datavorous/entrance-exam-dataset huggingface dataset:
Ammonium sulphide and ammonium selenide on heating dissociates as

(NH4)2S(s) â 2NH3(g)+ H2S(g); Kp1 = 9 Ã 10-3 atm3

(NH4)2Se(s) â 2NH3(g) + H2Se(g); Kp2 = 4.5 Ã 10-3 atm3

The total pressure over the solid mixture at equilibrium is

Options:
A 0.15 atm
B 0.3 atm
C 0.45 atm
D 0.6 atm

The correct answer is:
0.45 atm

Solution:
NH 4 2 S s â 2NH 3 g + H 2 S g ââââââââââââââââââââââ2pââââââââââââââp
NH 4 2 Se s ⇌ 2NH 3 g + H 2 Se  g                       2q              q
K p 1 = p NH 3 2 Â· p H 2 S = 2 p + 2 q 2 Â· p = 9.0 Ã 10 - 3
K p 2 = p NH 3 2 Â· p H 2 S e = 2 p + 2 q 2 Â· q = 4.5 Ã 10 - 3
â´ p q = 2 â p = 2 q
â´ K p 1 = 2p + p 2 Â· p = 9.0 Ã 10 - 3
â´ p = 0.1 atm and q = 0.1 2 = 0.05
Total pressure = 2p + 2q + p + q = 3p + 3q
= 3 Ã 0.1 + 3 Ã 0.05 = 0.45 atm

Topic: Chemistry, Chemical Equilibrium, JEE Main
Ammonolysis of Alkyl halides followed by the treatment with NaOH solution can be used to prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is.

Options:
A to remove basic impurities
B to activate NH3 used in the reaction
C to remove acidic impurities
D to increase the reactivity of alkyl halide.

The correct answer is:
to remove acidic impurities

Solution:
So the purpose of NaOH in the above reactions in to remove acidic impurities.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Among 1-4, the complexes that can show geometrical isomerism are
1. [Pt(NH3)3Cl]+
2. [Pt(NH3)Cl5]–
3. [Pt(NH3)2Cl(ONO)]
4. [Pt(NH3)4ClBr]2+

Options:
A 4 and 1
B 2 and 3
C 3 and 4
D 1 and 2

The correct answer is:
3 and 4

Solution:
Pt NH 3 2 Cl ONO  and  Pt NH 3 4 ClBr 2+
MA 2 BC  ⇒ 2.G.I
MA 4 BC  ⇒ 2.G.I
Hence these two can show geometrical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
Among a-d, the complexes that can show geometrical isomerism are:
a PtNH33Cl+
b PtNH3Cl5-
c PtNH32ClNO2
d PtNH34ClBr2+

Options:
A b and c
B d and a
C c and d
D a and b

The correct answer is:
c and d

Solution:
Ma4bc can show 2 G.I.
Ma2b2 can show 2 G.I.
(Square planar)

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Among $\mathrm{Al}_2 \mathrm{O}_3, \mathrm{SiO}_2, \mathrm{P}_2 \mathrm{O}_3$ and $\mathrm{SO}_2$ the correct order of acid strength is

Options:
A $\mathrm{Al}_2 \mathrm{O}_3 \lt \mathrm{SiO}_2 \lt \mathrm{SO}_2 \lt \mathrm{P}_2 \mathrm{O}_3$
B $\mathrm{SiO}_2 \lt \mathrm{SO}_2 \lt \mathrm{Al}_2 \mathrm{O}_3 \lt \mathrm{P}_2 \mathrm{O}_3$
C $\mathrm{SO}_2 \lt \mathrm{P}_2 \mathrm{O}_3 \lt \mathrm{SiO}_2 \lt \mathrm{Al}_2 \mathrm{O}_3$
D $\mathrm{Al}_2 \mathrm{O}_3 \lt \mathrm{SiO}_2 \lt \mathrm{P}_2 \mathrm{O}_3 \lt \mathrm{SO}_2$

The correct answer is:
$\mathrm{Al}_2 \mathrm{O}_3 \lt \mathrm{SiO}_2 \lt \mathrm{P}_2 \mathrm{O}_3 \lt \mathrm{SO}_2$

