RBSE Class 12th 2019 Chemistry-SS-41-2019-With Solution Previous Year Papers
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| Board | RBSE |
|---|---|
| Class | Class 12th |
| Exam year | 2019 |
| Subject | Chemistry-SS-41-2019-With Solution |
| Resource type | Previous Year Papers |
| Category | RBSE Previous Year Question Papers |
| Website | RBSE Solution |
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RBSE Class 12th 2019 Chemistry-SS-41-2019-With Solution
Scroll through the Previous Year Papers pages for Chemistry-SS-41-2019-With Solution (2019).
Rajasthan Board Class 12th Chemistry-SS-41-2019-With Solution 2019 solved Previous Year Question Papers
RBSE XII Examination – 2019
Series: 333 | Code No.: 41/Chem.
Roll No.: [TTT TEL | | | | | | | ]
Candidates must write the Code on the title page of the answer-book.
CHEMISTRY (Theory) & SOLUTION
Time allowed: 3¼ hours | Maximum Marks: 56
General Instructions to the Examinees:
- Candidate must write first his/her Roll No. on the question paper compulsorily.
- All the questions are compulsory.
- Write the answer to each question in the given answer-book only.
- For questions having more than one part, the answers to those parts are to be written together in continuity.
- If there is any error / difference / contradiction in Hindi & English versions of the question paper, the question of Hindi version should be treated valid.
-
Mark per question:
Q. Nos. Marks per question 1 – 23 1 24 2 25 – 27 3 28 – 30 4
Page content could not be transcribed.
SECTION A
4. Write any one example of network solid. [1 Mark]
Ans. Example of network solid is diamond. It is the network solid of carbon atoms.
2. Write definition of azeotropic mixture. [1 Mark]
Ans. Azeotropes are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation.
3. Rate constant of a chemical reaction is 7.2 × 10-5. Calculate the order of reaction. [1 Mark]
Ans.
k = Rate / [Reactant]n
Unit of k = (concentration / time) / (concentration)n = (conc.)1-n / time
Given unit of k = 7.2 × 10-5 (which is s-1 for first order)
So, (conc.)1-n / time = (conc.)0 / time
Therefore, 1 - n = 0, so n = 1
Order of reaction is (n) = 1
4. Define threshold energy. [1 Mark]
Ans. Threshold energy — The minimum amount of energy which must be associated with the molecules, so that their mutual collisions result in product formation.
5. Give any one example of bidentate ligand. [1 Mark]
Ans. Example of bidentate ligand is Ethylene diamine [H2N—CH2—CH2—NH2].
6. Write IUPAC name of Diethyl ether. [1 Mark]
Ans. IUPAC name of Diethyl ether is ethoxy ethane.
7. Draw the resonating structures of phenoxide ion.
Ans. Resonating structures of phenoxide ion are:
I
:O-
C6H5
II
:O:
C6H5
III
:O:
C6H5
IV
:O:
C6H5
V
:O:
C6H5
8. Write chemical equation of carbylamine reaction.
Ans. Carbylamine reaction:
R-NH2 + CHCl3 + 3KOH → R-NC + 3KCl + 3H2O
Class-XII / (RBSE) | Chemistry
9. Write name of the hormone secreted by thyroid gland.
Answer: The hormone secreted by the thyroid gland is thyroxine.
40. Write any one example of biodegradable polymer.
Answer: An example of a biodegradable polymer is Poly(3-hydroxybutyrate-co-3-hydroxyvalerate) (PHBV).
Its formation from monomers:
3-Hydroxybutanoic acid + 3-Hydroxypentanoic acid → PHBV
[–O–CH(CH₃)–CH₂–CO–O–CH(CH₂CH₃)–CH₂–CO–]ₙ
41. Write monomer units of polymer Buna-N.
Answer: The monomer units of Buna-N are:
- 1,3-Butadiene (CH₂=CH–CH=CH₂)
- Acrylonitrile (CH₂=CH–CN)
42. What is polydispersity index for a polymer?
Answer: Polydispersity index (PDI) is the ratio of weight average molecular weight (M̅w) to number average molecular weight (M̅n) of a polymer.
