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

Step 1: Definition of Colligative Properties:
Colligative properties depend on the number of solute particles in a solution and include:

- Osmotic Pressure
- Boiling Point Elevation
- Freezing Point Depression
- Vapor Pressure Lowering

Step 2: Why not other options?
- Surface tension and density depend on the nature of the substance, not the number of solute particles. Quick Tip: Colligative properties depend only on the number of solute particles, not their identity.

Step 1: Calculate oxidation number:
\[ Let oxidation number of Mn be x \]
\[ K_2MnO_4: 2(+1) + x + 4(-2) = 0 \]
\[ 2 - 8 + x = 0 \Rightarrow x = +6 \]

Thus, the oxidation number of Mn in \( K_2MnO_4 \) is +6. Quick Tip: Oxidation number is calculated by balancing charge contributions from all atoms in a compound.

Step 1: Definition of Ambidentate Ligand:
Ambidentate ligands can coordinate through two different donor atoms.

Step 2: Why is \( NO_2^- \) ambidentate?
- It can bind via Nitrogen (\( NO_2^- \) as Nitro) or Oxygen (\( ONO^- \) as Nitrito). Quick Tip: Ambidentate ligands can bond through different atoms, providing isomerism in complexes.

Step 1: Reaction Mechanism:
Rosenmund’s reduction involves catalytic hydrogenation of acyl chlorides to aldehydes.
\[ RCOCl + H_2 \xrightarrow{Pd/BaSO_4} RCHO + HCl \]

Step 2: Why aldehyde?
- The catalyst palladium on barium sulfate prevents over-reduction to alcohol. Quick Tip: Rosenmund’s reduction selectively converts acyl chlorides to aldehydes without over-reduction.

Step 1: Carbylamine Reaction:
Primary amines react with chloroform and alcoholic KOH to give foul-smelling isocyanides.
\[ R-NH_2 + CHCl_3 + 3KOH \rightarrow R-NC + 3KCl + 3H_2O \]

Step 2: Why not secondary or tertiary amines?
- Secondary and tertiary amines lack the necessary NH\(_2\) group. Quick Tip: The isocyanide test distinguishes primary amines due to their ability to form carbylamines.

Step 1: Definition:
Nucleotides in nucleic acids (DNA/RNA) are linked by phosphodiester bonds between sugar and phosphate groups.
\[ DNA Backbone: -Sugar-Phosphate-Sugar- \]

Step 2: Why not other options?
- Peptide bond: Found in proteins.
- Glycosidic bond: Links sugars in carbohydrates.
- Hydrogen bond: Stabilizes base pairing in DNA. Quick Tip: Phosphodiester bonds form the backbone of DNA, linking nucleotides through phosphate groups.

Step 1: Formula for Molality:
\[ Molality (m) = \frac{Moles of solute}{Mass of solvent (kg)} \]

Step 2: Calculate Moles of Glucose:
\[ Moles = \frac{2.82}{180} = 0.01567 mol \]

Step 3: Convert Solvent Mass to kg:
\[ 30 g = 0.030 kg \]

Step 4: Calculate Molality:
\[ m = \frac{0.01567}{0.030} = 0.522 mol/kg \]

Thus, the molality of the solution is 0.522 mol/kg. Quick Tip: Molality is independent of temperature because it is based on mass, unlike molarity which depends on volume.

Step 1: Preparation Steps:
\[ 4 FeCr_2O_4 + 8 Na_2CO_3 + 7 O_2 \rightarrow 8 Na_2CrO_4 + 2 Fe_2O_3 + 8 CO_2 \]
\[ 2 Na_2CrO_4 + H_2SO_4 \rightarrow Na_2Cr_2O_7 + Na_2SO_4 + H_2O \]

Step 2: Structural Formula of Dichromate Ion:
\[ [Cr_2O_7]^{2-} \] Quick Tip: Potassium dichromate is obtained by oxidation of chromite ore in an alkaline medium followed by acidification.

