NEB Class 11 Science Chemistry Question Paper 2078 Nepal

Each water molecule ($H_2O$) has 3 atoms, so two molecules have $2\times3=6$ atoms.

$N_2+3H_2\rightarrow 2NH_3$. 1 mol $N_2$ gives 2 mol $NH_3$, so 0.5 mol $N_2$ gives $2\times0.5=1$ mol $NH_3$.

Ammonia dissolves readily in water because both can form hydrogen bonds (N–H···O and O–H···N), so the answer is hydrogen bonding.

$SO_2$ oxidises $H_2S$: $SO_2+2H_2S\rightarrow 3S+2H_2O$. Here $H_2S$ is oxidised to sulphur (and $SO_2$ is reduced to sulphur). The matching option is that $H_2S$ is reduced to Sulphur — note: strictly $H_2S$ is oxidised to S; as printed, the intended best option is (b).

Isomorphism is the property by which substances of similar chemical composition/formula type crystallise in the same crystal form. (Note: options (a) and (b) are both printed as 'Isotopism'/'Isotopism' — likely typos for isomerism/isotopy.)

The forward reaction is exothermic and decreases moles of gas (3 → 2). By Le Chatelier's principle the reverse reaction is favoured by high temperature (endothermic direction) and low pressure (more moles of gas). Answer: low pressure and high temperature.

Thermal stability of Group II carbonates increases down the group, so the ease of decomposition (CO₂ release) decreases down the group: $BeCO_3>MgCO_3>CaCO_3>BaCO_3$ (option c).

A batch process involves a sequence of steps carried out in a specific order on a fixed quantity of material (not continuous). Answer: 'Involves sequence of steps followed in specific order'.

The sodium-glucose cotransporter moves glucose against its gradient using the energy stored in the Na⁺ gradient (which is itself maintained by the primary Na⁺/K⁺ ATPase). It is therefore a secondary active transport protein.

In urea manufacture, $2NH_3+CO_2\rightarrow NH_2COONH_4$ (ammonium carbamate), which then dehydrates to urea. The intermediate is ammonium carbamate, $NH_2COONH_4$.

In the Solvay process, $CO_2$ (recovered from heating $NaHCO_3$ and from limestone) and $NH_3$ (recovered by treating $NH_4Cl$ with $Ca(OH)_2$) are recycled. Answer: $CO_2$ and $NH_3$.

(i) Electronic configuration 2,8,5 → total 15 electrons → atomic number 15 → element is Phosphorus (P), with 15 protons and 15 electrons.

(ii) In $X^{3-}$ (i.e. $P^{3-}$), three extra electrons are added while the nuclear charge (protons) stays the same; increased electron–electron repulsion and reduced effective nuclear charge per electron make the ion larger than the neutral atom.

(iii) A compound with P in $-3$ state: $PH_3$ (phosphine) or $Na_3P$ (sodium phosphide).

OR

(i) First ionization energy is the minimum energy required to remove the most loosely bound (outermost) electron from one mole of gaseous atoms to form one mole of gaseous unipositive ions.

(ii) Factors affecting IE: nuclear charge, atomic size/radius, screening (shielding) effect, or electronic configuration stability (any one).

(iii) Neon (Ne, 2080 kJ/mol) — highest IE, hardest to ionize.

(iv) Nitrogen has a half-filled, extra-stable $2p^3$ configuration; removing an electron from this stable configuration needs more energy, whereas carbon's $2p^2$ is less stable, so IE rises steeply from C to N.

(i) An electrolytic cell is a device in which electrical energy is used to drive a non-spontaneous chemical (redox) reaction (electrolysis).

(ii) Charge $Q=It=0.1\times(30\times60)=180\,\text{C}$. Mass $=\dfrac{E\,Q}{96500}$ where equivalent mass of Na $=23$. $m=\dfrac{23\times180}{96500}=0.0429\,\text{g}$. At 75% efficiency: $m=0.75\times0.0429=0.0322\,\text{g}\approx 0.032\,\text{g}$.

(iii) Although CaCl₂ is added (to lower the melting point), sodium is preferentially discharged because, under the molten-salt conditions and concentrations used (Downs cell), Na⁺ is reduced more readily; Ca²⁺ requires higher reduction potential/energy, so Ca is not deposited.

(iv) In aqueous NaCl, water is more easily reduced than Na⁺, so hydrogen gas (not sodium metal) is liberated at the cathode; hence aqueous solution cannot give sodium metal.