PDI = M̅w / M̅n
43. Write the value of axis of symmetry (Cn) present in H₂O molecule.
Answer: Water (H₂O) has a C₂ axis of symmetry. Rotating the molecule by 180° brings it back to its original position.
Calculation: Cn = 360° / n = 360° / 180° = 2
SECTION B
44. (A) Write any two differences between Schottky and Frenkel defects.
| Schottky defect | Frenkel defect |
|---|---|
| Equal number of cationic and anionic vacancies are present. | Some cations are displaced from normal lattice sites to interstitial sites. |
| Density of the crystal is lowered. | Density of the crystal remains unaffected. |
44. (B) Calculate the packing efficiency in simple cubic lattice.
Answer: Packing efficiency (P.E.) is the percentage of total volume of a unit cell occupied by atoms, ions, or molecules.
For a simple cubic lattice:
- Number of atoms per unit cell = 1
- Volume of one atom = (4/3)πr³
- Edge length (a) = 2r
- Volume of unit cell = a³ = (2r)³ = 8r³
P.E. = (Volume of atoms / Volume of unit cell) × 100
= [(4/3)πr³ / 8r³] × 100
= (π/6) × 100
= (3.1416/6) × 100
= 52.36%
Thus, the packing efficiency of a simple cubic lattice is approximately 52.4%.
45. 0.05M solution of K₄[Fe(CN)₆] at 300K is 92% dissociated. Calculate the osmotic pressure of the solution. (R = 0.0824 atm. L K⁻¹ mol⁻¹)
Solution:
K₄[Fe(CN)₆] dissociates as:
K₄[Fe(CN)₆] ⇌ 4K⁺ + [Fe(CN)₆]⁴⁻
Number of ions (n) after dissociation = 5
Given: α = 92% = 0.92, n = 5
Van't Hoff factor (i) = 1 + α(n - 1)
i = 1 + 0.92(5 - 1)
i = 1 + 0.92 × 4
i = 1 + 3.68
i = 4.68
Osmotic pressure (π) = iCRT
Given: C = 0.05 mol L⁻¹, R = 0.0824 atm L K⁻¹ mol⁻¹, T = 300 K
π = 4.68 × 0.05 × 0.0824 × 300
π = 4.68 × 0.05 × 24.72
π = 4.68 × 1.236
π = 5.78448 atm
Correct Answer: 5.78 atm (approximately)
46. (A) Write any two factors which affect the conductance of electrolysis. [2 Marks]
Answer:
Factors which affect the conductance of electrolysis are:
- The nature of the electrolyte added
- The size of the ions produced and their solvation
46. (B) Corrosion is an electrochemical phenomenon. Explain. [2 Marks]
Explanation:
Corrosion is a redox process by which metals are oxidized by oxygen in the presence of moisture. For example, rusting of iron is an electrochemical phenomenon.
Oxidation: Fe(s) → Fe²⁺ + 2e⁻
Reduction: O₂ + 4H⁺ + 4e⁻ → 2H₂O
In the atmosphere:
4Fe²⁺ + O₂ + 4H₂O → 2Fe₂O₃ + 8H⁺
Further oxidation:
Fe₂O₃ + xH₂O → Fe₂O₃·xH₂O (Rust)
Thus, corrosion involves both oxidation (loss of electrons by metal) and reduction (gain of electrons by oxygen), making it an electrochemical process.
7. The conductivity of 0.40 M solution of KCl at 298 K is 0.0129 S cm⁻¹. Calculate its molar conductivity.
Solution:
Given: Conductivity (κ) = 0.0129 S cm⁻¹, Concentration (C) = 0.40 mol L⁻¹
Molar conductivity (Λm) is given by:
Λm = (κ × 1000) / C
Λm = (0.0129 × 1000) / 0.40
Λm = 12.9 / 0.40
Λm = 32.25 S cm² mol⁻¹
48. A first order reaction takes 40 minutes for 20% decomposition. Calculate half life (log₁₀ 4 = 0.602, log₁₀ 2 = 0.301).
