Introduction

The HISSAN Central Examination is one of the most closely watched preparatory board exams for Grade XII students across Nepal, run under the coordination of HISSAN (Higher Secondary Schools’ Association of Nepal). For Chemistry students preparing for the NEB board examination, the HISSAN paper functions almost like a rehearsal — it mirrors the actual board exam’s structure, marking scheme, and difficulty level closely enough that a strong performance here is genuinely predictive of board exam readiness. The 2081 (2025) Chemistry XII paper, worth 75 full marks split across an 11-mark objective section and a 64-mark subjective section, covers physical chemistry, inorganic chemistry, and organic chemistry in roughly the proportions students should expect on their final NEB exam.

What makes this particular paper valuable for revision is how it blends conceptual questions (definitions, distinguishing tests, mechanism-based reasoning) with numerical problem-solving (molarity calculations, entropy of vaporization, solubility product, electrode potential) and structural/sequential organic chemistry problems (multi-step conversions, reaction sequences, and compound identification from clues). This mix is exactly what NEB Chemistry demands, and working through this paper carefully — not just checking answers but understanding why each answer is correct — builds the kind of exam-day confidence that memorization alone cannot provide.

Group A of this paper tests eleven core concepts in just twenty-five minutes, forcing students to think fast without sacrificing accuracy. Group B and Group C, on the other hand, slow things down and demand real depth: tracing a secondary alcohol through three transformation steps to identify an unknown alkene, calculating the entropy of vaporization from boiling point and enthalpy data, or working out whether a galvanic cell reaction proceeds spontaneously based on standard electrode potentials. These aren’t questions you can answer by memorizing a textbook paragraph the night before — they require students to actually understand how the underlying chemistry works.

This article breaks the paper down chapter by chapter, estimates where the weightage concentrates, points out concepts that show up again and again across multiple HISSAN and NEB papers, and gives concrete preparation advice for students heading into their own board exam. Rather than simply reprinting the question paper, the aim here is to help students actually use it as a study tool — understanding not just what the correct answer is, but why it’s correct, and which related questions are likely to reappear in slightly different form on the actual NEB board exam.

For teachers and self-study students alike, this kind of question-paper breakdown is often more useful than working through a textbook chapter in isolation, since it shows exactly how examiners tend to frame questions, which topics get tested repeatedly, and where the real marks-scoring opportunities lie.

Quick Answer Key (Group A — Multiple Choice Questions)

Full Question Paper

Brief explanation of each MCQ:

1. A pH of 0 means an extremely high [latex]H^+[/latex] concentration ([latex][H^+] = 1,M[/latex]), making it strongly acidic, not neutral or amphoteric.
2. Rate = k[A]^n. When rate equals the rate constant itself, [A]^n must equal 1, which only happens when n = 0 (zero order), since anything raised to power zero is 1.
3. A secondary reference electrode is one whose potential is known relative to a primary standard (like the hydrogen electrode) and is used for practical convenience. The calomel electrode ([latex]Hg/Hg_2Cl_2[/latex]) is the classic example used in labs.
4. The given equation forms 2 moles of [latex]NH_3[/latex] with [latex]\Delta H = -M[/latex]. Enthalpy of formation is defined per mole of product, so dividing by 2 gives [latex]-\frac{M}{2}[/latex].
5. In the 3d transition series, manganese (Mn) shows the highest oxidation state of +7 (as in [latex]KMnO_4[/latex]), because up to five 3d and 4s electrons can be involved in bonding.
6. Corrosive sublimate is the common name for mercuric chloride, [latex]HgCl_2[/latex], historically used as a disinfectant and preservative.
7. Alkyl halides undergo nucleophilic substitution ([latex]S_N1[/latex] or [latex]S_N2[/latex]) most commonly because the C–X bond is polarized, making the carbon electrophilic and susceptible to attack by nucleophiles.
8. Grignard reagents react with [latex]CO_2[/latex] (carbonation) to form a magnesium carboxylate salt, which on acidification gives a carboxylic acid with one more carbon than the original Grignard reagent.
9. X is oxidized by acidified [latex]KMnO_4[/latex] to give Y, and Y reacts with chloroform in the presence of alkali to give a sleep-inducing drug (chloretone). This chloroform + alkali reaction is characteristic of methyl ketones, so Y must be propan-2-one (acetone). Since a secondary alcohol oxidizes to a ketone, X is propan-2-ol.
10. Among the four acids, formic acid (HCOOH) has no alkyl or aryl group to donate electron density, so its O–H bond is more polarized and it ionizes most readily, making it the strongest acid of the set.
11. The Mendius reaction reduces alkyl cyanides (nitriles) to primary amines using sodium metal in ethanol (Na/[latex]C_2H_5OH[/latex]), a classic reduction method distinct from catalytic hydrogenation.

