🔥🌟 MDCAT/ECAT/FSC/XII Grand Chemistry MCQs Interactive Quiz # 6 🧪💡 | Learn & Practice!

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1. Which of the following has maximum bond angle?

a) H₂O

b) CCl₄

c) BF₃

d) NH₃

✅ Correct Answer: (c) BF₃
🔎📌 Short Reason
Among the given molecules, BF₃ has the maximum bond angle (120°) because of its trigonal planar geometry. The others have smaller bond angles due to lone pair repulsion or tetrahedral arrangement.
🔎🧠 Step by Step Reasoning
🔎 (a) H₂O (Water):
🔥 Geometry: Bent (due to 2 lone pairs). Bond angle ≈ 104.5°. ❌
🔎 (b) CCl₄ (Carbon tetrachloride):
🔥 Geometry: Tetrahedral. Bond angle ≈ 109.5°. ❌
🔎 (c) BF₃ (Boron trifluoride):
🔥 Geometry: Trigonal planar. Bond angle ≈ 120°. ✅ 👈
🔎 (d) NH₃ (Ammonia):
🔥 Geometry: Trigonal pyramidal (due to lone pair). Bond angle ≈ 107°. ❌

2. Which of the following aqueous solution of salt is either acidic or basic?

a) NH₄CN

b) CH₃COONH₄

c) AgSH

d) All of these

✅ Correct Answer: (d) All of these
🔎📌 Short Reason
Each of the salts NH₄CN, CH₃COONH₄, and AgSH undergoes hydrolysis in water, producing either acidic or basic solutions. Therefore, the correct option is All of these.
🔎🧠 Step by Step Reasoning
🔎 (a) NH₄CN (Ammonium cyanide):
🔥 Contains NH₄⁺ (weak base conjugate → acidic) and CN⁻ (weak acid conjugate → basic).
🔥 Both ions hydrolyze → solution can be acidic or basic depending on relative strengths. So, not neutral. ✅
🔎 (b) CH₃COONH₄ (Ammonium acetate):
🔥 Contains NH₄⁺ (acidic) and CH₃COO⁻ (basic).
🔥 Both ions hydrolyze → solution may be slightly acidic or basic depending on Ka vs Kb. So, not neutral. ✅
🔎 (c) AgSH (Silver hydrosulfide):
🔥 Contains Ag⁺ (cation of weak base → acidic hydrolysis) and SH⁻ (anion of weak acid → basic hydrolysis).
🔥 Again, solution is not neutral. ✅
👉 Since all three salts undergo hydrolysis and give either acidic or basic solutions, the correct choice is All of these. ✅

3. If a≠ b ≠ c and α = γ = 90° and β ≠ 90°, then the crystal structure is:

a) Orthorhombic

b) Monoclinic

c) Triclinic

d) Hexagonal

✅ Correct Answer: (b) Monoclinic
🔎 📌 Short Reason
👉 Because in the monoclinic crystal system, the unit cell has three unequal axes (a ≠ b ≠ c), with two angles equal to 90° (α = γ = 90°) and the third angle different from 90° (β ≠ 90°).
🔎🧠 Step by Step Reasoning
🔎Crystal system classification depends on unit cell parameters (axial lengths a, b, c and interfacial angles α, β, γ):
🔎Orthorhombic: a≠b≠c, α=β=γ=90°
🔎Monoclinic: a≠b≠c, α=γ=90°, β≠90° ✅
🔎Triclinic: a≠b≠c, α≠β≠γ≠90°
🔎Hexagonal:a=b≠c, α=β=90°,γ=120°
👉 Given condition matches Monoclinic system exactly.

4. Which of the following is the correct electronic configuration of copper according to Hund’s rule?

a) [Ar] 4s² 3d⁹

b) [Ar] 4s¹ 3d⁹

c) [Ar] 4s¹ 3d¹⁰

d) [Ar] 4s² 3d¹⁰

✅ Correct Answer: (c) [Ar] 4s¹ 3d¹⁰
🔎📌Short Reason
👉 Copper is an exception to the Aufbau principle. Instead of [Ar] 4s² 3d⁹, it adopts the more stable configuration [Ar] 4s¹ 3d¹⁰, making option (c) correct.
🔎🧠 Step by Step Reasoning
🔎Expected configuration (Aufbau principle): After [Ar], we expect: 4s² 3d⁹. That corresponds to option (a).
🔎But copper is an exception.Copper prefers a completely filled 3d subshell (3d¹⁰) for extra stability.
To achieve this, one electron from 4s orbital shifts into 3d orbital.
🔎Actual stable configuration: Cu=[Ar]  4s¹ 3d¹⁰ 👈
🔎Hund’s rule consistency: Hund’s rule states that orbitals of equal energy are filled singly before pairing.
🔎In copper, the 3d orbitals are fully filled (stable), and 4s has one electron.

