PSEB 11TH CLASS CHEMISTRY FULLY SOLVED REAL QUESTION PAPER WITH PDF
PSEB 11TH CLASS CHEMISTRY FULLY SOLVED REAL QUESTION PAPER WITH PDF
Sure, I will systematically solve the paper by first writing the complete question with MCQ options (if applicable) and then providing the answer.
SECTION A: MCQ
Q1. (i) Mass of 2.6 gram molecule of SO₂
(a) 166.4 g
(b) 162.4 g
(c) 166.6 g
(d) 166 g
Answer:
The molar mass of SO₂ = 32 (S) + 2 × 16 (O) = 64 g/mol
Mass of 2.6 gram molecules of SO₂ = 2.6 × 64 = 166.4 g
Correct Option: (a) 166.4 g
Q1. (ii) The moles of sodium chloride in 250 cm³ of 0.50 M NaCl are
(a) 0.250 mol
(b) 2 mol
(c) 0.125 mol
(d) 1.0 mol
Answer:
Moles = Molarity × Volume (in liters)
Volume = 250 cm³ = 0.25 L
Moles of NaCl = 0.50 × 0.25 = 0.125 mol
Correct Option: (c) 0.125 mol
Q1. (iii) Neutrons are present in all atoms except
(a) He
(b) C
(c) H
(d) Ne
Answer:
Hydrogen (H) has no neutrons in its most common isotope (¹H).
Correct Option: (c) H
Q1. (iv) Which of the following orbitals does not make sense?
(a) 6s
(b) 3p
(c) 2d
(d) 4f
Answer:
The 2d orbital does not exist because the d-subshell starts from the third energy level (n=3).
Correct Option: (c) 2d
Q1. (v) Isotopes are atoms showing same
(a) mass number
(b) atomic mass
(c) number of neutrons
(d) atomic number
Answer:
Isotopes have the same atomic number but different mass numbers.
Correct Option: (d) atomic number
Q1. (vi) Maximum number of bonds between two atoms of a covalent bond can be
(a) 4
(b) 2
(c) 3
(d) 1
Answer:
The maximum number of bonds between two atoms is 3 (triple bond).
Correct Option: (c) 3
Q1. (vii) Which of the following bonds will be non-polar?
(a) N-H
(b) C-H
(c) F-F
(d) O-H
Answer:
F-F is non-polar because both atoms have the same electronegativity.
Correct Option: (c) F-F
Q1. (viii) A thermodynamic state function is a quantity
(a) used to determine heat change
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only
Answer:
A state function is independent of the path taken.
Correct Option: (b) whose value is independent of path
Q1. (ix) The heat of solution depends upon
(a) Nature of solute
(b) Nature of solvent
(c) Concentration of solution
(d) All of these
Answer:
The heat of solution depends on the nature of solute, solvent, and concentration.
Correct Option: (d) All of these
Q1. (x) According to Law of Mass Action, the rate of reaction is directly proportional to
(a) Volume of container
(b) Equilibrium constant
(c) Nature of reactants
(d) Molar concentration of reactants
Answer:
The rate of reaction is directly proportional to the molar concentration of reactants.
Correct Option: (d) Molar concentration of reactants
TRUE/FALSE
Q1. (xi) Phosphorous (Z=15) has three unpaired electrons
Answer:
True. Phosphorous has three unpaired electrons in its 3p orbital.
Q1. (xii) Diamond is more stable than graphite at room temperature and pressure
Answer:
False. Graphite is more stable than diamond at room temperature and pressure.
Q1. (xiii) A catalyst increases the speed of equilibrium reaction and hence the amount of products formed increases
Answer:
False. A catalyst speeds up the reaction but does not change the amount of products formed at equilibrium.
Q1. (xiv) Oxidation number of an element can be zero but valency is never zero
Answer:
True. Oxidation number can be zero (e.g., in H₂), but valency is never zero.
