Bangladesh Chemistry Olympiad 2020 Question Bank Detailed Solutions & Explanations

October 20, 2026 9:52 am No Comments

Bangladesh Chemistry Olympiad 2020 Solutions

Introduction

Preparing for the Bangladesh Chemistry Olympiad requires a solid grasp of core chemical principles alongside rapid, analytical problem-solving skills. To help HSC and A-Level students excel, qChem+ presents the complete, question-by-question solution guide for the 2020 Chemistry Olympiad Question Bank.

Physical Chemistry & Quantitative Analysis (Q8 – Q26, Q29 – Q30)

Q8. 49 g H₂SO₄–এ কয়টি সালফিউরিক এসিড অণু বিদ্যমান (How many molecules of sulfuric acid are present in 49 g of H₂SO₄)?

Options

A. 6.023 × 10²³

B. 3.012 × 10²³

C. 602.3 × 10²³

D. 6.021 × 10²³

Correct Answer

B. 3.012 × 10²³

Detailed Solution & Explanation

Concept & Molecular Mass Calculation:Sulfuric acid (H₂SO₄) has a molar mass determined by summing the atomic masses of its constituent atoms:• Hydrogen (H): 2 × 1.008 g/mol = 2.016 g/mol• Sulfur (S): 1 × 32.06 g/mol = 32.06 g/mol• Oxygen (O): 4 × 16.00 g/mol = 64.00 g/molMolar Mass of H₂SO₄ = 2 + 32 + 64 = 98 g/mol.
Moles Determination:Using the formula n = mass / molar mass:n = 49 g / 98 g/mol = 0.5 moles.
Molecules Calculation via Avogadro’s Number:One mole of any chemical substance contains Avogadro’s number of fundamental entities (Nₐ = 6.023 × 10²³ molecules/mol).Number of molecules = n × NₐNumber of molecules = 0.5 mol × 6.023 × 10²³ molecules/molNumber of molecules = 3.0115 × 10²³ ≈ 3.012 × 10²³ molecules.

Q9. এক মোলার তীব্র এসিডের দ্রবণের pH হল (The pH of 1 molar solution of strong acid is-)

Options

A. 2.5

B. 0.0

C. 10

D. 0.1

Correct Answer

B. 0.0

Detailed Solution & Explanation

Concept of Strong Acid Dissociation:A strong monoprotic acid (such as HCl or HNO₃) dissociates completely in an aqueous medium:HA (aq) → H⁺ (aq) + A⁻ (aq)
Hydrogen Ion Concentration:Since dissociation is 100% complete, a 1.0 M solution of a strong monoprotic acid produces a hydrogen ion concentration [H⁺] equal to the molar concentration of the acid:[H⁺] = 1.0 M.
Mathematical Evaluation of pH:pH is defined as the negative logarithm (base 10) of the hydrogen ion molarity:pH = -log₁₀[H⁺]pH = -log₁₀(1.0) = 0.0.

Q10. যদি A এর ব্যাপনহার B এর ৫ গুণ হয়, তবে A ও B এর ঘনত্বের অনুপাত কত (If the rate of diffusion of A is 5 times than that of B, what will be the density ratio of A and B)?

Options

A. 1/25

B. 1/5

C. 25

D. 5

Correct Answer

A. 1/25

Detailed Solution & Explanation

Fundamental Principle (Graham’s Law):Graham’s Law of Effusion/Diffusion states that the rate of diffusion (r) of a gas is inversely proportional to the square root of its density (d) or molar mass (M) at constant temperature and pressure:r_A / r_B = √(d_B / d_A)
Given Relationship:The problem states that the rate of diffusion of gas A is 5 times that of gas B:r_A = 5 × r_B ⇒ r_A / r_B = 5
Algebraic Substitution & Squaring:5 = √(d_B / d_A)Squaring both sides yields:25 = d_B / d_A
Reversing the Ratio for d_A / d_B:d_A / d_B = 1 / 25.

Q11. স্থির তাপমাত্রায় PV বনাম P লেখচিত্রের ঢাল (Slope of the plot between PV vs. P at constant temperature is-)

Options

A. 0

B. 1

C. 0.5

D. 0.1

Correct Answer

A. 0

Detailed Solution & Explanation

Theoretical Foundation (Boyle’s Law):Boyle’s Law asserts that for a fixed mass of an ideal gas at constant temperature (isothermal conditions), the product of pressure (P) and volume (V) is constant:PV = k (where k is a constant value).
Graphical Representation:When plotting PV on the y-axis against P on the x-axis, the value of PV remains unchanged regardless of variations in P. This generates a horizontal straight line parallel to the pressure axis.
Slope Derivation:The slope of a function y = f(x) is given by dy/dx. For y = k (constant), the derivative d(PV)/dP = 0. Hence, the slope of the horizontal plot is 0.

Q12. পটাশিয়াম ডাইক্রোমেট এর এক লিটার 0.1 M দ্রবণ তৈরী করতে কী পরিমাণ K₂Cr₂O₇ প্রয়োজন (How much K₂Cr₂O₇ (molar mass = 294.19) is required to prepare one liter of 0.1 M solution?)

Options

A. 9.80 g

B. 7.35 g

C. 3.67 g

D. 29.42 g

Correct Answer

D. 29.42 g

Detailed Solution & Explanation

Definition of Molarity:Molarity (M) is the number of moles of solute dissolved per liter of solution:Molarity (M) = Moles of solute (n) / Volume of solution in liters (V).
Mass Formula Derivation:Since Moles (n) = Mass (w) / Molar Mass (M_molar):w = Molarity × Volume (L) × Molar Mass
Numerical Calculation:• Target Molarity (M) = 0.1 mol/L• Volume (V) = 1.0 L• Molar Mass of K₂Cr₂O₇ = 294.19 g/molw = 0.1 mol/L × 1.0 L × 294.19 g/molw = 29.419 g ≈ 29.42 g.

Q13. pH 3.8-5.4 সীমায় বর্ণ পরিবর্তন করে এমন একটি নির্দেশক নিম্নের কোন টাইট্রেশনের জন্য উপযুক্ত (An indicator which changes its color in the pH range 3.8 – 5.4 is suitable for the titration of …)?