5. The maximum number of unpaired electrons in 3d energy level is:

a) 3

b) 5

c) 7

d) 9

✅ Correct Answer: (b) 5
🔎🧠 Step by Step Reasoning
📌 The 3d subshell has 5 orbitals (dxy, dyz, dzx, dx²–y², dz²).
📌 Each orbital can hold 2 electrons → maximum of 10 electrons in 3d.
📌 According to Hund’s rule, electrons occupy orbitals singly with parallel spins before pairing.
📌 So, when the 3d subshell is half filled (3d⁵):
📌 Each of the 5 orbitals has 1 electron.
📌 That gives the maximum number of unpaired electrons = 5. 👈
📌 If we go beyond 3d⁵ (like 3d⁶, 3d⁷…), pairing begins, so the number of unpaired electrons decreases.

6. If a catalyst is added in a chemical system at equilibrium, the value of Kc:

a) Will increase

b) Will decrease

c) Will become zero

d) Will not be changed

✅ Correct Answer: (d) Will not be changed
👉 A catalyst only lowers activation energy, speeding up the attainment of equilibrium. It does not alter the relative concentrations of reactants and products, hence the equilibrium constant (Kc) remains unchanged. 👈

7. One mole of H₂O contains this number of hydrogen atoms:

a) 1.204 × 10²⁴

b) 1.204 × 10²³

c) 6.02 × 10²³

d) 3.01 × 10²³

✅ Correct Answer: (a) 1.204 × 10²⁴
🔎📌 Short Reason
👉Each H₂O molecule has 2 hydrogen atoms. One mole of H₂O = 6.02 × 10²³ molecules. So hydrogen atoms = 2 × 6.02 × 10²³ = 1.204 × 10²⁴. Hence one mole of H₂O has twice Avogadro’s number of hydrogen atoms.
🔎🧠 Step by Step Reasoning
🔎1 mole of H₂O = 6.02 × 10²³ molecules (Avogadro’s number).
🔎Each molecule of H₂O has 2 hydrogen atoms.
🔎 Therefore, total hydrogen atoms in 1 mole: 2× 6.02 × 10²³ = 1.204 × 10²⁴.✅ 👈

8. The bond energy of this molecule is maximum:

a) N₂

b) O₂

c) CH₄

d) Cl₂

✅ Correct Answer: (a) N₂
👉The N≡N triple bond in nitrogen is one of the strongest bonds known, giving N₂ the highest bond energy among the listed molecules.

9. In James Chadwick's experiment alpha particles were bombarded on:

a) Beryllium

b) Boron

c) Bromine

d) Carbon

✅ Correct Answer: (c) Beryllium
👉In Chadwick’s experiment, alpha particles bombarded beryllium nuclei, releasing neutrons. This was the key step in the discovery of the neutron.

10. In the hydrogen atom spectrum, the series of lines obtained when electron jumps from higher orbits to the first orbit is called:

a) Balmer Series

b) Lyman Series

c) Brackett Series

d) Paschen Series

✅ Correct Answer: (b) Lyman Series
🔎📌 Short Reason
The Lyman Series corresponds to transitions ending at the first orbit (n = 1) in hydrogen, producing spectral lines in the ultraviolet region.
🔎🧠 Step by Step Reasoning
🔎 Hydrogen spectrum series:
📌 Lyman Series: electron falls to n = 1 (first orbit). Region: ultraviolet. ✅ 👈
📌 Balmer Series: electron falls to n = 2. Region: visible.
📌 Paschen Series: electron falls to n = 3. Region: infrared.
📌 Brackett Series: electron falls to n = 4. Region: infrared.
👉 Since the question specifies electron jumps to the first orbit (n = 1), the correct series is Lyman Series.

11. The range of bond energy of hydrogen bond is:

a) 10–20 kJ/mol

b) 20–40 kJ/mol

c) 40–50 kJ/mol

d) 50–60 kJ/mol

✅ Correct Answer: (b) 20–40 kJ/mol
🔎📌 Short Reason
👉 Hydrogen bonds are intermolecular forces with bond energies usually between 20–40 kJ/mol, strong enough to influence physical properties but weaker than covalent bonds.
🔎🧠 Step by Step Reasoning
📌 Hydrogen bonds are weaker than covalent bonds but stronger than van der Waals forces.
📌 Typical bond energy of a hydrogen bond lies in the range of 20–40 kJ/mol.
📌 This is why hydrogen bonds are strong enough to stabilize structures (like DNA double helix, water’s high boiling point), but not as strong as covalent bonds (~200–400 kJ/mol).

