Introduction

The HISSAN Central Examination has become one of the most reliable practice grounds for Grade XII students preparing for their NEB board exams in Nepal, and the 2079 (2023) Chemistry paper (Set B) is a strong example of why. Carrying 75 full marks across an 11-mark objective section and a 64-mark subjective section, this paper leans noticeably into applied problem-solving — ppm calculations, thermochemistry numericals, rate law derivation from experimental data, and multi-step organic reaction sequences that require students to work both forward and backward through a synthesis pathway.

What stands out in this particular set is how many questions test whether students can actually apply a concept rather than just recall it. Question 21, for instance, gives raw experimental rate data and asks students to derive the rate law themselves — a skill that goes well beyond memorizing formulas. Similarly, the reaction sequence questions (Q17, Q20-OR) require identifying five or more compounds in a chain, each depending correctly on the one before it. This style of testing rewards students who have built genuine conceptual maps of how organic transformations connect to each other, rather than students who have only memorized isolated named reactions.

The paper also carries a noticeably heavier physical chemistry component this year — chemical kinetics, thermochemistry, electrochemistry (fuel cells, standard hydrogen electrode), and ionic equilibrium together make up a large share of the subjective marks. Inorganic chemistry stays lighter, mostly touching transition metals and metallurgy of mercury, while organic chemistry still carries substantial weight through alcohols, carbonyl compounds, amines, and polymer chemistry.

This article walks through the paper chapter by chapter, works out brief solutions to every numerical and conceptual question, estimates chapter-wise weightage, and lays out a focused preparation strategy — so that instead of just checking your answers, you leave with a clearer sense of exactly how NEB-style Chemistry papers are built and where your revision time is best spent.

Quick Answer Key (Group A — Multiple Choice Questions)

Full Question Paper

Brief explanation of each MCQ:

1. In 2,2,3-trimethylpentan-1-ol, working through the carbon skeleton (C1–OH, C2 with two methyl branches, C3 with one methyl branch, C4, C5), only C4 is attached to exactly two other carbon atoms, making it the sole secondary carbon in the structure.
2. The Oxo process (hydroformylation) adds CO and H₂ across an alkene double bond to produce an aldehyde, which is fundamentally a carbonylation reaction since a carbonyl group is being introduced into the molecule.
3. When sodium reacts with ethanol to liberate hydrogen gas and form sodium ethoxide, it demonstrates that the O–H bond in alcohol can act as a weak acid, releasing H⁺ (as H₂ after reduction by Na), which is the basis of the acidic nature of alcohols.
4. This question needs a closer look — see the note directly below the table.
5. Under strongly acidic conditions, electrolytic reduction of nitrobenzene proceeds through phenylhydroxylamine, which undergoes the Bamberger rearrangement to give p-aminophenol as the major product.
6. Zinc has an electronic configuration ending in a completely filled 3d¹⁰4s² arrangement; since the d-subshell is completely filled and doesn’t participate in bonding beyond the 4s electrons, zinc shows only the +2 oxidation state.
7. Corrosive sublimate refers to mercuric chloride, HgCl₂, historically used as an antiseptic and wood preservative before its toxicity became well understood.
8. ppm = (mass of solute / mass of sample) × 10⁶ = (0.2/500) × 10⁶ = 400 ppm.
9. A stronger acid has a higher Ka (or lower pKa). Benzoic acid’s Ka is 6.3×10⁻⁵. Compound C with pKa = 4 has Ka = 10⁻⁴, which is greater than 6.3×10⁻⁵, making it the only option stronger than benzoic acid.
10. For a first order reaction, t½ = 0.693/k, which contains no concentration term at all — meaning half-life is entirely independent of the initial concentration of reactants, unlike zero or second order reactions.
11. KNO₃ is chosen for salt bridges because K⁺ and NO₃⁻ ions migrate at nearly identical speeds through the solution, which prevents the buildup of a liquid junction potential that would otherwise interfere with accurate EMF measurement.

A note on Question 4: The reaction shows propanoic acid (CH₃CH₂COOH) treated with Cl₂/red phosphorus (the Hell-Volhard-Zelinsky reaction), which halogenates the alpha carbon — the carbon directly attached to the carboxyl group — to give an alpha-chloro acid. Based on strict HVZ regiochemistry, the expected product A should be CH₃CHClCOOH (chlorine on the alpha carbon), which doesn’t exactly match any listed option as typically formatted. Depending on how your textbook or teacher labels the mechanism, you may see this recorded as option d in some official answer keys; this is worth clarifying with your chemistry teacher directly, since the correct regiochemistry of HVZ (alpha-substitution) is the more exam-critical concept to walk away with here, regardless of which lettered option is marked correct.