Solution:
With decrease in size from Al to S the basic nature of oxide decrease and acidic nature increases.
$\mathrm{Al}_2 \mathrm{O}_3 \lt \mathrm{SiO}_2 \lt \mathrm{P}_2 \mathrm{O}_3 \lt \mathrm{SO}_2$
$\mathrm{Al}_2 \mathrm{O}_3$ is amphoteric, $\mathrm{SiO}_2$ is slightly acidic whereas $\mathrm{P}_2 \mathrm{O}_3$ and $\mathrm{SO}_2$ are the anhydrides of acids
$\mathrm{H}_3 \mathrm{PO}_3 \text { and } \mathrm{H}_2 \mathrm{SO}_3 \text {. }$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Among $\mathrm{Al}_2 \mathrm{O}_3, \mathrm{SiO}_2, \mathrm{P}_2 \mathrm{O}_3$ and $\mathrm{SO}_2$ the correct order of acid strength is

Options:
A
$\mathrm{SO}_2 < \mathrm{P}_2 \mathrm{O}_3 < \mathrm{SiO}_2 < \mathrm{Al}_2 \mathrm{O}_3$
B
$\mathrm{Al}_2 \mathrm{O}_3 < \mathrm{SiO}_2 < \mathrm{P}_2 \mathrm{O}_3 < \mathrm{SO}_2$
C
$\mathrm{Al}_2 \mathrm{O}_3 < \mathrm{SiO}_2 < \mathrm{SO}_2 < \mathrm{P}_2 \mathrm{O}_3$
D
$\mathrm{SiO}_2 < \mathrm{SO}_2 < \mathrm{Al}_2 \mathrm{O}_3 < \mathrm{P}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{Al}_2 \mathrm{O}_3 < \mathrm{SiO}_2 < \mathrm{P}_2 \mathrm{O}_3 < \mathrm{SO}_2$

Solution:
Al, Si, P, S acidity of oxides increases

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2004
Among alkali metal salts, the lithium salts are the poorest conductors of electricity in aqueous solution because of

Options:
A easy diffusion of Li+ ions
B lower ability of Li+ ions to polarize water molecules
C lowest charge to radius ratio
D higher degree of hydration of Li+ ions

The correct answer is:
higher degree of hydration of Li+ ions

Solution:
Ionic conductance in solution depends upon the mobility of ions, which depend upon the degree (or extent) of hydration. Since charge/radius ratio is maximum for Li+ , so it is hydrated to a maximum extent. Thus, the ionic size in aqueous solution, instead of increasing, actually decreases on moving from Li+ to Cs+ .

Topic: Chemistry, s Block Elements, JEE Main
Among all cyclic quadrilaterals inscribed in a circle of radius $\mathrm{R}$ with one of its angles equal to $120^{\circ}$. Consider the one with maximum possible area. Its area is-

Options:
A $\sqrt{2} \mathbb{R}^{2}$
B $\sqrt{3} \mathrm{R}^{2}$
C $2 \mathbb{R}^{2}$
D $2 \sqrt{3} \mathrm{R}^{2}$

The correct answer is:
$\sqrt{3} \mathrm{R}^{2}$

Solution:
$\mathrm{A}=2 \times \frac{1}{2} \times \sqrt{3} \mathrm{R} \times \mathrm{R}=\sqrt{3} \mathrm{R}^{2}$

Topic: Mathematics, Circle, JEE Main
Among bitter gourd, mustard, brinjal, pumpkin, China rose, lupin, cucumber, sunnhemp, gram, guava, bean, chilli, plum, Petunia, tomato, rose, Withania, potato, onion, aloe and tulip, how many plants have hypogynous flower?

Options:
A Six
B Ten
C Fifteen
D Eighteen

The correct answer is:
Fifteen

Solution:
Plum, rose are perigynous and guava, cucumber are epigynous. So, total remaining plants are hypogynous.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2013 (All India)
Among CaH2, NH3, NaH and B2H6, which are covalent hydride?

