datavorous/entrance-exam-dataset huggingface dataset:
The incorrect statement about B2H6 is :

Options:
A All B-H-B angles are of 120°
B The two B-H-B bonds (bridged and terminal) are not of same length.
C Terminal B-H bonds have less p -character when compared to bridging bonds.
D Its fragment, BH3, behaves as a Lewis acid.

The correct answer is:
All B-H-B angles are of 120°

Solution:
θ2>θ1, ∴B-H (terminal) having less p
character as compare to bridge bond.

Both B-H-B (bridge bond and terminal bond) do not have same bond length.

B-H-B bond angle is ≈90°

BH3 is electron deficient species and therefore acts as Lewis acid

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2021 (25 Feb Shift 1)
The incorrect statement about the oxoacids, HClO4 and ClO , is

Options:
A The central atom in both HClO4 and HClO is sp3 hybridized
B HClO4 is formed in the reaction between Cl2 and H2O
C The conjugate base of HClO4 is weaker base than H2O
D HClO4 is more acidic than HClO because of the resonance stabilization of its anion

The correct answer is:
HClO4 is formed in the reaction between Cl2 and H2O

Solution:
Cl2+H2O→HCl+HOCl
Acidic: HClO4acid-1>H3O+acid-2
Basic: ClO4-C.B.-1<H2OC.B.-2
Resonance stabilization

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The incorrect statement about the property of a Zener diode is:

Options:
A Zener voltage remains constant at breakdown
B It is designed to operate under reverse bias
C Depletion region formed is very wide
D $\quad p$ and $n$ regions of zener diode are heavily doped

The correct answer is:
Depletion region formed is very wide

Solution:
Zener diode is highly doped $p$ - $n$ junction, and zener voltage is constant. Its depletion region is very thin.

Topic: Physics, Semiconductors, NEET, NEET 2022 (Phase 2)
The incorrect statement among the following is

Options:
A α-D-glucose and β-D- glucose are enantiomers.
B

the penta acetate of glucose does not react with hydroxyl amine.

C α-D-glucose and β-D-glucose are anomers.
D cellulose is a straight chain polysaccharide made up of only β-D-glucose units.

The correct answer is:
α-D-glucose and β-D- glucose are enantiomers.

Solution:
α-D-Glucose and β-D-glucose are anomer not enantiomer.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2017 (09 Apr Online)
The incorrect statement among the following is:

Options:
A Cr2O3 is an amphoteric oxide
B VOSO4 is a reducing agent
C RuO4 is an oxidizing agent
D Red colour of ruby is due to the presence of Co3+

The correct answer is:
Red colour of ruby is due to the presence of Co3+

Solution:
(i) InVOSO4,V'' is in +4 oxidation state.
So it act as oxidising agent.
(ii) Cr2O3 is an amphoteric oxide.
(iii) InRuO4, 'Ru' is in +8 oxidation state. So it act as oxidising agent.
(iv) Red colour of ruby is due to the presence of Cr+3 ions in Al2O2.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The incorrect statement among the following is

Options:
A α-D-glucose and β-D- glucose are enantiomers.
B

the penta acetate of glucose does not react with hydroxyl amine.

C α-D-glucose and β-D-glucose are anomers.
D cellulose is a straight chain polysaccharide made up of only β-D-glucose units.

The correct answer is:
α-D-glucose and β-D- glucose are enantiomers.

Solution:
α-D-Glucose and β-D-glucose are anomer not enantiomer.

Topic: Chemistry, Biomolecules, JEE Main
The incorrect statement among the following is:

Options:
A The colour of [Ni(H2O)6]2+ changes on adding ethylene diamine.
B Crystal field theory consider the covalent character of bonding between ligand and central atoms.
C The magnitude of CFSE of octahedral complexes is higher than that of corresponding tetrahedral complexes.
D [Co(C2O4)3]3– does not contain unpaired electrons.

The correct answer is:
Crystal field theory consider the covalent character of bonding between ligand and central atoms.

