🎉 Welcome to Inamjazbi Learn Chemistry! 🎉 🚀 Whether you're preparing for MDCAT, ECAT, or FSC, this MCQ Quiz is YOUR key to success! 🧪💡
✅ Perfect for exam preparation – designed to help you master the toughest topics in chemistry 🏆
✅ Boost your scores with smart tips and tricks 🔥
✅ Learn unique solving strategies and quick hacks 💯
💥 This isn’t just another quiz – it’s a game‑changer!
🧠 Related Posts:
🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 1 🧪💡 | Learn & Practice!
MDCAT/ECAT/FSC Chemistry Grand Quiz # 1 🧬 | High-Yield Chemistry MCQs💥 Exam-Focused, 🎯 💯 Exam Booster🚀 Practice & Win 190+! 📚 Topper’s Choice
🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 2 🧪💡 | Learn & Practice!
🔥🌟 Grand Chemistry MCQs Test for Class 12 | Interactive Quiz 🧪💡 | Learn & Practice!
🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 3 🧪💡 | Learn & Practice!
🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 4 🧪💡 | Learn & Practice!
🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 5 🧪💡 | Learn & Practice!
1. Which of the following is the correct order of increasing orbital energy?
a) 2s 2p, 1s, 3s 3p, 3d, 4s, 4p
b) 1s, 2s 2p, 3s 3p, 4s, 3d, 4p
c) 1s, 3s 3p, 3d, 4p, 4s, 2s 2p
d) 1s, 2s 3p, 3s 2p, 3d, 4s, 4p
✅ Correct Answer: (b) 1s, 2s 2p, 3s 3p, 4s, 3d, 4p
🔎📌 Short Reason
By the (n + l) rule: orbitals with lower n+l fill first; if equal, lower n comes first. So the sequence is: 1s → 2s,2p → 3s,3p → 4s → 3d → 4p.
🔎🧠 Reasoning (Aufbau Principle + (n + l) Rule)
📌The order of orbital energies is determined by the Aufbau principle and the (n + l) rule. According to this rule, orbitals with lower n+l values fill first; if two orbitals have the same n+l, the one with lower n is lower in energy. Applying this:
🔥1s (n+l=1) is the lowest.
🔥2s (2), 2p (3) come next.
🔥3s (3), 3p (4) follow.
🔥4s (4) is slightly lower than 3d (5).
🔥Finally, 4p (5) is higher than 3d because of its larger n.
📌Applying this:
🔥1s → n=1,l=0→n+l=1 → lowest energy.
🔥2s (n=2,l=0 → 2) and 2p (n=2,l=1 → 3) → next.
🔥3s (n=3,l=0 → 3) and 3p (n=3,l=1 → 4) → next.
🔥4s (n=4,l=0 → 4) → comes before 3d.
🔥3d (n=3,l=2 → 5) → higher than 4s.
🔥4p (n=4,l=1 → 5) → similar n+l but higher n, so after 3d.
📌 👉 Thus, the correct increasing energy sequence is 1s → 2s,2p → 3s,3p → 4s → 3d → 4p (1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p), which matches option (b).
🔎📌 Short Reason
By the (n + l) rule: orbitals with lower n+l fill first; if equal, lower n comes first. So the sequence is: 1s → 2s,2p → 3s,3p → 4s → 3d → 4p.
🔎🧠 Reasoning (Aufbau Principle + (n + l) Rule)
📌The order of orbital energies is determined by the Aufbau principle and the (n + l) rule. According to this rule, orbitals with lower n+l values fill first; if two orbitals have the same n+l, the one with lower n is lower in energy. Applying this:
🔥1s (n+l=1) is the lowest.
🔥2s (2), 2p (3) come next.
🔥3s (3), 3p (4) follow.
🔥4s (4) is slightly lower than 3d (5).
🔥Finally, 4p (5) is higher than 3d because of its larger n.
📌Applying this:
🔥1s → n=1,l=0→n+l=1 → lowest energy.
🔥2s (n=2,l=0 → 2) and 2p (n=2,l=1 → 3) → next.
🔥3s (n=3,l=0 → 3) and 3p (n=3,l=1 → 4) → next.
🔥4s (n=4,l=0 → 4) → comes before 3d.
🔥3d (n=3,l=2 → 5) → higher than 4s.
🔥4p (n=4,l=1 → 5) → similar n+l but higher n, so after 3d.
📌 👉 Thus, the correct increasing energy sequence is 1s → 2s,2p → 3s,3p → 4s → 3d → 4p (1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p), which matches option (b).
2. Which of the following pairs of gases have exactly double the rate of diffusion as each other?
a) N₂ and SO₃
b) He and CH₄
c) CO and SO₂
d) O₃ and F₂
✅ Correct Answer: (b) He and CH₄
🔎📌 Short Reason
By Graham’s Law, the diffusion rate ratio depends on the square root of molar mass ratio. Only He (4 g/mol) and CH₄ (16 g/mol) give a perfect square root ratio of 2, meaning helium diffuses exactly twice as fast as methane.
🔎 📝 Detailed Reasoning
According to Graham’s Law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass i.e. r ∝ 1/√M
So, for two gases A and B: rᴀ/rв =√Mв/√Mᴀ
🔎 Step by Step Check
📌 (a) N₂ (28 g/mol) and SO₃ (80 g/mol): rN₂/rSO₃ = √80/√28 ≈1.69 → Not exactly 2. ❌
📌 (b) He (4 g/mol) and CH₄ (16 g/mol): rʜₑ/rсʜ₄ = √16/√4 = 4 = 2 →✅ Exactly double.
📌 (c) CO (28 g/mol) and SO₂(64 g/mol): rCO/rSO₂ = √64/√28 ≈ 1.51 → Not exactly 2. ❌
📌 (d) O₃ (48 g/mol) and F₂ (38 g/mol): rF₂/rό₃ = √48/√38 ≈ 1.12 → Not exactly 2. ❌
🔎📌 Short Reason
By Graham’s Law, the diffusion rate ratio depends on the square root of molar mass ratio. Only He (4 g/mol) and CH₄ (16 g/mol) give a perfect square root ratio of 2, meaning helium diffuses exactly twice as fast as methane.
🔎 📝 Detailed Reasoning
According to Graham’s Law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass i.e. r ∝ 1/√M
So, for two gases A and B: rᴀ/rв =√Mв/√Mᴀ
🔎 Step by Step Check
📌 (a) N₂ (28 g/mol) and SO₃ (80 g/mol): rN₂/rSO₃ = √80/√28 ≈1.69 → Not exactly 2. ❌
📌 (b) He (4 g/mol) and CH₄ (16 g/mol): rʜₑ/rсʜ₄ = √16/√4 = 4 = 2 →✅ Exactly double.
