Series : WXYZ/S | SET-1 | Q.P.Code: 56/S/1
General Instructions :
Read the following instructions carefully and follow them :
- (i) This question paper contains 33 questions. All questions are compulsory
- (ii) This question paper is divided into five sections – Section A, B, C, D and E.
- (iii) Section A – questions number 1 to 16 are multiple choice type questions. Each question carries 1 mark.
- (iv) Section B – questions number 17 to 21 are very short answer type questions. Each question carries 2 marks.
- (v) Section C – questions number 22 to 28 are short answer type questions. Each question carries 3 marks.
- (vi) Section D – questions number 29 and 30 are case-based questions. Each question carries 4 marks.
- (vii) Section E – questions number 31 to 33 are long answer type questions. Each question carries 5 marks.
- (viii) There is no overall choice given in the question paper. However, an internal choice has been provided in few questions in all the sections except Section A.
- (ix) Kindly note that there is a separate question paper for Visually Impaired candidates.
- (x) Use of calculator is not allowed.
SECTION A
Questions no. 1 to 16 are Multiple Choice type Questions, carrying 1 mark each.16 × 1 = 16
1. A conductivity cell contains electrodes made up of:
(A) Silver(B) Copper
(C) Platinum
(D) Zinc
View Answer
Option (C) Platinum is the correct answer.
Explanation:
Conductivity cell electrodes are made of platinum sheets coated with finely divided platinum black. Platinum is chosen because it is chemically inert, meaning it will not react with the electrolyte solution during measurement. The platinum black coating increases the surface area of the electrodes, which helps reduce polarization effects and ensures accurate resistance readings.
2. The value of KH at 298 K for Ar(g), CO2(g), HCHO(g) and CH4(g) are 40.32, 1.68, 1.85 × 10–5 and 0.416 kbar respectively. When these gases are arranged in increasing order of solubility, the correct order is :
(A) HCHO < CH4 < CO2 < Ar(B) HCHO < CO2 < CH4 < Ar
(C) Ar < CO2 < CH4 < HCHO
(D) Ar < CH4 < CO2 < HCHO
View Answer
Option (C) Ar < CO2 < CH4 < HCHO is the correct answer.
Explanation:
According to Henry's Law, the solubility of a gas in a liquid is inversely proportional to its Henry's Law constant (KH) at a given pressure ($p = K_H \cdot x$, where $x$ is the mole fraction of the gas in solution). Therefore, a higher KH value means lower solubility. Comparing the given values:
• Ar: 40.32 kbar (Highest KH → Lowest solubility)
• CO2: 1.68 kbar
• CH4: 0.416 kbar
• HCHO: 1.85 × 10–5 kbar (Lowest KH → Highest solubility)
Thus, the increasing order of solubility is Ar < CO2 < CH4 < HCHO.
3. Out of the following organic compounds, the one which will react with Lucas reagent at room temperature is :
(A) CH2 = CH – CH2OH(B) CH3CH2CH2OH
(C) C6H5CH2OH
(D) (CH3)3COH
View Answer
Option (D) (CH3)3COH is the correct answer.
Explanation:
Lucas reagent (a mixture of concentrated HCl and anhydrous ZnCl2) reacts with alcohols via an SN1 mechanism involving a carbocation intermediate. Tertiary (3°) alcohols like (CH3)3COH (tert-butyl alcohol) form highly stable tertiary carbocations, allowing them to react immediately at room temperature to produce a characteristic cloudiness (turbidity). Primary (1°) alcohols do not react at room temperature.
4. Identify the Freon from the following compounds :
(A) CCl2F2(B) CCl2Br2
(C) CH2Cl– CHCl2
(D) CHCl2– CHCl2
View Answer
Option (A) CCl2F2 is the correct answer.
Explanation:
Freons are chlorofluorocarbon (CFC) compounds of methane and ethane. They are extremely stable, unreactive, non-toxic, non-corrosive, and easily liquefiable gases. Dichlorodifluoromethane (CCl2F2) is famously known as Freon-12 and is one of the most common Freon compounds widely used in refrigeration and air conditioning.
5. If the half-life period of a first order reaction is 1386 s, then the rate constant of this reaction is :
(A) 0.5 × 10–2s–1(B) 0.5 × 10–3s–1
(C) 5 × 10–2s–1
(D) 5 × 10–3s–1
View Answer
Option (B) 0.5 × 10–3s–1 is the correct answer.