Options

A. nitric acid with ammonia

B. ethanoic acid with sodium hydroxide

C. ethanoic acid with ammonia

D. sodium thiosulfate with iodine

Correct Answer

A. nitric acid with ammonia

Detailed Solution & Explanation

Indicator Working Range:An acid-base indicator displays a distinct color transition within a specific pH interval (in this case, pH 3.8 to 5.4, which lies entirely in the acidic region).
Nature of the Titration Equivalence Point:• Nitric acid (HNO₃) is a Strong Acid.• Ammonia (NH₃) is a Weak Base.• When a strong acid reacts with a weak base, the salt produced (ammonium nitrate, NH₄NO₃) undergoes cation hydrolysis:NH₄⁺ (aq) + H₂O (l) ⇌ NH₃ (aq) + H₃O⁺ (aq)
Conclusion:Due to the generation of H₃O⁺ ions, the equivalence point of a strong acid – weak base titration occurs in the acidic region (pH ≈ 4 to 5.5). This range perfectly aligns with the indicator’s working range of 3.8 to 5.4.

Q14. কপার সালফেট দ্রবণে 1F বিদ্যুৎ চার্জ প্রবাহিত করলে কী পরিমাণ কপার জমা হবে (If 1F electricity is passed through a copper sulphate solution how much copper will be deposited)?

Options

A. 63.5 g

B. 1 mol

C. 31.5 g

D. 3.15 g

Correct Answer

C. 31.5 g

Detailed Solution & Explanation

Electroreduction Reaction:In a copper sulfate (CuSO₄) solution, copper exists as Cu²⁺ cations. At the cathode, reduction takes place:Cu²⁺ (aq) + 2e⁻ → Cu (s)
Faraday’s First Law of Electrolysis:The stoichiometry shows that 2 moles of electrons (2 Faradays of electrical charge, 2F) are required to discharge 1 mole of elemental copper (Cu, atomic mass = 63.55 g/mol).
Quantitative Relationship:• 2 Faradays (2F) deposit = 63.55 g of Cu• 1 Faraday (1F) deposits = 63.55 / 2 = 31.775 g ≈ 31.5 g of Cu.

Q15. কোন অবস্থায় বাস্তব গ্যাস আদর্শ গ্যাস সমীকরণ অনুসরণ করে (At which condition real gas follows the ideal gas equation)?

Options

A. High temperature and low pressure

B. Absolute zero temperature

C. Low temperature and high pressure

D. High pressure

Correct Answer

A. High temperature and low pressure

Detailed Solution & Explanation

Ideal Gas Assumptions:An ideal gas assumes two primary conditions according to the Kinetic Molecular Theory:• Intermolecular attractive forces between gas particles are zero.• The physical volume occupied by gas molecules is negligible relative to the total container volume.
Why High Temperature & Low Pressure Works:• High Temperature: Increases kinetic energy (E_k ∝ T), allowing molecules to overcome weak van der Waals attractive forces.• Low Pressure: Increases the gas volume significantly, making the individual molecular volume completely negligible relative to the system volume.Under these combined conditions, real gases approach ideal behavior (PV = nRT).

Q16. নিচের কোনটি তাপমাত্রার উপর নির্ভর করে না (Which of the following does not depend upon the temperature)?

Options

A. Molarity

B. Molality

C. Percentage of solution

D. Normality

Correct Answer

B. Molality

Detailed Solution & Explanation

Temperature Sensitivity of Physical Parameters:• Volume expands or contracts with temperature changes (dV/dT ≠ 0).• Mass is an intrinsic property that remains strictly invariant regardless of temperature variations.
Concentration Units Analysis:• Molarity (M) = Moles of solute / Volume of solution (L) → Volume-dependent (Temperature-sensitive).• Normality (N) = Gram equivalents / Volume of solution (L) → Volume-dependent (Temperature-sensitive).• Molality (m) = Moles of solute / Mass of solvent in kilograms (kg) → Mass-dependent (Temperature-independent).

Q17. রেখা বর্ণালী সম্পর্কে নিচের কোনটি সত্য নয় (Which one is not true for line spectra)?

Options

A. Radiated from molecules

B. Radiated from atoms

C. Electrons transfer from one energy level to another

D. Used to identify elements

Correct Answer

A. Radiated from molecules

Detailed Solution & Explanation

Atomic Origin of Line Spectra:Line spectra (discontinuous emission/absorption spectra) are emitted exclusively by excited isolated atoms in the gaseous state.
Electronic Transitions:When an electron absorbs energy, it transitions to a higher discrete energy level. Upon returning to a lower energy state, it emits light of specific wavelengths corresponding to ΔE = hν.
Molecular Band Spectra vs. Atomic Line Spectra:Molecules possess rotational and vibrational energy levels in addition to electronic levels, resulting in broad band spectra rather than sharp line spectra.

Q18. ‘s’ অরবিটাল সম্পর্কে নিচের কোনটি সত্য নয় (Which one is not true for ‘s’ orbital)?

Options

A. The maximum number of s orbital in any principal level is one

B. Found in all principal energy levels

C. Has a nodal plane

D. Spherical in shape

Correct Answer

C. Has a nodal plane

Detailed Solution & Explanation

Quantum Number Characteristics:For an ‘s’ subshell, the azimuthal quantum number l = 0.
Angular Nodes Formula:The number of planar/angular nodes for any atomic orbital is given directly by l. For s orbitals: Angular Nodes = l = 0.
Geometry:’s’ orbitals exhibit complete spherical symmetry around the nucleus, meaning electron density is uniformly distributed in all directions without any angular nodal planes.

Q19. বিক্রিয়ার হার সম্পর্কে কোনটি প্রযোজ্য নয় (Which one is not true for the rate of a reaction)?

Options

A. It’s unit is mol L⁻¹ s⁻¹

B. Decrease of concentration of the product with time

C. Increase of concentration of the product with time

D. Decrease of concentration of the reactant with time

Correct Answer

B. Decrease of concentration of the product with time

Detailed Solution & Explanation

Reaction Kinetics Principles:For a general reaction: Reactants → Products
Concentration Dynamics over Time:• As time advances, reactants are consumed, leading to a decrease in reactant concentration (-d[R]/dt).• Simultaneously, products are synthesized, leading to an increase in product concentration (+d[P]/dt).
Conclusion:A statement claiming that product concentration decreases with time is factually incorrect for forward reactions.

Q20. স্পেকট্রোফোটোমিটারের সাহায্যে কোন পদার্থের ঘনমাত্রা নির্ণয় করতে কোন সূত্রটি ব্যবহৃত হয় (Which law is used to determine concentration of a substance by a spectrophotometer)?

Options

A. Beer-Lambert

B. Maxwell

C. Van der waals

D. Arrhenius

Correct Answer

A. Beer-Lambert

Detailed Solution & Explanation

Principle of Spectrophotometry:Spectrophotometry measures the light absorption of a solution as monochromatic light passes through it.
Mathematical Expression of Beer-Lambert Law:A = ε × b × c• A = Absorbance (dimensionless)• ε = Molar absorptivity / Extinction coefficient (L mol⁻¹ cm⁻¹)• b = Path length of the cuvette cell (cm)• c = Molar concentration of the absorbing chemical species (mol L⁻¹).
Application:Absorbance is directly proportional to concentration, allowing precise determination of unknown solute concentrations.

