Which of the following statements regarding interstitial compounds is incorrect?

Which of the following statements regarding interstitial compounds is incorrect? +

Correct Answer

Incorrect statement: Interstitial compounds are always non-stoichiometric.

Detailed Explanation

Interstitial compounds are formed when small atoms such as H, B, C or N occupy the interstitial spaces (holes) in the crystal lattice of transition metals. Common examples include:

Fe₃C   |   TiC   |   VN   |   Fe₄N

Their important characteristics are:

• They are generally hard and rigid.
• They often have high melting points.
• They generally retain the metallic conductivity of the parent metal.
• They are usually formed by small atoms occupying interstitial sites in a metal lattice.
• They may be non-stoichiometric, but they are not necessarily always non-stoichiometric.

Therefore, saying that all interstitial compounds are necessarily non-stoichiometric is incorrect.

Why are they called Interstitial Compounds?

The small atoms do not normally replace metal atoms in the crystal lattice. Instead, they occupy the empty spaces between the metal atoms.

Metal lattice + small atom in interstitial site → Interstitial compound

For example, carbon atoms can occupy the spaces in the iron lattice to form iron carbide.

JEE / NEET Quick Revision

Interstitial compounds → small atoms in lattice holes → hard + high m.p. + metallic conductivity → often non-stoichiometric, but not always.

30 Related MCQs with Solutions

1. Interstitial compounds are formed when small atoms occupy:

A. Substitutional sites only

B. Interstitial sites

C. Nuclei of metal atoms

D. Valence shells only

Answer: B. Interstitial sites

2. Which of the following atoms commonly forms interstitial compounds?

A. Na

B. K

C. C

D. Cs

Answer: C. Carbon

3. Interstitial compounds are mainly associated with:

A. Alkali metals

B. Transition metals

C. Noble gases

D. Halogens

Answer: B. Transition metals

4. Which is an example of an interstitial compound?

A. NaCl

B. TiC

C. H₂O

D. CO₂

Answer: B. TiC

5. Interstitial compounds are generally:

A. Soft

B. Hard

C. Gaseous

D. Volatile liquids

Answer: B. Hard

6. Interstitial compounds generally have:

A. Very low melting points

B. High melting points

C. No melting point

D. Melting points below room temperature

Answer: B. High melting points

7. Which property is usually retained by interstitial compounds?

A. Metallic conductivity

B. Molecular volatility

C. Ionic conductivity only

D. Gas-like behaviour

Answer: A. Metallic conductivity

8. Which statement is incorrect?

A. Small atoms occupy lattice holes

B. They are generally hard

C. They always have fixed stoichiometry

D. They can retain metallic character

Answer: C.
Interstitial compounds can often show variable composition.