Explanation:
For a first-order reaction, the relationship between half-life (t1/2) and the rate constant (k) is given by the formula:
t1/2 = 0.693/k
Rearranging to solve for k:
k = 0.693/1386 s = 0.0005s-1 = 0.5 × 10-3 s-1
6. The standard electrode potential of ‘A’, ‘B’ and ‘C’ are + 0.68 V, – 2.54 V and – 0.50 V respectively. The order of their reducing power is :
(A) A > B > C(B) A > C > B
(C) B > C > A
(D) C > B > A
View Answer
Option (C) B > C > A is the correct answer.
Explanation:
The reducing power of an element is inversely proportional to its standard reduction potential (E°). A lower (more negative) electrode potential means the substance loses electrons more easily, making it a stronger reducing agent. Comparing the values:
• B (–2.54 V) has the lowest value → Strongest reducing power.
• C (–0.50 V) has the intermediate value.
• A (+0.68 V) has the highest value → Weakest reducing power.
Therefore, the correct decreasing order of reducing power is B > C > A.
7. Glucose on reaction with Br2 water gives :
(A) Gluconic acid(B) Hexanoic acid
(C) Saccharic acid
(D) Glycolic acid
View Answer
Option (A) Gluconic acid is the correct answer.
Explanation:
Bromine water (Br2/H2O) is a mild oxidizing agent. It selectively oxidizes the aldehyde group (-CHO) at the C-1 position of glucose into a carboxylic acid group (-COOH), yielding gluconic acid. It is not strong enough to oxidize the primary alcohol group at the C-6 position (which would yield saccharic acid).
8. A compound CaCl2. 6H2O undergoes complete dissociation in water. The Van't Hoff factor (i) is :
(A) 9(B) 6
(C) 4
(D) 3
View Answer
Option (D) 3 is the correct answer.
Explanation:
When CaCl2.6H2O dissolves in water, the hydration water molecules simply merge with the solvent, while the salt dissociates completely into its constituent ions:
CaCl2 → Ca2+ + 2Cl-
Since one formula unit dissociates to produce 1 calcium ion and 2 chloride ions, the total number of particles after dissociation is 1 + 2 = 3. For complete dissociation, the Van't Hoff factor (i) equals the total number of ions produced, which is 3.
9. The transition element which shows both +1 and +2 oxidation states is :
(A) Sc(B) Mn
(C) Cu
(D) Zn
View Answer
Option (C) Cu is the correct answer.
Explanation:
Copper (Cu, Z=29) has an outer electronic configuration of [Ar] 3d10 4s1. Due to the stability of a fully filled d-orbital, it can easily lose the single 4s electron to show a +1 oxidation state (Cu+, cuprous). It can also lose an additional electron from the 3d subshell to form the more hydration-stabilized +2 oxidation state (Cu2+, cupric). Scandium typically only shows +3, Manganese ranges from +2 to +7, and Zinc uniquely exhibits +2.
10. Which one of the following halides contains Csp2 – X bond ?
(A) Alkyl halide(B) Allyl halide
(C) Benzyl halide
(D) Vinyl halide
View Answer
Option (D) Vinyl halide is the correct answer.
Explanation:
In a vinyl halide (such as vinyl chloride, CH2=CHCl, the halogen atom X is directly attached to one of the carbon atoms forming the carbon-carbon double bond. Because that carbon forms a double bond and two single bonds, it is sp2 hybridized. In contrast, alkyl, allyl, and benzyl halides all feature a halogen bonded to an sp3 hybridized carbon atom.
11. Which of the following amines has the highest boiling point ?
(A) 2,2-Dimethyl propanamine(B) 3-Methylbutanamine
(C) 2-Methylbutanamine
(D) Pentanamine
View Answer
Option (D) Pentanamine is the correct answer.
Explanation:
All four given compounds are isomeric primary amines containing five carbon atoms. Boiling points of isomeric amines depend heavily on molecular branching. Pentanamine is a straight-chain molecule, giving it a larger surface area and stronger intermolecular Van der Waals forces compared to its branched isomers. Branching makes a molecule more spherical, reducing its surface area and lowering its boiling point.
12. When phenyl methyl ether is heated with HI, it produces :
(A) Methyl chloride and Iodobenzene(B) Benzene and Methanol
(C) Iodobenzene and Methanol
(D) Phenol and Methyl iodide
View Answer
Option (D) Phenol and Methyl iodide is the correct answer.
