Depletion region in an unbiased semiconductor diode is a region consisting of
No formula involved
Understand formation of depletion region
Consider recombination of charge carriers
Region is depleted of mobile charges
Carrier distribution
Think about charge carriers
No formula involved
Understand formation of depletion region
Consider recombination of charge carriers
Region is depleted of mobile charges
Carrier distribution
The upper level of valence band and lower level of conduction band overlap in case of
No formula involved
Compare band structures of different materials
Identify that conductors have overlapping bands
Copper shows band overlap as a conductor
Band theory
Consider material properties
No formula involved
Compare band structures of different materials
Identify that conductors have overlapping bands
Copper shows band overlap as a conductor
Band theory
In diagram shown, Zener diode has reverse breakdown voltage Vz. The current through the laod resistance RL IS IL. The current through the zener diode is
I=Iz+IL
Apply Kirchhoff's current law at junction
Express current through Zener: Iz=I-IL
Calculate I=(V₀-Vz)/Rs, giving Iz=(V₀-Vz)/Rs-IL
Current distribution
Consider current division
I=Iz+IL
Apply Kirchhoff's current law at junction
Express current through Zener: Iz=I-IL
Calculate I=(V₀-Vz)/Rs, giving Iz=(V₀-Vz)/Rs-IL
Current distribution
A p-n junction diode connected to battery 5.7V in series with resistant 5kΩ such that it is forward biased. If the barrier potential of the diode is 0.7 V, neglecting the diode resistance, the current in the circuit is,
I=(V-Vd)/R
Consider forward bias voltage drop 0.7V
Calculate effective voltage: 5.7-0.7=5V
Current=5V/5kΩ=1mA
Voltage drop
Consider voltage drop
I=(V-Vd)/R
Consider forward bias voltage drop 0.7V
Calculate effective voltage: 5.7-0.7=5V
Current=5V/5kΩ=1mA
Voltage drop
An athlete runs along circular track of diameter 80m. The distance travelled and the magnitude of displacement of the athelete when he covers 3/4th of the circle is (in m)
s=rθ
Calculate distance=(3/4)×πd=60π m
Calculate displacement=2R=40√2 m
Distance > Displacement verified
Arc vs chord
Consider arc length
s=rθ
Calculate distance=(3/4)×πd=60π m
Calculate displacement=2R=40√2 m
Distance > Displacement verified
Arc vs chord
A point charge A of +10µC and another point charge B of +20µC are kept 1m apart in free space. The electrostatic force on A due to B is F₁ and the electrostatic force on B due to A is F₂. Then
F=kq₁q₂/r²
Apply Coulomb's law to find force on A due to B: F₁=k(10×20)×10⁻¹²/1² N
Apply Coulomb's law to find force on B due to A: F₂=k(20×10)×10⁻¹²/1² N
By Newton's third law, these forces are equal in magnitude but opposite in direction, so F₁=-F₂
Forgetting direction
Apply Newton's third law
F=kq₁q₂/r²
Apply Coulomb's law to find force on A due to B: F₁=k(10×20)×10⁻¹²/1² N
Apply Coulomb's law to find force on B due to A: F₂=k(20×10)×10⁻¹²/1² N
By Newton's third law, these forces are equal in magnitude but opposite in direction, so F₁=-F₂
Forgetting direction
Among given pair of vectors, the resultant of two vectors can never be 3 units. The vectors are
R²=a²+b²+2abcosθ
Apply triangle inequality: a-b ≤R≤ a+b
For 4,8 units: 4≤R≤12
3 units not possible in range
Misunderstanding Vector Addition Rules. Forgetting that angle between vectors affects resultant magnitude
Consider vector addition
R²=a²+b²+2abcosθ
Apply triangle inequality: a-b ≤R≤ a+b
For 4,8 units: 4≤R≤12