9. Which element can occupy interstitial positions in transition-metal lattices?

A. Hydrogen

B. Carbon

C. Nitrogen

D. All of these

Answer: D. All of these

10. The small atoms in interstitial compounds are generally:

A. Larger than metal atoms

B. Smaller than metal atoms

C. Equal in size to metal atoms

D. Always noble gases

Answer: B. Smaller than metal atoms

11. Which compound is associated with iron and carbon?

A. FeCl₃

B. Fe₃C

C. FeSO₄

D. Fe(NO₃)₃

Answer: B. Fe₃C

12. Which of the following is a carbide-type interstitial compound?

A. TiC

B. NaCl

C. MgO

D. CaCO₃

Answer: A. TiC

13. Interstitial compounds are generally formed by:

A. Large non-metals entering lattice sites

B. Small atoms entering lattice holes

C. Complete replacement of metal atoms

D. Noble gases entering the nucleus

Answer: B

14. Which is NOT a typical property of interstitial compounds?

A. High hardness

B. High melting point

C. Metallic conductivity

D. High volatility

Answer: D. High volatility

15. Interstitial compounds may be:

A. Only stoichiometric

B. Only non-stoichiometric

C. Stoichiometric or non-stoichiometric

D. Always molecular

Answer: C

16. The term "interstitial" refers to:

A. Surface of a crystal

B. Space between atoms in a lattice

C. Nucleus of an atom

D. Electron cloud

Answer: B

17. Which element is particularly small and can enter metal lattice holes?

A. H

B. Cs

C. Rb

D. Ba

Answer: A. Hydrogen

18. Interstitial compounds are usually:

A. Non-metallic insulators

B. Metallic or metal-like in conductivity

C. Molecular gases

D. Ionic liquids

Answer: B

19. Which statement about stoichiometry is correct?

A. All interstitial compounds must be non-stoichiometric

B. Some interstitial compounds have definite stoichiometry

C. None have definite composition

D. Stoichiometry is impossible

Answer: B

20. Which pair contains interstitial compounds?

A. TiC and VN

B. NaCl and KBr

C. H₂O and NH₃

D. CO₂ and SO₂

Answer: A. TiC and VN

21. Interstitial atoms mainly cause:

A. Weakening of the metallic lattice

B. Distortion and strengthening of the lattice

C. Complete destruction of lattice

D. Formation of a gas

Answer: B

22. Which is an important industrial feature of many interstitial compounds?

A. High hardness

B. Extreme volatility

C. Low thermal stability

D. Low melting point

Answer: A. High hardness

23. Carbon in iron lattice can form:

A. Fe₃C

B. FeCl₃

C. FeO₄

D. Fe(NO₃)₂

Answer: A. Fe₃C

24. Which small atoms are commonly involved in interstitial compounds?

A. H, B, C, N

B. Na, K, Rb, Cs

C. Cl, Br, I, At

D. Xe, Kr, Ar, Ne

Answer: A

25. Which statement is true?

A. Interstitial compounds are always ionic

B. Interstitial compounds are generally associated with metallic lattices

C. Interstitial compounds are always molecular

D. They contain no metals

Answer: B

26. Which property results from strong interaction between the interstitial atom and metal lattice?

A. Increased hardness

B. Increased volatility

C. Gas formation

D. Loss of all conductivity

Answer: A

27. Which statement is incorrect regarding interstitial compounds?

A. Small atoms occupy lattice holes

B. They are generally hard

C. They are always non-stoichiometric

D. They may have high melting points

Answer: C. They are always non-stoichiometric

28. VN is an example of an interstitial compound involving:

A. Vanadium and nitrogen

B. Vanadium and sodium

C. Vanadium and neon

D. Vanadium and chlorine

Answer: A. Vanadium and nitrogen

29. The electrical conductivity of many interstitial compounds is:

A. Metallic

B. Always zero

C. Only ionic

D. Molecular

Answer: A. Metallic

30. The best description of interstitial compounds is:

A. Small atoms occupy interstitial sites of a metal lattice

B. Large atoms replace all metal atoms

C. Only non-metals form them

D. They are always gaseous

Answer: A

Advertisement

Derive Raoult's law for a solution containing volatile components.

📘 Derivation of Raoult's Law for a Solution of Volatile Components

🔵 Step 1: Consider a Binary Solution

Consider a binary solution containing two volatile components, A and B.

x_A + x_B = 1

Let:

• P_A° = Vapour pressure of pure A
• P_B° = Vapour pressure of pure B
• x_A = Mole fraction of A in liquid phase
• x_B = Mole fraction of B in liquid phase
• P_A = Partial vapour pressure of A
• P_B = Partial vapour pressure of B

🟢 Step 2: Vapour Pressure of Component A

According to Raoult's law, the partial vapour pressure of a volatile component is directly proportional to its mole fraction in the solution.

P_A ∝ x_A

Introducing the proportionality constant, which is the vapour pressure of pure A:

P_A = x_AP_A°

🟠 Step 3: Vapour Pressure of Component B

Similarly, for component B:

P_B = x_BP_B°

🟣 Step 4: Total Vapour Pressure

The total vapour pressure of the solution is the sum of the partial vapour pressures of A and B.

P = P_A + P_B

Substituting the expressions:

P = x_AP_A° + x_BP_B°

⭐ Raoult's Law for a Binary Solution

P = x_AP_A° + x_BP_B°

🔴 Step 5: Special Case

If component B is absent, then x_A = 1. Therefore:

P = P_A°

Thus, the vapour pressure of pure A is obtained when the mole fraction of A is unity.

🎬 3D-Style Raoult's Law Animation

Volatile molecules A and B escape from the liquid surface into the vapour phase.

A B A B A B

A → P_A = x_AP_A°

B → P_B = x_BP_B°

Total → P = P_A + P_B

📌 General Form

For a solution containing several volatile components:

P_total = Σ x_iP_i°

where x_i is the mole fraction and P_i° is the vapour pressure of the pure component.

✅ Final Result

P_A = x_AP_A°

P_B = x_BP_B°

P_total = x_AP_A° + x_BP_B°

🎯 Exam Point

Raoult's law: P_i = x_iP_i°

For a binary volatile solution: P_total = x_AP_A° + x_BP_B°

20 g of a non-volatile solute is dissolved in 180 g of water. The vapour pressure of the solution is 23.5 mm Hg while that of pure water is 24 mm Hg. Calculate the molar mass of the solute.

📘 Complete Step-by-Step Solution

📌 Given Data

Mass of solute = 20 g

Mass of water = 180 g

Vapour pressure of pure water, P° = 24 mm Hg

Vapour pressure of solution, P = 23.5 mm Hg

Molar mass of water = 18 g mol⁻¹

🔵 Step 1: Apply Raoult's Law

For a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

(P° − P) / P° = X_solute

X_solute = (24 − 23.5) / 24

= 0.5 / 24

= 0.020833

🟢 Step 2: Calculate Moles of Water

Moles of water = Mass / Molar mass

= 180 / 18

= 10 mol

🟠 Step 3: Find Moles of Solute

We know:

X_solute = n_solute / (n_solute + n_water)

Let the moles of solute be n.

0.020833 = n / (n + 10)

Cross-multiplying:

0.020833(n + 10) = n

0.020833n + 0.20833 = n

0.20833 = 0.979167n

n ≈ 0.2128 mol

🟣 Step 4: Calculate Molar Mass of Solute

Molar mass = Mass / Number of moles

M = 20 / 0.2128

M ≈ 94.0 g mol⁻¹

🎬 3D-Style Vapour Pressure Animation

The non-volatile solute reduces the escaping tendency of water molecules.

H₂O H₂O Solute H₂O H₂O

Pure water: 24 mm Hg

Solution: 23.5 mm Hg

Lowering = 0.5 mm Hg

✅ Final Answer

Molar Mass of Solute ≈ 94 g mol⁻¹

🎯 Important Formula

(P° − P) / P° = X_solute

X_solute = n_solute / (n_solute + n_solvent)

Molar Mass = Mass / Moles

A solution is prepared by dissolving 10 g of urea in 90 g of water. Calculate the mole fraction of urea.

📘 Complete Step-by-Step Solution

📌 Given Data

Mass of urea = 10 g

Mass of water = 90 g

Molar mass of urea = 60 g mol⁻¹

Molar mass of H₂O = 18 g mol⁻¹

🔵 Step 1: Calculate Moles of Urea

Moles = Mass ÷ Molar Mass

Moles of urea = 10 ÷ 60

= 0.1667 mol

🟢 Step 2: Calculate Moles of Water

Moles of water = 90 ÷ 18

= 5 mol

🟠 Step 3: Calculate Total Moles

Total moles = Moles of urea + Moles of water

= 0.1667 + 5

= 5.1667 mol

🟣 Step 4: Calculate Mole Fraction of Urea

Mole fraction of urea is the ratio of moles of urea to the total moles of the solution.