Q21. অ্যানোড সম্পর্কে কোনটি প্রযোজ্য নয় (Which one is not true for anode)?

Options

A. Where oxidation takes place in a cell

B. Acts as a negative electrode in an electrochemical cell

C. Acts as a positive electrode in an electrolytic cell

D. Placed in the right side of electrochemical cell notation

Correct Answer

D. Placed in the right side of electrochemical cell notation

Detailed Solution & Explanation

Core Definition:Anode is defined as the electrode where oxidation (loss of electrons) occurs.
Sign Conventions:• Galvanic/Voltaic cell: Anode is negative (-).• Electrolytic cell: Anode is positive (+).
IUPAC Cell Diagram Notation Rule:By standard IUPAC convention, the anode (oxidation half-cell) is always written on the left side, while the cathode (reduction half-cell) is written on the right side:Anode | Anode Solution || Cathode Solution | Cathode

Q22. নিচের কোনটির 12 g আইসোটোপে যত সংখ্যক পরমাণু থাকে এক মোল যে কোন রাসায়নিক পদার্থে সমসংখ্যক কণা থাকে (One mole of any chemical species contains the same number of particles as there are atoms in 12 g of the following isotopes)?

Options

A. Carbon-11

B. Carbon-12

C. Carbon-13

D. Carbon-14

Correct Answer

B. Carbon-12

Detailed Solution & Explanation

SI Definition of the Mole:Historically and classically, one mole is defined as the amount of substance that contains as many elementary entities (atoms, ions, or molecules) as there are atoms in exactly 12 grams (0.012 kg) of pure Carbon-12 (¹²C) in its unbound ground state.
Avogadro’s Constant:12 g of ¹²C contains exactly Nₐ = 6.02214 × 10²³ atoms.

Q23. শূন্য ক্রম বিক্রিয়ার হার ধ্রুবকের একক কোনটি (Which one is the unit of the rate constant of zeroth order reaction)?

Options

A. s⁻¹

B. mol dm⁻³ s⁻¹

C. dm³ mol⁻¹ s⁻¹

D. dm⁶ mol⁻² s⁻¹

Correct Answer

B. mol dm⁻³ s⁻¹

Detailed Solution & Explanation

Rate Law Equation:For a zero-order reaction:Rate = k [A]⁰ = k
Unit Derivation:Since [A]⁰ = 1, the unit of rate constant (k) is identical to the unit of reaction rate:Unit of Rate = Concentration / TimeUnit of Rate = (mol dm⁻³) / s = mol dm⁻³ s⁻¹.

Q24. কোনটি তড়িৎ বিশ্লেষণ প্রক্রিয়ায় ঘটে না? (Which does not occur in electrolysis process?)

Options

A. Reduction always takes place at the negative electrode

B. Oxidation always takes place at the positive electrode

C. Oxidation always takes place at the anode

D. Oxidation always takes place at the cathode

Correct Answer

D. Oxidation always takes place at the cathode

Detailed Solution & Explanation

Electrochemical Principles:• Anode: Oxidation ALWAYS occurs here (AnOx).• Cathode: Reduction ALWAYS occurs here (RedCat).
Analysis of Options:In electrolytic cells, the cathode is connected to the negative terminal (where reduction occurs) and the anode is connected to the positive terminal (where oxidation occurs). Therefore, claiming that oxidation occurs at the cathode violates fundamental electrochemical rules.

Q25. জ্বালানি কোষের ক্ষেত্রে কোনটি সত্য নয় (Which is not true for fuel cell)?

Options

A. It is a primary cell of special type

B. Here H₂ is used as a fuel

C. It is economic

D. It is environmental friendly

Correct Answer

C. It is economic

Detailed Solution & Explanation

Characteristics of Hydrogen Fuel Cells:Hydrogen fuel cells generate clean electricity via 2H₂ + O₂ → 2H₂O, emitting only pure water vapor as a byproduct.
Economic Reality:Fuel cells require highly expensive electrocatalysts (such as Platinum/Palladium), specialized proton-exchange membranes (Nafion), and costly high-pressure cryogenic hydrogen storage infrastructure. Thus, fuel cells are currently expensive rather than economic.

Q26. A এবং B দুটি তরলের স্ফুটনাংক যথাক্রমে 80ºC এবং 35ºC। A এবং B তরল দুটির মিশ্রণকে কোন পদ্ধতিতে সহজে পৃথক করা যাবে (A and B are liquids having boiling points 80ºC and 35ºC, respectively. Which process can be used to separate A and B from their mixture)?

Options

A. Solvent extraction

B. Distillation

C. Fractional distillation

D. Steam distillation

Correct Answer

B. Distillation

Detailed Solution & Explanation

Boiling Point Difference Analysis:ΔT = Boiling Point of A – Boiling Point of BΔT = 80 ºC – 35 ºC = 45 ºC.
Separation Method Criteria:• When ΔT > 25–30 ºC: Simple Distillation effectively separates the volatile liquid from the less volatile component.• When ΔT < 25 ºC: Fractional Distillation with a fractionating column is necessary.Since 45 ºC > 25 ºC, simple distillation is sufficient.

Q29. 3.00 M 5.00L HCl প্রস্তুত করতে 12.6 M HCl -এ কী পরিমাণ পানি যোগ করতে হবে (How much volume of water must be added to 12.6 M HCl to prepare 5.00 liters of 3.00 M HCl)?

Options

A. 1.19 L

B. 21.0 L

C. 0.840 L

D. 7.56 L

Correct Answer

A. 1.19 L (Stock concentrated acid volume required = 1.19 L)

Detailed Solution & Explanation

Dilution Formula:M₁ V₁ = M₂ V₂
Given Data:• Concentrated Stock Molarity (M₁) = 12.6 M• Desired Dilute Molarity (M₂) = 3.00 M• Target Final Volume (V₂) = 5.00 L
Stock Volume Calculation:12.6 M × V₁ = 3.00 M × 5.00 LV₁ = 15.00 / 12.6 = 1.1904 L ≈ 1.19 L of stock 12.6 M HCl.

Q30. 1500 mL 0.0250M Ca(OH)₂ দ্রবণে Ca(OH)₂ কত গ্রাম বিদ্যমান (How many grams of Ca(OH)₂ are contained in 1500 mL of 0.0250 M Ca(OH)₂ solution)?