3 units not possible in range
Misunderstanding Vector Addition Rules. Forgetting that angle between vectors affects resultant magnitude
A block of certain mass is placed on a rough inclined plane. The angle between the plane and the horizontal is 30°. The coefficients of static and kinetic frictions 0.6 and 0.5 respectively. Then the magnitude ofthe acceleration of the block is [take g=10 m/s2],
fs≤μsN
Calculate limiting friction: fs=μsmgcosθ=0.6mg×0.866=0.52mg
Compare with component down plane: mgsinθ=0.5mg
As fs>mgsinθ, block remains at rest, a=0
Force resolution
Consider force balance
fs≤μsN
Calculate limiting friction: fs=μsmgcosθ=0.6mg×0.866=0.52mg
Compare with component down plane: mgsinθ=0.5mg
As fs>mgsinθ, block remains at rest, a=0
Force resolution
A particle of mass 500g at rest. It is free to move along a straight line. The power delivered to the particle varies with time according to graph: THE MOMENTUM OF THE PARTICLE AT t=5s is;
W=∫Pdt
Work done = Area under P-t graph = 25J
Use W=ΔKE=(1/2)mv²
Calculate v from p=mv=5Ns
Graph integration
Consider power-time relation
W=∫Pdt
Work done = Area under P-t graph = 25J
Use W=ΔKE=(1/2)mv²
Calculate v from p=mv=5Ns
Graph integration
Dimensional formula for activity of radioactive substance is
A=dN/dt
Activity = Rate of decay = dN/dt
Analyze dimensions: N is dimensionless
Therefore dimensions are [T⁻¹]
Dimension analysis
Consider time rate
A=dN/dt
Activity = Rate of decay = dN/dt
Analyze dimensions: N is dimensionless
Therefore dimensions are [T⁻¹]
Dimension analysis
A ceiling fan rotating around fixed axle as shown.The direction of angular velocity is along____
ω perpendicular to plane
Apply right-hand rule for rotation
Direction points out of plane of rotation
For clockwise rotation, ω points downward (-k̂)
Direction convention
Use right-hand rule
ω perpendicular to plane
Apply right-hand rule for rotation
Direction points out of plane of rotation
For clockwise rotation, ω points downward (-k̂)
Direction convention
A body of mass 1kg suspended by weightless string which passes over a frictionless pulley of mass 2kg as shown in the figure. The mass is releasedfrom a height of 1.6m from the ground. With what velocity does is strike the ground ?
KEtotal=KEtrans+KErot
Apply conservation of energy: mgh=½mv²+½Iω²
For disk I=½MR², ω=v/R
Solve for v=4ms⁻¹
Energy distribution
Consider rotational KE
KEtotal=KEtrans+KErot
Apply conservation of energy: mgh=½mv²+½Iω²
For disk I=½MR², ω=v/R
Solve for v=4ms⁻¹
Energy distribution
what is the value of acceleration due to gravity at a height equal to half radius of Earth, from its surface ?
g'=gR²/(R+h)²
Use g'=gR²/(R+h)²
Substitute h=R/2
Calculate g'=(9.8)(4/9)=4.4ms⁻²
Distance relation
Consider inverse square
g'=gR²/(R+h)²
Use g'=gR²/(R+h)²
Substitute h=R/2
Calculate g'=(9.8)(4/9)=4.4ms⁻²
Distance relation
A thick metal wire of density ρ and length 'L' is hung from rigid support. The increase in length of thr wire due to its own wieght is (Y= Young's Modulus of the material of the wire),
ΔL=∫(F/AY)dl
Consider varying force along length
Integrate strain over length
Get extension=½ρgL²/Y
Force variation
Consider variable force
ΔL=∫(F/AY)dl
Consider varying force along length
Integrate strain over length
Get extension=½ρgL²/Y
Force variation
Water flows through horizontal pipe of varying cross section at a rate 0.314m³s⁻¹. The velocity of water at a point where the radius of the pipe is 10cm is ...