::contentReference[oaicite:0]{index=0}

χ_urea = Moles of urea ÷ Total moles

= 0.1667 ÷ 5.1667

≈ 0.03226

🎬 3D-Style Mole Fraction Animation

Urea molecules are present in a larger number of water molecules.

Urea H₂O H₂O H₂O H₂O H₂O

Urea = 0.1667 mol
Water = 5 mol

Total = 5.1667 mol

✅ Final Answer

Mole fraction of urea ≈ 0.0323

🎯 Important Formula

χ_solute = Moles of solute ÷ Total moles of solution

χ_urea + χ_water = 1

Calculate the mole fraction of ethanol and water in a solution containing 46 g ethanol and 54 g water.

📘 Complete Step-by-Step Solution

📌 Given Data

Mass of ethanol = 46 g

Mass of water = 54 g

Molar mass of ethanol = 46 g mol⁻¹

Molar mass of water = 18 g mol⁻¹

🔵 Step 1: Calculate Moles of Ethanol

Moles = Mass ÷ Molar Mass

Moles of ethanol = 46 ÷ 46

= 1 mol

✅ Ethanol = 1 mol

🟢 Step 2: Calculate Moles of Water

Moles of water = 54 ÷ 18

= 3 mol

✅ Water = 3 mol

🟠 Step 3: Calculate Total Number of Moles

Total moles = Moles of ethanol + Moles of water

= 1 + 3

= 4 mol

🟣 Step 4: Mole Fraction of Ethanol

The mole fraction of a component is:

::contentReference[oaicite:0]{index=0}

χ_ethanol = Moles of ethanol ÷ Total moles

= 1 ÷ 4

= 0.25

✅ Mole fraction of ethanol = 0.25

🔴 Step 5: Mole Fraction of Water

χ_water = Moles of water ÷ Total moles

= 3 ÷ 4

= 0.75

✅ Mole fraction of water = 0.75

🎬 3D-Style Mole Fraction Animation

The solution contains 1 mole of ethanol and 3 moles of water.

EtOH H₂O H₂O H₂O

1 mol ethanol + 3 mol water

Total = 4 mol

χ_ethanol = 1/4 = 0.25
χ_water = 3/4 = 0.75

🟢 Step 6: Verify the Answer

χ_ethanol + χ_water

= 0.25 + 0.75

= 1

✅ Sum of mole fractions = 1

✅ Final Answer

Mole fraction of ethanol = 0.25

Mole fraction of water = 0.75

🎯 Quick Revision

46 g ethanol ÷ 46 = 1 mol
54 g water ÷ 18 = 3 mol
Total moles = 4 mol
χ_ethanol = 1/4 = 0.25
χ_water = 3/4 = 0.75

Calculate the molality of a solution containing 18 g of glucose dissolved in 180 g of water.

📘 Complete Step-by-Step Solution

🔵 Step 1: Identify the Given Data

Mass of glucose = 18 g

Mass of water (solvent) = 180 g

Molar mass of glucose = 180 g mol⁻¹

🟢 Step 2: Calculate Moles of Glucose

The formula for number of moles is:

Moles = Mass ÷ Molar Mass

Moles of glucose

= 18 ÷ 180

= 0.10 mol

🟠 Step 3: Convert Mass of Solvent into kg

Molality uses the mass of solvent in kilograms.

180 g = 180 ÷ 1000 kg

= 0.180 kg

🟣 Step 4: Apply the Molality Formula

Molality (m)

= Moles of solute ÷ Mass of solvent (kg)

m = 0.10 ÷ 0.180

= 0.5556 mol kg⁻¹

🎬 3D-Style Solution Animation

Glucose molecules are dissolved in water.

Glu H₂O H₂O Glu H₂O H₂O

18 g glucose → 0.10 mol glucose

180 g water → 0.180 kg solvent

✅ Final Answer

Molality = 0.5556 mol kg⁻¹

Approximately:
m = 0.556 mol kg⁻¹

🎯 Important Formula

Molality (m) = Moles of solute ÷ Mass of solvent in kg

m = n / W(kg)

Define molarity, molality, mole fraction and mass percentage. Derive the relationship between molarity and molality.

📘 Complete Explanation & Derivation

🔵 1. Molarity (M)

Molarity is defined as the number of moles of solute present in one litre (1000 mL) of solution.

M = Moles of solute ÷ Volume of solution in litre

M = n / V

Molarity depends on the volume of solution, so it can change with temperature.

Unit: mol L⁻¹ or M

🟢 2. Molality (m)

Molality is defined as the number of moles of solute present in 1 kg of solvent.

m = Moles of solute ÷ Mass of solvent in kg

m = n / W

Unit: mol kg⁻¹ or m

Molality does not depend on volume and is therefore essentially independent of temperature.

🟣 3. Mole Fraction (χ)

Mole fraction is the ratio of the number of moles of one component to the total number of moles of all components in the solution.

χ_A = n_A ÷ (n_A + n_B + ...)

For a binary solution:

χ_A + χ_B = 1

Mole fraction has no unit.

🟠 4. Mass Percentage

Mass percentage represents the mass of solute present in 100 g of solution.

Mass % = (Mass of solute ÷ Mass of solution) × 100

Mass % is expressed as a percentage (%).

