Section – A
Q.1 — MCQs (10 marks) | Q.2 — Very Short Answer (8 marks)
Q.1 Select and write the correct answer for the following multiple-choice questions. [10]
(1) Geometry of PCl₅ (VSEPR)
✔ (a) Trigonal Bipyramidal
PCl₅ has 5 bond pairs, 0 lone pairs → trigonal bipyramidal shape.
(2) Molecule with bent shape
✔ (b) SO₂
SO₂ has 2 bond pairs + 1 lone pair on S → bent (V-shaped). CO₂ is linear; BeBr₂ is linear; BF₃ is trigonal planar.
(3) Oxidation number of O in OF₂
✔ (c) +2
F is more electronegative (F = −1). OF₂: x + 2(−1) = 0 → x = +2.
(4) Stock notation for MnO₂
✔ (d) Mn(IV)O₂
In MnO₂, O = −2 each; 2×(−2) + Mn = 0 → Mn = +4 → Mn(IV).
(5) Not a redox reaction
✔ (d) BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
This is a double displacement/precipitation reaction; no change in oxidation state of any element.
(6) Elements in Mendeleev's periodic table
✔ (c) 63
Mendeleev's original periodic table (1869) contained 63 known elements.
(7) Products of auto-photolysis of water
✔ (c) H₃O⁺ and OH⁻
2H₂O ⇌ H₃O⁺ + OH⁻ (self-ionization / auto-photolysis of water).
(8) White phosphorus discrete molecules
✔ (b) P₄
White phosphorus exists as discrete tetrahedral P₄ molecules.
(9) Colloidal molecule — hydrophilic head + hydrophobic tail
✔ (b) Detergent
Detergent molecules are amphiphilic — ionic/polar head (hydrophilic) + long hydrocarbon tail (hydrophobic).
(10) Identical atoms on same side of double bond
✔ (a) cis isomer
When identical groups are on the same side of the double bond, the isomer is called cis.
Q.2 Answer the following questions. [8]
(1) Three common scales of temperature measurement:
- Celsius scale (°C) — water freezes at 0 °C and boils at 100 °C.
- Fahrenheit scale (°F) — water freezes at 32 °F and boils at 212 °F.
- Kelvin scale (K) — SI unit of temperature; absolute zero = 0 K. K = °C + 273.
(2) Define: Atomic mass unit (amu)
One atomic mass unit (amu or u) is defined as 1/12th of the mass of one carbon-12 (¹²C) atom.
1 amu = 1.66 × 10⁻²⁷ kg
(3) TRUE or FALSE: Inorganic molecules are often represented by molecular formulae indicating their elemental composition.
TRUE. Inorganic molecules are generally represented by molecular formulae (e.g., H₂O, NaCl, H₂SO₄) which indicate the types and number of atoms (elements) present.
(4) If QC < KCC, the reaction will proceed from right to left… — TRUE or FALSE?
FALSE. If QC < KCC, the reaction proceeds from left to right (forward direction), consuming more reactants to produce more products until equilibrium is reached.
(5) Give example of mirror nuclei.
Mirror nuclei are pairs of nuclei in which the number of protons in one equals the number of neutrons in the other, and vice versa.
- Example 1: ¹³N (Z=7, N=6) and ¹³C (Z=6, N=7)
- Example 2: ¹⁵O (Z=8, N=7) and ¹⁵N (Z=7, N=8)
- Example 3: ³H (Z=1, N=2) and ³He (Z=2, N=1)
Mirror nuclei have the same mass number (A) but their proton and neutron numbers are interchanged.
(6) What are the different types of radiations emitted by a radioactive element?
A radioactive element emits three types of radiations:
| Radiation | Nature | Charge | Mass | Penetrating Power |
|---|
| Alpha (α) rays | Helium nucleus (₂⁴He) | +2 | 4 u | Least (stopped by paper) |
| Beta (β) rays | Fast-moving electrons (e⁻) | −1 | Negligible | Moderate (stopped by Al sheet) |
| Gamma (γ) rays | High energy electromagnetic radiation | 0 (neutral) | 0 | Highest (needs thick Pb/concrete) |
(7) TRUE or FALSE: N-13 and C-13 are mirror nuclei.