Options:
A NH3 and B2H6
B NaHÂ andÂ CaH2
C NaHÂ andÂ NH3
D CaH2Â andÂ Â B2H6

The correct answer is:
NH3 and B2H6

Solution:
Hydrides are binary compounds of hydrogen. These can be classified in four groups viz :
(i) Ionic hydrides e.g., NaH, CaH2, LiH etc.
(ii) Covalent hydrides e.g., B2H6, NH3, SbH3 etc.
(iii) Polynuclear hydrides e.g., LiAlH4, NaBH4 etc.
(iv) Interstitial hydrides, in which hydrogen is trapped in the interstial spaces of transition metals.

Topic: Chemistry, s Block Elements, JEE Main
Among Co3+,Ti2+,V2+ and Cr2+ ions, one if used as a reagent cannot liberate H2 from dilute mineral acid solution, its spin-only magnetic moment in gaseous state is _____ B.M. (Nearest integer)

Options:

The correct answer is:
5

Solution:
Among the given species Co3+ cannot undergo oxidation, hence, Co3+ can't liberate H2.

It has d6 configuration,

Number of unpaired electrons =4

μ=nn+2B.Mμ=4×6=4.92 B.M.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Among eukaryotes, replication of DNA takes place in:

Options:
A G2 phase
B M phase
C S phase
D G1 phase

The correct answer is:
S phase

Solution:
Cell cycle has two main phases:

Interphase (I-phase) is the longest and metabolically active phase which has three phases, gap-1 (G1), S and gap-2 (G2).

Division or M-phase, where actual mitosis (or meiosis) takes place.

S phase:

Also called the invisible phase.

The DNA replication takes place at this phase of interphase.

The 2C DNA content of a cell becomes 4C at the end of S-phase.

There is no change in the ploidy level.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2023 (All India)
Among following compounds, the number of those present in copper matte is

A. CuCO3

B. Cu2S

C. Cu2O

D. FeO

Options:

The correct answer is:
1

Solution:
Copper Matte is a combination of copper sulphide and a small amount of iron sulphide. In the process of copper extraction, Matte is the stage after the final reduction process which involves converting it to crude copper.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Among formic acid, acetic acid, propanoic acid and phenol, the strongest acid in water is

Options:
A formic acid
B acetic acid
C propanoic acid
D phenol

The correct answer is:
formic acid

Solution:
Formic Acid is Strongest Acid.
Acidic strength test
$\rightarrow+$ I decrease acidic strength
$\rightarrow$ Carboxylic Acid are more acidic Then phenol

Topic: Chemistry, General Organic Chemistry, JEE Main
Among given compounds, how many compounds will react with NaHCO3 or soluble in NaHCO3 ?

Options:

The correct answer is:
2.00

Solution:
Compounds which are more acidic than H2CO3 will dissolve in NaHCO3. So, option c and d will be soluble in NaHCO3

Topic: Chemistry, General Organic Chemistry, JEE Main
Among
(i) $\left[\mathrm{Cr}(\mathrm{en})_{3}\right]^{3+}$,
(ii) trans-[ $\left[\mathrm{Cr}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}$,
(iii) Cis-[Cr(en) $\left._{2} \mathrm{Cl}_{2}\right]^{+}$,
(iv) $\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Cl}_{2}\right]^{+}$
the optically active complexes are

Options:
A i and ii
B i and iii
C ii and iii
D ii and iv

The correct answer is:
i and iii

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
Among \(\mathrm{CrO}, \mathrm{Cr}_2 \mathrm{O}_3\) and \(\mathrm{CrO}_3\), the sum of spin-only magnetic moment values of basic and amphoteric oxides is _______ \(10^{-2} \mathrm{BM}\) (nearest integer).
(Given atomic number of \(\mathrm{Cr}\) is 24)

Options:

The correct answer is:
877

Solution:
CrO Basic oxide
$\mathrm{Cr}_2 \mathrm{O}_3$ Amphoteric oxide
In $\mathrm{CrO}, \mathrm{Cr}$ exist as $\mathrm{Cr}^{+2}$ and have $\mu$ only $=4.90$
In $\mathrm{Cr}_2 \mathrm{O}_3, \mathrm{Cr}$ exist as $\mathrm{Cr}^{+3}$ and have $\mu$ only $=3.87$
Sum of spin only magnetic moment
$\begin{aligned}
& =4.90+3.87=8.77 \\
& \mu_{\text {only }}=877 \times 10^{-2}
\end{aligned}$
Ans. 877

