UP Board Class 12 Chemistry Question Paper 2024 (Code 347 GE) Available- Download Here with Solution PDF

Step 1: Colligative properties (such as boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure lowering) depend only on the number of solute particles, not their identity.

Step 2: These properties are a function of solute concentration, not the chemical nature of solute or solvent. Quick Tip: Colligative properties are independent of the type of solute and depend only on the concentration of solute particles.

Step 1: Let the oxidation number of Mn be \( x \) in \( KMnO_4 \).

Step 2: Since \( K = +1 \) and \( O_4 = -8 \), the equation is:
\[ 1 + x + (-8) = 0 \]
\[ x = +7 \]

Thus, the oxidation state of Mn in \( KMnO_4 \) is +7. Quick Tip: The oxidation state of Mn in \( KMnO_4 \) can be calculated using the sum rule: \( Sum of oxidation states = 0 \).

Step 1: The coordination number is determined by counting the number of ligands directly attached to the central metal atom.

Step 2: In \([Co(NH_3)_4Cl_2]Cl\), cobalt is surrounded by four \( NH_3 \) ligands and two \( Cl^- \) ligands, making a total coordination number of 6. Quick Tip: Coordination number is the number of donor atoms directly attached to the metal center.

Step 1: In Etard’s reaction, toluene is selectively oxidized to benzaldehyde using chromyl chloride (\( CrO_2Cl_2 \)).

Step 2: The reaction proceeds as:
\[ C_6H_5CH_3 + CrO_2Cl_2 \rightarrow C_6H_5CHO \] Quick Tip: Etard’s reaction is a mild oxidation method used to convert toluene to benzaldehyde.

Step 1: Benzene diazonium chloride reacts with phenol in basic medium to form p-hydroxy azobenzene, an example of azo coupling.

Step 2: The reaction is:
\[ C_6H_5N_2Cl + C_6H_5OH \xrightarrow{NaOH} C_6H_5N=N-C_6H_4OH \] Quick Tip: Azo coupling is an electrophilic substitution reaction between diazonium salts and phenols.

Step 1: Night blindness (nyctalopia) is caused due to the deficiency of Vitamin A.

Step 2: Vitamin A is essential for the production of rhodopsin, a pigment necessary for low-light vision.

Step 3: Deficiency can be treated by increasing intake of vitamin A-rich foods like carrots, fish oil, and leafy greens. Quick Tip: Vitamin A is crucial for vision and immune function. Its deficiency causes night blindness.

Step 1: Definition

- Molarity (M): Number of moles of solute per liter of solution.
\[ M = \frac{moles of solute}{volume of solution in liters} \]

- Molality (m): Number of moles of solute per kg of solvent.
\[ m = \frac{moles of solute}{mass of solvent in kg} \]

Step 2: Effect of Temperature

- Molarity depends on volume, which changes with temperature due to thermal expansion. Thus, molarity decreases with an increase in temperature.
- Molality depends on the mass of the solvent, which remains unaffected by temperature, making it temperature-independent. Quick Tip: Molarity changes with temperature due to volume expansion, whereas molality remains constant as it depends only on mass.

Step 1: Potassium permanganate (\( KMnO_4 \)) is prepared from pyrolusite (\( MnO_2 \)) via the following steps:

Step 2: Oxidation of Pyrolusite:
\[ 2 MnO_2 + 4 KOH + O_2 \rightarrow 2 K_2MnO_4 + 2 H_2O \]

Step 3: Disproportionation of Manganate:
\[ 3 K_2MnO_4 + 2 H_2O \rightarrow 2 KMnO_4 + MnO_2 + 4 KOH \]

Step 4: The structural formula of the permanganate ion (\( MnO_4^- \)):
\[ A tetrahedral structure with Mn in +7 oxidation state. \] Quick Tip: Potassium permanganate is synthesized via oxidation of manganese dioxide followed by disproportionation in acidic medium.

