NEB Grade XI Chemistry Set 1 (New Course) — Answer Key, Important Questions & Free PDF

Chemistry under the New Course for NEB Grade XI takes a noticeably different shape from the syllabus many students expect — instead of being purely an academic subject confined to reactions and formulas, this Set 1 paper weaves together atomic structure, electrochemistry, chemical equilibrium, industrial processes, organic chemistry, and even applied topics like cell membrane transport (the sodium-potassium pump) and the economics of chemical industries in Nepal. For a Grade XI student encountering this breadth for the first time, the sheer range can feel overwhelming — but it also means that strong performance here rewards genuinely broad preparation rather than narrow memorization of a handful of favorite chapters.

This paper, Subject Code 3011, follows the standard 75-mark NEB structure: Group A carries 11 marks of multiple choice questions, Group B carries 40 marks across eight short-answer questions (several with OR alternatives), and Group C carries 24 marks across three long-answer questions built around real chemical reactions and industrial processes. What stands out immediately is how application-driven the questions are — rather than asking students to simply state Avogadro’s number, Q1 asks them to calculate the number of water molecules in 1 mL using it; rather than just defining limiting reagent, Q20 walks through a complete stoichiometry problem with percentage yield; and Q19 connects Haber’s process directly to Nepal’s urea fertilizer dependency, asking students to think about industrial economics, not just chemistry.

Below, you’ll find the complete MCQ answer key with worked explanations for the calculation-based questions, a breakdown of the major topic areas this paper draws from across the New Course syllabus, a curated list of important questions worth prioritizing for your final exam, FAQs with fully worked solutions to the trickiest numericals and reaction-identification problems, and the complete downloadable question paper. Whether you’re working through this as exam practice, using it to identify gaps in your preparation across physical, inorganic, and organic chemistry, or building a revision plan around the industrial and applied chemistry sections that often get neglected, this guide should give you a clear and thorough starting point.

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• MCQ Answer Key
• Topic Areas Covered in This Paper
• Important Questions for Your Final Exam
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• How to Use This Paper Effectively

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Get the complete NEB Grade XI Chemistry Set 1 (New Course) paper as a PDF — includes the reaction scheme for Q18(b) and any structural diagrams referenced in the questions below.

MCQ Answer Key — NEB Chemistry Grade XI, Set 1

All 11 MCQs from Group A, 1 mark each (11 × 1 = 11 marks). Questions 1, 3, 4, and 6 involve calculation or careful structural reasoning — see the FAQ for full working.

Quick working for the trickier ones: Q1 — moles of water in 1 mL (= 1 g, since density ≈ 1 g/mL) = 1/18 mol. Number of molecules = (1/18) × 6.022×10^23 ≈ 3.35×10^22, matching option (d). Q3 — the question as printed says “hybridization of C in BF3,” but BF3 contains no carbon; this almost certainly should read “hybridization of B in BF3.” Boron in BF3 has three bonding pairs and no lone pairs, giving sp2 hybridization (trigonal planar geometry) — answer (a). Q4 — in OF2, fluorine is more electronegative than oxygen (the only common compound where this is true), so oxygen takes a positive oxidation state. With F at −1 each (total −2), oxygen must be +2 to balance to zero overall. Q6 — for the reaction PCl5 ⇌ PCl3 + Cl2, Δn (moles of gaseous products − moles of gaseous reactants) = 2 − 1 = +1, which is positive. Since Kp = Kc(RT)^Δn, a positive Δn with RT > 1 (in atm-L units, typically) gives Kp > Kc.

Topic Areas Covered in This Paper

This Set 1 paper draws from across the full breadth of the New Course Grade XI Chemistry syllabus. Here’s how the major areas break down:

