Question 1
Which of the following acts as an ambidentate ligand in coordination compounds?
✅ Correct Answer: $SCN^-$
Explanation: An ambidentate ligand can ligate through two different atoms. The thiocyanate ion ($SCN^-$) can bind to the central metal atom/ion through either the sulfur atom or the nitrogen atom. Water and ammonia are strictly unidentate, while oxalate ($C_2O_4^{2-}$) is a bidentate ligand.
Question 2
According to NCERT, which of the following pairs of nitrogenous bases is present in DNA but NOT entirely in RNA?
✅ Correct Answer: Cytosine and Thymine
Explanation: DNA contains four bases: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T). RNA contains A, G, C, but Thymine is exclusively replaced by Uracil (U). Therefore, the pair containing Thymine is unique to DNA.
Question 3
In the Hinsberg test for amines, which observation accurately corresponds to a secondary amine?
✅ Correct Answer: Forms a precipitate that is insoluble in aqueous $NaOH$.
Explanation: Secondary amines react with Hinsberg's reagent (benzenesulphonyl chloride) to form N,N-dialkylbenzenesulphonamide. Because this product has no acidic hydrogen attached to the nitrogen atom, it is completely insoluble in an alkali medium like $NaOH$.
Question 4
Transition metals form a large number of interstitial compounds. Which of the following statements regarding these compounds is INCORRECT as per NCERT?
✅ Correct Answer: They are highly chemically reactive.
Explanation: Interstitial compounds are formed when small atoms like H, C, or N are trapped inside the crystal lattices of metals. According to NCERT, one of their primary characteristics is that they are chemically inert, not reactive. They also retain their metallic conductivity and become extremely hard.
Question 5
Arrange the following molecules in the decreasing order of their dipole moments: $NH_3$, $NF_3$, $BF_3$.
✅ Correct Answer: $NH_3 > NF_3 > BF_3$
Explanation: $BF_3$ is highly symmetrical (trigonal planar) resulting in a net dipole moment of exactly zero. In $NH_3$, the orbital dipole (lone pair) and the N-H bond dipoles are in the same direction, adding up constructively. In $NF_3$, the highly electronegative F atoms create N-F bond dipoles that oppose the orbital dipole, resulting in a significantly lower net dipole moment than $NH_3$.
Question 6
Which of the following alkyl halides undergoes $S_N2$ reaction the fastest?
✅ Correct Answer: $CH_3-Br$
Explanation: The rate of $S_N2$ reactions is predominantly governed by steric hindrance around the electrophilic carbon. The less hindered the carbon, the faster the nucleophilic attack. The order of reactivity is Methyl > Primary > Secondary > Tertiary. Hence, Methyl bromide ($CH_3-Br$) reacts the fastest.
Question 7
What is the hybridization and molecular geometry of $XeF_4$?
✅ Correct Answer: $sp^3d^2$, Square planar
Explanation: Xenon in $XeF_4$ has 8 valence electrons. It forms 4 sigma bonds with 4 Fluorine atoms, leaving 2 lone pairs. Total electron pairs = 6, yielding $sp^3d^2$ hybridization. The octahedral electron geometry with 2 lone pairs arranged at $180^\circ$ to minimize repulsion results in a square planar molecular geometry.
Question 8
For the equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $\Delta H = -92.4\text{ kJ/mol}$, what will be the effect of adding an inert gas at constant pressure?
✅ Correct Answer: The equilibrium will shift in the backward direction.
Explanation: Adding an inert gas at constant pressure increases the total volume of the system. According to Le Chatelier's principle, the equilibrium will shift towards the direction with a larger number of gaseous moles to counteract the volume increase. Here, reactants have 4 moles and products have 2 moles, so it strictly shifts backward.
Question 9
Assertion (A): The molar conductivity ($\Lambda_m$) of a weak electrolyte like acetic acid increases steeply with dilution.
Reason (R): The degree of dissociation ($\alpha$) of weak electrolytes increases significantly upon dilution.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: For weak electrolytes, the number of ions in solution is small at high concentrations. Upon dilution, the degree of dissociation ($\alpha$) increases drastically, leading to a steep rise in the total number of mobile ions, thereby sharply increasing the overall molar conductivity. Strong electrolytes only show a slight increase due to decreased interionic forces.
Question 10
Assertion (A): The complex $[Fe(H_2O)_6]^{3+}$ is strongly paramagnetic, whereas $[Fe(CN)_6]^{3-}$ is weakly paramagnetic.
Reason (R): $H_2O$ is a weak field ligand causing high spin complexes, while $CN^-$ is a strong field ligand causing pairing of electrons against Hund's rule.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: Both complexes contain $Fe^{3+}$ ($3d^5$). $H_2O$ is a weak field ligand ($\Delta_o < P$), leading to no pairing (5 unpaired electrons, strongly paramagnetic). $CN^-$ is a strong field ligand ($\Delta_o > P$), forcing electron pairing and leaving only 1 unpaired electron (weakly paramagnetic).
Question 11
Statement I: The boiling points of alkyl halides follow the order $RI > RBr > RCl > RF$.
Statement II: For isomeric haloalkanes, the boiling point increases with an increase in branching.
✅ Correct Answer: Statement I is correct, but Statement II is incorrect.
Explanation: Statement I is true because as the size and mass of the halogen atom increase, the van der Waals dispersion forces increase, raising the boiling point. Statement II is false; boiling point actively decreases with an increase in branching because the surface area decreases, leading to weaker overall van der Waals forces.
