datavorous/entrance-exam-dataset huggingface dataset:
The complex number z=i-1cosπ3+isinπ3 is equal to:

Options:
A 2icos5π12-isin5π12
B cosπ12-isinπ12
C 2cosπ12+isinπ12
D 2cos5π12+isin5π12

The correct answer is:
2cos5π12+isin5π12

Solution:
Given,

z=i-1cosπ3+isinπ3

⇒z=2-12+i2cosπ3+isinπ3

Now argument of complex number -12+i2 is 3π4

So, by euler's formula eiθ=cosθ+isinθ we get,

⇒z=2ei3π/4eiπ/3

⇒z=2ei5π/12

⇒z=2cos5π12+isin5π12

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The complex number z, satisfying the equation z3=z- and arg⁡z+1=π4 simultaneously, is (where, i2=-1)

Options:
A i
B 1+2i
C 2+3i
D 3+4i

The correct answer is:
i

Solution:
z3=z-⇒z=0 or 1
z=0⇒z=0
z=1⇒z4=1
⇒z=±1,±1
Only z=i satisfies arg⁡z+1=π4

Topic: Mathematics, Complex Number, JEE Main
The complex $\left[\mathrm{PtCl} \mathrm{C}_{2}(\mathrm{en})_{2}\right]^{2+}$ ion shows

Options:
A structural isomerism only
B optical isomerism only
C geometrical and optical isomerism
D geometrical isomerism only

The correct answer is:
geometrical and optical isomerism

Solution:
$\left[\mathrm{PtCl}_{2}(\mathrm{en})_{2}\right]^{2+}$ complex forms geometrical isomers (cis and trans). Trans-isomer does not show optical isomerism, since it is symmetrical while cis-isomer shows optical isomerism as it is unsymmetrical.

Topic: Chemistry, Coordination Compounds, JEE Main
The complex that can show fac - and mer -isomers is:

Options:
A CoNH34Cl2+
B PtNH32Cl2
C CoCl2en2
D CoNH33NO23

The correct answer is:
CoNH33NO23

Solution:
Ma3b3 type complex shows facial and meridional isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The complex that can show \( f a c \) - and mer -isomers is:

Options:
A \( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+} \)
B \( \left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right] \)
C \( \left[\mathrm{CoCl}_{2}(\mathrm{en})_{2}\right] \)
D \( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right] \)

The correct answer is:
\( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{3}\left(\mathrm{NO}_{2}\right)_{3}\right] \)

Solution:
Ma3b3 type complex shows facial and meridional isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
The complex that exists as a pair of enantiomers is

Options:
A trans- CoH2NCH2CH2NH22Cl2+
B cis- CoNH34Cl2+
C Pt PPh 3 Cl Br CN −
D CoH2NCH2CH2NH233+

The correct answer is:
CoH2NCH2CH2NH233+

Solution:
Complexes of the type Maa3,Maa2b2,Maab2c2 exhibit optical isomerism and exist as a pair of enantiomers.
Thus, CoH2NCH2CH2NH233+ exists as a pair of enantiomers. The structure of its entiomers are as follows

Topic: Chemistry, Coordination Compounds, JEE Main
The complex that has highest crystal field splitting energy ∆ , is:

Options:
A K3CoCN6
B CoNH35Cl Cl2
C CoNH35H2OCl3
D K2CoCl4

The correct answer is:
K3CoCN6

Solution:
Strong field ligands have more CFSE value while weak field ligand have less CFSE value.
CN- is strongest ligand among all so K3CoCN6 has maximum CFSE value.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The complexes $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Cr}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$ and $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_6\right]\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$

Options:
A Linkage isomerism
B Geometrical isomerism
C Coordination isomerism
D Ionisation isomerism

The correct answer is:
Coordination isomerism

Solution:
Co-ordination isomerism is possible when both +ve and -ve ions of a salt are complex ions and the two isomers differ in the distribution of ligands in the cation and the anion

Topic: Chemistry, Coordination Compounds, JEE Main
The composite mapping $f \circ g$ of the map $f: R \rightarrow R, f(x)=\sin x, g: R \rightarrow R, g(x)=x^2$ is

Options:
A $\sin x+x^2$
B $(\sin x)^2$
C $\sin x^2$
D $\frac{\sin x}{x^2}$

The correct answer is:
$\sin x^2$

Solution:
$\begin{aligned} & f: R \rightarrow R, f(x)=\sin x \text { and } g: R \rightarrow R, g(x)=x^2 \\ & \Rightarrow f \circ g(x)=f(g(x))=f\left(x^2\right)=\sin x^2 .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
The composition of glomerular filtrate and urine is not same. Comment.