Solution:
Crystal field theory considers the interaction between ligand and central metal atom/ion to be purely electrostatic.

Topic: Chemistry, Coordination Compounds, JEE Main
The incorrect statement from the following for borazine is:

Options:
A It contains banana bonds
B

It can react with water

C

It is a cyclic compound

D

It has electronic delocalization

The correct answer is:
It contains banana bonds

Solution:
The statement "It contains a banana bond" is incorrect about Borazine.

Borazine, also known as inorganic benzene or borazole, is a chemical compound with the formula B3N3H6. Borazine is a hexagonal planar molecule with alternating boron and nitrogen atoms, and it exhibits some structural and chemical similarities to benzene. Borazine is an aromatic compound with electron delocalization. It does not contain banana bond.

On heating, borazine hydrolyses slowly.B3 N3H6+9H2O→3NH3+3H3BO3+3H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2023 (13 Apr Shift 1)
The incorrect statement is:

Options:
A Cu2+ ion gives chocolate coloured precipitate with potassium ferrocyanide solution
B Cu2+ and Ni2+ ions give black precipitate with H2S in presence of HCl solution
C Ferric ion gives blood red colour with potassium thiocyanate
D Cu2+ salts give red coloured borax bead test in reducing flame

The correct answer is:
Cu2+ and Ni2+ ions give black precipitate with H2S in presence of HCl solution

Solution:
Cu+2 gives black precipitate in presence of dil. HCl Ni+2 ions gives black precipitate in presence of NH4Cl +NH3
with HCl S2- concentration is not just sufficient to overcome the solubility product of NiS.
Cu+2+H2S→dil. HClCu2S↓(black)+2H+
Ni+2+H2S→NH4ClNH4OHNiS↓(black)+2H+

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2018 (16 Apr Online)
The incorrect statement is.

Options:
A The spin-only magnetic moments of FeH2O62+ and CrH2O62+ are nearly similar.
B The spin-only magnetic moment of NiNH34H2O22+ is 2.83 BM.
C The gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.
D The color of CoClNH352+ is violet as it absorbs the yellow light.

The correct answer is:
The gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.

Solution:
A FeH2O62+, Fe2+→3d6→4 unpaired electron
CrH2O62+, Cr2+→3d4→4 unpaired electron
B NiNH34H2O22+=Ni2+→3d8
→2 unpaired electron
μm=2.83 B.M
C In gemstone, ruby has Cr3+ ion occupying the octahedral sites of aluminium oxide Al2O3 normally occupied by Al3+ ion.
D Complimentary color of violet is yellow.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The incorrect statement is:

Options:
A Manganate and permanganate ions are tetrahedral
B In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of maganese
C Manganate and permanganate ions are paramagnetic
D Manganate ion is green in colour and permanganate ion is purple in colour

The correct answer is:
Manganate and permanganate ions are paramagnetic

Solution:
Manganate

MnO42-

Paramagnetic, green in colour,
Tetrahedral & contains pπ-dπ bond

Permanganate

MnO4Θ

Diamagnetic, purple in colour,
Tetrahedral & contains pπ-dπ bond

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The INCORRECT statement is :

Options:
A bronze is an alloy of copper and tin.
B cast iron is used to manufacture wrought iron.
C german silver is an alloy of zinc, copper and nickel.
D brass is an alloy of copper and nickel.

The correct answer is:
brass is an alloy of copper and nickel.

Solution:
S.No. Alloy Composition 1. Bronze Cu(75-90%)+Sn(10,25%)2. Brass Cu(60-80%)+Zn(20-40%)3. GermanSilver Cu(50-62%)+Zn(17-19%)+Ni(21-30%)

Cast iron is used to manufacture wrought iron.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The incorrect statement is:

Options:
A F2 is a stronger oxidizing agent than Cl2 in aqueous solution.
B On hydrolysis ClF forms HOCl and HF.
C Cl2 is more reactive than ClF.
D F2 is more reactive than ClF.

