Haryana Board (HBSE) Class 11 Chemistry SAT-1 Question Paper 2026 Answer Key. Get the Students Assessment Test (SAT) Class 11 Chemistry question paper with complete solution, accurate answer key, and expert preparation tips. The Haryana Board of School Education (HBSE) conducts SAT as an important assessment for Class 11 students. Best resource for Haryana Board Class 11 SAT Chemistry exam practice, quick revision, and scoring better marks.
HBSE Class 11 Chemistry SAT-1 Question Paper 2026 Answer Key
Instructions :
• All questions are compulsory.
• Questions (1-6) carry 1 mark each.
• Questions (7) carry 2 marks.
• Questions (8) carry 3 marks.
• Questions (9) case study, carry 4 marks.
• Questions (10) carry 5 marks.
1. Which of the following properties of matter is independent of temperature?
(A) Volume
(B) Density
(C) Molarity
(D) Molality
Answer – (D) Molality
2. Which of the following contains equal number of atoms as 20 g of calcium?
(A) 24 g of magnesium
(B) 8 g of oxygen gas
(C) 12 g Carbon
(D) 4 g Helium
Answer – (B) 8 g of oxygen gas
3. Assertion (A) : Across a period, the atomic radius generally decreases from left to right.
Reason (R) : As electrons are added to orbitals in the same principal quantum level, the effective nuclear charge increases, pulling the electrons closer to the nucleus.
(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) Both Assertion and Reason are false.
Answer – (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
4. At STP, one mole of a gas in grams is called its …………….
Answer – molar mass
5. The elements belonging to groups 13 to 18 in the Periodic Table are collectively known as the …………….
Answer – p-block elements
6. 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.
7. State Heisenberg’s Uncertainty Principle. Why is it significant only for microscopic objects like electrons and negligible for macroscopic objects?
Answer – According to Heisenberg’s Uncertainty Principle, it is impossible to determine simultaneously the exact position and momentum of a moving particle like an electron. It is significant only for microscopic particles due to their very small mass, where measurement causes noticeable uncertainty, while for macroscopic objects the mass is large, making the uncertainty negligible.
8. A compound is found to contain 6.0 g of carbon and 8.0 g of oxygen.
(i) 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
(ii) 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
9. CASE STUDY : Ionization enthalpy, also known as ionization energy, is a fundamental periodic property of elements. It is defined as the minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. The first ionization enthalpy (∆H1) is the energy required to remove the first electron. The second ionization enthalpy (∆H2) is the energy required to remove the second electron from the resultant monovalent cation.
(i) What happens to ionisation enthalpy as atomic size increases?
Answer – Ionization enthalpy decreases as atomic size increases because the outermost electron is farther from the nucleus and experiences weaker attraction.
(ii) Which factor leads to a higher ionization enthalpy due to stronger electron-nucleus attraction?
Answer – Higher effective nuclear charge (more protons in the nucleus) increases electron-nucleus attraction, leading to a higher ionization enthalpy.
(iii) Why do noble gases have very high ionization enthalpies?
Answer – Noble gases have completely filled electron shells, making their electrons tightly bound and difficult to remove, resulting in very high ionization enthalpies.
(iv) How does the shielding effect influence ionization enthalpy?
Answer – Greater shielding by inner electrons reduces the effective nuclear charge on outer electrons, making them easier to remove and thus lowering ionization enthalpy.
10. (i) Explain the photoelectric effect. List any two observations from the photoelectric effect experiment that could not be explained by classical wave theory.
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.
Observations not explained by Classical Wave Theory are Existence of Threshold Frequency and Independence of Kinetic Energy from Intensity.
(ii) The work function (W) for a certain metal is 2.0 eV. If light of wavelength 400 nm is incident on the metal, calculate the kinetic energy of the ejected electrons in Joules.
Answer – We know, 1 eV = 1.602 × 10–19
W = 2.0 eV = 2 × (1.602 × 10–19) = 3.204 × 10–19 J
Energy (E) = hc/λ = (6.626×10–34Js)(3×108m/s) / (400×10–9) = 4.97 × 10–19 J
K.E. = E – W = (4.97 × 10–19) – (3.204 × 10–19) = 1.77 × 10–19 J
OR
(i) For a given principal quantum number ‘n’, what are the possible values of the azimuthal (I) and magnetic (m) quantum numbers? How many orbitals are possible for a given ‘n’ value?
Answer – For a given principal quantum number ‘n’, the possible values of the azimuthal quantum number (l) are integers from 0 to (n–1). The possible values of the magnetic quantum number m are integers from –l to +l, including 0. The total number of orbitals for a given ‘n’ value is n2.
(ii) An electron in a hydrogen atom undergoes a transition from an excited state with n = 5 to the ground state (n = 1).
(a) Calculate the wavelength of the emitted radiation in nanometers.
Answer – Using Rydberg formula,
1/λ = R(1/n12 – 1/n22) = 1.097 × 107 (1/12 – 1/52) = 1.097 × 107 (1 – 1/25) = 1.097 × 107 × 24/25
λ = 9.495 × 10–8 m = 94.95 nm
(b) To which series does this transition belong?
Answer – Lyman series