Step 1: Main Postulates of Werner’s Theory:

1. Metals exhibit primary valency (oxidation state) and secondary valency (coordination number).
2. Secondary valency is directed towards specific ligands in a fixed geometry.
3. Example: In \([Co(NH_3)_6]Cl_3\), Co has:
- Primary valency: +3 (satisfied by Cl\(^-\))
- Secondary valency: 6 (satisfied by NH\(_3\)) Quick Tip: Werner’s theory explains the geometry of coordination complexes, distinguishing between oxidation state and coordination number.

Step 1: Definition of \( S_N2 \) Reaction:

The \( S_N2 \) reaction is a one-step nucleophilic substitution mechanism where the nucleophile attacks while the leaving group departs simultaneously.

Step 2: Mechanism:
\[ R-Br + OH^- \rightarrow R-OH + Br^- \]

- Rate Law: \( Rate = k [R-X] [Nu^-] \)
- Stereochemistry: Inversion of configuration (Walden Inversion). Quick Tip: \( S_N2 \) reactions are favored by primary alkyl halides and strong nucleophiles.

Step 1: Definition:
Osmotic pressure is the pressure required to stop the osmotic flow of solvent molecules across a semipermeable membrane.
\[ \pi = CRT \]

Step 2: Difference Between Osmosis and Diffusion:




Quick Tip: Osmosis occurs in biological systems, while diffusion applies to all states of matter.

Step 1: Resonance Stabilization:
\[ C_6H_5OH \rightleftharpoons C_6H_5O^- + H^+ \]

Phenol ionizes to release H\(^+\), stabilized by resonance.

Step 2: Stronger Acid than Alcohols:
- Phenol (\( pK_a = 9.95 \)) is more acidic than alcohols because of resonance stabilization. Quick Tip: Phenols are weak acids due to delocalization of the negative charge on the oxygen in the conjugate base.

Step 1: Definition:
Tollen’s reagent is an ammoniacal silver nitrate solution that oxidizes aldehydes to carboxylates, forming a silver mirror.

Step 2: Reaction:
\[ R-CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow R-COO^- + 2Ag + 4NH_3 + H_2O \] Quick Tip: Tollen’s test is used to distinguish aldehydes from ketones as only aldehydes undergo oxidation.

Denaturation disrupts the secondary and tertiary structures of proteins due to heat, acids, or chemicals, causing loss of function.
\[ Example: Cooking eggs denatures ovalbumin in egg whites. \] Quick Tip: Denatured proteins lose biological activity, but their primary structure remains intact.

Step 1: Formula for Freezing Point Depression:
\[ \Delta T_f = i \cdot K_f \cdot m \]

Step 2: Calculate Moles of Ethylene Glycol:
\[ Moles = \frac{25}{62} = 0.403 mol \]

Step 3: Convert Solvent Mass to kg:
\[ 300 g = 0.300 kg \]

Step 4: Calculate Molality:
\[ m = \frac{0.403}{0.300} = 1.343 mol/kg \]

Step 5: Calculate Depression in Freezing Point:
\[ \Delta T_f = 1.86 \times 1.343 = 2.50 K \]

Step 6: Calculate Freezing Point of Solution:
\[ T_f = 273.15 - 2.50 = 270.65 K \]

Thus, the freezing point of the solution is 270.65 K. Quick Tip: Molal depression constant (\(K_f\)) is a colligative property, depending only on the number of solute particles.

Step 1: Kohlrausch’s Law:
The molar conductivity of an electrolyte at infinite dilution is the sum of the conductivities of its individual ions.
\[ \Lambda_m^\circ = \lambda^+ + \lambda^- \]

Step 2: Applications:

1. Determination of Limiting Molar Conductivity of weak electrolytes.
2. Calculation of Degree of Dissociation (\(\alpha\)).
3. Determination of Solubility of Sparingly Soluble Salts. Quick Tip: Kohlrausch’s law helps determine ionic conductivities and solubilities of weak electrolytes.

Step 1: Difference Between Molecularity and Order:








Step 2: First-Order Reaction Derivation:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]

For three-fourth completion:
\[ t_{3/4} = \frac{2.303}{k} \log \frac{4}{1} = 2 t_{1/2} \]

Thus, the time taken for 75% completion is twice the half-life. Quick Tip: Molecularity is theoretical while order is experimentally determined. First-order reactions follow exponential decay.