The correct answer is:
Cl2 is more reactive than ClF.

Solution:
In general, interhalogen compounds are more reactive than halogens (except fluorine). This is because X-X' bond in interhalogens is weaker than X-X bond in halogens except F-F bond. All these undergo hydrolysis giving halide ion derived from the smaller halogen and a hypohalite (when XX'), halite (when XX3'), halate (when XX5') and perhalate (whenXX7' anion derived from the larger halogen.

XX'+H2O→HX'+HOX

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (26 Aug Shift 1)
The incorrect statement is

Options:
A The first ionization enthalpy of K is less than that of Na and Li
B Xe does not have the lowest first ionization enthalpy in its group
C The first ionization enthalpy of element with atomic number 37 is lower than that of the element with atomic number 38.
D The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

The correct answer is:
The first ionization enthalpy of Ga is higher than that of the d-block element with atomic number 30.

Solution:
Ionisation enthalpy of alkali metals decreases down the group from Li to Cs. This is because of increase in size which outweighs the increasing nuclear charge, and the outermost electron is very well screened from the nuclear charge.

Li>Na>K

Due to stable electronic configuration noble gases exhibit very high ionisation enthalpy. However, it decreases down the group with increase in atomic size.

He>Ne>Ar>Kr>Xe>Rn

Alkaline earths have higher ionization enthalpy values than alkali metals. This is because the atomic size of alkaline earths is smaller than that of alkali metals.

Sr>Rb

Because of completely filled configuration, zinc is stable and requires higher ionization energy whereas gallium is yet to fill its p-orbitals, so it has lower ionization energy compared to zinc.

Zn>Ga

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2022 (27 Jul Shift 1)
The incorrect statement is

Options:
A Be2+ cation has largest hydration energy among alkaline earth metal
B The second ionisatio enthalpies of alkaline earth metal are smaller than those of the corresponding alkali metals
C Li is the strongest reducing agent among all elements
D LiCl and MgCl2 are most ionic in their groups

The correct answer is:
LiCl and MgCl2 are most ionic in their groups

Solution:
As LiCl and MgCl2 are least ionic in their groups so option D is incorrect. Rest other statements are correct here.

Topic: Chemistry, s Block Elements, JEE Main
The incorrect statement regarding above reactions is

Options:
A \( \mathrm{Al} \) shows amphoteric character
B Gas \( P \) and \( Q \) are different
C Both \( \mathrm{X} \) and \( \mathrm{Y} \) are water soluble
D Gas \( \mathrm{Q} \) in inflammable

The correct answer is:
Gas \( P \) and \( Q \) are different

Solution:
These reactions represent amphoteric nature of aluminium and both P and Q are hydrogen and the salts X(AlCl3) and Y(Na[Al(OH)4] are water-soluble. Hydrogen is flammable gas

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The incorrect statement regarding above reactions is

Options:
A Al shows amphoteric character
B Gas 'P' and 'Q' are different
C Both X and Y are water soluble
D Gas Q in inflammable

The correct answer is:
Gas 'P' and 'Q' are different

Solution:
These reactions represents amphoteric nature of aluminium and both P and Q are hydrogen and the salts X(AlCl3) and Y(Na[Al(OH)4] are water soluble. Hydrogen is flammable gas

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The incorrect statement regarding conformations of ethane is:

Options:
A Ethane has infinite number of conformations
B The dihedral angle in staggered conformation is 60°
C Eclipsed conformation is the most stable conformation.
D The conformations of ethane are interconvertible to one-another

The correct answer is:
Eclipsed conformation is the most stable conformation.

Solution:
In staggered conformation arrangement, the hydrogens of the two carbon atoms are staggered with respect to one another. As a result, they are at maximum distance apart and have minimum repulsion between the hydrogen atoms attached to the two carbon atoms.