Brief Solutions to Group B & C Numerical and Descriptive Questions

Q12(a): Molarity of 38% HCl (sp. gr. 1.19):

[latex]M = \frac{10 \times d \times %}{Molar,mass} = \frac{10 \times 1.19 \times 38}{36.5} \approx 12.39,M[/latex]

Q12(b): To neutralize 1 L of 0.1 M NaOH, moles of NaOH = 0.1 mol, so moles of HCl required = 0.1 mol.
[latex]Volume = \frac{0.1}{12.39} \approx 8.07,mL[/latex] of conc. HCl.

Q12 OR: Rate of reaction depends directly on catalyst (lowers activation energy, increases rate), surface area (more exposed particles increase collision frequency), temperature (raises kinetic energy and collision frequency per Arrhenius equation), and concentration (higher concentration means more frequent collisions, per rate law).

Q13(b): Entropy of vaporization:
[latex]\Delta S = \frac{\Delta H_{vap}}{T} = \frac{42400,J/mol}{351.4,K} \approx 120.6,J,mol^{-1}K^{-1}[/latex]
(Boiling point converted to Kelvin: 78.4°C + 273 = 351.4 K)

Q14: Blue vitriol is copper(II) sulfate pentahydrate, [latex]CuSO_4 \cdot 5H_2O[/latex]. It’s prepared industrially by dissolving copper scrap or copper oxide in dilute sulfuric acid and crystallizing. It’s blue due to hydrated [latex]Cu^{2+}[/latex] ions, loses water on heating (turns white/anhydrous), and is used as a fungicide, in electroplating, and in Fehling’s/Benedict’s reagents.

Q15: Metal M (group IIB, used for coating iron sheets to prevent rusting via galvanization) is zinc.

• Ore: zinc blende, [latex]ZnS[/latex]
• Extraction involves roasting ZnS to ZnO, then reduction in a reverberatory/blast furnace using coke.
• Reduction reaction: [latex]ZnO + C \rightarrow Zn + CO[/latex]

Q16: Working backward from the ozonolysis product (ethanal + methanal mixture means the alkene was but-2-ene, [latex]CH_3-CH=CH-CH_3[/latex]… actually giving ethanal and methanal suggests a terminal alkene arrangement). Z = but-1-ene ([latex]CH_2=CH-CH_2-CH_3[/latex]) fits better with one aldehyde being methanal (from the terminal =CH2) — students should verify the exact carbon count against dehydrohalogenation of Y. Y is the alkyl halide from PCl5 treatment of secondary alcohol X (butan-2-ol), giving 2-chlorobutane, which on dehydrohalogenation gives but-2-ene or but-1-ene depending on the direction of elimination; ozonolysis of but-2-ene gives two molecules of ethanal (not ethanal + methanal), so students should re-check X as butan-1-ol type possibilities — this question is a good candidate for classroom discussion since multiple elimination products are possible.

Q17:
i. Benzene diazonium chloride hydrolyzed with water (warm, in presence of dilute [latex]H_2SO_4[/latex]) gives phenol and [latex]N_2[/latex] gas.
ii. Phenol is more acidic than aliphatic alcohols because the phenoxide ion formed after loss of [latex]H^+[/latex] is resonance-stabilized by the benzene ring, spreading the negative charge, while alkoxide ions have no such stabilization.
iii. (a) Phenol + [latex]CH_3I[/latex]/NaOH → anisole (Williamson ether synthesis). (b) Phenol treated with conc. [latex]HNO_3[/latex] gives picric acid (2,4,6-trinitrophenol).
iv. Laboratory test: phenol gives a violet color with neutral [latex]FeCl_3[/latex] solution.

Q18: A = propan-1-ol (3-carbon alcohol, no iodoform test since it’s primary and doesn’t have the [latex]CH_3-CHOH-[/latex] pattern). B = 1-bromopropane (via [latex]PBr_3[/latex]). C = butanenitrile (via KCN, chain extension by one carbon). D = butanoic acid (via acid hydrolysis, [latex]H_2O/H^+[/latex]). E = the corresponding nitrile/anhydride from D with [latex]P_2O_5[/latex] (dehydration). D converted to propane requires decarboxylation (soda-lime) followed by consideration of carbon count adjustments — a good discussion point for chain-length tracking with students.

Q19: Reaction examples:
a. Benzoin condensation: two benzaldehyde molecules + KCN catalyst → benzoin.
b. Claisen condensation: two ethyl acetate molecules + NaOEt → ethyl acetoacetate.
c. Reimer-Tiemann: phenol + chloroform + NaOH → salicylaldehyde.
d. Cannizzaro reaction: benzaldehyde (no alpha-H) + conc. NaOH → benzyl alcohol + sodium benzoate.
e. Hoffmann’s ammonolysis: alkyl halide + excess [latex]NH_3[/latex] → primary amine.