📌 (c) CO (28 g/mol) and SO₂(64 g/mol): rCO/rSO₂ = √64/√28 ≈ 1.51 → Not exactly 2. ❌
📌 (d) O₃ (48 g/mol) and F₂ (38 g/mol): rF₂/rό₃ = √48/√38 ≈ 1.12 → Not exactly 2. ❌
3. Which of the following is not an example of oxidation?
a) Liberating bromine from a bromide
b) Converting calcium atoms into calcium ions
c) Converting iron (III) salts into iron (II) salts
d) Dissolving of a copper anode during electrolysis
✅ Correct Answer: (c) Converting iron (III) salts into iron (II) salts
📝 📌 Quick Recap
Oxidation means loss of electrons. In option (a), Fe³⁺ is reduced to Fe²⁺ (gain of electron), so it is not oxidation. All other processes involve electron loss, hence they are oxidation.
🔎🧠 Step by Step Reasoning
🔥Oxidation = loss of electrons / increase in oxidation state.
🔥Reduction = gain of electrons / decrease in oxidation state.
🔎 (a) Liberating Br₂ from Br⁻
🔥Br⁻ → Br₂ involves loss of electrons. That is oxidation ✅.
🔎 (b) Converting Ca → Ca²⁺
🔥Calcium atom loses 2 electrons to form Ca²⁺. That is oxidation ✅.
🔎 (c) Converting Fe³⁺ → Fe²⁺
🔥Iron(III) to Iron(II) means gain of electron (Fe³⁺ + ē → Fe²⁺). That is reduction, not oxidation.
🔎 (d) Dissolving copper anode in electrolysis
🔥Cu → Cu²⁺ + 2e⁻. Copper loses electrons. That is oxidation ✅.
📝 📌 Quick Recap
Oxidation means loss of electrons. In option (a), Fe³⁺ is reduced to Fe²⁺ (gain of electron), so it is not oxidation. All other processes involve electron loss, hence they are oxidation.
🔎🧠 Step by Step Reasoning
🔥Oxidation = loss of electrons / increase in oxidation state.
🔥Reduction = gain of electrons / decrease in oxidation state.
🔎 (a) Liberating Br₂ from Br⁻
🔥Br⁻ → Br₂ involves loss of electrons. That is oxidation ✅.
🔎 (b) Converting Ca → Ca²⁺
🔥Calcium atom loses 2 electrons to form Ca²⁺. That is oxidation ✅.
🔎 (c) Converting Fe³⁺ → Fe²⁺
🔥Iron(III) to Iron(II) means gain of electron (Fe³⁺ + ē → Fe²⁺). That is reduction, not oxidation.
🔎 (d) Dissolving copper anode in electrolysis
🔥Cu → Cu²⁺ + 2e⁻. Copper loses electrons. That is oxidation ✅.
4. Which of the following statement is true?
a) Samples of same element have different spectra
b) When an electron makes a transition continuous spectra is obtained.
c) Under similar conditions elements produce some wavelength.
d) The continuous spectrum is called spectral signature of elements.
✅ Correct Answer: (c) Under similar conditions elements produce some wavelength.
🔎📌Short Reason
👉 Each element has a unique set of spectral lines due to fixed electronic transitions. These characteristic wavelengths are reproducible under similar conditions and serve as the spectral signature of the element.
🔎🧠 Step by Step Reasoning
(a) ❌ Wrong → Samples of the same element always give the same line spectrum, because the electronic transitions are fixed for that element.
(b) ❌ Wrong → Electron transitions produce line spectra (discrete wavelengths), not continuous spectra. Continuous spectra come from hot solids or dense gases.
(c) ✅ Correct → Under similar conditions, each element produces the same characteristic wavelengths (its spectral lines). This is why spectra are used to identify elements.
(d) ❌ Wrong → The line spectrum (not continuous spectrum) is the spectral signature of elements.
🔎📌Short Reason
👉 Each element has a unique set of spectral lines due to fixed electronic transitions. These characteristic wavelengths are reproducible under similar conditions and serve as the spectral signature of the element.
🔎🧠 Step by Step Reasoning
(a) ❌ Wrong → Samples of the same element always give the same line spectrum, because the electronic transitions are fixed for that element.
(b) ❌ Wrong → Electron transitions produce line spectra (discrete wavelengths), not continuous spectra. Continuous spectra come from hot solids or dense gases.
(c) ✅ Correct → Under similar conditions, each element produces the same characteristic wavelengths (its spectral lines). This is why spectra are used to identify elements.
(d) ❌ Wrong → The line spectrum (not continuous spectrum) is the spectral signature of elements.
5. Which of the following sets of light waves is the correct order of increasing energy?
a) Infra-red radiation < visible light < Ultraviolet radiation < X-rays
b) X-rays < Ultraviolet radiation < visible light < infra-red radiation
c) X-rays < visible light < infra-red radiation < Ultraviolet radiation
d) Visible light < X-rays < Ultraviolet radiation < infra-red radiation
✅ Correct Answer: (a) Infrared radiation < Visible light < Ultraviolet radiation < X rays
🔎 Energy of electromagnetic waves increases with frequency (shorter wavelength → higher energy).
👉 Correct scientific order (low → high energy): Infrared < Visible < Ultraviolet < X rays
🔎 Energy of electromagnetic waves increases with frequency (shorter wavelength → higher energy).
👉 Correct scientific order (low → high energy): Infrared < Visible < Ultraviolet < X rays
6. The ions P³⁻, S²⁻ and Cl⁻ have radii 0.212 nm, 0.184 nm and 0.181 nm respectively. Which one of the following correctly explains the decrease in radius in going from P³⁻ to Cl⁻?
a) An increase in the total number of electrons and in the nuclear charge
b) An increase in the total number of electrons with the nuclear charge remaining constant
c) A decrease in the total number of electrons with the nuclear charge remaining constant
d) A constant number of electrons and an increase in the nuclear charge
✅ Correct Answer: (d) A constant number of electrons and an increase in the nuclear charge.
🔎📌 Short Reason
All three ions are isoelectronic (same electrons). As we move from P³⁻ to Cl⁻, the nuclear charge increases while electrons remain constant, pulling the electron cloud inward and decreasing the radius.
🔎🧠 Step by Step Reasoning
📌 P³⁻, S²⁻, Cl⁻ are all isoelectronic ions (they each have 18 electrons).
📌 Since the number of electrons is constant, the difference in size must be due to the nuclear charge (protons).
📌 Going from P (Z=15) → S (Z=16) → Cl (Z=17), the nuclear charge increases while electrons remain the same.
📌 Greater nuclear charge pulls the same electron cloud closer, reducing the radius.
📌 That explains why radius decreases from 0.212 nm (P³⁻) → 0.184 nm (S²⁻) → 0.181 nm (Cl⁻).
🔎📌 Short Reason
All three ions are isoelectronic (same electrons). As we move from P³⁻ to Cl⁻, the nuclear charge increases while electrons remain constant, pulling the electron cloud inward and decreasing the radius.
🔎🧠 Step by Step Reasoning
📌 P³⁻, S²⁻, Cl⁻ are all isoelectronic ions (they each have 18 electrons).
📌 Since the number of electrons is constant, the difference in size must be due to the nuclear charge (protons).
📌 Going from P (Z=15) → S (Z=16) → Cl (Z=17), the nuclear charge increases while electrons remain the same.
📌 Greater nuclear charge pulls the same electron cloud closer, reducing the radius.