TRUE.
- ¹³N has Z = 7 protons and N = 6 neutrons.
- ¹³C has Z = 6 protons and N = 7 neutrons.
- The proton number of one equals the neutron number of the other → they are mirror nuclei.
- Both have the same mass number A = 13.
(8) Why is food stored for a long time?
Food is stored for a long time by the following methods and reasons:
- Refrigeration / Low temperature: Low temperature slows down the growth of microorganisms (bacteria, fungi) and reduces the rate of enzymatic reactions that cause food spoilage.
- Addition of preservatives: Chemical preservatives like sodium benzoate, salt, sugar, and vinegar (acetic acid) inhibit microbial growth.
- Canning and vacuum packing: Removes oxygen, preventing oxidation and aerobic bacterial growth.
- Dehydration (drying): Removing moisture prevents microbial growth since microbes need water to survive.
- Pasteurization: Heating food to a specific temperature destroys harmful microorganisms.
In summary, food is stored to prevent spoilage by microorganisms, delay oxidation, and maintain nutritional value for longer periods.
Section – B
Answer in one or two sentences — Any EIGHT from Q.3 to Q.14 | [16 marks]
Q.3
What is semi-microanalysis?
Semi-microanalysis is a type of qualitative chemical analysis in which the amount of substance used for testing is between 0.01 g to 0.1 g (10 mg to 100 mg).
- It requires smaller quantities of reagents compared to macro analysis.
- Tests are performed on a smaller scale, making it economical and faster.
- It is more sensitive than macro analysis.
Q.4
What does classical qualitative analysis method include?
Classical qualitative analysis includes:
- Dry tests: Flame test, charcoal cavity test, cobalt nitrate test, borax bead test — performed without dissolving the substance.
- Wet tests: Systematic group analysis using group reagents (H₂S, NH₄OH, etc.) to identify cations (acid radical) and anions (basic radical) in solution.
Q.5
Subtract 5.8 × 10⁻³ from 3.5 × 10⁻² and express in scientific notation.
3.5 × 10⁻² − 5.8 × 10⁻³
= 35.0 × 10⁻³ − 5.8 × 10⁻³
= 29.2 × 10⁻³
∴ = 2.92 × 10⁻²
Q.6
Give the names of quantum numbers.
- Principal quantum number (n)
- Azimuthal (Angular momentum) quantum number (l)
- Magnetic quantum number (ml)
- Spin quantum number (ms)
Q.7
Complete the table of Functional Groups, Prefix and Suffix.
| S.No. | Functional Group | Prefix | Suffix |
|---|
| (I) | >C=O (Ketone) | oxo– | –one |
| (II) | –CONH₂ (Amide) | carboxamido– | –amide |
| (III) | –COCl (Acid chloride) | chlorocarbonyl– | –oyl chloride |
| (IV) | –COOR (Ester) | alkoxycarbonyl– | –oate |
Q.8
What are alkynes? Write their general formula.
Alkynes are unsaturated hydrocarbons that contain a carbon–carbon triple bond (C≡C).
General formula: CₙH₂ₙ₋₂ (n ≥ 2)
e.g., Ethyne: HC≡CH | Propyne: CH₃–C≡CH
Q.9
How will you convert phenol to benzene?
Phenol is converted to benzene by reduction with zinc dust (Zn) on heating:
C₆H₅OH + Zn → C₆H₆ + ZnO
Phenol reacts with zinc dust on heating; zinc removes the –OH group. Benzene and zinc oxide are formed.
Q.10
What are meta directing groups? Enlist a few of them.
Meta directing groups are substituents already present on a benzene ring that direct the incoming electrophile (during electrophilic aromatic substitution) to the meta position (3rd or 5th carbon).
These groups are generally electron-withdrawing groups.
Examples:
–NO₂, –CN, –CHO, –COOH, –SO₃H, –COOR, –COR, –CCl₃
Q.11
Convert 1,2-dichloropropane to propyne.
