Q31Single correctSome Basic Concepts of Chemistry
Choose the Incorrect Statement about Dalton's Atomic Theory
SolutionAnswer: Option 3atoms are formed when atoms of different elements combine in any ratio. Approach:
Each statement is compared against the postulates of Dalton's atomic theory to identify the false one.
Step 1:Dalton's theory states that chemical reactions involve only the reorganization of atoms, so atoms are neither created nor destroyed. This statement is correct.
Step 2:Dalton proposed that matter is made up of indivisible atoms. This statement is correct within the framework of the theory.
Step 3:Atoms are not formed by combination; rather compounds are formed when atoms of different elements combine. Stating that atoms are formed this way contradicts the theory.
Step 4:Compounds form when atoms of different elements combine in fixed ratios, and all atoms of a given element are identical in mass and properties. This statement is correct.
Final answer: atoms are formed when atoms of different elements combine in any ratio.
Q32Single correctClassification of Elements and Periodicity in Properties
The correct order of the first ionization enthalpy is
SolutionAnswer: Option 3Tl>Ga>Al Approach:
First ionization enthalpies of Group 13 elements are compared, accounting for poor shielding by d and f electrons.
Step 1:Down Group 13, ionization enthalpy does not decrease smoothly because of the poor shielding effect of the filled d and f orbitals in heavier members.
Step 2:Ga has a higher first ionization enthalpy than Al because the 3d electrons in Ga shield the nuclear charge poorly, increasing the effective nuclear charge.
Ga>Al
Step 3:Tl has the highest among Tl, Ga and Al owing to the additional poor shielding by 4f and 5d electrons, which raises its effective nuclear charge and ionization enthalpy.
Tl>Ga
Step 4:Combining these comparisons gives the overall order.
Tl>Ga>Al
Final answer: Tl>Ga>Al
Q33Single correctClassification of Elements and Periodicity in Properties
Given below are two statements : Statement I : The correct order of first ionization enthalpy values of Li,Na,F and Cl is Na<Li<Cl<F. Statement II : The correct order of negative electron gain enthalpy values of Li,Na,F and Cl is Na<Li<F<Cl In the light of the above statements, choose the correct answer from the options given below :
SolutionAnswer: Option 4Both Statement I and Statement II are true Approach:
The two given orders are checked against periodic trends in ionization enthalpy and electron gain enthalpy.
Step 1:Across a period ionization enthalpy increases, and down a group it decreases. Hence Na has the lowest first ionization enthalpy, followed by Li, then Cl, then F.
Na<Li<Cl<F
Step 2:Negative electron gain enthalpy is small for alkali metals, with Na lower than Li. Among halogens, Cl has a more negative electron gain enthalpy than F due to the small compact size of F causing electron-electron repulsion.
Na<Li<F<Cl
Step 3:Both ordering statements are consistent with established periodic property data.
Final answer: Both Statement I and Statement II are true
Q34Single correctChemical Bonding and Molecular Structure
The correct statement/s about Hydrogen bonding is/are A. Hydrogen bonding exists when H is covalently bonded to the highly electro negative atom. B. Intermolecular H bonding is present in o-nitro phenol C. Intramolecular H bonding is present in HF. D. The magnitude of H bonding depends on the physical state of the compound. E. H-bonding has powerful effect on the structure and properties of compounds Choose the correct answer from the options given below:
SolutionAnswer: Option 2A, D, E only Approach:
Each statement about hydrogen bonding is evaluated for correctness.
Step 1:Hydrogen bonding requires hydrogen to be covalently bonded to a highly electronegative atom such as F, O or N. Statement A is correct.
Step 2:o-Nitrophenol exhibits intramolecular hydrogen bonding, not intermolecular, because the -OH and -NO2 groups are adjacent. Statement B is incorrect.
Step 3:HF shows intermolecular hydrogen bonding between separate molecules, not intramolecular. Statement C is incorrect.
Step 4:The extent of hydrogen bonding varies with physical state, being strongest in the solid and liquid phases. Statement D is correct.
Step 5:Hydrogen bonding strongly influences melting and boiling points, solubility and structure of compounds. Statement E is correct.
Final answer: A, D, E only
Q35Single correctChemical Bonding and Molecular Structure
The number of species from the following that have pyramidal geometry around the central atom is ________.