In eclipsed conformation, the hydrogens of one carbon atom are directly behind those of the other. Consequently, the repulsion in between these hydrogen atoms is maximum. Therefore, staggered conformation of ethane is more stable than eclipsed conformation.

Eclipsed conformation is the least stable conformation of ethane.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The incorrect statement regarding ethyne is

Options:
A The \(\mathrm{C}-\mathrm{C}\) bonds in ethyne is shorter than that in ethene
B Both carbons are sp hybridised
C Ethyne is linear
D The corbon - carbon bonds in ethyne is weaker than that in ethene

The correct answer is:
The corbon - carbon bonds in ethyne is weaker than that in ethene

Solution:
The carbon-carbon bonds in ethyne is stronger than that in ethene.
$(\mathrm{H}-\mathrm{C} \equiv \mathrm{C}-\mathrm{H})$ Ethyne is linear and carbon atoms are SP hybridised.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (09 Apr Shift 2)
The incorrect statement regarding S N 2 reaction is

Options:
A The reaction is favoured by strong nucleophile
B An enantiomer of the reactant is formed
C A single stereoisomer is formed during the reaction
D The group which stabilize transition state increase the rate of reaction

The correct answer is:
An enantiomer of the reactant is formed

Solution:
Product is not always an enantiomer. It may be a completely different compound.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The incorrect statement regarding the geometrical isomers of 2-butene is :

Options:
A cis-2-butene and trans-2-butene are not interconvertible at room temperature.
B cis-2-butene and trans-2-butene are stereoisomers.
C cis-2-butene has less dipole moment than trans-2-butene.
D trans-2-butene is more stable than cis-2-butene.

The correct answer is:
cis-2-butene has less dipole moment than trans-2-butene.

Solution:
Cis-but-2-ene has higher Dipole moment than trans-but-2-ene.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The incorrect statement regarding the given structure is

Options:
A can be oxidized to a dicarboxylic acid with \(\mathrm{Br}_2\) water
B will coexist in equilibrium with 2 other cyclic structure
C despite the presence of - \(\mathrm{CHO}\) does not give Schiff's test
D has 4 asymmetric carbon atom

The correct answer is:
can be oxidized to a dicarboxylic acid with \(\mathrm{Br}_2\) water

Solution:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The INCORRECT statement(s) about heavy water is (are)
(A) used as a moderator in nuclear reactor
(B) obtained as a by-product in fertilizer industry.
(C) used for the study of reaction mechanism
(D) has a higher dielectric constant than water
Choose the correct answer from the options given below:

Options:
A (B) only
B (C) only
C (D) only
D (B) and (D) only

The correct answer is:
(D) only

Solution:
The dielectric constant of H2O is greater than heavy water.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The incorrect statement(s) among (a) - (c) is (are) :

(a) W(VI) is more stable than Cr(VI).

(b) in the presence of HCl, permanganate titrations provide satisfactory results.

(c) some lanthanoid oxides can be used as phosphors

Options:
A (b) and (c) only
B (a) and (b) only
C (b) only
D (a) only

The correct answer is:
(b) only

Solution:
(a) In transition metals on moving down the group higher oxidation states are more stable due to smaller size of atoms, due to lanthanide and actinide contractions.

(b) KMnO4 can oxidise chloride into chlorine, so it will give incorrect results

(c) its a fact

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The incorrect statement(s) among (a) – (d) regarding acid rain is (are):

a) It can corrode water pipes

(b) It can damage structures made up of stone.

(c) It cannot cause respiratory ailments in animals

(d) It is not harmful for trees

Options:
A (a), (c) and (d)
B (c) only
C (a), (b) and (d)
D (c) and (d)

The correct answer is:
(c) and (d)

Solution:
Acid rain is harmful for trees and plants. It causes breathing problem in human being and animals.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The INCORRECT statements below regarding colloidal solutions is :

Options:
A A colloidal solution shows colligative properties.
B An ordinary filter paper can stop the flow of colloidal particles.
C The flocculation power of Al3+ is more than that of Na+
D A colloidal solution shows Brownian motion of colloidal particles.

