Question 1
According to NCERT, what happens to the structure of a protein during denaturation (e.g., coagulation of egg white on boiling)?
✅ Correct Answer: Secondary and tertiary structures are destroyed, but the primary structure remains intact.
Explanation: During denaturation, the physical and biological properties of proteins change. The delicate hydrogen bonds are disturbed, uncoiling the protein. NCERT strictly notes that secondary and tertiary structures are destroyed, but the primary structure (the exact sequence of amino acids linked by strong covalent peptide bonds) remains completely intact.
Question 2
In the formation of metal carbonyls, the synergic bonding effect involves:
✅ Correct Answer: Donation of lone pair from carbon to metal $\sigma$-orbital and back-donation from filled metal d-orbital to empty $\pi^*$ orbital of carbon monoxide.
Explanation: According to NCERT's explicit diagram on synergic bonding, the $M-C$ $\sigma$-bond is formed by the donation of a lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The $M-C$ $\pi$-bond is simultaneously formed by the back-donation of a pair of electrons from a filled d-orbital of the metal directly into the vacant antibonding $\pi^*$ orbital (LUMO) of the $CO$ ligand.
Question 3
In the collision theory of chemical kinetics, the rate equation is modified to $\text{Rate} = P \cdot Z_{AB} \cdot e^{-E_a/RT}$. What does the factor '$P$' strictly account for?
✅ Correct Answer: The probability or steric factor ensuring molecules collide with proper orientation.
Explanation: The factor $P$ is called the probability or steric factor. It is introduced to account for the fact that not all collisions with sufficient threshold energy ($>E_a$) lead to a product. For an effective reaction to occur, the reacting molecules must also physically collide with the proper spatial orientation.
Question 4
The brown ring test for nitrates depends on the ability of $Fe^{2+}$ to reduce nitrates to nitric oxide. What is the true oxidation state of Iron ($Fe$) in the brown ring complex $[Fe(H_2O)_5(NO)]^{2+}$?
✅ Correct Answer: +1
Explanation: In the brown ring complex $[Fe(H_2O)_5(NO)]^{2+}$, the $NO$ ligand exceptionally behaves as a positively charged nitrosyl ion ($NO^+$). Therefore, applying the standard algebraic formula: $x + 5(0) + 1 = +2 \implies x = +1$. Iron exists in a very rare +1 oxidation state here.
Question 5
During the electrolysis of an aqueous solution of $NaCl$ using platinum electrodes, which gas is liberated at the anode, and what is the core conceptual reason?
✅ Correct Answer: $Cl_2$ gas, due to the high overpotential (overvoltage) required for the oxidation of water to $O_2$.
Explanation: At the anode, both $Cl^-$ and $H_2O$ compete for oxidation. Thermodynamically, the oxidation of water to $O_2$ is favored (it has a lower $E^\circ$ requirement). However, kinetically, the process of forming $O_2$ bubbles requires a high activation energy, leading to an "overpotential" barrier. Because of this overpotential, the oxidation of $Cl^-$ to $Cl_2$ becomes faster and practically preferred.
Question 6
Which of the following chemical species is aromatic according to Hückel's rule?
✅ Correct Answer: Cyclopentadienyl anion
Explanation: For a species to be aromatic, it must be cyclic, planar, completely conjugated, and possess $(4n+ 2)\pi$ electrons. The cyclopentadienyl anion has $6\pi$ electrons ($4\pi$ from the two double bonds + $2$ from the delocalized lone pair on the negative charge), fulfilling Hückel's rule ($n=1$). The cation has $4\pi$ (anti-aromatic), and cycloheptatriene lacks full conjugation.
Question 7
The correct order of ionic radii for the isoelectronic species $N^{3-}$, $O^{2-}$, $F^-$, $Na^+$, $Mg^{2+}$ is:
✅ Correct Answer: $N^{3-} > O^{2-} > F^- > Na^+ > Mg^{2+}$
Explanation: For isoelectronic species (all possessing exactly 10 electrons here), the ionic size steadily decreases as the nuclear charge (atomic number, $Z$) increases. Magnesium ($Z=12$) pulls the 10 electrons strongest towards its nucleus, making it the smallest. Nitrogen ($Z=7$) pulls them weakest, allowing the electron cloud to expand, making it the largest.