Step 1 — Dehydrohalogenation (1st mol alc. KOH)
CH₃–CHCl–CH₂Cl + KOH(alc.) → CH₃–CH=CHCl + KCl + H₂O
Step 2 — Dehydrohalogenation (2nd mol alc. KOH)
CH₃–CH=CHCl + KOH(alc.) → CH₃–C≡CH + KCl + H₂O
Net reaction:
CH₃–CHCl–CH₂Cl + 2KOH(alc.) → CH₃–C≡CH + 2KCl + 2H₂O
1,2-dichloropropane → Propyne
Q.12
State the action of HBr on acetylene and methyl acetylene.
Action on Acetylene (Ethyne, HC≡CH):
HC≡CH + HBr → CH₂=CHBr (vinyl bromide) [1 mol HBr]
CH₂=CHBr + HBr → CH₃–CHBr₂ (1,1-dibromoethane) [2 mol HBr]
The addition follows Markovnikov's rule in the second step.
Action on Methyl Acetylene (Propyne, CH₃–C≡CH):
CH₃–C≡CH + HBr → CH₃–CBr=CH₂ (2-bromopropene) [1 mol, Markovnikov]
CH₃–C≡CH + 2HBr → CH₃–CBr₂–CH₃ (2,2-dibromopropane) [2 mol HBr]
Q.13
Differentiate between: Saturated and Unsaturated fats.
| Property | Saturated Fats | Unsaturated Fats |
|---|
| Bonds | Only single bonds (C–C) in fatty acid chains | One or more double bonds (C=C) in chains |
| Physical state | Solid at room temperature | Liquid at room temperature (oils) |
| Sources | Animal sources (butter, ghee, lard) | Plant sources (olive oil, sunflower oil) |
| Health effect | Excess intake raises cholesterol; unhealthy | Considered healthier; contain essential fatty acids |
| Example | Palmitic acid, stearic acid | Oleic acid, linoleic acid |
Q.14
What happens when proteins and carbohydrates present in foods are digested in the presence of enzymes?
Proteins:
Enzymes called proteases (e.g., pepsin, trypsin) catalyze the hydrolysis of peptide bonds (–CO–NH–) in proteins.
Proteins + H₂O → Polypeptides → Peptides → Amino acids
Final products: Amino acids
Carbohydrates:
Enzymes called amylases (e.g., salivary amylase, pancreatic amylase) catalyze the hydrolysis of glycosidic bonds.
Carbohydrates + H₂O → Disaccharides → Monosaccharides (Glucose)
Final products: Simple sugars (Glucose, Fructose)
Section – C
Answer the following in brief — Any EIGHT from Q.15 to Q.26 | [24 marks]
Q.15
Which of the following techniques is used for purification of solid organic compounds? (a) Crystallisation (b) Distillation
(a) Crystallisation is used for purification of solid organic compounds.
- The impure solid is dissolved in a minimum volume of a suitable hot solvent.
- The hot solution is filtered to remove insoluble impurities.
- On cooling, pure crystals of the compound separate out.
- Crystals are filtered, washed, and dried.
Distillation is used for purification of liquids, not solids.
Q.16
Write a short note on magnetic orbital quantum number (ml).
- The magnetic quantum number (ml) describes the orientation of an orbital in space relative to an external magnetic field.
- It can have integral values from −l to +l including zero, giving (2l + 1) values in total.
- For s subshell (l = 0): ml = 0 → 1 orbital
- For p subshell (l = 1): ml = −1, 0, +1 → 3 orbitals
- For d subshell (l = 2): ml = −2, −1, 0, +1, +2 → 5 orbitals
- It explains the splitting of spectral lines in a magnetic field (Zeeman effect).
Q.17
What is oxidation? Which one of the following pairs of species is in its oxidized state? (a) Mg/Mg²⁺ (b) Cu/Cu²⁺ (c) O₂/O²⁻ (d) Cl₂/Cl⁻
Oxidation is defined as the loss of electrons or an increase in oxidation number of a species.
The oxidized species in each pair:
- (a) Mg²⁺ — Mg has lost 2 electrons (Mg → Mg²⁺ + 2e⁻), so Mg²⁺ is in the oxidized state.