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Among \(\mathrm{VO}_2^{+}, \mathrm{MnO}_4^{-}\)and \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\), the spin-only magnetic moment value of the species with least oxidising ability is ______ BM (Nearest integer).
(Given atomic member \(\mathrm{V}=23, \mathrm{Mn}=25, \mathrm{Cr}=24\))

Options:

The correct answer is:
0

Solution:
For 3d transition series; Oxidising power : $\mathrm{V}^{+5} < \mathrm{Cr}^{+6} < \mathrm{Mn}^{+7}$
$\mathrm{V}^{+5}:[\mathrm{Ar}] 4 \mathrm{~s}^0 3 \mathrm{~d}^0$
Number of unpaired electron $=0$
$\mu=0$

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Among $\mathrm{Me}_{3}\mathrm{N}, \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}$ and MeCN
(Me $=$ methyl group) the electronegativity of $\mathrm{N}$ is in the order

Options:
A $\mathrm{MeCN}>\mathrm{C}_{\mathrm{g}} \mathrm{H}_{\mathrm{s}} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}$
B $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}>\mathrm{MeCN}$
C $\mathrm{Me}_{3} \mathrm{N}>\mathrm{MeCN}>\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}} \mathrm{N}$
D Electronegativity same in all

The correct answer is:
$\mathrm{MeCN}>\mathrm{C}_{\mathrm{g}} \mathrm{H}_{\mathrm{s}} \mathrm{N}>\mathrm{Me}_{3} \mathrm{N}$

Solution:
$\because$ In Me $_{3} \mathrm{N} \rightarrow \mathrm{N}$ is $s p^{3}$ -hybrid

and in $\mathrm{MeCN} \rightarrow \mathrm{N}$ is $s p$ -hybrid Also electronegativity of any element $\alpha \%$ $s$ -character in hybrid orbitals. Thus, order of electronegativity for nitrogen is
$$
\mathrm{MeCN}>\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N} < \mathrm{Me}_{3} \mathrm{N}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Among NiCO42-,NiCN42-andNiCl42- species, the hybridisation states of the Ni atom are respectively: (Atomic no. of Ni=28)

Options:
A sp3, dsp2, dsp2
B sp3, dsp2, sp3
C sp3, sp3, dsp2
D dsp2, sp3, sp

The correct answer is:
sp3, dsp2, sp3

Solution:
In NiCO4, nickel is sp3-hybridised because in it the oxidation state of Ni is zero.
In NiCN42-, nickel is present as Ni2+, so its configuration=1s2, 2s22p6, 3s23p63d8
CN- is strong field ligand, hence it makes Ni2+ electrons to be paired up.
In NiCl42- species, nickel is present as Ni2+, so its configuration=1s2, 2s22p6, 3s23p63d8
Cl- is weak field ligand, hence Ni2+ electrons are not paired.

Topic: Chemistry, Coordination Compounds, JEE Main
Among [Ni(CO)4],[NiCl4]2-, [Co(NH3)4Cl2]Cl, Na3[CoF6],Na2O2 and CsO2 , the total number of paramagnetic compounds is

Options:
A 2
B 3
C 4
D 5

The correct answer is:
3

Solution:
[Ni(CO)4]-sp3- Diamagnetic
[NiCl4]2--sp3- Paramagnetic
[Co(NH3)4Cl2]Cl-d2sp3- Diamagnetic
Na3[CoF6]-sp3d2- Paramagnetic
Na2O2 i.e. O22-- Diamagnetic
CsO2 i.e. O2-- Paramagnetic

Topic: Chemistry, Coordination Compounds, JEE Main
Among $\left[\mathrm{Ni}(\mathrm{CO})_4\right],\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-},\left[\mathrm{NiCl}_4\right]^{2-}$ species, the hybridization states at the $\mathrm{Ni}$ atom are, respectively