Brief Solutions to Group B & C Numerical and Descriptive Questions

Q12:
a. Oxalic acid is used as a primary standard because it is available in high purity, is stable and non-hygroscopic, doesn’t change composition on storage, and its exact equivalent weight can be calculated directly from its formula.
b. Redox reaction: [latex]5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O[/latex]. Manganese is reduced from +7 to +2 while oxalate is oxidized to carbon dioxide, confirming this is a redox titration.
c. The solution is warmed (to about 60°C) because at room temperature the reaction between oxalate and permanganate ions is slow; heating increases the rate so the initial drops of permanganate decolorize promptly instead of the titration stalling before the autocatalytic effect of Mn²⁺ kicks in.
d. Decinormal = 0.1 N. Equivalents needed for 500 cc = 0.1 × 0.5 = 0.05 eq. Equivalent weight of hydrated oxalic acid ([latex]H_2C_2O_4 \cdot 2H_2O[/latex], molar mass 126, basicity 2) = 126/2 = 63. Weight required = 0.05 × 63 = 3.15 g.

Q13:
a. Enthalpy of formation is the heat change when one mole of a compound forms from its elements in their standard states, whereas enthalpy of combustion is the heat released when one mole of a substance is completely burnt in excess oxygen.
b. Combustion of ammonia: [latex]4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O[/latex], with [latex]\Delta H = 4 \times (-9.6) = -38.4,kcal[/latex] for 4 moles.
Using [latex]\Delta H_{comb} = \Sigma \Delta H_f(products) – \Sigma \Delta H_f(reactants)[/latex]:
[latex]-38.4 = [6(-68)] – [4 \times \Delta H_f(NH_3)][/latex]
[latex]-38.4 = -408 – 4\Delta H_f(NH_3)[/latex]

[latex]\Delta H_f(NH_3) = \frac{-408+38.4}{4} = -92.4,kcal/mol[/latex]

Q13 OR:
a. Ostwald’s dilution law: for a weak electrolyte, [latex]\alpha = \sqrt{\frac{K}{C}}[/latex], where α is degree of dissociation, K is the dissociation constant, and C is concentration. Limitation: it only holds for weak electrolytes with low degree of dissociation, since it assumes ideal solution behavior and breaks down for strong electrolytes where ionic interactions can’t be ignored.
b. Mixing equal volumes halves each concentration: [latex][Ba^{2+}] = 1\times10^{-3},M[/latex], [latex][SO_4^{2-}] = 1\times10^{-4},M[/latex]. Ionic product = [latex]1\times10^{-3} \times 1\times10^{-4} = 1\times10^{-7}[/latex], which is far greater than [latex]K_{sp} = 1\times10^{-10}[/latex], so precipitation will occur.

Q14: Hoffmann’s method uses diethyl oxalate: primary amines form solid oxalamides, secondary amines form liquid oxalamides, and tertiary amines don’t react at all (separated by simple distillation). The solid and liquid oxalamides are separated physically, then each is hydrolyzed with acid or alkali to regenerate the original primary or secondary amine.

Q15: Metal M (atomic number 80, “quicksilver”) is mercury. Its sulphide ore is cinnabar (HgS). Since mercury is a fairly unreactive/noble metal, simple roasting in air is sufficient for extraction: [latex]HgS + O_2 \rightarrow Hg + SO_2[/latex]. The mercury vapor produced is condensed and collected as the pure liquid metal — no separate chemical reducing agent is needed.

Q16:
a. Transition metals are elements with partially filled d-orbitals in their atomic or common ionic state; they occupy groups 3–12 of the periodic table (the d-block).
b. Variable oxidation states arise because the energy gap between (n-1)d and ns orbitals is small, allowing electrons from both to be lost or shared in bonding.
c. Most transition metal ions are colored because their partially filled d-orbitals allow d-d electronic transitions — electrons absorb specific wavelengths of visible light to jump between split d-orbital energy levels (crystal field splitting), and the complementary color is what we observe.

Q17: Working backward from D: D is obtained from benzene + CH₃Cl/AlCl₃ (Friedel-Crafts alkylation), so D = toluene. C → D uses ZnHg/HCl (Clemmensen reduction, reduces C=O to CH₂), so C = benzaldehyde (C₆H₅CHO). B → C uses H₂/Pd-BaSO₄ (Rosenmund reduction, converts acid chloride to aldehyde), so B = benzoyl chloride (C₆H₅COCl). A → B uses PCl₃ (converts carboxylic acid to acid chloride), so A = benzoic acid (C₆H₅COOH).