The correct answer is:
An ordinary filter paper can stop the flow of colloidal particles.

Solution:
* Colloidal solution exhibits colligative properties

* An ordinary filter can not stop the flow of colloidal particles.

* Flocculation power increases with increase in the opposite charge of electrolyte.

* Colloidal particles show Brownian motion.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 2)
The incorrect statements regarding enzymes are :
(A) Enzymes are biocatalysts.
(B) Enzymes are non-specific and can catalyse different kinds of reactions.
(C) Most Enzymes are globular proteins.
(D) Enzyme - oxidase catalyses the hydrolysis of maltose into glucose.
Choose the correct answer from the option given below :

Options:
A (B), (C) and (D)
B (B) and (D)
C (A), (B) and (C)
D (B) and (C)

The correct answer is:
(B) and (D)

Solution:
Direct NCERT Based

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (06 Apr Shift 2)
The increase in osmolarity from outer to inner medullary interstitium is maintained due to :
(i) Close proximity between Henle's loop and vasa recta
(ii) Counter current mechanism
(iii) Selective secretion of HCO3- and hydrogen ions in PCT
(iv) Higher blood pressure in glomerular capillaries

Options:
A Only(ii)
B (iii) and (iv)
C (i), (ii) and (iii)
D (i) and (ii)

The correct answer is:
(i) and (ii)

Solution:
In the outer part of the medulla, the osmotic concentration of interstitial fluid is 300 mosmol/litre. It gradually increases and reaches a value of 1200 mosmol/litre in the inner part of the medulla. This gradient is maintained for the concentration of urine as it passes through the collecting tubule. The vasa recta contribute to the maintenance of the hyperosmotic condition of ECF in the medulla by a mechanism known as counter-current exchange. The counter-current mechanism works by exchange of water and electrolytes between Loop of Henle and vasa recta.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2020 (Phase 2)
The increase in pressure required to decrease the $200 \mathrm{~L}$ volume of a liquid by $0.008 \%$ in $\mathrm{kPa}$ is (Bulk modulus of the liquid $=2100 \mathrm{MPa}$ is )

Options:
A $8.4$
B 84
C $92.4$
D 168

The correct answer is:
84

Solution:
Bulk modulus $K=\frac{\Delta p}{\Delta V} \cdot V$
$$
\begin{array}{l}
\Delta p=\frac{K \Delta V}{V} \\
\Delta p=\frac{2100 \times 10^{6} \times 0.008}{200}
\end{array}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The increase of pressure on ice ⇌water system at constant temperature will lead to:

Options:
A A decrease in the entropy of the system
B An increase in the Gibbs energy of the system
C A shift of the equilibrium in the forward direction
D No effect on the equilibrium

The correct answer is:
A shift of the equilibrium in the forward direction

Solution:
H2Os⇌H2O(l)
At melting point of H2O at (273 K = T)
dG = VdP - SdT
At constant temperature
d∆G=∆VdP=MρL-MρSdP
ρL=Density of Liq. H2O
ρS=Density of ice
d∆G=MρS - ρLρL ⋅ ρSdP
Integrating
∆G=MρS - ρLρL ⋅ ρS(P-1)
Pressure increased from 1 atm to P atm
ρS<ρL , P>1
∆G is -ve, so melting in forward direction spontaneous.
Method 2 :
Ice ⇌ Water
On Increasing P : Molecule Come closer & density increasing; water is formed;
Ice → Water;
Since S ice <   S water
∵Process From Ice to water is spontaneous;
ΔG<0

In physical equilibrium, on increasing pressure, equilibrium shift toward more density side.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2015 (11 Apr Online)
The increasing basicity order of the following compounds is:

Options:
A (D) < (C) < (B) < (A)
B (D) < (C) < (A) < (B)
C (A) < (B) < (C) < (D)
D (A) < (B) < (D) < (C)

The correct answer is:
(D) < (C) < (A) < (B)