Options:

None

Solution:
Glomerular filtrate contains all the content of the blood plasma except proteins. About 180 litres of glomerular filtrate like water, glucose, nutrients ions etc. occurs. As a result, now the composition of urine is quite different from that of the glomerular filtrate. Some ions are also added to this fluid by tubules i.e. tubular secretion to maintain ionic and acid base balance of body fluids. Thus the composition of glomerular filtrate andd urine is not same.

Topic: Biology, Excretory Products and their Elimination
The composition of 'Golden spangles' is

Options:
A $\mathrm{PbCr}_4$
B $\mathrm{Pbl}_2$
C $\mathrm{As}_2 \mathrm{~S}_3$
D $\mathrm{BaCrO}_4$

The correct answer is:
$\mathrm{Pbl}_2$

Solution:
The correct option is $(2) \mathrm{Pbl}_2$
$\mathrm{Pbl}_2$ is yellow and is called as 'Golden spangles'. When a solution of $\mathrm{KI}$ is added to a solution containing $\mathrm{Pb}^{2+}$, $\mathrm{Pbl}_2$ is formed which is golden yellow in colour and it is insoluble at room temperature.

Topic: Chemistry, Practical Chemistry, JEE Main
The composition of the equilibrium mixture for the equilibrium Cl2⇌2Cl at 1470 K diffuses 1.16 times as fast as Krypton (84) under the same conditions. Find the degree of dissociation of Cl2 at equilibrium.

Options:
A 0.14
B 0.41
C 0.91
D 0.24

The correct answer is:
0.14

Solution:
Cl21-x⇌2Cl2x
Mmix=(1-x)71+2x×35.5(1-x)+2x=711+x
RmixRKr=MKrMmix
1.16= 84 71 1+x
On solving x=0.14

Topic: Chemistry, States of Matter, JEE Main
The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi$ results in the displacement of the particle along

Options:
A Straight line
B Circle
C Ellipse
D Figure of eight

The correct answer is:
Straight line

Solution:
If $y_1=a_1 \sin \omega t$ and $y_2=a_2 \sin (\omega t+\pi)$ $\Rightarrow \frac{y_1}{a_1}+\frac{y_2}{a_2}=0 \Rightarrow y_2=-\frac{a_2}{a_1} y_1$
This is the equation of straight line.

Topic: Physics, Oscillations, JEE Main
The composition of wustite is Fe0.95O1 . What is the percentage of iron as Fe3+ ions?

Options:
A 10.5
B 25
C 35
D 45

The correct answer is:
10.5

Solution:
Fe0.95O1
Let there are x ions of Fe3+ then 0.95-x ions will be Fe2+ . Compound is electrically neutral.
+3x+2(0.95-x)-2=0
3x+1.90-2x-2=0
x=0.1
%Fe3+=0.10.95×100
=10.5

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The Compound A and B in the equation given above are

Options:
A CH3COOH, CH3CH3
B DCH2-COOD, CH4
C DCH2-COOH, CH2D2
D CH3-COOD, CH3D

The correct answer is:
DCH2-COOH, CH2D2

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The compound 'A' having formula $C_8 H_{10}$ (aromatic) which gives 1 mononitro substitute and 3 nitrosubstitute compound is

Options:
A m-Xylene
B p-Xylene
C o-Xylene
D Ethyl benzene

The correct answer is:
p-Xylene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The compound A in the following reactions is:

A →(ii) Conc. H2SO4/∆(i) CH3MgBr/H2OB →(ii) Zn/H2O(i) O3C+D

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The compound 'A' is a complementary base of .......... in DNA strands.

