FY 3025 Chemistry - Answer Key
Note: This answer key provides solutions for selected questions from the First Year Higher Secondary Second Terminal Examination, December 2025. Calculations and logical steps are shown where necessary.
Section A: 1-Score Questions (Answer any 4 from 1 to 5)
Each question carries 1 score
1. The mass of 1.2 moles of H2O is ______ g.
Molar mass of H2O = (2×1) + 16 = 18 g/mol
Mass = Number of moles × Molar mass = 1.2 × 18 = 21.6 g
Mass = Number of moles × Molar mass = 1.2 × 18 = 21.6 g
2. The number of electrons present in a metal M2+ ion is 28, its atomic number is ______
M2+ ion has 28 electrons. The neutral atom M would have 28 + 2 = 30 electrons.
Atomic number = Number of protons = Number of electrons in neutral atom = 30.
Correct option: c) 30
Atomic number = Number of protons = Number of electrons in neutral atom = 30.
Correct option: c) 30
3. The most electronegative element in the periodic table is ______
Fluorine (F)
4. Which of the following molecules has the highest dipole moment?
CO2 and BF3 are non-polar. CCl4 is symmetrical and non-polar. H2S is bent and polar.
a) H2S
a) H2S
5. The conjugate acid of NH3 is ______
NH4+ (Ammonium ion)
Section B: 2-Score Questions (Answer any 8 from 6 to 15)
Each question carries 2 scores
6. State and explain the law of Conservation of Mass.
Law: Mass can neither be created nor destroyed in a chemical reaction.
Explanation: The total mass of the reactants equals the total mass of the products. This is based on the principle that atoms are merely rearranged during a chemical change.
Explanation: The total mass of the reactants equals the total mass of the products. This is based on the principle that atoms are merely rearranged during a chemical change.
9. Write the values of n, l, m and s for the valence electron of sodium atom (atomic number of Na = 11).
Electronic configuration of Na (11): 1s2 2s2 2p6 3s1
The valence electron is in the 3s orbital.
n (principal quantum number) = 3
l (azimuthal quantum number for s-orbital) = 0
m (magnetic quantum number for l=0) = 0
s (spin quantum number) = +1/2 or -1/2 (either is acceptable)
The valence electron is in the 3s orbital.
n (principal quantum number) = 3
l (azimuthal quantum number for s-orbital) = 0
m (magnetic quantum number for l=0) = 0
s (spin quantum number) = +1/2 or -1/2 (either is acceptable)
12. 2Cl(g) → Cl2(g). What are the signs of ΔH and ΔS?
Bond formation is exothermic. ∴ ΔH is negative (ΔH < 0).
Two gaseous atoms combining to form one gaseous molecule reduces disorder. ∴ ΔS is negative (ΔS < 0).
Two gaseous atoms combining to form one gaseous molecule reduces disorder. ∴ ΔS is negative (ΔS < 0).
13. ΔH and ΔS of a reaction are 45.84 and 0.999 kilo joules per mol respectively at 1 atm pressure. Calculate the temperature at which the reaction is in equilibrium.
At equilibrium, ΔG = 0 = ΔH - TΔS
Therefore, T = ΔH / ΔS
T = (45.84 kJ/mol) / (0.999 kJ mol-1 K-1)
T ≈ 45.89 K (Note: Units of ΔS must be kJ/mol·K for this calculation. The given value 0.999 is assumed to be in correct units).
Therefore, T = ΔH / ΔS
T = (45.84 kJ/mol) / (0.999 kJ mol-1 K-1)
T ≈ 45.89 K (Note: Units of ΔS must be kJ/mol·K for this calculation. The given value 0.999 is assumed to be in correct units).