Solution:
For a first order reaction: t = (2.303 / k) log [a / (a - x)]
Given: t = 40 min, a = 100%, x = 20% (decomposition), so (a - x) = 80%
40 = (2.303 / k) log (100 / 80)
40 = (2.303 / k) log (10 / 8)
40 = (2.303 / k) [log 10 - log 8]
40 = (2.303 / k) [1 - log 2³]
40 = (2.303 / k) [1 - 3 log 2]
40 = (2.303 / k) [1 - 3 × 0.301]
40 = (2.303 / k) [1 - 0.903]
40 = (2.303 / k) × 0.097
k = (2.303 × 0.097) / 40
k = 0.2234 / 40
k = 0.005585 min⁻¹
Half life for first order: t1/2 = 0.693 / k
t1/2 = 0.693 / 0.005585
t1/2 ≈ 124.1 min
49. (A) What is the role of graphite rod in the electrometallurgy of aluminium?
Answer: In the electrometallurgy of aluminium (Hall-Héroult process), the graphite rod acts as the anode. It helps in the reduction of aluminium oxide to aluminium metal. The overall reaction is:
2Al2O3 + 3C → 4Al + 3CO2
The graphite anode is consumed during the process as it reacts with oxygen to form carbon dioxide.
49. (B) Draw a labelled diagram of reverberatory furnace.
Answer: A labelled diagram of a reverberatory furnace is shown below:
┌─────────────────────────────┐
│ Charge Hopper │
│ (Ore) │
└─────────────┬───────────────┘
│
┌─────────────▼───────────────┐
│ Hangers │
│ (for support) │
└─────────────┬───────────────┘
│
┌─────────────▼───────────────┐
│ Charge Hopper │
│ (Ore) │
└─────────────┬───────────────┘
│
┌─────────────▼───────────────┐
│ ┌───────────────┐ │
│ │ Reverberatory│ │
│ │ Furnace │ │
│ │ (Hearth) │ │
│ └───────────────┘ │
│ ▲ │
│ │ │
│ ┌────┴────┐ │
│ │ Fuel │ │
│ │ Burner │ │
│ └─────────┘ │
└─────────────────────────────┘
Key parts: Charge hopper (for feeding ore), hangers (for support), reverberatory furnace hearth (where ore is heated), and fuel burner (for heating). The flame reflects (reverberates) from the roof onto the charge.
20. Give the oxidation state and coordination number of the central metal ion in the following complexes:
(A) [Co(en)3]3+ (B) K4[Fe(CN)6]
Calculation of oxidation state of metal in complex ion:
(A) [Co(en)3]3+
Let oxidation state of Co = x.
en (ethylenediamine) is a neutral ligand, so charge = 0.
x + 3(0) = +3
x = +3
Oxidation state = +3
Coordination number = 6 (en is bidentate, 3 × 2 = 6)
(B) K4[Fe(CN)6]
K+ has charge +1 each, so 4K = +4.
CN− has charge −1 each, so 6CN = −6.
Let oxidation state of Fe = x.
+4 + x + (−6) = 0
x − 2 = 0
x = +2
Oxidation state = +2
Coordination number = 6 (CN is monodentate, 6 × 1 = 6)
24. (A) Why are phenols more acidic than alcohols? Explain.
Phenols are stronger acids (Ka ≈ 10−10) than alcohols (Ka ≈ 10−16 to 10−18) because the phenoxide ion (C6H5O−) is stabilized by resonance delocalization of the negative charge into the benzene ring, whereas the alkoxide ion (RO−) has no such resonance stabilization and the negative charge is localized on oxygen.
(B) Arrange the following alcohols in increasing order of their reactivity towards esterification reaction.
CH3–CH2–OH, (CH3)2CH–OH, (CH3)3C–OH
Increasing order of reactivity towards esterification:
(CH3)3C–OH < (CH3)2CH–OH < CH3–CH2–OH
Reason: Reactivity decreases with increasing steric hindrance around the hydroxyl group. Primary alcohols are most reactive, followed by secondary, and tertiary alcohols are least reactive.