Explanation:
Phenyl methyl ether (anisole, C6H5-OCH3) is protonated by HI to form an oxonium ion. The bond between the oxygen and the phenyl ring (OC6H5) possesses partial double-bond character due to resonance stabilization with the aromatic ring, making it incredibly strong and difficult to break. Consequently, the nucleophile I- attacks the less sterically hindered methyl group via an SN2 pathway, yielding phenol (C6H5OH) and methyl iodide (CH3I).
13. Assertion (A): Actinoids show a wide range of oxidation states.
Reason (R): Actinoids are radioactive in nature.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
View Answer
Option (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). is the correct answer.
Explanation:
• Assertion (A) is true: Actinoids exhibit a broad variety of oxidation states (ranging up to +7 for elements like Np and Pu). This happens because the energy gap between the 5f, 6d, and 7s subshells is very small, allowing electrons from all three shells to participate in bond formation.
• Reason (R) is true: Most actinoids have unstable nuclei and are indeed radioactive.
• Conclusion: While both statements are scientifically factual, radioactivity is a nuclear property and has nothing to do with the chemical sharing/loss of valence electrons that dictates oxidation states. Therefore, R is not the correct explanation for A.
14. Assertion (A): Presence of –NO2 group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution reactions.
Reason (R): Nitro group decreases the electron density over the benzene ring.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
View Answer
Option (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). is the correct answer.
Explanation:
• Assertion (A) is true: Nitro groups (-NO2) situated at the ortho or para positions strongly activate haloarenes toward nucleophilic aromatic substitution.
• Reason (R) is true: The nitro group is a powerful electron-withdrawing group via both resonance (-M effect) and induction (-I effect), which directly decreases the electron density of the aromatic ring.
• Conclusion: Because the ring's overall electron density drops, it becomes less susceptible to repelling arriving nucleophiles. Furthermore, when placed specifically at ortho or para positions, the negative charge of the intermediate carbanion can be directly delocalized onto the electronegative oxygen atoms of the nitro group. Hence, R is the correct and accurate explanation for A.
15. Assertion (A): The molecularity of the reaction H + Br2 → HBr + Br appears to be 2.
Reason (R): Two molecules of the reactants are involved in the given elementary reaction.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
View Answer
Option (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). is the correct answer.
Explanation:
• Assertion (A) is true: Molecularity is defined as the number of reacting species (atoms, ions, or molecules) taking part in an elementary chemical reaction that must collide simultaneously. For this specific step, the molecularity is exactly 2 (bimolecular).
• Reason (R) is true: There are exactly two reacting species on the left side of this elementary step: one hydrogen atom and one bromine molecule.
• Conclusion: Since molecularity is fundamentally defined by counting the number of reactant particles colliding in a single elementary step, Reason (R) directly explains why Assertion (A) is true.
16. Assertion (A): p-methoxyphenol is a stronger acid than p-nitrophenol.
Reason (R): Methoxy group is electron-donating group whereas nitro group is electron-withdrawing.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
View Answer
Option (D) Assertion (A) is false, but Reason (R) is true. is the correct answer.
Explanation:
• Assertion (A) is false: p-nitrophenol is actually a much stronger acid than p-methoxyphenol. Acidity depends on the stability of the conjugate base (phenoxide ion).
• Reason (R) is true: The methoxy group (-OCH3) acts as an electron-donating group through resonance (+M effect), which destabilizes the phenoxide ion charge. Conversely, the nitro group (-NO2) is an electron-withdrawing group (-M and -I effects) that highly stabilizes the phenoxide charge by pulling it away from the ring.
• Conclusion: Because electron-withdrawing groups increase acidity and electron-donating groups decrease it, p-nitrophenol is more acidic than p-methoxyphenol, making the initial assertion incorrect.
SECTION B
17. Give one point of difference between the following : 2
(a) Starch and Cellulose
(b) Primary and Secondary structure of protein
Answer:
Starch: Made of α-glucose units connected by α-1,4 and α-1,6 linkages. This creates a coiled, helical structure that is easy for enzymes to break down, making it ideal for energy storage in plants.
Cellulose: Made of β-glucose units connected by β-1,4 linkages. This creates straight, rigid, unbranched chains that form tough microfibrils, making it ideal for providing structural support in plant cell walls.
Primary Structure: This is the linear, one-dimensional sequence of amino acids in a polypeptide chain. It is held together exclusively by strong, covalent peptide bonds between adjacent amino acids.
Secondary Structure: This is the local, two-dimensional folding or coiling of that linear chain into specific shapes (most commonly an α-helix or β-pleated sheet). It is stabilized by weaker, non-covalent hydrogen bonds formed between the oxygen and hydrogen atoms of the peptide backbone.