Pentan-2-one has a terminal methyl group adjacent to the carbonyl, so it would give a positive test. Pentan-3-one does not. Therefore, $X$ is Pentan-3-one.
Question 20
The total number of stereoisomers for the octahedral coordination complex $[Co(en)_2Cl_2]^+$ is:
✅ Correct Answer: 3
Explanation: This complex of type $[M(AA)_2a_2]$ exhibits geometrical isomerism, forming cis and trans isomers. The trans-isomer possesses a plane of symmetry and is optically inactive (meso form). The cis-isomer lacks a plane of symmetry and is optically active, existing as a pair of non-superimposable enantiomers (d- and l- forms). Total stereoisomers = 1 (trans) + 2 (cis enantiomers) = 3.
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