OR
(A) Why are the boiling points of alcohols higher than hydrocarbons and ethers of comparable molecular mass? Explain.
The high boiling points of alcohols are mainly due to the presence of intermolecular hydrogen bonding between alcohol molecules, which requires extra energy to break. Hydrocarbons and ethers of comparable molecular mass lack such hydrogen bonding (ethers have only weak dipole-dipole interactions, hydrocarbons have only van der Waals forces), so they have lower boiling points.
(B) Arrange the following alcohols in increasing order of their reactivity towards dehydration reaction.
CH3–CH2–OH, (CH3)2CH–OH, (CH3)3C–OH
Increasing order of reactivity towards dehydration:
CH3–CH2–OH < (CH3)2CH–OH < (CH3)3C–OH
Reason: Dehydration follows carbocation stability order. Tertiary alcohols form the most stable tertiary carbocation, secondary forms a less stable carbocation, and primary forms the least stable carbocation. Hence, reactivity increases with increasing alkyl substitution.
22. Identify [A] and [B] in the following chemical reactions:
C6H5N2Cl → (with Cu powder) → [A] → (with Sn/HCl) → [B]
Step 1: C6H5N2Cl (benzenediazonium chloride) reacts with Cu powder (or Cu2O) in the presence of water to give nitrobenzene (C6H5NO2). This is a substitution reaction where the diazonium group is replaced by a nitro group.
[A] = Nitrobenzene (C6H5NO2)
Step 2: Nitrobenzene is reduced by Sn/HCl (tin and hydrochloric acid) to aniline (C6H5NH2).
[B] = Aniline (C6H5NH2)
23. (A) What happens when glucose reacts with concentrated HNO3? Give chemical equation.
When glucose reacts with concentrated HNO3, it undergoes oxidation. The aldehyde group (−CHO) at C1 and the primary alcohol group (−CH2OH) at C6 are both oxidized to carboxylic acid groups (−COOH), forming saccharic acid (glucaric acid).
Chemical equation:
CHO–(CHOH)4–CH2OH + 2HNO3 → (conc.) → HOOC–(CHOH)4–COOH + 2NO2 + 2H2O
Glucose → Saccharic acid
(B) Draw the structure of β-D-ribose sugar.
The structure of β-D-ribose (a pentose sugar) in its cyclic furanose form is:
HOH₂C O OH
\ / \
C C
/ \ / \
H C C
| |
OH OH
Note: In β-D-ribose, the anomeric carbon (C1) has the −OH group on the same side as the −CH2OH group (up in the Haworth projection). The sugar is a furanose (five-membered ring) with the formula C5H10O5.
24. (A) Differentiate between configurational and conformational isomers.
| Configurational Isomers | Conformational Isomers |
|---|---|
| 1. Arise due to presence of one or more chiral carbon atoms (optical isomers) or restricted rotation of C=C (geometrical isomers). | 1. Arise due to free rotation of C—C single bond. |
| 2. Position of atoms or groups in space is fixed. | 2. Position of atoms or groups in space changes. |
| 3. One form of configuration is not converted into another form under normal conditions. | 3. At room temperature, one form of conformation is converted into another. |
24. (B) The chair conformer of cyclohexane is more stable than boat conformer. Explain.
The boat conformer is free of angle strain. However, it is less stable than the chair conformer because some of the bonds in the boat conformer are eclipsed, giving it torsional strain. The boat conformer is further destabilized by the close proximity of the flagpole hydrogens (the hydrogens at the "bow" and "stern" of the boat), which causes steric strain. Torsional strain and flagpole interactions cause the boat conformation to have considerably higher energy than the chair conformation. The chair form is more stable than the boat form by 44 kJ mol−1.
SECTION C
25. Read the given paragraph and answer the following questions.
The rare earth elements of the modern periodic table are known as lanthanoids. They have a separate block in the periodic table. The lanthanoid series consists of fourteen elements starting from Cerium (atomic number 58) to Lutetium (atomic number 71). All lanthanoids generally exhibit +3 oxidation state. In addition, some lanthanoids show +2 and +4 oxidation states also. As we move from left to right in the lanthanoid series, there is a regular decrease in the size of an atom. This is known as lanthanoid contraction. There are many industrial applications such as formation of mischmetal and production of parts of jet engines.
