The energy associated with the first orbit of He+ is:
En = -2.18 ร 10^-18 Z^2/n^2.
Use the formula En = -2.18 ร 10^-18 Z^2/n^2, where Z = 2 for He+ and n = 1.
Substituting values: En = -2.18 ร 10^-18 ร 4 = -8.72 ร 10^-18.
Final Answer: Energy = -8.72 ร 10^-18 J; Option B is correct.
Misplacing Z or n values in the formula for hydrogen-like species.
Apply the energy equation for hydrogen-like species.
En = -2.18 ร 10^-18 Z^2/n^2.
Use the formula En = -2.18 ร 10^-18 Z^2/n^2, where Z = 2 for He+ and n = 1.
Substituting values: En = -2.18 ร 10^-18 ร 4 = -8.72 ร 10^-18.
Final Answer: Energy = -8.72 ร 10^-18 J; Option B is correct.
Misplacing Z or n values in the formula for hydrogen-like species.
For which one of the following mixtures is composition uniform throughout?
No formula involved.
Homogeneous mixtures have uniform composition throughout.
Aqueous sugar solution dissolves uniformly without phase separation.
Final Answer: Option D; sugar solution is homogeneous.
Assuming phase-separated mixtures like oil and water to be uniform.
Differentiate between homogeneous and heterogeneous mixtures.
No formula involved.
Homogeneous mixtures have uniform composition throughout.
Aqueous sugar solution dissolves uniformly without phase separation.
Final Answer: Option D; sugar solution is homogeneous.
Assuming phase-separated mixtures like oil and water to be uniform.
NOโ gas is:
No formula involved.
NOโ is a reddish-brown gas that is acidic in nature.
The brown color arises from dimerization of NOโ at lower temperatures, and its acidic property comes from reaction with water.
Final Answer: Brown and acidic; Option C is correct.
Confusing NOโ with other nitrogen oxides (e.g., NO, NโO).
Recall the properties of nitrogen oxides.
No formula involved.
NOโ is a reddish-brown gas that is acidic in nature.
The brown color arises from dimerization of NOโ at lower temperatures, and its acidic property comes from reaction with water.
Final Answer: Brown and acidic; Option C is correct.
Confusing NOโ with other nitrogen oxides (e.g., NO, NโO).
Identify the wrong relation for real gases:
Van der Waals equation corrections.
Compressibility factor Z = pV/nRT, not directly Vreal/Videal.
Real gas behavior is modeled using Van der Waals corrections, including p and V adjustments.
Final Answer: Option A is incorrect as Z involves pressure, volume, and temperature.
Confusing Z with a simplistic ratio of real and ideal volumes.
Recall the real gas compressibility factor Z.
Van der Waals equation corrections.
Compressibility factor Z = pV/nRT, not directly Vreal/Videal.
Real gas behavior is modeled using Van der Waals corrections, including p and V adjustments.
Final Answer: Option A is incorrect as Z involves pressure, volume, and temperature.
Confusing Z with a simplistic ratio of real and ideal volumes.
From the diagram ฮHโ = +10J A โ 2B, ฮHโ = +25J B โ C, ฮH for the reaction CโA is:
ฮH(reverse) = -ฮH(forward)
Use Hess's Law for cyclic process
Calculate ฮH for AโC: ฮHโ + ฮHโ = +35J
Reverse reaction for CโA: -(ฮHโ + ฮHโ) = -35J
Wrong sign conversion
Consider sign convention in thermochemistry
ฮH(reverse) = -ฮH(forward)
Use Hess's Law for cyclic process
Calculate ฮH for AโC: ฮHโ + ฮHโ = +35J
Reverse reaction for CโA: -(ฮHโ + ฮHโ) = -35J
Wrong sign conversion
A pair of isoelectronic species with bond order of 1 is:
Bond order = 1/2 (Bonding electrons - Antibonding electrons).
Isoelectronic species O2- and F2 both have 18 electrons and bond order of 1.
Compare electronic configuration and bond order of the given species.
Final Answer: O2- and F2 have bond order 1; Option C is correct.
Overlooking isoelectronic nature or miscalculating bond order for given species.
Use bond order formulas and molecular configurations.
Bond order = 1/2 (Bonding electrons - Antibonding electrons).
Isoelectronic species O2- and F2 both have 18 electrons and bond order of 1.
Compare electronic configuration and bond order of the given species.
Final Answer: O2- and F2 have bond order 1; Option C is correct.
Overlooking isoelectronic nature or miscalculating bond order for given species.
A metalloid is:
None required
Identify elements lying on the border between metals and non-metals
Check electronic configuration and properties
Verify metalloid characteristics: semiconducting properties, intermediate electronegativity
Confusing with transition metals
Consider periodic table position
None required
Identify elements lying on the border between metals and non-metals
Check electronic configuration and properties
Verify metalloid characteristics: semiconducting properties, intermediate electronegativity
Confusing with transition metals
Which statement about lanthanoids is incorrect?
Color arises from f-f transitions.
Lanthanoid colors arise from f-f transitions, not d-d transitions.
Recognize f-orbitals as the source of color for lanthanoid ions in solution.
Final Answer: Option D is incorrect; f-f transitions cause colors in lanthanoids.
Assuming d-d transitions occur in lanthanoids instead of f-f.
Understand the electronic transitions in lanthanoids.
Color arises from f-f transitions.
Lanthanoid colors arise from f-f transitions, not d-d transitions.
Recognize f-orbitals as the source of color for lanthanoid ions in solution.
Final Answer: Option D is incorrect; f-f transitions cause colors in lanthanoids.
Assuming d-d transitions occur in lanthanoids instead of f-f.