Step 1: Definition:
Lanthanide contraction refers to the gradual decrease in atomic and ionic radii of lanthanides across the series due to poor shielding by f-electrons.

Step 2: Reason:
- Poor shielding increases effective nuclear charge, pulling electrons inward. Quick Tip: Lanthanide contraction affects the properties of transition elements, reducing size variation.

Step 1: Electronic Configurations:

- Zn\(^{2+}\) = \(3d^{10}\) (fully filled, no d-d transitions)
- Cu\(^{2+}\) = \(3d^9\) (one unpaired electron, d-d transitions possible)

Step 2: Reason for Colour:
- Zn\(^{2+}\) has no unpaired d-electrons → No d-d transitions → Appears white.
- Cu\(^{2+}\) has d-d transitions absorbing red-orange light, reflecting blue. Quick Tip: Transition metal colour is due to d-d transitions; fully filled orbitals prevent absorption of visible light.

Step 1: Introduction to Lead Storage Battery
- The lead storage battery is a secondary battery that can be recharged.
- It consists of lead (Pb) as the anode and lead dioxide (PbO\(_2\)) as the cathode, immersed in sulfuric acid (H\(_2\)SO\(_4\)) electrolyte.

Step 2: Cell Reactions During Discharge

Anode Reaction (Oxidation): \[ Pb (s) + SO_4^{2-} (aq) \rightarrow PbSO_4 (s) + 2e^- \]

Cathode Reaction (Reduction): \[ PbO_2 (s) + 4H^+ (aq) + SO_4^{2-} (aq) + 2e^- \rightarrow PbSO_4 (s) + 2H_2O (l) \]

Overall Cell Reaction: \[ Pb (s) + PbO_2 (s) + 4H^+ (aq) + 2SO_4^{2-} (aq) \rightarrow 2PbSO_4 (s) + 2H_2O (l) \]

Step 3: Working Principle
- During discharge, Pb and PbO\(_2\) get converted into PbSO\(_4\), releasing electrical energy.
- During charging, the reaction is reversed, regenerating Pb and PbO\(_2\). Quick Tip: The lead storage battery is rechargeable, making it widely used in automobiles and inverters.

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

Step 2: Standard E.M.F. Calculation:
\[ E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 \]
\[ = 0.80 - 0.34 = 0.46V \]

Thus, the e.m.f. of the cell is 0.46V. Quick Tip: The Nernst equation helps calculate cell potential under non-standard conditions.

Step 1: First-Order Rate Law:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]

Step 2: Calculate Rate Constant \( k \):

For 40% completion:

t₄₀% = 50 = (2.303 / k) * log(100 / 60)

\[ k = \frac{2.303}{50} \times 0.2218 = 0.0102 min⁻¹ \]

Step 3: Calculate Time for 80% Completion:

t₈₀% = (2.303 / 0.0102) * log(100 / 20)

\[ = \frac{2.303}{0.0102} \times 0.7782 = 176 min \]

Thus, the time for 80% completion is 176 minutes. Quick Tip: The rate of a first-order reaction depends only on the concentration of one reactant, making calculations straightforward.

Step 1: Ionisation Isomerism:
Occurs when compounds produce different ions in solution.

Example:
\[ [Co(NH_3)_5SO_4]Br and [Co(NH_3)_5Br]SO_4 \]

Step 2: Coordination Isomerism:
Occurs when the ligands exchange between cation and anion parts.

Example:
\[ [Cr(NH_3)_6][Co(CN)_6] and [Co(NH_3)_6][Cr(CN)_6] \] Quick Tip: Ionisation isomerism changes the counter-ion, while coordination isomerism swaps ligands between metal centers.

(x) \([CoCl_2(en)_2]Cl\)


Dichlorobis(ethylenediamine)cobalt(C) chloride



(y) \(Fe_4[Fe(CN)_6]_3\)


Tetrairon(C) hexacyanoferrate(B)
Quick Tip: IUPAC naming follows metal oxidation state, ligand count, and anionic charge.