Q1. (xv) Hexane has four structural isomers
Answer:
False. Hexane has five structural isomers.
COMPREHENSION
Q1. (xvi) Who observed the X-rays’ characteristics?
(a) Henry Moseley
(b) Mendeleev
(c) Pauli
(d) Newland
Answer:
Henry Moseley observed the X-rays’ characteristics.
Correct Option: (a) Henry Moseley
Q1. (xvii) The physical and chemical properties of an element are the periodic function of its
(a) Atomic mass
(b) Element behaviour
(c) Number of electrons
(d) Atomic number
Answer:
The properties are a periodic function of the atomic number.
Correct Option: (d) Atomic number
Q1. (xviii) What does (Z) represent in this graph?
(a) Atomic mass
(b) Atomic number
(c) Frequency
(d) Wavelength
Answer:
Z represents the atomic number.
Correct Option: (b) Atomic number
Q1. (xix) Name the ore of uranium
(a) Bauxite
(b) Cinnabar
(c) Pitch blende
(d) Zinc blende
Answer:
The ore of uranium is pitch blende.
Correct Option: (c) Pitch blende
Q1. (xx) What is the fundamental property of an element?
(a) Atomic mass
(b) Atomic number
(c) Neutrons
(d) Nuclear charge
Answer:
The fundamental property of an element is its atomic number.
Correct Option: (b) Atomic number
SECTION B: TWO MARKS QUESTIONS
Q2. Classify the following substances as elements, compounds, and mixtures: Water, tea, silver, steel, carbon dioxide, and platinum.
Answer:
- Elements: Silver (Ag), Platinum (Pt)
- Compounds: Water (H₂O), Carbon dioxide (CO₂)
- Mixtures: Tea, Steel
Q3. Out of sigma and pi bonds, which one is stronger and why?
Answer:
- Sigma bond is stronger than a pi bond.
- Reason: Sigma bonds are formed by head-on overlapping of orbitals, which results in a stronger bond. Pi bonds are formed by side-on overlapping, which is weaker.
Q4. The equilibrium constant for a reaction is 10. What will be the value of ΔG°? (R = 8.314 J K⁻¹ mol⁻¹, T = 300 K)
Answer:
Using the formula:
ΔG° = -RT ln K
ΔG° = -8.314 × 300 × ln(10)
ΔG° = -8.314 × 300 × 2.303
ΔG° = -5744.1 J/mol
ΔG° = -5.744 kJ/mol
Q5. State and explain the first law of thermodynamics. Give its mathematical form.
Answer:
- First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or converted from one form to another.
- Mathematical Form: ΔU = q – W
Where ΔU = change in internal energy, q = heat added to the system, W = work done by the system.
Q6. What is Kc for the following equilibrium when the equilibrium concentration of each substance is [SO₂] = 0.6 M, [O₂] = 0.82 M, and [SO₃] = 1.9 M?
Answer:
The reaction is:
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Kc = [SO₃]² / ([SO₂]² [O₂])
Kc = (1.9)² / ((0.6)² × 0.82)
Kc = 3.61 / (0.36 × 0.82)
Kc = 3.61 / 0.2952
Kc ≈ 12.23
Q7. (a) State the law of mass action.
Answer:
The rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Q7. (b) Define equilibrium constant.
Answer:
The equilibrium constant (Kc) is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficients at equilibrium.
Q8. The compound AgF₂ is unstable. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer:
AgF₂ is unstable because silver (Ag) is in a +2 oxidation state, which is not common. However, if formed, it acts as a strong oxidizing agent because it can easily gain electrons to reduce Ag²⁺ to Ag⁺ or Ag⁰.
Q9. Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg, and Zn.
Answer:
The reactivity order is:
Mg > Al > Zn > Fe > Cu
Q10. What are electrophiles and nucleophiles? Explain with examples.
Answer:
- Electrophiles: Electron-deficient species that accept electrons. Example: H⁺, BF₃.
- Nucleophiles: Electron-rich species that donate electrons. Example: OH⁻, NH₃.
Q11. What effect does branching of an alkane chain have on its boiling point?
Answer:
Branching decreases the boiling point of an alkane because it reduces the surface area and weakens the van der Waals forces between molecules.
Q12. State the following laws of chemical combination and give one example in each case:
(a) Law of constant composition
(b) Law of multiple proportions
Answer:
- (a) Law of Constant Composition: A compound always contains the same elements in the same proportion by mass. Example: Water (H₂O) always has a 1:8 ratio of hydrogen to oxygen by mass.
- (b) Law of Multiple Proportions: When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. Example: CO and CO₂ (carbon combines with oxygen in a 1:2 ratio).
Q13. Write the favourable factors for the formation of an ionic bond.
Answer:
- High electronegativity difference between the atoms.
- Low ionization energy of the metal.
- High electron affinity of the non-metal.
- Lattice energy released during the formation of the ionic compound.
Q14. Define Hess’s law of constant heat summation.
Answer:
Hess’s law states that the total enthalpy change for a reaction is the same, regardless of the number of steps in which the reaction is carried out.
Q15. Name two tests to test the presence of a double bond in a compound.
Answer:
- Bromine Water Test: The compound decolorizes bromine water.
- Baeyer’s Test: The compound reacts with alkaline KMnO₄, turning it from purple to brown.
SECTION C: THREE MARKS QUESTIONS
Q16. Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
Answer:
- Moles of Au = 1 / 197 ≈ 0.0051 mol
- Moles of Na = 1 / 23 ≈ 0.0435 mol
- Moles of Li = 1 / 7 ≈ 0.1429 mol
Since Li has the highest number of moles, it will have the largest number of atoms.
Correct Option: (iii) 1 g Li (s)
Q17. List some postulates of Bohr’s model of an atom.
Answer:
- Electrons revolve in discrete orbits around the nucleus.
- Electrons do not radiate energy while in stable orbits.
- Energy is emitted or absorbed when electrons jump between orbits.
- The angular momentum of electrons is quantized.
Q18. Discuss the following terms and account for the variation in groups and periods of the periodic table:
(i) Electron gain enthalpy
(ii) Ionization enthalpy
Answer:
- (i) Electron Gain Enthalpy: It is the energy released when an electron is added to a neutral atom. It generally increases across a period and decreases down a group.
- (ii) Ionization Enthalpy: It is the energy required to remove an electron from a neutral atom. It generally increases across a period and decreases down a group.
Q19. Explain four types of organic reactions with examples.
Answer:
- Substitution Reaction: CH₄ + Cl₂ → CH₃Cl + HCl
- Addition Reaction: CH₂=CH₂ + H₂ → CH₃-CH₃
- Elimination Reaction: CH₃-CH₂Br → CH₂=CH₂ + HBr
- Rearrangement Reaction: CH₃-CH₂-CH₂⁺ → CH₃-CH⁺-CH₃
SECTION D: FIVE MARKS QUESTIONS
Q20. Write short notes on:
(i) Friedel-Crafts reaction
(ii) Markovnikov’s and anti-Markovnikov’s rule
Answer:
- (i) Friedel-Crafts Reaction: It involves the alkylation or acylation of an aromatic ring using an alkyl or acyl halide in the presence of a Lewis acid catalyst (e.g., AlCl₃).
- (ii) Markovnikov’s Rule: In the addition of HX to an alkene, the hydrogen attaches to the carbon with more hydrogen atoms.
- Anti-Markovnikov’s Rule: In the presence of peroxides, the addition of HBr to an alkene follows the opposite pattern.
Q21. Explain the difference between sigma bond and pi bond.
Answer:
- Sigma Bond: Formed by head-on overlapping of orbitals. It is stronger and allows free rotation.
- Pi Bond: Formed by side-on overlapping of orbitals. It is weaker and restricts rotation.