Match the following: I. Znยฒโบ II. Cuยฒโบ III. Niยฒโบ with i. dโธ configuration ii. Colourless iii. ฮผ=1.73 BM
ฮผ = โn(n+2) BM
Write electronic configurations: Znยฒโบ(3dยนโฐ), Cuยฒโบ(3dโน), Niยฒโบ(3dโธ)
Calculate magnetic moments
Match properties with ions
Wrong electron counting
Consider d-orbital filling
ฮผ = โn(n+2) BM
Write electronic configurations: Znยฒโบ(3dยนโฐ), Cuยฒโบ(3dโน), Niยฒโบ(3dโธ)
Calculate magnetic moments
Match properties with ions
Wrong electron counting
The transition element (~5%) present with lanthanoid metal in Misch metal is:
No formula involved.
Misch metal contains ~95% lanthanide metals and ~5% iron.
Identify the additional transition element (iron, Fe) in Misch metal composition.
Final Answer: Iron (Fe) is the transition element in Misch metal.
Misinterpreting "transition element" as any metal instead of Fe.
Recall the composition of Misch metal.
No formula involved.
Misch metal contains ~95% lanthanide metals and ~5% iron.
Identify the additional transition element (iron, Fe) in Misch metal composition.
Final Answer: Iron (Fe) is the transition element in Misch metal.
Misinterpreting "transition element" as any metal instead of Fe.
The synthetically produced radioactive noble gas by collision of 249Cf98 with 48Ca20 is:
Reaction formula: 249Cf + 48Ca.
Use the fusion reaction: 249Cf + 48Ca โ 294Og + 3n.
Recognize that Oganesson (Og, Z=118) is the product of this synthetic reaction.
Final Answer: Oganesson (Og) is synthetically produced in this reaction.
Confusing other noble gases with the synthesized element (Og).
Relate atomic structure and fusion reactions.
Reaction formula: 249Cf + 48Ca.
Use the fusion reaction: 249Cf + 48Ca โ 294Og + 3n.
Recognize that Oganesson (Og, Z=118) is the product of this synthetic reaction.
Final Answer: Oganesson (Og) is synthetically produced in this reaction.
Confusing other noble gases with the synthesized element (Og).
Identify the incorrect statement from the following:
No formula involved.
Ozone absorbs ultraviolet (UV) radiation, not infrared (IR) radiation.
IR absorption is more characteristic of greenhouse gases like COโ.
Final Answer: Option D is incorrect because ozone does not absorb IR radiation.
Assuming ozone absorbs IR radiation like greenhouse gases.
Differentiate between ozone's UV protection and greenhouse effects.
No formula involved.
Ozone absorbs ultraviolet (UV) radiation, not infrared (IR) radiation.
IR absorption is more characteristic of greenhouse gases like COโ.
Final Answer: Option D is incorrect because ozone does not absorb IR radiation.
Assuming ozone absorbs IR radiation like greenhouse gases.
The correct decreasing order of boiling point of hydrogen halides is:
No formula involved.
Analyze boiling points based on intermolecular forces (hydrogen bonding, van der Waals forces).
HF has the highest boiling point due to strong hydrogen bonding, followed by halides with increasing molar masses.
Final Answer: HF > HI > HBr > HCl; Option C is correct.
Overlooking the significant impact of hydrogen bonding in HF.
Focus on the role of hydrogen bonding and molecular mass in boiling points.
No formula involved.
Analyze boiling points based on intermolecular forces (hydrogen bonding, van der Waals forces).
HF has the highest boiling point due to strong hydrogen bonding, followed by halides with increasing molar masses.
Final Answer: HF > HI > HBr > HCl; Option C is correct.
Overlooking the significant impact of hydrogen bonding in HF.
The incorrect statement about Hall-Heroult process is:
No formula involved.
Analyze the role of oxygen and carbon anode in the process.
Oxygen reacts at the carbon anode, but its oxidation state remains unchanged.
Final Answer: Oxygen's oxidation state does not change; Option D is incorrect.
Assuming oxygen changes oxidation states during the process.
Understand the reactions occurring at the anode and cathode.
No formula involved.
Analyze the role of oxygen and carbon anode in the process.
Oxygen reacts at the carbon anode, but its oxidation state remains unchanged.
Final Answer: Oxygen's oxidation state does not change; Option D is incorrect.
Assuming oxygen changes oxidation states during the process.
Select the correct statement:
No formula involved.
Recognize the definition and purpose of calcination, roasting, and smelting.
Calcination of CaCOโ releases COโ, requiring heat (endothermic).
Final Answer: Calcination of CaCOโ is endothermic; Option D is correct.
Misunderstanding the conditions of calcination and roasting.
Focus on the nature of heat transfer during calcination.
No formula involved.
Recognize the definition and purpose of calcination, roasting, and smelting.
Calcination of CaCOโ releases COโ, requiring heat (endothermic).
Final Answer: Calcination of CaCOโ is endothermic; Option D is correct.
Misunderstanding the conditions of calcination and roasting.
Gold sol is not a :
No formula involved.
The oxidation state of oxygen does not change in the Hall-Heroult process.
Carbon serves as the anode and reacts with oxygen to produce CO/COโ.
Final Answer: Option D is incorrect.
Assuming oxygen changes oxidation state during electrolysis.
Focus on the role of oxygen in the process.
No formula involved.
The oxidation state of oxygen does not change in the Hall-Heroult process.
Carbon serves as the anode and reacts with oxygen to produce CO/COโ.
Final Answer: Option D is incorrect.
Assuming oxygen changes oxidation state during electrolysis.
Which one of the following does not represent Arrhenius equation?
k = Ae^(-Ea/RT) and related variations.
Recognize the valid forms of Arrhenius equation.
Compare all forms of the equation. The correct forms involve negative exponential or logarithmic terms.
Final Answer: Option D does not match the equation.
Misinterpreting positive exponents as valid forms.
Recall valid mathematical forms of Arrhenius equation.
k = Ae^(-Ea/RT) and related variations.
Recognize the valid forms of Arrhenius equation.
Compare all forms of the equation. The correct forms involve negative exponential or logarithmic terms.
Final Answer: Option D does not match the equation.
Misinterpreting positive exponents as valid forms.
For the coagulation of positively charged hydrated ferric oxide sol, flocculating power of the ions is in order is:
Flocculation โ ion charge.