📌 That explains why radius decreases from 0.212 nm (P³⁻) → 0.184 nm (S²⁻) → 0.181 nm (Cl⁻).
7. The oxidation number of oxygen in OF₂ and O₂F₂ is respectively :
a) +1 and +2
b) +2 and +1
c) –1 and –2
d) –2 and +1
✅ Correct Answer: (b) +2 and +1
👉 In compounds with fluorine, oxygen takes positive oxidation states. In OF₂, oxygen is +2, and in O₂F₂, oxygen is +1.
👉 Hence option (b) is correct.
👉 In compounds with fluorine, oxygen takes positive oxidation states. In OF₂, oxygen is +2, and in O₂F₂, oxygen is +1.
👉 Hence option (b) is correct.
8. During electrolysis of 18 g of acidified water, H₂ released at cathode at s.t.p. is:
a) 22.4 L
b) 11.2 L
c) 5.6 L
d) 44.8 L
✅ Correct Answer: (a) 22.4 liter
🔎📌 Short Reason
👉Electrolysis of 18 g (1 mole) of water produces 1 mole of H₂ gas. At STP, 1 mole of H₂ occupies 22.4 L, so the correct answer is option (a).
🔎🧠 Stepwise Explanation
📌 Molar mass of water (H₂O) = 18 g/mol. → So 18 g of water = 1 mole of H₂O.
📌 Electrolysis reaction: 2H₂O → electrolysis 2H₂ + O₂
📌 2 moles of water produce 2 moles of H₂ gas.
📌 Therefore, 1 mole of water (18g) produces 1 mole of H₂ gas.
📌 At STP: 1 mole of any gas occupies 22.4 L.
👉 So, 1 mole of H₂ = 22.4 L.
🔎📌 Short Reason
👉Electrolysis of 18 g (1 mole) of water produces 1 mole of H₂ gas. At STP, 1 mole of H₂ occupies 22.4 L, so the correct answer is option (a).
🔎🧠 Stepwise Explanation
📌 Molar mass of water (H₂O) = 18 g/mol. → So 18 g of water = 1 mole of H₂O.
📌 Electrolysis reaction: 2H₂O → electrolysis 2H₂ + O₂
📌 2 moles of water produce 2 moles of H₂ gas.
📌 Therefore, 1 mole of water (18g) produces 1 mole of H₂ gas.
📌 At STP: 1 mole of any gas occupies 22.4 L.
👉 So, 1 mole of H₂ = 22.4 L.
9. The order of rate of diffusion of gases PH₃, SO₃, F₂, & CO is:
a) PH₃ > SO₃ > F₂ > CO
b) F₂ > SO₃ > CO > PH₃
c) CO > PH₃ > F₂ > SO₃
d) PH₃ > CO > F₂ > SO₃
✅ Correct Answer: (c) CO > PH₃ > F₂ > SO₃
🔎📌 Short Reason
By Graham’s Law, diffusion rate is inversely proportional to the square root of molar mass. Since CO has the lowest molar mass, it diffuses fastest, followed by PH₃, then F₂, and finally SO₃ as the slowest. Hence option (c) is correct.
🔎🧠 Step by Step Reasoning
🔎Graham’s Law: r ∝ 1/√M → Lighter gases diffuse faster.
🔎Order of diffusion (lowest molar mass → fastest):
📌CO (28 g/mol) → fastest (least molar mass)
📌PH₃ (34 g/mol) → next
📌F₂ (38 g/mol) → next
📌SO₃ (80 g/mol) → slowest (largest molar mass)
👉So the correct order is: CO > PH₃ > F₂ > SO₃✅
🔎📌 Short Reason
By Graham’s Law, diffusion rate is inversely proportional to the square root of molar mass. Since CO has the lowest molar mass, it diffuses fastest, followed by PH₃, then F₂, and finally SO₃ as the slowest. Hence option (c) is correct.
🔎🧠 Step by Step Reasoning
🔎Graham’s Law: r ∝ 1/√M → Lighter gases diffuse faster.
🔎Order of diffusion (lowest molar mass → fastest):
📌CO (28 g/mol) → fastest (least molar mass)
📌PH₃ (34 g/mol) → next
📌F₂ (38 g/mol) → next
📌SO₃ (80 g/mol) → slowest (largest molar mass)
👉So the correct order is: CO > PH₃ > F₂ > SO₃✅
10. In which of the following substances does sulphur exhibit its highest oxidation state?
a) S₈
b) SO₂Cl₂
c) Na₂S₄O₆
d) Na₂S₂O₃
✅ Correct Answer: (b) SO₂Cl₂
🔎📌 Short Reason
👉 Sulphur reaches its maximum oxidation state of +6 in SO₂Cl₂ (sulphuryl chloride). In S₈ it is 0, in Na₂S₄O₆ and Na₂S₂O₃ it has mixed lower oxidation states. Hence option (b) is correct.
🔎 🧠 Stepwise Explanation
📌 (a) S₈ → Elemental sulphur. Oxidation state = 0.
📌 (b) SO₂Cl₂ (Sulphuryl chloride) → S+2(–2)+2(–1)=0 ⇒ S–4–2=0 ⇒ S=+6. ✅ 👈
📌 (c) Na₂S₄O₆ (Sodium tetrathionate) → This has mixed oxidation states of sulphur (average +2.5). Not the highest.
📌 (d) Na₂S₂O₃ (Sodium thiosulphate) → Contains sulphur in +2 and +6 states (average +2). Not the highest overall.
🔎📌 Short Reason
👉 Sulphur reaches its maximum oxidation state of +6 in SO₂Cl₂ (sulphuryl chloride). In S₈ it is 0, in Na₂S₄O₆ and Na₂S₂O₃ it has mixed lower oxidation states. Hence option (b) is correct.
🔎 🧠 Stepwise Explanation
📌 (a) S₈ → Elemental sulphur. Oxidation state = 0.
📌 (b) SO₂Cl₂ (Sulphuryl chloride) → S+2(–2)+2(–1)=0 ⇒ S–4–2=0 ⇒ S=+6. ✅ 👈
📌 (c) Na₂S₄O₆ (Sodium tetrathionate) → This has mixed oxidation states of sulphur (average +2.5). Not the highest.
📌 (d) Na₂S₂O₃ (Sodium thiosulphate) → Contains sulphur in +2 and +6 states (average +2). Not the highest overall.
11. 18 kg of water contains?
a) 18 moles
b) 180 moles
c) 1800 moles
d) 1000 moles
✅ Correct Answer: (d) 1000 moles
🔎📌 Short Reason
👉 18 kg of water = 18,000 g. Since 1 mole of water = 18 g, dividing gives 1000 moles. Hence option (d) is correct.
🔎 🧠 Stepwise Explanation
📌 Molar mass of water (H₂O) = 18 g/mol.
📌 Given mass of water = 18 kg = 18,000 g.
📌 Number of moles = mass/molar mass = 18,000/18 = 1000
🔎📌 Short Reason
👉 18 kg of water = 18,000 g. Since 1 mole of water = 18 g, dividing gives 1000 moles. Hence option (d) is correct.