- (b) Cu²⁺ — Cu has lost 2 electrons; Cu²⁺ is in the oxidized state.
- (c) O₂ — O₂ (ox. no. = 0) is more oxidized than O²⁻ (ox. no. = −2).
- (d) Cl₂ — Cl₂ (ox. no. = 0) is more oxidized than Cl⁻ (ox. no. = −1).
Pairs (a) and (b) show the ionic form (Mg²⁺, Cu²⁺) as the oxidized state.
Q.18
What is the action of heat on crystalline sodium carbonate (washing soda)?
Crystalline sodium carbonate is Na₂CO₃·10H₂O (washing soda).
On gentle heating — Efflorescence
Na₂CO₃·10H₂O → Na₂CO₃·H₂O + 9H₂O↑
It loses 9 molecules of water of crystallization.
On strong heating
Na₂CO₃·H₂O → Na₂CO₃ + H₂O↑
Forms anhydrous sodium carbonate (soda ash).
Anhydrous Na₂CO₃ does NOT decompose further on heating as it is thermally stable.
Q.19
GeCl₄ is more stable than GeCl₂, while PbCl₂ is more stable than PbCl₄. Explain.
This is explained by the Inert Pair Effect.
- In Group 14, as we go down the group, the ns² electrons become increasingly reluctant to participate in bond formation due to the poor shielding by d and f electrons (relativistic effects).
- Ge (Period 4): The 4s electrons are readily available for bonding → higher oxidation state (+4) is more stable. Hence GeCl₄ > GeCl₂.
- Pb (Period 6): The 6s² pair is highly stabilized (inert pair effect is maximum) and does not participate in bonding easily → lower oxidation state (+2) is more stable. Hence PbCl₂ > PbCl₄.
This trend shows the increasing stability of the lower oxidation state down Group 14.
Q.20
Write physical properties of diamond. Also, state its uses.
Physical Properties of Diamond:
- It is the hardest natural substance (hardness = 10 on Mohs scale).
- Has a very high melting point (~3550 °C).
- It is a poor conductor of electricity (all four valence electrons of C are used in covalent bonding — no free electrons).
- Has a high refractive index (2.42), giving it brilliance and sparkle.
- It is transparent and colorless.
- High density = 3.5 g/cm³.
Uses of Diamond:
- Used as a gemstone in jewellery.
- Used for cutting, drilling, and grinding hard materials (drill bits, cutting tools).
- Used for glass cutting.
- Used in precision scientific instruments.
Q.21
Ammonia is a good complexing agent. Explain.
- Ammonia (NH₃) has a lone pair of electrons on the nitrogen atom.
- It acts as a Lewis base / ligand and can donate this lone pair to transition metal ions (Lewis acids).
- It forms coordinate (dative) bonds with metal ions, creating stable complex compounds (coordination complexes).
- The N–H bonds are polar, making NH₃ a good donor of the lone pair.
Examples:
[Cu(NH₃)₄]²⁺ — Tetraamminecopper(II) complex (deep blue colour)
[Ag(NH₃)₂]⁺ — Diamminesilver(I) complex (used in Tollens' reagent)
Q.22
Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.
Given
P = 4.7 atm, V = 2.32 L, T = 32 °C = 32 + 273 = 305 K, R = 0.0821 L·atm/(mol·K)
Using Ideal Gas Equation: PV = nRT
n = PV / RT
n = (4.7 × 2.32) / (0.0821 × 305)
n = 10.904 / 25.04
∴ n ≈ 0.435 mol
Q.23
At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10 °C and volume of the gas is 640 mL?
Given
P₁ = 760 mm Hg, V₁ = 600 mL, T₁ = 25 + 273 = 298 K
V₂ = 640 mL, T₂ = 10 + 273 = 283 K, P₂ = ?
Using Combined Gas Law:
P₁V₁/T₁ = P₂V₂/T₂
P₂ = (P₁ × V₁ × T₂) / (T₁ × V₂)
P₂ = (760 × 600 × 283) / (298 × 640)
P₂ = 129,048,000 / 190,720
∴ P₂ ≈ 676.8 mm Hg
Q.24
How does a bicycle pump work?