Flocculating power is proportional to the charge of ions (Schulze-Hardy rule).
High-charge ions (e.g., [Fe(CN)โ]ยณโป) exhibit greater coagulating power than low-charge ions.
Final Answer: Option B is correct.
Overlooking the impact of ion charge on coagulation.
Recall the relationship between charge and coagulation.
Flocculation โ ion charge.
Flocculating power is proportional to the charge of ions (Schulze-Hardy rule).
High-charge ions (e.g., [Fe(CN)โ]ยณโป) exhibit greater coagulating power than low-charge ions.
Final Answer: Option B is correct.
Overlooking the impact of ion charge on coagulation.
Identify the incorrect statement regarding colloidal solutions:
No formula involved.
Brownian movement is caused by unbalanced molecular bombardment.
Correct the misunderstanding about balanced versus unbalanced forces in colloids.
Final Answer: Brownian movement is unbalanced, not balanced.
Confusing the reason for Brownian motion (balanced vs. unbalanced forces).
Review molecular behavior in colloids.
No formula involved.
Brownian movement is caused by unbalanced molecular bombardment.
Correct the misunderstanding about balanced versus unbalanced forces in colloids.
Final Answer: Brownian movement is unbalanced, not balanced.
Confusing the reason for Brownian motion (balanced vs. unbalanced forces).
For the reaction A โ B, Ea = 50 kJ molโปยน and ฮH = -20 kJ molโปยน. After addition of catalyst, decreases Ea by 10 kJ molโปยน, find Ea (backward)
ฮH = Ea(f) - Ea(b).
Use the relation ฮH = Ea(f) - Ea(b).
Ea(forward) = 50 - 10 = 40 kJ/mol. Hence Ea(backward) = ฮH + Ea(f) = -20 + 40 = 60 kJ/mol.
Final Answer: Ea(backward) = 60 kJ/mol.
Incorrectly assuming ฮH = Ea without considering catalyst effect.
Understand catalyst effects on Ea.
ฮH = Ea(f) - Ea(b).
Use the relation ฮH = Ea(f) - Ea(b).
Ea(forward) = 50 - 10 = 40 kJ/mol. Hence Ea(backward) = ฮH + Ea(f) = -20 + 40 = 60 kJ/mol.
Final Answer: Ea(backward) = 60 kJ/mol.
Incorrectly assuming ฮH = Ea without considering catalyst effect.
For the reaction PClโ โ PClโ + Clโ, rate and rate constant are 1.02 ร 10โปโด mol/L/s and 3.4 ร 10โปโต sโปยน respectively. What is [PClโ ]?
Rate = k[Reactant].
Use the formula Rate = k[PClโ ].
Rearrange to find [PClโ ]: [PClโ ] = Rate / k = (1.02 ร 10โปโด) / (3.4 ร 10โปโต).
Final Answer: [PClโ ] = 3.0 mol/L.
Forgetting to divide rate by rate constant or misplacing units.
Relate rate and rate constant to molarity.
Rate = k[Reactant].
Use the formula Rate = k[PClโ ].
Rearrange to find [PClโ ]: [PClโ ] = Rate / k = (1.02 ร 10โปโด) / (3.4 ร 10โปโต).
Final Answer: [PClโ ] = 3.0 mol/L.
Forgetting to divide rate by rate constant or misplacing units.
The value of 'A' in the equation ฮป_m = ฮป_0_m - A โC is same for the pair...
ฮป_m = ฮป_0 - A โC (empirical relation).
A depends on ion type and interactions in solution.
NaCl and KBr have similar ionic types and interaction effects.
Result: Value of A is identical for NaCl and KBr.
Assuming all salts affect A equally regardless of charge or type.
Relate ionic properties to conductivity coefficients.
ฮป_m = ฮป_0 - A โC (empirical relation).
A depends on ion type and interactions in solution.
NaCl and KBr have similar ionic types and interaction effects.
Result: Value of A is identical for NaCl and KBr.
Assuming all salts affect A equally regardless of charge or type.
A current of 3A is passed through molten calcium salt for 1 hr 47 min 13 sec. The mass of calcium deposited is...
Mass = (E ร I ร t) / 96500.
Use Faradayโs law: Mass deposited = (E ร I ร t) / 96500.
Substituting values: E = 40, I = 3A, t = 6432 s. Result: (40 ร 3 ร 6432) / 96500 = 3.99 g โ 4.0 g.
Final Mass: 4.0 g.
Errors in time conversion or forgetting valency of calcium during calculations.
Apply Faraday's Law to time-dependent mass calculations.
Mass = (E ร I ร t) / 96500.
Use Faradayโs law: Mass deposited = (E ร I ร t) / 96500.
Substituting values: E = 40, I = 3A, t = 6432 s. Result: (40 ร 3 ร 6432) / 96500 = 3.99 g โ 4.0 g.
Final Mass: 4.0 g.
Errors in time conversion or forgetting valency of calcium during calculations.
How many Coulombs are required to oxidise 0.1 mole of H2O to oxygen?
q = n ร F (Faraday constant).
Oxidation of water to oxygen involves 4 electrons per molecule.
For 0.1 mole of H2O: (0.1 ร 2 ร 96500 = 1.93 ร 10^4 C).
Final Answer: 1.93 ร 10^4 C.
Mixing up electron moles or Faraday's constant during calculations.
Use Faraday's Laws of Electrolysis.
q = n ร F (Faraday constant).
Oxidation of water to oxygen involves 4 electrons per molecule.
For 0.1 mole of H2O: (0.1 ร 2 ร 96500 = 1.93 ร 10^4 C).
Final Answer: 1.93 ร 10^4 C.
Mixing up electron moles or Faraday's constant during calculations.
The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is...
d = (Z ร M) / (Na ร aยณ).
Use the formula d = (Z ร M) / (Na ร aยณ) to find the density of the unit cell.
Substituting values: d = 10, a = 300 pm, Na = 6.022 ร 10^23, Z = 4, M = 40.5 g/mol.