(A) The basic nature of hydroxides of lanthanoid elements decreases moving from left to right. Explain.
With an increase in atomic number, the basic strength of the oxides and hydroxides decreases. This contraction causes a decrease in the size of lanthanoid cations and, therefore, the polarizing power of the cations increases. This further decreases the ionic character of the oxides and hydroxides. Thus, Ce(OH)3 is most basic and Lu(OH)3 is least basic.
(B) Write the name of two lanthanoid elements used in the formation of mischmetal.
Ce (Cerium) and La (Lanthanum).
(C) Write the name of one lanthanoid element exhibiting +4 oxidation state.
Ce (Cerium) forms Ce4+.
26. Read the given paragraph and write answer of the following questions.
Chemicals have special importance in various fields of daily life such as in foods, soaps, and detergents. Chemicals are used in food materials for preservation, to enhance appeal, and to increase their nutritive quality. Chemical substances which are added to food materials to prevent their spoilage and retain nutritive value for long times are called food preservatives. Artificial sweeteners are those chemical compounds which are used to give sweetening effect to food materials. Diabetic patients are advised to use saccharin in place of sugar.
(A) Why are chemicals added to food materials?
Chemicals are added to food for (i) their preservation, (ii) enhancing their appeal, and (iii) adding nutritive value to them.
(B) Write the name of any two food preservatives.
Sodium benzoate (C6H5COONa) and salts of sorbic acid and propanoic acid.
(C) Why diabetic patients are advised to use saccharin?
Answer: Saccharin is about 550 times sweeter than sugar on a mass-to-mass basis. It is not biodegradable (or is not metabolized in the body) and does not have any calorific value as food. It is excreted as such in urine. Therefore, it is primarily used as a sweetening agent by diabetic patients.
27. (A) Write the structural formula of oxalic acid.
Answer: The structural formula of oxalic acid is:
HOOC – COOH
Or, in expanded form:
O O
|| ||
HO – C – C – OH
(B) Explain the mechanism of Aldol condensation.
Answer: The mechanism of Aldol condensation involves the following steps:
- Formation of enolate ion: In the presence of a base (OH⁻), the aldehyde (e.g., acetaldehyde) loses a proton from the α-carbon to form a resonance-stabilized enolate ion.
Reaction: CH₃CHO + OH⁻ ⇌ :CH₂CHO + H₂O - Nucleophilic attack: The enolate ion (nucleophile) attacks the carbonyl carbon of another unreacted aldehyde molecule, forming a new carbon-carbon bond.
Reaction: CH₃CHO + :CH₂CHO → CH₃–CH(–O⁻)–CH₂–CHO (alkoxide intermediate) - Protonation: The alkoxide intermediate accepts a proton from water to form the aldol product (β-hydroxy aldehyde) and regenerates the hydroxide ion.
Reaction: CH₃–CH(–O⁻)–CH₂–CHO + H₂O → CH₃–CH(OH)–CH₂–CHO + OH⁻
When the aldol product is heated, it undergoes dehydration to form an α,β-unsaturated aldehyde or ketone.
OR
(A) Write the structural formula of Diethyl ketone.
Answer: The structural formula of diethyl ketone (also known as 3-pentanone) is:
O
||
CH₃–CH₂–C–CH₂–CH₃
(B) Explain the mechanism of Kolbe electrolysis.
Answer: Kolbe electrolysis is an electrochemical reaction used to produce alkanes (or alkenes) from the electrolysis of aqueous solutions of alkali metal salts of carboxylic acids. The mechanism involves the following steps:
- Ionization: The sodium salt of the carboxylic acid (e.g., sodium acetate) dissociates in water to give carboxylate ions (RCOO⁻) and Na⁺ ions.
- Anodic oxidation: At the anode, the carboxylate ion loses an electron to form a carboxyl radical (RCOO•).
- Decarboxylation: The carboxyl radical is unstable and rapidly loses CO₂ to form an alkyl radical (R•).