Q=Av
Use continuity equation Q=Av
A=πr²=π(0.1)²=0.0314m²
v=Q/A=0.314/0.0314=10ms⁻¹
Area calculation
Consider flow conservation
Q=Av
Use continuity equation Q=Av
A=πr²=π(0.1)²=0.0314m²
v=Q/A=0.314/0.0314=10ms⁻¹
Area calculation
A solid cube of mass m at temperature θ₀ is heated at a constant rate. It becomes liquid at temperature θ₁ and vapour at temperature θ₂. Lets S1 and S2 be specific heats in its solid and liquid states respectively. If Lf and Lv are latent heats of fusion and vaporisation respectively, then the minimum heat energy supplied to the cube until it vaporises is,
Q=mcΔθ+mL
Consider heating in solid phase
Add latent heat of fusion
Add heating in liquid phase and vaporization
Phase transition
Account all phase changes
Q=mcΔθ+mL
Consider heating in solid phase
Add latent heat of fusion
Add heating in liquid phase and vaporization
Phase transition
A uniform electric field E=3×10⁵NC⁻¹ is acting along the positive Y-axis. The electric flux through a rectangle of area 10cm×30cm whose plane is parallel to the Z-X plane is
Φ=E·A
Convert dimensions to meters: 10cm×30cm = 0.1m×0.3m = 0.03m²
Apply electric flux formula Φ=E·A. Since plane is parallel to Z-X plane, it's perpendicular to field (Y-axis)
Calculate Φ=EA=(3×10⁵)(0.03)=9×10³Vm
Area unit conversion errors
Consider orientation of surface
Φ=E·A
Convert dimensions to meters: 10cm×30cm = 0.1m×0.3m = 0.03m²
Apply electric flux formula Φ=E·A. Since plane is parallel to Z-X plane, it's perpendicular to field (Y-axis)
Calculate Φ=EA=(3×10⁵)(0.03)=9×10³Vm
Area unit conversion errors
One mole ideal monoatomic gas is taken round the cyclic process MNOM. The work done by the gas is,
W=∮PdV
Calculate area of cycle in P-V diagram
Work done = Area enclosed
W=½(3V₀)(2P₀-P₀)=2P₀V₀
PV diagram
Consider cycle area
W=∮PdV
Calculate area of cycle in P-V diagram
Work done = Area enclosed
W=½(3V₀)(2P₀-P₀)=2P₀V₀
PV diagram
Which of the following set of polymers are used as fiber?
No specific formula.
Terylene and Orlon are classified as synthetic fibers due to their high tensile strength and elasticity.
Starch is a natural polymer and does not behave as a fiber.
Final result: Polymers used as fibers are Terylene and Orlon.
Confusing synthetic polymers like Teflon with fiber-forming ones.
Recall fiber properties of synthetic polymers.
No specific formula.
Terylene and Orlon are classified as synthetic fibers due to their high tensile strength and elasticity.
Starch is a natural polymer and does not behave as a fiber.
Final result: Polymers used as fibers are Terylene and Orlon.
Confusing synthetic polymers like Teflon with fiber-forming ones.
The biodegradable polymer obtained by polymerization of Glycine and Aminocaproic acid is...
Polymerization reactions basics.
Nylon 2–Nylon 6 is a biodegradable copolymer formed by condensation polymerization of Glycine and Aminocaproic acid.
Reaction: Glycine + Aminocaproic acid → Nylon 2–Nylon 6.
Final result: Biodegradable polymer is Nylon 2–Nylon 6.
Confusing Nylon-6 with biodegradable polymers like PHBV.
Recall examples of biodegradable synthetic polymers.
Polymerization reactions basics.
Nylon 2–Nylon 6 is a biodegradable copolymer formed by condensation polymerization of Glycine and Aminocaproic acid.
Reaction: Glycine + Aminocaproic acid → Nylon 2–Nylon 6.
Final result: Biodegradable polymer is Nylon 2–Nylon 6.
Confusing Nylon-6 with biodegradable polymers like PHBV.
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.
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.
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 3′ hydroxyl group of one sugar and the 5′ 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 3′ hydroxyl group of one sugar and the 5′ 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.
α-D-(+)-glucose and β-D-(+)-glucose are...
No specific formula.
α-D-(+)-glucose and β-D-(+)-glucose are anomers differing at the anomeric carbon (C1), where the –OH group orientation varies.
Both belong to the D-glucose family but differ in their C1 position (α/β).
Final result: They are Anomers.