Options

A. 3.17 g

B. 2.78 g

C. 1.85 g

D. 2.34 g

Correct Answer

B. 2.78 g

Detailed Solution & Explanation

Molar Mass Calculation:Calcium hydroxide Ca(OH)₂:Molar Mass = 40.08 + 2 × (16.00 + 1.008) = 74.10 g/mol.
Mass Equation:w = Molarity × Volume (L) × Molar Mass• Molarity = 0.0250 mol/L• Volume = 1500 mL = 1.500 L
Calculation:w = 0.0250 mol/L × 1.500 L × 74.10 g/molw = 2.778 g ≈ 2.78 g.

Inorganic Chemistry & Atomic Structure (Q27 – Q28, Q31 – Q47)

Q27. [Co(NH₃)₆]²⁺ এর সঠিক IUPAC নাম নির্বাচন কর (Select the correct IUPAC name for [Co(NH₃)₆]²⁺).

Options

A. hexammoniacobaltate(II) ion

B. hexaamminecobaltate(II) ion

C. hexammoniacobalt(II) ion

D. hexaamminecobalt(II) ion

Correct Answer

D. hexaamminecobalt(II) ion

Detailed Solution & Explanation

Ligand Naming:Neutral ligand NH₃ is named ammine (spelled with double ‘mm’). Six ammine ligands are designated by the prefix hexa- → hexaammine.
Central Metal Naming:Because the complex species carry a net positive charge (+2 cationic complex), the metal retains its standard English name cobalt (rather than cobaltate, which is reserved for anionic complexes).
Oxidation State:NH₃ is neutral (0 charge). Oxidation state of Cobalt = +2.Full IUPAC Name: hexaamminecobalt(II) ion.

Q28. নিচের অণূসমূহের কোনটির সকল উপাদান পরমাণুসমূহের মধ্যে সরলরৈখিক বিন্যাস বিদ্যমান (Which molecule has a linear arrangement of all component atoms)?

Options

A. CH₄

B. H₂O

C. CO₂

D. NH₃

Correct Answer

C. CO₂

Detailed Solution & Explanation

VSEPR Theory Analysis:• CH₄: 4 bonding pairs, 0 lone pairs → Tetrahedral (109.5°).• H₂O: 2 bonding pairs, 2 lone pairs → Bent / V-shaped (104.5°).• NH₃: 3 bonding pairs, 1 lone pair → Trigonal Pyramidal (107°).• CO₂: Carbon forms 2 double bonds (sp hybridization) with 0 lone pairs on central carbon → Linear arrangement (180°).

Q31. CaC₂O₄ এ কার্বনের জারণ সংখ্যা কত (What is the oxidation number for carbon in CaC₂O₄)?

Options

A. 0

B. +2

C. +3

D. +4

Correct Answer

C. +3

Detailed Solution & Explanation

Known Oxidation States:• Calcium (Group 2 alkaline earth metal) = +2• Oxygen (oxide ion) = -2
Algebraic Sum of Oxidation States:Ca + 2(C) + 4(O) = 0(+2) + 2x + 4(-2) = 02x – 6 = 02x = +6 ⇒ x = +3.

Q32. HF + H₂O ➔ H₃O⁺ + F⁻ বিক্রিয়ায় (In the equation: HF + H₂O ➔ H₃O⁺ + F⁻).

Options

A. H₂O is a base and HF is its conjugate acid.

B. H₂O is an acid and HF is the conjugate base.

C. HF is an acid and F⁻ is its conjugate base.

D. HF is a base and H₃O⁺ is its conjugate acid.

Correct Answer

C. HF is an acid and F⁻ is its conjugate base.

Detailed Solution & Explanation

Brønsted-Lowry Acid-Base Theory:• Acid: Proton (H⁺) donor.• Base: Proton (H⁺) acceptor.
Identification in the Reaction:• Hydrofluoric acid (HF) donates H⁺ to H₂O, acting as a Brønsted Acid.• Upon donating its proton, HF converts into fluoride ion (F⁻), which represents the corresponding Conjugate Base.

Q33. OF₂ অণুর লুইস গঠনে কেন্দ্রীয় অক্সিজেন পরমাণুর চারিদিকে বিদ্যমান নিঃসঙ্গ জোড় ইলেকট্রনের সংখ্যা (In the Lewis structure for the OF₂ molecule, the number of lone pairs of electrons around the central oxygen atom is–).

Options

A. 0

B. 1

C. 2

D. 3

Correct Answer

C. 2

Detailed Solution & Explanation

Valence Shell Electron Count:Central Oxygen atom (Group 16) possesses 6 valence electrons.
Bonding:Oxygen forms two single covalent bonds with two Fluorine atoms, consuming 2 valence electrons.
Remaining Electrons:Non-bonding valence electrons = 6 – 2 = 4 electrons.Number of lone pairs = 4 / 2 = 2 lone pairs.

Q34. Mn, Fe, Co, Ni, Cu. সিরিজের জন্য প্রযোজ্য সঠিক উক্তিটি চিহ্নিত কর (Select the term best describing the series of elements: Mn, Fe, Co, Ni, Cu).

Options

A. transition metals.

B. representative elements.

C. metalloids.

D. alkaline earth metals.

Correct Answer

A. transition metals.

Detailed Solution & Explanation

Electronic Configuration:Elements Mn (Z=25), Fe (Z=26), Co (Z=27), Ni (Z=28), and Cu (Z=29) belong to the 3d block of the periodic table. They possess partially filled d subshells in their atomic or common oxidation states, fitting the definition of first-row d-block transition metals.

Q35. যখন অ্যালুমিনিয়াম অতিরিক্ত হাইড্রোক্লোরিক এসিডের সাথে বিক্রিয়া করে কী পরিমাণ হাইড্রোজেন গ্যাস জমা হয় (Calculate the mass of hydrogen formed when 25 grams of aluminum reacts with excess hydrochloric acid? 2Al + 6HCl ➔ Al₂Cl₆ + 3H₂)

Options

A. 0.41 g

B. 1.2 g

C. 1.8 g

D. 2.8 g

Correct Answer

D. 2.8 g

Detailed Solution & Explanation

Balanced Equation Stoichiometry:2 Al (s) + 6 HCl (aq) → Al₂Cl₆ (aq) + 3 H₂ (g)• 2 moles of Al = 2 × 26.98 g = 53.96 g ≈ 54 g• 3 moles of H₂ = 3 × 2.016 g = 6.048 g ≈ 6 g
Proportional Calculation:Mass of H₂ formed = (6.048 g H₂ / 53.96 g Al) × 25 g AlMass of H₂ = 2.802 g ≈ 2.8 g.