Result: Number of atoms = 6.6 ร 10^22.
Miscalculating Z or misinterpreting the given edge length (pm to cm).
Focus on unit cell dimensions and atomic packing.
d = (Z ร M) / (Na ร aยณ).
Use the formula d = (Z ร M) / (Na ร aยณ) to find the density of the unit cell.
Substituting values: d = 10, a = 300 pm, Na = 6.022 ร 10^23, Z = 4, M = 40.5 g/mol.
Result: Number of atoms = 6.6 ร 10^22.
Miscalculating Z or misinterpreting the given edge length (pm to cm).
Which of the following pairs will show positive deviation from Raoult's Law?
None (Conceptual).
Positive deviation occurs when intermolecular forces are weaker in the solution than in pure components.
Benzene and methanol have weak interactions compared to their pure molecular interactions.
Final Answer: Benzene-methanol shows positive deviation.
Confusing positive deviation as stronger interactions between solute and solvent.
Recall weaker molecular interactions.
None (Conceptual).
Positive deviation occurs when intermolecular forces are weaker in the solution than in pure components.
Benzene and methanol have weak interactions compared to their pure molecular interactions.
Final Answer: Benzene-methanol shows positive deviation.
Confusing positive deviation as stronger interactions between solute and solvent.
A mixture of phenol and aniline shows negative deviation from Raoult's law. This is because of:
No formula involved.
The negative deviation is due to strong intermolecular hydrogen bonding between phenol and aniline.
Phenol and aniline form stronger intermolecular bonds than with their individual molecules.
Result: Negative deviation due to intermolecular hydrogen bonding.
Assuming polar or non-polar covalent bonds cause deviation.
Recall the nature of molecular interactions.
No formula involved.
The negative deviation is due to strong intermolecular hydrogen bonding between phenol and aniline.
Phenol and aniline form stronger intermolecular bonds than with their individual molecules.
Result: Negative deviation due to intermolecular hydrogen bonding.
Assuming polar or non-polar covalent bonds cause deviation.
Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at 100ยฐC is...
P_solution = P_water ร (1 - ฯ_solute).
Use Raoultโs law: P_solution = P_water ร (1 - ฯ_solute).
Calculate ฯ_solute = moles of solute / (moles of solute + moles of solvent).
Result: P_solution = 760 - 7.6 = 752.4 torr.
Incorrect mole calculation for glucose or water.
Understand Raoult's law application.
P_solution = P_water ร (1 - ฯ_solute).
Use Raoultโs law: P_solution = P_water ร (1 - ฯ_solute).
Calculate ฯ_solute = moles of solute / (moles of solute + moles of solvent).
Result: P_solution = 760 - 7.6 = 752.4 torr.
Incorrect mole calculation for glucose or water.
MnO exhibits:
No formula involved.
MnO exhibits antiferromagnetism due to its arrangement of magnetic moments in opposite directions.
Magnetic moments in MnO cancel out, resulting in no net magnetic moment.
Final conclusion: MnO is antiferromagnetic.
Assuming MnO has paramagnetic properties like Mn ions.
Recall magnetic properties and arrangements of magnetic moments.
No formula involved.
MnO exhibits antiferromagnetism due to its arrangement of magnetic moments in opposite directions.
Magnetic moments in MnO cancel out, resulting in no net magnetic moment.
Final conclusion: MnO is antiferromagnetic.
Assuming MnO has paramagnetic properties like Mn ions.
Which of the following crystals has the unit cell such that a = b โ c and ฮฑ = ฮฒ = 90ยฐ, ฮณ = 120ยฐ?
No formula involved.
The unit cell described corresponds to a hexagonal lattice structure.
Graphite has a layered hexagonal structure, meeting the given conditions.
Graphite satisfies a = b โ c and ฮฑ = ฮฒ = 90ยฐ, ฮณ = 120ยฐ.
Confusing between unit cell parameters for hexagonal and tetragonal structures.
Recall the geometry of different crystal lattices.
No formula involved.
The unit cell described corresponds to a hexagonal lattice structure.
Graphite has a layered hexagonal structure, meeting the given conditions.
Graphite satisfies a = b โ c and ฮฑ = ฮฒ = 90ยฐ, ฮณ = 120ยฐ.
Confusing between unit cell parameters for hexagonal and tetragonal structures.
First chlorinated organic insecticide prepared is...
DDT = Dichlorodiphenyltrichloroethane.
DDT (Dichlorodiphenyltrichloroethane) was the first chlorinated organic insecticide prepared.
Recognize historical usage of DDT as an insecticide.
Final answer: DDT.
Confusing DDT with other chlorinated compounds.
Recall DDT synthesis and significance.
DDT = Dichlorodiphenyltrichloroethane.
DDT (Dichlorodiphenyltrichloroethane) was the first chlorinated organic insecticide prepared.
Recognize historical usage of DDT as an insecticide.
Final answer: DDT.
Confusing DDT with other chlorinated compounds.
Sequence of reactions yielding Br-substituted products is...
Addition-elimination pathways for Br substitution.
Recognize reaction steps: addition, substitution, and elimination pathways involving Brโ and NaNHโ.
Follow sequential mechanism: CHโ=CHโ โ Br-substituted products.
Final products involve multiple bromination and alkane formation.
Misunderstanding intermediate steps in bromination sequence.
Carefully track intermediates and reagents.
Addition-elimination pathways for Br substitution.
Recognize reaction steps: addition, substitution, and elimination pathways involving Brโ and NaNHโ.
Follow sequential mechanism: CHโ=CHโ โ Br-substituted products.
Final products involve multiple bromination and alkane formation.
Misunderstanding intermediate steps in bromination sequence.
0.48g of an organic compound on complete combustion produced 0.22g of COโ .The percentage of C in the given organic compound is;
%C = (mass of C in COโ / mass of compound) ร 100.
Calculate %C using the relation: (mass of C in COโ / mass of compound) ร 100.
Use data: (12/44) ร (0.22/0.48) ร 100 = 12.5%.
Carbon percentage = 12.5%.