📊 Quick Comparison

Molarity

moles / L solution

Unit: M

Molality

moles / kg solvent

Unit: m

Mole Fraction

moles of component / total moles

No unit

Mass %

mass solute / mass solution × 100

Unit: %

🔴 Derivation: Relationship Between Molarity and Molality

Let:

Molarity = M
Molality = m
Molar mass of solute = M_B g mol⁻¹
Mass percentage of solution = w%
Density of solution = d g mL⁻¹

Step 1: Consider 1 L of solution

For 1 litre (1000 mL) of solution:

Mass of solution = Volume × Density

= 1000 × d

= 1000d g

Step 2: Calculate mass of solute

If the solution contains w% solute:

Mass of solute = (w/100) × 1000d

= 10dw g

Step 3: Calculate mass of solvent

Mass of solvent = Mass of solution − Mass of solute

= 1000d − 10dw

= 10d(100 − w) g

Step 4: Calculate moles of solute

Moles of solute = Mass of solute ÷ Molar mass

= 10dw ÷ M_B

But for 1 L of solution, the number of moles of solute is equal to molarity M.

M = 10dw / M_B

Step 5: Calculate molality

m = Moles of solute ÷ Mass of solvent in kg

= (10dw / M_B) ÷ [10d(100 − w) / 1000]

Therefore:

m = 1000M ÷ [1000d − M M_B]

⭐ Final Relationship

m = 1000M / [1000d − M M_B]

📌 Common Form Using Mass Percentage

Since:

M = 10dw / M_B

The relationship can also be written as:

M = 1000d m ÷ [1000 + mM_B]

where d is density in g mL⁻¹ and M_B is molar mass of solute.

🎬 3D-Style Concentration Animation

Solute particles are distributed throughout the solvent.

Solute Solvent Solute Solvent Solvent Solute

Molarity → Volume of Solution

Molality → Mass of Solvent

Mole Fraction → Number of Moles

Mass % → Mass of Solution

✅ Final Summary

M = Moles of solute / Volume of solution (L)

m = Moles of solute / Mass of solvent (kg)

χ = Moles of component / Total moles

Mass % = (Mass of solute / Mass of solution) × 100

m = 1000M / [1000d − MM_B]

🎯 Exam Point

Molarity depends on volume and can change with temperature.
Molality depends on mass of solvent and is independent of temperature.
Mole fraction has no unit.
Mass percentage is expressed as %.

Class 12 Chemistry – Chapter 2: Solutions

 

Class 12 Chemistry – Chapter 2: Solutions

Competitive / JEE Level – Questions Only

1. Define molarity, molality, mole fraction and mass percentage. Derive the relationship between molarity and molality.
2. Calculate the molality of a solution containing 18 g of glucose dissolved in 180 g of water.
3. A solution contains 5.85 g of NaCl in 500 g of water. Calculate its molality.
4. Calculate the mole fraction of ethanol and water in a solution containing 46 g ethanol and 54 g water.
5. A solution is prepared by dissolving 10 g of urea in 90 g of water. Calculate the mole fraction of urea.
6. 20 g of a non-volatile solute is dissolved in 180 g of water. The vapour pressure of the solution is 23.5 mm Hg while that of pure water is 24 mm Hg. Calculate the molar mass of the solute.
7. Derive Raoult's law for a solution containing volatile components.
8. Explain the difference between ideal and non-ideal solutions with suitable examples.
9. A solution containing two volatile liquids A and B obeys Raoult's law. Derive the expression for its total vapour pressure.
10. 10 g of a non-volatile solute is dissolved in 90 g of water. Calculate the relative lowering of vapour pressure if the molar mass of the solute is 100 g mol⁻¹.
11. Explain positive and negative deviations from Raoult's law.
12. What is an azeotrope? Explain the formation of minimum-boiling and maximum-boiling azeotropes.
13. Calculate the elevation in boiling point when 18 g glucose is dissolved in 500 g water.
    (Kb​=0.52 Kkgmol−1)
14. A solution containing 5 g of a non-electrolyte solute in 100 g water boils at 100.52∘C. Calculate the molar mass of the solute.
    (Kb​=0.52 Kkgmol−1)
15. Calculate the depression in freezing point when 9 g glucose is dissolved in 100 g water.
    (Kf​=1.86 Kkgmol−1)
16. A solution freezes at −0.372∘C. Calculate the molality of the solution.
    (Kf​=1.86 Kkgmol−1)
17. Explain the phenomenon of osmosis and define osmotic pressure.
18. Derive the equation for osmotic pressure of a dilute solution.
19. A solution contains 2 g of a polymer dissolved in 100 mL solution. Its osmotic pressure at 27°C is 0.246 atm. Calculate the molar mass of the polymer.
20. Calculate the osmotic pressure of a 0.1 M glucose solution at 27°C.
    (R=0.0821 LatmK−1mol−1)
21. Why is osmotic pressure preferred over other colligative properties for determining the molar mass of proteins and polymers?
22. Explain reverse osmosis and mention its practical application.
23. What are colligative properties? Explain why they depend only on the number of solute particles.
24. Derive the expression for depression in freezing point.
25. Derive the expression for elevation in boiling point.
26. A solution containing 1.8 g of glucose in 100 g water shows a depression in freezing point of 0.186∘C. Calculate the molar mass of glucose.
27. 1 g of a non-electrolyte solute dissolved in 100 g benzene produces a depression in freezing point of 0.5 K. Calculate the molar mass of the solute.
    (Kf​ for benzene=5.12 Kkgmol−1)
28. Explain abnormal molar mass with suitable examples.
29. Define the van't Hoff factor and derive its relation with observed and calculated colligative properties.
30. A 0.1 molal solution of CaCl2​ shows 80% ionisation. Calculate its effective molality.
31. Calculate the van't Hoff factor for Na2​SO4​ if its degree of dissociation is 60%.
32. A solution of K2​SO4​ is 50% dissociated. Calculate its van't Hoff factor.
33. A solute undergoes association to form dimers. Derive the expression for van't Hoff factor in terms of degree of association.
34. A 0.1 molal solution of an electrolyte has a freezing point depression of 0.279 K. Calculate its van't Hoff factor.
    (Kf​=1.86 Kkgmol−1)
35. A compound AB2​ dissociates into three ions with 70% dissociation. Calculate its van't Hoff factor.
36. A solute associates to form trimers. If the degree of association is 60%, calculate the van't Hoff factor.
37. Derive the relation between molar mass and elevation in boiling point.
38. Derive the relation between molar mass and depression in freezing point.
39. Explain why the addition of a non-volatile solute lowers the vapour pressure of a solvent.
40. JEE Level: A solution contains equal masses of glucose and sucrose dissolved in equal masses of water. Compare their osmotic pressures at the same temperature.
41. JEE Level: Two solutions have the same osmotic pressure at the same temperature. What relationship exists between their molar concentrations?
42. JEE Level: 0.1 M NaCl and 0.1 M glucose solutions are compared at the same temperature. Which has greater osmotic pressure? Explain using van't Hoff factor.
43. JEE Level: A solution contains NaCl with 90% ionisation. Calculate the van't Hoff factor and compare its freezing point depression with that of an ideal 0.1 m glucose solution.
44. JEE Level: A 1 g sample of a polymer is dissolved in enough water to make 100 mL solution. At 27°C, its osmotic pressure is 0.0821 atm. Calculate its molar mass.
45. JEE Level: A solution containing a non-volatile solute has a vapour pressure of 22.4 mm Hg at 25°C, while pure solvent has vapour pressure 24 mm Hg. Calculate the mole fraction of the solute. 