14. i) Write the equation for the equilibrium constant Kc for the reaction.
ii) 4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(l). What happens to Kc when the reaction is reversed?
ii) 4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(l). What happens to Kc when the reaction is reversed?
i) For a general reaction: aA + bB ⇌ cC + dD
Kc = ([C]c[D]d) / ([A]a[B]b)
ii) For the given reaction:
Kc(forward) = ([NO]4[H2O]6) / ([NH3]4[O2]5)
When reversed: 4 NO(g) + 6 H2O(l) ⇌ 4 NH3(g) + 5 O2(g)
Kc(reverse) = ([NH3]4[O2]5) / ([NO]4[H2O]6) = 1 / Kc(forward)
Kc = ([C]c[D]d) / ([A]a[B]b)
ii) For the given reaction:
Kc(forward) = ([NO]4[H2O]6) / ([NH3]4[O2]5)
When reversed: 4 NO(g) + 6 H2O(l) ⇌ 4 NH3(g) + 5 O2(g)
Kc(reverse) = ([NH3]4[O2]5) / ([NO]4[H2O]6) = 1 / Kc(forward)
15. Calculate the solubility (S) of MgSO4 at 298 K, if its Ksp at this temperature is 9 × 10-6.
MgSO4(s) ⇌ Mg2+(aq) + SO42-(aq)
Ksp = [Mg2+][SO42-] = S × S = S2
S2 = 9 × 10-6
S = √(9 × 10-6) = 3 × 10-3 mol/L
Solubility S = 3 × 10-3 M
Ksp = [Mg2+][SO42-] = S × S = S2
S2 = 9 × 10-6
S = √(9 × 10-6) = 3 × 10-3 mol/L
Solubility S = 3 × 10-3 M
Section C: 3-Score Questions (Answer any 8 from 16 to 26)
Each question carries 3 scores
16. i) Define one mole. (1)
ii) How much copper can be obtained from 79.5 g of CuSO4? (Given atomic mass of Cu, S and O are 63.5, 32, and 16 g respectively) (2)
ii) How much copper can be obtained from 79.5 g of CuSO4? (Given atomic mass of Cu, S and O are 63.5, 32, and 16 g respectively) (2)
i) One mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 g of carbon-12.
ii) Molar mass of CuSO4 = 63.5 + 32 + (4×16) = 159.5 g/mol
Moles of CuSO4 = 79.5 g / 159.5 g/mol = 0.4984 mol
1 mol CuSO4 contains 1 mol Cu.
∴ Moles of Cu = 0.4984 mol
Mass of Cu = 0.4984 mol × 63.5 g/mol = 31.65 g
ii) Molar mass of CuSO4 = 63.5 + 32 + (4×16) = 159.5 g/mol
Moles of CuSO4 = 79.5 g / 159.5 g/mol = 0.4984 mol
1 mol CuSO4 contains 1 mol Cu.
∴ Moles of Cu = 0.4984 mol
Mass of Cu = 0.4984 mol × 63.5 g/mol = 31.65 g
20. The first ionization enthalpy of sodium is lower than that of magnesium but the second ionization enthalpy is higher than that of magnesium. Explain.
First Ionization: Na: [Ne]3s1, Mg: [Ne]3s2. Removing an electron from Mg's filled 3s orbital requires more energy than from Na's half-filled orbital. Hence, I1(Na) < I1(Mg).
Second Ionization: Na+: [Ne] (stable noble gas configuration), Mg+: [Ne]3s1. Removing an electron from the stable Na+ configuration requires a very high energy jump compared to removing the lone 3s electron from Mg+. Hence, I2(Na) > I2(Mg).
Second Ionization: Na+: [Ne] (stable noble gas configuration), Mg+: [Ne]3s1. Removing an electron from the stable Na+ configuration requires a very high energy jump compared to removing the lone 3s electron from Mg+. Hence, I2(Na) > I2(Mg).
22. The oxygen atom in water molecule is sp3 hybridized but the water molecule has no tetrahedral shape. Why?
Oxygen in H2O undergoes sp3 hybridization, resulting in four sp3 hybrid orbitals with a tetrahedral electron pair geometry. Two orbitals form O-H bonds, and the other two contain lone pairs.