Miscalculating %C using incorrect ratios.
Carefully compute based on given mass ratios.
%C = (mass of C in COโ / mass of compound) ร 100.
Calculate %C using the relation: (mass of C in COโ / mass of compound) ร 100.
Use data: (12/44) ร (0.22/0.48) ร 100 = 12.5%.
Carbon percentage = 12.5%.
Miscalculating %C using incorrect ratios.
Halide that cannot be hydrolyzed is...
Hydrolysis needs d-orbitals, absent in carbon.
CClโ cannot be hydrolyzed due to the absence of d-orbitals in carbon.
Recognize that hydrolysis requires availability of d-orbitals.
Final result: CClโ cannot hydrolyze.
Assuming all halides undergo hydrolysis.
Understand the role of d-orbitals in hydrolysis reactions.
Hydrolysis needs d-orbitals, absent in carbon.
CClโ cannot be hydrolyzed due to the absence of d-orbitals in carbon.
Recognize that hydrolysis requires availability of d-orbitals.
Final result: CClโ cannot hydrolyze.
Assuming all halides undergo hydrolysis.
Which one of the following Nitrate will decompose to give NOโ on heating...
LiNOโ decomposition: 4LiNOโ โ 2LiโO + 4NOโ + Oโ.
LiNOโ decomposes to NOโ due to its unique thermal decomposition properties.
Recognize that LiNOโ decomposition differs from other nitrates.
Final product: 4LiNOโ โ 2LiโO + 4NOโ + Oโ.
Confusing LiNOโ decomposition with other alkali nitrates.
Focus on differences in thermal decomposition of nitrates.
LiNOโ decomposition: 4LiNOโ โ 2LiโO + 4NOโ + Oโ.
LiNOโ decomposes to NOโ due to its unique thermal decomposition properties.
Recognize that LiNOโ decomposition differs from other nitrates.
Final product: 4LiNOโ โ 2LiโO + 4NOโ + Oโ.
Confusing LiNOโ decomposition with other alkali nitrates.
Property not applicable to ionic hydrides is...
Ionic hydrides: Non-volatile, crystalline, non-conducting.
Ionic hydrides are generally non-volatile, crystalline, and non-conducting in solid state.
Focus on physical properties of ionic hydrides.
Volatility is not applicable to ionic hydrides.
Mistaking ionic hydrides for covalent hydrides.
Recall high melting points and crystalline nature of ionic hydrides.
Ionic hydrides: Non-volatile, crystalline, non-conducting.
Ionic hydrides are generally non-volatile, crystalline, and non-conducting in solid state.
Focus on physical properties of ionic hydrides.
Volatility is not applicable to ionic hydrides.
Mistaking ionic hydrides for covalent hydrides.
Solubility product of CaCโOโ at a given temperature in pure water is 4 ร 10โปโน(moL-1)2. Its solubility at same temperature is ...
Ksp = Sยฒ โ Solve for S.
Solubility (S) is calculated from Ksp = Sยฒ for salts dissociating into 1:1 cations and anions.
Ksp = Sยฒ โ Solve for S: โ(4 ร 10โปโน) = 6.3 ร 10โปโต.
Solubility = 6.3 ร 10โปโต mol/L.
Miscalculating solubility for salts dissociating into 1:1 ions.
Use Ksp formula correctly for solubility.
Ksp = Sยฒ โ Solve for S.
Solubility (S) is calculated from Ksp = Sยฒ for salts dissociating into 1:1 cations and anions.
Ksp = Sยฒ โ Solve for S: โ(4 ร 10โปโน) = 6.3 ร 10โปโต.
Solubility = 6.3 ร 10โปโต mol/L.
Miscalculating solubility for salts dissociating into 1:1 ions.
Reaction between moist SOโ and acidified permanganate...
Oxidation: SOโ โ SOโยฒโป; Reduction: MnOโโป โ Mnยฒโบ.
Reaction involves oxidation of SOโ to SOโยฒโป and reduction of MnOโโป to Mnยฒโบ in acidic medium.
Balance the reaction in acidic medium: MnOโโป + SOโ โ Mnยฒโบ + SOโยฒโป.
Final result: SOโ โ SOโยฒโป; MnOโโป โ Mnยฒโบ.
Misidentifying oxidized and reduced species in redox reaction.
Focus on oxidation states and redox balancing.
Oxidation: SOโ โ SOโยฒโป; Reduction: MnOโโป โ Mnยฒโบ.
Reaction involves oxidation of SOโ to SOโยฒโป and reduction of MnOโโป to Mnยฒโบ in acidic medium.
Balance the reaction in acidic medium: MnOโโป + SOโ โ Mnยฒโบ + SOโยฒโป.
Final result: SOโ โ SOโยฒโป; MnOโโป โ Mnยฒโบ.
Misidentifying oxidized and reduced species in redox reaction.
A haloalkane undergoes SN1 or SN2 reaction depending on:
Polar protic solvents stabilize SN1 carbocations; aprotic favor SN2 nucleophiles.
Reaction mechanism depends on the polarity and protic/aprotic nature of the solvent.
Classify solvent as polar protic (favoring SN1) or polar aprotic (favoring SN2).
Correct: SN1 or SN2 determined by solvent type.
Overlooking solvent role in mechanism choice.
Consider solvent polarity and its effect on carbocation stability.
Polar protic solvents stabilize SN1 carbocations; aprotic favor SN2 nucleophiles.
Reaction mechanism depends on the polarity and protic/aprotic nature of the solvent.
Classify solvent as polar protic (favoring SN1) or polar aprotic (favoring SN2).
Correct: SN1 or SN2 determined by solvent type.
Overlooking solvent role in mechanism choice.
2-Methylpropane can be prepared by the Wurtz reaction. The haloalkanes taken along metallic sodium and dry ether are...
RCl + Na โ Hydrocarbon (alkane) formation.
Wurtz reaction involves metallic sodium in dry ether reacting with haloalkanes to produce alkanes.