A bicycle pump works on the basis of Boyle's Law (P ∝ 1/V at constant T).
- When the piston is pushed inward, the volume of the air inside the cylinder decreases.
- According to Boyle's Law, as volume decreases, pressure increases.
- This increased pressure forces air through the valve and into the tyre.
- When the piston is pulled outward, the volume increases and pressure decreases, allowing fresh air to enter the cylinder through the inlet valve.
Thus, a bicycle pump converts mechanical energy into pressure energy of air using Boyle's Law.
Q.25
Write resonance structures of H–COO⁻ and comment on their relative stability.
H–COO⁻ is the formate ion.
Structure I
O
‖
H–C
|
O⁻
⟺
Structure II
O⁻
|
H–C
‖
O
Comment on relative stability:
- Both resonance structures are equivalent (identical in bond lengths, energies, and charge distribution).
- Since they are mirror images of each other, they have equal stability.
- In the actual resonance hybrid, the negative charge is equally distributed on both oxygen atoms.
- Both C–O bonds are equivalent, with bond length intermediate between a C–O single bond and a C=O double bond.
Q.26
Identify primary, secondary, tertiary, and quaternary carbon in the following compounds.
(a) CH₃–C(CH₃)(CH₃)–CH(CH₃)–CH₂–CH₂–CH₃
CH₃
|
CH₃ – C – CH – CH₂ – CH₂ – CH₃
| |
CH₃ CH₃
| Carbon | Type | Reason |
|---|
| CH₃ (top branch), CH₃ (left), terminal CH₃'s | Primary (1°) | Bonded to only 1 other carbon |
| CH₂–CH₂ (in chain) | Secondary (2°) | Bonded to 2 other carbons |
| –CH– (with one CH₃ branch) | Tertiary (3°) | Bonded to 3 other carbons |
| –C– (central, no H; bonded to 4 C groups) | Quaternary (4°) | Bonded to 4 other carbons, no H attached |
(b) Cage structure (Adamantane — C₁₀H₁₆)
The diamond-shaped cage structure represents adamantane.
| Type | Count | Reason |
|---|
| Tertiary (3°) — CH | 4 carbon atoms | Each CH is bonded to 3 carbon atoms |
| Secondary (2°) — CH₂ | 6 carbon atoms | Each CH₂ is bonded to 2 carbon atoms |
There are no primary or quaternary carbons in adamantane.
Section – D
Long answer questions — Any THREE from Q.27 to Q.31 | [12 marks]
Q.27
Differentiate between: Isotopes and Isobars.
| Property | Isotopes | Isobars |
|---|
| Definition | Atoms of the same element with same atomic number but different mass numbers | Atoms of different elements with same mass number but different atomic numbers |
| Atomic number (Z) | Same | Different |
| Mass number (A) | Different | Same |
| No. of protons | Same | Different |
| No. of neutrons | Different | Different |
| No. of electrons | Same | Different |
| Chemical properties | Same (same electronic configuration) | Different |
| Physical properties | Different (different mass) | Different |
| Position in periodic table | Same position | Different positions |
| Examples | ¹H, ²H (deuterium), ³H (tritium)
¹²C and ¹⁴C | ¹⁴C and ¹⁴N (A = 14)
⁴⁰Ar, ⁴⁰Ca, ⁴⁰K (A = 40) |
Q.28
Dipole moment in case of BeF₂ is zero. Explain.
- BeF₂ (Beryllium fluoride) has sp hybridization of the central Be atom.
- This gives it a linear geometry with a F–Be–F bond angle of 180°.
- F is more electronegative than Be, so each Be–F bond is polar (bond dipole ≠ 0).
- However, the two Be–F bond dipoles are equal in magnitude but point in exactly opposite directions (180° apart).
- As a result, the two bond dipoles cancel each other out.
← (δ⁻)F ← Be → F(δ⁻) →
Bond dipoles cancel → Net μ = 0
Therefore, despite having polar bonds, BeF₂ is a non-polar molecule with dipole moment = 0 D.
Q.29
Write a short note on dipole moment.