🔎 🧠 Stepwise Explanation
📌 Molar mass of water (H₂O) = 18 g/mol.
📌 Given mass of water = 18 kg = 18,000 g.
📌 Number of moles = mass/molar mass = 18,000/18 = 1000
12. At zero kelvin, a gas:
a) Is super cooled
b) Has zero volume
c) Freezes
d) Liquefies
✅ Correct Answer: (b) Has zero volume
👉 At absolute zero (0 K), the ideal gas law predicts that gases would have zero volume, since molecular motion ceases. Hence option (a) is correct.
👉 At absolute zero (0 K), the ideal gas law predicts that gases would have zero volume, since molecular motion ceases. Hence option (a) is correct.
13. Which of the following compounds has or have sp³ hybridization?
a) CO₂
b) C₂H₄
c) C₂H₆
d) C₂H₂
✅ Correct Answer: (c) C₂H₆ (Ethane)
🔎 📌 Short Reason
👉 Only C₂H₆ (ethane) has carbons with sp³ hybridization, because each carbon makes 4 sigma bonds. CO₂ is sp, C₂H₄ is sp², and C₂H₂ is sp.
🔎 🧠 Step by Step Reasoning
🔎 (a) CO₂
Structure: O=C=O (linear).
Carbon forms two double bonds → hybridization = sp. ❌ Not sp³.
🔎 (b) C₂H₄ (Ethene)
Each carbon forms 3 sigma bonds + 1 π bond.
Hybridization = sp². ❌ Not sp³.
🔎 (c) C₂H₆ (Ethane)
Each carbon forms 4 sigma bonds (single bonds only).
Hybridization = sp³. ✅ Correct. 👈
🔎 (d) C₂H₂ (Ethyne)
Each carbon forms 2 sigma bonds + 2 π bonds.
Hybridization = sp. ❌ Not sp³.
🔎 📌 Short Reason
👉 Only C₂H₆ (ethane) has carbons with sp³ hybridization, because each carbon makes 4 sigma bonds. CO₂ is sp, C₂H₄ is sp², and C₂H₂ is sp.
🔎 🧠 Step by Step Reasoning
🔎 (a) CO₂
Structure: O=C=O (linear).
Carbon forms two double bonds → hybridization = sp. ❌ Not sp³.
🔎 (b) C₂H₄ (Ethene)
Each carbon forms 3 sigma bonds + 1 π bond.
Hybridization = sp². ❌ Not sp³.
🔎 (c) C₂H₆ (Ethane)
Each carbon forms 4 sigma bonds (single bonds only).
Hybridization = sp³. ✅ Correct. 👈
🔎 (d) C₂H₂ (Ethyne)
Each carbon forms 2 sigma bonds + 2 π bonds.
Hybridization = sp. ❌ Not sp³.
14. Which one of the following is independent of temperature?
a) Reaction order
b) Molality
c) Kc
d) Kw
✅ Correct Answer: (b) Molality
👉 Molality is based on the ratio of moles of solute to mass of solvent. Since mass does not change with temperature, molality is independent of temperature, unlike Kc and Kw which vary with temperature. Hence option (b) is correct.
👉 Molality is based on the ratio of moles of solute to mass of solvent. Since mass does not change with temperature, molality is independent of temperature, unlike Kc and Kw which vary with temperature. Hence option (b) is correct.
15. Which one of the following is an endothermic change?
a) Condensation
b) Electrolysis
c) Freezing
d) Fermentation
✅ Correct Answer: (b) Electrolysis
👉 Condensation, freezing and fermentation release heat (exothermic), while electrolysis requires energy input to break chemical bonds, making it endothermic. Hence option (b) is correct.
👉 Condensation, freezing and fermentation release heat (exothermic), while electrolysis requires energy input to break chemical bonds, making it endothermic. Hence option (b) is correct.
16. The oxidation number of oxygen in a molecule is not –2 when it is bonded to a ___________________ atom.
a) N
b) S
c) F
d) Br
✅ Correct Answer: (c) F
👉 Oxygen usually has –2 oxidation state, but when bonded to fluorine (the most electronegative element), oxygen takes positive oxidation states (+2 in OF₂, +1 in O₂F₂). Hence option (b) is correct.
👉 Oxygen usually has –2 oxidation state, but when bonded to fluorine (the most electronegative element), oxygen takes positive oxidation states (+2 in OF₂, +1 in O₂F₂). Hence option (b) is correct.
17. In hydrates, water of crystallization combine with positive metal ions by:
a) Ionic bond
b) Hydrogen bond
c) Co ordinate covalent bond
d) Covalent bond
✅ Correct Answer: (c) Co ordinate covalent bond
👉 In hydrates, water of crystallization binds to positive metal ions through co ordinate covalent bonds, where the oxygen atom donates its lone pair to the metal ion. Hence option (c) is correct.
👉 In hydrates, water of crystallization binds to positive metal ions through co ordinate covalent bonds, where the oxygen atom donates its lone pair to the metal ion. Hence option (c) is correct.
18. If it took 10 ml of 1 M HCl to titrate 20 ml of NaOH solution of unknown strength to its end point, what was the concentration of NaOH?
a) 1.5 M
b) 0.5 M
c) 2.0 M
d) 2.5 M
✅ Correct Answer: (b) 0.5 M
🔎It took 10 mL of 1 M HCl to neutralize 20 mL of NaOH solution. We want the concentration of NaOH.
🔎By applying the simple titration formula M₁V₁=M₂V₂, we find that the NaOH solution has a concentration of 0.5 M.
🔎M₁V₁ = M₂V₂
🔎M₁ = concentration of HCl = 1 M
🔎V₁ = volume of HCl = 10 mL
🔎M₂ = concentration of NaOH = ? 🔎V₂ = volume of NaOH = 20 mL
🔎1 × 10 = M₂ × 20
👉 M₂ =10/20 = 0.5 M✅
🔎It took 10 mL of 1 M HCl to neutralize 20 mL of NaOH solution. We want the concentration of NaOH.
🔎By applying the simple titration formula M₁V₁=M₂V₂, we find that the NaOH solution has a concentration of 0.5 M.
🔎M₁V₁ = M₂V₂
🔎M₁ = concentration of HCl = 1 M
🔎V₁ = volume of HCl = 10 mL
🔎M₂ = concentration of NaOH = ? 🔎V₂ = volume of NaOH = 20 mL
🔎1 × 10 = M₂ × 20
👉 M₂ =10/20 = 0.5 M✅
19. The value of ionic product (Kw) of water at 25ºC is equal to:
a) 1 × 10¹⁴
b) 1 × 10⁻¹⁴
c) 1 × 10⁻⁷
d) 1 × 10⁻¹⁰
✅ Correct Answer: (b) 1 × 10⁻¹⁴
📌 Kw = ionic product of water = [H⁺][OH⁻].
📌 At 25 °C, pure water has: [H⁺] = [OH⁻] = 1 × 10⁻⁷ M.
👉 Therefore: Kw = (1 × 10⁻⁷)(1 × 10⁻⁷)=1 × 10⁻¹⁴ ✅
📌 At 25 °C, the ionic product of water is Kw = 1 × 10⁻¹⁴, since both [H⁺] and [OH⁻] are 1×10⁻⁷ M. Hence option (b) is correct.