Q18:
a. Aldol condensation: [latex]2CH_3CHO \xrightarrow{dil.,NaOH} CH_3CH(OH)CH_2CHO[/latex] (aldol)
b. Reimer-Tiemann: phenol + CHCl₃ + NaOH → salicylaldehyde
c. Sandmeyer’s reaction: benzene diazonium chloride + CuCl/HCl → chlorobenzene + N₂
d. Diazotization: aniline + NaNO₂ + HCl (0–5°C) → benzene diazonium chloride
e. Decarbonylation: an aldehyde loses CO on heating with a catalyst, e.g., [latex]CH_3CHO \rightarrow CH_4 + CO[/latex]

Q19:
a. From C₄H₁₀O: primary alcohol — butan-1-ol, [latex]CH_3CH_2CH_2CH_2OH[/latex]; secondary alcohol — butan-2-ol, [latex]CH_3CH_2CH(OH)CH_3[/latex].
b. Victor Meyer’s method converts the alcohol to an alkyl iodide, then to a nitro compound, then treats it with nitrous acid and KOH: primary alcohols give a blood-red color, secondary alcohols give a blue color, and tertiary alcohols give no color at all — allowing all three classes to be distinguished in one test.

Q19 OR: The sweet-smelling compound P that oxidizes in sunlight to a poisonous gas, reacts with acetone to give a sleep-inducing drug, and was historically used as anesthesia is chloroform.
a. P = chloroform (CHCl₃); Q = phosgene (COCl₂), formed by aerial oxidation of chloroform in sunlight; R = chloretone, formed when chloroform reacts with acetone in the presence of alkali.
b. Chloroform heated with concentrated nitric acid forms chloropicrin: [latex]CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O[/latex].
c. Laboratory preparation: bleaching powder reacts with ethanol (or acetone) in the haloform reaction to give chloroform.

Q20:
a. Aldehydes and ketones are called carbonyl compounds because both classes share the C=O (carbonyl) functional group as their defining structural feature.
b. Carbonyl compounds are tested using 2,4-dinitrophenylhydrazine (2,4-DNP), which forms a yellow-orange precipitate with both aldehydes and ketones.
c. Propanal (an aldehyde) gives a positive Tollens’ test (silver mirror) and Fehling’s test (brick-red precipitate), while propanone (a ketone) gives negative results for both — this distinguishes the two directly.
d. Converting propanal to propanone requires a change in carbon skeleton arrangement: propanal is first oxidized to propanoic acid, which is then converted (via its calcium salt, using standard ketonic decarboxylation methods taught in your textbook) into a ketone — this step is worth confirming against your specific textbook’s method since exact reagent conditions vary between NEB textbook editions.
e. Ozonolysis of 2-methylpent-2-ene ([latex]CH_3-C(CH_3)=CH-CH_2CH_3[/latex]) cleaves the double bond to give propanone (latex_2C=O[/latex]) from one side and propanal ([latex]CH_3CH_2CHO[/latex]) from the other — producing both target compounds from a single starting alkene.

Q20 OR: Reaction sequence: A –CO/Δ–> CH₃COONa –(B)–> CH₃COOH –P₂O₅–> C, and CH₃COOH –(D)–> CH₃COCl –(E)–> CH₃COOCH₃.
a. A = sodium methoxide (CH₃ONa), which undergoes carbonylation under heat and pressure to give sodium acetate. B = dilute mineral acid (e.g., HCl), used to convert sodium acetate to acetic acid. C = acetic anhydride (latex_2O[/latex]), formed by dehydration of acetic acid with P₂O₅. D = PCl₅ (or SOCl₂), which converts acetic acid to acetyl chloride. E = methanol (CH₃OH), which reacts with acetyl chloride to form methyl acetate.
b. Acetic anhydride (C) heated with LiAlH₄ is reduced to give two moles of ethanol.
c. Converting CH₃COOH to CH₃NH₂: acetic acid is first converted to acetamide (CH₃CONH₂) by reaction with ammonia, and acetamide then undergoes Hofmann bromamide degradation (Br₂/KOH) to give methylamine (CH₃NH₂), losing one carbon in the process.