CBSE Class 10 Science Sample Paper 10 - 2026-27

www.careerinformationportal.in

APNA CAREER PORTAL

www.careerinformationportal.in

CBSE Sample Practice Paper — 10  (Session 2026–27)

Class: X (10th)

Subject: Science — Code: 086

Time: 3 Hours

Max. Marks: 80

Student Name: ______________________

Roll No.: __________

Section: __________

General Instructions:

1. This question paper is divided into three parts — Part A (Chemistry, 25 Marks), Part B (Biology, 30 Marks) and Part C (Physics, 25 Marks).

2. Each part is further divided into four sections: Section 1 — Very Short Answer (including objective-type and Assertion–Reason questions), Section 2 — Short Answer, Section 3 — Long Answer, and Section 4 — Case/Source-based integrated question.

3. Internal choices have been provided in approximately one-third of the questions across the paper. Attempt only one option where a choice is given.

4. Draw neat, labelled diagrams wherever necessary.

5. Write balanced chemical equations wherever applicable.

6. Marks for each question are indicated against it.

PART A — CHEMISTRY  (25 Marks)

Section 1 — Very Short Answer  (1 Mark each) — Q1 to Q8

Q1. Name the type of reaction in which a more reactive element takes the place of a less reactive element in a compound. 1

Q2. Write the chemical formula of caustic soda. 1

Q3. State the colour change observed when a few drops of phenolphthalein are added to a basic solution. 1

Q4. Name the functional group present in aldehydes. 1

Q5. Identify the substance that gets oxidised in the reaction: ZnO + C → Zn + CO. 1

Q6. Write the common name and chemical formula of washing soda. 1

Q7. Assertion (A): Copper vessels develop a green coating when exposed to moist air for a long time.
Reason (R): Copper reacts slowly with moisture, carbon dioxide, and oxygen present in air to form a green basic copper carbonate layer on its surface. 1

(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is NOT the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Q8. Assertion (A): Diamond does not conduct electricity, while graphite does.
Reason (R): In diamond, each carbon atom is bonded to four other carbon atoms with no free electrons, whereas in graphite, each carbon atom is bonded to only three others, leaving one free electron per atom. 1

(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is NOT the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Section 2 — Short Answer  (2 Marks each) — Q9 to Q12

Q9. Write a balanced chemical equation for the reaction of iron with steam. State the type of reaction involved. 2

Q10. What is meant by a "precipitation reaction"? Give one example with a balanced chemical equation. 2

Q11. State any two reasons why gold and platinum are used for making jewellery rather than more reactive metals. 2

— OR —

Explain, with reason, why the decomposition of silver chloride in sunlight is considered a photochemical reaction. 2

Q12. Draw the electron dot structure of ethyne (C₂H₂), and state the type of bond present between the two carbon atoms. 2

Section 3 — Long Answer  (5 Marks) — Q13

Q13. (a) Describe the process of obtaining metals of medium reactivity (like iron and zinc) from their oxides through reduction with carbon, with a suitable example. (3)
(b) Why must metal ores be first converted to oxides (through roasting or calcination) before reduction with carbon? (2) 5

— OR —

(a) Define homologous series. State any two characteristic features of a homologous series with reference to alkanes. (3)
(b) Why does the melting and boiling point of carbon compounds in a homologous series generally increase with increasing molecular mass? (2)

Section 4 — Case-based Question  (4 Marks) — Q14

Q14. Read the following passage and answer the questions that follow.4

Most metals occur in nature combined with other elements in the form of compounds called minerals, and a mineral from which a metal can be profitably extracted is called an ore. The method used to extract a metal from its ore depends largely on the metal's position in the reactivity series. Highly reactive metals require electrolytic reduction, moderately reactive metals are usually extracted by reduction with carbon, and the least reactive metals often occur in their free (native) state and require little or no chemical processing to obtain the pure metal.

