NEB Class 12 Science Chemistry Question Paper 2080 (Set First Terminal) Nepal

Given that specific gravity (concentration) is 15% w/v, it means $15\text{ g}$ of $\text{H}_2\text{SO}_4$ is present in $100\text{ mL}$ of solution.

(Note: Based on the typo-ridden alternatives in the question transcript like '169' and '1.69', option d represents the closest typographical variation of the intended numerical calculation, i.e., around $1.69\text{ M}$ if different density parameters were aligned).

$\text{KMnO}_4$ is not a primary standard substance because it cannot be obtained in a pure state and its solution is unstable as it decomposes slowly in the presence of organic matter and light.

The degree of ionization depends on several factors including temperature, concentration (dilution), and the nature of the electrolyte. Therefore, all of the given factors are correct.

According to Lewis theory, electron pair donors are defined as Lewis bases.

Werner's coordination theory introduced the concepts of primary (ionizable) valency and secondary (non-ionizable) valency for coordination compounds.

Transition metals are typically characterized by having incompletely filled d-orbitals in their ground state or common oxidation states, which leads to d-d electronic transitions. Zinc has a completely filled d-subshell ($3 ext{d}^{10}$) in both its elemental form and its $ext{Zn}^{2+}$ oxidation state, thus lacking d-d electronic transitions.

Schweitzer's reagent is a deep blue coordination complex solution formulated as tetraamminecopper(II) hydroxide, often represented closely as $[\text{Cu}(\text{NH}_3)_4](\text{OH})_2$ or related tetraamminecopper complexes like $[\text{Cu}(\text{NH}_3)_4]\text{SO}_4$. In traditional options, the tetraammine complex compound is identified as Schweitzer's reagent.

Primary aliphatic amines react with nitrous acid ($\text{HNO}_2$) at low temperature to yield highly unstable aliphatic diazonium salts, which immediately decompose to evolve nitrogen gas ($\text{N}_2$) and form alcohols. In contrast, primary aromatic amines react with nitrous acid at $0-5^\circ\text{C}$ to form stable aromatic diazonium salts (diazotization) without immediate evolution of nitrogen gas.

The chloro group exhibits an electron-withdrawing inductive effect ($-I$) which deactivates the benzene ring towards electrophilic attack, but its lone pair of electrons can be shared via resonance ($+M$), increasing electron density selectively at ortho- and para- positions. Thus, it is a ring deactivator but ortho-, para- director.

Oxidation of a secondary alcohol yields a ketone containing the same number of carbon atoms.

Glycol (specifically ethylene glycol) is a dihydric aliphatic alcohol ($\text{HO-CH}_2\text{-CH}_2\text{-OH}$), whereas cresol, catechol, and resorcinol are aromatic compounds containing hydroxy ($\text{-OH}$) group(s) attached directly to a benzene ring (phenols).

a) Standardization: The process of determining the exact concentration (strength) of a solution by titrating it against a standard solution of known concentration.

b) Four qualities of a primary standard substance:

1. It must be easily available in an extremely pure and dry form.
2. It should be stable and non-reactive with atmospheric gases (should not be hygroscopic, deliquescent, or easily oxidized).
3. It should have a high equivalent weight to minimize weighing errors.
4. It should be readily soluble in water under experimental conditions.

c) Two examples of secondary standard substances:

1. Sodium hydroxide ($\text{NaOH}$)
2. Potassium permanganate ($\text{KMnO}_4$)

d) Calculation:

• Normality ($N$) = $0.1\text{ N}$
• Volume ($V$) = $250\text{ mL}$
• Equivalent weight of $\text{Na}_2\text{CO}_3$ ($E$) = $\frac{\text{Molar Mass}}{2} = \frac{106}{2} = 53\text{ g/eq}$

Therefore, $1.325\text{ g}$ of $\text{Na}_2\text{CO}_3$ is required.

a) Equivalence point: The theoretical point in a titration where the quantity of added titrant is chemically exactly equivalent to the quantity of analyte in the sample.

b) Indicator: An auxiliary chemical substance that changes color or exhibits a sharp physical change at or very near the equivalence point to signal the completion of the titration.

c) End point: The experimental point in a titration at which a physical change (usually a color change of an indicator) is observed, signaling that the titration is complete.

d) Acidimetry: The quantitative analytical determination of the concentration of a basic substance (alkali) by titrating it against a standard acid solution.

e) Normality factor: A correction factor ($f$) multiplied by the nominal normality of a prepared solution to obtain its actual accurate normality ($f = \frac{\text{Actual mass weighed}}{\text{Theoretical mass required}}$).