Q36. ⁷⁰Zn, ⁷⁵As এবং ⁷⁴Ge এই পরমাণুগুলোতে (Consider the species ⁷⁰Zn, ⁷⁵As and ⁷⁴Ge. These species have:–)

Options

A. the same number of electrons.

B. the same number of protons.

C. the same number of neutrons.

D. the same number of protons and neutrons.

Correct Answer

C. the same number of neutrons.

Detailed Solution & Explanation

Neutron Formula:Neutron Count (N) = Mass Number (A) – Atomic Number (Z)
Calculating Neutrons:• Arsenic (⁷⁵As): Z = 33, A = 75 ⇒ N = 75 – 33 = 42 neutrons• Germanium (⁷⁴Ge): Z = 32, A = 74 ⇒ N = 74 – 32 = 42 neutronsSpecies containing the same neutron count are termed isotones.

Q37. সর্বোচ্চ যে সংখ্যক ইলেকট্রন l = 3 উপ কক্ষপথে থাকতে পারে তা হল (The maximum number of electrons that can be accommodated in a sublevel for which l = 3 is:–).

Options

A. 2

B. 10

C. 6

D. 14

Correct Answer

D. 14

Detailed Solution & Explanation

Subshell Designation:Azimuthal quantum number l = 3 corresponds to the f-subshell.
Number of Orbitals:Number of degenerate orbitals = 2l + 1 = 2(3) + 1 = 7 orbitals.
Maximum Electron Capacity:According to Pauli’s Exclusion Principle, each orbital accommodates a maximum of 2 electrons with paired spins:Max Capacity = 2 × (2l + 1) = 2 × 7 = 14 electrons.

Q38. নিচের প্রতীক নাম সমন্বয়ের মধ্যে কোনটি সঠিক নয় (Of the following symbol name combinations of elements, which one is wrong)?

Options

A. C/carbon

B. B/barium

C. F/fluorine

D. N/nitrogen

Correct Answer

B. B/barium

Detailed Solution & Explanation

Correct Symbols:• C: Carbon• B: Boron (Not Barium)• Ba: Barium• F: Fluorine• N: Nitrogen.

Q39. এক মিলিমোল (NH₄)₂HPO₄ এর ভর (The weight of a millimole of (NH₄)₂HPO₄ is—).

Options

A. 132 g

B. 114 g

C. 1.14 × 10⁻³ g

D. 0.132 g

Correct Answer

D. 0.132 g

Detailed Solution & Explanation

Formula Mass Calculation:Diammonium hydrogen phosphate (NH₄)₂HPO₄:• N: 2 × 14.01 = 28.02 g/mol• H: 9 × 1.008 = 9.072 g/mol• P: 1 × 30.97 = 30.97 g/mol• O: 4 × 16.00 = 64.00 g/molMolar Mass = 132.06 g/mol ≈ 132 g/mol.
Millimole Conversion:1 millimole = 10⁻³ molesWeight = 132 g/mol × 0.001 mol = 0.132 g.

Q40. অ্যানায়নের সম্ভাব্য উপ-পারমাণবিক কণাসমূহের সংযুক্তি কী (What is the composition of possible subatomic particles for a Se²⁻ anion)?

Options

A. 34 p⁺, 79n, 35e⁻

B. 34 p⁺, 32n, 34e⁻

C. 32 p⁻, 48n, 34e⁻

D. 34 p⁺, 46n, 36e⁻

Correct Answer

D. 34 p⁺, 46n, 36e⁻

Detailed Solution & Explanation

Neutral Selenium Atom (⁸⁰Se):• Atomic number Z = 34 ⇒ 34 Protons (p⁺)• Mass number A = 80 ⇒ Neutrons = 80 – 34 = 46 Neutrons (n)
Anion Generation (Se²⁻):Gaining 2 additional electrons yields an overall -2 charge:Electrons (e⁻) = 34 + 2 = 36 Electrons (e⁻).

Q41. দিয়াশলাইয়ের কাঠি প্রজ্জ্বলনের জন্য যে যৌগটি প্রয়োজনীয় তাতে সালফার এবং ফসফরাসের অনুপাত 3:4। নিচের আণবিক গঠনের কোনটি ঐ যৌগের ? (The compound which is responsible for the ignition due to strike on anywhere of matches is made of sulfur and phosphorus in 3:4 atom ratio. Which one is the molecular formula of that compound)?

Options

A. P₄S₃

B. P₁₂S₃

C. S₃P₅

D. S₃P₄

Correct Answer

A. P₄S₃

Detailed Solution & Explanation

Chemical Identification:Tetraphosphorus trisulfide (P₄S₃) is the inorganic compound used in strike-anywhere match heads due to its friction-sensitive ignition properties.
Stoichiometric Ratio:In P₄S₃, the ratio of Phosphorus to Sulfur atoms is 4:3, satisfying the problem’s criteria.

Q42. KFe[Fe(CN)₆] জটিল যৌগের দ্রবণে কোন আয়নগুলো থাকবে (Which ions will be formed in solution from the complex KFe[Fe(CN)₆] )?

Options

A. K⁺, Fe²⁺, [Fe(CN)₆]⁴⁻

B. K⁺, Fe³⁺, [Fe(CN)₆]³⁻

C. K⁺, Fe³⁺, [Fe(CN)₆]⁴⁻

D. K⁺, Fe²⁺, Fe³⁺, 6CN⁻

Correct Answer

C. K⁺, Fe³⁺, [Fe(CN)₆]⁴⁻

Detailed Solution & Explanation

Nature of Prussian Blue Complex:Soluble Prussian Blue has the molecular composition KFeIII[FeII(CN)₆].
Dissociation Behavior:When dissolved in water, outer coordination cations separate while the inner coordination complex sphere remains intact:KFe[Fe(CN)₆] (s) → K⁺ (aq) + Fe³⁺ (aq) + [Fe(CN)₆]⁴⁻ (aq).

Q43. Fe³⁺-আয়নের ভিত্তিতর ইলেকট্রন বিন্যাস কোনটি (What is the ground state electronic configuration of Fe³⁺)?

Options

A. [Ar] 4s¹ 3d⁴

B. [Ar] 4s² 3d³

C. [Ar] 4s⁰ 3d⁵

D. [Ar] 4s⁰ 3d⁶

Correct Answer

C. [Ar] 4s⁰ 3d⁵

Detailed Solution & Explanation

Neutral Iron Configuration:Fe (Z = 26): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶ = [Ar] 4s² 3d⁶.
Ionization Sequence:Electrons are always removed first from the outermost principal quantum level (4s orbital before 3d orbital):• Fe → Fe²⁺ + 2e⁻ ⇒ [Ar] 4s⁰ 3d⁶• Fe²⁺ → Fe³⁺ + 1e⁻ ⇒ [Ar] 4s⁰ 3d⁵ (a highly stable half-filled d-subshell).

