Question 1
Which of the following nitrogenous bases is found strictly in DNA and NOT in RNA?
✅ Correct Answer: Thymine
Explanation: According to NCERT, DNA contains four bases: Adenine, Guanine, Cytosine, and Thymine. RNA also contains four bases, but Thymine is strictly replaced by Uracil. Therefore, Thymine is an exclusive marker for DNA.
Question 2
Which of the following statements regarding the interstitial compounds formed by transition metals is INCORRECT?
✅ Correct Answer: They are highly reactive chemically due to trapped non-metal atoms.
Explanation: Interstitial compounds are formed when small atoms like $H$, $C$, or $N$ are trapped inside the crystal lattices of transition metals. According to NCERT, a key characteristic of these compounds is that they are chemically inert, not highly reactive. They also remain extremely hard and retain their metallic conductivity.
Question 3
When primary aliphatic amines are treated with cold nitrous acid ($HNO_2$), the major observable outcome is:
✅ Correct Answer: Evolution of nitrogen gas and formation of alcohols.
Explanation: Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts, which are highly unstable even at low temperatures. They immediately decompose to yield a mixture of alcohols, liberating nitrogen gas ($N_2$) as a visible effervescence. (Aromatic amines form stable salts at $0-5^\circ\text{C}$).
Question 4
Despite having similar geometries, $NH_3$ has a significantly higher dipole moment ($1.47\text{ D}$) than $NF_3$ ($0.23\text{ D}$). This is primarily because:
✅ Correct Answer: In $NH_3$, the orbital dipole of the lone pair and the bond dipoles are in the same direction, whereas in $NF_3$, they are in opposite directions.
Explanation: In $NH_3$, nitrogen is more electronegative than hydrogen, so the $N-H$ bond dipoles point towards nitrogen (upwards, identically aligned with the lone pair orbital dipole). In $NF_3$, fluorine is more electronegative, so the $N-F$ bond dipoles point downwards, opposing the lone pair dipole and drastically reducing the net sum.
Question 5
In metal carbonyls, the synergic bonding effect strengthens the metal-ligand bond. The $\pi$-back bonding is formed by the donation of electron density from:
✅ Correct Answer: A filled $d$-orbital of the metal into the empty antibonding $\pi^*$ orbital of $CO$.
Explanation: Synergic bonding involves two concerted steps: a $\sigma$-bond forms by the donation of a lone pair from Carbon to the metal, and a $\pi$-back bond forms by the robust donation of electrons from the filled metal $d$-orbitals directly into the empty antibonding $\pi^*$ orbitals (LUMO) of the carbon monoxide ligand.
Question 6
Among the isoelectronic species $O^{2-}$, $F^-$, $Na^+$ and $Mg^{2+}$ the correct decreasing order of ionic radii is:
✅ Correct Answer: $O^{2-} > F^- > Na^+ > Mg^{2+}$
Explanation: For isoelectronic species (all possessing exactly 10 electrons here), the ionic radius decreases as the nuclear charge (atomic number $Z$) increases because the electron cloud is pulled more strongly towards the nucleus. The atomic numbers are O (8), F (9), Na (11), Mg (12). Thus, the largest is $O^{2-}$ and the smallest is $Mg^{2+}$.
Question 7
Arrange the following compounds in decreasing order of their acidic strength:
I. Benzoic acid
II. 2-Methylbenzoic acid
III. 4-Methylbenzoic acid
IV. 4-Nitrobenzoic acid
✅ Correct Answer: IV > II > I > III
Explanation: 1. Electron-withdrawing groups ($-NO_2$) strongly increase acidity, making 4-Nitrobenzoic acid (IV) the absolute strongest.
2. Due to the "ortho effect", ortho-substituted benzoic acids (even with electron-donating groups like $-CH_3$) are generally stronger acids than plain benzoic acid. Therefore, II > I.
3. 4-Methylbenzoic acid (III) is the weakest due to the full $+I$ and hyperconjugation effects of the methyl group at the para position weakening the acid.
Final Order: IV > II > I > III.
Question 8
For a general elementary reaction $2A+B \rightarrow 3C$, the rate of appearance of $C$ is found to be $1.5 \times 10^{-3}\text{ mol L}^{-1}\text{s}^{-1}$. What is the rate of disappearance of $A$?
✅ Correct Answer: $1.0 \times 10^{-3}\text{ mol L}^{-1}\text{s}^{-1}$
Explanation: For the reaction $2A+B \rightarrow 3C$, the rate of reaction is written as:
$\text{Rate} = -\frac{1}{2}\frac{d[A]}{dt} = +\frac{1}{3}\frac{d[C]}{dt}$.
We need to find $-\frac{d[A]}{dt}$ (the rate of disappearance of $A$).