The molecular shape is determined by the positions of atoms, not electron pairs. The repulsion between two lone pairs is greater than between a bond pair and a lone pair, which compresses the H-O-H bond angle from the ideal 109.5° to about 104.5°. This gives a bent or V-shape, not a perfect tetrahedron.
The molecular shape is determined by the positions of atoms, not electron pairs. The repulsion between two lone pairs is greater than between a bond pair and a lone pair, which compresses the H-O-H bond angle from the ideal 109.5° to about 104.5°. This gives a bent or V-shape, not a perfect tetrahedron.
26. Justify that the following reactions are redox reactions.
i) Fe2O3 + 3CO → 2Fe + 3CO2 (1½)
ii) 4NH3 + 5O2 → 4NO + 6H2O (1½)
i) Fe2O3 + 3CO → 2Fe + 3CO2 (1½)
ii) 4NH3 + 5O2 → 4NO + 6H2O (1½)
A redox reaction involves both oxidation (loss of electrons/increase in oxidation number) and reduction (gain of electrons/decrease in oxidation number).
i) In Fe2O3 + 3CO → 2Fe + 3CO2:
• Oxidation state of Fe changes from +3 (in Fe2O3) to 0 (in Fe). Reduction.
• Oxidation state of C changes from +2 (in CO) to +4 (in CO2). Oxidation.
∴ It is a redox reaction.
ii) In 4NH3 + 5O2 → 4NO + 6H2O:
• Oxidation state of N changes from -3 (in NH3) to +2 (in NO). Oxidation.
• Oxidation state of O changes from 0 (in O2) to -2 (in H2O and NO). Reduction.
∴ It is a redox reaction.
i) In Fe2O3 + 3CO → 2Fe + 3CO2:
• Oxidation state of Fe changes from +3 (in Fe2O3) to 0 (in Fe). Reduction.
• Oxidation state of C changes from +2 (in CO) to +4 (in CO2). Oxidation.
∴ It is a redox reaction.
ii) In 4NH3 + 5O2 → 4NO + 6H2O:
• Oxidation state of N changes from -3 (in NH3) to +2 (in NO). Oxidation.
• Oxidation state of O changes from 0 (in O2) to -2 (in H2O and NO). Reduction.
∴ It is a redox reaction.
Section D: 4-Score Questions (Answer any 4 from 27 to 31)
Each question carries 4 scores
27. i) What is mass percent? (1)
ii) A compound contains 4.07% H, 24.27% C, and 71.65% Cl. Its molar mass is 96.96. What is the empirical and molecular formula? (3)
ii) A compound contains 4.07% H, 24.27% C, and 71.65% Cl. Its molar mass is 96.96. What is the empirical and molecular formula? (3)
i) Mass percent (or weight percent) is the mass of a particular component divided by the total mass of the mixture/compound, multiplied by 100%.
ii) Assume 100 g of compound.
Empirical formula mass = 12 + 2×1 + 35.5 = 49.5 g/mol
n = Molar mass / Empirical mass = 96.96 / 49.5 ≈ 1.96 ≈ 2
Molecular Formula = (CH2Cl)2 = C2H4Cl2
ii) Assume 100 g of compound.
• Moles of H = 4.07 g / 1 g/mol = 4.07 mol
• Moles of C = 24.27 g / 12 g/mol = 2.0225 mol
• Moles of Cl = 71.65 g / 35.5 g/mol ≈ 2.0183 mol
Find simplest mole ratio (divide by smallest ~2.018):
• Moles of C = 24.27 g / 12 g/mol = 2.0225 mol
• Moles of Cl = 71.65 g / 35.5 g/mol ≈ 2.0183 mol
H: 4.07 / 2.018 ≈ 2.02 ≈ 2
C: 2.0225 / 2.018 ≈ 1.00 ≈ 1
Cl: 2.0183 / 2.018 ≈ 1.00 ≈ 1
Empirical Formula = CH2ClC: 2.0225 / 2.018 ≈ 1.00 ≈ 1
Cl: 2.0183 / 2.018 ≈ 1.00 ≈ 1
Empirical formula mass = 12 + 2×1 + 35.5 = 49.5 g/mol
n = Molar mass / Empirical mass = 96.96 / 49.5 ≈ 1.96 ≈ 2
Molecular Formula = (CH2Cl)2 = C2H4Cl2
28. ii) Electrons are emitted with zero velocity from a metal surface when exposed to radiation of wavelength 6800 Å. Calculate the threshold frequency and work function.