• Atomic structure and periodicity (Q1, Q12): mole concept and Avogadro’s number, Rutherford’s gold foil experiment and its conclusions, Hund’s rule, quantum numbers, electronic configuration, atomic and ionic radii trends across groups and periods.
• Chemical bonding (Q3): hybridization and molecular geometry (sp2, sp, dsp2, sp3).
• Stoichiometry and chemical laws (Q2, Q20): law of reciprocal proportions, limiting reagent calculations, percentage yield, gas volume calculations using the ideal gas equation.
• Oxidation-reduction and electrochemistry (Q4, Q13): oxidation states, oxidizing and reducing agents, Faraday’s laws of electrolysis (coulombs required for electrodeposition, mass of metal plated).
• Gaseous state (Q5, Q14 OR): real vs. ideal gas behavior, Boyle’s and Charles’s laws, derivation of PV=nRT, Graham’s law of diffusion and gas density calculations.
• Chemical equilibrium (Q6, Q14): active mass, equilibrium constant (Kp, Kc) for heterogeneous equilibria, Le Chatelier’s principle (effect of temperature and pressure on equilibrium position).
• s-block and p-block chemistry / industrial chemistry (Q7, Q8, Q15, Q16, Q19, Q21): amphoteric oxides, amalgams, Haber’s process for ammonia manufacture, ammonia reactions with copper sulfate solution, metallurgy terms (ore, calcination, roasting, flux, electrometallurgy), urea manufacturing flow sheets, washing soda (Solvay process) manufacture.
• Chemical industry and economics (Q9, Q19): cost structures in chemical industries (fixed vs. variable costs), challenges of establishing chemical industries in developing countries like Nepal.
• Biomolecules and applied chemistry (Q10): active transport mechanisms in cell membranes (sodium-potassium pump).
• Agricultural chemistry / fertilizers (Q11, Q19): NPK fertilizers and their nitrogen/phosphorus/potassium content, urea as a nitrogen source.
• Organic chemistry — hydrocarbons and nomenclature (Q17, Q22): homologous series, general formulas (alkenes), structural isomerism (C2H6O isomers and IUPAC naming), electrophiles and nucleophiles, dehydrohalogenation, Wurtz reaction, peroxide effect (anti-Markovnikov addition), ozonolysis.
• Aromatic chemistry (Q18): Hückel’s rule of aromaticity, reaction sequences involving sodium benzoate derivatives, directing effects of substituents (meta-directing nature of -NO2) in electrophilic aromatic substitution.
• Distinguishing organic compounds (Q22): chemical tests to differentiate ethene and ethyne.

Important Questions for Your Final Exam

These topics from Set 1 are high-value because they combine conceptual definitions with calculations or reaction-writing — exactly the format NEB consistently favors.

• Mole concept and Avogadro’s number — calculating molecules/atoms from mass or volume
• Rutherford’s gold foil experiment — observations and conclusions about atomic structure
• Hund’s rule, quantum numbers (n, l, m) for given subshells, and electronic configurations
• Atomic radius trends across periods and groups, and comparing isoelectronic species (O2-, F-, Na+, Mg2+)
• Faraday’s laws of electrolysis — coulombs required and mass deposited calculations
• Equilibrium constant expressions (Kp, Kc) for heterogeneous reactions
• Le Chatelier’s principle — effect of temperature, pressure, and concentration on equilibrium
• Derivation of the ideal gas equation PV=nRT from Boyle’s and Charles’s laws
• Graham’s law of diffusion — relating rates of diffusion to molar mass and density
• Haber’s process — principle, conditions, and flow diagram for ammonia manufacture
• Reactions of ammonia with copper sulfate solution (formation of complex ions)
• Metallurgy terminology — ore, calcination, roasting, flux, electrometallurgy
• Solvay process (washing soda manufacture) — principle, steps, and why it doesn’t work for potassium carbonate
• Homologous series, general formulas of alkanes/alkenes/alkynes
• Isomerism — structural isomers of C2H6O with IUPAC names (ethanol and dimethyl ether)
• Electrophiles vs. nucleophiles — definitions and examples
• Hückel’s rule of aromaticity and identifying aromatic compounds
• Reaction sequences involving sodium salts of carboxylic acids (decarboxylation, Friedel-Crafts alkylation)
• Directing effects of substituents in electrophilic aromatic substitution (why -NO2 is meta-directing)
• Limiting reagent identification and stoichiometric calculations
• Percentage yield calculations
• Dehydrohalogenation reactions — examples and mechanism
• Wurtz reaction — definition and identifying starting materials from products
• Peroxide effect (Kharasch effect) in addition reactions to alkenes
• Ozonolysis of alkenes — predicting carbonyl products
• Chemical tests to distinguish alkenes from alkynes (e.g., Tollens’ test, bromine water, ammoniacal silver nitrate)
• Urea manufacturing flow sheet and challenges of chemical industry development in Nepal

Frequently Asked Questions

What conclusions can be drawn from Rutherford’s gold foil experiment observations in Q12(a-b)?