Identify 2-methylpropane as the desired product formed from chloromethane and 2-chloropropane.
Final product is 2-methylpropane.
Misinterpreting haloalkane pairing for Wurtz reaction.
Recall Wurtz reaction and pairing mechanism.
RCl + Na โ Hydrocarbon (alkane) formation.
Wurtz reaction involves metallic sodium in dry ether reacting with haloalkanes to produce alkanes.
Identify 2-methylpropane as the desired product formed from chloromethane and 2-chloropropane.
Final product is 2-methylpropane.
Misinterpreting haloalkane pairing for Wurtz reaction.
Purpose of adding NHโCl to NHโOH in the analysis of Group III basic radicals...
NHโCl addition suppresses NHโOH dissociation.
NHโCl suppresses NHโOH dissociation due to the common ion effect.
Recognize the role of NHโCl in suppressing OHโป ion increase, facilitating selective precipitation.
NHโCl suppresses NHโOH dissociation due to the common ion effect.
Confusing role of NHโCl in Group III analysis.
Focus on the effect of common ions in suppressing dissociation.
NHโCl addition suppresses NHโOH dissociation.
NHโCl suppresses NHโOH dissociation due to the common ion effect.
Recognize the role of NHโCl in suppressing OHโป ion increase, facilitating selective precipitation.
NHโCl suppresses NHโOH dissociation due to the common ion effect.
Confusing role of NHโCl in Group III analysis.
True statements about [CoFโ]ยณโป ion include:
Hybridization: spยณdยฒ for high spin weak field ligands like Fโป.
Analyze hybridization, geometry, and spin state based on ligand field strength and oxidation state of cobalt.
Recognize weak-field ligand (Fโป) leads to high spin complex and spยณdยฒ hybridization.
True: (I) Octahedral geometry; (III) spยณdยฒ hybridized; (IV) High spin complex.
Confusion in hybridization of weak vs. strong field ligands.
Correlate weak-field ligands with spin and hybridization.
Hybridization: spยณdยฒ for high spin weak field ligands like Fโป.
Analyze hybridization, geometry, and spin state based on ligand field strength and oxidation state of cobalt.
Recognize weak-field ligand (Fโป) leads to high spin complex and spยณdยฒ hybridization.
True: (I) Octahedral geometry; (III) spยณdยฒ hybridized; (IV) High spin complex.
Confusion in hybridization of weak vs. strong field ligands.
The complex compounds Co(NHโ)โ SOโ]Br and [Co(NHโ)โ Br]SOโ are:
Ionization isomers: different ions outside the coordination sphere.
Complexes differ in anion association, leading to ionization isomerism.
Recognize that the same ligands but different counterions lead to ionization isomers.
The complexes are ionization isomers.
Confusing ionization isomers with coordination isomers.
Focus on anion association vs. coordination sphere.
Ionization isomers: different ions outside the coordination sphere.
Complexes differ in anion association, leading to ionization isomerism.
Recognize that the same ligands but different counterions lead to ionization isomers.
The complexes are ionization isomers.
Confusing ionization isomers with coordination isomers.
PCC is:
PCC: Cโ Hโ NHโบCrOโClโป
PCC refers to pyridinium chlorochromate, a mild oxidizing agent used to oxidize alcohols to aldehydes or ketones.
Recognize PCC as a complex containing CrOโ, pyridine, and HCl.
PCC structure is pyridinium chlorochromate (Cโ Hโ NHโบCrOโClโป).
Misinterpreting PCC's composition and function.
Recall mild oxidation reagents like PCC.
PCC: Cโ Hโ NHโบCrOโClโป
PCC refers to pyridinium chlorochromate, a mild oxidizing agent used to oxidize alcohols to aldehydes or ketones.
Recognize PCC as a complex containing CrOโ, pyridine, and HCl.
PCC structure is pyridinium chlorochromate (Cโ Hโ NHโบCrOโClโป).
Misinterpreting PCC's composition and function.
A complex CrClโยท6HโO gives 2.86 g AgCl with AgNOโ. Identify the complex.
Precipitation stoichiometry: moles AgCl = moles ionizable Clโป ions.
Calculate moles of AgCl formed to infer the number of chloride ions available for precipitation.
Analyze the coordination and ionization in Cr complex from given AgCl precipitation data.
The complex is [Cr(HโO)โClโ]Clยท2HโO based on moles of precipitated AgCl.
Misinterpreting ionizable vs. bonded chloride ions.
Carefully relate ionizable Clโป with AgCl precipitation.
Precipitation stoichiometry: moles AgCl = moles ionizable Clโป ions.
Calculate moles of AgCl formed to infer the number of chloride ions available for precipitation.
Analyze the coordination and ionization in Cr complex from given AgCl precipitation data.
The complex is [Cr(HโO)โClโ]Clยท2HโO based on moles of precipitated AgCl.
Misinterpreting ionizable vs. bonded chloride ions.
8.8 g of monohydric alcohol added to ethyle magnesium iodide in ether liberates 2240 cmยณ of ethane at STP. This monohydric alcohol when oxidised using pyridium-chloromate, forms a carbonyl compound that answers silver mirrior test( Tollenโs test). The monohydric alcohol is,
Use STP gas laws: 1 mol gas = 22400 cmยณ; calculate molar ratios and products.
8.8 g alcohol reacts to produce 2240 cmยณ ethane, indicating a primary alcohol (oxidized to aldehyde).
Use weight-volume relations to deduce molar mass and identity of alcohol.
Final alcohol is 2,2-dimethylpropan-1-ol (primary alcohol).
Misidentifying alcohol type (primary/secondary).
Recognize characteristics of primary alcohols under oxidation.
Use STP gas laws: 1 mol gas = 22400 cmยณ; calculate molar ratios and products.
8.8 g alcohol reacts to produce 2240 cmยณ ethane, indicating a primary alcohol (oxidized to aldehyde).
Use weight-volume relations to deduce molar mass and identity of alcohol.
Final alcohol is 2,2-dimethylpropan-1-ol (primary alcohol).