Step 1: Ionisation Isomerism:

- Occurs when two coordination compounds produce different ions in solution but have the same molecular formula.
- Example:
\[ [Co(NH_3)_5Br]SO_4 \quad and \quad [Co(NH_3)_5SO_4]Br \]

Step 2: Linkage Isomerism:

- Occurs when a ligand can bind through two different donor atoms.
- Example:
\[ [Co(NO_2)(NH_3)_5]^{2+} \quad (Nitro complex, N-bound) \]
\[ [Co(ONO)(NH_3)_5]^{2+} \quad (Nitrito complex, O-bound) \] Quick Tip: Ionisation isomerism occurs due to exchangeable ions, while linkage isomerism occurs when a ligand bonds through different atoms.

Step 1: Definition:

- The \( S_N1 \) (unimolecular nucleophilic substitution) reaction follows a two-step mechanism.
- It occurs in tertiary haloalkanes due to carbocation stability.

Step 2: Mechanism:

Step 2.1: Formation of Carbocation (Rate Determining Step):
\[ R-X \rightarrow R^+ + X^- \]

Step 2.2: Nucleophilic Attack:
\[ R^+ + Nu^- \rightarrow R-Nu \]

Step 3: Example:
\[ (CH_3)_3CBr \xrightarrow{H_2O} (CH_3)_3C^+ + Br^- \]
\[ (CH_3)_3C^+ + H_2O \rightarrow (CH_3)_3COH + H^+ \] Quick Tip: \( S_N1 \) reactions involve carbocation formation and follow first-order kinetics: rate \( \propto [R-X] \).

Step 1: Statement of Raoult’s Law:

Raoult’s law states that the partial vapor pressure of a solvent in a solution is directly proportional to its mole fraction.
\[ P_A = X_A P_A^0 \]

where \( P_A \) is the partial vapor pressure of the solvent, \( X_A \) is its mole fraction, and \( P_A^0 \) is the vapor pressure of the pure solvent.

Step 2: Mathematical Derivation:

For a solution containing a non-volatile solute:
\[ X_A + X_B = 1 \]
\[ X_A = 1 - X_B \]

Substituting in Raoult’s law:
\[ P_A = (1 - X_B) P_A^0 \]

Step 3: Relative Lowering of Vapor Pressure:
\[ \frac{P_A^0 - P_A}{P_A^0} = X_B \]

Thus, the relative lowering of vapor pressure is equal to the mole fraction of the solute. Quick Tip: Raoult’s law helps explain colligative properties like boiling point elevation and freezing point depression.

Step 1: Definition:

Dehydration of ethanol is an acid-catalyzed elimination reaction, producing ethene.

Step 2: Mechanism:

Step 2.1: Protonation of Ethanol:
\[ C_2H_5OH + H^+ \rightarrow C_2H_5OH_2^+ \]

Step 2.2: Formation of Carbocation:
\[ C_2H_5OH_2^+ \rightarrow C_2H_5^+ + H_2O \]

Step 2.3: Elimination of Proton:
\[ C_2H_5^+ \rightarrow C_2H_4 + H^+ \]

Step 3: Overall Reaction:
\[ C_2H_5OH \xrightarrow{H_2SO_4} C_2H_4 + H_2O \] Quick Tip: Dehydration of alcohols follows an E1 mechanism in acid medium, favoring alkene formation.

Step 1: Fehling’s Test:

- Aldehydes reduce Fehling’s solution (blue) to a red precipitate of Cu\(_2\)O.
- Ketones do not react.
\[ R-CHO + 2Cu^{2+} + 5OH^- \rightarrow R-COO^- + Cu_2O + 3H_2O \]

Step 2: Schiff’s Test:

- Aldehydes turn Schiff’s reagent pink/magenta.
- Ketones do not react.

