Question 1
According to the NCERT text on electrochemistry, what is the approximate efficiency of a galvanic cell like the hydrogen-oxygen fuel cell in converting the heat of combustion into electrical energy?
✅ Correct Answer: 70%
Explanation: NCERT explicitly states that conventional thermal plants have an efficiency of about 40%, whereas fuel cells, such as the $H_2-O_2$ fuel cell (used in the Apollo space program), operate at a highly impressive efficiency of approximately 70%.
Question 2
Which of the following statements about potassium permanganate ($KMnO_4$) is strictly accurate as per NCERT?
✅ Correct Answer: It is prepared by the fusion of $MnO_2$ with an alkali metal hydroxide and an oxidizing agent like $KNO_3$.
Explanation: $KMnO_4$ is commercially prepared by fusing $MnO_2$ with KOH and an oxidizing agent ($KNO_3$ or $O_2$). Permanganate ($MnO_4^-$) is diamagnetic ($d^0$), while manganate ($MnO_4^{2-}$) is paramagnetic ($d^1$). $HCl$ is not used in titrations because it gets oxidized to $Cl_2$. In strongly acidic media, it reduces to $Mn^{2+}$, not $MnO_2$.
Question 3
Read the following statements regarding nucleic acids. Which statement represents the exact correct structural linkage as per NCERT?
✅ Correct Answer: In DNA and RNA, the phosphodiester linkage is formed between the $5^{\prime}$ and $3^{\prime}$ carbon atoms of the pentose sugar.
Explanation: The backbone of nucleic acids is formed by a phosphodiester linkage between the 5' and $3^{\prime}$ carbon atoms of the pentose sugar. DNA strands are complementary and antiparallel, not identical and parallel. A base + sugar = nucleoside, not nucleotide. The sequence of nucleotides is called the primary structure, not secondary.
Question 4
For the inversion of cane sugar, which is a pseudo first-order reaction, what is the exact kinetic description according to NCERT?
✅ Correct Answer: The rate equation is $\text{Rate} = k'[C_{12}H_{22}O_{11}][H_2O]$, where water is in large excess and its concentration remains practically constant.
Explanation: The reaction is inherently bimolecular (molecularity = 2) but behaves as first-order experimentally because water is present in such a massive excess that its concentration barely changes. The true rate law includes water, but the effective rate law absorbs it into a new pseudo constant $k = k'[H_2O]$.
Question 5
Which of the following species has a perfectly tetrahedral geometry and a zero dipole moment ($\mu=0$)?
✅ Correct Answer: $NH_4^+$
Explanation: $NH_4^+$ has 4 bond pairs and 0 lone pairs ($sp^3$ hybridization), resulting in a perfectly symmetrical tetrahedral shape where all four N-H bond dipoles cancel out completely ($\mu=0$). $CHCl_3$ is tetrahedral but asymmetrical. $NH_3$ is pyramidal, and $SF_4$ is see-saw shaped.
Question 6
The stabilization of coordination compounds due to chelation is called the chelate effect. Which of the following complexes is thermodynamically the most stable?
✅ Correct Answer: $[Fe(C_2O_4)_3]^{3-}$
Explanation: The oxalate ion ($C_2O_4^{2-}$) is a bidentate ligand that binds and forms stable 5-membered ring structures with the central metal ion. This ring formation (chelation) greatly increases the thermodynamic stability of the complex through an enormous increase in entropy compared to monodentate ligands like $CN^-$ or $NH_3$.
Question 7
For an isolated system undergoing a spontaneous process, which of the following sets of thermodynamic conditions must be satisfied?
✅ Correct Answer: $\Delta S_{\text{system}} > 0$ and $\Delta U = 0$
Explanation: An isolated system cannot exchange heat ($q=0$) or work ($w=0$) with its surroundings. By the First Law of Thermodynamics, $\Delta U=0$. For any spontaneous process, the total universe entropy must increase. Since the surroundings are completely unaffected, the entropy of the system itself must increase ($\Delta S_{\text{system}} > 0$).
Question 8
When a mixture of benzaldehyde and formaldehyde is heated with concentrated aqueous NaOH, the major organic products formed are:
✅ Correct Answer: Benzyl alcohol and sodium formate
Explanation: This is a classic Crossed Cannizzaro reaction. Formaldehyde ($HCHO$) is significantly more reactive towards nucleophilic attack by $OH^-$ due to lack of steric hindrance and $+I$ groups. It acts as the reducing agent, easily getting oxidized to the formate ion ($HCOO^-$). Benzaldehyde is correspondingly forced to reduce to benzyl alcohol ($C_6H_5CH_2OH$).