12. Principal and azimuthal quantum number values for 3d orbital are:

a) n = 2, l = 1

b) n = 2, l = 3

c) n = 3, l = 3

d) n = 3, l = 2

✅ Correct Answer: (d) n = 3, l = 2
🔎 📌 Short Reason
Because the 3d orbital belongs to the 3rd shell (n = 3) and the d subshell corresponds to l = 2.
🔎 🧠 Step by Step Reasoning (shortest, best logic)
🔎 Principal quantum number (n): indicates the shell. For 3d orbital, n = 3.
🔎 Azimuthal quantum number (l): indicates the subshell type. For 3d orbital, d → l = 2
📌 s → l = 0
📌 p → l = 1
📌 d → l = 2
📌 f → l = 3
👉 For 3d orbital, n = 3 and l = 2.

13. If the Kelvin temperature of an ideal gas is increased to double and pressure is reduced to one half, the volume of gas will:

a) Remain same

b) Double

c) Half

d) Four times

✅ Correct Answer: (d) Four times
🔎 📌 Short Reason
By the ideal gas law, V ∝ T/P. Doubling T and halving P multiplies volume by 4.
🔎 🧠 Step by Step Reasoning
📌 Ideal gas law: PV=nRT
📌 Rearranging for volume: V = nRT/P
📌 Changes:
👉 Temperature T → doubled (×2).
👉 Pressure P → reduced to half (÷2).
👉 New volume: Vnе⍵ = nR(2T)/(P/2) = 2nRTP/2 = (2nRT/P) × 2 = 4⋅nRTP
👉 The volume becomes four times the original.

14. The energy corresponds to the given thermochemical process is labeled as: Li⁺(g) + Cl⁻(g) → LiCl(s)

a) Ionization energy

b) Lattice energy

c) Enthalpy of atomization

d) Enthalpy of combustion

✅ Correct Answer: (b) Lattice energy
🔎 📌 Short Reason
👉 Because the reaction represents gaseous ions combining to form a solid ionic lattice, the energy released is lattice energy. 🔎 🧠 Step by Step Reasoning 📌 Ionization energy: energy required to remove an electron from an atom in gaseous state → not this case. 📌 Enthalpy of atomization: energy required to break bonds in an element to form gaseous atoms → not this case. 📌 Enthalpy of combustion: energy released when a substance burns in oxygen → not this case. 📌 Lattice energy: energy released when gaseous ions combine to form a solid ionic crystal. 👉 The given process shows Li⁺(g) + Cl⁻(g) → LiCl(s), which is exactly the definition of lattice energy.

15. For a hypothetical reaction x + y → z, if the concentration of x is doubled, the rate increases by square, and if the concentration of y is doubled, the rate increases by twice. The experimental rate law of this reaction is:

a) R = K [x]¹ [y]¹

b) R = K [x]² [y]¹

c) R = K [x][y]

d) R = K [x][y]

✅ Correct Answer: (b) R = K [x]² [y]¹
🔎 📌 Short Reason
👉 Because the rate depends on the square of [x] and first power of [y], the experimental rate law is R = K [x]² [y]¹.
🔎 🧠 Step by Step Reasoning
📌 Rate law form: R = K[x] ᵐ[y]ⁿ
📌 If [x] is doubled → rate increases by square (×4) → exponent m = 2.
📌 If [y] is doubled → rate increases by twice (×2) → exponent n = 1.
👉 Therefore: R = K [x]² [y]¹✅ Correct Answer

16. The unit of rate constant (K) for the zero order reaction is:

a) s⁻¹

b) conc·s⁻¹

c) conc⁻¹·s

d) conc⁻¹·s⁻¹

✅ Correct Answer: (b) conc·s⁻¹
🔎 📌 Short Reason
For a zero order reaction, rate is independent of concentration, so the rate constant has units of concentration per time (mol·L⁻¹·s⁻¹).
🔎 🧠 Step by Step Reasoning
📌 General rate law: R=k[A]ⁿ (where n = order of reaction)
📌 For zero order reaction (n = 0): R = k[A]⁰ = k
📌 Rate R has units of concentration per time (e.g., mol·L⁻¹·s⁻¹).
📌 Since R = k, the unit of k must be the same: mol·L⁻¹·s⁻¹
👉 In short form: conc·s⁻¹

17. The overall order of reaction to which the rate law is R=K:

a) First order

b) Zero order

c) Second order

d) Third order

✅ Correct Answer: (b) Zero order
🔎 📌 Short Reason
👉 Since the rate law has no dependence on reactant concentration, the overall order is zero.
🔎 🧠 Step by Step Reasoning
📌 General rate law: R = K[x] ᵐ[y]ⁿ
📌 Overall order = m + n.
📌 Here, given: R = K → No concentration terms appear.
📌 That means the reaction rate is independent of concentration. Such reactions are classified as zero order reactions.