Step 3: Conclusion:

- Aldehydes give positive tests in both reactions.
- Ketones give negative results. Quick Tip: Fehling’s test detects reducing aldehydes, while Schiff’s reagent gives a color change for aldehydes but not ketones.

Step 1: Definition of Carbohydrates:

Carbohydrates are biomolecules consisting of carbon, hydrogen, and oxygen, generally following the formula \( C_n(H_2O)_n \). They serve as an energy source and structural material.

Step 2: Differences Between Glucose and Fructose:
Quick Tip: Glucose is an aldose with a six-membered ring, while fructose is a ketose with a five-membered ring.

Step 1: Definition:
The molal elevation constant (\( K_b \)) is the increase in boiling point when 1 mole of a non-volatile solute is dissolved in 1 kg of solvent.

Step 2: Formula for Boiling Point Elevation:
\[ \Delta T_b = K_b \times m \]

where,
\[ m = \frac{moles of solute}{mass of solvent (kg)} \]

Step 3: Calculate Molality:
\[ Moles of urea = \frac{0.6}{60} = 0.01 mol \]
\[ m = \frac{0.01}{0.1} = 0.1 mol/kg \]

Step 4: Calculate Boiling Point Elevation:
\[ \Delta T_b = 0.52 \times 0.1 = 0.052 K \]

Step 5: Final Boiling Point:
\[ T_b = 373.15 + 0.052 = 373.202 K \]

Thus, the boiling point of the solution is 373.20 K. Quick Tip: The increase in boiling point is proportional to the molality of the solute, and molal elevation constant (\( K_b \)) is solvent-specific.

Step 1: Difference Between Electrochemical and Electrolytic Cells:



Step 2: Kohlrausch’s Law:

Kohlrausch’s law states:
\[ \Lambda_m^0 = \lambda^0_+ + \lambda^0_- \]

where \( \lambda^0_+ \) and \( \lambda^0_- \) are the limiting molar conductivities of cations and anions, respectively.

Step 3: Applications:

- Used to calculate the molar conductivity of weak electrolytes.
- Helps in determining dissociation constants of weak acids and bases. Quick Tip: Electrochemical cells generate electricity via redox reactions, whereas electrolytic cells drive non-spontaneous reactions using electrical energy.

Step 1: Definitions:

- Molecularity: The number of reactant molecules participating in an elementary reaction.
- Order of Reaction: The sum of powers of reactant concentrations in the rate law.

Step 2: Derivation of First-Order Rate Equation:
\[ Rate = k[A] \]

Step 3: Separating Variables:
\[ \frac{d[A]}{[A]} = -k dt \]

Step 4: Integrating Both Sides:
\[ \int \frac{d[A]}{[A]} = - \int k dt \]
\[ \ln [A] = -kt + C \]

Step 5: Applying Initial Conditions: \( [A] = [A]_0 \) at \( t = 0 \):
\[ \ln [A]_0 = C \]
\[ \ln [A] = -kt + \ln [A]_0 \]
\[ [A] = [A]_0 e^{-kt} \]

Thus, the first-order rate equation is:
\[ [A] = [A]_0 e^{-kt} \] Quick Tip: Molecularity is a theoretical concept for elementary steps, while reaction order is experimentally determined.

Step 1: Basicity and Lanthanoid Contraction:

- As we move across the lanthanide series, atomic size decreases due to lanthanoid contraction.
- \( La(OH)_3 \) has a larger ionic radius, making it more basic than \( Lu(OH)_3 \).

Step 2: Explanation:

- Larger ionic size in \( La^{3+} \) leads to weaker bond strength in \( La(OH)_3 \), making hydroxide ions (\( OH^- \)) more available.
- \( Lu^{3+} \) has a smaller radius, leading to a stronger attraction between \( Lu^{3+} \) and \( OH^- \), reducing basicity. Quick Tip: Lanthanoid contraction reduces atomic size across the series, affecting basicity trends.