$K_p$ and $K_c$: For a gaseous equilibrium $aA+bB \rightleftharpoons cC+dD$, using $PV=nRT$ so $P=\dfrac{n}{V}RT=[\,]RT$ (partial pressure = concentration × RT). Substituting into $K_p=\dfrac{P_C^c P_D^d}{P_A^a P_B^b}$ gives $K_p=K_c(RT)^{\Delta n}$ where $\Delta n=(c+d)-(a+b)$.

Example with $K_p>K_c$ ($\Delta n>0$): $N_2O_4(g)\rightleftharpoons 2NO_2(g)$, $\Delta n=+1$, so $K_p=K_c(RT)>K_c$ (since $RT>1$ at usual T).

OR — Ideal gas equation: Combine Boyle's law ($V\propto 1/P$), Charles's law ($V\propto T$) and Avogadro's law ($V\propto n$): $V\propto \dfrac{nT}{P}$, so $V=\dfrac{nRT}{P}$, i.e. $PV=nRT$ (R = universal gas constant). Conditions for ideal behaviour: high temperature and low pressure (so molecular volume and intermolecular forces become negligible).

(i) Concentrated $H_2SO_4$ is a strong oxidising agent; it oxidises the HI formed to iodine ($2HI+H_2SO_4\rightarrow I_2+SO_2+2H_2O$), so pure HI cannot be obtained this way (a non-oxidising acid like phosphoric acid is used instead).

(ii) Hydrogen chloride gas (HCl gas) is the pure covalent gaseous compound and does not conduct electricity or turn litmus red when dry; hydrochloric acid is the aqueous solution of HCl gas in water, which ionises ($HCl\rightarrow H^++Cl^-$), is acidic, conducts electricity and turns blue litmus red.

(iii) Bring a glass rod dipped in ammonia (or open a bottle of ammonia) near the bottle: HCl gas reacts with $NH_3$ to give dense white fumes of ammonium chloride ($NH_3+HCl\rightarrow NH_4Cl$), confirming HCl.

(i) Roasting is heating a (usually sulphide) ore strongly in a regular supply of air below its melting point to convert it into oxide and drive off volatile impurities. Calcination is heating a (usually carbonate or hydrated) ore strongly in the absence or limited supply of air to convert it into oxide and remove moisture/CO₂/volatile matter.

(ii) Roasting is carried out in a reverberatory furnace (or a roaster).

(iii) Impurities removed during roasting: sulphur (as $SO_2$), arsenic (as $As_2O_3$), antimony, phosphorus, and moisture (any two — e.g. sulphur and arsenic).

(i) A homologous series is a family of organic compounds having the same general formula, similar chemical properties and the same functional group, in which successive members differ by a $-CH_2-$ unit (14 mass units) and show a gradual gradation in physical properties.

(ii) Successive members differ by $CH_2 = 14$. X is the third member (propan-1-ol, $CH_3CH_2CH_2OH$, $C_3H_8O$). Mass of X: methanol = 32, ethanol = 46, so X = $46+14=60$, and butan-1-ol = 74 (consistent, $60+14=74$). Molecular mass of X = 60.

(iii) The fourth member (butan-1-ol) has the highest molecular mass, so it has the strongest van der Waals (and similar H-bonding) forces among the small members → highest boiling point. Its solubility is least because the larger non-polar hydrocarbon chain dominates, reducing the proportion that can hydrogen-bond with water → least soluble.

D is $C_6H_{14}$ formed by Wurtz reaction (Na/ether) coupling two C of equal carbons → C is a 3-carbon halide ($C_3H_7Br$). So B = propene.

(i) $CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3$ (Markovnikov) [B → C]. $2CH_3CHBrCH_3 + 2Na \xrightarrow{ether} CH_3CH(CH_3)CH(CH_3)CH_3 + 2NaBr$... Wurtz on 2-bromopropane gives 2,3-dimethylbutane ($C_6H_{14}$) [C → D]. (If C is 1-bromopropane, D is n-hexane.)

(ii) C = 2-bromopropane (propan-2-yl bromide); D = 2,3-dimethylbutane (or n-hexane if C is 1-bromopropane).

(iii) Positional isomer of C ($C_3H_7Br$): 1-bromopropane $CH_3CH_2CH_2Br$.

(i) Flow-sheet (described): purified $N_2$ (from air) and $H_2$ (from natural gas/water gas) in 1:3 ratio are compressed to ~200 atm, passed over a finely divided iron catalyst (with $Mo$/promoters) at ~450–500 °C in the catalytic chamber; $N_2+3H_2\rightleftharpoons 2NH_3$. The hot gas mixture is cooled/condensed to liquefy ammonia, which is removed; unreacted $N_2$ and $H_2$ are recycled back to the compressor.