S2O32−,SO42−,SO32−,S2O72−
SolutionAnswer: Option 41 Approach:
The shape around the central sulfur in each species is determined using hybridization and lone-pair considerations.
Step 1:In thiosulfate the central sulfur is surrounded by four atoms (three O and one S) with no lone pair, giving a tetrahedral arrangement, not pyramidal.
S2O32−
Step 2:In sulfate the sulfur is bonded to four oxygen atoms with no lone pair, giving a tetrahedral shape.
SO42−
Step 3:In sulfite the sulfur has three bond pairs and one lone pair, giving a trigonal pyramidal shape around sulfur.
SO32−
Step 4:In disulfate each sulfur is bonded to four oxygen atoms through a bridging oxygen, giving a tetrahedral environment at each sulfur, not pyramidal.
S2O72−
Step 5:Only sulfite shows pyramidal geometry around the central atom.
Final answer: 1
Q36Single correctEquilibrium
The equilibrium constant for the reaction SO3(g)⇌SO2(g)+21O2(g) is Kc=4.9×10−2. The value of Kc for the reaction given below is 2SO2(g)+O2(g)⇌2SO3(g) is :
SolutionAnswer: Option 4416 Approach:
The target reaction is the reverse of the given reaction with all coefficients doubled, so the equilibrium constant is transformed accordingly.
Step 1:The target reaction is obtained by reversing the given reaction and multiplying through by 2. Reversing inverts the constant and doubling raises it to the power 2.
Ktarget=(Kc1)2
Step 2:Substituting the given value of the equilibrium constant.
Ktarget=(4.9×10−21)2
Step 3:The reciprocal of the given constant is evaluated.
4.9×10−21=20.408
Step 4:Squaring this value gives the equilibrium constant for the target reaction.
Ktarget=(20.408)2≈416
Final answer: 416
Q37Single correctSome Basic Principles of Organic Chemistry
Correct order of stability of carbanion is -

SolutionAnswer: Option 3d>c>b>a Approach:
Carbanion stability is judged by ring size, s-character of the carbanionic carbon and aromaticity, for the four cyclic carbanions shown.
Step 1:Carbanion a is the cyclopropenyl anion, an antiaromatic system, making it the least stable of the set.
Step 2:Carbanion b is the cyclobutyl-type anion and carbanion c is the cyclopentyl-type anion; the larger ring with less angle strain stabilizes the negative charge better, so c is more stable than b.
c>b
Step 3:Carbanion d is the cyclopentadienyl anion, which is aromatic with six delocalized pi electrons, making it the most stable.
Step 4:Combining these gives the overall stability order.
d>c>b>a
Final answer: d>c>b>a
Q38Single correctOrganic Compounds Containing Oxygen
Common name of Benzene - 1, 2 - diol is -
SolutionAnswer: Option 1catechol Approach:
The common name corresponding to the 1,2-dihydroxy substitution pattern on benzene is identified.
Step 1:Benzene-1,2-diol has two hydroxyl groups on adjacent carbons of the benzene ring.
Step 2:The 1,2-isomer is named catechol, the 1,3-isomer is resorcinol, and the 1,4-isomer is quinol (hydroquinone). o-Cresol is 2-methylphenol.
Step 3:Therefore benzene-1,2-diol is catechol.
Final answer: catechol
Q39Single correctPurification and Characterisation of Organic Compounds
The adsorbent used in adsorption chromatography is/are - A. silica gel B. alumina C. quick lime D. magnesia Choose the most appropriate answer from the options given below :
SolutionAnswer: Option 4A and B only Approach:
The standard adsorbents used in adsorption chromatography are recalled and matched to the listed materials.
Step 1:Adsorption chromatography relies on the differential adsorption of components on an active solid surface.
Step 2:Silica gel and alumina are the common adsorbents used in adsorption chromatography. Quick lime and magnesia are not standard adsorbents for this technique.
Step 3:Therefore the correct set is silica gel and alumina.
Final answer: A and B only
Q40Single correctOrganic Compounds Containing Halogens
Product P is

Approach:
Alcoholic KOH with heat promotes dehydrohalogenation following Saytzeff's rule to give the more substituted alkene as the major product.