18. Which of the following is not a buffer solution?

a) Na₂CO₃ / NaHCO₃

b) NaOH / HCl

c) CH₃COOH / CH₃COONa

d) NH₄OH / NH₄Cl

✅ Correct Answer: (b) NaOH / HCl
🔎 📌 Short Reason
👉 A buffer requires a weak acid/base with its salt. NaOH/HCl are strong acid and strong base, so they cannot act as a buffer solution.
🔎 🧠 Step by Step Reasoning
📌 Buffer solution definition: A buffer is a mixture of a weak acid + its salt (conjugate base), or a weak base + its salt (conjugate acid).
📌 Option (a): Na₂CO₃ / NaHCO₃ → weak acid (HCO₃⁻) + conjugate base (CO₃²⁻). ✅ Buffer.
📌 Option (c): CH₃COOH / CH₃COONa → weak acid (acetic acid) + conjugate base (acetate). ✅ Buffer.
📌 Option (d): NH₄OH / NH₄Cl → weak base (NH₄OH) + conjugate acid (NH₄⁺). ✅ Buffer.
📌 Option (b): NaOH / HCl → strong base + strong acid. ❌ Not a buffer, because strong acid/base neutralization does not resist pH change. 👈

19. The molar volume of oxygen gas is maximum at:

a) 0 °C and 2 atm

b) 25 °C and 2 atm

c) RTP

d) STP

✅ Correct Answer: (c) RTP
🔎 📌 Short Reason
👉 Since Vₘ ∝ T/P, the highest temperature with lowest pressure gives maximum molar volume. At RTP (25 °C, 1 atm), oxygen has the largest molar volume (~24.5 L/mol).
🔎 🧠 Step by Step Reasoning
📌 Molar volume (Vₘ) is given by the ideal gas law: Vₘ=RT/P
📌 At 0 °C, 2 atm: Pressure is doubled → volume halves → Vₘ ≈ 11.2 L.
📌 At 25 °C, 2 atm: Higher T but also double pressure → Vₘ ≈ 12.25 L.
📌 At RTP (25°C, 1 atm): Vₘ ≈ 24.5 L. (largest molar volume) 👈
📌 At STP (0°C, 1 atm): Vₘ ≈ 22.4 L.
👉 Clearly, the largest molar volume is at RTP (25 °C, 1 atm).

20. A non-polar molecule with bigger size will experience:

a) Hydrogen bonding

b) dipole-dipole interaction

c) London forces

d) All of these

✅ Correct Answer: (c) London forces
👉 A non polar molecule cannot form hydrogen bonds or dipole dipole interactions. Its intermolecular attraction arises only from London dispersion forces, which increase with molecular size.

21. If the bond angle in an AB₂ type molecule is 104.5°, its geometry should be:

a) Linear

b) Pyramidal

c) Bent

d) Planar Trigonal

✅ Correct Answer: (c) Bent
🔎 📌 Short Reason
👉 An AB₂ molecule with bond angle ≈ 104.5° (like H₂O) has lone pair repulsion, giving it a bent geometry. 🔎 🧠 Step by Step Reasoning
📌 Linear AB₂ → bond angle = 180°.
📌 Planar trigonal → bond angle = 120°.
📌 Pyramidal → typical for AB₃ with lone pair, bond angle ~107°.
📌 Bent AB₂ → occurs when central atom has two bonding pairs + lone pairs.
📌 Example: H₂O → bond angle ≈ 104.5°.
👉 Therefore, AB₂ with bond angle 104.5° must be Bent geometry.

22. The shape and hybridization of BeCl₂ molecule is:

a) Tetrahedral and sp³

b) Linear and sp

c) Planar trigonal and sp²

d) Angular & sp³

✅ Correct Answer: (b) Linear and sp
🔎 📌 Short Reason
👉 BeCl₂ has two bonding pairs and no lone pairs on Be, so it adopts linear geometry with sp hybridization.
🔎 🧠 Step by Step Reasoning
📌 Central atom = Be.
📌 Electronic configuration of Be: 1s² 2s².
📌 In BeCl₂, Be forms two covalent bonds with Cl atoms.
📌 To form two bonds, Be undergoes sp hybridization (mixing of 2s and one 2p orbital).
📌 sp hybridization → gives linear geometry with bond angle ≈ 180°.
👉 Therefore, BeCl₂ is linear and sp hybridized.