📌 Kw = ionic product of water = [H⁺][OH⁻].
📌 At 25 °C, pure water has: [H⁺] = [OH⁻] = 1 × 10⁻⁷ M.
👉 Therefore: Kw = (1 × 10⁻⁷)(1 × 10⁻⁷)=1 × 10⁻¹⁴ ✅
📌 At 25 °C, the ionic product of water is Kw = 1 × 10⁻¹⁴, since both [H⁺] and [OH⁻] are 1×10⁻⁷ M. Hence option (b) is correct.
20. The sum of pH and pOH of any solution at 25 ºC is:
a) 7
b) 9
c) 14
d) 3
✅ Correct Answer: (c) 14
👉 At 25 ºC, the sum of pH and pOH of any aqueous solution is always 14, because Kw = 1 × 10⁻¹⁴. Hence option (c) is correct.
🔎 🧠 Step by Step Reasoning
📌At 25 ºC, the ionic product of water is:
📌Kw = 1 × 10⁻¹⁴
📌Since: pKw = −log Kw = 14 And:
📌pH + pOH = pKw. Therefore:
👉 pH + pOH=14✅ Correct Answer
👉 At 25 ºC, the sum of pH and pOH of any aqueous solution is always 14, because Kw = 1 × 10⁻¹⁴. Hence option (c) is correct.
🔎 🧠 Step by Step Reasoning
📌At 25 ºC, the ionic product of water is:
📌Kw = 1 × 10⁻¹⁴
📌Since: pKw = −log Kw = 14 And:
📌pH + pOH = pKw. Therefore:
👉 pH + pOH=14✅ Correct Answer
21. The aqueous solution of sodium acetate is:
a) Neutral
b) Alkaline
c) Weakly acidic
d) Strongly acidic
✅ Correct Answer: (b) Alkaline
🔎 📌 Short Reason
👉 Sodium acetate solution is alkaline because the acetate ion hydrolyzes to produce hydroxide ions (anionic hydrolysis), raising the pH above 7. Hence option (b) is correct.
🔎 🧠 Step by Step Reasoning
📌Sodium acetate (CH₃COONa) is the salt of a strong base (NaOH) and a weak acid (CH₃COOH).
📌When dissolved in water:
📌Na⁺ ion → neutral (does not hydrolyze).
📌CH₃COO⁻ ion → undergoes hydrolysis:
📌CH₃COO⁻ + H₂O → CH₃COOH+OH⁻
👉 This reaction produces OH⁻ ions, making the solution basic (alkaline). ✅
🔎 📌 Short Reason
👉 Sodium acetate solution is alkaline because the acetate ion hydrolyzes to produce hydroxide ions (anionic hydrolysis), raising the pH above 7. Hence option (b) is correct.
🔎 🧠 Step by Step Reasoning
📌Sodium acetate (CH₃COONa) is the salt of a strong base (NaOH) and a weak acid (CH₃COOH).
📌When dissolved in water:
📌Na⁺ ion → neutral (does not hydrolyze).
📌CH₃COO⁻ ion → undergoes hydrolysis:
📌CH₃COO⁻ + H₂O → CH₃COOH+OH⁻
👉 This reaction produces OH⁻ ions, making the solution basic (alkaline). ✅
22. Photons of red color are:
a) More energetic than that of violet colour
b) Equal energetic than that of violet colour
c) Less energetic than that of violet colour
d) None of the above
✅ Correct Answer: (c) Less energetic than that of violet colour
👉 Photon energy decreases with increasing wavelength. Since red light has a longer wavelength than violet, red photons are less energetic. Hence option (b) is correct.
👉 Photon energy decreases with increasing wavelength. Since red light has a longer wavelength than violet, red photons are less energetic. Hence option (b) is correct.
23. Which one is larger in size?
a) Na⁺
b) Mg²⁺
c) Al³⁺
d) All are equal in size
✅ Correct Answer: (a) Na⁺
👉 Among Na⁺, Mg²⁺, and Al³⁺, the Na⁺ ion is largest because it has the lowest positive charge and hence the weakest nuclear pull on its electrons. The size decreases as the charge increases.
👉 Among Na⁺, Mg²⁺, and Al³⁺, the Na⁺ ion is largest because it has the lowest positive charge and hence the weakest nuclear pull on its electrons. The size decreases as the charge increases.
24. Which of the following has maximum bond angle?
a) H₂O
b) CCl₄
c) SO₂
d) NH₃
✅ Correct Answer: (c) SO₂
🔎 📌 Short Reason
👉 Among the given molecules, SO₂ has the largest bond angle (~119°) because of its bent structure derived from trigonal planar geometry.
🔎 🧠 Step by Step Reasoning
📌 (a) H₂O → Bent structure due to 2 lone pairs. → Bond angle ≈ 104.5°.
📌 (b) CCl₄ → Tetrahedral geometry. → Bond angle = 109.5°.
📌 (c) SO₂ → Bent structure with resonance. → Bond angle ≈ 119° (close to trigonal planar) → Maximum bond angle 👈
📌 (d) NH₃ → Trigonal pyramidal (1 lone pair). → Bond angle ≈ 107°.
🔎 📌 Short Reason
👉 Among the given molecules, SO₂ has the largest bond angle (~119°) because of its bent structure derived from trigonal planar geometry.
🔎 🧠 Step by Step Reasoning
📌 (a) H₂O → Bent structure due to 2 lone pairs. → Bond angle ≈ 104.5°.
📌 (b) CCl₄ → Tetrahedral geometry. → Bond angle = 109.5°.
📌 (c) SO₂ → Bent structure with resonance. → Bond angle ≈ 119° (close to trigonal planar) → Maximum bond angle 👈
📌 (d) NH₃ → Trigonal pyramidal (1 lone pair). → Bond angle ≈ 107°.
25. Which of the following aqueous solution of salt is acidic?
a) Na₂CO₃
b) KCN
c) NaOCl
d) CuSO₄
✅ Correct Answer: (d) CuSO₄
🔎 📌 Short Reason
👉 Among the given salts, CuSO₄ forms an acidic solution because it is derived from a weak base (Cu(OH)₂) and a strong acid (H₂SO₄). The other salts are derived from strong bases and weak acids, making their solutions basic.
🔎 🧠 Step by Step Reasoning
🔎 (a) Na₂CO₃ (Sodium carbonate)
📌 Salt of a strong base (NaOH) and weak acid (H₂CO₃). Hydrolysis produces OH⁻ ions → basic solution. ❌
🔎 (b) KCN (Potassium cyanide)
📌 Salt of strong base (KOH) and weak acid (HCN). Hydrolysis produces OH⁻ ions → basic solution. ❌
🔎 (c) NaOCl (Sodium hypochlorite)
📌 Salt of strong base (NaOH) and weak acid (HOCl). Hydrolysis produces OH⁻ ions → basic solution. ❌
🔎 (d) CuSO₄ (Copper(II) sulfate)
📌 Salt of weak base (Cu(OH)₂) and strong acid (H₂SO₄). Hydrolysis produces H⁺ ions → acidic solution. ✅ 👈
🔎 📌 Short Reason
👉 Among the given salts, CuSO₄ forms an acidic solution because it is derived from a weak base (Cu(OH)₂) and a strong acid (H₂SO₄). The other salts are derived from strong bases and weak acids, making their solutions basic.