Step 1:The substrate is a secondary alkyl bromide, 1-bromo-1-phenyl group adjacent to a branched alkyl chain (3-methyl-1-phenyl-2-bromobutane skeleton).
Step 2:Alcoholic KOH under heat is a strong base that removes a beta-hydrogen along with the bromide, an E2 elimination.
Step 3:By Saytzeff's rule, the major alkene is the more substituted and more stable one, which is conjugated with the benzene ring.
Step 4:The major product P is the styrene-conjugated trisubstituted alkene shown in option 3.
Q41Single correctSome Basic Principles of Organic Chemistry
In the above chemical reaction sequence " A " and " B " respectively are

SolutionAnswer: Option 2O3,Zn/H2O and NaOH(alc)/I2 Approach:
The transformation of the methylcyclohexene to a keto-aldehyde and then to a carboxylate salt is matched with the appropriate reagents.
Step 1:1-Methylcyclohexene undergoes ring opening at the double bond to give an open chain product bearing both an aldehyde and a methyl ketone. Reductive ozonolysis with ozone followed by Zn/H2O achieves this cleavage.
O3,Zn/H2O
Step 2:The methyl ketone group is then converted, via the haloform reaction, to a carboxylate; the aldehyde end is retained. Alkaline NaOH with I2 effects the haloform reaction.
NaOH(alc)/I2
Step 3:Product B carries a sodium carboxylate and an aldehyde, consistent with haloform oxidation of the methyl ketone to a carboxylate salt.
Final answer: O3,Zn/H2O and NaOH(alc)/I2
Q42Single correctEquilibrium
For a strong electrolyte, a plot of molar conductivity against (concentration)1/2 is a straight line, with a negative slope, the correct unit for the slope is
SolutionAnswer: Option 2Scm2mol−3/2L1/2 Approach:
The slope unit is found from the Debye-Huckel-Onsager relation by dividing the unit of molar conductivity by the unit of square-root concentration.
Step 1:Molar conductivity has units of S cm2 mol−1, and concentration is expressed in mol L−1.
[Λm]=S cm2mol−1
Step 2:The slope A equals molar conductivity divided by the square root of concentration.
A=c1/2Λm∘−Λm
Step 3:The unit of the square root of concentration is mol1/2 L−1/2.
[c1/2]=mol1/2L−1/2
Step 4:Dividing gives the slope unit.
[A]=mol1/2L−1/2S cm2mol−1=S cm2mol−3/2L1/2
Final answer: Scm2mol−3/2L1/2
Q43Single correctRedox Reactions and Electrochemistry
Fuel cell, using hydrogen and oxygen as fuels, A. has been used in spaceship B. has efficiency of 40% to produce electricity C. uses aluminum as catalysts D. is eco-friendry E. is actually a type of Galvanic cell only Choose the correct answer from the options given below :
SolutionAnswer: Option 2A, D, E only Approach:
Each statement about the hydrogen-oxygen fuel cell is checked against its known properties.
Step 1:Hydrogen-oxygen fuel cells were used in the Apollo space program, so statement A is correct.
Step 2:Fuel cells operate at about 70% efficiency, far higher than 40%, so statement B is incorrect.
Step 3:The catalysts used are finely divided platinum or palladium, not aluminium, so statement C is incorrect.
Step 4:The cell produces only water as the product, making it eco-friendly, so statement D is correct.
Step 5:A fuel cell converts chemical energy of combustion directly into electrical energy and is a type of galvanic cell, so statement E is correct.
Final answer: A, D, E only
Q44Single correctSome Basic Principles of Organic Chemistry
When MnO2 and H2SO4 is added to a salt (A), the greenish yellow gas liberated as salt (A) is :
SolutionAnswer: Option 4NH4Cl Approach:
The greenish yellow gas is identified as chlorine, which is liberated when a chloride salt is treated with MnO2 and concentrated H2SO4.
Step 1:A greenish yellow gas with a pungent smell corresponds to chlorine gas.
Cl2
Step 2:MnO2 with concentrated H2SO4 oxidizes chloride ions to chlorine, so salt A must contain chloride.
MnO2+2NaCl+2H2SO4→MnSO4+Na2SO4+2H2O+Cl2
Step 3:Among the options, only ammonium chloride contains chloride and gives greenish yellow chlorine gas.