23. Benzene has bond angles same as:

a) Ethene

b) BF₃

c) SO₃²⁻

d) All of these

✅ Correct Answer: (d) All of these
🔎 📌 Short Reason
👉 Benzene carbons are sp² hybridized with bond angles of 120°, identical to Ethene, BF₃, and SO₃²⁻. 🔎 🧠 Step by Step Reasoning 📌 Benzene (C₆H₆): Each carbon is sp² hybridized, bond angle = 120°. 📌 Ethene (C₂H₄): Each carbon is also sp² hybridized, bond angle ≈ 120°. 📌 BF₃: Boron is sp² hybridized, trigonal planar, bond angle = 120°. 📌 SO₃²⁻ (sulfite ion): Central S is sp² hybridized, bond angle ≈ 120°. 👉 All three species have bond angles ≈ 120°, same as benzene.

24. Bond length is largest in case of ………. overlapping:

a) sp²–s

b) sp²–sp²

c) sp³–sp³

d) sp–sp

✅ Correct Answer: (c) sp³–sp³
🔎 📌 Short Reason
👉 Bond length increases as s character decreases. Since sp³ has the least s character (25%), sp³–sp³ overlap gives the longest bond length.
🔎 🧠 Step by Step Reasoning
📌 Bond length depends on extent of orbital overlap.
📌 Greater s character → stronger overlap → shorter bond length.
📌 Lesser s character → weaker overlap → longer bond length.
⚙️ Order of s character:
📌 sp (50% s) → strongest overlap → shortest bond length.
📌 sp² (33% s) → intermediate.
📌 sp³ (25% s) → weakest overlap → longest bond length.
👉 Therefore, sp³–sp³ overlap produces the largest bond length. 👈

25. In PH₃, number of bond pairs and lone pairs of electrons are respectively:

a) 2, 8

b) 3, 1

c) 2, 2

d) 2, 4

✅ Correct Answer: (b) 3, 1
🔎 📌 Reason (short)
👉 Phosphorus in PH₃ uses 3 electrons to bond with H atoms (3 bond pairs) and keeps 1 lone pair.
🔎 🧠 Step by Step Reasoning
⚙️Step 1: Valence electrons
📌 P = 5 valence electrons
📌 H = 1 × 3 = 3
📌 Total = 8 valence electrons.
⚙️Step 2: Bonding
📌 P forms 3 single bonds with 3 H atoms → 3 bond pairs (6 electrons).
📌 Step 3: Lone pairs
📌 Remaining 2 electrons stay as 1 lone pair on phosphorus.
👉 So, PH₃ has 3 bond pairs and 1 lone pair.

26. Ice has vacant spaces in what percentage due to its open hydrogen bonded structure?

a) 4%

b) 9%

c) 12%

d) 15%

✅ Correct Answer: (b) 9%
🔎 📌 Reason (short)
👉 The 9% vacant space in ice’s lattice makes its density lower than liquid water, hence ice floats.
🔎 🧠 Step by Step Reasoning
📌 Ice is less dense than water because of its open hexagonal lattice structure formed by hydrogen bonds.
📌 This arrangement leaves vacant spaces (voids) in the solid structure.
👉 The percentage of these vacant spaces is about 9%, which explains why ice floats on water.

27. If the radius of Zn²⁺ ion is 0.74 Å and that of S²⁻ ion is 1.84 Å, the radius ratio of ZnS should be:

a) 2.48

b) 1.84

c) 0.74

d) 0.40

✅ Correct Answer: (d) 0.40
🔎 📌 Reason (short)
👉 Radius ratio is cation radius ÷ anion radius. For Zn²⁺ and S²⁻, it equals 0.40, which determines the stable crystal structure of ZnS. 🔎 🧠 Step by Step Reasoning 📌 Radius ratio is defined as: Radius ratio = ʀᴄₐₜᵢₒₙ/ ʀᴀₙᵢₒₙ 📌 Given: 📌 rZn²⁺ = 0.74 A˚ 📌 rS²⁻ = 1.84 A˚ 📌 Calculation: 0.74/1.84 = 0.4/02 ≈ 0.40 👉 So the radius ratio = 0.40.