🔎 🧠 Step by Step Reasoning
🔎 (a) Na₂CO₃ (Sodium carbonate)
📌 Salt of a strong base (NaOH) and weak acid (H₂CO₃). Hydrolysis produces OH⁻ ions → basic solution. ❌
🔎 (b) KCN (Potassium cyanide)
📌 Salt of strong base (KOH) and weak acid (HCN). Hydrolysis produces OH⁻ ions → basic solution. ❌
🔎 (c) NaOCl (Sodium hypochlorite)
📌 Salt of strong base (NaOH) and weak acid (HOCl). Hydrolysis produces OH⁻ ions → basic solution. ❌
🔎 (d) CuSO₄ (Copper(II) sulfate)
📌 Salt of weak base (Cu(OH)₂) and strong acid (H₂SO₄). Hydrolysis produces H⁺ ions → acidic solution. ✅ 👈
26. Which of the following molecule has non zero dipole moment?
a) BF₃
b) CCl₄
c) CO₂
d) SO₂
✅ Correct Answer: (d) SO₂
🔎 📌 Short Reason
👉 Among the given molecules, only SO₂ has a non zero dipole moment because of its bent geometry. BF₃, CCl₄, and CO₂ are symmetrical, so their bond dipoles cancel out.
🔎 🧠 Step by Step Reasoning
🔎 (a) BF₃
📌Geometry: Trigonal planar.
📌Symmetrical → dipoles cancel. Net dipole moment = 0. ❌
🔎 (b) CCl₄
📌Geometry: Tetrahedral.
📌Symmetrical → dipoles cancel. Net dipole moment = 0. ❌
🔎 (c) CO₂
📌Geometry: Linear (O=C=O).
📌Symmetrical → dipoles cancel. Net dipole moment = 0. ❌
🔎 (d) SO₂
Geometry: Bent (like H₂O).
📌Dipoles do not cancel. Net dipole moment ≠ 0. ✅👈
🔎 📌 Short Reason
👉 Among the given molecules, only SO₂ has a non zero dipole moment because of its bent geometry. BF₃, CCl₄, and CO₂ are symmetrical, so their bond dipoles cancel out.
🔎 🧠 Step by Step Reasoning
🔎 (a) BF₃
📌Geometry: Trigonal planar.
📌Symmetrical → dipoles cancel. Net dipole moment = 0. ❌
🔎 (b) CCl₄
📌Geometry: Tetrahedral.
📌Symmetrical → dipoles cancel. Net dipole moment = 0. ❌
🔎 (c) CO₂
📌Geometry: Linear (O=C=O).
📌Symmetrical → dipoles cancel. Net dipole moment = 0. ❌
🔎 (d) SO₂
Geometry: Bent (like H₂O).
📌Dipoles do not cancel. Net dipole moment ≠ 0. ✅👈
27. In which of the following substances chromium does not exhibit its highest oxidation state of +6?
a) Cr₂O₃
b) CrO₃
c) K₂Cr₂O₇
d) CrO₂Cl₂
✅ Correct Answer: (a) Cr₂O₃
🔎 📌 Short Reason
Chromium exhibits its highest oxidation state of +6 in CrO₃, K₂Cr₂O₇, and CrO₂Cl₂. But in Cr₂O₃, chromium is only in the +3 oxidation state, so this is the correct answer.
🔎 🧠 Step by Step Reasoning
🔎 (a) Cr₂O₃ (Chromium(III) oxide):
📌Oxygen = –2. Let Cr = x.
📌Equation: 2x + 3(–2) = 0 → 2x – 6 = 0 → x = +3. Oxidation state = +3. ❌ NOT +6. 👈
🔎 (b) CrO₃ (Chromium trioxide):
📌Oxygen = –2.
📌Equation: x + 3(–2) = 0 → x – 6 = 0 → x = +6. Oxidation state = +6. ✅
🔎 (c) K₂Cr₂O₇ (Potassium dichromate):
📌Oxygen = –2, K = +1.
📌Equation: 2(+1) + 2x + 7(–2) = 0 → 2 + 2x – 14 = 0 → 2x – 12 = 0 → x = +6. Oxidation state = +6. ✅
🔎 (d) CrO₂Cl₂ (Chromyl chloride):
📌Oxygen = –2, Cl = –1.
📌Equation: x + 2(–2) + 2(–1) = 0 → x – 4 – 2 = 0 → x = +6. Oxidation state = +6. ✅
🔎 📌 Short Reason
Chromium exhibits its highest oxidation state of +6 in CrO₃, K₂Cr₂O₇, and CrO₂Cl₂. But in Cr₂O₃, chromium is only in the +3 oxidation state, so this is the correct answer.
🔎 🧠 Step by Step Reasoning
🔎 (a) Cr₂O₃ (Chromium(III) oxide):
📌Oxygen = –2. Let Cr = x.
📌Equation: 2x + 3(–2) = 0 → 2x – 6 = 0 → x = +3. Oxidation state = +3. ❌ NOT +6. 👈
🔎 (b) CrO₃ (Chromium trioxide):
📌Oxygen = –2.
📌Equation: x + 3(–2) = 0 → x – 6 = 0 → x = +6. Oxidation state = +6. ✅
🔎 (c) K₂Cr₂O₇ (Potassium dichromate):
📌Oxygen = –2, K = +1.
📌Equation: 2(+1) + 2x + 7(–2) = 0 → 2 + 2x – 14 = 0 → 2x – 12 = 0 → x = +6. Oxidation state = +6. ✅
🔎 (d) CrO₂Cl₂ (Chromyl chloride):
📌Oxygen = –2, Cl = –1.
📌Equation: x + 2(–2) + 2(–1) = 0 → x – 4 – 2 = 0 → x = +6. Oxidation state = +6. ✅
28. Which of the following is the value of atmospheric pressure in SI unit?
a) 14.7 psi
b) 101325 Pa
c) 760 torr
d) 760 mm Hg
✅ Correct Answer: (b) 101325 Pa
🔎 📌 Reason (short)
👉 Atmospheric pressure at sea level equals 101325 Pa, which is the SI unit of pressure. The other values are correct equivalents but expressed in non SI units.
🔎 🧠 Step by Step Reasoning
📌Atmospheric pressure at sea level is commonly expressed as:
📌1 atm = 101325 Pa (SI unit). ✅
📌1 atm = 760 mm Hg.
📌1 atm = 760 torr.
📌1 atm ≈ 14.7 psi.
📌The SI unit of pressure is Pascal (Pa).
📌Therefore, the correct value in SI units is 1 atm = 101325 Pa✅👈
🔎 📌 Reason (short)
👉 Atmospheric pressure at sea level equals 101325 Pa, which is the SI unit of pressure. The other values are correct equivalents but expressed in non SI units.