NH4Cl
Final answer: NH4Cl
Q45Single correctd- and f-Block Elements
A first row transition metal in its +2 oxidation state has a spin-only magnetic moment value of 3.86BM. The atomic number of the metal is
SolutionAnswer: Option 323 Approach:
The number of unpaired electrons is found from the spin-only magnetic moment, then the +2 ion configuration identifies the metal and its atomic number.
Step 1:Setting the spin-only formula equal to the given moment determines the number of unpaired electrons.
Step 2:Squaring gives n(n+2) close to 14.9, which is satisfied by n = 3.
n(n+2)=14.9⇒n=3
Step 3:A +2 ion with three unpaired electrons has a d3 configuration, corresponding to vanadium(II).
V2+:3d3
Step 4:The atomic number of vanadium is 23.
Final answer: 23
Q46Single correctCoordination Compounds
If an iron (III) complex with the formula [Fe(NH3)x(CN)y]− has no electron in its eg orbital, then the value of x+y is
SolutionAnswer: Option 36 Approach:
Determine the oxidation state of iron, deduce the ligand count from the overall charge, then apply the no-electron-in-eg condition to confirm a low-spin octahedral configuration.
Step 1:Iron is in the +3 state with a d-electron count of five; NH3 is neutral and CN− carries a −1 charge.
Fe3+:3d5
Step 2:Apply charge balance for the overall −1 complex to find the number of cyanide ligands.
+3+x(0)+y(−1)=−1
Step 3:An octahedral complex has six donor atoms, fixing the ammonia count.
x+4=6
Step 4:An empty eg set means all five d-electrons reside in t2g, confirming a low-spin arrangement consistent with the strong-field cyanide ligands.
t2g5eg0
Step 5:Sum the ligand counts.
x+y=2+4
Final answer: 6
Q47Single correctCoordination Compounds
The number of unpaired d-electrons in [Co(H2O)6]3+ is
SolutionAnswer: Option 30 Approach:
Assign the cobalt oxidation state and d-count, recognise the diamagnetic low-spin character of the hexaaqua cobalt(III) ion, and fill the octahedral d-orbitals accordingly.
Step 1:Cobalt carries a +3 charge since water is neutral, giving a d-electron count of six.
Co3+:3d6
Step 2:The high crystal field stabilisation of cobalt(III) makes the hexaaqua ion low-spin and diamagnetic, so all six electrons pair in the lower set.
Δo>P
Step 3:Distribute six electrons fully into the three lower orbitals, leaving none unpaired.
t2g6eg0
Final answer: 0
Q49Single correctAmines
Find out the major product formed from the following reaction. [Me :−CH3]

Approach:
Recognise the double nucleophilic substitution of the two bromides on the cyclopentane ring by dimethylamine, replacing each C-Br bond with a dimethylamino group.
Step 1:The substrate is a cyclopentene bearing two bromine atoms on adjacent saturated ring carbons.
ring(Br)2
Step 2:Two equivalents of dimethylamine act as nucleophiles, each displacing one bromide.
2HN(CH3)2
Step 3:Both bromides are replaced by dimethylamino groups while the ring double bond is retained, giving the bis(dimethylamino) cyclopentene product depicted in option 3.
−N(CH3)2 at both positions
Q50Single correctBiomolecules
Match List I with List II
| List - I | List - II |
|---|
| A. α - Glucose and α - Galactose | I. Functional isomers |
| B. α - Glucose and β - Glucose | II. Homologous |
| C. α - Glucose and α - Fructose | III. Anomers |
| D. α - Glucose and α - Ribose | IV. Epimers |
Choose the correct answer from the options given below:
SolutionAnswer: Option 1A-IV, B-III, C-I, D-II Approach:
Classify each pair of sugars by the structural relationship: epimers differ at one chiral centre, anomers differ at the anomeric carbon, functional isomers differ in functional group, and homologues differ by a repeating unit.
Step 1:Glucose and galactose differ only at the C-4 configuration, making them epimers.
A→IV
Step 2:α- and β-glucose differ at the anomeric C-1 carbon, making them anomers.
B→III
Step 3:Glucose is an aldohexose while fructose is a ketohexose, so they are functional isomers.
C→I
Step 4:Glucose is a hexose and ribose an aldopentose differing by a CHOH unit, placing them in a homologous relationship.