Confusing anomers with epimers or enantiomers.
Differentiate anomers from other stereoisomers.
No specific formula.
α-D-(+)-glucose and β-D-(+)-glucose are anomers differing at the anomeric carbon (C1), where the –OH group orientation varies.
Both belong to the D-glucose family but differ in their C1 position (α/β).
Final result: They are Anomers.
Confusing anomers with epimers or enantiomers.
Propanone and Propanal are...
Functional group → Isomer identification
Propanone (ketone) and Propanal (aldehyde) are functional isomers due to different functional groups with the same molecular formula.
Reaction: CH₃COCH₃ (ketone) ↔ CH₃CH₂CHO (aldehyde).
Final result: They are Functional isomers.
Confusing functional isomers with chain or positional isomers.
Recall differences between functional group isomerism.
Functional group → Isomer identification
Propanone (ketone) and Propanal (aldehyde) are functional isomers due to different functional groups with the same molecular formula.
Reaction: CH₃COCH₃ (ketone) ↔ CH₃CH₂CHO (aldehyde).
Final result: They are Functional isomers.
Confusing functional isomers with chain or positional isomers.
Sodium ethanoate on heating with soda lime gives ‘X’. Electrolysis of aqueous solution of sodium ethanoate gives ‘Y’. Identify ‘X’ and ‘Y’ respectively are
X = Decarboxylation; Y = Kolbe’s electrolysis
Decarboxylation: Sodium ethanoate + NaOH → Methane (X). Kolbe’s electrolysis: 2CH₃COONa → Ethane (Y).
Decarboxylation: CO₂ released; Electrolysis: Dimerization of CH₃ radicals.
Final result: X = Methane, Y = Ethane.
Misinterpreting electrolysis products or decarboxylation intermediates.
Differentiate between reactions for decarboxylation and electrolysis.
X = Decarboxylation; Y = Kolbe’s electrolysis
Decarboxylation: Sodium ethanoate + NaOH → Methane (X). Kolbe’s electrolysis: 2CH₃COONa → Ethane (Y).
Decarboxylation: CO₂ released; Electrolysis: Dimerization of CH₃ radicals.
Final result: X = Methane, Y = Ethane.
Misinterpreting electrolysis products or decarboxylation intermediates.
But-1-yne on reaction with dil. H₂SO₄ in the presence of Hg²⁺ ions at 333k gives
Hydration reaction → Tautomerization
But-1-yne reacts with H₂SO₄ and Hg²⁺ to give Butanone after tautomerization (initially forms enol).
Reaction: CH≡CH-CH₃ → CH₂=CH-CH₂OH → CH₃COCH₃ (Butanone).
Final result: Product is Butanone.
Misunderstanding hydration mechanisms of alkynes.
Relate terminal alkyne reactions to hydration mechanisms.
Hydration reaction → Tautomerization
But-1-yne reacts with H₂SO₄ and Hg²⁺ to give Butanone after tautomerization (initially forms enol).
Reaction: CH≡CH-CH₃ → CH₂=CH-CH₂OH → CH₃COCH₃ (Butanone).
Final result: Product is Butanone.
Misunderstanding hydration mechanisms of alkynes.
The total electric flux through a closed spherical surface of radius 'r' enclosing an electric dipole of dipole moment 2aq is (Give ε₀=permittivity of free space)
Φ=Q/ε₀
Recall Gauss's law relates flux to enclosed charge: Φ=Q_enclosed/ε₀
For dipole, total charge enclosed is zero as positive and negative charges are equal
Therefore, total flux must be zero regardless of radius
Confusing with single charge
Use Gauss's law
Φ=Q/ε₀
Recall Gauss's law relates flux to enclosed charge: Φ=Q_enclosed/ε₀
For dipole, total charge enclosed is zero as positive and negative charges are equal
Therefore, total flux must be zero regardless of radius
Confusing with single charge
Biologically active adrenaline and ephedrine used to increase blood pressure contain...
Secondary amine = R-NH-R’
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-R’
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.
In the reaction: Anilin→ NaNO₂/dil.HCl-->(P)→Phenol/NaOH-->(Q).