Options:
A Uracil
B Guanine
C Adenine
D Cytosine

The correct answer is:
Adenine

Solution:
Given structure is Thymine and Thymine being paired with adenine

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The compound 'A' is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The compound $A$ on heating gives a colourless gas and a residue that is dissolved in water to obtain $B$. Excess of $\mathrm{CO}_2$ is bubbled through aqueous solution of $B, C$ is formed which is recovered in the solid form. Solid $C$ on gentle heating gives back $A$. The compound $A$ is

Options:
A $\mathrm{CaCO}_3$
B $\mathrm{Na}_2 \mathrm{CO}_3$
C $\mathrm{K}_2 \mathrm{CO}_3$
D $\mathrm{CaSO}_4 \cdot 2 \mathrm{H}_2 \mathrm{O}$

The correct answer is:
$\mathrm{CaCO}_3$

Solution:
Thus, the compound ' $A$ ' must be $\mathrm{CaCO}_3$ and the reactions are as follows

Topic: Chemistry, s Block Elements, NEET, NEET 2010 (Mains)
The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of CO2 is bubbled through aqueous solution of B, C is formed which
is recovered in the solid form. Solid C on gentle heating gives back A. The compound is

Options:
A CaCO3
B Na2CO3
C K2CO3
D CaSO4·2H2O

The correct answer is:
CaCO3

Solution:
The reactions can be summarised as follows :
A → Δ Colourless gas + residue
Residue + H 2 O → B → Excess CO 2 C → Δ A
This is possible only when A is CaCO3 . The reactions are as follows :
CaCO 3 → Δ CO 2 + CaO A Colourless gas Residue
CaO + H 2 O → Ca OH 2 → excess CO 2 Ca HCO 3 2 → Δ CaCO 3 B C A

Topic: Chemistry, s Block Elements, JEE Main
The compound 'A' when treated with $\mathrm{HNO}_3$ (in presence of $\mathrm{H}_2 \mathrm{SO}_4$ ) gives compound ' $\mathrm{B}$ ' which is then reduced with $\mathrm{Sn}$ and $\mathrm{HCl}$ to aniline. The compound ' $\mathrm{A}$ ' is

Options:
A Toulene
B Benzene
C Ethane
D Acetamide

The correct answer is:
Benzene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The compound $A$ with following sequence of reaction gave benzoic acid
$A \xrightarrow{\text { NeNO} O_2 / \mathrm{HCl}} B \xrightarrow{\mathrm{KCN}} C \xrightarrow{\mathrm{H}_3 0^{+}}$benzoic acid. The compound $A$ is

Options:
A Nitrobenzene
B Aniline
C Benzaldehyde
D Amides

The correct answer is:
Aniline

Solution:

Topic: Chemistry, Amines, JEE Main
The compound $\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}$ can show

Options:
A Metamerism
B Functional isomerism
C Positional isomerism
D All types

The correct answer is:
All types

Solution:
Position isomers :
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}$ and
Functional isomers:
$\underset{\text { Butanol }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{CH}_2-\mathrm{OH}} \underset{\text { and }}{\mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5}$

Topic: Chemistry, General Organic Chemistry, JEE Main
The compound C7H8 undergoes the following reactions:
The product 'C' is

Options:
A 3-bromo-2, 4, 6-trichlorotoluene
B o-bromotoluene
C m-bromotoluene
D p-bromotoluene

The correct answer is:
m-bromotoluene

Solution:
In electrophilic aromatic substitution, a substituent that favors electrophilic attack ortho or para to the substituent. Most ortho/para directors are also activators, except for the halogens, which are deactivators. A carboxylic acid is a meta director because it experiences resonance, a delocalization of electrons.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The compound 'D' is

Options:
A
B
C
D

The correct answer is:

Solution:
DU in (A) = 1. So, (A) is unsaturated alcohol and it is converted into cyclopentanone.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The compound formed as a result of oxidation of ethyl benzene by $\mathrm{KMnO}_4$ is

Options:
A
benzophenone

B
acetophenone

C
benzoic acid

D
benzyl alcohol

The correct answer is:
benzoic acid

Solution:
Any aliphatic carbon with hydrogen attached to it, in combination with benzene ring, will be oxidized to benzoic acid by $\mathrm{KMnO}_4 / \mathrm{H}^{+}$.
Hence, (C) is correct.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2007
The compound formed as a result of oxidation of ethyl benzene by $\mathrm{KMnO}_4$ is

Options:
A
benzophenone

B
acetophenone

C
benzoic acid

D
benzyl alcohol

The correct answer is:
benzoic acid

Solution:
Any aliphatic carbon with hydrogen attached to it, in combination with benzene ring, will be oxidized to benzoic acid by $\mathrm{KMnO}_4 / \mathrm{H}^{+}$.
Hence, (C) is correct.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The compound formed by the reaction of ethanal with semicarbazide contains _______ number of nitrogen atoms.