Q44. ডিউটেরিয়াম (²H) কে —পদ্ধতিতে প্রোটিয়াম (¹H) থেকে পৃথক করা হয় (Deuterium (²H) was separated from the protium (¹H) by—).

Options

A. fractional distillation

B. solvent extraction

C. chromatography

D. evaporation

Correct Answer

A. fractional distillation

Detailed Solution & Explanation

Isotopic Difference:Deuterium oxide (Heavy Water, D₂O) has a boiling point of 101.4 ºC, whereas ordinary water (H₂O) boils at 100.0 ºC.
Industrial Separation:Due to this slight 1.4 ºC boiling point disparity, precision fractional distillation of liquid hydrogen or water mixtures is used to isolate deuterium.

Q45. নিচের প্রজাতিগুলোর আয়নিক ব্যাসার্ধের সঠিক ক্রম নির্বাচন কর (Choose the correct order of ionic radius for the following species).

Options

A. Cl⁻ > I⁻ > Te²⁻ > Ar

B. Te²⁻ > I⁻ > Cl⁻ > Ar

C. Te²⁻ > I⁻ > Ar > Cl⁻

D. Te²⁻ > Ar > I⁻ > Cl⁻

Correct Answer

B. Te²⁻ > I⁻ > Cl⁻ > Ar

Detailed Solution & Explanation

Principal Shell Count:• Te²⁻ and I⁻ have 5 filled electron shells (Kr core + 5s² 5p⁶ = 54 electrons).• Cl⁻ and Ar have 3 filled electron shells (Ne core + 3s² 3p⁶ = 18 electrons).Naturally, 5-shell species are vastly larger than 3-shell species.
Isoelectronic Pair Analysis:Within an isoelectronic series, radius decreases as nuclear charge (Z) increases due to stronger electrostatic pull:• For 54 e⁻ pair: Te²⁻ (Z=52) > I⁻ (Z=53).• For 18 e⁻ pair: Cl⁻ (Z=17) > Ar (Z=18).Overall Sequence: Te²⁻ > I⁻ > Cl⁻ > Ar.

Q46. নিচের কোন আকরিকটি Ca এবং Mg এর উভয়ের আকরিক রূপে বিবেচনা করা যায় (Which of the following ores can be considered as an ore of both Ca and Mg)?

Options

A. Dolomite

B. Calcite

C. Magnesite

D. Magnetite

Correct Answer

A. Dolomite

Detailed Solution & Explanation

Mineral Formulas:• Dolomite: CaMg(CO₃)₂ (Double carbonate of Calcium and Magnesium)• Calcite: CaCO₃ (Calcium carbonate)• Magnesite: MgCO₃ (Magnesium carbonate)• Magnetite: Fe₃O₄ (Iron oxide).

Q47. ক্রায়োস্কোপিক পদ্ধতিতে সুপারকন্ডাকটিভিটি গবেষণায় নিচের কোনটি ব্যবহৃত হয় (Which of the following elements is used in cryoscopic study for superconductivity)?

Options

A. Liquid mercury

B. liquid bromine

C. Liquid butane

D. liquid nitrogen

Correct Answer

D. liquid nitrogen

Detailed Solution & Explanation

High-Temperature Superconductors:High-temperature ceramic superconductors (such as YBCO) display zero electrical resistance below critical temperatures (T_c ≈ 93 K).
Cryogenic Medium:Liquid nitrogen boils at 77 K (-196 ºC), providing an accessible, economically viable cryogenic refrigerant to achieve superconductivity.

Organic Chemistry & Spectroscopy (Q48 – Q67)

Q48. IUPAC পদ্ধতিতে CH₃― CBr ═ CH ― CO ―CH₂Cl যৌগটির নাম কী (What is the IUPAC name of the compound CH₃― CBr ═ CH ― CO ―CH₂Cl)?

Options

A. 1-Chloro-4-bromo-pent-3-in-2-on

B. 4-Bromo-1-chloro-pent-3-in-2-on

C. 2-Bromo-5-chloro-pent-3-in-2-on

D. 4-Bromo-1-chloro-pent-3-in-2-al

Correct Answer

B. 4-Bromo-1-chloro-pent-3-in-2-on

Detailed Solution & Explanation

Priority Functional Group:Ketone (-CO-) takes precedence over alkene (=) and halo substituents.
Carbon Chain Numbering:Numbering starts from the right end to give the carbonyl group the lowest possible locant (C2):• C1: -CH₂Cl (Chloro attached)• C2: -CO- (Ketone)• C3: =CH- (Double bond)• C4: -CBr- (Bromo attached)• C5: -CH₃
Alphabetical Order of Substituents:4-Bromo-1-chloro-pent-3-en-2-one (spelled ‘in/en’ depending on translation).

Q49. কোনটি আলোক সক্রিয় (Which one is optically active)?

Options

A. CH₂(OH)― CH(OH) ― CH₂(OH)

B. (CO₂H)₂

C. (CH₃)₂CH(OH)

D. CH₃ ― CH(OH) ― CH₂―CH₃

Correct Answer

D. CH₃ ― CH(OH) ― CH₂―CH₃

Detailed Solution & Explanation

Optical Activity Requirement:A molecule is optically active if it lacks an internal plane of symmetry, which requires the presence of at least one chiral carbon center (sp³ carbon bonded to 4 non-identical groups).
Analysis of Butan-2-ol:In CH₃-C*H(OH)-CH₂-CH₃, the C2 carbon is bonded to:• -H• -OH• -CH₃• -CH₂CH₃Since all 4 attached groups are distinct, it exists as non-superimposable enantiomeric pairs that rotate plane-polarized light.

Q50. C₆H₅OH + B + 3NaOH ──Δ──> C₆H₄(OH)CHO + 3NaCl + 2H₂O (What is B for the above reaction)?

Options

A. H―CHO

B. CHCl₃

C. CCl₄

D. CH₃―CHO

Correct Answer

B. CHCl₃

Detailed Solution & Explanation

Reaction Name:This is the classical Reimer-Tiemann Reaction used to formylating phenols.
Mechanism:Phenol reacts with chloroform (CHCl₃) in the presence of hot aqueous sodium hydroxide to generate dichlorocarbene (:CCl₂) as an electrophilic intermediate, yielding salicylaldehyde (2-hydroxybenzaldehyde).

Q51. এসিডের শক্তির কোন ক্রমটি সঠিক (Which order of strength is correct for acids)?