Step 1: Definition:
Carbohydrates are biomolecules composed of C, H, and O, serving as energy sources.

Step 2: Difference Between Glucose and Fructose:




Quick Tip: Glucose is an aldose, while fructose is a ketose; both are monosaccharides.

Step 1: Vitamin A:
- Sources: Carrots, milk, fish liver oil
- Deficiency Disease: Night blindness

Step 2: Vitamin C:
- Sources: Citrus fruits, green vegetables
- Deficiency Disease: Scurvy (bleeding gums, weakness) Quick Tip: Vitamins are essential for metabolism; deficiencies cause specific diseases.

(A) Sulphonation of chlorobenzene
\[ C_6H_5Cl + H_2SO_4 \xrightarrow{SO_3/H_2SO_4} C_6H_4ClSO_3H \]

(B) Nitration of chlorobenzene
\[ C_6H_5Cl + HNO_3 \xrightarrow{H_2SO_4} p-C_6H_4ClNO_2 + o-C_6H_4ClNO_2 \]

(C) Reaction of ethyl bromide with sodium ethoxide
\[ C_2H_5Br + C_2H_5ONa \rightarrow C_2H_5OC_2H_5 + NaBr \]

(D) Wurtz-Fittig reaction
\[ C_6H_5Cl + 2Na + CH_3Cl \rightarrow C_6H_5CH_3 + 2NaCl \]

(v) Fittig reaction
\[ C_6H_5Cl + 2Na + C_6H_5Cl \rightarrow C_6H_5C_6H_5 + 2NaCl \] Quick Tip: Electrophilic aromatic substitution reactions follow regioselectivity; ortho- and para-directing groups influence product formation.

Step 1: Electrophilic Substitution in Chlorobenzene:
Chlorine has a \(-I\) effect (electron-withdrawing) and a \(+M\) effect (electron-donating via resonance), making the ortho and para positions more reactive.

Step 2: Example - Nitration of Chlorobenzene:
\[ C_6H_5Cl + HNO_3 \xrightarrow{H_2SO_4} o-C_6H_4ClNO_2 + p-C_6H_4ClNO_2 \] Quick Tip: Halogens are deactivating but ortho/para-directing due to resonance.

Step 1: Freons:
- Used as refrigerants and aerosol propellants.
- Environmental Effect: Depletes ozone layer.

Step 2: DDT:
- Used as an insecticide.
- Environmental Effect: Accumulates in the food chain, causing toxicity. Quick Tip: Freons destroy ozone, while DDT bioaccumulates, harming ecosystems.

(A) Preparation of Primary Alcohol:
\[ RMgX + HCHO \xrightarrow{H_3O^+} RCH_2OH \]

(B) Preparation of Secondary Alcohol:
\[ RMgX + R'CHO \xrightarrow{H_3O^+} RCH(OH)R' \]

(C) Preparation of Tertiary Alcohol:
\[ RMgX + R'COR'' \xrightarrow{H_3O^+} RCR'(OH)R'' \] Quick Tip: Grignard reagents react with aldehydes/ketones to form alcohols.

Step 1: Reaction Mechanism:
\[ R-O^- + R'X \rightarrow R-O-R' + X^- \]

Step 2: Example:
\[ C_2H_5ONa + C_6H_5Cl \rightarrow C_6H_5OC_2H_5 + NaCl \] Quick Tip: Williamson’s synthesis is useful for the preparation of ethers.

(x) Anisole reacts with acetyl chloride in the presence of anhydrous AlCl\(_3\)?
\[ C_6H_5OCH_3 + CH_3COCl \xrightarrow{AlCl_3} p-CH_3COC_6H_4OCH_3 + HCl \]

(y) Phenol reacts with chloroform in the presence of aqueous NaOH?
\[ C_6H_5OH + CHCl_3 + 3NaOH \rightarrow o-C_6H_4CHO + 3NaCl + 2H_2O \] Quick Tip: Phenol reacts with chloroform under basic conditions to form salicylaldehyde.