Options:

The correct answer is:
03

Solution:
Semicarbazide is the chemical compound with the formula OC(NH2)(N2H3). It is a water-soluble white solid. It is a derivative of urea.

Acetaldehyde reacts with semicarbazide give acetaldehyde semicarbazone, the reaction will be

Hence the product formed has three nitrogen atoms in it.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The compound formed in the positive test for nitrogen with the Lassaigne solution of an organic compound is

Options:
A
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

B
$\mathrm{Na}_4\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NOS}\right]$

C
$\mathrm{Fe}(\mathrm{CN})_3$

D
$\mathrm{Na}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

The correct answer is:
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2004
The compound formed on heating chlorobenzene with chloral in the presence concentrated sulphuric acid is

Options:
A
gammexene

B
hexachloroethane

C
Freon

D
DDT

The correct answer is:
DDT

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2004
The compound formed when Ethyl bromide is heated with dry Silver oxide is

Options:
A dimethylether
B diethylether
C methylalcohol
D ethylalcohol

The correct answer is:
diethylether

Solution:
2 C 2 H 5 Br+ Ag 2 O Dry → C 2 H 5 −O− C 2 H 5 +2AgBr
If we take moist Ag 2 O then alcohol is formed
Ag 2 O+ H 2 O→2AgOH
C 2 H 5 Br+AgOH→ C 2 H 5 OH+AgBr

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The compound formed when ethyl bromide is heated with dry silver oxide is

Options:
A Dimethyl ether
B Diethyl ether
C Methyl alcohol
D Ethyl alcohol

The correct answer is:
Diethyl ether

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The compound in which the number of $d \pi$ - $p \pi$ bonds are equal to those present in $\mathrm{ClO}_4^{-}$

Options:
A $\mathrm{XeF}_4$
B $\mathrm{XeO}_3$
C $\mathrm{XeO}_4$
D $\mathrm{XeF}_6$

The correct answer is:
$\mathrm{XeO}_3$

Solution:
The structure of $\mathrm{ClO}_4^{-}$is
Thus, it contains $3 d \pi-p \pi$ bonds. $\mathrm{XeO}_3$ also contains $3 d \pi-p \pi$ bonds as

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The compound
is used as

Options:
A
antiseptic

B
antibiotic

C
analgesic

D
pesticide

The correct answer is:
analgesic

Solution:
The $E^0$ of cell will be zero

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2002
The compound most likely to decolourize a solution of potassium permanganate is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_3$
B

C $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{CH}_3$
D

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_2 \mathrm{CH}_3$

Solution:
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_3$ it decolourizes $\mathrm{KMnO}_4$ solution because double bond is present.

Topic: Chemistry, Hydrocarbons, JEE Main
The compound of copper which turns green on keeping in air is

Options:
A Copper sulphate
B Copper nitrate
C Cupric chloride
D Cuprous chloride

The correct answer is:
Cuprous chloride

Solution:
Cuprous chloride slowly oxidises to green basic cupric chloride.
$\mathrm{CuCl} \xrightarrow{\text { air }} 3 \mathrm{CuO} . \mathrm{CuCl}_2 \cdot 3 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, d and f Block Elements, JEE Main
The compound of xenon with zero dipole moment is

Options:
A
$\mathrm{XeO}_3$

B
$\mathrm{XeF}_4$

C
$\mathrm{XeOF}_4$

D
$\mathrm{XeO}_2$

The correct answer is:
$\mathrm{XeF}_4$

Solution:
$\mathrm{XeF}_4$ has zero dipole moment. It has square planar structure due to which the bond moments of $\mathrm{Xe}-\mathrm{F}$ cancel each other.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (19 May Online)
The compound
on reaction with $\mathrm{NaIO}_4$ in the presence of $\mathrm{KMnO}_4$ gives

Options:
A $\mathrm{CH}_3 \mathrm{CHO}+\mathrm{CO}_2$
B $\mathrm{CH}_3 \mathrm{COCH}_3$
C $\mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CH}_3 \mathrm{COOH}$
D $\mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CH}_3 \mathrm{CHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{COCH}_3+\mathrm{CH}_3 \mathrm{COOH}$

Solution:
The given compound is isopentylene. When it reacts with \(\mathrm{NaIO}_4\) in presence of potassium permanganate it undergoes oxidation to form acetone and acetic acid respectively.