Diazonium ion formation: ArNH₂ + NaNO₂ + HCl → ArN₂⁺Cl⁻
Reaction involves diazotization of aniline, coupling with phenol, and further treatment with NaOH to form para-hydroxyazobenzene.
Diazotization: C₆H₅NH₂ → NaNO₂, dil. HCl → C₆H₅N₂Cl → 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: ArNH₂ + NaNO₂ + HCl → ArN₂⁺Cl⁻
Reaction involves diazotization of aniline, coupling with phenol, and further treatment with NaOH to form para-hydroxyazobenzene.
Diazotization: C₆H₅NH₂ → NaNO₂, dil. HCl → C₆H₅N₂Cl → 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.
In the reactions; C₂H₅Cl--X--> C₂H₅F , C₂H₅Cl--Y--> CH₂=CH , C₂H₅Cl --Z-->C₄H₁₀, the reagents X,Y,Z respectively are:
HgF₂ for halide exchange; Alcoholic KOH for dehydrohalogenation; Na in dry ether for coupling.
Understand the conversion of C₂H₅Cl to C₂H₅F (by HgF₂), then to CH₂=CH₂ (dehydrohalogenation with alcoholic KOH), followed by Wurtz reaction.
Follow reaction mechanisms stepwise: halogen exchange → elimination → coupling via Wurtz reaction.
Reactions yield butane (C₄H₁₀) via Wurtz reaction.
Misinterpreting reagent functionality for fluorination or Wurtz reaction.
Carefully trace reagents and sequence of steps.
HgF₂ for halide exchange; Alcoholic KOH for dehydrohalogenation; Na in dry ether for coupling.
Understand the conversion of C₂H₅Cl to C₂H₅F (by HgF₂), then to CH₂=CH₂ (dehydrohalogenation with alcoholic KOH), followed by Wurtz reaction.
Follow reaction mechanisms stepwise: halogen exchange → elimination → coupling via Wurtz reaction.
Reactions yield butane (C₄H₁₀) via Wurtz reaction.
Misinterpreting reagent functionality for fluorination or Wurtz reaction.
8.8 g of monohydric alcohol added to ethyle magnesium iodide in ether liberates 2240 cm³ of ethane at STP. This monohydric alcohol when oxidised using pyridium-chloromate, forms a carbonyl compound that answers silver mirrior test( Tollen’s test). The monohydric alcohol is,
Use STP gas laws: 1 mol gas = 22400 cm³; calculate molar ratios and products.
8.8 g alcohol reacts to produce 2240 cm³ 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 cm³; calculate molar ratios and products.
8.8 g alcohol reacts to produce 2240 cm³ 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' (C₄H₁₀O) reacts with 20% H₃PO₄ at 358 K to yield a major product 'B' (C₄H₈). The IUPAC name of the compound.
Dehydration of alcohol: R–C(OH)–R → R=CH₂ + H₂O
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 (C₄H₈).
Confusion in determining major alkene product (Zaitsev's rule).
Recognize dehydration patterns for tertiary alcohols.
Dehydration of alcohol: R–C(OH)–R → R=CH₂ + H₂O
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 (C₄H₈).
Confusion in determining major alkene product (Zaitsev's rule).
PCC is:
PCC: C₅H₅NH⁺CrO₃Cl⁻
PCC refers to pyridinium chlorochromate, a mild oxidizing agent used to oxidize alcohols to aldehydes or ketones.
Recognize PCC as a complex containing CrO₃, pyridine, and HCl.
PCC structure is pyridinium chlorochromate (C₅H₅NH⁺CrO₃Cl⁻).
Misinterpreting PCC's composition and function.
Recall mild oxidation reagents like PCC.
PCC: C₅H₅NH⁺CrO₃Cl⁻
PCC refers to pyridinium chlorochromate, a mild oxidizing agent used to oxidize alcohols to aldehydes or ketones.
Recognize PCC as a complex containing CrO₃, pyridine, and HCl.
PCC structure is pyridinium chlorochromate (C₅H₅NH⁺CrO₃Cl⁻).