- Dimerization: Two alkyl radicals combine at the anode to form a stable alkane (R–R).
Example: Electrolysis of sodium acetate (CH₃COONa) yields ethane (CH₃–CH₃) at the anode, along with CO₂ and H₂ gas at the cathode.
28. (A) Write definition of adsorption.
Ans. The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed adsorption. The molecular species or substance, which concentrates or accumulates at the surface is termed adsorbate and the material on the surface of which the adsorption takes place is called adsorbent.
28. (B) What happens when an electric current is passed through colloidal solution?
Ans. The colloidal particles move towards oppositely charged electrodes, get discharged and precipitated.
28. (C) Why is alum added for purification of water?
Ans. The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.
28. (D) Draw a labelled diagram of electro-dialysis method for purification of colloidal solutions.
Ans. A labelled diagram of electro-dialysis is shown below:
+ -
| |
| |
┌────┤ ├────┐
│ │ │ │
│ ┌─┴─┐ ┌─┴─┐ │
│ │ │ │ │ │
│ │ S │ │ C │ │
│ │ o │ │ r │ │
│ │ l │ │ y │ │
│ │ │ │ s │ │
│ │ p │ │ t │ │
│ │ a │ │ a │ │
│ │ r │ │ l │ │
│ │ t │ │ l │ │
│ │ i │ │ o │ │
│ │ c │ │ i │ │
│ │ l │ │ d │ │
│ │ e │ │ │ │
│ │ │ │ │ │
│ └───┘ └───┘ │
│ ↑ ↑ │
│ │ │ │
└────┴───────┴────┘
Electro-dialysis
Labels: Sol particle, Crystalloid, Electrodes (+ and -).
OR
28. (A) Write definition of chemical adsorption.
Ans. When the forces of attraction existing between adsorbate and adsorbent are strong chemical bonds, the adsorption is called chemical adsorption. In chemical adsorption, the adsorbate forms a product by reaction at the surface of the adsorbent.
29. (A) Write oxidation state of nitrogen in nitric acid.
Ans. In HNO₃, let the oxidation state of N be x.
H = +1, O = –2 (for 3 O atoms, total –6).
Equation: (+1) + x + (–6) = 0 → x = +5.
Thus, the oxidation state of nitrogen in nitric acid is +5.
(B) What happens when sulphur reacts with concentrated H₂SO₄? Give chemical equation.
Ans. Sulphur is oxidised by concentrated sulphuric acid to sulphur dioxide.
Chemical equation:
S + 2H₂SO₄ (conc.) → 3SO₂ + 2H₂O
(C) Why is helium used as a diluent for oxygen in modern diving apparatus?
Ans. Helium is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood. This reduces the risk of decompression sickness (bends) and allows divers to breathe more easily at high pressures.
(D) Draw a labelled diagram of Cottrell smoke precipitator.
Diagram of Cottrell smoke precipitator:
┌─────────────────────────────┐
│ Electrode (30,000 volts │
│ or more) │
│ │ │
│ │ │
│ ┌─────┴─────┐ │
│ │ │ │
│ │ Smoke │ │
│ │ inlet │ │
│ │ │ │
│ └─────┬─────┘ │
│ │ │
│ ▼ │
│ ┌───────────┐ │
│ │ │ │
│ │ Gases │ │
│ │ free │ │
│ │ from │ │
│ │ carbon │ │
│ │ particles│ │
│ │ │ │
│ └───────────┘ │
│ │ │
│ ▼ │
│ Precipitated carbon │
└─────────────────────────────┘
Labels: Electrode (high voltage), Smoke inlet, Gases free from carbon particles, Precipitated carbon.
28. (B) What happens when a beam of light is passed through a colloidal solution?
Ans. The Tyndall effect is observed. The Tyndall effect is due to the fact that colloidal particles scatter light in all directions in space. This scattering of light illuminates the path of the beam in the colloidal dispersion.
(C) Why does the colour of the sky appear blue?
Ans. Dust particles along with water suspended in air scatter blue light (shorter wavelength) more than other colours. This scattered blue light reaches our eyes, and the sky appears blue to us.