Q21:
a. In an energy profile diagram, both catalyzed and uncatalyzed pathways start and end at the same energy levels (same reactants and products), but the catalyzed pathway shows a lower activation energy peak, meaning the reaction proceeds through an alternative, lower-energy transition state.
b. Glucose, being a simpler monosaccharide with a less complex crystal lattice and weaker intermolecular hydrogen-bonding network than sucrose (a disaccharide), breaks apart and hydrates faster in water, giving it a quicker apparent dissolution rate.
c. Rate law is the experimentally determined mathematical expression relating the rate of a reaction to the concentrations of reactants, each raised to a power called its order: [latex]Rate = k[A]^m[B]^n[/latex].
d. Comparing experiments 1 and 2 ([M] constant, [N] doubles, rate doubles) shows the reaction is first order in N. Comparing experiments 2 and 3 ([N] constant, [M] doubles, rate unchanged) shows the reaction is zero order in M. So [latex]Rate = k[N][/latex]. Using experiment 1: [latex]1.6\times10^{-4} = k \times 0.5[/latex], giving [latex]k = 3.2\times10^{-4}[/latex]. At [M] = 2.00, [N] = 4.00: [latex]Rate = 3.2\times10^{-4} \times 4.00 = 1.28\times10^{-3},mol,L^{-1}s^{-1}[/latex].

Q21 OR:
a. In a hydrogen-oxygen fuel cell, hydrogen gas is fed to the anode and oxygen to the cathode, both over porous electrodes dipped in KOH electrolyte.
ii. Anode: [latex]2H_2 + 4OH^- \rightarrow 4H_2O + 4e^-[/latex]; Cathode: [latex]O_2 + 2H_2O + 4e^- \rightarrow 4OH^-[/latex]; Net: [latex]2H_2 + O_2 \rightarrow 2H_2O[/latex].
iii. Major use: generating electricity aboard spacecraft, with drinking water as a useful byproduct (famously used in the Apollo missions).
b. The standard hydrogen electrode consists of hydrogen gas at 1 atm bubbled over a platinum electrode immersed in a 1M H⁺ solution.
ii. It’s called the reference electrode because its potential is arbitrarily fixed at 0V, and all other electrode potentials are measured relative to it.

Q22:
a. Addition polymers form when monomer units (typically containing double bonds) join together repeatedly without losing any atoms, e.g., polyethylene from ethylene; condensation polymers form when monomers with different functional groups link together with the elimination of a small molecule like water, e.g., nylon or terylene.
b. Paper production stages: raw material preparation (wood/bamboo chipping) → pulping (mechanical or chemical, e.g., kraft/sulfite process) → bleaching → beating and refining → sizing → sheet formation on a wire mesh → pressing → drying → calendering and finishing into rolls or sheets.

Exam Overview

• Examination: HISSAN Central Examination 2079 (2023), Set B
• Class: XII
• Subject: Chemistry (Code 3021 B)
• 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” alternatives on some), Group C (3 long-answer questions, 8 marks each, with “OR” alternatives)

Question Pattern Analysis

This paper’s MCQ section leans more heavily on applied calculation (ppm concentration, Ka/pKa comparison) and mechanistic reasoning (HVZ reaction, Bamberger rearrangement) than on pure definitional recall, making Group A noticeably more demanding than a typical “just remember the fact” section. Group B mixes conceptual short-answer parts with quantitative sub-parts almost every time — enthalpy calculations, solubility product checks, and normality-based weight calculations all appear as sub-parts of larger questions. Group C is where the paper gets most rigorous: Question 21 in particular gives raw tabulated experimental data and expects students to independently derive the order of reaction and the rate constant, rather than simply plugging numbers into a pre-stated formula.

Topics Covered

• Carbon classification and IUPAC nomenclature
• Oxo process / hydroformylation (carbonylation reactions)
• Acidic nature of alcohols
• Hell-Volhard-Zelinsky (HVZ) reaction
• Reduction of nitrobenzene (aniline, p-aminophenol)
• d-block elements (zinc’s fixed valency, transition metal properties, color of transition metal ions)
• Mercury compounds and metallurgy (extraction of mercury from cinnabar)
• Concentration units (ppm)
• Acid strength comparison (Ka, pKa)
• Chemical kinetics (half-life, rate law, order of reaction from experimental data)
• Electrochemistry (salt bridge, standard hydrogen electrode, hydrogen-oxygen fuel cell)
• Redox titrations (oxalic acid-permanganate titration)
• Thermochemistry (enthalpy of formation, enthalpy of combustion)
• Ionic equilibrium (Ostwald’s dilution law, solubility product)
• Amines (Hoffmann’s method of separation)
• Alcohols (Victor Meyer’s test)
• Carbonyl compounds (aldehydes, ketones, distinguishing tests)
• Named reactions (Aldol condensation, Reimer-Tiemann, Sandmeyer’s, diazotization, decarbonylation)
• Carboxylic acid derivatives (acid chlorides, anhydrides, esters, amides)
• Polymers (addition vs condensation) and paper manufacturing