(ii) Major challenge: lack of capital/investment and energy/raw-material supply, plus inadequate infrastructure and skilled manpower (and landlocked transport costs). Strategy to overcome: government incentives and investment-friendly policy, attracting foreign direct investment, developing hydropower/energy and infrastructure, technology transfer and human-resource training, and public-private partnership.

(i) Ratio $NH_3:O_2=4:5$. For 2 mol $NH_3$ we need $2\times5/4=2.5$ mol $O_2$, but only 2 mol $O_2$ available → $O_2$ is limiting. From 2 mol $O_2$: $NO=\dfrac{4}{5}\times2=1.6$ mol. Mass NO $=1.6\times30=48\,\text{g}$.

(ii) The limiting reactant is the reactant completely consumed first; it determines (limits) the maximum amount of product and the extent of reaction — all yield calculations are based on it.

(iii) From $4NH_3\rightarrow6H_2O$: 2 mol $NH_3$ should give $\dfrac{6}{4}\times2=3$ mol $H_2O=3\times18=54\,\text{g}$ (theoretical). Actual = 50 g. Percentage yield $=\dfrac{50}{54}\times100=92.6\%$.

(iv) $NH_3+HCl\rightarrow NH_4Cl$ (1:1), so 2 mol $NH_3$ needs 2 mol HCl. Using $PV=nRT$: $V=\dfrac{nRT}{P}$. $P=750/760=0.987\,\text{atm}$, $T=293\,K$, $R=0.0821$. $V=\dfrac{2\times0.0821\times293}{0.987}=\dfrac{48.11}{0.987}\approx 48.7\,\text{L}$.

(i) Acidic oxide: $SO_2$; Basic oxide: $Na_2O$; Neutral oxide: $CO$; Mixed oxide: $Fe_3O_4$ (= $FeO\cdot Fe_2O_3$). ($Al_2O_3$ is amphoteric; $H_2O_2$ is a peroxide.)

(ii) $Al_2O_3$ is amphoteric: with acid $Al_2O_3+6HCl\rightarrow 2AlCl_3+3H_2O$; with base $Al_2O_3+2NaOH\rightarrow 2NaAlO_2+H_2O$.

(iii) CO is harmful because it binds haemoglobin (forming carboxyhaemoglobin) about 200–300× more strongly than oxygen, preventing oxygen transport in blood and causing suffocation/poisoning.

(iv) Oxygen is used industrially in steel making (Bessemer/oxygen converters), oxy-acetylene welding/cutting, or as an oxidiser/rocket propellant (any one).

OR — Contact process from FeS₂: (1) $4FeS_2+11O_2\rightarrow 2Fe_2O_3+8SO_2$; (2) $2SO_2+O_2 \xrightarrow{V_2O_5} 2SO_3$; (3) $SO_3+H_2SO_4\rightarrow H_2S_2O_7$ (oleum); (4) $H_2S_2O_7+H_2O\rightarrow 2H_2SO_4$. Precipitant: $BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl$. Dehydrating agent: $C_{12}H_{22}O_{11}\xrightarrow{H_2SO_4}12C+11H_2O$. Fertilizer: $2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4$. Oxidising agent: hot conc. $H_2SO_4$ readily gives up oxygen ($H_2SO_4\rightarrow H_2O+SO_2+[O]$), so it acts as an oxidising agent.

Ozonolysis gives two $C_3H_6O$ fragments that are functional isomers (one aldehyde, one ketone): propanal $CH_3CH_2CHO$ and propanone (acetone) $CH_3COCH_3$. The alkene X is therefore $CH_3CH_2-CH=C(CH_3)-CH_3$? — to give propanal + acetone, X = 2-methyl-2-pentene: $CH_3CH_2CH=C(CH_3)_2$.

(i) $X + O_3 \rightarrow$ ozonide; ozonide $+ H_2O/Zn \rightarrow CH_3CH_2CHO + CH_3COCH_3$.

(ii) Y = propanal $CH_3CH_2CHO$ (aldehyde); Z = propanone $CH_3COCH_3$ (ketone). They have the same molecular formula $C_3H_6O$ but different functional groups (–CHO vs >C=O), hence functional isomers.

(iii) $H_2/Ni$ adds across the double bond (hydrogenation), giving the saturated alkane 2-methylpentane.

(iv) Ozonolysis is used to locate the position of the carbon–carbon double bond in an alkene (the carbonyl fragments reveal where the C=C was).

(v) X decolourises bromine water (or alkaline KMnO₄ — Baeyer's reagent), proving unsaturation (presence of C=C).