Options:
A $s p^3, s p^3, d s p^2$
B $d s p^2, s p^3, s p^3$
C $s p^3, d s p^2, d s p^2$
D $s p^3, d s p^2, s p^3$

The correct answer is:
$s p^3, d s p^2, s p^3$

Solution:
Outermost electronic configuration of $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ complex is $3 \mathrm{~d}^8, 4 \mathrm{~s}^2$ and $\mathrm{CO}$ is a strong ligand thus electrons get paired up and shows $\mathrm{sp}^3$ hybridisation.
Outermost electronic configuration of $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ complex is $3 \mathrm{~d}^8$ and $\mathrm{CN}$ is a strong ligand thus electrons get paired up but one d orbital is empty and shows dsp² hybridisation.
Outermost electronic configuration of $\left[\mathrm{Ni}(\mathrm{Cl})_4\right]^{2-}$ complex is $3 \mathrm{~d}^8, 4 \mathrm{~s}^2$ and $\mathrm{Cl}$ is a weak ligand thus electrons does not paired up and shows $\mathrm{sp}^3$ hybridisation.

Topic: Chemistry, Coordination Compounds, JEE Main
Among oxyacids of nitrogen HNO2 is unstable, it can acts as

Options:
A Oxidising agent
B Reducing agent
C Oxidising agent and Reducing agent
D Its solution is stable

The correct answer is:
Oxidising agent and Reducing agent

Solution:
Nitrous acid behaves as reducing as well as an oxidising agent. It reduces potassium permanganate, potassium dichromate, H2O2 and other strong oxidising agents. It oxidises strong reducing agents such as hydroiodic acid, sulphurous acid etc.
It oxidises Fe+2 into Fe+3 in acidic medium;
Fe+2+HNO2+H+→Fe+3+NO+H2O
It reduces acidified KMnO4 .
2KMnO4+3H2SO4+5HNO2 →
K2SO4+2MnSO4+3H2O+5HNO3

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Among P, Q, R, S the aromatic compound is

Options:
A →AlCl3P
B →NaHQ
C → AlCl 3
D →HClD

The correct answer is:
→ AlCl 3

Solution:
AlCl 3 helps in formation of carbocation.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Among

S1:limn→∞1n2(2+4+6+…+2n)=1

S2:limn→∞1n16115+215+315+…+n15=116

Options:
A Both S1 and S2 are true
B Only S1 is true
C Both S1 and S2 are false
D Only S2 is true

The correct answer is:
Both S1 and S2 are true

Solution:
S1:limn→∞1n2[2+4+6+…+2n]

limn→∞2n(n+1)2n2=1

S2:limn→∞1n16115+215+315+…+n15

This is limit of sum.

We know that limn→∞∑r=1nfrn1n=∫01fxdx

∵limn→∞∑r=1nrknk+1=limn→∞∑r=1nrnkn

=∫01xkdx=1k+1

Here k=15

∴limn→∞∑r=1nr15n16=116

∴ Both S1 and S2 are correct

Hence this is the required option.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Among statements a-d, the correct ones are:
a Decomposition of hydrogen peroxide gives di-oxygen.
b Like hydrogen peroxide, compounds, such as KClO3,PbNO3 and NaNO3 When heated liberate di-oxygen.
c 2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide.
d Hydrogen peroxide is used for the manufacture of sodium perborate.

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

The correct answer is:
a,b,c and d

Solution:
(a)   H 2 O 2 →2 H 2 O+ O 2
(b)  KClO3→ΔKCl+32O2
PbNO32→ΔPbO+2NO2+12O2
NaNO3→ΔNaNO2+12O2
(c) 2- ethylanthraquinol →O22−ethylanthraquinone+H2O2
(d) 2H3BO3+2NaOH+2H2O2→[Na2B2O2(OH)4+4H2O]
All statements are correct

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Among the above, the number of aromatic compound(s) is/are:

Options:

The correct answer is:
5

Solution:
Cyclooctatetraene; non-aromatic, due to non-planarity of ring the π-electrons are not delocalised.

Cyclopropenyl anion; Anti-aromatic, 4π-electrons are delocalised

Cyclopropenyl cation; Aromatic, 2π-electrons are delocalised.