Question 9
Assertion (A): The order of a reaction can be zero or a fractional number, but molecularity is always a positive whole number.
Reason (R): Order of a reaction is an experimentally determined quantity, whereas molecularity represents the number of reacting species colliding simultaneously in an elementary step.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: Molecularity is a theoretical collision concept; you cannot have 'half' a molecule colliding, so it must be positive integers like 1, 2, or 3. Order is derived experimentally from rate laws and can mathematically be fractions or zero. The reason accurately explains the fundamental definitions causing the assertion.
Question 10
Assertion (A): The boiling point of a 0.1 M aqueous solution of NaCl is higher than that of a 0.1 M aqueous solution of urea.
Reason (R): Elevation in boiling point is a colligative property which depends strictly on the number of solute particles in the solution.
✅ Correct Answer: Both (A) and (R) are true and (R) is the correct explanation of (A).
Explanation: NaCl is a strong electrolyte that dissociates into two distinct ions ($Na^+$ and $Cl^-$), providing a van't Hoff factor $i \approx 2$. Urea is a non-electrolyte and does not dissociate ($i=1$). Since boiling point elevation directly depends on the total number of particles (colligative property $\Delta T_b = i \cdot K_b \cdot m$), the NaCl solution will have double the elevation.
Question 11
Read the following statements:
Statement I: The number of radial nodes for a 3p orbital is 1.
Statement II: The total number of nodes for any atomic orbital is given by the formula $(n-1)$.
✅ Correct Answer: Both Statement I and Statement II are correct.
Explanation: For a 3p orbital, $n=3$ and $l=1$. The number of radial nodes = $n-l-1 = 3-1-1 = 1$. The total number of nodes (radial + angular) is always equal to $(n-l-1) + l = n-1$. Both statements mathematically align with the quantum mechanical definitions.
Question 12
Given below are two statements regarding electronic displacement effects:
Statement I: The inductive effect is a permanent electron displacement effect, whereas the electromeric effect is a temporary effect.
Statement II: In a carbon-chlorine bond ($C-Cl$), chlorine exhibits a $+I$ effect due to the presence of lone pairs of electrons.
✅ Correct Answer: Statement I is true, Statement II is false.
Explanation: Statement I is correct; the inductive effect is inherent to the polar bond, while the electromeric effect requires the presence of an attacking reagent. Statement II is completely false; Chlorine is much more electronegative than carbon and pulls the $\sigma$-electrons towards itself, exhibiting a $-I$ (electron-withdrawing) effect, not a $+I$ effect.
Question 13
How many moles of electrons weigh exactly one kilogram?
(Given: Mass of an electron = $9.1 \times 10^{-31}\text{ kg}$, Avogadro's number $N_A = 6.02 \times 10^{23}$)
✅ Correct Answer: $\frac{1}{9.1 \times 10^{-31} \times 6.02 \times 10^{23}}$
Explanation: Mass of one single electron = $m_e$.
Mass of 1 full mole of electrons = $N_A \times m_e$.
Therefore, the number of moles contained in $1\text{ kg} = \frac{\text{Total Mass}}{\text{Mass of 1 mole}} = \frac{1}{N_A \times m_e}$.
Substituting the given values yields $\frac{1}{6.02 \times 10^{23} \times 9.1 \times 10^{-31}}$.
Question 14
50 mL of a 0.2 M weak acid HA ($K_a=1 \times 10^{-5}$) is mixed with 50 mL of 0.1 M NaOH. What is the approximate pH of the resulting solution?
✅ Correct Answer: 5
Explanation: Initial millimoles of HA = $50 \times 0.2 = 10\text{ mmol}$.
Initial millimoles of NaOH = $50 \times 0.1 = 5\text{ mmol}$.
NaOH acts as the limiting reagent and reacts completely: $HA + NaOH \rightarrow NaA + H_2O$.
After reaction: HA remaining = $10 - 5 = 5\text{ mmol}$. Salt (NaA) formed = $5\text{ mmol}$.
The solution forms an ideal acidic buffer. Using Henderson-Hasselbalch equation:
$pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) = -\log(10^{-5}) + \log\left(\frac{5}{5}\right) = 5 + 0 = 5$.