$-\frac{d[A]}{dt} = \frac{2}{3} \times \frac{d[C]}{dt} = \frac{2}{3} \times (1.5 \times 10^{-3}) = 1.0 \times 10^{-3}\text{ mol L}^{-1}\text{s}^{-1}$.
Question 9
Assertion (A): A $0.1\text{ M}$ aqueous solution of $KCl$ has a lower freezing point than a $0.1\text{ M}$ aqueous solution of Glucose.
Reason (R): $KCl$ dissociates into ions in water, increasing the van't Hoff factor ($i$) and thus increasing the magnitude of freezing point depression.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: Freezing point depression ($\Delta T_f$) is a colligative property, depending heavily on the number of particles. Glucose is a non-electrolyte ($i=1$), whereas $KCl$ dissociates into $K^+$ and $Cl^-$ ($i=2$). Since $KCl$ produces more particles, its $\Delta T_f$ is significantly larger, meaning the actual freezing point ($T_f = T_f^\circ - \Delta T_f$) is pushed to a lower temperature than Glucose.
Question 10
Assertion (A): Reaction of Anisole (methoxybenzene) with concentrated $HI$ gives Phenol and Methyl iodide, and not Iodobenzene and Methanol.
Reason (R): The $O-C_{aryl}$ bond has partial double bond character due to resonance, making it stronger and harder to cleave than the $O-C_{alkyl}$ bond.
✅ Correct Answer: Both A and R are true and R is the correct explanation of A.
Explanation: In Anisole, the lone pair on the oxygen atom is in conjugation with the benzene ring, imparting partial double bond character to the $Phenyl-Oxygen$ bond. When protonated by $HI$, the iodide ion ($I^-$) favors attacking the less hindered, weaker $O-CH_3$ bond via an $S_N2$ mechanism, resulting in Methyl iodide and leaving behind Phenol.
Question 11
Statement I: The principal quantum number ($n$) determines the main energy shell and largely determines the size of the orbital.
Statement II: The magnetic quantum number ($m_l$) defines the three-dimensional shape of the orbital.
✅ Correct Answer: Statement I is correct, but Statement II is incorrect.
Explanation: Statement I is true; $n$ determines the energy and spatial size. Statement II is false because the azimuthal quantum number ($l$) defines the structural shape of the orbital (s, p, d, f). The magnetic quantum number ($m_l$) determines the spatial orientation of the orbital in a uniform magnetic field.
Question 12
Statement I: For any exothermic reaction, the enthalpy change ($\Delta H$) is negative.
Statement II: All exothermic reactions are strictly spontaneous at all temperatures.
✅ Correct Answer: Statement I is correct, but Statement II is incorrect.
Explanation: Statement I is correct by standard thermodynamic definition. Statement II is incorrect because spontaneity is determined by Gibbs Free Energy ($\Delta G = \Delta H - T\Delta S$). If an exothermic reaction has a large decrease in entropy (negative $\Delta S$), the term $-T\Delta S$ becomes heavily positive. At high temperatures, this positive term can completely overcome the negative $\Delta H$, making the reaction non-spontaneous ($\Delta G > 0$).
Question 13
Calculate the mass of Ascorbic acid ($C_6H_8O_6$) that must be dissolved in $75\text{ g}$ of acetic acid to lower its melting point by $1.5^\circ\text{C}$. ($K_f \text{ for acetic acid} = 3.9\text{ K kg mol}^{-1}$)
✅ Correct Answer: $5.08\text{ g}$
Explanation: Molar mass of Ascorbic acid ($C_6H_8O_6$) = $(12 \times 6) + (1 \times 8) + (16 \times 6) = 176\text{ g/mol}$.
Formula: $\Delta T_f = K_f \times m = K_f \times \frac{w_2 \times 1000}{M_2 \times w_1}$
$1.5 = 3.9 \times \frac{w_2 \times 1000}{176 \times 75}$
$w_2 = \frac{1.5 \times 176 \times 75}{3.9 \times 1000} = \frac{19800}{3900} = 5.076\text{ g} \approx 5.08\text{ g}$.
Question 14
At $298\text{ K}$, the solubility product ($K_{sp}$) of $CaF_2$ is $3.2 \times 10^{-11}$. What is its molar solubility in a $0.1\text{ M } NaF$ aqueous solution?
✅ Correct Answer: $3.2 \times 10^{-9}\text{ M}$
Explanation: $CaF_2 \rightleftharpoons Ca^{2+} + 2F^-$. The solution already contains $0.1\text{ M } F^-$ from the $NaF$ matrix (Common Ion Effect). Let solubility be $s$. Thus, $[Ca^{2+}] = s$, and $[F^-] \approx 0.1\text{ M}$ (ignoring the tiny $2s$ from $CaF_2$).
$K_{sp} = [Ca^{2+}][F^-]^2$
$3.2 \times 10^{-11} = (s) \times (0.1)^2$
$s = \frac{3.2 \times 10^{-11}}{10^{-2}} = 3.2 \times 10^{-9}\text{ M}$.