Dipole Moment (μ) is a measure of the polarity of a molecule.
Definition: Dipole moment is defined as the product of the magnitude of charge (q) and the distance (d) between the two charges.
μ = q × d
- Unit: Debye (D) in CGS system; SI unit is C·m (coulomb·metre).
1 D = 3.336 × 10⁻³⁰ C·m
- It is a vector quantity, directed from the positive charge (δ⁺) to the negative charge (δ⁻).
- In a diatomic molecule: μ = 0 if bond is nonpolar (e.g., H₂, Cl₂), μ ≠ 0 if bond is polar (e.g., HCl, HF).
- Non-polar molecules — net dipole moment = 0 (e.g., CO₂, BF₃, CCl₄, BeF₂).
- Polar molecules — net dipole moment ≠ 0 (e.g., H₂O, NH₃, SO₂, HCl).
- Dipole moment helps determine the shape and geometry of molecules and predicts their polarity, solubility, and boiling point.
| Molecule | Shape | Dipole Moment |
|---|
| H₂O | Bent (V-shaped) | 1.85 D (polar) |
| CO₂ | Linear | 0 D (non-polar) |
| NH₃ | Pyramidal | 1.46 D (polar) |
| BF₃ | Trigonal planar | 0 D (non-polar) |
Q.30
Explain: (a) Reversible reaction (b) Rate of reaction
(a) Reversible Reaction:
- A reversible reaction is one that can proceed in both forward and reverse directions under the same conditions.
- In the forward reaction, reactants form products; in the reverse reaction, products reform reactants.
- The reaction does not go to completion. Instead, it reaches a state of dynamic equilibrium when the rate of forward reaction equals the rate of reverse reaction.
- Represented by a double arrow (⇌) symbol.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
H₂(g) + I₂(g) ⇌ 2HI(g)
- Conditions: The reaction must be carried out in a closed system at constant temperature.
(b) Rate of Reaction:
- The rate of reaction is defined as the change in concentration of reactants or products per unit time.
- For a reaction: A → B
Rate = −d[A]/dt = +d[B]/dt
- Unit: mol L⁻¹ s⁻¹ (mol per litre per second).
- Average rate: Change in concentration over a finite time interval: Δ[conc.]/Δt
- Instantaneous rate: Rate at a specific instant (slope of the tangent to the concentration–time graph).
- Factors affecting rate of reaction:
- Temperature (rate increases with temperature)
- Concentration of reactants
- Presence of a catalyst
- Pressure (for gaseous reactions)
- Surface area (for solid reactions)
Q.31
Draw structures of cis–trans isomers for the following:
(a) CH₃–CH₂–CH=CH–CH₂–CH₃ (Hex-3-ene)
cis-Hex-3-ene:
CH₃CH₂ CH₂CH₃
\ /
C = C
/ \
H H
Both C₂H₅ on same side
trans-Hex-3-ene:
CH₃CH₂ H
\ /
C = C
/ \
H CH₂CH₃
C₂H₅ on opposite sides
(b) CH₃–CH₂–C(CH₃)=CH–CH₃ (2-methylpent-2-ene)
cis: CH₂CH₃ and CH₃ on same side
CH₃CH₂ CH₃
\ /
C = C
/ \
CH₃ H
trans: CH₂CH₃ and CH₃ on opposite sides
CH₃CH₂ H
\ /
C = C
/ \
CH₃ CH₃
(c) C₂H₅–C(CH₃)=C(CH₃)–C₂H₅ (3,4-dimethylhex-3-ene)
cis: Both CH₃ on same side
CH₃ CH₃
| |
C₂H₅–C = C–C₂H₅
(same side)
trans: CH₃ groups on opposite sides
CH₃CH₂ CH₃
\ /
C = C
/ \
CH₃ CH₂CH₃
(d) CH₃–CH=CH–C₆H₅ (1-phenylprop-1-ene / β-methylstyrene)
cis: CH₃ and C₆H₅ on same side
CH₃ C₆H₅
\ /
C = C
/ \
H H
trans: CH₃ and C₆H₅ on opposite sides
CH₃ H
\ /
C = C
/ \
H C₆H₅