Zero velocity means kinetic energy (KE) of emitted electron is 0.
Photoelectric equation: hν = hν0 + KE, so hν = hν0.
Given λ = 6800 Å = 6800 × 10-10 m = 6.8 × 10-7 m.
ν = c / λ = (3 × 108 m/s) / (6.8 × 10-7 m) ≈ 4.41 × 1014 Hz.
Since ν = ν0 at KE=0, Threshold frequency ν0 = 4.41 × 1014 Hz.
Work function φ = hν0 = (6.626 × 10-34 Js) × (4.41 × 1014 s-1) ≈ 2.92 × 10-19 J.
(In eV: φ ≈ 2.92×10-19 / 1.6×10-19 ≈ 1.83 eV).
Photoelectric equation: hν = hν0 + KE, so hν = hν0.
Given λ = 6800 Å = 6800 × 10-10 m = 6.8 × 10-7 m.
ν = c / λ = (3 × 108 m/s) / (6.8 × 10-7 m) ≈ 4.41 × 1014 Hz.
Since ν = ν0 at KE=0, Threshold frequency ν0 = 4.41 × 1014 Hz.
Work function φ = hν0 = (6.626 × 10-34 Js) × (4.41 × 1014 s-1) ≈ 2.92 × 10-19 J.
(In eV: φ ≈ 2.92×10-19 / 1.6×10-19 ≈ 1.83 eV).
29. Write the molecular orbital electronic configuration and calculate the bond order of N2 and F2.
N2 (14 electrons):
MO Configuration: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2px)2 (π2py)2 (σ2pz)2
Bond order = (Number of bonding electrons - Number of antibonding electrons)/2
= (10 - 4)/2 = 3
F2 (18 electrons):
MO Configuration: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2 (π2py)2 (π*2px)2 (π*2py)2
Bond order = (10 - 8)/2 = 1
MO Configuration: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (π2px)2 (π2py)2 (σ2pz)2
Bond order = (Number of bonding electrons - Number of antibonding electrons)/2
= (10 - 4)/2 = 3
F2 (18 electrons):
MO Configuration: (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2 (π2py)2 (π*2px)2 (π*2py)2
Bond order = (10 - 8)/2 = 1
31. ii) Write the Henderson-Hasselbalch equation for an acidic buffer. Calculate the pH of an acidic buffer containing 0.1 M CH3COOH and 0.5 M CH3COONa. [Ka for CH3COOH is 1.8 × 10-6].
Henderson-Hasselbalch equation for acidic buffer:
pH = pKa + log([Salt]/[Acid])
Given: [Acid] = 0.1 M, [Salt] = 0.5 M, Ka = 1.8 × 10-6
pKa = -log(1.8 × 10-6) = -(-5.7447) ≈ 5.745
pH = 5.745 + log(0.5 / 0.1) = 5.745 + log(5)
log(5) ≈ 0.6990
pH ≈ 5.745 + 0.699 = 6.444
pH = pKa + log([Salt]/[Acid])
Given: [Acid] = 0.1 M, [Salt] = 0.5 M, Ka = 1.8 × 10-6
pKa = -log(1.8 × 10-6) = -(-5.7447) ≈ 5.745
pH = 5.745 + log(0.5 / 0.1) = 5.745 + log(5)
log(5) ≈ 0.6990
pH ≈ 5.745 + 0.699 = 6.444
End of Answer Key. This key provides solutions for a representative selection of questions from the paper. For complete solutions, all parts of each attempted question must be answered as per the exam instructions.