Question 15
The approximate number of total atoms present in $4.25\text{ g}$ of Ammonia ($NH_3$) gas is:
✅ Correct Answer: $6.02 \times 10^{23}$
Explanation: Molar mass of $NH_3 = 14 + 3 = 17\text{ g/mol}$.
Moles of $NH_3 = 4.25 / 17 = 0.25\text{ mol}$.
Number of $NH_3$ molecules = $0.25 \times N_A$.
Since one molecule of $NH_3$ contains 4 distinct atoms (1 N + 3 H), the Total atoms = $4 \times 0.25 \times N_A = 1.0 \times N_A = 6.02 \times 10^{23}$.
Question 16
Identify the major product formed when 2-bromo-2-methylbutane is treated with a bulky base, potassium tert-butoxide, in tert-butanol.
✅ Correct Answer: 2-methylbut-1-ene
Explanation: Normally, elimination strictly follows Saytzeff's rule to give the more highly substituted alkene (2-methylbut-2-ene). However, when a highly sterically bulky base like potassium tert-butoxide is used, it struggles immensely to abstract the hindered internal proton. Instead, it abstracts a less hindered terminal proton on the edge, leading to the less substituted Hofmann product (2-methylbut-1-ene) as the overwhelming major product.
Question 17
Consider an octahedral coordination complex where the crystal field splitting energy ($\Delta_o$) is $18000\text{ cm}^{-1}$ and the pairing energy ($P$) is $20000\text{ cm}^{-1}$. The Crystal Field Stabilization Energy (CFSE) for a $d^5$ metal ion in this complex will be:
✅ Correct Answer: $0.0 \Delta_o$
Explanation: Since $\Delta_o < P$ ($18000 < 20000$), it is energetically more favorable for the electrons to occupy the higher energy $e_g$ orbitals rather than forcibly pairing up in the lower tier. This creates a weak field (high spin) scenario. The electronic configuration for $d^5$ becomes $t_{2g}^3 e_g^2$.
CFSE = $[-0.4 \times (n_{t2g}) + 0.6 \times (n_{eg})]\Delta_o = [-0.4 \times 3 + 0.6 \times 2]\Delta_o = [-1.2 + 1.2]\Delta_o = 0.0 \Delta_o$.
Question 18
An organic compound 'A' with the molecular formula $C_3H_6O$ forms an addition product with sodium hydrogen sulfite but does NOT reduce Tollens' reagent. When 'A' undergoes reduction with Zinc amalgam and concentrated $HCl$, it yields compound 'B'. Identify 'B'.
✅ Correct Answer: Propane
Explanation: 1. Forms addition product with $NaHSO_3$: It is a functional carbonyl compound (aldehyde or ketone).
2. Does NOT reduce Tollens' reagent: It must be a ketone. For $C_3H_6O$, 'A' is Acetone (Propan-2-one).
3. Reduction with $Zn(Hg) / \text{conc. } HCl$ is the famous Clemmensen reduction, which completely strips carbonyl groups ($>C=O$) to methylene groups ($-CH_2-$). Acetone is thus reduced completely to Propane ('B').
Question 19
An unknown alkene 'X' upon reductive ozonolysis gives a mixture of propan-2-one and ethanal. When 'X' is treated with $HBr$ in the presence of an organic peroxide, the major product formed is:
✅ Correct Answer: 2-Bromo-3-methylbutane
Explanation: 1. Retro-ozonolysis mapping: Propan-2-one ($CH_3-CO-CH_3$) + Ethanal ($CH_3-CHO$) means alkene 'X' is 2-Methylbut-2-ene ($CH_3-C(CH_3)=CH-CH_3$).
2. Addition of $HBr$ with peroxide mandates anti-Markovnikov (Kharasch effect) addition via a free radical mechanism.
3. The $Br^\bullet$ radical attacks the double bond to form the most stable alkyl radical intermediate (attacking C3 forms a more stable tertiary radical at C2).
4. The final product has $Br$ at C3 and $H$ at C2: $CH_3-CH(CH_3)-CH(Br)-CH_3$. Standard IUPAC numbering gives 2-Bromo-3-methylbutane.
Question 20
Among the following halogens, which one has the highest bond dissociation enthalpy?
✅ Correct Answer: $Cl_2$
Explanation: Generally, bond dissociation enthalpy decreases as atomic size increases. However, $F_2$ is a drastic exception. Due to the extremely small volume of the fluorine atom, there is high interelectronic repulsion between the non-bonding electron lone pairs packed on the two $F$ atoms, which physically weakens the $F-F$ bond. Therefore, $Cl_2$ ends up having the highest overall bond dissociation enthalpy. Order: $Cl_2 > Br_2 > F_2 > I_2$.
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