Topic: Chemistry, Hydrocarbons, JEE Main
The compound showing complete stereochemical inversion during an \( S_{N} 2 \) reaction is:

Options:
A
B
C \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl} \)
D \( \mathrm{CH}_{3} \mathrm{Cl} \)

The correct answer is:
\( \mathrm{CH}_{3} \mathrm{Cl} \)

Solution:
CH3Cl being most reactive towards SN2 reaction undergoes complete inversion.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The compound statement P∨Q∧~P⇒Q equivalent to:

Options:
A P∨Q
B P∧~Q
C ~P⇒Q
D ~P⇒Q⇔P∧~Q

The correct answer is:
~P⇒Q⇔P∧~Q

Solution:
P	Q	P∨Q	~P	P∨Q∧~P	P∨Q∧~P⇒Q

T	T	T	F	F	T

T	F	T	F	F	T

F	T	T	T	T	T

F	F	F	T	F	T

P	Q	~Q	P∧~Q	P⇒Q	~P⇒Q

T	T	F	F	T	F

T	F	T	T	F	T

F	T	F	F	T	F

F	F	T	F	T	F

~P⇒Q	P∧~Q	~P⇒Q⇔P∧~Q

F	F	T

T	T	T

F	F	T

F	F	T

Hence, P∨Q∧~P⇒Q equivalent to ~P⇒Q⇔P∧~Q

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The compound statement p→∼q∨p∧q is logically equivalent to

Options:
A p↔q
B p∧q
C Tautology
D Contradiction

The correct answer is:
Tautology

Solution:
We know that ∼p→∼q≡p∧∼∼q≡p∧q
∴ Given statement is p→∼q∨∼p→∼q
which is a tautology as p∨~p≡t

Topic: Mathematics, Mathematical Reasoning, JEE Main
The compound statement p↔q∨p↔~q is logically equivalent to

Options:
A p↔q
B p∨q
C tautology
D contradiction

The correct answer is:
tautology

Solution:
We know that ~p↔q≡p↔~q
∴ given statement is p↔q∨~p↔q
which is a tautology ∵ p∨~p≡t

Topic: Mathematics, Mathematical Reasoning, JEE Main
The compound statement p→q∨p∧~q is logically equivalent to

Options:
A p↔q
B p∧q
C tautology
D contradiction

The correct answer is:
tautology

Solution:
We know that ~p→q≡p∧~q
∴ given statement is p→q∨~p→q
Which is a tautology as, p∨~p≡t

Topic: Mathematics, Mathematical Reasoning, JEE Main
The compound statement (~(P∧Q))∨((~P)∧Q)⇒((~P)∧(~Q)) is equivalent to

Options:
A

((~P)∨Q)∧((~Q)∨P)

B (~Q)∨P
C ((~P)∨Q)∧(~Q)
D (~P)∨Q

The correct answer is:

((~P)∨Q)∧((~Q)∨P)

Solution:
We have,

P	Q	~P∧Q	~P∧Q	~P∧Q∨~P∧Q	~P∧~Q	~P∧Q∨~P∧Q⇒~P∧~Q

T	T	F	F	F	F	T

T	F	T	F	T	F	F

F	T	T	T	T	F	F

F	F	T	F	T	T	T

Also,

P	Q	~P	~Q	~P∨Q	~Q∨P	~P∧Q∧~Q∨P

T	T	F	F	T	T	T

T	F	F	T	T	F	F

F	T	T	F	F	F	F

F	F	T	T	T	T	T

Hence, ~P∧Q∨~P∧Q⇒~P∧~Q≡~P∧Q∧~Q∨P.

Topic: Mathematics, Mathematical Reasoning, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The compound that cannot act both as oxidizing and reducing agent is

Options:
A H3PO4
B HNO2
C H2SO3
D H2O2

The correct answer is:
H3PO4

Solution:
In HNO2, Nitrogen has oxidation state +3, nitrogen has maximum oxidation state +5 and minimum -3, hence nitrous acid can act both as oxidizing and reducing agent.

Sulphur has maximum oxidation state +6 and minimum is -2, in H2SO3 sulphur has oxidation state +4, hence it can act as both oxidizing and reducing agent.