14.1 Define the term "ore". (1)

14.2 Name one metal that is typically extracted by electrolytic reduction because of its high reactivity. (1)

14.3 Explain why the method of extraction of a metal depends on its position in the reactivity series. (2)

--- Continued on next page ---

Page 1 of 5  |  www.careerinformationportal.in

www.careerinformationportal.in

PART B — BIOLOGY  (30 Marks)

Section 1 — Very Short Answer  (1 Mark each) — Q15 to Q22

Q15. Name the hormone secreted by the thyroid gland, and state the mineral required for its synthesis. 1

Q16. Define "tropic movement" in plants. 1

Q17. Name the process by which pollen grains are transferred from the anther to the stigma of a flower. 1

Q18. What is the term used for the physical and chemical features that are transmitted from parents to offspring? 1

Q19. In pea plants, a cross between a plant with round, yellow seeds (RrYy) and a plant with wrinkled, green seeds (rryy) is called a: 1

Q20. Name one ozone-depleting substance commonly released from refrigerators and air-conditioners. 1

Q21. Assertion (A): Phloem tissue can transport food materials both upward and downward in a plant.
Reason (R): Unlike xylem, phloem transport of food (translocation) occurs using energy from ATP, and can move materials in the direction required by different parts of the plant. 1

(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is NOT the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Q22. Assertion (A): A person who has undergone vasectomy can no longer produce sperm.
Reason (R): Vasectomy involves cutting and sealing the vas deferens, which blocks the passage of sperm from the testes but does not stop sperm production itself. 1

(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is NOT the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Section 2 — Short Answer  (2 Marks each) — Q23 to Q26

Q23. Name the hormone responsible for regulating the level of calcium and phosphate in the blood, and name the gland that secretes it. 2

Q24. Differentiate between spore formation in Rhizopus (bread mould) and budding in Hydra as modes of asexual reproduction. 2

— OR —

Explain why a large number of sperms are produced by a healthy human male, even though only one is needed for fertilisation. 2

Q25. What is the role of the gastric glands present in the stomach lining during digestion? 2

Q26. Name any two sexually transmitted infections, and state one way of preventing their spread. 2

Section 3 — Long Answer  (5 Marks each) — Q27 to Q28

Q27. (a) With reference to the human respiratory system, describe the pathway of air from the nostrils to the alveoli, naming each structure in sequence. (3)
(b) Why are alveoli well-suited for efficient exchange of gases? (2) 5

— OR —

(a) Describe the structure of the human excretory system, explaining the roles of the kidney, ureter, urinary bladder and urethra. (3)
(b) What is the role of ADH (antidiuretic hormone) in regulating the water content of urine? (2)

Q28. (a) Explain how a cross between a pure-breeding round-seeded (RR) pea plant and a pure-breeding wrinkled-seeded (rr) pea plant results in an entirely round-seeded F1 generation. What does this tell us about dominance? (3)
(b) If the F1 plants are self-pollinated, predict the phenotypic ratio expected in the F2 generation. (2) 5

— OR —

(a) Describe the mechanism of coordination between the nervous system and the endocrine system, using the example of the body's response to a sudden fright. (3)
(b) Why is hormonal (chemical) coordination generally slower but longer-lasting compared to nervous coordination? (2)

Section 4 — Case-based Question  (4 Marks) — Q29

Q29. Read the following passage and answer the questions that follow.4

A stable ecosystem depends on a balanced flow of energy and nutrients between its living and non-living components. Producers, such as green plants, capture solar energy and convert it into chemical energy, which is then passed on to various consumers through food chains. Decomposers, including bacteria and fungi, play an equally important role by breaking down the remains of dead organisms and waste products, releasing nutrients back into the soil so that they can be reused by producers, thereby completing the natural nutrient cycle.

29.1 Name the group of organisms that capture solar energy and convert it into chemical energy. (1)

29.2 What would happen to the nutrient cycle if decomposers were absent from an ecosystem? (1)

29.3 Explain, with an example, how decomposers help maintain the fertility of soil. (2)

--- Continued on next page ---

Page 2 of 5  |  www.careerinformationportal.in

www.careerinformationportal.in

PART C — PHYSICS  (25 Marks)

Section 1 — Very Short Answer  (1 Mark each) — Q30 to Q37

Q30. Define the "principal focus" of a concave mirror. 1

Q31. Name the phenomenon responsible for the sun being visible for a short while even after it has actually set below the horizon. 1

Q32. State one cause of presbyopia. 1

Q33. Write the formula relating electric power, resistance, and current. 1

Q34. Calculate the potential difference required to pass a current of 2 A through a resistor of 5 Ω. 1

Q35. Name the device used to detect the presence of a small current in a circuit, based on the deflection of a magnetic needle. 1

Q36. Assertion (A): A convex mirror always forms a virtual, erect and diminished image of an object, regardless of the object's distance from the mirror.
Reason (R): A convex mirror is a diverging mirror, so the reflected rays always appear to meet behind the mirror, no matter where the object is placed. 1

(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is NOT the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Q37. Assertion (A): Electrical appliances with a metallic body are always earthed.
Reason (R): Earthing provides a low-resistance path for leakage current to flow safely into the ground, protecting the user from electric shock in case of a fault. 1

(A) Both A and R are true, and R is the correct explanation of A. (B) Both A and R are true, but R is NOT the correct explanation of A. (C) A is true, but R is false. (D) A is false, but R is true.