Misidentifying alcohol type (primary/secondary).
When a Tertiary alcohol 'A' (CโHโโO) reacts with 20% HโPOโ at 358 K to yield a major product 'B' (CโHโ). The IUPAC name of the compound.
Dehydration of alcohol: RโC(OH)โR โ R=CHโ + HโO
Reaction involves dehydration of tertiary alcohol to form an alkene (major product).
Dehydration: tertiary alcohol loses water to form 2-methylpropene as the major alkene product.
Final product is 2-methylpropene (CโHโ).
Confusion in determining major alkene product (Zaitsev's rule).
Recognize dehydration patterns for tertiary alcohols.
Dehydration of alcohol: RโC(OH)โR โ R=CHโ + HโO
Reaction involves dehydration of tertiary alcohol to form an alkene (major product).
Dehydration: tertiary alcohol loses water to form 2-methylpropene as the major alkene product.
Final product is 2-methylpropene (CโHโ).
Confusion in determining major alkene product (Zaitsev's rule).
In the reactions; CโHโ Cl--X--> CโHโ F , CโHโ Cl--Y--> CHโ=CH , CโHโ Cl --Z-->CโHโโ, the reagents X,Y,Z respectively are:
HgFโ for halide exchange; Alcoholic KOH for dehydrohalogenation; Na in dry ether for coupling.
Understand the conversion of CโHโ Cl to CโHโ F (by HgFโ), then to CHโ=CHโ (dehydrohalogenation with alcoholic KOH), followed by Wurtz reaction.
Follow reaction mechanisms stepwise: halogen exchange โ elimination โ coupling via Wurtz reaction.
Reactions yield butane (CโHโโ) via Wurtz reaction.
Misinterpreting reagent functionality for fluorination or Wurtz reaction.
Carefully trace reagents and sequence of steps.
HgFโ for halide exchange; Alcoholic KOH for dehydrohalogenation; Na in dry ether for coupling.
Understand the conversion of CโHโ Cl to CโHโ F (by HgFโ), then to CHโ=CHโ (dehydrohalogenation with alcoholic KOH), followed by Wurtz reaction.
Follow reaction mechanisms stepwise: halogen exchange โ elimination โ coupling via Wurtz reaction.
Reactions yield butane (CโHโโ) via Wurtz reaction.
Misinterpreting reagent functionality for fluorination or Wurtz reaction.
Sodium ethanoate on heating with soda lime gives โXโ. Electrolysis of aqueous solution of sodium ethanoate gives โYโ. Identify โXโ and โYโ respectively are
X = Decarboxylation; Y = Kolbeโs electrolysis
Decarboxylation: Sodium ethanoate + NaOH โ Methane (X). Kolbeโs electrolysis: 2CHโCOONa โ Ethane (Y).
Decarboxylation: COโ released; Electrolysis: Dimerization of CHโ radicals.
Final result: X = Methane, Y = Ethane.
Misinterpreting electrolysis products or decarboxylation intermediates.
Differentiate between reactions for decarboxylation and electrolysis.
X = Decarboxylation; Y = Kolbeโs electrolysis
Decarboxylation: Sodium ethanoate + NaOH โ Methane (X). Kolbeโs electrolysis: 2CHโCOONa โ Ethane (Y).
Decarboxylation: COโ released; Electrolysis: Dimerization of CHโ radicals.
Final result: X = Methane, Y = Ethane.
Misinterpreting electrolysis products or decarboxylation intermediates.
In the reaction: Anilinโ NaNOโ/dil.HCl-->(P)โPhenol/NaOH-->(Q).
Diazonium ion formation: ArNHโ + NaNOโ + HCl โ ArNโโบClโป
Reaction involves diazotization of aniline, coupling with phenol, and further treatment with NaOH to form para-hydroxyazobenzene.
Diazotization: CโHโ NHโ โ NaNOโ, dil. HCl โ CโHโ NโCl โ coupling with phenol forms the desired product.
The final product is para-hydroxyazobenzene.
Confusion between isomers ortho/para in coupling reactions.
Recall diazotization and coupling reactions.
Diazonium ion formation: ArNHโ + NaNOโ + HCl โ ArNโโบClโป
Reaction involves diazotization of aniline, coupling with phenol, and further treatment with NaOH to form para-hydroxyazobenzene.
Diazotization: CโHโ NHโ โ NaNOโ, dil. HCl โ CโHโ NโCl โ coupling with phenol forms the desired product.
The final product is para-hydroxyazobenzene.
Confusion between isomers ortho/para in coupling reactions.
The female sex hormone which is responsible for the development of secondary female characteristics and participates in the control of menstrual cycle is:
Hormonal roles: Estrogen (Estradiol) responsible for secondary characteristics.
The hormone responsible is estradiol, which is a key estrogen in female reproductive development.
Recognize that estradiol contributes to female secondary sexual traits and the menstrual cycle.
Correct hormone is Estradiol.
Confusing estradiol with other steroidal hormones.
Focus on key hormones in female reproduction.
Hormonal roles: Estrogen (Estradiol) responsible for secondary characteristics.
The hormone responsible is estradiol, which is a key estrogen in female reproductive development.
Recognize that estradiol contributes to female secondary sexual traits and the menstrual cycle.
Correct hormone is Estradiol.
Confusing estradiol with other steroidal hormones.
ฮฑ-D-(+)-glucose and ฮฒ-D-(+)-glucose are...
No specific formula.
ฮฑ-D-(+)-glucose and ฮฒ-D-(+)-glucose are anomers differing at the anomeric carbon (C1), where the โOH group orientation varies.
Both belong to the D-glucose family but differ in their C1 position (ฮฑ/ฮฒ).
Final result: They are Anomers.
Confusing anomers with epimers or enantiomers.
Differentiate anomers from other stereoisomers.
No specific formula.
ฮฑ-D-(+)-glucose and ฮฒ-D-(+)-glucose are anomers differing at the anomeric carbon (C1), where the โOH group orientation varies.
Both belong to the D-glucose family but differ in their C1 position (ฮฑ/ฮฒ).