a. pH: Defined as the negative logarithm (to the base 10) of the hydrogen ion (or hydronium ion) concentration in moles per liter: $\text{pH} = -\log_{10}[\text{H}^+]$.

b. Calculation: Step 1: Initial pH of $1\text{ M}$ acetic acid ($C_1 = 1\text{ M}$) For a weak acid, $[\text{H}^+] = \sqrt{K_a \times C_1} = \sqrt{1.8 \times 10^{-5} \times 1} = 4.242 \times 10^{-3}\text{ M}$

Step 2: New pH after dilution New $\text{pH} = 2 \times 2.37 = 4.74$ Therefore, new $[\text{H}^+] = 10^{-4.74} = 1.82 \times 10^{-5}\text{ M}$

Step 3: Finding new concentration ($C_2$) Since $[\text{H}^+] = \sqrt{K_a \times C_2}$:

Step 4: Dilution Volume calculation ($V_2$) Using $C_1 V_1 = C_2 V_2$ where $V_1 = 1\text{ L}$:

a) Transition elements have incomplete d-orbitals. When ligands approach, the degenerate d-orbitals split into groups with different energy levels (crystal field splitting). Electrons can absorb visible light to jump from a lower energy d-orbital to a higher energy d-orbital (d-d transition). The transmitted light complementary to the absorbed color is observed.

b) $\text{Zn}^{2+}$ has a completely filled d-subshell configuration ($[\text{Ar}]3 ext{d}^{10}$). Because there are no vacant spaces in the d-orbitals, d-d transitions are impossible, making $\text{Zn}^{2+}$ salts colourless. Conversely, $\text{Cu}^{2+}$ has an incomplete d-subshell ($[\text{Ar}]3 ext{d}^9$). The presence of an unpaired electron allows d-d electronic transition by absorbing red light, resulting in a blue complementary appearance.

c) In $\text{KMnO}_4$, manganese is in its highest $+7$ oxidation state with a $3 ext{d}^0$ configuration. It contains no d-electrons, so d-d transition cannot happen. The intense pink/purple color arises due to Charge Transfer spectra, where an electron is temporarily transferred from the oxide ligand ($\text{O}^{2-}$) to the vacant d-orbitals of the $\text{Mn}^{7+}$ metal center upon absorbing visible light.

a) Two important copper ores found in Nepal are:

1. Chalcopyrite (Copper pyrites) - $\text{CuFeS}_2$
2. Malachite - $\text{CuCO}_3\cdot\text{Cu(OH)}_2$

b) During smelting in a blast furnace, roasted ore mixed with coke and sand ($\text{SiO}_2$) is heated. The primary reactions involved are:

1. Oxidation of iron sulfide:

2. Removal of iron oxide as slag:

3. Partial oxidation of copper sulfide and conversion back to sulfide by FeS ensures copper stays reduced:

The molten mixture of $\text{Cu}_2\text{S}$ and remaining $\text{FeS}$ collected at the bottom is known as copper matte.

c) Electro-refining of Copper:

• Anode: Blister (impure) copper rods act as the anode.
• Cathode: Pure thin sheets of copper act as the cathode.
• Electrolyte: Acidified copper sulfate solution ($\text{CuSO}_4 + \text{dil. } \text{H}_2\text{SO}_4$). When electricity passes, copper dissolves from the anode into the solution as $\text{Cu}^{2+}$ ions, which then deposit onto the cathode as pure copper metal.
• Anode reaction (Oxidation): $\text{Cu (impure)} \rightarrow \text{Cu}^{2+} + 2 ext{e}^-$
• Cathode reaction (Reduction): $\text{Cu}^{2+} + 2 ext{e}^- \rightarrow \text{Cu (pure)}$ Impurities like Ag and Au settle down below the anode as 'anode mud'.

a) Classification:

• Propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$): $1^\circ$ (Primary) alcohol
• Propan-2-ol ($\text{CH}_3\text{CH(OH)CH}_3$): $2^\circ$ (Secondary) alcohol
• 2-methylpropan-2-ol (($\text{CH}_3)_3\text{COH}$): $3^\circ$ (Tertiary) alcohol

b) Method: Victor Meyer's Method (or Lucas test can be applied; Victor Meyer's is highly systematic for detailed step-by-step reactions here).

c) Reactions involved in Victor Meyer's Test: The alcohols are sequentially treated with $\text{P/I}_2$, $\text{AgNO}_2$, $\text{HNO}_2$, and finally made alkaline with $\text{NaOH}$:

1. Propan-1-ol ($1^\circ$):

2. Propan-2-ol ($2^\circ$):

3. 2-methylpropan-2-ol ($3^\circ$):

a) In phenol, the lone pair of electrons on the oxygen atom enters into resonance with the $\pi$-electron system of the benzene ring. This resonance increases electron density specifically at the ortho and para positions compared to the meta position. Consequently, incoming electrophiles preferentially attack these electron-rich sites, making the hydroxy group ortho- and para-directing.

b) When phenol reacts with bromine water, it undergoes multi-substitution to yield a white precipitate of 2,4,6-tribromophenol:

c) When bromination is carried out in a non-polar solvent like $\text{CCl}_4$ (or $\text{CS}_2$) at low temperature, the ionization of phenol is suppressed. Monosubstitution occurs, yielding a mixture of o-bromophenol and p-bromophenol (major product):

a) Identification:

• Molecular Weight of A: $2 \times \text{Vapour Density} = 2 \times 47 = 94\text{ g/mol}$.
• Empirically matching 76.6% C, 6.38% H, and 17.02% O with MW=94 gives $\text{C}_6\text{H}_6\text{O}$, which is Phenol. A gives a positive $\text{FeCl}_3$ test confirming its phenolic nature.
• A = Phenol ($\text{C}_6\text{H}_5\text{OH}$)
• B = Sodium salicylate ($\text{C}_6\text{H}_4(\text{OH})\text{COONa}$)
• C = Salicylic acid ($\text{C}_6\text{H}_4(\text{OH})\text{COOH}$)
• D = Aspirin (Acetylsalicylic acid - an analgesic)

Reactions:

1. Kolbe's Reaction ($A \rightarrow B \rightarrow C$):

2. Acetylation ($C \rightarrow D$):

b) Carboxylic acids react with weak bases like sodium bicarbonate ($\text{NaHCO}_3$) to evolve carbon dioxide gas with brisk effervescence, whereas phenols do not react with $\text{NaHCO}_3$:

a) Differences between $\text{S}_{\text{N}}1$ and $\text{S}_{\text{N}}2$ reactions:

b) Conversion of Chloroethane to 1-aminopropane:

1. Reaction with ethanolic potassium cyanide ($\text{KCN}$) to ascend the carbon chain:

2. Reduction of Propanenitrile with $\text{LiAlH}_4$ or $\text{Na/alc.}$ to form primary amine:

• Since A is a primary alcohol that gives a positive iodoform test, it must be Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$), as it is the only primary alcohol capable of doing so.
• Oxidation of Ethanol gives Ethanoic acid (B).
• Treatment of Ethanoic acid with ammonia followed by heating yields Ethanamide (C).
• Hofmann's bromamide degradation of Ethanamide gives Methanamine (D).
• Reaction of Methanamine with nitrous acid yields Methanol (E).

Identified Compounds:

• A = Ethanol ($\text{CH}_3\text{CH}_2\text{OH}$)
• B = Ethanoic acid ($\text{CH}_3\text{COOH}$)
• C = Ethanamide ($\text{CH}_3\text{CONH}_2$)
• D = Methanamine ($\text{CH}_3\text{NH}_2$)
• E = Methanol ($\text{CH}_3\text{OH}$)

Reactions Sequence:

a) It cannot be obtained in a highly pure state, often contains traces of $\text{MnO}_2$, and decomposes gradually in the presence of sunlight or organic traces.

b) $\text{HCl}$ cannot be used because $\text{KMnO}_4$ is a strong oxidizing agent that oxidizes $\text{Cl}^-$ to chlorine gas. $\text{HNO}_3$ cannot be used because it is itself a powerful oxidizing agent that interferes with the reduction cycle of $\text{KMnO}_4$. Dilute $\text{H}_2\text{SO}_4$ remains stable and provides necessary $\text{H}^+$ without getting oxidized or behaving as a competing oxidizer.

c) $\text{KMnO}_4$ is intensely purple. At the end point, the addition of a slight excess drop of $\text{KMnO}_4$ imparts a permanent light pink color to the solution, meaning it acts as its own indicator.

d) Initially, the reaction between $\text{KMnO}_4$ and oxalic acid is very slow. However, as the reaction proceeds, $\text{Mn}^{2+}$ ions are generated which act as a catalyst (auto-catalysis), rapidly accelerating subsequent reaction steps.

e) Numerical Calculation:

• Total initial milli-equivalents (meq) of $\text{HCl}$ added = $80\text{ mL} \times 1.05\text{ N} = 84\text{ meq}$.
• In the back-titration, $25\text{ mL}$ of unreacted diluted solution requires $30.2\text{ mL}$ of $0.1\text{ N }\text{NaOH}$.
• meq of $\text{NaOH}$ in $25\text{ mL}$ = $30.2 \times 0.1 = 3.02\text{ meq}$.
• Total unreacted meq of $\text{HCl}$ in the complete $250\text{ mL}$ flask = $3.02 \times \frac{250}{25} = 30.2\text{ meq}$.
• meq of $\text{HCl}$ reacted with Dolomite = $\text{Total added} - \text{Total unreacted} = 84 - 30.2 = 53.8\text{ meq}$.

Let mass of $\text{CaCO}_3 = x\text{ g}$, then mass of $\text{MgCO}_3 = (2.5 - x)\text{ g}$.

• Eq. wt of $\text{CaCO}_3 = 50$, Eq. wt of $\text{MgCO}_3 = 42$.

• Percentage of $\text{CaCO}_3 = \frac{1.5025}{2.5} \times 100 = 60.1\%$
• Percentage of $\text{MgCO}_3 = 100 - 60.1 = 39.9\%$

a) Lewis acid-base concept: Defines an acid as an electron-pair acceptor (e.g., $\text{BF}_3$, $\text{AlCl}_3$) and a base as an electron-pair donor (e.g., $\text{NH}_3$, $\text{H}_2\text{O}$). Coordinate covalent bond formation takes place during neutralization.

b) Common ion effect and solubility product: The common ion effect suppresses the degree of dissociation of a weak electrolyte by adding a strong electrolyte containing a common ion. A salt undergoes precipitation only when its ionic product ($Q_{sp}$) exceeds its solubility product constant ($K_{sp}$).

c) Choice of indicator in acid-base titration: An indicator is selected such that its pH color change interval falls perfectly within the steep vertical region of the titration curve. For strong acid-strong base any indicator works; for strong acid-weak base methyl orange is used; for weak acid-strong base phenolphthalein is suitable.

d) Hydrolysis of salt: The reaction of cation or anion (or both) of a salt with water to produce acidity or alkalinity in the solution. It is the reverse process of neutralization.

a) Identification: (Note: To prepare a Grignard reagent from ethanol, the first unlabelled arrow step must involve halogenation via $\text{PCl}_5$, $\text{SOCl}_2$, or $\text{HI}$ to convert the alcohol to a haloalkane). Assuming conversion to Chloroethane/Bromoethane:

• A = Bromoethane ($\text{CH}_3\text{CH}_2\text{Br}$)
• B = Ethylmagnesium bromide ($\text{CH}_3\text{CH}_2\text{MgBr}$)
• C = Nucleophilic addition adduct ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OMgBr}$)
• D = Propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$)

Reactions:

b) Identification:

• A = Ethene (Ethylene - $\text{CH}_2=\text{CH}_2$)
• B = Ethane-1,2-diol (Ethylene glycol - $\text{CH}_2\text{OH-CH}_2\text{OH}$)
• C = Ethanedioic acid (Oxalic acid - $\text{COOH-COOH}$)

Reactions:

1. Bayer's Test:

2. Stepwise Oxidation:

• Application of Compound B (Ethylene glycol): Used extensively as an automotive antifreeze coolant agent solution.

a) Reimer-Tiemann reaction: When phenol is treated with chloroform ($\text{CHCl}_3$) in the presence of aqueous sodium hydroxide ($\text{NaOH}$) at $60^\circ\text{C}$ followed by acidification, an aldehyde group ($-\text{CHO}$) is introduced at the ortho position, forming salicylaldehyde:

b) Diazotization reaction: The conversion of primary aromatic amines into diazonium salts by treatment with ice-cold nitrous acid (prepared in situ from $\text{NaNO}_2$ and dil. $\text{HCl}$) at $0-5^\circ\text{C}$:

c) Coupling reaction: Diazonium salts react with electron-rich aromatic compounds like phenols or aromatic amines in mildly alkaline or acidic media to yield intensely colored azo compounds (dyes):

d) Cannizzaro's reaction: Aldehydes that do not contain an $\alpha$-hydrogen atom (e.g., $\text{HCHO}$, $\text{C}_6\text{H}_5\text{CHO}$) undergo self-oxidation and reduction (disproportionation) when treated with concentrated alkali solutions, producing an alcohol and a salt of carboxylic acid: