🎉 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!
1: The px, py and pz orbitals are called degenerate orbitals as they have
a) Same size
b) Same spatial orientation
c) Same energy
d) Same orbit
✅ Correct Answer: (c) Same energy
🔎📌 Short Reason
✨ In fact, degeneracy means that multiple orbitals (like px, py, pz) have equal energy levels even though their orientations differ.
🔎🧠 Step by Step Reasoning
📌Orbitals like px, py, pz belong to the same subshell (the p subshell).
📌They differ in spatial orientation (x, y, z axes), so option (b) is not correct.
📌Their size depends on the principal quantum number (n), not on the orientation, so option (a) is not correct.
📌They are not the same orbit (orbit ≠ orbital; orbit is Bohr’s model, orbital is quantum mechanical), so option (d) is not correct.
👉🏼What makes them degenerate is that they all have the same energy in the absence of an external field.
🔎📌 Short Reason
✨ In fact, degeneracy means that multiple orbitals (like px, py, pz) have equal energy levels even though their orientations differ.
🔎🧠 Step by Step Reasoning
📌Orbitals like px, py, pz belong to the same subshell (the p subshell).
📌They differ in spatial orientation (x, y, z axes), so option (b) is not correct.
📌Their size depends on the principal quantum number (n), not on the orientation, so option (a) is not correct.
📌They are not the same orbit (orbit ≠ orbital; orbit is Bohr’s model, orbital is quantum mechanical), so option (d) is not correct.
👉🏼What makes them degenerate is that they all have the same energy in the absence of an external field.
2: The total values of magnetic quantum for a given value of azimuthal quantum number is
a) 2𝓵
b) 2𝓵 – 1
c) 2𝓵 + 1
d) 2𝓵 – 2
✅ Correct Answer: (c) 2𝓵 + 1
🔎📌 Short Reason
For a given azimuthal quantum number ℓ, the magnetic quantum number mℓ ranges from −ℓ to +ℓ. That gives a total of 2ℓ+1 values.
🔎🧠 Step by Step Reasoning
📌The azimuthal quantum number (ℓ) defines the subshell (s, p, d, f …).
📌For a given ℓ, the magnetic quantum number (mℓ) can take integer values from –ℓ to +ℓ.
📌That means the possible values are: mℓ = −ℓ, (−ℓ+1), …, 0, …, (+ℓ−1), +ℓ
👉🏼The total number of values is: (2ℓ+1)
🔎Example:
📌ℓ = 1 (p subshell) → mℓ = –1, 0, +1 → 3 values = 2(ℓ)+1.
📌ℓ = 2 (d subshell) → mℓ = –2, –1, 0, +1, +2 → 5 values = 2(2)+1.
🔎📌 Short Reason
For a given azimuthal quantum number ℓ, the magnetic quantum number mℓ ranges from −ℓ to +ℓ. That gives a total of 2ℓ+1 values.
🔎🧠 Step by Step Reasoning
📌The azimuthal quantum number (ℓ) defines the subshell (s, p, d, f …).
📌For a given ℓ, the magnetic quantum number (mℓ) can take integer values from –ℓ to +ℓ.
📌That means the possible values are: mℓ = −ℓ, (−ℓ+1), …, 0, …, (+ℓ−1), +ℓ
👉🏼The total number of values is: (2ℓ+1)
🔎Example:
📌ℓ = 1 (p subshell) → mℓ = –1, 0, +1 → 3 values = 2(ℓ)+1.
📌ℓ = 2 (d subshell) → mℓ = –2, –1, 0, +1, +2 → 5 values = 2(2)+1.
3: The number of unpaired electrons in oxygen and nitrogen is respectively:
a) 3 and 2
b) 2 and 3
c) 2 and 2
d) 3 and 3
✅ Correct Answer: (b) 2 and 3
🔎📌 Short Reason
✨ Short reason: Oxygen has 2 unpaired electrons in its 2p orbitals, while nitrogen has 3 unpaired electrons in its 2p orbitals.
🔎🧠 Step by Step Reasoning
📌Nitrogen (Z = 7): Electronic configuration = 1s² 2s² 2p³ → In the 2p subshell, 3 electrons occupy 3 orbitals singly (Hund’s rule). → Unpaired electrons = 3
📌Oxygen (Z = 8): Electronic configuration = 1s² 2s² 2p⁴ → In the 2p subshell, 4 electrons: three fill singly, the fourth pairs up in one orbital. → Unpaired electrons = 2
🔎📌 Short Reason
✨ Short reason: Oxygen has 2 unpaired electrons in its 2p orbitals, while nitrogen has 3 unpaired electrons in its 2p orbitals.
🔎🧠 Step by Step Reasoning
📌Nitrogen (Z = 7): Electronic configuration = 1s² 2s² 2p³ → In the 2p subshell, 3 electrons occupy 3 orbitals singly (Hund’s rule). → Unpaired electrons = 3
📌Oxygen (Z = 8): Electronic configuration = 1s² 2s² 2p⁴ → In the 2p subshell, 4 electrons: three fill singly, the fourth pairs up in one orbital. → Unpaired electrons = 2
4: The number of unpaired electrons in oxygen and nitrogen molecules is respectively:
a) 2 and 0
b) 0 and 3
c) 2 and 3
d) 0 and 0
✅ Correct Answer: (a) 2 and 0
🔎📌 Short Reason
📌O₂ molecule: Molecular orbital configuration (up to 16 electrons): σ(1s)² σ(1s)² σ(2s)² σ(2s)² σ(2px)² π(2py)² π(2pz)² π(2py)¹ π(2pz)¹ → Two electrons remain unpaired in the π* antibonding orbitals. → Unpaired electrons = 2.
📌 N₂ molecule: Molecular orbital configuration (up to 14 electrons): σ(1s)² σ(1s)² σ(2s)² σ(2s)² σ(2px)² π(2py)² π(2pz)² → All orbitals fully paired. → Unpaired electrons = 0
🔎📌 Short Reason
📌O₂ molecule: Molecular orbital configuration (up to 16 electrons): σ(1s)² σ(1s)² σ(2s)² σ(2s)² σ(2px)² π(2py)² π(2pz)² π(2py)¹ π(2pz)¹ → Two electrons remain unpaired in the π* antibonding orbitals. → Unpaired electrons = 2.
📌 N₂ molecule: Molecular orbital configuration (up to 14 electrons): σ(1s)² σ(1s)² σ(2s)² σ(2s)² σ(2px)² π(2py)² π(2pz)² → All orbitals fully paired. → Unpaired electrons = 0
5: The number of unpaired electrons in chromium atom (Z=24) is
a) 1
b) 6
c) 5
d) 2
✅ Correct Answer: (b) 6
🔎📌 Short Reason
📌 Chromium’s ground state configuration is [Ar] 3d⁵ 4s¹ (due to extra stability of half filled d orbitals).
📌 In 3d⁵ → all 5 d orbitals singly occupied → 5 unpaired electrons.
📌 In 4s¹ → 1 electron, also unpaired.
📌 Total unpaired electrons = 5 + 1 = 6 👈
🔎📌 Short Reason
📌 Chromium’s ground state configuration is [Ar] 3d⁵ 4s¹ (due to extra stability of half filled d orbitals).
📌 In 3d⁵ → all 5 d orbitals singly occupied → 5 unpaired electrons.
📌 In 4s¹ → 1 electron, also unpaired.
📌 Total unpaired electrons = 5 + 1 = 6 👈
6: The sum of all quantum numbers of the electron of hydrogen atom is
a) - ½
b) 1
c) + ½
d) ³/₂
✅ Correct Answer: (d) ³/₂
🔎📌 Short Reason
For hydrogen’s ground state electron (1s¹), the quantum numbers add up to (1 +0 +0 + ½) ³/₂
🔎🧠 Stepwise Reasoning
📌 The 🔥 Hydrogen atom has 1 electron in the ground state. Its quantum numbers are: ⚡Principal quantum number n =1
⚡Azimuthal quantum number ℓ = 0 (s orbital)
⚡Magnetic quantum number mℓ = 0
⚡Spin quantum number ms = + ½
🔥 Now, sum of all quantum numbers: n+ℓ+mℓ+ms = 1 + 0 + 0 + ½ = ³/₂ ✅👈
🔎📌 Short Reason
For hydrogen’s ground state electron (1s¹), the quantum numbers add up to (1 +0 +0 + ½) ³/₂
🔎🧠 Stepwise Reasoning
📌 The 🔥 Hydrogen atom has 1 electron in the ground state. Its quantum numbers are: ⚡Principal quantum number n =1
⚡Azimuthal quantum number ℓ = 0 (s orbital)
⚡Magnetic quantum number mℓ = 0
⚡Spin quantum number ms = + ½
🔥 Now, sum of all quantum numbers: n+ℓ+mℓ+ms = 1 + 0 + 0 + ½ = ³/₂ ✅👈
7: The outermost electronic configuration of manganese (Z = 25) is:
a) 3d⁶ 4s¹
b) 3d⁵ 4s²
c) 3d⁵ 4s¹
d) 3d⁶ 4s²
✅ Correct Answer: (b) 3d⁵ 4s²
🔎📌 Short Reason
Manganese (Z = 25) has a half filled 3d⁵ subshell and a filled 4s² subshell, giving stability.
🔎🧠 Step by Step Reasoning
📌 Manganese has atomic number 25.
📌 Its ground state electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
📌 The outermost configuration (valence shell) is therefore: 3d⁵ 4s²
🔎📌 Short Reason
Manganese (Z = 25) has a half filled 3d⁵ subshell and a filled 4s² subshell, giving stability.
🔎🧠 Step by Step Reasoning
📌 Manganese has atomic number 25.
📌 Its ground state electronic configuration is:
1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s²
📌 The outermost configuration (valence shell) is therefore: 3d⁵ 4s²
8: The subshell which does not exist, has the quantum numbers:
a) n = 2, ℓ = 0
b) n = 2, ℓ = 2
c) n = 2, ℓ = 1
d) n = 3, ℓ = 0
✅ Correct Answer: (b) n = 2, ℓ = 2
🔎📌 Short Reason
✨ Short reason: For a given principal quantum number n, the azimuthal quantum number ℓ can take values from 0 to (n – 1).
If ℓ=n, that subshell does not exist.
👉 For n = 2, allowed ℓ values are 0 and 1 only. ℓ = 2 is not possible, so the 2d subshell does not exist.
🔎📌 Short Reason
✨ Short reason: For a given principal quantum number n, the azimuthal quantum number ℓ can take values from 0 to (n – 1).
If ℓ=n, that subshell does not exist.
👉 For n = 2, allowed ℓ values are 0 and 1 only. ℓ = 2 is not possible, so the 2d subshell does not exist.
9: The ground state electronic configuration of carbon atom has ________ paired and _________ unpaired electrons.
a) 1,2
b) 2,1
c) 2,3
d) 2,2
✅ Correct Answer: (d) 2,2
🔎📌 Short Reason
✨ Short reason: Carbon’s ground state (1s² 2s² 2p²) has 2 paired electrons in 1s and 2s, and 2 unpaired electrons in 2p.
🔎📌 Short Reason
✨ Short reason: Carbon’s ground state (1s² 2s² 2p²) has 2 paired electrons in 1s and 2s, and 2 unpaired electrons in 2p.
10: The second electron affinity of an element is always:
a) Negative
b) Zero
c) Positive
d) Infinity
✅ Correct Answer: (c) Positive
🔎📌 Short Reason
✨ The second electron affinity is always positive because energy is required to add an electron to a negatively charged ion due to repulsion.6
🔎📌 Short Reason
✨ The second electron affinity is always positive because energy is required to add an electron to a negatively charged ion due to repulsion.6
11: An element M has an atomic mass 19 and atomic number 9, its ion is represented by:
a) M⁺
b) M²⁺
c) M²⁻
d) M⁻
✅ Correct Answer: (d) M⁻
🔎📌 Short Reason
✨ Fluorine (Z = 9, mass = 19) gains one electron to achieve stable noble gas configuration, so its ion is M⁻.
🔎📌 Short Reason
✨ Fluorine (Z = 9, mass = 19) gains one electron to achieve stable noble gas configuration, so its ion is M⁻.
12: The total number of electron pairs in a nitrogen and oxygen molecules are respectively:
a) 7,7
b) 4 and 4
c) 7 and 8
d) 9 and 7
✅ Correct Answer: (c) 7 and 8
🔎📌 Short Reason
✨ N₂ has 14 electrons = 7 pairs, O₂ has 16 electrons = 8 pairs.
🔎📌 Short Reason
✨ N₂ has 14 electrons = 7 pairs, O₂ has 16 electrons = 8 pairs.
13: The compounds which contain both ionic and covalent bonds are:
a) CHCl₃ and CCl₄
b) KCN and NaOH
c) KCl and AlCl₃
d) H₂ and CH₄
✅ Correct Answer: (b) KCN and NaOH
🔎📌 Short Reason
✨ KCN and NaOH each have an ionic bond (metal–nonmetal) and a covalent bond within the polyatomic ion (CN⁻, OH⁻).
🔎📌 Short Reason
✨ KCN and NaOH each have an ionic bond (metal–nonmetal) and a covalent bond within the polyatomic ion (CN⁻, OH⁻).
14: The colour of universal indicator in neutral solution is:
a) Red
b) Green
c) Blue
d) Pink
✅ Correct Answer: (b) Green
🔎📌 Short Reason
Universal indicator turns green in a neutral solution (pH = 7). Acidic solutions give red/orange, while basic solutions give blue/purple.
🔎📌 Short Reason
Universal indicator turns green in a neutral solution (pH = 7). Acidic solutions give red/orange, while basic solutions give blue/purple.
15: Which of the following processes occurs at anode in electrochemical cell?
a) Hydrolysis
b) Redox
c) Reduction
d) Oxidation
✅ Correct Answer: (d) Oxidation
🔎📌 Short Reason
In any electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode. This is a fundamental rule of electrochemistry.
🔎🧠 Step by Step Reasoning
🔎 In an electrochemical cell (galvanic or electrolytic):
📌 Anode is the electrode where oxidation takes place.
📌 Cathode is the electrode where reduction takes place.
🔎 Easy way to remember:
📌 AnOx → Anode = Oxidation
📌 RedCat → Reduction = Cathode.
👉🏼Therefore, the correct process at the anode is oxidation.
🔎📌 Short Reason
In any electrochemical cell, oxidation occurs at the anode and reduction occurs at the cathode. This is a fundamental rule of electrochemistry.
🔎🧠 Step by Step Reasoning
🔎 In an electrochemical cell (galvanic or electrolytic):
📌 Anode is the electrode where oxidation takes place.
📌 Cathode is the electrode where reduction takes place.
🔎 Easy way to remember:
📌 AnOx → Anode = Oxidation
📌 RedCat → Reduction = Cathode.
👉🏼Therefore, the correct process at the anode is oxidation.
16: The n + l value for 4d orbital is:
a) 4
b) 5
c) 7
d) 6
✅ Correct Answer: (d) 6
🔎📌 Short Reason
The n + l value for the 4d (since n=4, l=2) orbital is 6 (4+2).
🔎📌 Short Reason
The n + l value for the 4d (since n=4, l=2) orbital is 6 (4+2).
17: The number of orbitals in each energy level is given by the formula:
a) (2ℓ + 1)
b) n²
c) 2(2ℓ + 1)
d) 2n²
✅ Correct Answer: (b) n²
🔎📌 Short Reason
✨The total number of orbitals in the nth energy level is always n².
🔎📌 Short Reason
✨The total number of orbitals in the nth energy level is always n².
18: If the volume of gas increases twice on increasing temperature 4 times, pressure will be:
a) Increased 4 times
b) Increased twice
c) Decreased by one half
d) Remains same
✅ Correct Answer: (b) Increased twice
🔎📌 Short Reason
✨ By the ideal gas law, pressure ∝ T/V. If T increases 4× and V increases 2×, pressure increases 2×.
🔎📌 Short Reason
✨ By the ideal gas law, pressure ∝ T/V. If T increases 4× and V increases 2×, pressure increases 2×.
19: Which of the following pair of transformations described below take place with the absorption of heat?
a) Hydration and dissolution
b) Condensation and combustion
c) Sublimation and fermentation
d) Fusion and Vaporization
✅ Correct Answer: (d) Fusion and Vaporization
🔎📌 Short Reason
✨ Both fusion (melting) and vaporization (boiling) are endothermic processes that occur with absorption of heat.
🔎📌 Short Reason
✨ Both fusion (melting) and vaporization (boiling) are endothermic processes that occur with absorption of heat.
20: The incompleteness in Bohr’s atomic theory is explained by
a) Uncertainty principle
b) Rutherford’s model
c) de-Broglie’s concept
d) Quantum theory
✅ Correct Answer: (a) Uncertainty principle
🔎📌 Short Reason
Bohr’s atomic model assumes that an electron moves in a fixed circular orbit with known position and momentum at the same time.
According to Heisenberg’s Uncertainty Principle, it is impossible to determine simultaneously the exact position and momentum of an electron.
👉 This contradiction explains the incompleteness of Bohr’s atomic theory.
🔎📌 Short Reason
Bohr’s atomic model assumes that an electron moves in a fixed circular orbit with known position and momentum at the same time.
According to Heisenberg’s Uncertainty Principle, it is impossible to determine simultaneously the exact position and momentum of an electron.
👉 This contradiction explains the incompleteness of Bohr’s atomic theory.
21: In which compound, nitrogen is present in least oxidation state?
a) N₂H₄
b) NH₂OH
c) NH₄⁺
d) HNO₃
✅ Correct Answer: (c) NH₄⁺
🔎📌 Short Reason
✨ Nitrogen reaches its lowest oxidation state (–3) in NH₄⁺, while in N₂H₄ it is –2, in NH₂OH it is –1 and in HNO₃ it is +5.
🔎🧠 Step by Step Reasoning
Let oxidation state of N = x.
📌N₂H₄ (Hydrazine): 2x+4(+1) = 0 ⇒ 2x+4 = 0 ⇒ x = −2. → Nitrogen = –2
📌NH₄⁺ (Ammonium ion): x + 4(+1)=+1 ⇒ x+4 = 1 ⇒ x =−3. → Nitrogen = –3 → least oxidation state (most negative) ✅👈
📌NH₂OH (hydroxylamine): x+(3×+1) + (−2) = 0 ⇒ x+3−2=0 ⇒ x + 1= 0 ⇒ x = −1 → Nitrogen = –1
📌HNO₃ (Nitric acid): x+(+1) + 3(−2) = 0 ⇒ x+1−6=0 ⇒ x = +5. → Nitrogen = +5
🔎📌 Short Reason
✨ Nitrogen reaches its lowest oxidation state (–3) in NH₄⁺, while in N₂H₄ it is –2, in NH₂OH it is –1 and in HNO₃ it is +5.
🔎🧠 Step by Step Reasoning
Let oxidation state of N = x.
📌N₂H₄ (Hydrazine): 2x+4(+1) = 0 ⇒ 2x+4 = 0 ⇒ x = −2. → Nitrogen = –2
📌NH₄⁺ (Ammonium ion): x + 4(+1)=+1 ⇒ x+4 = 1 ⇒ x =−3. → Nitrogen = –3 → least oxidation state (most negative) ✅👈
📌NH₂OH (hydroxylamine): x+(3×+1) + (−2) = 0 ⇒ x+3−2=0 ⇒ x + 1= 0 ⇒ x = −1 → Nitrogen = –1
📌HNO₃ (Nitric acid): x+(+1) + 3(−2) = 0 ⇒ x+1−6=0 ⇒ x = +5. → Nitrogen = +5
22: Powdered marble reacts more vigorously with HCl than its chucks due to its greater surface area because it
a) Increases pace of colliding molecules
b) Increases number of effective collisions
c) Increases sites for effective collisions
d) Provides alternative reaction pathway
✅ Correct Answer: (c) Increases sites for effective collisions
🔎📌 Short Reason
✨Powdered marble has a larger surface area, which exposes more reactive sites to hydrochloric acid.
📌This allows more effective collisions between CaCO₃ particles and HCl molecules per unit time speeding up the reaction.
👉 More exposed sites → more effective collisions → faster reaction. ✅
🔎❌ Why other options are incorrect?
📌 (a) ❌ No change in reaction pathway (that would be catalysis)
📌 (b) ❌ Does not increase molecular speed (that depends on temperature)
📌 (d) ❌ Incomplete; the correct idea is more sites for collisions
🔎📌 Short Reason
✨Powdered marble has a larger surface area, which exposes more reactive sites to hydrochloric acid.
📌This allows more effective collisions between CaCO₃ particles and HCl molecules per unit time speeding up the reaction.
👉 More exposed sites → more effective collisions → faster reaction. ✅
🔎❌ Why other options are incorrect?
📌 (a) ❌ No change in reaction pathway (that would be catalysis)
📌 (b) ❌ Does not increase molecular speed (that depends on temperature)
📌 (d) ❌ Incomplete; the correct idea is more sites for collisions
23: Rain drops have spherical shape because a sphere has the least
a) Surface to volume ratio
b) Length
c) Volume
d) Area
✅ Correct Answer: (a) Surface to volume ratio
🔎📌 Short Reason
✨ Short reason: Raindrops are spherical because a sphere has the least surface area for a given volume, minimizing energy due to surface tension. 📌Rain drops form due to surface tension of water.
📌Surface tension tends to minimize the surface area for a given volume.
📌Among all 3D shapes, a sphere has the minimum surface area for a given volume.
This is why raindrops naturally adopt a spherical shape.
🔎📌 Short Reason
✨ Short reason: Raindrops are spherical because a sphere has the least surface area for a given volume, minimizing energy due to surface tension. 📌Rain drops form due to surface tension of water.
📌Surface tension tends to minimize the surface area for a given volume.
📌Among all 3D shapes, a sphere has the minimum surface area for a given volume.
This is why raindrops naturally adopt a spherical shape.
24: The quantitative relationship between the substances according to balanced equation describes:
a) Stoichiometry
b) Reversible reaction
c) Percentage compound
d) Limiting reactant
✅ Correct Answer: (a) Stoichiometry
🔎📌 Short Reason
✨ Balanced equations describe the mole ratios of reactants and products, which is the essence of stoichiometry.
🔎📌 Short Reason
✨ Balanced equations describe the mole ratios of reactants and products, which is the essence of stoichiometry.
25: How many significant figures are there in 87.00900?
a) 7
b) 5
c) 3
d) 4
✅ Correct Answer: (a) 7
🔎📌 Short Reason
✨ 87.00900 has 7 significant figures because the captive zeroes (between 7 and 9) and trailing zeros after the decimal (two zeros after 9) are significant.
🔎📌 Short Reason
✨ 87.00900 has 7 significant figures because the captive zeroes (between 7 and 9) and trailing zeros after the decimal (two zeros after 9) are significant.
26: S²⁻ (Z = 16) is isoelectronic with:
a) ₁₁Na⁺
b) ₉F⁻
c) ₁₉K⁺
d) ₁₇Cl
✅ Correct Answer: (c) ₁₉K⁺
🔎📌 Short Reason
✨ S²⁻ has 18 electrons (sulphur atom gains 2 electrons), which matches K⁺ (19 protons, 18 electrons).
🔎📌 Short Reason
✨ S²⁻ has 18 electrons (sulphur atom gains 2 electrons), which matches K⁺ (19 protons, 18 electrons).
27: The most of the radiations coming out from pitch blend were
a) Proton
b) Electron
c) X-rays
d) Neutron
✅ Correct Answer: (b) Electron
🔎📌 Short Reason
✨ Pitchblende is a uranium ore, and its radioactivity is mainly due to emission of β particles (electrons).
🔎📌 Short Reason
✨ Pitchblende is a uranium ore, and its radioactivity is mainly due to emission of β particles (electrons).
28: Which one is correct statement?
(I) Ethane (C₂H₆) contains 7 sigma bonds and 1 pi bond.
(II) Ethene (C₂H₄) contains 5 sigma bonds and 1 pi bond.
(III) Ethyne (C₂H₂) contains 3 sigma bonds and 2 pi bonds.
(IV) Benzene (C₆H₆) contains 12 sigma bonds and 3 pi bonds.
(I) Ethane (C₂H₆) contains 7 sigma bonds and 1 pi bond.