Final result: They are Anomers.
Confusing anomers with epimers or enantiomers.
Biologically active adrenaline and ephedrine used to increase blood pressure contain...
Secondary amine = R-NH-Rโ
Adrenaline and Ephedrine have secondary amines (-NH-) that increase blood pressure and act as stimulants.
Reaction mechanisms align with secondary amine reactivity in biochemistry.
Final result: Amino group is Secondary.
Confusing primary/tertiary amines with secondary ones.
Recall functional groups in biologically active amines.
Secondary amine = R-NH-Rโ
Adrenaline and Ephedrine have secondary amines (-NH-) that increase blood pressure and act as stimulants.
Reaction mechanisms align with secondary amine reactivity in biochemistry.
Final result: Amino group is Secondary.
Confusing primary/tertiary amines with secondary ones.
But-1-yne on reaction with dil. HโSOโ in the presence of Hgยฒโบ ions at 333k gives
Hydration reaction โ Tautomerization
But-1-yne reacts with HโSOโ and Hgยฒโบ to give Butanone after tautomerization (initially forms enol).
Reaction: CHโกCH-CHโ โ CHโ=CH-CHโOH โ CHโCOCHโ (Butanone).
Final result: Product is Butanone.
Misunderstanding hydration mechanisms of alkynes.
Relate terminal alkyne reactions to hydration mechanisms.
Hydration reaction โ Tautomerization
But-1-yne reacts with HโSOโ and Hgยฒโบ to give Butanone after tautomerization (initially forms enol).
Reaction: CHโกCH-CHโ โ CHโ=CH-CHโOH โ CHโCOCHโ (Butanone).
Final result: Product is Butanone.
Misunderstanding hydration mechanisms of alkynes.
Propanone and Propanal are...
Functional group โ Isomer identification
Propanone (ketone) and Propanal (aldehyde) are functional isomers due to different functional groups with the same molecular formula.
Reaction: CHโCOCHโ (ketone) โ CHโCHโCHO (aldehyde).
Final result: They are Functional isomers.
Confusing functional isomers with chain or positional isomers.
Recall differences between functional group isomerism.
Functional group โ Isomer identification
Propanone (ketone) and Propanal (aldehyde) are functional isomers due to different functional groups with the same molecular formula.
Reaction: CHโCOCHโ (ketone) โ CHโCHโCHO (aldehyde).
Final result: They are Functional isomers.
Confusing functional isomers with chain or positional isomers.
The type of linkage present between nucleotides is...
Phosphodiester linkage = Sugar-phosphate backbone
Nucleotides in DNA and RNA are joined by phosphodiester bonds between the 3โฒ hydroxyl group of one sugar and the 5โฒ phosphate group of another.
Linkage ensures the formation of the sugar-phosphate backbone.
Final result: Linkage between nucleotides is Phosphodiester linkage.
Misidentifying amide or glycosidic bonds in nucleotides.
Recall the structure of DNA and RNA linkages.
Phosphodiester linkage = Sugar-phosphate backbone
Nucleotides in DNA and RNA are joined by phosphodiester bonds between the 3โฒ hydroxyl group of one sugar and the 5โฒ phosphate group of another.
Linkage ensures the formation of the sugar-phosphate backbone.
Final result: Linkage between nucleotides is Phosphodiester linkage.
Misidentifying amide or glycosidic bonds in nucleotides.
Which one of the following is a cationic detergent?
No specific formula.
Cetyltrimethylammonium bromide is a cationic detergent due to its positively charged quaternary ammonium group.
Other options are anionic detergents or non-detergent molecules.
Final result: Cationic detergent is Cetyltrimethylammonium bromide.
Confusing cationic detergents with anionic or neutral compounds.
Relate cationic detergents to their functional groups.
No specific formula.
Cetyltrimethylammonium bromide is a cationic detergent due to its positively charged quaternary ammonium group.
Other options are anionic detergents or non-detergent molecules.
Final result: Cationic detergent is Cetyltrimethylammonium bromide.
Confusing cationic detergents with anionic or neutral compounds.
One mole ideal monoatomic gas is taken round the cyclic process MNOM. The work done by the gas is,
W=โฎPdV
Calculate area of cycle in P-V diagram
Work done = Area enclosed
W=ยฝ(3Vโ)(2Pโ-Pโ)=2PโVโ
PV diagram
Consider cycle area
W=โฎPdV
Calculate area of cycle in P-V diagram
Work done = Area enclosed
W=ยฝ(3Vโ)(2Pโ-Pโ)=2PโVโ
PV diagram
The compound is...
No specific formula.
Saccharin is an artificial sweetener, and its given structure matches the functional groups of saccharin.
Confirmed by analyzing structural formula (no misalignment).
Final result: Compound is Saccharin.
Misidentifying functional groups in sweeteners.
Identify functional groups for artificial sweeteners.
No specific formula.
Saccharin is an artificial sweetener, and its given structure matches the functional groups of saccharin.
Confirmed by analyzing structural formula (no misalignment).
Final result: Compound is Saccharin.
Misidentifying functional groups in sweeteners.
The biodegradable polymer obtained by polymerization of Glycine and Aminocaproic acid is...
Polymerization reactions basics.
Nylon 2โNylon 6 is a biodegradable copolymer formed by condensation polymerization of Glycine and Aminocaproic acid.
Reaction: Glycine + Aminocaproic acid โ Nylon 2โNylon 6.
Final result: Biodegradable polymer is Nylon 2โNylon 6.
Confusing Nylon-6 with biodegradable polymers like PHBV.
Recall examples of biodegradable synthetic polymers.
Polymerization reactions basics.
Nylon 2โNylon 6 is a biodegradable copolymer formed by condensation polymerization of Glycine and Aminocaproic acid.
Reaction: Glycine + Aminocaproic acid โ Nylon 2โNylon 6.
Final result: Biodegradable polymer is Nylon 2โNylon 6.
Confusing Nylon-6 with biodegradable polymers like PHBV.
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