Step 1: Paramagnetic Behaviour:

- Transition metals exhibit paramagnetism due to the presence of unpaired electrons in d-orbitals.
- Magnetic moment:
\[ \mu = \sqrt{n(n+2)} BM \]

where \( n \) is the number of unpaired electrons.

Step 2: Formation of Coloured Ions:

- Colour in transition metal ions arises due to d-d transitions when electrons absorb visible light and jump between d-orbitals.
- Example:
\[ [Cu(H_2O)_6]^{2+} (Blue) \] Quick Tip: The paramagnetic nature of transition elements depends on unpaired electrons, while colour formation is due to d-d transitions.

Step 1: Definition of Fuel Cells:

Fuel cells are electrochemical devices that convert chemical energy into electrical energy through a redox reaction between fuel (e.g., hydrogen) and an oxidizing agent (e.g., oxygen).

Step 2: Advantages of Hydrogen-Oxygen Fuel Cell:

1. Higher Efficiency: Fuel cells have higher efficiency compared to conventional electrochemical cells.
2. Eco-friendly: Produces only water as a byproduct, making it environmentally friendly. Quick Tip: Fuel cells are highly efficient and environmentally friendly as they produce water instead of harmful emissions.

Step 1: Cell Reaction:
\[ Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu \]

Step 2: Standard E.M.F. Calculation:
\[ E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 \]
\[ = 0.34 - (-2.37) = 2.71V \]

Step 3: Nernst Equation:
\[ E_{cell} = E_{cell}^0 - \frac{0.0591}{n} \log \frac{[Mg^{2+}]}{[Cu^{2+}]} \]
\[ E_{cell} = 2.71 - \frac{0.0591}{2} \log \frac{0.1}{0.001} \]
\[ E_{cell} = 2.71 - \frac{0.0591}{2} \times 2 \]
\[ E_{cell} = 2.71 - 0.0591 = 2.6509V \]

Thus, the e.m.f. of the cell is 2.65V. Quick Tip: The Nernst equation helps calculate the actual cell potential under non-standard conditions.

Step 1: Effect of Temperature:

- Higher temperature increases reaction rate by providing activation energy.
- Follows Arrhenius equation:
\[ k = A e^{-E_a/RT} \]

Step 2: First-Order Reaction Calculation:

Given: 20% completion in 10 minutes.
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \]
\[ k = \frac{2.303}{10} \log \frac{100}{80} \]
\[ k = \frac{2.303}{10} \times 0.0969 = 0.0223 min^{-1} \]

For 80% completion:
\[ t = \frac{2.303}{k} \log \frac{100}{20} \]
\[ t = \frac{2.303}{0.0223} \times 0.6989 \]
\[ t = 72.2 minutes \]

Thus, the time required for 80% completion is 72.2 minutes. Quick Tip: Reaction rate increases with temperature due to increased kinetic energy and collision frequency.

Step 1: Electronic Configuration:

- \( Cr^{3+} \): \( 3d^3 \), unpaired electrons → Paramagnetic
- \( Ni^{2+} \): \( 3d^8 \), paired electrons due to strong ligand effect → Diamagnetic

Step 2: Explanation:

- \( NH_3 \) is a weak field ligand, so \( Cr^{3+} \) retains unpaired electrons.
- \( CN^- \) is a strong field ligand, causing pairing in \( Ni^{2+} \). Quick Tip: Ligand strength affects electron pairing, influencing magnetic properties of complexes.

(a) \( K[Cr(H_2O)_2(C_2O_4)_2] \)
Potassium diaquadioxalatochromate(C)

(b) \( [PtCl(NO_2)(NH_3)_4]SO_4 \)
Tetraamminechloronitroplatinum(D) sulfate Quick Tip: IUPAC naming follows sequence: ligands (alphabetical) → metal with oxidation state → counter ion.

Step 1: Definition:
Nucleic acids (DNA & RNA) are biomolecules composed of nucleotides, responsible for genetic information storage.