Misinterpreting PCC's composition and function.
A complex CrCl₃·6H₂O gives 2.86 g AgCl with AgNO₃. Identify the complex.
Precipitation stoichiometry: moles AgCl = moles ionizable Cl⁻ 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(H₂O)₄Cl₂]Cl·2H₂O based on moles of precipitated AgCl.
Misinterpreting ionizable vs. bonded chloride ions.
Carefully relate ionizable Cl⁻ with AgCl precipitation.
Precipitation stoichiometry: moles AgCl = moles ionizable Cl⁻ 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(H₂O)₄Cl₂]Cl·2H₂O based on moles of precipitated AgCl.
Misinterpreting ionizable vs. bonded chloride ions.
The complex compounds Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄ 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.
True statements about [CoF₆]³⁻ ion include:
Hybridization: sp³d² for high spin weak field ligands like F⁻.
Analyze hybridization, geometry, and spin state based on ligand field strength and oxidation state of cobalt.
Recognize weak-field ligand (F⁻) leads to high spin complex and sp³d² hybridization.
True: (I) Octahedral geometry; (III) sp³d² hybridized; (IV) High spin complex.
Confusion in hybridization of weak vs. strong field ligands.
Correlate weak-field ligands with spin and hybridization.
Hybridization: sp³d² for high spin weak field ligands like F⁻.
Analyze hybridization, geometry, and spin state based on ligand field strength and oxidation state of cobalt.
Recognize weak-field ligand (F⁻) leads to high spin complex and sp³d² hybridization.
True: (I) Octahedral geometry; (III) sp³d² hybridized; (IV) High spin complex.
Confusion in hybridization of weak vs. strong field ligands.
Under electrostatic condition of a charged conductor, which among the following statements is true?
E=σ/2ε₀
Understand that in electrostatic equilibrium, charges must be stationary
In a conductor, free electrons move until reaching equilibrium where internal field is zero
All excess charge must reside on surface to maintain zero internal field
Confusing field directions
Consider charge distribution
E=σ/2ε₀
Understand that in electrostatic equilibrium, charges must be stationary
In a conductor, free electrons move until reaching equilibrium where internal field is zero
All excess charge must reside on surface to maintain zero internal field
Confusing field directions
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 → 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 → 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 NH₄Cl to NH₄OH in the analysis of Group III basic radicals...
NH₄Cl addition suppresses NH₄OH dissociation.
NH₄Cl suppresses NH₄OH dissociation due to the common ion effect.
Recognize the role of NH₄Cl in suppressing OH⁻ ion increase, facilitating selective precipitation.
NH₄Cl suppresses NH₄OH dissociation due to the common ion effect.
Confusing role of NH₄Cl in Group III analysis.
Focus on the effect of common ions in suppressing dissociation.
NH₄Cl addition suppresses NH₄OH dissociation.
NH₄Cl suppresses NH₄OH dissociation due to the common ion effect.
Recognize the role of NH₄Cl in suppressing OH⁻ ion increase, facilitating selective precipitation.
NH₄Cl suppresses NH₄OH dissociation due to the common ion effect.
Confusing role of NH₄Cl in Group III analysis.
Solubility product of CaC₂O₄ at a given temperature in pure water is 4 × 10⁻⁹(moL-1)2. Its solubility at same temperature is ...
Ksp = S² → Solve for S.
Solubility (S) is calculated from Ksp = S² for salts dissociating into 1:1 cations and anions.
Ksp = S² → Solve for S: √(4 × 10⁻⁹) = 6.3 × 10⁻⁵.
Solubility = 6.3 × 10⁻⁵ mol/L.
Miscalculating solubility for salts dissociating into 1:1 ions.
Use Ksp formula correctly for solubility.
Ksp = S² → Solve for S.
Solubility (S) is calculated from Ksp = S² for salts dissociating into 1:1 cations and anions.
Ksp = S² → Solve for S: √(4 × 10⁻⁹) = 6.3 × 10⁻⁵.
Solubility = 6.3 × 10⁻⁵ mol/L.
Miscalculating solubility for salts dissociating into 1:1 ions.