For oxygen maximum oxidation state is +2 and minimum is -2, in H2O2 Oxygen has oxidation state -1, hence it can act bot as oxidizing and reducing agent.

As in H3PO4, Phosphorous is present in it’s maximum oxidation state i.e., +5, hence it cannot act as reducing agent.

Topic: Chemistry, Redox Reactions, JEE Main
The compound that cannot act both as oxidizing and reducing agent is

Options:
A H3PO4
B HNO2
C H2SO3
D H2O2

The correct answer is:
H3PO4

Solution:
In HNO2, Nitrogen has oxidation state +3, nitrogen has maximum oxidation state +5 and minimum -3, hence nitrous acid can act both as oxidizing and reducing agent.

Sulphur has maximum oxidation state +6 and minimum is -2, in H2SO3 sulphur has oxidation state +4, hence it can act as both oxidizing and reducing agent.

For oxygen maximum oxidation state is +2 and minimum is -2, in H2O2 Oxygen has oxidation state -1, hence it can act bot as oxidizing and reducing agent.

As in H3PO4, Phosphorous is present in it’s maximum oxidation state i.e., +5, hence it cannot act as reducing agent.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The compound that does not produce nitrogen gas by thermal decomposition is:

Options:
A NH42SO4
B BaN32
C NH42Cr2O7
D NH4NO2

The correct answer is:
NH42SO4

Solution:
(i) NH42Cr2O7 →Δ Cr2O3+N2+4H2O
(ii) NH4NO2 →Δ N2+2H2O
(iii) NH42SO4 →Δ 2NH3+H2SO4
(iv) BaN32 →Δ Ba+3N2
In reaction (iii) NH3 is evolved whereas in reaction 1, 2 and 4 N2 is evolved.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2018 (08 Apr)
The compound that forms a yellow oily liquid with nitrous acid is

Options:
A 2 -methyl aniline
B methyl amine
C Benzyl amine
D diethyl amine

The correct answer is:
diethyl amine

Solution:
Secondary amines (-NH group containing amines), when reacts with nitrous acid $\left(\mathrm{HNO}_{2}\right)$, a yellow oily nitroso compound is formed.

Topic: Chemistry, Amines, JEE Main
The compound that has the largest H–M–H bond angle (M = N, O, S, C), is :

Options:
A H2O
B NH3
C H2S
D CH4

The correct answer is:
CH4

Solution:
Comparison of H2O and NH3

H2O :

sp3,ℓ·p=2, B. A.=104°30°
H2S→ No hybridisation [Drago's rule], bond angle =92°

NH3 : sp3 hybridisation ,ℓ.p=1,B.A.=107°

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The compound that inhibits the growth of tumors is:

Options:
A trans-PtCl2NH32
B cis- PtCl2NH32
C cis- PdCl2NH32
D trans-PdCl2NH32

The correct answer is:
cis- PtCl2NH32

Solution:
Cis-chloroplatin =Cis [Pt (NH3)2Cl2]

Cis-chloroplatin inhibits the growth of tumors Hence it is used for cancer treatment.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The compound that is most reactive towards electrophilic nitration is

Options:
A Toluene
B Benzene
C Benzoic acid
D Nitrobenzene

The correct answer is:
Toluene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The compound that is NOT a common component of photochemical smog is:

Options:
A O3
B CF2Cl2
C

D CH2=CHCHO

The correct answer is:
CF2Cl2

Solution:
Photochemical smog consists of nitrogen oxides, volatile organic compounds, tropospheric ozone, etc., CF2Cl2 , a chlorofluorocarbon, is not present in photochemical smog, but is a major cause of ozone-layer depletion.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The compound that is white in color is

Options:
A ammonium sulphide
B lead sulphate
C lead iodide
D ammonium arsinomolybdate

The correct answer is:
lead sulphate

Solution:
Lead sulphate-white

Lead sulphate is a white, somewhat crystalline, very heavy powder, its specific gravity being about 6.3. It is only slightly soluble in water, insoluble in dilute acids and alcohol, but soluble in solutions of ammonium salts and strong sulphuric acid.

Ammonium sulphide-soluble

Lead iodide-bright yellow

Ammonium arsinomolybdate-yellow

Hence, the answer is option B.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The compound that undergoes decarboxylation most readily under mild condition is

Options:
A
B
C
D

The correct answer is:

Solution:
β-keto acids undergoes decarboxylation easily.
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Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main