Question 15
A steady current of 2.0 A is passed for 5.0 hours through a molten metal salt, depositing 22.2 g of the metal (Atomic mass = 177 u) at the cathode. The oxidation state of the metal in the salt is approximately: (Given $1\text{ F} = 96500\text{ C}$)
✅ Correct Answer: +3
Explanation: Using Faraday's First Law equation: $W = ZIt = \left(\frac{M}{n \times 96500}\right) \times I \times t$
Convert time to seconds: $t = 5 \times 3600 = 18000\text{ s}$.
$22.2 = \frac{177}{n \times 96500} \times 2 \times 18000$
$n = \frac{177 \times 2 \times 18000}{22.2 \times 96500} = \frac{6372000}{2142300} \approx 2.97$
Since '$n$' represents the valency factor (number of electrons), the nearest integer oxidation state is +3.
Question 16
Following IUPAC nomenclature conventions for elements with atomic number > 100, the official name and symbol of the element with atomic number 117 is:
✅ Correct Answer: Tennessine (Ts)
Explanation: Based on the standard NCERT periodic table updates:
$Z=115$ is Moscovium (Mc)
$Z=116$ is Livermorium (Lv)
$Z=117$ is Tennessine (Ts) (Group 17 halogen equivalent)
$Z=118$ is Oganesson (Og).
Question 17
An alkyl halide 'P' reacts with alcoholic KOH to yield an alkene 'X'. The reductive ozonolysis of 'X' yields a mixture of propan-2-one and methanal. The original alkyl halide 'P' is:
✅ Correct Answer: 1-Bromo-2-methylpropane
Explanation: The final ozonolysis products are propan-2-one ($CH_3-CO-CH_3$) and methanal ($HCHO$).
Reassembling them synthetically by removing the oxygens gives the alkene 'X' as 2-methylpropene ($CH_3-C(CH_3)=CH_2$).
The alkyl halide 'P' undergoing dehydrohalogenation to give 2-methylpropene must be 1-Bromo-2-methylpropane (or 2-Bromo-2-methylpropane, but matching the options, 1-Bromo-2-methylpropane is structurally correct).
Question 18
In which of the following organic chlorides will the $C-Cl$ bond length be the shortest?
✅ Correct Answer: $CH_2=CH-Cl$
Explanation: In vinyl chloride ($CH_2=CH-Cl$), the lone pair of electrons on the chlorine atom sits in direct conjugation with the $\pi$-bond of the adjacent double bond. This $+R$ resonance imparts a partial double bond character to the $C-Cl$ bond, pulling the chlorine atom closer to the carbon, making the bond exceptionally short and strong compared to alkyl chlorides.
Question 19
The freezing point depression ($\Delta T_f$) of a $0.001\text{ M } K_3[Fe(CN)_6]$ aqueous solution is $7.10 \times 10^{-3}\text{ K}$. If $K_f$ for water is $1.86\text{ K kg mol}^{-1}$ and assuming molarity roughly equals molality, the percentage dissociation of the coordination complex is approximately:
✅ Correct Answer: 94%
Explanation: Step 1: Calculate the observed van't Hoff factor ($i$).
$\Delta T_f = i \cdot K_f \cdot m \implies 7.10 \times 10^{-3} = i \times 1.86 \times 0.001 \implies i = 3.817$
Step 2: Relate $i$ to the degree of dissociation ($\alpha$).
$K_3[Fe(CN)_6]$ dissociates into $3~K^+$ and $1~[Fe(CN)_6]^{3-}$ ion. So, $n=4$.
Using formula: $i = 1 + \alpha(n-1)$
$3.817 = 1 + \alpha(4-1) \implies 3\alpha = 2.817 \implies \alpha \approx 0.939$, which equates to 93.9% or roughly ~94%.
Question 20
Among the halogens, which element has the most negative electron gain enthalpy ($\Delta_{eg}H$), and what is the core reason for this exception?
✅ Correct Answer: Chlorine, due to its larger size preventing the high inter-electronic repulsions seen in the compact 2p subshell.
Explanation: Normally, electron gain enthalpy becomes less negative down a group. However, Fluorine is so small that its 2p electrons are tightly packed, creating massive inter-electronic repulsion for any incoming electron. Chlorine's 3p orbital is significantly larger and accommodates the extra electron far more easily, making its $\Delta_{eg}H$ more negative than Fluorine's.
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