Reaction between moist SO₂ and acidified permanganate...
Oxidation: SO₂ → SO₄²⁻; Reduction: MnO₄⁻ → Mn²⁺.
Reaction involves oxidation of SO₂ to SO₄²⁻ and reduction of MnO₄⁻ to Mn²⁺ in acidic medium.
Balance the reaction in acidic medium: MnO₄⁻ + SO₂ → Mn²⁺ + SO₄²⁻.
Final result: SO₂ → SO₄²⁻; MnO₄⁻ → Mn²⁺.
Misidentifying oxidized and reduced species in redox reaction.
Focus on oxidation states and redox balancing.
Oxidation: SO₂ → SO₄²⁻; Reduction: MnO₄⁻ → Mn²⁺.
Reaction involves oxidation of SO₂ to SO₄²⁻ and reduction of MnO₄⁻ to Mn²⁺ in acidic medium.
Balance the reaction in acidic medium: MnO₄⁻ + SO₂ → Mn²⁺ + SO₄²⁻.
Final result: SO₂ → SO₄²⁻; MnO₄⁻ → Mn²⁺.
Misidentifying oxidized and reduced species in redox reaction.
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.
Which one of the following Nitrate will decompose to give NO₂ on heating...
LiNO₃ decomposition: 4LiNO₃ → 2Li₂O + 4NO₂ + O₂.
LiNO₃ decomposes to NO₂ due to its unique thermal decomposition properties.
Recognize that LiNO₃ decomposition differs from other nitrates.
Final product: 4LiNO₃ → 2Li₂O + 4NO₂ + O₂.
Confusing LiNO₃ decomposition with other alkali nitrates.
Focus on differences in thermal decomposition of nitrates.
LiNO₃ decomposition: 4LiNO₃ → 2Li₂O + 4NO₂ + O₂.
LiNO₃ decomposes to NO₂ due to its unique thermal decomposition properties.
Recognize that LiNO₃ decomposition differs from other nitrates.
Final product: 4LiNO₃ → 2Li₂O + 4NO₂ + O₂.
Confusing LiNO₃ decomposition with other alkali nitrates.
Halide that cannot be hydrolyzed is...
Hydrolysis needs d-orbitals, absent in carbon.
CCl₄ cannot be hydrolyzed due to the absence of d-orbitals in carbon.
Recognize that hydrolysis requires availability of d-orbitals.
Final result: CCl₄ cannot hydrolyze.
Assuming all halides undergo hydrolysis.
Understand the role of d-orbitals in hydrolysis reactions.
Hydrolysis needs d-orbitals, absent in carbon.
CCl₄ cannot be hydrolyzed due to the absence of d-orbitals in carbon.
Recognize that hydrolysis requires availability of d-orbitals.
Final result: CCl₄ cannot hydrolyze.
Assuming all halides undergo hydrolysis.
0.48g of an organic compound on complete combustion produced 0.22g of CO₂ .The percentage of C in the given organic compound is;
%C = (mass of C in CO₂ / mass of compound) × 100.
Calculate %C using the relation: (mass of C in CO₂ / mass of compound) × 100.
Use data: (12/44) × (0.22/0.48) × 100 = 12.5%.
Carbon percentage = 12.5%.
Miscalculating %C using incorrect ratios.
Carefully compute based on given mass ratios.
%C = (mass of C in CO₂ / mass of compound) × 100.
Calculate %C using the relation: (mass of C in CO₂ / mass of compound) × 100.
Use data: (12/44) × (0.22/0.48) × 100 = 12.5%.
Carbon percentage = 12.5%.
Miscalculating %C using incorrect ratios.
Sequence of reactions yielding Br-substituted products is...
Addition-elimination pathways for Br substitution.
Recognize reaction steps: addition, substitution, and elimination pathways involving Br₂ and NaNH₂.
Follow sequential mechanism: CH₂=CH₂ → 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 Br₂ and NaNH₂.
Follow sequential mechanism: CH₂=CH₂ → Br-substituted products.
Final products involve multiple bromination and alkane formation.
Misunderstanding intermediate steps in bromination sequence.