Q20(b): Solubility of AgCl in 0.1 M NaCl (common ion effect):
[latex]K_{sp} = [Ag^+][Cl^-][/latex], with [latex][Cl^-] \approx 0.1,M[/latex] from NaCl,
[latex][Ag^+] = \frac{9.965 \times 10^{-11}}{0.1} = 9.965 \times 10^{-10},M[/latex] = solubility of AgCl.

Q20 OR:

[latex]E^{\circ}{cell} = E^{\circ}{cathode} – E^{\circ}_{anode} = 0.77 – (-0.76) = 1.53,V[/latex]

. Since Fe³⁺/Fe²⁺ has the higher reduction potential, it acts as cathode and Zn as anode in the spontaneous direction (Zn oxidized, Fe³⁺ reduced). The reaction as written in the question ([latex]Zn^{2+} + 2Fe^{2+} \rightarrow Zn + 2Fe^{3+}[/latex]) runs opposite to this spontaneous direction, so it will not occur spontaneously.

Q21(a): Hoffmann’s method separates primary, secondary, and tertiary amines using diethyl oxalate: primary amines form solid oxalamide, secondary amines form liquid oxalamide, tertiary amines don’t react and can be separated by distillation; further hydrolysis regenerates the original amines.

Q21(b): i. Lucas test or oxidation test: propan-2-ol (secondary) turns acidified [latex]K_2Cr_2O_7[/latex] orange-to-green faster than propan-1-ol; propan-2-ol also gives a positive iodoform test while propan-1-ol doesn’t.
ii. Sucrose fermentation (using yeast/zymase enzyme) first hydrolyzes into glucose and fructose (invertase), which then ferment further into ethanol and [latex]CO_2[/latex].

Q21 OR: Compound A gives white precipitate with bromine water and couples with benzene diazonium chloride — this is characteristic of phenol.
i. Phenol reacts with [latex]Br_2/H_2O[/latex] to give 2,4,6-tribromophenol (white ppt).
ii. Phenol is prepared industrially from cumene (cumene process) or from benzene diazonium chloride hydrolysis.
iii. B = salicylaldehyde (from Reimer-Tiemann on phenol), C = salicyl alcohol (after LiAlH4 reduction of the aldehyde).
iv. Phenol → converted to benzene first (via diazonium salt route: aniline → benzene diazonium chloride → treat with [latex]H_3PO_2[/latex] to remove the group and get benzene), then Friedel-Crafts alkylation of benzene with [latex]CH_3Cl/AlCl_3[/latex] gives toluene.

Q22:
a. Paper production flow: wood chips → pulping (chemical/mechanical) → bleaching → beating → sheet formation → pressing/drying → finished paper.
b. Portland cement composition: mainly calcium silicates ([latex]C_3S, C_2S[/latex]), tricalcium aluminate ([latex]C_3A[/latex]), and tetracalcium aluminoferrite ([latex]C_4AF[/latex]), made by heating limestone and clay.
c. i. Antipyretic drug example: paracetamol. ii. Analgesic drug example: aspirin.
d. Carbon dating uses the known decay rate of radioactive [latex]^{14}C[/latex] in organic remains to estimate their age; it’s widely used to date archaeological artifacts and fossils.

Exam Overview

• Examination: HISSAN Central Examination 2081 (2025)
• Class: XII
• Subject: Chemistry
• Full Marks: 75 (11 marks objective + 64 marks subjective)
• Time: 3 hours
• Structure: Group A (11 MCQs, 1 mark each, 25 minutes), Group B (8 short-answer questions, 5 marks each, with “OR” alternative choices on several), Group C (8 long-answer questions, 3 marks each, again with alternative choices on some)

Question Pattern Analysis

The paper follows NEB’s standard three-tier format. Group A tests rapid recall and conceptual clarity across physical, inorganic, and organic chemistry without giving away much room for partial credit — students either know the concept or they don’t. Group B questions are more elaborate, frequently combining a short definitional or conceptual sub-part with a numerical or descriptive sub-part, and nearly every Group B question offers an “OR” alternative, meaning students can choose the topic they’re stronger in. Group C questions lean heavily into applied problem-solving — electrode potentials, solubility equilibria, and multi-step organic transformations — and reward students who can trace a logical sequence from clue to answer rather than just recalling isolated facts.

A notable pattern across Groups B and C is the recurring use of “identify the compound” style questions in organic chemistry, where students are given a set of reactions or clues and must work backward or forward to name specific compounds. This style tests genuine understanding of reaction mechanisms rather than memorized name-reactions in isolation.