28. Which of the following enthalpy change is always negative?

a) ∆Hcₒₘ°

b) ∆Hᵣ°

c) ∆Hf°

d) ∆Hₐ°

✅ Correct Answer: (a) ∆Hcₒₘ°
👉 Combustion always releases heat (exothermic), so ∆Hc° is always negative, unlike reaction, formation, or atomization enthalpies which may vary.
🔎 ∆Hc° (enthalpy of combustion):
📌 Combustion is always exothermic → heat released → ∆H is negative. ✅ 👈
🔎 ∆Hr° (enthalpy of reaction):
📌 Can be positive or negative depending on whether the reaction is endothermic or exothermic.
🔎 ∆Hf° (enthalpy of formation):
📌 Can be negative (stable compounds like CO₂) or positive (unstable compounds like NO).
🔎 ∆Ha° (enthalpy of atomization):
📌 Always positive, since energy is required to break bonds into atoms.
👉 The only enthalpy change that is always negative is enthalpy of combustion (∆Hc°).

29. This principle involves in the liquefaction of gas:

a) Joule Thomson effect

b) Henry’s Law

c) Le Chatelier’s Principle

d) Raoult’s Law

✅ Correct Answer: (a) Joule Thomson effect
🔎 📌 Reason (short)
👉 Liquefaction of gases is achieved by cooling during expansion, explained by the Joule Thomson effect.
🔎 Step by Step Reasoning
📌 Joule Thomson effect: Real gases cool upon expansion without external work → basis of liquefaction. ✅
📌 Henry’s Law: Solubility of gases in liquids → unrelated to liquefaction.
📌 Le Chatelier’s Principle: Explains equilibrium shifts → not directly about liquefaction.
📌 Raoult’s Law: Deals with vapor pressure lowering in solutions → not about liquefaction.
👉 The principle used in liquefaction of gases is the Joule Thomson effect.

30. The geometry of NH₄⁺ and CrO₄²⁻ ions is:

a) Trigonal

b) Pyramidal

c) Tetrahedral

d) Square planar

✅ Correct Answer: (c) Tetrahedral
🔎 📌 Short Reason
👉 Both NH₄⁺ and CrO₄²⁻ have 4 bond pairs and no lone pairs on the central atom, giving them tetrahedral geometry.
🔎 🧠 Step by Step Reasoning
🔎 NH₄⁺ (ammonium ion):
📌 Central atom = N.
📌 N has 5 valence electrons, but in NH₄⁺ it forms 4 bonds with H atoms (positive charge removes lone pair).
📌 So, 4 bond pairs, no lone pair → tetrahedral geometry.
🔎 CrO₄²⁻ (chromate ion):
📌 Central atom = Cr.
📌 Surrounded by 4 O atoms, all equivalent due to resonance.
📌 4 bond pairs, no lone pair → tetrahedral geometry.
👉 Both ions have tetrahedral geometry. ✅ 👈

31. Which of the following atoms has the highest electron affinity?

a) F

b) Cl

c) N

d) O

✅ Correct Answer: (b) Cl
🔎 📌 Short Reason
👉 Although fluorine is more electronegative, chlorine has the highest electron affinity among the given atoms because its larger size reduces repulsion, making electron addition more favorable.
🔎 🧠 Step by Step Reasoning
💡 Electron affinity (EA): Energy released when an atom gains an electron.
💡 General trend: Increases across a period (left → right) and decreases down a group.
💡 But there are exceptions due to atomic size and repulsions.
🔎 Fluorine (F):
📌 Very electronegative, but small size → strong electron–electron repulsion in compact 2p orbital.
📌 EA is high, but not the maximum among halogens.
🔎 Chlorine (Cl):
📌 Larger size than F, so incoming electron faces less repulsion.
📌 EA of Cl is actually greater than F (≈ –349 kJ/mol vs –328 kJ/mol). ✅ 👈
🔎 Nitrogen (N):
📌 Half filled 2p³ configuration → adding an electron causes extra repulsion.
📌 EA is very low (close to zero). ❌
🔎 Oxygen (O):
📌 EA is higher than N, but still less than F and Cl. ❌