Step 2: Structure of DNA:
- Double-helix model by Watson and Crick.
- Consists of phosphate backbone and nitrogenous bases (A, T, C, G).
- Complementary base pairing: A-T, C-G.

Step 3: Diagram:
\[ (Include diagram of DNA double-helix) \] Quick Tip: DNA follows Chargaff’s rule: A pairs with T, and C pairs with G via hydrogen bonds.

- Primary Structure: Linear sequence of amino acids in a polypeptide chain.
- Secondary Structure: Regular coiling or folding due to hydrogen bonds (e.g., α-helix, β-pleated sheet). Quick Tip: Protein structure determines its function: primary structure defines sequence, while secondary gives stability.

Step 1: Resonance Effect in Haloarenes:

Haloarenes (e.g., chlorobenzene) show resonance due to delocalization of lone pair electrons on the halogen. This results in partial double bond character, reducing reactivity towards nucleophilic substitution.
\[ C_6H_5Cl \leftrightarrow Various Resonating Structures \]

Step 2: Wurtz-Fittig Reaction:

- This reaction involves the coupling of aryl halides with alkyl halides using sodium in dry ether.
- Reaction:
\[ C_6H_5Cl + 2Na + C_2H_5Cl \xrightarrow{Dry Ether} C_6H_5C_2H_5 + 2NaCl \]

Step 3: Friedel-Crafts Reaction:

- Aromatic compounds undergo electrophilic substitution with alkyl halides in the presence of a Lewis acid catalyst (\( AlCl_3 \)).
- Example:
\[ C_6H_6 + CH_3Cl \xrightarrow{AlCl_3} C_6H_5CH_3 + HCl \] Quick Tip: Haloarenes resist nucleophilic substitution due to resonance stabilization. Wurtz-Fittig and Friedel-Crafts reactions introduce alkyl groups.

(A) Ethyl bromide reacts with sodium ethoxide?
\[ C_2H_5Br + C_2H_5ONa \rightarrow C_2H_5-O-C_2H_5 + NaBr \]

(B) Ethyl chloride reacts with alcoholic KCN?
\[ C_2H_5Cl + KCN \xrightarrow{alc.} C_2H_5CN + KCl \]

(C) Chlorobenzene reacts with nitric acid?
\[ C_6H_5Cl + HNO_3 \xrightarrow{H_2SO_4} o-C_6H_4ClNO_2 + p-C_6H_4ClNO_2 \]

(D) \( n \)-butyl chloride reacts with alcoholic KOH?
\[ C_4H_9Cl + KOH \xrightarrow{alc.} C_4H_8 + H_2O + KCl \]

(v) Ethyl bromide reacts with magnesium metal in the presence of dry ether?
\[ C_2H_5Br + Mg \xrightarrow{Dry Ether} C_2H_5MgBr \] Quick Tip: Reactions with alkyl halides depend on nucleophile strength and solvent polarity, affecting substitution vs. elimination outcomes.

(A) Preparation of ethanol from fermentation of sugar
\[ C_6H_{12}O_6 \xrightarrow{Yeast} 2C_2H_5OH + 2CO_2 \]

(B) Reimer-Tiemann Reaction
\[ C_6H_5OH + CHCl_3 + NaOH \rightarrow o-hydroxybenzaldehyde + p-hydroxybenzaldehyde \]

(C) Williamson’s Synthesis
\[ R-O^- + R'-X \rightarrow R-O-R' + X^- \] Quick Tip: Fermentation produces ethanol biologically, Reimer-Tiemann introduces -CHO into phenol, and Williamson’s synthesis forms ethers.