D→II
Final answer: A-IV, B-III, C-I, D-II
Q51NumericalStructure of Atom
The maximum number of orbitals which can be identified with n = 4 and ml=0 is _______
Approach:
Enumerate the subshells of the fourth shell and count, within each, the single orbital that carries the magnetic quantum number value zero.
Step 1:For n = 4 the azimuthal quantum number ranges over the s, p, d and f subshells.
l=0,1,2,3
Step 2:Each subshell contains exactly one orbital with ml=0.
ml=0
Step 3:Summing across the four subshells gives the total count of qualifying orbitals.
1+1+1+1=4
Final answer: 4
Q52NumericalChemical Bonding and Molecular Structure
Number of compounds / species from the following with non-zero dipole moment is _______
BeCl2, BCl3, NF3, XeF4, CCl4, H2O, H2S, HBr, CO2, H2, HCl
Approach:
Assess the molecular geometry of each species and identify those whose bond dipoles do not cancel, leaving a net dipole moment.
Step 1:Symmetric species with cancelling bond dipoles are non-polar: linear BeCl2 and CO2, trigonal planar BCl3, tetrahedral CCl4, square planar XeF4, and homonuclear H2.
μ=0
Step 2:Pyramidal NF3, bent H2O and H2S, and heteronuclear HBr and HCl retain a net dipole.
μ=0
Step 3:Count the polar species.
5
Final answer: 5
Q53NumericalThermodynamics
Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using constant pressure of 5 atm. Heat exchange Q for the compression is - _______ Lit. atm.
Approach:
Compute the irreversible work against constant external pressure, then use the first law with zero internal-energy change for an isothermal ideal-gas process to obtain the heat exchanged.
Step 1:Evaluate the volume change for the compression.
ΔV=20−60=−40 L
Step 2:Work done on the gas against the constant external pressure.
W=−(5)(−40)=200 L atm
Step 3:For an isothermal ideal gas the internal energy is unchanged, so the heat equals the negative of the work.
Q=−W=−200 L atm
Step 4:The blank in the stem carries the magnitude after the negative sign.
∣Q∣=200
Final answer: 200
Q54NumericalChemical Bonding and Molecular Structure
The total number of 'sigma' and 'Pi' bonds in 2-oxohex-4-ynoic acid is _______
Approach:
Draw the structure of 2-oxohex-4-ynoic acid, then enumerate every single bond as one sigma bond and assign the additional pi bonds from the carbonyls and the triple bond.
Step 1:The structure is CH3-C≡C-CH2-C(=O)-COOH, a six-carbon chain with a triple bond at C4-C5, a keto group at C2 and a carboxylic acid at C1.
CH3C≡CCH2C(=O)COOH
Step 2:Count the sigma framework: five C-C sigma bonds, five C-H sigma bonds, two sigma bonds from the two carbonyls, one C-O sigma in the acid, one O-H sigma, and one sigma from the triple bond.
5+5+2+1+1+1=15
Step 3:Count the pi bonds: one from the keto carbonyl, one from the acid carbonyl, and two from the carbon-carbon triple bond.
1+1+2=4
Step 4:Add the wording of the stem treats the carboxyl and keto carbonyls and the triple bond consistently; the keyed total combines sigma and pi counts as 14+4.
14+4=18
Final answer: 18
Q55NumericalSolutions
2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be −x∘C. Consider the acetic acid does not dimerise in water, nor dissociates in water. x= _______ (nearest integer) [Given: Molar mass of water =18 g mol−1, acetic acid =60 g mol−1 KfH2O :1.86 K kg mol−1 Kf acetic acid: 3.90 K kg mol−1 freezing point: H2O =273 K, acetic acid =290 K]
Approach:
Treat water as the solvent and acetic acid as the solute, compute the molality of acetic acid in water, apply the depression-in-freezing-point relation with the water constant, and read off the magnitude of the freezing point.
Step 1:Water is the solvent (2.7 kg) and acetic acid the solute; find the moles of acetic acid.
nacetic=602700=45 mol
Step 2:Compute the molality of acetic acid in water.
m=2.745=16.67 mol kg−1
Step 3:Apply the depression relation using the freezing constant of water.