Options

A. CH₃― CH₂ ― OH > (CH₃)₂CH(OH) > (CH₃)₃C(OH) > C₆H₅OH

B. (CH₃)₃C(OH) > (CH₃)₂CH(OH) > CH₃― CH₂ ― OH > C₆H₅OH

C. (CH₃)₂CH(OH) > (CH₃)₃C(OH) > C₆H₅OH > CH₃― CH₂ ― OH

D. C₆H₅OH > CH₃― CH₂ ― OH > (CH₃)₂CH(OH) > (CH₃)₃C(OH)

Correct Answer

D. C₆H₅OH > CH₃― CH₂ ― OH > (CH₃)₂CH(OH) > (CH₃)₃C(OH)

Detailed Solution & Explanation

Phenol Acidity:Phenol (C₆H₅OH, pK_a ≈ 10) is exponentially more acidic than aliphatic alcohols because the phenoxide anion is resonance-stabilized across the aromatic ring.
Aliphatic Alcohol Acidity Trend:Alkyl groups exhibit electron-donating inductive (+I) effects, which destabilize the alkoxide negative charge:Primary (1º) > Secondary (2º) > Tertiary (3º)Overall Trend: Phenol > Ethanol > Isopropanol > tert-Butanol.

Q52. কোন কার্যকরীমূলক IR বর্ণালীতে 1700-1750 cm⁻¹-এ বৈশিষ্ট্যমূলক ব্যান্ড দেখায় (Which functional group shows characteristic band at 1700 – 1750 cm⁻¹ in IR spectrum)?

Options

A. ―C≡C―

B. ―OH

C. ═C═O

D. ―NH₂

Correct Answer

C. ═C═O

Detailed Solution & Explanation

Infrared Spectroscopy Absorption Ranges:• Carbonyl (C=O): Strong, sharp stretching vibration at 1700–1750 cm⁻¹.• Alcohol (-OH): Broad stretching at 3200–3600 cm⁻¹.• Alkyne (-C≡C-): Weak stretching at 2100–2260 cm⁻¹.• Amine (-NH₂): Doublet absorption at 3300–3500 cm⁻¹.

Q53. কোনটি নিউক্লিওফিলিক সংযোজন বিক্রিয়ায় অধিক সক্রিয় (Which one is more reactive to Nucleophilic addition reaction)?

Options

A. H-CHO

B. CH₃-CHO

C. CH₃-CO-CH₃

D. CH₃-COOH

Correct Answer

A. H-CHO

Detailed Solution & Explanation

Electronic Factor:Alkyl groups donate electron density (+I effect) toward the carbonyl carbon, diminishing its partial positive charge (δ+) and reducing its electrophilicity.
Steric Factor:Bulky alkyl groups sterically hinder the nucleophile’s trajectory. Formaldehyde (HCHO) has 0 alkyl groups, making it both sterically unhindered and highly electrophilic compared to acetaldehyde or acetone.

Q54. CH₃― CHCl― CH₃ + NaOH(alc) ──Δ──> “A” + 3NaCl + H₂O (Of the above reaction compound A is – i) unsymmetrical alkene ii) alcohol iii) shows geometrical isomerism. Which one is correct?)

Options

A. i

B. ii

C. iii

D. i, ii & iii

Correct Answer

A. i

Detailed Solution & Explanation

Elimination Reaction:Dehydrohalogenation of 2-chloropropane using alcoholic NaOH proceeds via an E2 mechanism to produce propene:CH₃-CHCl-CH₃ + NaOH (alc) → CH₃-CH=CH₂ (Propene) + NaCl + H₂O
Evaluation of Statements:• i. Unsymmetrical alkene: Correct, because different groups are attached to either side of the double bond (CH₃-CH= vs =CH₂).• ii. Alcohol: Incorrect.• iii. Geometrical isomerism: Incorrect, because terminal C1 has two identical hydrogen atoms.

Q55. Benzene + CH₃Cl ──AlCl₃/Δ──> A ──KMnO₄──> B (Number of delocalised π electrons in compound B are–).

Options

A. 6

B. 8

C. 10

D. 12

Correct Answer

A. 6

Detailed Solution & Explanation

Reaction Sequence:• Step 1: Friedel-Crafts alkylation of Benzene with CH₃Cl/AlCl₃ gives Toluene (A).• Step 2: Oxidation of Toluene with alkaline KMnO₄ converts the side chain to Benzoic Acid (B, C₆H₅COOH).
Aromatic π-Electron Count:The carboxyl group (-COOH) is non-aromatic. The aromatic ring itself retains Huckel’s rule count of 6 delocalized π-electrons.

Q56. কোনটি ক্যানিজারো বিক্রিয়ায় সাড়া দেয় (Which one responses to Cannizzaro reaction)?

Options

A. CH₃― CH₂―CHO

B. CH₃―CHO

C. CH₃ ― CO ― CH₃

D. C₆H₅― CHO

Correct Answer

D. C₆H₅― CHO

Detailed Solution & Explanation

Cannizzaro Requirement:The Cannizzaro reaction is a base-induced disproportionation limited strictly to aldehydes that lack α-hydrogen atoms.
Analysis:• CH₃CHO and CH₃CH₂CHO possess α-hydrogens and undergo Aldol condensation instead.• Benzaldehyde (C₆H₅CHO) has no α-hydrogen, undergoing self-redox to form benzyl alcohol and benzoate ion.

Q57. C ― C বন্ধন দৈর্ঘ্য অনুসারে কোন ক্রমটি সঠিক (Which order is correct according to C ― C bond length)?

Options

A. C₂H₂ > C₂H₄ > C₂H₆ > C₆H₆

B. C₂H₆ > C₆H₆ > C₂H₄ > C₂H₂

C. C₆H₆ > C₂H₆ > C₂H₄ > C₂H₂

D. C₂H₂ > C₂H₄ > C₆H₆ > C₂H₆

Correct Answer

B. C₂H₆ > C₆H₆ > C₂H₄ > C₂H₂

Detailed Solution & Explanation

Bond Multiplicity vs Length:Higher bond order increases electron density between nuclei, pulling them closer together:• Single Bond (Ethane, C₂H₆): 1.54 Å• Aromatic Resonance Bond (Benzene, C₆H₆): 1.39 Å• Double Bond (Ethene, C₂H₄): 1.34 Å• Triple Bond (Ethyne, C₂H₂): 1.20 Å.

Q58. গ্রিগনার্ড বিকারক হতে 3⁰ অ্যালকোহল তৈরিতে কোনটি ব্যবহৃত হয় (Which one is used to prepare 3⁰ alcohol from Grignard reagent)?