32. The shape of crystal of graphite is:

a) Cubic

b) Monoclinic

c) Hexagonal

d) Orthorhombic

✅ Correct Answer: (c) Hexagonal
👉 Among the given options, the shape of graphite crystal is hexagonal. 👈
📌 Cubic → typical of diamond, not graphite.
📌 Monoclinic and Orthorhombic → belong to other minerals, not graphite.
👉 Hexagonal → correct for graphite. 👈
🔎 Graphite is a crystalline allotrope of carbon.
🔎 Its atoms are arranged in layers of hexagonal rings of carbon atoms.
🔎 Each carbon atom is bonded to three others in a plane, forming a two dimensional sheet of hexagons.
🔎 These sheets are stacked parallel to each other, held together by weak van der Waals forces, which makes graphite soft and slippery.
🔎 Crystallographically, graphite belongs to the hexagonal crystal system 👈

33. Which one of the following is NOT an isoelectronic triad?

a) Na⁺, Ne, F⁻

b) Ca²⁺, K⁺, P³⁻

c) K⁺, Ar, Cl⁻

d) O⁻, Ne, Al³⁺

✅ Correct Answer: (d) O⁻, Ne, Al³⁺
🔎 📌 Short Reason
👉 Isoelectronic species must have the same number of electrons. Options (a), (b), and (c) all match, but in option (d), O⁻ has 9 electrons while Ne and Al³⁺ have 10. Therefore, (d) is not an isoelectronic triad.
🔎 🧠 Step by Step Reasoning
🔎 Isoelectronic species: Atoms/ions having the same number of electrons.
🌀 (a) Na⁺, Ne, F⁻
⚛️ Na (Z=11) → Na⁺ has 10 ē.
⚛️ Ne (Z=10) → 10 ē.
⚛️ F (Z=9) → F⁻ has 10 ē.
⚛️ All have 10 ē → Isoelectronic ✅
🌀 (b) Ca²⁺, K⁺, P³⁻
⚛️ Ca (Z=20) → Ca²⁺ has 18 ē.
⚛️ K (Z=19) → K⁺ has 18 ē.
⚛️ P (Z=15) → P³⁻ has 18 ē.
⚛️ All have 18 ē → Isoelectronic ✅
🌀 (c) K⁺, Ar, Cl⁻
⚛️ K (Z=19) → K⁺ has 18 ē.
⚛️ Ar (Z=18) → 18 ē.
⚛️ Cl (Z=17) → Cl⁻ has 18 ē.
⚛️ All have 18 ē → Isoelectronic ✅
🌀 (d) O⁻, Ne, Al³⁺
⚛️ O (Z=8) → O⁻ has 9 ē.
⚛️ Ne (Z=10) → 10 ē.
⚛️ Al (Z=13) → Al³⁺ has 10 ē.
⚛️ Not all equal (9 vs 10 vs 10) → Not isoelectronic ❌ ✅ Correct Answer 👈

34. The volume of 1 g of hydrogen gas at STP is:

a) 11.2 cm³

b) 22.4 dm³

c) 11200 cm³

d) 0.0224 m³

✅ Correct Answer: (c) 11200 cm³
🔎 🧠 Step by Step Reasoning
⚛️Molar mass of H₂: M = 2 g/mol
⚛️Number of moles in 1 g H₂: n = mass/molar mass = ½ = 0.5 mol
⚛️Molar volume at STP: 1 mol gas = 22.4 L = 22.4 dm³
⚛️Volume of 0.5 mol H₂: V = 0.5 × 22.4 = 11.2 L 💙 👈
⚛️Unit conversions:
⚡11.2 L = 11.2 dm³
⚡11.2 L = 11200 cm³ 💙 👈
⚡11.2 L = 0.0112 m³
🔎 📌 Short Reason
👉 At STP, 1 g of hydrogen gas corresponds to 0.5 mol, which occupies 11.2 L. Converting to cm³ gives 11200 cm³, so the correct option is (c).

35. Which of the following compounds has/have sp² hybridization?

a) C₆H₆

b) C₂H₄

c) CH₃⁺

d) All of these

✅ Correct Answer: (d) All of these
🔎 📌 Short Reason
👉 C₆H₆ (benzene) → sp² hybridized carbons (planar hexagonal ring).
👉 C₂H₄ (ethene) → sp² hybridized carbons (planar double bond).
👉 CH₃⁺ (methyl cation) → sp² hybridized carbon (trigonal planar). Thus, all of them exhibit sp² hybridization.
🔎 🧠 Step by Step Reasoning
🌀 (a) C₆H₆ (Benzene):
📌 Each carbon atom in benzene forms 3 sigma bonds (2 with C, 1 with H) and has delocalized π electrons.
📌 Steric number = 3 → sp² hybridization. ✅
🌀 (b) C₂H₄ (Ethene):
📌 Each carbon forms 3 sigma bonds (2 with H, 1 with C) and 1 π bond.
📌 Steric number = 3 → sp² hybridization. ✅
🌀 (c) CH₃⁺ (Methyl cation):
📌 Carbon forms 3 sigma bonds with H, no lone pairs.
📌 Steric number = 3 → sp² hybridization. ✅
🌀 (d) All of these:
👉 Since all three (C₆H₆, C₂H₄, CH₃⁺) are sp² hybridized, the correct choice is All of these. ✅ 💜👈