(II) Ethene (C₂H₄) contains 5 sigma bonds and 1 pi bond.
(III) Ethyne (C₂H₂) contains 3 sigma bonds and 2 pi bonds.
(IV) Benzene (C₆H₆) contains 12 sigma bonds and 3 pi bonds.
a) I and II only
b) II, III and IV only
c) III and IV only
d) II and IV only
✅ Correct Answer: (b) II, III and IV only
🔎📌 Short Reason
✨ Ethane has 7 sigma only, not a pi bond. Ethene, Ethyne, and Benzene statements are correct.
🔎🧠 Step by Step Reasoning
🔎Ethane (C₂H₆):
📌Structure: H₃C–CH₃
📌Bonds: 6 C–H sigma + 1 C–C sigma = 7 sigma bonds.
📌No double/triple bonds → 0 pi bonds.
📌Statement (I) says 7 sigma + 1 pi → ❌ Incorrect.
🔎Ethene (C₂H₄):
📌Structure: H₂C=CH₂
📌Bonds: 4 C–H sigma + 1 C–C sigma = 5 sigma.
📌Double bond has 1 pi → 5 sigma + 1 pi.
📌Statement (II) → ✅ Correct.
🔎Ethyne (C₂H₂):
📌Structure: HC≡CH
📌Bonds: 2 C–H sigma + 1 C–C sigma = 3 sigma.
📌Triple bond has 2 pi → 3 sigma + 2 pi.
📌Statement (III) → ✅ Correct.
🔎Benzene (C₆H₆):
📌Structure: Hexagon with alternating double bonds.
📌Bonds: 6 C–C sigma + 6 C–H sigma = 12 sigma.
📌3 delocalized double bonds → 3 pi.
📌Statement (IV) → ✅ Correct.
✅ Correct Answer (b) II, III and IV only 👈
🔎📌 Short Reason
✨ Ethane has 7 sigma only, not a pi bond. Ethene, Ethyne, and Benzene statements are correct.
🔎🧠 Step by Step Reasoning
🔎Ethane (C₂H₆):
📌Structure: H₃C–CH₃
📌Bonds: 6 C–H sigma + 1 C–C sigma = 7 sigma bonds.
📌No double/triple bonds → 0 pi bonds.
📌Statement (I) says 7 sigma + 1 pi → ❌ Incorrect.
🔎Ethene (C₂H₄):
📌Structure: H₂C=CH₂
📌Bonds: 4 C–H sigma + 1 C–C sigma = 5 sigma.
📌Double bond has 1 pi → 5 sigma + 1 pi.
📌Statement (II) → ✅ Correct.
🔎Ethyne (C₂H₂):
📌Structure: HC≡CH
📌Bonds: 2 C–H sigma + 1 C–C sigma = 3 sigma.
📌Triple bond has 2 pi → 3 sigma + 2 pi.
📌Statement (III) → ✅ Correct.
🔎Benzene (C₆H₆):
📌Structure: Hexagon with alternating double bonds.
📌Bonds: 6 C–C sigma + 6 C–H sigma = 12 sigma.
📌3 delocalized double bonds → 3 pi.
📌Statement (IV) → ✅ Correct.
✅ Correct Answer (b) II, III and IV only 👈
29: Which molecule has linear structure?
a) CH₄
b) C₂H₂
c) BF₃
d) NH₃
✅ Correct Answer: (b) C₂H₂
🔎📌 Short Reason
✨ Ethyne (C₂H₂) has a linear structure due to sp hybridization of carbon atoms, giving a bond angle of 180°.
🔎📌 Short Reason
✨ Ethyne (C₂H₂) has a linear structure due to sp hybridization of carbon atoms, giving a bond angle of 180°.
30: Which of the following is intensive property of system?
a) Density
b) Energy
c) Volume
d) Entropy
✅ Correct Answer: (a) Density
🔎📌 Short Reason
✨ Density is an intensive property because it does not depend on the amount of substance, unlike energy, volume, or entropy. 👉🏼Ratio properties (mass/volume, etc.) are usually intensive. 👈
🔎📌 Short Reason
✨ Density is an intensive property because it does not depend on the amount of substance, unlike energy, volume, or entropy. 👉🏼Ratio properties (mass/volume, etc.) are usually intensive. 👈
31: For the reaction:
2NH₃ ⇌ N₂+3H₂
the relationship between Kc and Kₚ is:
2NH₃ ⇌ N₂+3H₂
the relationship between Kc and Kₚ is:
a) Kc > Kₚ
b) Kₚ = Kc
c) Kₚ = Kc²
d) Kc < Kₚ
✅ Correct Answer: (d) Kc < Kₚ
🔎📌 Short Reason
✨ Because Δn = 2, Kₚ = Kc(RT) ², making Kₚ greater than Kc.
🔎🧠 Step by Step Reasoning
General relation: Kₚ = Kc(RT)▵ⁿ 🔎Larger Kₐ → stronger acid.
where (Δn = moles of gaseous products - moles of gaseous reactants]
🔎For this reaction:
Reactants: 2 NH₃ (g) → 2 moles
Products: 1 N₂ (g) + 3 H₂ (g) → 4 moles
So, Δn = 4−2 = 2.
Therefore: Kₚ = Kc(RT) ²
Since RT>1 (positive temperature and gas constant), Kₚ > Kc or Kc < Kₚ✅
🔎📌 Short Reason
✨ Because Δn = 2, Kₚ = Kc(RT) ², making Kₚ greater than Kc.
🔎🧠 Step by Step Reasoning
General relation: Kₚ = Kc(RT)▵ⁿ 🔎Larger Kₐ → stronger acid.
where (Δn = moles of gaseous products - moles of gaseous reactants]
🔎For this reaction:
Reactants: 2 NH₃ (g) → 2 moles
Products: 1 N₂ (g) + 3 H₂ (g) → 4 moles
So, Δn = 4−2 = 2.
Therefore: Kₚ = Kc(RT) ²
Since RT>1 (positive temperature and gas constant), Kₚ > Kc or Kc < Kₚ✅
32: Which one of the following solution is basic?
a) NH₄Cl
b) KCl
c) NaCl
d) Na₂CO₃
✅ Correct Answer: (d) Na₂CO₃
🔎📌 Short Reason
A negative catalyst (inhibitor) slows down the rate of a reaction.
✨ 👉 Sodium carbonate solution is basic because carbonate ions hydrolyze to form OH⁻ ions, while the others are either acidic or neutral.
🔎📌 Short Reason
A negative catalyst (inhibitor) slows down the rate of a reaction.
✨ 👉 Sodium carbonate solution is basic because carbonate ions hydrolyze to form OH⁻ ions, while the others are either acidic or neutral.
33: The oxidation number of Cr in Cr₂O₇²⁻ and in CrO₄²⁻ ions is respectively:
a) +6, +6
b) +3, +3
c) +12, +6
d) +6, +12
✅ Correct Answer: (a) +6, +6
🔎📌 Short Reason
✨ Both dichromate (Cr₂O₇²⁻) and chromate (CrO₄²⁻) ions have chromium in the +6 oxidation state.
🔎📌 Short Reason
✨ Both dichromate (Cr₂O₇²⁻) and chromate (CrO₄²⁻) ions have chromium in the +6 oxidation state.
34: If the rate of reaction is independent of concentration of the reactant, the reaction is of:
a) 1st order
b) Zero order
c) 2nd order
d) 3rd order
✅ Correct Answer: (b) Zero order
🔎📌 Short Reasoning
✨ In a zero order reaction, the rate remains constant regardless of reactant concentration.👈
🔎📌 Short Reasoning
✨ In a zero order reaction, the rate remains constant regardless of reactant concentration.👈
35: The product of pressure and volume, PV has the dimension
a) Pressure
b) Energy
c) Volume
d) Temperature
✅ Correct Answer: (b) Energy
🔎📌 Short Reason
✨ The product PV has the same dimension as work or energy (ML²T⁻²).
🔎📌 Step by Step Reasoning
📌Pressure (P): Dimension = Force/Area = MLT⁻²/L² = ML⁻¹T⁻².
📌Volume (V): Dimension = L³.
Multiply:
PV = (ML⁻¹T⁻²)(L³) = ML²T⁻² This is the dimension of energy (work).
Recall: Work = Force × Distance = MLT⁻²⋅L = ML²T⁻².
🔎📌 Short Reason
✨ The product PV has the same dimension as work or energy (ML²T⁻²).
🔎📌 Step by Step Reasoning
📌Pressure (P): Dimension = Force/Area = MLT⁻²/L² = ML⁻¹T⁻².
📌Volume (V): Dimension = L³.
Multiply:
PV = (ML⁻¹T⁻²)(L³) = ML²T⁻² This is the dimension of energy (work).
Recall: Work = Force × Distance = MLT⁻²⋅L = ML²T⁻².
36: The penetration power of β particle in air as compared to α particle is:
a) 2 times
b) 100 times
c) 1000 times
d) 200 times
✅ Correct Answer: (b) 100 times
🔎📌 Short Reason
✨ β particles have much greater penetration in air than α particles, roughly 100 times more.
🔎📌 Short Reason
✨ β particles have much greater penetration in air than α particles, roughly 100 times more.
37: A gas at zero kelvin:
a) Has zero molecular motion
b) Represents absolute zero temperature
c) has zero Volume (for an ideal gas)
d) All of these
✅ Correct Answer: (d) All of these
🔎📌 Short Reason
At 0 K, molecular motion stops, entropy approaches zero, internal energy is minimized, and volume tends to zero for an ideal gas. But in reality, absolute zero cannot be achieved, and gases do not obey the ideal gas law perfectly at such extremes.
🔎📌 Short Reason
At 0 K, molecular motion stops, entropy approaches zero, internal energy is minimized, and volume tends to zero for an ideal gas. But in reality, absolute zero cannot be achieved, and gases do not obey the ideal gas law perfectly at such extremes.