A cube of side 1 cm contains 100 molecules each having an induced dipole moment of 0.2×10⁻⁶ C-m in an external electric field of 4NC⁻¹. The electric susceptibility of the materials is _____ C²N⁻¹m⁻²
χₑ=P/E
Convert dimensions to SI units: volume=(0.01m)³=10⁻⁶m³. Calculate number density n=100/10⁻⁶=10⁸ molecules/m³
Use relation P=χₑE where P=np (n=number density, p=dipole moment)
Calculate χₑ=(0.2×10⁻⁶×100)/(10⁻⁶×4)=5 C²N⁻¹m⁻²
Unit conversion errors
Check units carefully
χₑ=P/E
Convert dimensions to SI units: volume=(0.01m)³=10⁻⁶m³. Calculate number density n=100/10⁻⁶=10⁸ molecules/m³
Use relation P=χₑE where P=np (n=number density, p=dipole moment)
Calculate χₑ=(0.2×10⁻⁶×100)/(10⁻⁶×4)=5 C²N⁻¹m⁻²
Unit conversion errors
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.
Which of the following crystals has the unit cell such that a = b ≠ c and α = β = 90°, γ = 120°?
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 ≠ c and α = β = 90°, γ = 120°.
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 ≠ c and α = β = 90°, γ = 120°.
Confusing between unit cell parameters for hexagonal and tetragonal structures.
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.
The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is...
d = (Z × M) / (Na × a³).
Use the formula d = (Z × M) / (Na × a³) to find the density of the unit cell.
Substituting values: d = 10, a = 300 pm, Na = 6.022 × 10^23, Z = 4, M = 40.5 g/mol.
Result: Number of atoms = 6.6 × 10^22.
Miscalculating Z or misinterpreting the given edge length (pm to cm).
Focus on unit cell dimensions and atomic packing.
d = (Z × M) / (Na × a³).
Use the formula d = (Z × M) / (Na × a³) to find the density of the unit cell.
Substituting values: d = 10, a = 300 pm, Na = 6.022 × 10^23, Z = 4, M = 40.5 g/mol.
Result: Number of atoms = 6.6 × 10^22.
Miscalculating Z or misinterpreting the given edge length (pm to cm).
Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at 100°C is...
P_solution = P_water × (1 - χ_solute).
Use Raoult’s law: P_solution = P_water × (1 - χ_solute).
Calculate χ_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 × (1 - χ_solute).
Use Raoult’s law: P_solution = P_water × (1 - χ_solute).
Calculate χ_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.
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.
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.
How many Coulombs are required to oxidise 0.1 mole of H2O to oxygen?
q = n × F (Faraday constant).
Oxidation of water to oxygen involves 4 electrons per molecule.
For 0.1 mole of H2O: (0.1 × 2 × 96500 = 1.93 × 10^4 C).
Final Answer: 1.93 × 10^4 C.
Mixing up electron moles or Faraday's constant during calculations.
Use Faraday's Laws of Electrolysis.
q = n × F (Faraday constant).
Oxidation of water to oxygen involves 4 electrons per molecule.
For 0.1 mole of H2O: (0.1 × 2 × 96500 = 1.93 × 10^4 C).
Final Answer: 1.93 × 10^4 C.
Mixing up electron moles or Faraday's constant during calculations.
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 × I × t) / 96500.
Use Faraday’s law: Mass deposited = (E × I × t) / 96500.
Substituting values: E = 40, I = 3A, t = 6432 s. Result: (40 × 3 × 6432) / 96500 = 3.99 g ≈ 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 × I × t) / 96500.
Use Faraday’s law: Mass deposited = (E × I × t) / 96500.
Substituting values: E = 40, I = 3A, t = 6432 s. Result: (40 × 3 × 6432) / 96500 = 3.99 g ≈ 4.0 g.
Final Mass: 4.0 g.
Errors in time conversion or forgetting valency of calcium during calculations.
The value of 'A' in the equation λ_m = λ_0_m - A √C is same for the pair...
λ_m = λ_0 - A √C (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.
λ_m = λ_0 - A √C (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.