Class 11 Chemistry Half Yearly Question Paper 2025 Answer Key

Class 11 Chemistry Half Yearly Question Paper 2025 Answer Key (NCERT Based)

Instructions :
• All questions are compulsory.
• Questions (1-9) carry 1 mark each.
• Questions (10-12) carry 2 marks each.
• Questions (13-14) carry 3 marks each.
• Question (15) case study, carry 4 marks.
• Questions (16-17) carry 5 marks each.

1. Bond lengths are lower in elements having :
(A) Double Bonds
(B) Triple Bonds
(C) Crystal Structure
(D) Single Bonds
Answer – (B) Triple Bonds

2. The molar mass of C₆H₁₂O₆ is :
(A) 1.80 amu
(B) 180 u
(C) 180 g
(D) 18.0 g
Answer – (C) 180 g
Atomic mass of C = 12 u, H = 1 u, O = 16 u
Molecular mass = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 u
Molar mass = 180 g/mol or 180 g (for one mole)

3. Which of the following properties of matter is independent of temperature?
(A) Volume
(B) Density
(C) Molarity
(D) Molality
Answer – (D) Molality

4. A Reaction A + B → C + D + q is found to have a positive entropy change. The reaction will be :
(A) Possible at high temperature
(B) Possible at low temperature
(C) Not possible at any temperature
(D) Possible at any temperature
Answer – (D) Possible at any temperature

5. Assertion (A) : For the balmer series of hydrogen spectrum the value of n₁ = 2 and n₂ = 3, 4, 5, ………
Reason (R) : The value of ‘n’ for spectral lines in the balmer series of hydrogen spectrum having the highest wavelength is 4 and 5.
(A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(B) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(C) Assertion is true but Reason is false.
(D) Assertion is false but Reason is true.
Answer – (C) Assertion is true but Reason is false.

6. Fe²⁺ is ………….. in size in comparison to F³⁺.
Answer – larger

7. The maximum ionization enthalpy is shown by ……………
Answer – Helium (He)

8. The amount of heat energy required to raise the temperature of a given quantity of matter by one degree Celsius is known as ……………
Answer – specific heat capacity

9. Define Orbit and Orbital.
Answer – An Orbit is a fixed circular path around the nucleus, while an Orbital is a region around the nucleus where the probability of finding an electron is maximum.

10. If the density of methanol is 0.793 g/ml, what is its volume required for making 2.5 L of its 0.25 M solution?
Answer – Molecular mass of methanol (CH₃OH) = 12 + 4 + 16 + 1 = 32 g/mol
Molarity = no. of moles / Volume (litres)
M = n/V
n = M × V = 0.25 × 2.5 = 0.625 mol
Mass = n × Molar mass = 0.625 × 32 = 20 g
Volume = Mass / Density = 20 / 0.793 = 25.22 ml

11. Chlorophyll present in green leaves of plants absorb light at 4.620 × 10¹⁴ Hz. Calculate the wavelength of radiation in nanometres.
Answer – Wavelength = Speed / Frequency
λ = c/𝛎 = (3.0 × 10⁸ m/s) / (4.620 × 10¹⁴ Hz) = 0.6494 × 10^–6 m = 649.4 nm

12. Which hybrid orbitals are used by carbon atom in the following molecules :
(a) CH ≡ CH
Answer – sp hybridized

(b) CH₃CHO
Answer – CH3 is sp³ hybridized and CHO is sp² hybridized.

13. A compound is found to contain 6.0 g of carbon and 8.0 g of oxygen.
(a) Determine the empirical formula of the compound.
Answer – Moles of C = 6.0/12 = 0.5 mol
Moles of O = 8.0/16 = 0.5 mol
Ratio = C : O = 0.5 : 0.5 = 1 : 1
Empirical formula = CO

(b) If the molar mass of the compound is 28.0 g/mol, what is its molecular formula?
Answer – Empirical formula mass = 12 + 16 = 28 g/mol
n = Molar mass / Empirical formula mass = 28/28 = 1
Molecular formula = n × Empirical formula
Molecular formula = CO

14. Predict in which of the following, entropy increases / decreases :
(a) A liquid crystallizes into a solid.
Answer – Entropy decreases

(b) The temperature of a crystalline solid is raised from 0 K to 115 K.
Answer – Entropy increases

(c) 2NaHCO₃ (s) → Na₂CO₃ (s) + CO₂ (g) + H₂O (g)
Answer – Entropy increases

(d) H₂ (g) → 2 H (g)
Answer – Entropy increases

15. CASE STUDY : A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity. Unlike ionization enthalpy and electron gain enthalpy, it is not a measurable quantity. However, a number of numerical scales of electronegativity of elements viz. Pauling Scale, Mulliken Jaffe Scale, Allred Rochow Scale have been developed. In 1992, assigned arbitrarily a value of 4.0 to Fluorine, the element considered to have the greatest ability to attract electrons. The electronegativity of any given element is not constant, it varies depending on the element to which it is bound. Though it is a not measurable quantity, it does provide a means, strong tendency to gain electrons. Therefore, electronegativity is directly related to the non-metallic properties of elements. Thus, the increase in electronegativities across a period is accompanied by an increase in non-metallic properties (or decrease in metallic properties) of elements.
Questions :
(a) What is the increasing order of electronegativities of N, C, P, Si?
Answer – Si < P < C < N

(b) What is the increasing order of atomic radii of Li, Na, K, Rb, Cs?
Answer – Li < Na < K < Rb < Cs

(c) Why does oxygen have lower ionization enthalpy than Nitrogen?
Answer – Oxygen has lower ionization enthalpy than nitrogen due to repulsion in paired electrons.

(d) What happens to metallic character down the group?
Answer – Metallic character increases down the group.

16. (a) Define ionic bond.
Answer – An ionic bond is a type of chemical bond formed by the electrostatic attraction between positively charged ions (cations) and negatively charged ions (anions). It usually occurs when one atom donates electrons and another atom accepts electrons to achieve a stable electron configuration. For example, in sodium chloride (NaCl), sodium donates one electron to chlorine, forming Na⁺ and Cl^– ions that are held together by an ionic bond.

(b) Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer – NH₃ has a higher dipole moment than NF₃ because in NH₃, the N–H bond dipoles and the lone pair dipole on nitrogen point in the same general direction, strengthening the overall dipole. In NF₃, the N–F bond dipoles point opposite to the lone pair dipole, which causes partial cancellation and results in a much lower net dipole moment.

(c) Compare relative stability of the following species and indicate their magnetic properties on the basis of Molecular Orbital Theory : O², O²⁺, O^2–
Answer – Bond order of O²⁺ is 2.5, bond order of O² is 2, bond order of O^2– is 1.5.
Order of the stability is O²⁺ > O² > O^2–

17. (a) Why line spectrum is also called fingerprint spectrum.
Answer – A line spectrum is called a fingerprint spectrum because each element produces a unique pattern of spectral lines that is different from every other element, just like every person has a unique fingerprint. These distinct lines arise from specific electronic transitions within the atoms, which depend on the element’s unique energy levels. Therefore, by studying the line spectrum, we can identify the element present in a sample just as fingerprints are used to identify individuals.

(b) What were the weaknesses or limitations of Bohr’s model of atom?
Answer – Limitations of Bohr’s Model of Atom :
• It explains only hydrogen and hydrogen-like (one-electron) species, not multi-electron atoms.
• It fails to explain fine spectral lines and the Zeeman effect (splitting in magnetic field).
• It does not consider the wave nature of electrons.
• It violates the Heisenberg uncertainty principle by assuming definite electron orbits.
• It cannot explain the shapes and orientations of orbitals in atoms.

(c) Explain the photoelectric effect.
Answer – The photoelectric effect is the process in which electrons are released from a metal surface when light of a certain frequency falls on it. The emitted electrons are called photoelectrons, and the electric current formed due to these electrons is called photoelectric current.