🔎 🧠 Step by Step Reasoning
📌Atmospheric pressure at sea level is commonly expressed as:
📌1 atm = 101325 Pa (SI unit). ✅
📌1 atm = 760 mm Hg.
📌1 atm = 760 torr.
📌1 atm ≈ 14.7 psi.
📌The SI unit of pressure is Pascal (Pa).
📌Therefore, the correct value in SI units is 1 atm = 101325 Pa✅👈
29. For what value of Kc, the forward reaction is almost complete?
a) 0
b) 1
c) 10⁻³⁰
d) 10³⁰
✅ Correct Answer: (d) 10³⁰
🔎 📌 Short Reason
👉 A very large equilibrium constant (like 10³⁰) means the reaction strongly favors products, so the forward reaction is almost complete.
🔎 🧠 Step by Step Reasoning
📌 Equilibrium constant (Kc): Kc = [Products]/[Reactants]
📌 If Kc ≈ 0 → Reaction hardly proceeds forward (mostly reactants). ❌
📌 If Kc ≈ 1 → Significant amounts of both reactants and products present (reaction is at balance). ❌
📌 If Kc ≈ 10⁻³⁰ → Extremely small → reaction does not proceed forward (almost only reactants). ❌
📌 If Kc ≈ 10³⁰ → Extremely large → reaction goes almost completely forward (almost only products). ✅
🔎 📌 Short Reason
👉 A very large equilibrium constant (like 10³⁰) means the reaction strongly favors products, so the forward reaction is almost complete.
🔎 🧠 Step by Step Reasoning
📌 Equilibrium constant (Kc): Kc = [Products]/[Reactants]
📌 If Kc ≈ 0 → Reaction hardly proceeds forward (mostly reactants). ❌
📌 If Kc ≈ 1 → Significant amounts of both reactants and products present (reaction is at balance). ❌
📌 If Kc ≈ 10⁻³⁰ → Extremely small → reaction does not proceed forward (almost only reactants). ❌
📌 If Kc ≈ 10³⁰ → Extremely large → reaction goes almost completely forward (almost only products). ✅
30. If 28 g of nitrogen gas comprises of y molecules; how many molecules are there in 96 g of ozone?
a) y
b) 2y
c) ½y
d) ¼y
✅ Correct Answer: (b) 2y
🔎 📌 Short Reason
👉 28 g of N₂ corresponds to y molecules (1 mole). 96 g of O₃ corresponds to 2 moles = 2y molecules. Hence option (a) 2y is correct.
🔎 🧠 Step by Step Reasoning
🔎 Molecular mass of N₂:
📌 M(N₂) = 14 × 2 = 28 g/mol
📌 Given: 28 g of N₂ → 1 mole.
📌 Molecules in 1 mole = Avogadro’s number (Nₐ).
📌 So, 28 g N₂ = y molecules = Nₐ molecules.
📌 Hence, y = Na.
🔎 Molecular mass of O₃ (ozone):
📌 M(O₃)=16×3 = 48 g/mol
📌 Given: 96 g of O₃.
📌 Moles =96/48 =2 moles
📌 Molecules in 2 moles:
📌 2 × Na = 2y👈
🔎 📌 Short Reason
👉 28 g of N₂ corresponds to y molecules (1 mole). 96 g of O₃ corresponds to 2 moles = 2y molecules. Hence option (a) 2y is correct.
🔎 🧠 Step by Step Reasoning
🔎 Molecular mass of N₂:
📌 M(N₂) = 14 × 2 = 28 g/mol
📌 Given: 28 g of N₂ → 1 mole.
📌 Molecules in 1 mole = Avogadro’s number (Nₐ).
📌 So, 28 g N₂ = y molecules = Nₐ molecules.
📌 Hence, y = Na.
🔎 Molecular mass of O₃ (ozone):
📌 M(O₃)=16×3 = 48 g/mol
📌 Given: 96 g of O₃.
📌 Moles =96/48 =2 moles
📌 Molecules in 2 moles:
📌 2 × Na = 2y👈
31. Which of the following molecules form sigma bond by the of s-p overlap?
a) O₂
b) CH₄
c) HF
d) F₂
✅ Correct Answer: (c) HF
👉 Among the given molecules, HF forms its sigma bond by s–p overlap (H 1s orbital with F 2p orbital). The others involve either p–p overlap or hybrid orbital overlaps.
👉 Among the given molecules, HF forms its sigma bond by s–p overlap (H 1s orbital with F 2p orbital). The others involve either p–p overlap or hybrid orbital overlaps.
32. Which of the following pairs of gases have exactly the same rate of diffusion as each other?
a) N₂ and SO₃
b) CO and SO₂
c) He and CH₄
d) CO₂ and C₃H₈
✅ Correct Answer: (d) CO₂ and C₃H₈
👉 According to Graham’s Law, gases with equal molar masses diffuse at the same rate. Only CO₂ (44 g/mol) and C₃H₈ (44 g/mol) have identical molar masses, so they diffuse at the same rate.
👉 According to Graham’s Law, gases with equal molar masses diffuse at the same rate. Only CO₂ (44 g/mol) and C₃H₈ (44 g/mol) have identical molar masses, so they diffuse at the same rate.
33. Which one of the following compounds in a 1 mol/dm3 solution has the lowest pH value?
a) Ammonia
b) Sodium hydroxide
c) Hydrochloric acid
d) Ethanoic acid
✅ Correct Answer: (c) Hydrochloric acid
👉 Among the given compounds, 1 M HCl has the lowest pH (~0)
👉 Among the given compounds, 1 M HCl has the lowest pH (~0)
34. Which one of the following is a molecular solid?
a) Carborundum
b) Dry ice
c) Iron
d) KCl
✅ Correct Answer: (b) Dry ice
👉 Among the given options, dry ice (solid CO₂) is a molecular solid, since it consists of discrete molecules held together by weak intermolecular forces. The others are ionic, metallic, or covalent network solids.
👉 Among the given options, dry ice (solid CO₂) is a molecular solid, since it consists of discrete molecules held together by weak intermolecular forces. The others are ionic, metallic, or covalent network solids.
35. Which of the following is the correct electronic configuration of carbon according to Hund’s rule?
a) 1s², 2s² 2px¹ 2py¹
b) 1s², 2s¹ 2px¹ 2py¹ 2pz¹
c) 1s², 2s² 2px² 2py⁰ 2pz⁰
d) 1s², 2s² 2px¹ 2py¹ 2pz¹
✅ Correct Answer: (a) 1s², 2s² 2px¹ 2py¹
👉 Carbon’s correct electronic configuration according to Hund’s rule is 1s², 2s², 2px¹, 2py¹. This ensures maximum multiplicity by keeping the two 2p electrons unpaired in separate orbitals.
👉 Carbon’s correct electronic configuration according to Hund’s rule is 1s², 2s², 2px¹, 2py¹. This ensures maximum multiplicity by keeping the two 2p electrons unpaired in separate orbitals.