Primary Alcohol:
\[ R-MgX + HCHO \xrightarrow{H_2O} R-CH_2OH + Mg(OH)X \]

Secondary Alcohol:
\[ R-MgX + CH_3CHO \xrightarrow{H_2O} R-CHOH-CH_3 + Mg(OH)X \]

Tertiary Alcohol:
\[ R-MgX + R'CO-R'' \xrightarrow{H_2O} R-C(OH)-R'R'' + Mg(OH)X \] Quick Tip: Grignard reagents react with aldehydes/ketones to produce alcohols of different types.

(a) Picric acid from phenol
\[ C_6H_5OH + 3HNO_3 \xrightarrow{H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O \]

(b) Methanol from ethanol
\[ C_2H_5OH \xrightarrow{Cu, 300^\circ C} CH_3OH + CO \] Quick Tip: Nitration of phenol produces picric acid, while methanol can be synthesized by oxidation of ethanol.

Step 1: Definition:
Cannizzaro’s reaction is a redox reaction where non-enolizable aldehydes undergo disproportionation in a strong base.

Step 2: Reaction:
\[ 2 C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]

One aldehyde molecule is reduced to an alcohol, and the other is oxidized to a carboxylate.


Cannizzaro’s reaction occurs with aldehydes lacking \(\alpha\)-hydrogen atoms.
(B) Gattermann-Koch reaction

Step 1: Definition:
Gattermann-Koch reaction is a method for formylation of benzene.

Step 2: Reaction:
\[ C_6H_6 + CO + HCl \xrightarrow{AlCl_3, CuCl} C_6H_5CHO \]

The reaction introduces a formyl (-CHO) group into benzene.


The Gattermann-Koch reaction is useful for direct formylation of aromatic rings.


(C) Aldol condensation

Step 1: Definition:
Aldol condensation is an organic reaction in which an enolate ion reacts with a carbonyl compound, forming a β-hydroxy ketone.

Step 2: Reaction:
\[ CH_3CHO + CH_3CHO \xrightarrow{NaOH} CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O \] Quick Tip: Aldol condensation is crucial in carbon-carbon bond formation and occurs under basic or acidic conditions.

(A) Ethyl acetate from acetic acid
\[ CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O \]

(B) Acetone hydrazone from acetone
\[ CH_3COCH_3 + NH_2NH_2 \rightarrow CH_3C=NNH_2CH_3 + H_2O \]

(C) Acetyl chloride from acetic acid
\[ CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl \]

(D) Benzaldehyde from benzyl chloride
\[ C_6H_5CH_2Cl + H_2O \xrightarrow{MnO_2} C_6H_5CHO + HCl \]

(v) Benzoic acid from benzamide
\[ C_6H_5CONH_2 + H_2O \xrightarrow{H^+} C_6H_5COOH + NH_3 \] Quick Tip: Functional group conversions rely on specific reagents, such as oxidation for aldehydes and acid hydrolysis for amides.

(A) Gabriel phthalimide synthesis
\[ C_6H_4(CO)_2N^-K^+ + RX \rightarrow C_6H_4(CO)_2NR + KX \]

It produces primary amines selectively.

(B) Hoffmann bromamide reaction
\[ RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O \]

Converts amides to primary amines.

(C) Carbylamine reaction
\[ R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O \]

Produces isocyanides (carbylamines). Quick Tip: Gabriel synthesis selectively forms primary amines, Hoffmann’s reaction removes carbonyls, and Carbylamine reaction identifies primary amines.

Preparation of Benzene Diazonium Chloride:
\[ C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{0^\circ C} C_6H_5N_2Cl + 2H_2O \]

Conversions:

(A) Benzene
\[ C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl \]

(B) Chlorobenzene
\[ C_6H_5N_2Cl + CuCl \rightarrow C_6H_5Cl + N_2 \]

(C) Nitrobenzene
\[ C_6H_5N_2Cl + CuNO_2 \rightarrow C_6H_5NO_2 + N_2 \] Quick Tip: Diazonium salts are versatile intermediates, enabling substitution with hydrogen, halogens, and functional groups.