38: E + PV is called:
a) Internal Energy
b) Entropy
c) Free Energy
d) Enthalpy
✅ Correct Answer: d) Enthalpy
🔎📌 Short Reason
✨ The thermodynamic function defined as E+PV is Enthalpy (H).
🔎📌 Short Reason
✨ The thermodynamic function defined as E+PV is Enthalpy (H).
39: The quantum number values for 3p orbitals are:
a) n = 2, l = 1
b) n = 3, l = 0
c) n = 3, l = 1
d) n = 2, l = 2
✅ Correct Answer: c) n = 3, l = 1
🔎📌 Short Reason
✨ The 3p orbital belongs to the third shell (n=3) and has l=1 because it is a p orbital.
🔎📌 Short Reason
✨ The 3p orbital belongs to the third shell (n=3) and has l=1 because it is a p orbital.
40: Which of the following compound does not contain hydrogen bonding?
a) H₂O
b) NH₃
c) CH₄
d) HF
✅ Correct Answer: (c) CH₄
🔎📌 Short Reason
Hydrogen bonding requires a highly electronegative atom (like O, N, or F) directly bonded to hydrogen.
H₂O, NH₃, HF → all have O–H, N–H, or F–H bonds, so they form hydrogen bonds.
CH₄ → carbon is not electronegative enough, so no hydrogen bonding occurs.
👉 Therefore, CH₄ does not contain hydrogen bonding.
🔎📌 Short Reason
Hydrogen bonding requires a highly electronegative atom (like O, N, or F) directly bonded to hydrogen.
H₂O, NH₃, HF → all have O–H, N–H, or F–H bonds, so they form hydrogen bonds.
CH₄ → carbon is not electronegative enough, so no hydrogen bonding occurs.
👉 Therefore, CH₄ does not contain hydrogen bonding.
🔥💥 جون ایلیا غزل 🔥💥
🌙 حال یہ ہے کہ خواہشِ پُرسشِ حال بھی نہیں
💭 اُس کا خیال بھی نہیں، اپنا خیال بھی نہیں
💭 اُس کا خیال بھی نہیں، اپنا خیال بھی نہیں
🍂 اے شجرِ حیاتِ شوق، ایسی خزاں رسیدگی؟
🌿 پوششِ برگ و گُل تو کیا، جسم پہ چھال بھی نہیں
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📉 مُجھ میں وہ شخص ہو چکا جس کا کوئی حساب تھا
📊 سُود ہی کیا، زیاں ہے کیا، اس کا سوال بھی نہیں
📊 سُود ہی کیا، زیاں ہے کیا، اس کا سوال بھی نہیں
💔 مست ہیں اپنے حال میں دل زدگان و دلبراں
⚔️ صُلح و سلام تو کُجا، بحث و جدال بھی نہیں
⚔️ صُلح و سلام تو کُجا، بحث و جدال بھی نہیں
🧭 تُو مرا حوصلہ تو دیکھ، داد تو دے کہ اب مجھے
🎯 شوقِ کمال بھی نہیں، خوفِ زوال بھی نہیں
🎯 شوقِ کمال بھی نہیں، خوفِ زوال بھی نہیں
🏕️ خیمہ گہِ نگاہ کو لوٹ لیا گیا ہے کیا؟
🌫️ آج افق کے دوش پر گرد کی شال بھی نہیں
🌫️ آج افق کے دوش پر گرد کی شال بھی نہیں
🌬️ اف یہ فضا سے احتیاط تا کہیں اڑ نہ جائیں ہم
🌪️ بادِ جنوب بھی نہیں، بادِ شمال بھی نہیں
🌪️ بادِ جنوب بھی نہیں، بادِ شمال بھی نہیں
🧾 وجہ معاشِ بے دلاں، یاس ہے اب مگر کہاں
🙏 اس کے درود کا گماں، فرضِ محال بھی نہیں
🙏 اس کے درود کا گماں، فرضِ محال بھی نہیں
⏳ غارتِ روز و شب کو دیکھ، وقت کا یہ غضب تو دیکھ
🕰️ کل تو نڈھال بھی تھا میں، آج نڈھال بھی نہیں
🕰️ کل تو نڈھال بھی تھا میں، آج نڈھال بھی نہیں
🗣️ میرے زبان و ذات کا ہے یہ معاملہ کہ اب
🌄 صبحِ فراق بھی نہیں، شامِ وصال بھی نہیں
🌄 صبحِ فراق بھی نہیں، شامِ وصال بھی نہیں
🎨 پہلے ہمارے ذہن میں حسن کی اک مثال تھی
🧠 اب تو ہمارے ذہن میں کوئی مثال بھی نہیں
🧠 اب تو ہمارے ذہن میں کوئی مثال بھی نہیں
🤯 میں بھی بہت عجیب ہوں اتنا عجیب ہوں کہ بس
💣 خود کو تباہ کر لیا اور ملال بھی نہیں
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