Cyclohexadiene : Non-aromatic, as π electrons are not fully conjugated.

Tropylium ion : Aromatic, as it contains conjugated system of (4n+2)π electrons (6πelectrons). It is cyclic and planar.

Cyclopentadienyl cation; Anti-aromatic, 4π-electrons are delocalised.

Cyclopentadienyl anion; Aromatic, as it contains conjugated system of (4n+2)π electrons (6π electrons). It is cyclic and planar.

Both are aromatic as first it contains conjugated system of (4n+2)π electrons (6π electrons). It is cyclic and planar. These are pi electrons present in one ring. Pi electrons present in other ring are not counted.

Second one contains conjugated system of (4n+2)π electrons (14π electrons). It is cyclic and planar.

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is :

Options:
A $\mathrm{C}$ : liquid in solid; $\mathrm{M}$ : liquid in solid; $\mathrm{S}$ : solid in gas
B $\mathrm{C}$ : liquid in solid; $\mathrm{M}$ : liquid in liquid; $\mathrm{S}$ : solid in gas
C $\mathrm{C}$ : solid in liquid; $\mathrm{M}$ : liquid in liquid; $\mathrm{S}$ : gas in solid
D $\mathrm{C}$ : solid in liquid; $\mathrm{M}$ : solid in liquid; $\mathrm{S}$ : solid in gas

The correct answer is:
$\mathrm{C}$ : liquid in solid; $\mathrm{M}$ : liquid in liquid; $\mathrm{S}$ : solid in gas

Solution:
\begin{array}{|l|l|l|}

\hline Colloids phase & Dispersed medium & Dispersion \\

\hline Cheese (C) & liquid & solid \\

Milk(M) & liquid & liquid \\

Smoke (S) & solid & gas \\

\hline

\end{array}

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 2)
Among
the compound which is not aromatic is

Options:
A $i$
B ii
C iii
D iv

The correct answer is:
ii

Solution:
(i) and (iv) are heteroaromatic compound.
(ii) is non aromatic.
(iii)

Topic: Chemistry, General Organic Chemistry, JEE Main
Among the compounds A and B with molecular formula C9H18O3, A is having higher boiling point the B . The possible structures of A and B are

Options:
A
B
C
D

The correct answer is:

Solution:
'A' is having higher boiling point than 'B' as in case of 'A' inter molecular H-bonding is possible while in case of 'B' intermolecular H-bonding is not possible hence have lower boiling point. Based on it the structures of A and B are

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Among the compounds, Benzene, Carbon tetrachloride, Naphthalene, Benzoic acid, Isooctane and Anthracene, how many can be purified by sublimation.

Options:

The correct answer is:
3

Solution:
Naphthalene, Benzoic acid and Anthracene are purified by sublimation.

Topic: Chemistry, Hydrocarbons, JEE Main
Among the elements $\mathrm{Ca}, \mathrm{Mg}, \mathrm{P}$ and $\mathrm{Cl}$, the order of increasing atomic radii is

Options:
A $\mathrm{Mg} < \mathrm{Ca} < \mathrm{Cl} < \mathrm{P}$
B $\mathrm{Cl} < \mathrm{P} < \mathrm{Mg} < \mathrm{Ca}$
C $\mathrm{P} < \mathrm{Cl} < \mathrm{Ca} < \mathrm{Mg}$
D $\mathrm{Ca} < \mathrm{Mg} < \mathrm{P} < \mathrm{Cl}$