Section 2 — Short Answer  (2 Marks each) — Q38 to Q41

Q38. Define "refractive index" of a medium with respect to air. 2

Q39. Why do we see a spectrum of colours when white light passes through a glass prism? 2

— OR —

An object is placed at the focus of a convex lens. State the nature and position of the image formed. 2

Q40. Calculate the electrical energy consumed by an appliance of power 1000 W used for 3 hours, expressing your answer in kilowatt-hours. 2

Q41. State any two factors on which the strength of the magnetic field produced at the centre of a current-carrying circular loop depends. 2

Section 3 — Long Answer  (5 Marks) — Q42

Q42. (a) With a labelled ray diagram, explain the formation of the image of an object placed between the pole and the focus of a concave mirror. (3)
(b) An object is placed 15 cm from a convex mirror of focal length 10 cm. Using the mirror formula, find the position of the image formed. (2) 5

— OR —

(a) With the help of a circuit diagram, explain how you would experimentally verify Ohm's law in a laboratory. (3)
(b) A wire of resistance 6 Ω is bent in the form of a circle. Calculate the effective resistance between two points diametrically opposite to each other on the circle. (2)

Section 4 — Case-based Question  (4 Marks) — Q43

Q43. Read the following passage and answer the questions that follow.4

Every current-carrying conductor is surrounded by a magnetic field, whose strength and pattern depend on the shape of the conductor and the amount of current flowing through it. This basic relationship between electricity and magnetism, discovered by Hans Christian Oersted, forms the basis for a wide range of devices that use magnetic fields produced by electric currents, from simple electromagnets used to sort scrap metal, to more complex instruments used in hospitals and industries.

43.1 Who discovered that a current-carrying conductor produces a magnetic field around it? (1)

43.2 Name one everyday device (other than a motor or generator) that makes use of an electromagnet. (1)

43.3 State two factors on which the strength of the magnetic field produced by a straight current-carrying conductor depends. (2)

--- End of Question Paper --- All the best! ---

Page 3 of 5  |  www.careerinformationportal.in

www.careerinformationportal.in

APNA CAREER PORTAL

www.careerinformationportal.in

Design of Question Paper — Blueprint & Competency Analysis

1. Part-wise & Section-wise Blueprint

Part	Section 1 (VSA + AR)	Section 2 (SA)	Section 3 (LA)	Section 4 (Case-based)	Total
A — Chemistry	8 × 1 = 8	4 × 2 = 8	1 × 5 = 5	1 × 4 = 4	25
B — Biology	8 × 1 = 8	4 × 2 = 8	2 × 5 = 10	1 × 4 = 4	30
C — Physics	8 × 1 = 8	4 × 2 = 8	1 × 5 = 5	1 × 4 = 4	25
Total	24	24	20	12	80

2. Unit-wise Weightage (as per given specification)

Unit	Topic Focus	Marks
Unit I — Chemical Substances	Chemical reactions/equations, acids/bases/salts, metals & non-metals, carbon compounds	25
Unit II — World of Living	Life processes, control & coordination, reproduction, heredity	25
Unit V — Natural Resources	Ecosystems, energy flow, decomposers, environmental issues	5
Unit III — Natural Phenomena	Reflection/refraction, human eye and its defects	12
Unit IV — Effects of Current	Ohm's law, resistance, magnetic effects (excluding motor/generator/EMI)	13
Total		80

3. Competency-Level Distribution

Competency Level	Approx. Marks	Approx. %
Knowledge & Understanding (state, define, list, describe, identify)	40	50%
Application (calculate, illustrate, show, explain, distinguish)	24	30%
Analyse, Evaluate & Create (interpret, analyse, compare, contrast, examine)	16	20%
Total	80	100%

4. Internal Choice Summary

Part A: Internal choice provided in Q11 (SA) and Q13 (LA) — 2 of 5 non-VSA questions (~40%).

Part B: Internal choice provided in Q24 (SA), Q27 (LA) and Q28 (LA) — 3 of 6 non-VSA questions (50%).

Part C: Internal choice provided in Q39 (SA) and Q42 (LA) — 2 of 5 non-VSA questions (~40%).

Overall, internal choices are provided in roughly one-third to one-half of the non-VSA questions across the paper, consistent with the CBSE 2026-27 design intent.

5. Exclusions Confirmed

✔ No questions from Periodic Classification of Elements.

✔ No questions from Evolution (speciation, fossils) — only Heredity concepts included.

✔ No questions on Electric Motor, Electromagnetic Induction, or Electric Generator.

✔ No content drawn from "Do You Know" / informational boxes in NCERT textbooks.

Page 4 of 5 (Blueprint & Analysis)  |  www.careerinformationportal.in

www.careerinformationportal.in

Answer Key & Marking Scheme (Key Points)

PART A — CHEMISTRY

Q1. Displacement reaction.

Q2. NaOH.

Q3. Colourless solution turns pink.

Q4. –CHO.

Q5. Carbon (C) gets oxidised to CO.

Q6. Washing soda; Na₂CO₃·10H₂O.

Q7. (A)   Q8. (A)

Q9. 3Fe + 4H₂O → Fe₃O₄ + 4H₂; a combination-type redox reaction (iron is oxidised, steam provides oxidising conditions).

Q10. A reaction in which an insoluble solid (precipitate) is formed when two solutions are mixed; e.g., BaCl₂ + Na₂SO₄ → BaSO₄↓ + 2NaCl.

Q11. Gold and platinum are highly unreactive, so jewellery made from them does not tarnish or corrode; they are also lustrous and malleable, making them suitable for intricate designs (any two points). OR: The decomposition of silver chloride occurs specifically due to exposure to light (sunlight), not heat; reactions that occur due to the absorption of light energy are called photochemical reactions.

Q12. Structure showing a C≡C triple bond, with each carbon also bonded to one hydrogen atom (HC≡CH); a triple covalent bond is present between the two carbon atoms.

Q13. (a) Metal oxides of medium-reactivity metals are reduced by heating with carbon, which is more reactive than the metal and takes away the oxygen; e.g., ZnO + C → Zn + CO. (b) Ores usually occur as sulphides or carbonates, not oxides; roasting (heating sulphide ores in air) or calcination (heating carbonate ores in limited air) converts them into oxides, which can then be more easily reduced by carbon. OR: A homologous series is a series of carbon compounds having the same general formula and similar chemical properties, where each successive member differs from the previous one by a –CH₂– unit; features: same general formula (e.g., CnH2n+2 for alkanes), same functional group, and gradual change in physical properties. (b) As molecular mass increases, the strength of intermolecular forces (van der Waals forces) between molecules increases, requiring more energy (higher temperature) to overcome them and change state.