36. If 200 cm³ of 1 M solution is diluted up to 2000 cm³, its molarity would be:

a) 10 M

b) 0.2 M

c) 0.1 M

d) 1 M

✅ Correct Answer: (c) 0.1 M
🔎 📌 Short Reason
👉 When 200 cm³ of a 1 M solution is diluted to 2000 cm³, the concentration decreases by a factor of 10. The new molarity is 0.1 M. 👈
🔎 🧠 Step by Step Reasoning
🌀 Dilution formula: M₁V₁ = M₂V₂
🌀 Given:
📌 M₁ = 1 M
📌 V₁ = 200 cm³
📌 V₂ = 2000 cm³
📌 M₂ = ?
🌀 Substitution:
📌 1 × 200 = M₂ × 2000
📌 M₂=200/2000 = 0.1 M 👈

37. What is the pOH of a solution whose pH is 8?

a) 10

b) 6

c) 4

d) 2

✅ Correct Answer: (b) 6
🔎 📌 Short Reason
👉 For a solution with pH = 8, the corresponding pOH = 6. This follows directly from the relation pH + pOH = 14.
🔎 🧠 Step by Step Reasoning
📌 The fundamental relation between pH and pOH: pH + pOH = 14 (at 25°C)
📌 Given: pH = 8
📌 So: pOH = 14−8 = 6 👈

38. When gaseous anions and cations are brought closer, the energy evolved:

a) Electron affinity

b) E.N. (Electronegativity)

c) I.P. (Ionization potential)

d) Lattice energy

✅ Correct Answer: (d) Lattice energy
🔎 📌 Short Reason
👉 When gaseous cations and anions approach each other and form an ionic lattice, the energy released is lattice energy. This stabilizes the ionic solid structure.
🔎 🧠 Step by Step Reasoning
🌀 Electron affinity (EA): Energy released when a neutral atom gains an electron. → Not correct here.
🌀 Electronegativity (EN): Tendency of an atom to attract electrons in a bond. → Not an energy term.
🌀 Ionization potential (IP): Energy required to remove an electron from an atom. → Not relevant here.
🌀 Lattice energy: Energy released when gaseous cations and anions combine to form an ionic solid.
👉 This exactly matches the situation described: gaseous ions coming together → energy evolved = lattice energy. ✅ 👈

39. The energy of activation of a chemical reaction is 75.0 kJ·mol⁻¹. What will be its value in presence of a positive catalyst?

a) 53.3 kJ·mol⁻¹

b) 75.0 kJ·mol⁻¹

c) 86.7 kJ·mol⁻¹

d) 109.1 kJ·mol⁻¹

✅ Correct Answer: (a) 53.3 kJ·mol⁻¹
🔎 📌 Short Reason
👉 A positive catalyst reduces the activation energy of a reaction. Since the uncatalyzed Eₐ is 75.0 kJ·mol⁻¹, the catalyzed value must be lower. The correct option is 53.3 kJ·mol⁻¹.
🔎 🧠 Step by Step Reasoning
📌 Activation energy (Eₐ): Minimum energy required for reactants to form products.
📌 Positive catalyst: Provides an alternative pathway with lower activation energy.
📌 The original Eₐ = 75.0 kJ·mol⁻¹.
📌 With a catalyst, Eₐ must be less than 75.0 kJ·mol⁻¹.
👉 Among the given options, only 53.3 kJ·mol⁻¹ is lower.

40. An orbital can have a maximum of:

a) 10 electrons

b) 6 electrons

c) 2 electrons

d) 14 electrons

✅ Correct Answer: (c) 2 electrons
🔎 🧠 Step by Step Reasoning
📌 An orbital is a region of space where the probability of finding an electron is high.
📌 According to the Pauli Exclusion Principle, each orbital can hold a maximum of 2 electrons, with opposite spins. 👈
📌 The other numbers (6, 10, 14) correspond to maximum electrons in subshells:
📌 p subshell → 3 orbitals × 2 = 6 electrons.
📌 d subshell → 5 orbitals × 2 = 10 electrons.
📌 f subshell → 7 orbitals × 2 = 14 electrons.
📌 But the question asks specifically about one orbital, not subshells.

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