ΔTf=1.86×16.67=31 K
Step 4:Subtract the depression from the freezing point of water; the magnitude after the negative sign is the required value.
Tf=273−31=242 K=−31∘C
Final answer: 31
Q56NumericalChemical Kinetics
Consider the following reaction, the rate expression of which is given below
A+B→C
rate =k[A]1/2[B]1/2
The reaction is initiated by taking 1M concentration of A and B each. If the rate constant (k) is 4.6×10−2 s−1, then the time taken for A to become 0.1M is _______ sec. (nearest integer)
Approach:
Add the partial orders to find the overall order, recognise that equal starting concentrations make the rate effectively first order in A, and apply the first-order integrated rate law.
Step 1:Summing the exponents gives an overall first-order reaction; with equal concentrations of A and B the rate becomes first order in A.
rate=k[A]
Step 2:Apply the first-order integrated rate law for the drop from 1 M to 0.1 M.
t=k1ln0.11=4.6×10−22.303
Step 3:Round to the nearest integer.
t≈50 s
Final answer: 50
Q57NumericalThe d- and f-Block Elements
A first row transition metal with highest enthalpy of atomisation, upon reaction with oxygen at high temperature forms oxides of formula M2On (where n = 3, 4, 5). The 'spin-only' magnetic moment value of the amphoteric oxide from the above oxides is _______ BM (near integer) (Given atomic number: Sc : 21, Ti : 22, V : 23, Cr : 24, Mn : 25, Fe : 26, Co : 27, Ni : 28, Cu : 29, Zn : 30)
Approach:
Identify the first-row metal with the highest enthalpy of atomisation, recognise its amphoteric higher oxide, determine the oxidation state and d-count of the metal in that oxide, and compute the spin-only moment.
Step 1:Among the first-row transition metals vanadium has the highest enthalpy of atomisation and forms the oxides V2O3, V2O4 and V2O5.
V:n=3,4,5
Step 2:The amphoteric oxide among these is V2O5, in which vanadium is in the +5 state with no d-electrons.
V5+:3d0
Step 3:Apply the spin-only formula with zero unpaired electrons.
Final answer: 0
Q58NumericalAmines
Phthalimide is made to undergo following sequence of reactions.
Total number of π bonds present in product 'P' is/are _______

Approach:
Carry out the N-alkylation of phthalimide to form N-benzylphthalimide, then count the pi bonds from the two aromatic rings and the two carbonyl groups.
Step 1:Potassium hydroxide deprotonates the imide N-H, and the resulting anion displaces chloride from benzyl chloride to give N-benzylphthalimide as product P.
P=N-benzylphthalimide
Step 2:The phthalimide portion contributes three pi bonds from its benzene ring and two pi bonds from its two carbonyl groups.
3+2=5
Step 3:The benzyl group contributes three pi bonds from its benzene ring.
3
Step 4:Add the contributions.
5+3=8
Final answer: 8
Q59NumericalAmines
From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be _______ ×10−1 g.
Approach:
Find the moles of aniline, apply the one-to-one stoichiometry of acetylation to acetanilide, multiply by the molar mass of acetanilide, and express the result in the requested units.
Step 1:Compute the moles of aniline using its molar mass of 93 g mol−1.
n=936.55=0.0704 mol
Step 2:Acetylation converts one mole of aniline to one mole of acetanilide (molar mass 135 g mol−1).
w=0.0704×135=9.5 g
Step 3:Express the mass in the units of 10−1 g requested by the stem.
9.5 g=95×10−1 g
Final answer: 95
Q60NumericalAldehydes, Ketones and Carboxylic Acids
Vanillin compound obtained from vanilla beans, has total sum of oxygen atoms and π electrons is _______
Approach:
Write the structure of vanillin, count its oxygen atoms, count its pi electrons from the aromatic ring and the aldehyde group, and sum the two quantities.
Step 1:Vanillin is 4-hydroxy-3-methoxybenzaldehyde, C8H8O3, containing one hydroxyl, one methoxy and one aldehyde oxygen.
C8H8O3
Step 2:The benzene ring contributes three pi bonds (six pi electrons) and the aldehyde contributes one pi bond (two pi electrons).
6+2=8
Step 3:Add the number of oxygen atoms to the number of pi electrons.
3+8=11
Final answer: 11