The fact that most alpha particles pass straight through the gold foil without deflection tells us that an atom is mostly empty space — the volume occupied by the nucleus is a tiny fraction of the total atomic volume. The fact that some alpha particles deflect through small angles tells us that there is a small, dense, positively charged region within the atom (the nucleus) that repels the positively charged alpha particles when they pass close to it. Together, these observations led Rutherford to propose the nuclear model of the atom: a small, dense, positively charged nucleus surrounded by mostly empty space through which electrons move.

For Q12(d), what are the quantum numbers for an electron in the 3p subshell?

For 3p: the principal quantum number n = 3 (from the “3”). The azimuthal (orbital angular momentum) quantum number l = 1 (since “p” corresponds to l = 1; s=0, p=1, d=2, f=3). The magnetic quantum number m can take any integer value from −l to +l, so for l = 1, m can be −1, 0, or +1 (corresponding to the three p orbitals: px, py, pz).

How do you find the atomic number and electron count in Q12(e) for valence configuration 4s1?

A valence shell configuration of 4s1 with no other electrons in the 4th shell or beyond, combined with a completely filled inner shell structure (1s2 2s2 2p6 3s2 3p6 3d10), describes an atom with total electrons = 2+2+6+2+6+10+1 = 29. This is copper (Cu, atomic number 29) — note that copper is an exception to the expected filling order, having configuration [Ar] 3d10 4s1 rather than [Ar] 3d9 4s2, due to the extra stability of a completely filled d-subshell. Counting s-electrons: 1s2, 2s2, 3s2, 4s1 = 2+2+2+1 = 7 s-electrons. Counting p-electrons: 2p6, 3p6 = 6+6 = 12 p-electrons.

For Q12 (OR, b), which is larger — K or K+, and why?

K (the neutral potassium atom) is larger than K+. When potassium loses one electron to form K+, the remaining electrons experience a stronger effective nuclear charge per electron (same nuclear charge, fewer electrons), pulling the remaining electron shells inward. Additionally, K+ has one fewer electron shell occupied compared to K (K has configuration [Ar]4s1, while K+ has configuration [Ar], the same as a noble gas) — losing the entire 4th shell causes a significant decrease in size.

For Q12 (OR, c), how do O2-, F-, Na+, and Mg2+ compare in size?

All four of these ions are isoelectronic — each has 10 electrons (the same electronic configuration as neon: 1s2 2s2 2p6). For isoelectronic species, size is determined entirely by nuclear charge: more protons means a stronger pull on the same number of electrons, resulting in a smaller ionic radius. The nuclear charges are O (8), F (9), Na (11), Mg (12). Therefore the order of decreasing size is: O2- > F- > Na+ > Mg2+. O2- is largest because it has the fewest protons (8) pulling on the 10 electrons, while Mg2+ is smallest because it has the most protons (12) pulling on the same 10 electrons.

How do you calculate the coulombs required in Q13(b)?

Note: the electrode reaction given (Mg2+ + 2e- → Mg) appears to be a typo for the aluminum deposition mentioned in the question — for aluminum, the relevant half-reaction is Al3+ + 3e- → Al. Using Faraday’s first law: moles of Al = mass/molar mass = 36/27 = 4/3 mol (using atomic mass of Al = 27). Each mole of Al requires 3 moles of electrons (3 Faradays). Total Faradays needed = (4/3) × 3 = 4 Faradays. Charge = 4 × 96500 C = 386000 C (or 3.86×10^5 C). If the question genuinely intends the Mg2+ + 2e- → Mg reaction with “36 grams of aluminum” being a wording error for magnesium (molar mass 24): moles of Mg = 36/24 = 1.5 mol, requiring 2 Faradays per mole, so 3 Faradays = 3 × 96500 = 289500 C. We recommend clarifying with your teacher which element (Al or Mg) and which corresponding atomic mass/electron count this question intends, as the printed reaction and the stated element don’t match.

How do you calculate the mass of silver plated in Q13(c)?

Charge passed Q = I × t = 5 A × (10 × 3600 s) = 5 × 36000 = 180000 C. Using Faraday’s first law, mass deposited = (Q × equivalent weight) / Faraday’s constant = (180000 × 108) / 96500. Calculating: 180000 × 108 = 19440000. Then 19440000 / 96500 ≈ 201.4 grams of silver.

For Q14(c), what’s the Kp expression for CaCO3(s) ⇌ CaO(s) + CO2(g)?

For heterogeneous equilibria, only gaseous species appear in the equilibrium constant expression — pure solids and liquids are assigned an activity of 1 and don’t appear. Since CO2 is the only gas, Kp = P(CO2) — the equilibrium constant simply equals the partial pressure of carbon dioxide gas at equilibrium.