Question 8
Which of the following sets contains only intensive thermodynamic properties?
✅ Correct Answer: Density, Molarity, Electromotive Force (EMF) of a cell
Explanation: Intensive properties do not depend on the physical quantity or bulk size of matter present. Density, Molarity, and EMF ($E_{cell}$) are purely intensive. Extensive properties (Enthalpy, Volume, Mass, Heat Capacity, Gibbs Free Energy) depend on the total mass of the system. Note: The ratio of two extensive properties (like Mass/Volume) yields an intensive property (Density).
Question 9
Assertion (A): Propanone (acetone) is less reactive than ethanal (acetaldehyde) towards nucleophilic addition reactions.
Reason (R): Alkyl groups are electron-releasing ($+I$ effect), which decreases the electrophilicity of the carbonyl carbon, and two methyl groups create more steric hindrance than one.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: The reactivity of carbonyl compounds towards nucleophilic addition directly depends on both steric and electronic factors. Ketones (like propanone) have two alkyl groups causing significant steric crowding and pushing electron density ($+I$) onto the carbonyl carbon, making it far less positively charged (less electrophilic) than an aldehyde.
Question 10
Assertion (A): Osmotic pressure is widely used to determine the molar masses of proteins, polymers, and other macromolecules.
Reason (R): Osmotic pressure is measured at room temperature, meaning biomacromolecules remain stable, and its magnitude is significantly large even for very dilute solutions.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: Proteins and polymers are highly unstable at elevated temperatures (they denature easily) and often possess very low solubilities. Traditional methods like boiling point elevation would instantly destroy them. Osmotic pressure ($\pi = CRT$) operates perfectly at room temperature and yields easily measurable values (in atm/bar) even for trace concentrations, making it the superior choice.
Question 11
Read the following statements regarding chemical equilibrium:
Statement I: A positive catalyst alters the equilibrium composition of a reaction mixture to yield more products.
Statement II: A catalyst lowers the activation energy of the forward reaction and the backward reaction by the exact same amount.
✅ Correct Answer: Statement I is incorrect but Statement II is correct.
Explanation: Statement I is strictly false; a catalyst rapidly helps the system reach equilibrium faster but never alters the final equilibrium constant ($K_c$) or the final product composition. Statement II is correct; it merely provides an alternate pathway, lowering the activation energy ($E_a$) equally for both the forward and backward directions.
Question 12
Given below are two statements regarding haloalkanes:
Statement I: $S_N1$ reactions of optically active alkyl halides are accompanied by complete inversion of spatial configuration.
Statement II: $S_N2$ reactions proceed via a planar carbocation intermediate.
✅ Correct Answer: Both Statement I and Statement II are false.
Explanation: Both statements are perfectly swapped/incorrect. $S_N1$ reactions proceed via a planar carbocation intermediate and result in racemization (both retention and inversion). $S_N2$ reactions proceed via a single-step transition state (no intermediate) and result in complete Walden inversion.
Question 13
The freezing point of a $0.1\text{ molal}$ aqueous solution of a weak acid ($HA$) is $-0.2046^\circ\text{C}$. If the cryoscopic constant ($K_f$) of water is $1.86\text{ K kg mol}^{-1}$, what is the degree of dissociation ($\alpha$) of the acid?
✅ Correct Answer: 10%
Explanation: 1. From $\Delta T_f = i \cdot K_f \cdot m \implies 0.2046 = i \times 1.86 \times 0.1 \implies i = \frac{0.2046}{0.186} = 1.1$.
2. For dissociation of $HA \rightleftharpoons H^+ + A^-$, $n=2$ ions are formed.
3. The formula linking $i$ and $\alpha$ is: $i = 1 + \alpha(n-1) \implies 1.1 = 1 + \alpha(2-1) \implies 1.1 = 1 + \alpha \implies \alpha = 0.1$.
4. Converting to percentage: $0.1 \times 100 = 10\%$.
Question 14
The half-life of a first-order reaction is 15 minutes. What fraction of the reactant will remain unreacted after 1 hour?
✅ Correct Answer: $\frac{1}{16}$
Explanation: 1 hour = 60 minutes.
Number of elapsed half-lives ($n$) = $\frac{\text{Total Time}}{\text{Half-life}} = \frac{60}{15} = 4$.