36. Which of the following is the SI unit of dipole moment?
a) debye
b) Coulomb meter
c) e.s.u. m
d) All of the above
✅ Correct Answer: (b) Coulomb meter
👉 The SI unit of dipole moment is coulomb meter (C·m). Debye and e.s.u. m are non SI units, though commonly used in chemistry. Hence option (b) is correct.
👉 The SI unit of dipole moment is coulomb meter (C·m). Debye and e.s.u. m are non SI units, though commonly used in chemistry. Hence option (b) is correct.
37. Which of the following has pyramidal geometry?
a) CH₄
b) SO₂
c) PCl₃
d) BF₃
✅ Correct Answer: (c) PCl₃
🔎 📌 Short Reason
Among the given molecules, PCl₃ has pyramidal geometry due to the presence of one lone pair on phosphorus, which pushes the three bonding pairs downward into a trigonal pyramidal arrangement.
🔎 🧠 Step by Step Reasoning
🔎 (a) CH₄ (Methane):
📌 Central atom: C. 4 bonding pairs, no lone pairs. Geometry: Tetrahedral (109.5°). ❌
🔎 (b) SO₂ (Sulfur dioxide):
📌 Central atom: S.2 bonding pairs, 1 lone pair. Geometry: Bent (≈119°). ❌
🔎 (c) PCl₃ (Phosphorus trichloride):
📌 Central atom: P. 3 bonding pairs, 1 lone pair. Geometry: Trigonal pyramidal. ✅ 👈
🔎 (d) BF₃ (Boron trifluoride):
📌 Central atom: B. 3 bonding pairs, no lone pairs. Geometry: Trigonal planar (120°). ❌
🔎 📌 Short Reason
Among the given molecules, PCl₃ has pyramidal geometry due to the presence of one lone pair on phosphorus, which pushes the three bonding pairs downward into a trigonal pyramidal arrangement.
🔎 🧠 Step by Step Reasoning
🔎 (a) CH₄ (Methane):
📌 Central atom: C. 4 bonding pairs, no lone pairs. Geometry: Tetrahedral (109.5°). ❌
🔎 (b) SO₂ (Sulfur dioxide):
📌 Central atom: S.2 bonding pairs, 1 lone pair. Geometry: Bent (≈119°). ❌
🔎 (c) PCl₃ (Phosphorus trichloride):
📌 Central atom: P. 3 bonding pairs, 1 lone pair. Geometry: Trigonal pyramidal. ✅ 👈
🔎 (d) BF₃ (Boron trifluoride):
📌 Central atom: B. 3 bonding pairs, no lone pairs. Geometry: Trigonal planar (120°). ❌
38. Which of the following compounds has/have sp³ hybridization?
a) SO₄²⁻
b) CH₃⁻
c) CH₄
d) All of them
✅ Correct Answer: (d) All of them
🔎 SO₄²⁻ → sp³ (tetrahedral arrangement).
🔎 CH₃⁻ → sp³ (trigonal pyramidal geometry due to lone pair).
🔎 CH₄ → sp³ (tetrahedral geometry).
📌 Thus, all of them exhibit sp³ hybridization. ✅ 👈
🔎 SO₄²⁻ → sp³ (tetrahedral arrangement).
🔎 CH₃⁻ → sp³ (trigonal pyramidal geometry due to lone pair).
🔎 CH₄ → sp³ (tetrahedral geometry).
📌 Thus, all of them exhibit sp³ hybridization. ✅ 👈
39. The oxidation number of oxygen in H₂O₂ and KO₂ is:
a) –1 and +2 respectively
b) –2 and +½ respectively
c) –1 and –½ respectively
d) –2 and +1 respectively
✅ Correct Answer: (c) –1 and –½ respectively
👉 In H₂O₂, oxygen has oxidation number –1.
👉 In KO₂, oxygen exists in the superoxide ion (O₂⁻), giving an average oxidation number of –½. Thus, the correct option is (c).
👉 In H₂O₂, oxygen has oxidation number –1.
👉 In KO₂, oxygen exists in the superoxide ion (O₂⁻), giving an average oxidation number of –½. Thus, the correct option is (c).
40. Which ion has the greatest charge density?
a) K⁺
b) Al³⁺
c) O²⁻
d) Mg²⁺
✅ Correct Answer: (b) Al³⁺
👉 Among the given ions, Al³⁺ has the greatest charge density because it combines a high charge (+3) with a small ionic radius, making it far stronger in polarizing power compared to K⁺, O²⁻, or Mg²⁺.
👉 Among the given ions, Al³⁺ has the greatest charge density because it combines a high charge (+3) with a small ionic radius, making it far stronger in polarizing power compared to K⁺, O²⁻, or Mg²⁺.
Your Score: 0 / 0
🔥 جونؔ ایلیاء 🔥
زندانیانِ شام و سحر خیریت سے ہیں 🌙
ہر لمحہ جی رہے ہیں مگر خیریت سے ہیں ⏳
شہرِ یقیں میںاب کوئی دم خم نہیں رہا 🏙️
دشتِ گماں کے خاک بسر خیریت سے ہیں 🏜️
آخر ہے کون جو کسی پل کہ سکے یہ بات 🙏
اللہ اور تمام بشر خیریت سے ہیں 🌍
ہے اپنے اپنے طور پہ ہر چیز اس گھڑی ⏱️
مژگانِ خشک و دامنِ تر خیریت سے ہیں 💧
اب فیصلوں کا کم نظروں پر مدار ہے 👀
یعنی تمام اہل نظر خیریت سے ہیں 🌟
پیروں سے آبلوں کا وہی ہے معاملہ 👣
سودائیانِ حال کے سر خیریت سے ہیں 🌀
ہم جن گھروں کو چھوڑ کے آئے تھے ناگہاں 🏠
شکوے کی بات ہے،وہ اگر خیریت سے ہیں 💭
لو چل رہی ہے،محو ہے اپنے میں دوپہر ☀️
خاک اڑ رہی ہے اور کھنڈر خیریت سے ہیں 🏚️
ہم اہلِ شہر اپنے جوانوں کے درمیاں 👥
جونؔ!ایک معجزہ ہے اگر خیریت سے ہیں ✨
برباد ہوچکا ہے ہنر اک ہنر کے ساتھ 🎭
اور اپنے صاحبانِ ہنر خیریت سے ہیں 🎨
شکرِ خدا شہید ہوئے اہلِ حق تمام ⚔️
برگستوان وتیغ و تبر خیریت سے ہیں 🛡️
اب اس کا قصرِ ناز کہاں اور وہ کہاں 🏰
بس در ہے اور بندئہ در خیریت سے ہیں 🚪
ہم ہیں کہ شاعری ہے ہمارے لئے عذاب 📜
ورنہ تمام جوشؔ و جگرؔ خیریت سے ہیں ❤️
شاعر تو دو ہیں میرؔ تقی اور میر جونؔ ✒️
باقی جوہیں وہ شام و سحر خیریت سے ہیں 🌅
Tags
1st Year
Aptitude test (MDCAT/ECAT)
Class XI chemistry MCQs
grand chemistry test
Interactive Chemistry Quiz
MDCAT chemistry MCQs
MDCAT ECAT FSC chemistry quiz
online chemistry practice