Q14. (i) A mineral from which a metal can be extracted profitably and conveniently is called an ore. (ii) Sodium (or potassium/calcium/magnesium — any highly reactive metal). (iii) Since different metals require different amounts of energy to be separated from their compounds depending on their reactivity, highly reactive metals need strong methods like electrolysis, while less reactive metals can be extracted using simpler methods like heating with carbon or, for very unreactive metals, may not need any chemical extraction at all.

PART B — BIOLOGY

Q15. Thyroxine; iodine is required for its synthesis.

Q16. Tropic movement is the directional growth movement of a plant part in response to an external stimulus, either towards or away from it.

Q17. Pollination.

Q18. Heredity.

Q19. Test cross.

Q20. Chlorofluorocarbons (CFCs).

Q21. (A)   Q22. (D)

Q23. Parathyroid hormone (parathormone); secreted by the parathyroid gland.

Q24. In Rhizopus, tiny spores are formed inside sporangia and are released to develop into new individuals under favourable conditions; in Hydra, a new individual develops as a bud/outgrowth from the parent's body and eventually detaches to become independent. OR: A large number of sperms are produced to increase the probability that at least one sperm will successfully reach and fertilise the egg, since most sperms are lost or fail to reach the egg during their journey.

Q25. Gastric glands secrete gastric juice, containing hydrochloric acid (which kills bacteria and provides an acidic medium) and the enzyme pepsin (which digests proteins).

Q26. HIV-AIDS and gonorrhoea (any two valid examples); prevention — use of condoms during intercourse.

Q27. (a) Air passes through the nostrils → pharynx → larynx → trachea → bronchi → bronchioles → alveoli. (b) Alveoli have thin walls (one-cell thick), are richly supplied with blood capillaries, and provide a very large surface area, all of which allow efficient diffusion of oxygen and carbon dioxide. OR: Kidneys filter blood and form urine; ureters carry urine from kidneys to the urinary bladder; the bladder stores urine until it is expelled through the urethra. (b) ADH increases the reabsorption of water by the kidney tubules when the body needs to conserve water, resulting in the production of a smaller volume of more concentrated urine.

Q28. (a) Since F1 is entirely round-seeded, the round-seed trait (R) is dominant over the wrinkled-seed trait (r), which is recessive and remains hidden in the F1 generation. (b) On self-pollination of F1 (Rr × Rr), the F2 phenotypic ratio expected is 3 round : 1 wrinkled. OR: On sudden fright, the nervous system triggers an immediate reflex response (e.g., jumping back), while the adrenal gland simultaneously releases adrenaline into the blood, increasing heart rate and preparing the body for prolonged "fight or flight" action. (b) Hormones travel through the bloodstream to reach target organs, which takes more time than the electrical impulses of the nervous system, but their effects tend to last longer since hormone levels change and act gradually over time.

Q29. (i) Producers (green plants). (ii) Dead organic matter and waste would accumulate indefinitely, and essential nutrients would remain locked up, disrupting the cycling of nutrients and eventually limiting the growth of new producers. (iii) Decomposers break down dead plant and animal matter into simpler substances, releasing nutrients like nitrogen and phosphorus back into the soil, which are then absorbed by plant roots, enriching soil fertility (e.g., composting of dead leaves by fungi and bacteria).

PART C — PHYSICS

Q30. The principal focus of a concave mirror is the point on its principal axis where parallel rays of light, after reflection from the mirror, actually converge (meet).

Q31. Atmospheric refraction (advance sunrise/delayed sunset effect).

Q32. Gradual weakening of the ciliary muscles and reduced flexibility of the eye lens with age.

Q33. P = I²R.

Q34. V = IR = 2 × 5 = 10 V.

Q35. Galvanometer.

Q36. (A)   Q37. (A)

Q38. The refractive index of a medium with respect to air is the ratio of the speed of light in air (or vacuum) to the speed of light in that medium (n = c/v).

Q39. Different colours of white light have different wavelengths and are refracted by different amounts as they pass through the glass prism, causing them to separate and form a visible spectrum (dispersion). OR: The image formed is real, inverted, highly diminished, and formed at the focus on the other side of the convex lens... (correction: an object at the focus of a convex lens actually produces an image at infinity, highly enlarged; state accordingly) — the image forms at infinity and is real, inverted, and highly magnified.

Q40. Energy = P × t = 1000 W × 3 h = 3000 Wh = 3 kWh.

Q41. The number of turns in the coil, and the magnitude of current flowing through the coil (also depends on the radius of the loop).

Q42. (a) Diagram showing a virtual, erect, and magnified image formed behind the mirror when the object is between the pole and focus of a concave mirror. (b) For a convex mirror, f = +10 cm, u = −15 cm; using 1/v + 1/u = 1/f: 1/v = 1/10 − 1/(−15)... correctly, 1/v = 1/f − 1/u = 1/10 − 1/(−15) = 1/10 + 1/15 = 5/30 = 1/6 ⟹ v = 6 cm (behind the mirror, virtual image). OR: Standard Ohm's law verification circuit description using ammeter, voltmeter, rheostat, and battery, plotting V vs I to get a straight line confirming Ohm's law. (b) Bending the wire into a circle creates two equal semicircular arcs of 3 Ω each between the diametrically opposite points, connected in parallel: 1/R = 1/3+1/3 = 2/3 ⟹ R = 1.5 Ω.

Q43. (i) Hans Christian Oersted. (ii) Electric bell (or electromagnetic crane used to lift scrap iron). (iii) The magnitude of the current flowing through the conductor, and the distance from the conductor at which the field is measured.

Page 5 of 5 (Answer Key)  |  www.careerinformationportal.in