Chapter-wise Weightage (Estimated)

Physical chemistry and organic chemistry are almost evenly weighted in this paper, which is slightly different from papers where organic chemistry dominates more clearly — a useful signal that strong numerical fluency in kinetics, thermochemistry, and electrochemistry is just as important as organic reaction mastery for scoring well this year.

Difficulty Analysis

The MCQ section is moderately challenging, with several questions (ppm calculation, Ka comparison, HVZ mechanism) requiring a calculation or a mechanistic step rather than pure recall. Group B numericals (oxalic acid weight calculation, enthalpy of formation, solubility product check) follow standard formula patterns that become very manageable with regular practice. Group C is the most demanding section this year — particularly Question 21, which tests whether students can genuinely derive a rate law from raw data rather than recognize a pre-labeled order, and Question 17/20-OR, which require tracking four to five compounds through a reaction sequence with correct reasoning at every step. Students who are comfortable working backward from a final product to identify starting materials will find this paper considerably more approachable than students relying purely on forward memorization of individual reactions.

Student Preparation Tips

1. Practice deriving rate laws from tabulated data, not just plugging into memorized formulas — compare pairs of experiments where only one concentration changes, and work out the order for each reactant independently before combining them.
2. Get comfortable working backward through reaction sequences. Questions like Q17 and Q20-OR give you the last step or an external clue and expect you to trace the entire sequence in reverse — practice this direction specifically, since forward-only practice won’t prepare you for it.
3. Revisit ppm, molarity, and normality calculations until they’re automatic. These conversions appear in some form on nearly every HISSAN and NEB paper and are easy marks once the formula is second nature.
4. Build clean example reactions for each named reaction (Aldol condensation, Reimer-Tiemann, Sandmeyer’s, diazotization, decarbonylation, HVZ) — these are quick 1-mark questions that only pay off if you can write the reaction correctly and completely.
5. Understand the reasoning behind common tests, not just the test names — why Tollens’ test distinguishes aldehydes from ketones, why Victor Meyer’s test gives different colors for different alcohol classes, and so on. Examiners frequently ask “give reason” alongside these tests.
6. Don’t neglect applied chemistry (polymers, paper production). These topics are short and predictable, and skipping them means giving up easy marks that many well-prepared students pick up without much extra effort.

Frequently Asked Questions

Q: How is this HISSAN paper different from a typical NEB board Chemistry paper?
It follows the same overall format and marking scheme as NEB, though this particular set places slightly more emphasis on numerical derivation (especially in kinetics) than some other years, making it excellent practice for calculation-heavy questions.

Q: Which topics should I prioritize given the weightage in this paper?
Organic chemistry (carbonyl compounds, alcohols, amines, carboxylic acid derivatives) and physical chemistry (kinetics, thermochemistry, electrochemistry) together make up over three-quarters of the marks and should be your top priorities.

Q: Is deriving a rate law from experimental data a common NEB exam pattern?
Yes, this is a recurring question type in recent years — examiners increasingly prefer testing genuine understanding of rate laws over simple recall, so practicing this data-based derivation is a valuable use of study time.

Q: How important are reaction sequence identification questions?
Very important — they typically carry 5 or more marks in Group B/C and test deep, connected understanding of organic chemistry rather than isolated facts, making them high-value, high-effort questions worth dedicated practice.

Q: Is this paper harder than the average HISSAN Chemistry set?
It’s on the more rigorous end, particularly due to the data-based kinetics question and the multi-step reaction sequences, but it remains fully within the syllabus and expected NEB difficulty range with consistent practice.

Conclusion

The HISSAN 2079 Chemistry XII Set B paper rewards students who can move beyond memorization into genuine application — deriving rate laws from raw data, tracing multi-step reaction sequences in both directions, and connecting named reactions to their underlying mechanisms. Working through this paper carefully, understanding not just the correct answers but the reasoning behind each one, will meaningfully strengthen your readiness for the NEB board exam. Use the chapter-wise weightage and preparation tips above to direct your remaining revision time toward the areas that will earn you the most marks.