30. (A) Write chemical equation of Finkelstein reaction.
Ans. R–Cl or R–Br + NaI → R–I + NaCl or NaBr (in dry acetone)
(Finkelstein reaction)
(B) Why aryl halides are less reactive towards nucleophilic substitution reactions? Explain.
Ans. In aryl halides, due to resonance, the C–Cl bond acquires a partial double bond character. This makes bond cleavage in haloarenes more difficult than in haloalkanes (where carbon is attached to halogen by a pure single bond). Hence, aryl halides are less reactive towards nucleophilic substitution reactions.
(C) Arrange the following alkyl halides in increasing order of their reactivity towards SN2 reaction.
CH3–CH2–Cl, (CH3)2CH–Cl, (CH3)3C–Cl
Ans. Increasing order of reactivity towards SN2 reaction:
(CH3)3C–Cl < (CH3)2CH–Cl < CH3–CH2–Cl
Reason: SN2 reactions are favoured by less steric hindrance around the carbon atom. Primary alkyl halides (CH3–CH2–Cl) have the least steric hindrance, while tertiary alkyl halides ((CH3)3C–Cl) have the most.
(D) Draw the orbital diagram of CH3Cl.
Ans. The orbital diagram of CH3Cl shows:
- Carbon is sp3 hybridised.
- Three sp3 hybrid orbitals of carbon overlap with 1s orbitals of three hydrogen atoms to form three C–H sigma bonds.
- One sp3 hybrid orbital of carbon overlaps with a 3p orbital of chlorine to form a C–Cl sigma bond.
- The chlorine atom also has three lone pairs of electrons in its 3p orbitals.
(A simple diagram would show a central carbon with four sp3 orbitals, three bonded to H and one to Cl, with Cl having lone pairs.)
OR
(A) Write chemical equation of Wurtz-Fitting reaction.
Ans. Wurtz-Fitting reaction:
C6H5–I + 2Na + I–CH3 → C6H5–CH3 + 2NaI
(Iodobenzene + Methyl iodide + Sodium → Toluene + Sodium iodide)
(B) The reaction of alkyl chloride with aqueous KOH leads to the formation of alcohols but in presence of alcoholic KOH, alkenes are major products. Explain.
Ans. Aqueous KOH provides OH– ions which act as a nucleophile and favour substitution reaction (SN2 or SN1) to form alcohols. Alcoholic KOH, on the other hand, provides a strong base (C2H5O–) which favours elimination reaction (E2 or E1) to form alkenes as major products. The solvent polarity and the nature of the base determine the pathway.
(C) Arrange the following halogen derivatives in increasing order of their reactivity towards nucleophilic substitution reaction.
R–Cl, R–Br, R–I
Answer: R–Cl < R–Br < R–I
Explanation: In nucleophilic substitution reactions, the reactivity of alkyl halides depends on the bond strength of the C–X bond. The bond dissociation energy decreases from C–Cl to C–I, making the C–I bond the weakest and easiest to break. Thus, alkyl iodides are most reactive, followed by bromides, and chlorides are least reactive.
(D) Draw a labelled diagram of the laboratory method of preparation of chloroform.
Answer: The laboratory preparation of chloroform (CHCl₃) involves heating ethanol (or acetone) with bleaching powder (CaOCl₂) and water. The reaction mixture is distilled, and chloroform vapours are condensed and collected.
Labelled diagram description:
- A round-bottom flask (RBF) containing ethanol, bleaching powder, and water is placed on a wire gauze over a Bunsen burner.
- A thermometer is inserted into the flask to monitor temperature (around 60–70°C).
- The flask is connected to a Liebig condenser via a delivery tube.
- Cold water enters the condenser from the bottom and exits from the top.
- The condensed chloroform (liquid) is collected in a receiver flask kept in an ice bath to prevent evaporation.
- Uncondensed gases (if any) are vented out.
Note: The diagram should show: RBF, thermometer, Liebig condenser, water inlet/outlet, receiver flask in ice bath, and labelled product as "Chloroform".
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