For Q14(d), how do temperature and pressure affect the SO2 + O2 ⇌ SO3 equilibrium?

The reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) + heat is exothermic (heat is released as the reaction proceeds forward). By Le Chatelier’s principle: increasing temperature shifts the equilibrium backward (toward SO2 and O2), since the system opposes the added heat by favoring the endothermic (reverse) direction, which absorbs heat. For pressure: the forward reaction has 3 moles of gas (2 SO2 + 1 O2) converting to 2 moles of gas (2 SO3) — fewer moles on the product side. Increasing pressure shifts the equilibrium forward (toward SO3), since the system favors the side with fewer gas molecules to reduce the pressure increase.

For Q14 (OR, b), how do you find the density of gas A?

By Graham’s law of diffusion, rate of diffusion is inversely proportional to the square root of molar mass (or density): rate_A / rate_B = √(M_B/M_A) = √(d_B/d_A). Given rate_A = 4 × rate_B: 4 = √(64/M_A), so 16 = 64/M_A, giving M_A = 4 g/mol. Since density is proportional to molar mass at the same conditions (d = M/22.4 at STP for ideal gases, or more simply, ratio of densities equals ratio of molar masses for gases at the same T and P): d_A/d_B = M_A/M_B = 4/64 = 1/16. If gas B’s molar mass is 64 (used directly as a proxy for density in this simplified ratio), then gas A’s “density” in the same relative units is 4/64 = 0.0625 times that of gas B, or M_A = 4 g/mol corresponds to a gas like helium or methane-range molar mass — the molar mass of gas A is 4 g/mol.

How do you identify X and Y in the reaction sequence in Q18(b)?

The starting material is sodium benzoate (the sodium salt of benzoic acid, shown as a benzene ring with a -COONa group). Step 1: heating sodium benzoate with NaOH and CaO (soda lime) causes decarboxylation — loss of the -COONa group as Na2CO3, leaving behind benzene as product X. Step 2: treating benzene (X) with CH3Cl in the presence of AlCl3 is a Friedel-Crafts alkylation reaction, which introduces a methyl group onto the benzene ring, giving toluene (methylbenzene) as product Y. So: X = Benzene, Y = Toluene (IUPAC name: methylbenzene).

Why is the -NO2 group meta-directing in electrophilic substitution (Q18c)?

The -NO2 group is strongly electron-withdrawing through both resonance (its resonance structures pull electron density from the ring toward the nitro group, especially from the ortho and para positions) and the inductive effect (the highly electronegative N and O atoms withdraw electron density through the sigma bonds). This withdrawal makes the ortho and para positions electron-deficient relative to the meta position. Since electrophilic substitution requires the ring to donate electron density to the incoming electrophile, the ring preferentially reacts at the position where electron density is least depleted — the meta position — making -NO2 a meta-directing (and overall deactivating) group.

How do you find the limiting reactant and product masses in Q20?

The reaction is 2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3. Molar mass of NH4Cl = 14+4(1)+35.5 = 53.5 g/mol. Molar mass of Ca(OH)2 = 40+2(16+1) = 40+34 = 74 g/mol. Moles of NH4Cl = 7/53.5 ≈ 0.1308 mol. Moles of Ca(OH)2 = 7/74 ≈ 0.0946 mol. The stoichiometric ratio requires 2 mol NH4Cl per 1 mol Ca(OH)2. For 0.0946 mol Ca(OH)2, we’d need 0.1892 mol NH4Cl — but we only have 0.1308 mol. Since NH4Cl is insufficient relative to what’s needed, NH4Cl is the limiting reactant.

Significance of limiting reactant: it determines the maximum amount of product that can form — the reaction stops once the limiting reactant is fully consumed, regardless of how much excess reactant remains. Mass of CaCl2 formed: from the stoichiometry, 2 mol NH4Cl produces 1 mol CaCl2 (molar mass = 40+2(35.5)=111 g/mol). Moles of CaCl2 = 0.1308/2 ≈ 0.0654 mol. Mass = 0.0654 × 111 ≈ 7.26 g.

Volume of NH3 at 27°C and 1.5 atm: moles of NH3 = moles of NH4Cl (1:1 ratio from the equation, since 2 mol NH4Cl gives 2 mol NH3) = 0.1308 mol. Using PV=nRT with T=300K, R=0.0821 L·atm/(mol·K): V = nRT/P = (0.1308 × 0.0821 × 300)/1.5 ≈ (3.222)/1.5 ≈ 2.15 L.