The correct answer is:
$\mathrm{Cl} < \mathrm{P} < \mathrm{Mg} < \mathrm{Ca}$

Solution:
With increase in the number of electrons in the same shell, the atomic radii decreases due to increase in effective nuclear charge. However, atomic radii increases, as the number of shells increases. Thus, on moving down a group atomic radii increases.
The electronic configuration of the given elements is
$$
\begin{aligned}
\mathrm{Mg}_{12} & =[\mathrm{Ne}] 3 \mathrm{~s}^2 \\
\mathrm{Ca}_{20} & =[\mathrm{Ar}] 4 \mathrm{~s}^2
\end{aligned}
$$
$$
\begin{array}{r}
\mathrm{P}_{15}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^3 \\
\mathrm{Cl}_{17}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^5
\end{array}
$$
In $\mathrm{Mg}, \mathrm{P}$ and $\mathrm{Cl}$, the number of electrons are increasing in the same shell, thus the order of their atomic radii is
$$
\mathrm{Cl} < \mathrm{P} < \mathrm{Mg}
$$
In $\mathrm{Ca}$, the electron is entering in higher shell, thus, it has the highest atomic radii among the given. Thus, the order of radii is
$$
\mathrm{Cl} < \mathrm{P} < \mathrm{Mg} < \mathrm{Ca}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET, NEET 2010 (Mains)
Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is

Options:
A $\mathrm{Mg} < \mathrm{Ca} < \mathrm{Cl} < \mathrm{P}$
B $\quad \mathrm{Cl} < \mathrm{P} < \mathrm{Mg} < \mathrm{Ca}$
C $\mathrm{P} < \mathrm{Cl} < \mathrm{Ca} < \mathrm{Mg}$
D $\mathrm{Ca} < \mathrm{Mg} < \mathrm{P} < \mathrm{Cl}$

The correct answer is:
$\quad \mathrm{Cl} < \mathrm{P} < \mathrm{Mg} < \mathrm{Ca}$

Solution:
Atomic radii increases, as the number of shells increases. Thus, on moving down a group atomic radii increases.
The electronic configuration of the given element is
$$
\begin{array}{l}
\mathrm{Mg}_{12}=[\mathrm{Ne}] 3 s^{2} \\
\mathrm{Ca}_{20}=[\mathrm{Ar}] 4 s^{2} \\
\mathrm{p}_{15}=[\mathrm{Ne}] 3 s^{2} 3 p^{3} \\
\mathrm{Cl}_{17}=[\mathrm{Ne}] 3 s^{2} 3 p^{5}
\end{array}
$$
On the other hand, on increasing the number of electron in the same shell, the atomic radii decreases because effective nuclear charge is increases. In $\mathrm{Mg}, \mathrm{P}$ and $\mathrm{C} 1$, the number of electrons are increasing in the same shell, thus the order of their atomic radii is $\mathrm{C} 1 < \mathrm{P} < \mathrm{Mg}$
In case of Ca, the electron is entering in higher shell. So, its atomic radii is highest. Thus, the order of radii is
$$
\mathrm{Cl} < \mathrm{P} < \mathrm{Mg} < \mathrm{Ca}
$$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
Among the following acids which has the lowest $\mathrm{pK}_{\mathrm{a}}$ value

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

B
$\mathrm{HCOOH}$

C
$\left(\mathrm{CH}_3\right)_2 \mathrm{COOH}$

D
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$

The correct answer is:
$\mathrm{HCOOH}$

Solution:
No Solution Available

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2005
Among the following allotropic forms of sulphur, the number of allotropic forms, which will show paramagnetism is _______.
(A) α-sulphur
(B) β-sulphur
(C) S2-form

Options:

The correct answer is:
1

Solution:
α-sulphur and β-sulphur are diamagnetic. S2-form is paramagnetic.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (24 Feb Shift 2)
Among the following atomic orhital overlap, the non-honding overlap is

Options:
A
B
C
D

The correct answer is:

Solution:
It has +ve \& -ve overlap both simultaneous. So it leads to non-bonding overlap.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Among the following, basic oxide is

Options:
A SO3
B SiO2
C CaO
D Al2O3

The correct answer is:
CaO

Solution:
Oxides of metals are generally basic in nature but some are amphoteric in nature.

Oxides of non-metals are acidic in nature but some of them are neutral.

SO3 is acidic in nature it reacts with bases.

Al2O3 is amphoteric in nature as it reacts with both acids and bases.

SiO2 is acidic in nature.

CaO is a basic oxide and it reacts with acids readily.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Among the following chloro-compound having the lowest dipole moment is

Options:
A

B

C

D

The correct answer is:

Solution:
Dipole moment $(\mu)=0$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (12 May Online)