Topics Covered

• Acid-base chemistry and pH
• Chemical kinetics (order of reaction, rate laws)
• Electrochemistry (reference electrodes, galvanic cells, standard electrode potential)
• Thermochemistry (enthalpy of formation, Hess’s law, entropy of vaporization)
• d-block elements (oxidation states)
• s and p-block/coordination compounds (mercury compounds)
• Ionic equilibrium (solubility product, common ion effect, strong vs weak electrolytes)
• Metallurgy (extraction of zinc, furnace design)
• Alkyl halides and nucleophilic substitution
• Alcohols, phenols, and ethers (acidity, tests, conversions)
• Grignard reagent reactions
• Aldehydes and ketones (condensation reactions: benzoin, Claisen, Reimer-Tiemann, Cannizzaro)
• Carboxylic acids (acidity comparison)
• Amines (Hoffmann’s ammonolysis, Mendius reaction, separation of 1°/2°/3° amines)
• Carbohydrates (fermentation of sucrose)
• Applied/environmental chemistry (paper production, cement composition, drugs, carbon dating)

Chapter-wise Weightage (Estimated)

Organic chemistry consistently carries the heaviest weightage in NEB-style Chemistry papers, and this exam is no exception — nearly two-fifths of the total marks trace back to organic reaction mechanisms, conversions, and identification problems. Physical chemistry follows closely, driven largely by numerical problems in electrochemistry and ionic equilibrium.

Difficulty Analysis

The MCQ section is moderately difficult — most questions test direct recall (definitions, standard values, named reactions) rather than multi-step reasoning, making them a good confidence-builder if students have covered the syllabus properly. Group B and C, however, escalate quickly in difficulty. The organic conversion questions (Q16, Q18, Q21-OR) require students to hold multiple reaction steps in mind simultaneously and correctly predict intermediate products, which is where most students tend to lose marks. The numerical questions (Q12, Q13b, Q20b, Q20-OR) are standard formula-based problems, so students who practice these formula types regularly should find them very manageable. Overall, this paper is fair but rewards depth of preparation over surface-level memorization, particularly in organic chemistry.

Student Preparation Tips

1. Master reaction sequences, not isolated reactions. Questions like Q16 and Q18 require you to track a compound through 3-4 transformation steps. Practice drawing out full reaction chains rather than memorizing reactions individually.
2. Drill the standard numericals until they’re automatic: molarity from percentage and density, entropy of vaporization, solubility product with common ion effect, and standard cell EMF calculations appear almost every year in some form.
3. Build a table of named reactions (Cannizzaro, Claisen, Reimer-Tiemann, Benzoin condensation, Hoffmann’s ammonolysis, Mendius reaction) with one clean example each — these show up as quick 1-mark identification questions regularly.
4. Revise tests to distinguish similar compounds: propan-1-ol vs propan-2-ol, phenol’s FeCl3 test, iodoform test conditions — these recur in both MCQs and short answers.
5. Don’t skip applied chemistry. Topics like cement composition, paper production, and carbon dating are short, low-effort, high-return topics that many students ignore until the last week.
6. Practice “identify the compound” style organic problems from multiple sources, since HISSAN and NEB both favor this format for testing deeper understanding over rote learning.

Frequently Asked Questions

Q: Is the HISSAN Chemistry paper a good indicator of the actual NEB board exam?
Yes. HISSAN papers are set to closely mirror NEB’s format, marking scheme, and difficulty distribution, so performing well here is a strong signal of board exam readiness.

Q: Which chapters should I prioritize if I’m short on time before the exam?
Organic chemistry (alcohols, phenols, alkyl halides, amines, carbonyl condensation reactions) and physical chemistry (electrochemistry and ionic equilibrium numericals) carry the most weightage and should be prioritized first.

Q: Are the numerical questions in this paper typical of what NEB usually asks?
Yes, the numerical patterns here — molarity/dilution, entropy of vaporization, solubility product with common ion effect, and standard cell EMF — are recurring numerical types across recent NEB and HISSAN papers.

Q: How important are named reactions like Cannizzaro or Reimer-Tiemann for the exam?
Very important. They typically appear as quick, low-time-cost marks (1 mark each) but only if you can recall a correct example reaction immediately, so they’re worth memorizing precisely.

Q: Is this paper harder or easier than a typical NEB board paper?
It’s comparable in difficulty, perhaps slightly favoring conceptual organic chemistry questions a bit more heavily than some past NEB papers, but well within the expected range for board-level preparation.

Conclusion

The HISSAN 2081 Chemistry XII paper offers a well-rounded snapshot of what NEB expects from Class XII Chemistry students — a mix of quick conceptual recall, standard numerical problem-solving, and layered organic chemistry reasoning. Students who work through this paper analytically, focusing on understanding reaction sequences and formula applications rather than just checking their answers against a key, will walk into their board exam considerably better prepared. Use the chapter-wise weightage and preparation tips above to guide your remaining study time toward the areas that matter most.