Percentage yield in Q20(e): moles of Ca(OH)2 = 3.7/74 = 0.05 mol, which theoretically produces 2 × 0.05 = 0.1 mol NH3. At STP, volume of NH3 expected = 0.1 × 22.4 = 2.24 L. Actual volume obtained = 2.18 L. Percentage yield = (actual/theoretical) × 100 = (2.18/2.24) × 100 ≈ 97.3%.

How do you identify X in the Wurtz reaction in Q22(b)?

The Wurtz reaction involves treating a haloalkane with sodium metal in dry ether, where two molecules of the haloalkane combine (with loss of two halide ions as NaX) to form a higher alkane with double the carbon count of each starting haloalkane fragment — specifically, 2 R-X + 2Na → R-R + 2NaX. To get 2,3-dimethylbutane (which has the structure (CH3)2CH-CH(CH3)2, i.e., two isopropyl groups joined together), the starting haloalkane X must be 2-chloropropane (isopropyl chloride, (CH3)2CHCl) or the corresponding bromide/iodide. The reaction: 2(CH3)2CHCl + 2Na → (CH3)2CH-CH(CH3)2 + 2NaCl. The Wurtz reaction is defined as the coupling of two alkyl halide molecules in the presence of sodium metal (in dry ether) to form a higher alkane with twice the number of carbons.

What is the peroxide effect mentioned in Q22(c)?

The peroxide effect (also called the Kharasch effect) refers to the anti-Markovnikov addition of HBr to unsymmetrical alkenes when the reaction is carried out in the presence of peroxides (like H2O2). Normally, HBr adds to alkenes following Markovnikov’s rule (H attaches to the carbon with more hydrogens already, Br to the carbon with fewer). However, in the presence of peroxides, the mechanism shifts from ionic to free radical, and bromine adds to the less substituted carbon first, reversing the usual product. Importantly, the peroxide effect is observed only with HBr — not with HCl or HI, due to the relative bond strengths and the thermodynamics of the radical chain steps involved.

What product forms from ozonolysis of 2-methyl but-1-ene in Q22(d)?

2-methylbut-1-ene has the structure CH2=C(CH3)-CH2-CH3, with the double bond between C1 and C2. Ozonolysis cleaves the C=C double bond and converts each carbon of the double bond into a carbonyl group (C=O). The C1 carbon (CH2=) becomes formaldehyde (HCHO), and the C2 carbon (=C(CH3)-CH2CH3) becomes butan-2-one (CH3-CO-CH2CH3, methyl ethyl ketone). So the products are formaldehyde and butan-2-one.

How can you distinguish ethene from ethyne using a chemical test (Q22e)?

The most direct test uses ammoniacal silver nitrate (Tollens’ reagent) or ammoniacal cuprous chloride. Ethyne (HC≡CH) has a terminal alkyne hydrogen (acidic due to the high s-character of the sp carbon), which reacts with ammoniacal AgNO3 to form a white precipitate of silver acetylide (AgC≡CAg), or with ammoniacal CuCl to form a red precipitate of copper acetylide (CuC≡CCu). Ethene (CH2=CH2), having no acidic hydrogen, gives no precipitate with these reagents. This single test cleanly distinguishes the two gases.

How to Use This Paper Effectively

This paper’s organic chemistry section (Q17, Q18, Q22) rewards being able to draw and manipulate structures, not just name reactions — when revising the Wurtz reaction, peroxide effect, ozonolysis, and aromatic substitution questions, practice drawing out the full structures and mechanisms by hand rather than just memorizing the named outcomes. NEB markers generally award more credit for a correctly drawn mechanism or structure than for a correct name alone.

The industrial chemistry questions (Q15, Q16, Q19, Q21) are often under-revised because they feel like “extra” content compared to core physical and organic chemistry — but together they account for nearly 20 marks across this paper. Make sure you can draw the flow diagrams for Haber’s process, the Solvay process (washing soda), and urea manufacture from memory, and that you understand the economic/strategic questions (Q9, Q19ii-iii) about chemical industries in developing countries, since these connect chemistry to real-world context in a way NEB increasingly favors.

Finally, the calculation-heavy questions (Q1, Q13, Q14 OR, Q20) all hinge on careful unit handling and correct molar mass calculations — practice writing out atomic masses and molecular formulas explicitly before plugging numbers into formulas, and double-check whether a question is asking for STP conditions (22.4 L/mol) versus conditions specified by a given temperature and pressure (requiring PV=nRT).