Solution:
In case of polar protic solvents, two factors decide the basic character. The positive inductive effect from alkyl groups which increases basicity.
3o>2o>1o Amine
& the Steric Hindrance. 1o>2 >3o
Thus, the resultant basic character is determined by taking the above two factors into account.
So order will be 2o>1o>3o if methyl group is present as alkyl group otherwise order will be 2°>3°>1°.
In aromatic & Aliphatic: Aliphatic Amine are more basic than aromatic because the lone pair of N is involved in delocalization.
So basic character is in the order 2o>1o>3o>Aromatic Amine
So (B) > (A) > (C) > (D)

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The increasing order of Ag+ ion concentration in

I. Saturated solution of AgCl

II. Saturated solution of Agl

III. 1 M Ag NH 3 2 + in 0.1 M NH3

IV. 1 M Ag CN 2 - in 0.1 M KCN

Given :

K sp of AgCl = 1 . 0 Ã 1 0 - 1 0 ,

K sp of AgI = 1 . 0 Ã 1 0 - 1 6

K d of Ag NH 3 2 + = 1 . 0 Ã 1 0 - 8

K d of Ag CN 2 - = 1 . 0 Ã 1 0 - 2 1

Options:
A I < II < III < IV
B IV < III < II < I
C IV < II < III < I
D IV < II < I < III

The correct answer is:
IV < II < III < I

Solution:
I Ag + = K sp AgCl = 1 Ã 1 0 - 1 0 = 1 0 - 5 M
II Ag + = K sp AgI = 1 Ã 1 0 - 1 6 = 1 0 - 8 M
III Ag NH 3 2 + â Ag + + 2 NH 3
Kd=1Ã10-8=Ag+NH32Ag NH32aq+=Ag+0.121.0âAg+=1Ã10-6M
IV AgCN2(aq)-1âAgaq++2CNaq-1
K CN 0 . 1 M â¶ K aq + + CN aq - 1
(0.1M)
K d = 1 Ã 1 0 - 2 1 = Ag + CN - 2 Ag CN 2 - 1 = Ag + 0 . 1 2 1 . 0
[Ag+ ] = 1 x 10- 19 M.

Topic: Chemistry, Ionic Equilibrium, JEE Main
The increasing order of basicity among the following is

Options:
A Y < X < Z
B Y < Z < X
C X < Z < Y
D X < Y < Z

The correct answer is:
Y < X < Z

Solution:
Electron releasing substituent like -CH3-OCH3 increase the basicity of amines by dispersing positive charge of the anilium ion. However, their presence at ortho position decreases the basicity due to ortho effect.
Thus, is most basic and compound Y is least basic.

Topic: Chemistry, General Organic Chemistry, JEE Main
The increasing order of basicity for the following intermediates is (from weak to strong)

Options:
A iii<i<ii<iv<v
B v<i<iv<ii<iii
C v<iii<ii<iv<i
D iii<iv<ii<i<v

The correct answer is:
v<iii<ii<iv<i

Solution:
Basicity is inversely proportional to electronegativity.

Higher the electron affinity or core charge of an atom, the less likely it is to donate its electrons to a proton.
Greater the delocalization of electrons that could potentially donate to a proton, the less able they are to donate.
Greater the electron-withdrawing effects in another part of a molecule, the less likely the electrons on a particular atom are to donate.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The increasing order of basicity of the following compounds is

(1)

(2)

(3)

(4)

Options:
A (4) < (2) < (1) < (3)
B (1) < (2) < (3) < (4)
C (2) < (1) < (3) < (4)
D (2) < (1) < (4) < (3)

The correct answer is:
(2) < (1) < (4) < (3)

Solution:
3 is most basic because its conjugate acid is stabilized by equivalent resonance.
Out of 1 and 4, 4 is more basic due to +I effect. 2 is less basic than 1 and 4 because N atom is sp2 hybridized (more s%).

Topic: Chemistry, General Organic Chemistry, JEE Main