Question 12
Statement I: $Cu^+$ ion is highly stable in aqueous solutions and does not undergo disproportionation.
Statement II: The hydration enthalpy of $Cu^{2+}$ is much more negative than that of $Cu^+$, which more than compensates for the second ionization enthalpy of Copper.
✅ Correct Answer: Statement I is incorrect, but Statement II is correct.
Explanation: Statement I is strictly incorrect because $Cu^+$ is actually unstable in aqueous solutions and rapidly undergoes disproportionation: $2Cu^+(aq) \rightarrow Cu^{2+}(aq) + Cu(s)$. Statement II is correct and is the exact NCERT thermodynamic reasoning for why $Cu^{2+}$ is vastly more stable in water despite needing more ionization energy to form.
Question 13
What is the mole fraction of the solute in a $1.00$ molal aqueous solution?
✅ Correct Answer: $0.0177$
Explanation: A $1.00$ molal aqueous solution signifies that exactly $1$ mole of solute is dissolved in $1000\text{ g}$ of the solvent (water).
Moles of water = $1000 / 18 = 55.5\text{ mol}$.
Total moles = $1 + 55.5 = 56.5\text{ mol}$.
Mole fraction of solute = $n_{\text{solute}} / n_{\text{total}} = 1 / 56.5 \approx 0.0177$.
Question 14
For a first-order reaction, what is the mathematical relationship between the time required for 99% completion ($t_{99\%}$) and the time required for 90% completion ($t_{90\%}$)?
✅ Correct Answer: $t_{99\%} = 2 \times t_{90\%}$
Explanation: Using the integrated rate law: $t = (2.303/k) \log (a/(a-x))$.
For 90%: $t_{90\%} = (2.303/k) \log (100/10) = (2.303/k) \log 10$.
For 99%: $t_{99\%} = (2.303/k) \log (100/1) = (2.303/k) \log 100 = 2 \times (2.303/k) \log 10$.
Hence, comparing the terms mathematically yields $t_{99\%} = 2 \times t_{90\%}$.
Question 15
How many Faradays of electricity are required to produce $20.0\text{ g}$ of Calcium from molten $CaCl_2$? (Atomic mass of Ca = $40\text{ g/mol}$)
✅ Correct Answer: 1 F
Explanation: The reduction half-reaction at the cathode is $Ca^{2+} + 2e^- \rightarrow Ca$. This shows that 2 Faradays (2 moles of electrons) deposit exactly 1 mole of Ca ($40\text{ g}$). To deposit $20.0\text{ g}$ (which is exactly $0.5\text{ moles}$), the required charge is $0.5 \times 2 = 1\text{ Faraday}$.
Question 16
What is the number of angular nodes and radial nodes for a $3p$ orbital, respectively?
✅ Correct Answer: 1 and 1
Explanation: For any atomic orbital, the number of angular nodes = $l$. For a p-orbital, $l = 1$.
The number of radial nodes = $n - l - 1$. For the $3p$ orbital ($n=3$, $l=1$), radial nodes = $3 - 1 - 1 = 1$.
Thus, it possesses 1 angular node and 1 radial node.
Question 17
In the Reimer-Tiemann reaction of Phenol with Chloroform and aqueous $NaOH$, the reactive electrophilic intermediate is:
✅ Correct Answer: Dichlorocarbene ($:CCl_2$)
Explanation: During the initial steps of the Reimer-Tiemann reaction, chloroform ($CHCl_3$) reacts with a strong base ($NaOH$) to lose a proton and a chloride ion, effectively generating Dichlorocarbene ($:CCl_2$). This electron-deficient species acts as the core electrophile that directly attacks the phenol ring.
Question 18
The solubility product ($K_{sp}$) of $AgCl$ is $1.0 \times 10^{-10}$. What will be its solubility in a $0.1\text{ M } NaCl$ solution?
✅ Correct Answer: $1.0 \times 10^{-9}\text{ M}$
Explanation: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$. Let the solubility be '$s$'. Due to the powerful common ion effect from the $NaCl$ solvent, $[Cl^-] \approx 0.1\text{ M}$.
$K_{sp} = [Ag^+][Cl^-] \implies 1.0 \times 10^{-10} = s \times (0.1)$.
$s = 1.0 \times 10^{-9}\text{ M}$.
Question 19
Compound 'A' on reaction with concentrated $NaOH$ undergoes a simultaneous oxidation and reduction to yield two products, one of which is benzyl alcohol. What is compound 'A'?
✅ Correct Answer: Benzaldehyde
Explanation: The reaction framework described is the classic Cannizzaro reaction, which specifically occurs in aldehydes lacking alpha-hydrogens when treated with concentrated alkali. Benzaldehyde ($C_6H_5CHO$) undergoes disproportionation to form benzyl alcohol (the reduction product) and sodium benzoate (the oxidation product).
Question 20
Which element among the following halogens has the most negative electron gain enthalpy?
✅ Correct Answer: Chlorine
Explanation: This is a standard periodic table exception. Although Fluorine is the most electronegative element, Chlorine actually possesses the most negative electron gain enthalpy. This is because the $2p$ orbitals in F are very small, leading to strong interelectronic repulsions when an extra electron is added. Chlorine's larger $3p$ orbital accommodates the incoming electron much more easily.
🔙 Back to Scorecard