Amount remaining after $n$ half-lives = $N_0 \times \left(\frac{1}{2}\right)^n$.
Fraction remaining = $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
Question 15
According to Bohr's theory, the radius of the second orbit ($n=2$) of a $Li^{2+}$ ion will be: (Given Bohr radius for hydrogen $a_0=0.529\text{ \AA}$)
✅ Correct Answer: $0.705\text{ \AA}$
Explanation: The radius of the $n^{\text{th}}$ Bohr orbit for hydrogen-like species is $r_n = 0.529 \times \frac{n^2}{Z}\text{ \AA}$.
For $Li^{2+}$ ion, atomic number $Z=3$. For the $2^{\text{nd}}$ orbit, $n=2$.
$r = 0.529 \times \frac{2^2}{3} = 0.529 \times \frac{4}{3} \approx 0.7053\text{ \AA}$.
Question 16
Which of the following represents the correct Crystal Field Stabilization Energy (CFSE) for a $d^6$ metal ion in an octahedral field acting with a strong field ligand? (Let $P=$ pairing energy)
✅ Correct Answer: $-2.4 \Delta_o + 3P$
Explanation: In a strong octahedral field, $\Delta_o > P$ so pairing occurs heavily before electrons go to the higher $e_g$ level. A $d^6$ ion will fill the lower $t_{2g}$ level completely: $t_{2g}^6 e_g^0$.
CFSE = $(-0.4 \times 6) \Delta_o = -2.4 \Delta_o$.
Since all 6 electrons are completely paired in 3 orbitals, there are 3 extra pairs formed relative to a free ion, so $+3P$ is added to the total energy stabilization expression.
Question 17
When 2-methylbut-2-ene is subjected to reductive ozonolysis ($O_3$ followed by $Zn/H_2O$), the products obtained are:
✅ Correct Answer: Propanone and Ethanal
Explanation: The structure of 2-methylbut-2-ene is $CH_3-C(CH_3)=CH-CH_3$.
In reductive ozonolysis, you cleanly cleave the $C=C$ double bond and assign a double-bonded Oxygen atom ($=O$) to both carbons.
Left fragment: $CH_3-C(CH_3)=O$ (Propanone/Acetone).
Right fragment: $O=CH-CH_3$ (Ethanal/Acetaldehyde).
Question 18
Identify the final major product 'C' in the following reaction sequence:
$\text{Benzene} \xrightarrow{Br_2/FeBr_3} A \xrightarrow{Mg / \text{dry ether}} B \xrightarrow{1. \text{CO}_2 \text{ (dry ice)}, \ 2. H_3O^+} C$
✅ Correct Answer: Benzoic acid
Explanation: 1. Electrophilic aromatic substitution of benzene gives Bromobenzene (A).
2. Bromobenzene reacts with Magnesium in dry ether to form Phenylmagnesium bromide (B), a highly reactive Grignard reagent.
3. Grignard reagents nucleophilically attack solid $CO_2$ (dry ice) to form a carboxylate salt intermediate, which on acidic hydrolysis yields Benzoic acid (C).
Question 19
Arrange the following isomers of Dichlorobenzene in decreasing order of their dipole moments ($\mu$):
(I) $o$-Dichlorobenzene
(II) $m$-Dichlorobenzene
(III) $p$-Dichlorobenzene
✅ Correct Answer: I > II > III
Explanation: Dipole moment is a vector sum of individual $C-Cl$ bond dipoles. In ortho ($60^\circ$ angle), the vectors add up most constructively. In meta ($120^\circ$ angle), the resultant is smaller. In para ($180^\circ$ angle), the two equal dipoles point in exact opposite directions and cancel out entirely, giving $\mu=0$. Therefore, the order is ortho > meta > para.
Question 20
What is the correct increasing order of boiling points for the group 15 hydrides?
✅ Correct Answer: $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$
Explanation: Usually, boiling point increases steadily down the group due to an increase in molecular mass and Van der Waals forces ($PH_3 < AsH_3 < SbH_3 < BiH_3$). However, $NH_3$ forms exceptionally strong intermolecular hydrogen bonds, raising its boiling point abnormally high. It sits directly between $AsH_3$ and $SbH_3$. Thus, the correct sequence is $PH_3 < AsH_3 < NH_3 < SbH_3 < BiH_3$.
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