Step 1: Reaction:
\[ R_2B-H + R'CH=CH_2 \rightarrow R'CH_2CH_2B(R_2) \]
\[ R'CH_2CH_2B(R_2) + H_2O_2/OH^- \rightarrow R'CH_2CH_2OH \]

Step 2: Characteristics:
- Anti-Markovnikov addition (OH group attaches to the least substituted carbon).
- Syn addition (H and OH add to the same face of the alkene). Quick Tip: Hydroboration-oxidation forms anti-Markovnikov alcohols in one step.

(A) Acetone reacts with sodium hypoiodite
\[ (CH_3)_2CO + 3NaOI \rightarrow (CH_3)_2COO^- Na^+ + I_2 + 2NaOH \]

(B) Acetaldehyde reacts with ammonia
\[ CH_3CHO + NH_3 \rightarrow CH_3CH=NH + H_2O \]

(C) Acetic acid reacts with ethyl alcohol in the presence of concentrated H\(_2\)SO\(_4\)
\[ CH_3COOH + C_2H_5OH \xrightarrow{H_2SO_4} CH_3COOC_2H_5 + H_2O \]

(D) Formaldehyde reacts with ammonia
\[ HCHO + NH_3 \rightarrow (CH_2)_6N_4 \]

(E) Benzene reacts with carbon monoxide and hydrogen chloride in the presence of anhydrous AlCl\(_3\)
\[ C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO \] Quick Tip: Reagents like NaOI and AlCl\(_3\) act as oxidizing and Lewis acid catalysts respectively, influencing the type of reaction.

Cross Aldol Condensation

- Aldol condensation occurs between two different aldehydes or ketones.
- Example:
\[ CH_3CHO + C_6H_5CHO \xrightarrow{NaOH} CH_3CH(OH)C_6H_5CHO \]

Etard’s Reaction

- Selective oxidation of toluene to benzaldehyde.
- Reaction:
\[ C_6H_5CH_3 + CrO_2Cl_2 \rightarrow C_6H_5CHO + CrCl_3 + HCl \]

Cannizzaro Reaction

- Aldehydes without α-hydrogen undergo disproportionation.
- Example:
\[ 2 HCHO + NaOH \rightarrow CH_3OH + HCOONa \] Quick Tip: Cross aldol reaction occurs in the presence of a base, while Cannizzaro reaction is unique to aldehydes without α-hydrogen.

Hinsberg Test:

- Used to distinguish primary, secondary, and tertiary amines using benzene sulfonyl chloride \((C_6H_5SO_2Cl)\).

Reactions:

Primary Amine: \[ RNH_2 + C_6H_5SO_2Cl \rightarrow R-NH-SO_2C_6H_5 + HCl \]
(Soluble in alkali)

Secondary Amine: \[ R_2NH + C_6H_5SO_2Cl \rightarrow R_2N-SO_2C_6H_5 + HCl \]
(Insoluble in alkali)

Tertiary Amine: \[ R_3N + C_6H_5SO_2Cl \nrightarrow No reaction \] Quick Tip: Hinsberg’s test differentiates amines based on their solubility in an alkali.

- Hydrogen bonding in primary amines increases boiling points.
- Tertiary amines lack hydrogen bonding due to the absence of an H atom attached to nitrogen.
\[ Primary amine: Strong H-bonding > Tertiary amine: No H-bonding \] Quick Tip: Hydrogen bonding significantly impacts boiling points of organic compounds.

(A) Diazotisation Reaction

- Primary aromatic amines react with NaNO\(_2\) and HCl at 0-5°C to form diazonium salts.
- Example:
\[ C_6H_5NH_2 + NaNO_2 + HCl \rightarrow C_6H_5N_2^+Cl^- + H_2O \]

(B) Gabriel Phthalimide Synthesis

- Used for synthesizing primary amines.
- Example:
\[ C_6H_4(CO)_2N^-K^+ + RBr \rightarrow C_6H_4(CO)_2NR + KBr \]

(C) Carbylamine Reaction

- Primary amines react with chloroform and alcoholic KOH to form isocyanides (foul-smelling).
- Example:
\[ RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O \] Quick Tip: Diazotisation forms diazonium salts, crucial for azo dye formation.