Options

A. H― CHO

B. CH₃ ― CHO

C. CH₃ ― CO ― CH₃

D. CH₃― CH₂ ―CHO

Correct Answer

C. CH₃ ― CO ― CH₃

Detailed Solution & Explanation

Grignard Reaction Trends:• Formaldehyde (HCHO) + RMgX → Primary (1º) Alcohol.• Other Aldehydes (RCHO) + RMgX → Secondary (2º) Alcohol.• Ketones (R-CO-R’, e.g., Acetone) + RMgX → Tertiary (3º) Alcohol upon acid hydrolysis.

Q59. কোনটিতে α―গ্লাইকোসাইড বন্ধন আছে (Which one contains α ― glycoside bond)?

Options

A. Cellulose

B. Fructose

C. Starch

D. Glucose

Correct Answer

C. Starch

Detailed Solution & Explanation

Polymer Linkages:• Starch consists of α-D-glucose monomers linked via α(1→4) and α(1→6) glycosidic bonds.• Cellulose consists of β-D-glucose monomers linked via β(1→4) glycosidic bonds.

Q60. Catechol (1,2-dihydroxybenzene) + HI ──> ? (What is the major product of the following reaction)?

Options

A. 1,2-diiodobenzene

B. 2-iodophenol

C. 2-hydroxybenzyl alcohol

D. Glucose

Correct Answer

B. 2-iodophenol

Detailed Solution & Explanation

Resistance of Phenolic Hydroxyls:Ar-OH bonds display partial double bond character due to p-π resonance with the aromatic ring, resisting substitution.
Severe Cleavage:Under forced reaction conditions with concentrated HI, mono-substitution occurs yielding 2-iodophenol.

Q61. (CH₃)₃C―Cl + I⁻ ──> (CH₃)₃C―I + Cl⁻ (Consider the Sₙ1 reaction of tert-butyl chloride with iodide ion, if the concentration of iodide ion is doubled, the rate of forming tert-butyl iodide will be – .)

Options

A. double

B. increase 4 times

C. remain the same

D. decrease

Correct Answer

C. remain the same

Detailed Solution & Explanation

Sₙ1 Kinetics:An Sₙ1 reaction proceeds via a two-step mechanism where the rate-determining step (RDS) is unimolecular heterolysis to form a carbocation:Rate = k [(CH₃)₃C-Cl]
Nucleophile Independence:The nucleophile (I⁻) only participates in the fast second step. Thus, doubling [I⁻] has zero effect on the reaction rate.

Q62. কোন উক্তিটি সঠিক (Which statement is correct)?

Options

A. trans-2-butene have higher boiling point with more polarity

B. trans-2-butene have lower boiling point with more polarity

C. cis-2-butene is more polar and have higher boiling point

D. cis-2-butene is more polar with lower boiling point

Correct Answer

C. cis-2-butene is more polar and have higher boiling point

Detailed Solution & Explanation

Polarity:In cis-2-butene, C-CH₃ bond dipole moments reinforce one another, producing a net dipole moment (μ > 0). In trans-2-butene, dipoles cancel out (μ = 0).
Boiling Point:Stronger dipole-dipole intermolecular attractions in polar cis-2-butene require more energy to vaporize, giving it a higher boiling point than non-polar trans-2-butene.

Q63. কার্বক্সিলিক এসিড সম্পর্কে কোন উক্তিটি সঠিক নয় (Which of the following statements is not true for the carboxylic acids)?

Options

A. Carboxylic acid with five or fewer carbon atoms is soluble in water completely

B. The carbon atom of formic acid is sp² hybridized

C. The carbon-oxygen bonds of the formate ion (HCOO⁻) are of the same length

D. Carboxylic acids are almost completed dissociated in water

Correct Answer

D. Carboxylic acids are almost completed dissociated in water

Detailed Solution & Explanation

Weak Acid Character:Carboxylic acids are weak organic acids that only partially ionize in water (typically < 1% dissociation, K_a ≈ 10⁻⁵). Claiming complete dissociation is factually incorrect.

Q64. Cyclohex-2-en-1-one ──[?]──> Cyclohex-2-en-1-ol (Choose the right reagent(s) for the following reaction).

Options

A. H₂/Ni

B. Hg(OAc)₂, H₂O

C. NaBH₄, THF

D. BH₃, THF and H₂O₂, NaOH

Correct Answer

C. NaBH₄, THF

Detailed Solution & Explanation

Chemoselectivity:Sodium borohydride (NaBH₄) in THF is a mild hydride donor that selectively reduces carbonyl groups (C=O) to allylic alcohols without reducing conjugated C=C double bonds.

Q65. সাইক্লোপেন্টিনের ব্রোমিনেশনে প্রত্যাশিত উৎপাদ কী (For the bromination of cyclopentene what will be the expected product/products)?

Options

A. trans-1,2-dibromocyclopentane

B. cis-1,2-dibromocyclopentane

C. 50% Cis-1,2-dibromocylcopentane and 50% trans-1,2-dibromocyclopentane

D. None

Correct Answer

A. trans-1,2-dibromocyclopentane

Detailed Solution & Explanation

Stereochemistry of Electrophilic Halogenation:Bromination proceeds through a cyclic bromonium ion intermediate. The nucleophilic bromide ion (Br⁻) attacks from the opposite face (anti-addition), forming exclusively the trans-1,2-dibromocyclopentane stereoisomer.

Q66. 2,5-dimethyltetrahydrofuran (How many stereogenic centers are there for the following compound)?

Options

A. 1

B. 3

C. 2

D. 4

Correct Answer

C. 2

Detailed Solution & Explanation

Chiral Center Identification:In 2,5-dimethyltetrahydrofuran, Carbon-2 and Carbon-5 each bear 4 distinct substituents:• -H• -CH₃• Ring Oxygen (-O-)• Ring Carbon (-CH₂-)Thus, there are 2 stereogenic centers.

Q67. CH₃― O― CH₂ ― CH₂ ― CH₃ ও CH₃― CH₂― O ― CH₂― CH₃ পরস্পর (CH₃―O―CH₂―CH₂―CH₃ and CH₃―CH₂―O―CH₂―CH₃ are ……… to each other)

Options

A. metamer

B. functional group isomer

C. tautomer

D. chain isomer

Correct Answer

A. metamer

Detailed Solution & Explanation

Metamerism Definition:Metamerism is a form of structural isomerism where compounds belonging to the same functional class (ethers, -O-) possess different alkyl groups attached to the central divalent heteroatom:• Compound 1: Methyl propyl ether (methyl + propyl)• Compound 2: Diethyl ether (ethyl + ethyl).