datavorous/entrance-exam-dataset huggingface dataset:
Which one is a wrong statement?

Options:
A Mucor has biflagellate zoospores
B Haploid endosperm is typical feature of gymnosperms
C Brown algae have chlorophyll a and c, and fucoxanthin
D Archegonia are found in Bryophyta, Pteridophyta and Gymnosperms

The correct answer is:
Mucor has biflagellate zoospores

Solution:
Zoospores: These are microscopic motile spores that move by means of flagella. They are formed in zoosporangia. They are mainly found in algae like Chlamydomonas that has biflagellated zoospores.

Mucor has non-motile spore i.e. sporangiospores or aplanospores.

Topic: Biology, Biological Classification, NEET, NEET 2015 (Phase 2)
Which one is an example of gel

Options:
A Soap
B Cheese
C Milk
D Fog

The correct answer is:
Cheese

Solution:
Correct option is 2. Cheese
Cheese is an example of gel because it's a colloid and the dispersed phase is liquid, and dispersion medium is solid.

Topic: Chemistry, Surface Chemistry, JEE Main
Which one is an INCORRECTLY matched pair?

Options:
A Phycomycetes - Mucor, Albugo
B Ascomycetes - Penicillium, Aspergillus
C Basidiomycetes - Puccinia, Agaricus
D Deuteromycetes - Ustilago, Colletotrichum

The correct answer is:
Deuteromycetes - Ustilago, Colletotrichum

Solution:
Fungi form a unique group of heterotrophic organisms showing wide diversity. Based various features like spore formation mode, fruiting bodies fungi is categorized into -
Phycomycetes - Mainly aquatic habitat, also seen growing on decaying wood. Their mycelium is aseptate and coenocytic. Zoospores or aplanosopres are formed during asexual reproduction. Mucor, Rhizopus, Albugo are included in this group.
Ascomycetes - Majorly multi cellular and also called sac-fungi as their reproductive spores are produced in sac like structure called ascus. For asexual reproduction, conidia are produced in specialized structure called conidiophore. This group includes Penicillium, Aspergillus, Claviceps.
Basidiomycetes - Filamentous fungi having hyphae and specialized club-shaped cells called basidia involved in sexual reproduction. Agaricus (mushroom), Puccinia, Ustilago fall in this group.
Deuteromycetes - Reproduce only via asexual mode through spores called conidia. The mycelium of this group is septate and branched. Trichoderma, Colletotrichum are included in this group.

Topic: Biology, Biological Classification, NEET
Which one is component of Ornithine cycle?

Options:
A Ornithine, citrulline and alanine
B Ornithine, citrulline and arginine
C Amino acid are not used
D Ornithine, citrulline and fumaric acid

The correct answer is:
Ornithine, citrulline and arginine

Solution:
Ornithine cycle or Krebs-Henseleit cycle or urea cycle was discovered by Hans Krebs andKurt Henseleit. It takes place in liver cells for synthesis of urea. The main component of ornithine cycle are arginine, ornithine and citrulline.

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2010
Which one is component of Ornithine cycle?

Options:
A Ornithine, citrulline and alanine
B Ornithine, citrulline and arginine
C Amino acid are not used
D Ornithine, citrulline and fumaric acid

The correct answer is:
Ornithine, citrulline and arginine

Solution:
Ornithine cycle or urea cycle or Krebs-Henseleit cycle was discovered by Hans Krebs and Kurt Henseleit. It takes place in liver cells. The main component of ornithine cycle are arginine, ornithine and citrulline.

Topic: Biology, Excretory Products and their Elimination, JIPMER, JIPMER 2009
Which one is correct about fission?

Options:
A Approx. $0.1 \%$ mass converts into energy.
B Most of energy of fission is in the form of heat.
C In a fission of $U^{235}$ about 200 eV energy is released.
D On an average, one neutron is released per fission of $\mathrm{U}^{235}$

The correct answer is:
Approx. $0.1 \%$ mass converts into energy.

Solution:
No Solution available

Topic: Physics, Nuclear Physics, JEE Main
Which one is known as oil of vitriol

Options:
A Hydrochloric acid
B Sulphuric acid
C Nitric acid
D None of these

The correct answer is:
Sulphuric acid

Solution:
Sulphuric acid is known as the oil of vitriol because it looks like a dense oil and is produced from green vitriol (ferrous sulphate). It is also known as the king of chemicals because of its wide range of uses.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which one is least ionic in the following compounds

Options:
A $\mathrm{AgCl}$
B $\mathrm{KCl}$
C $\mathrm{BaCl}_2$
D $\mathrm{CaCl}_2$

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

Solution:
Acoording to Fajan's rule ionic character is less.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which one is not an allylic halide?

Options:
A 3-chloro cyclo hex-1-ene
B 1-chloro but-1-ene
C 1-chloro but-2-ene
D 3-chloro prop-1-ene.

The correct answer is:
1-chloro but-1-ene

Solution:
1-chloro but-1-ene CH3-CH=CH-Cl is classified as a vinylic halide.

Topic: Chemistry, Hydrocarbons, JEE Main
Which one is not called a anode reaction from the following

Options:
A $\mathrm{Cl}^{-} \rightarrow \frac{1}{2} \mathrm{Cl}_2+\mathrm{e}^{-}$
B $\mathrm{Cu} \rightarrow \mathrm{Cu}^{++}+2 e^{-}$
C $\mathrm{Hg}^{+} \rightarrow \mathrm{Hg}^{++}+e$
D $\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}$

The correct answer is:
$\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}$

Solution:
Reduction
$\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}$ It shows reduction reaction.

Topic: Chemistry, Electrochemistry, JEE Main
Which one is not correct according to Langmuir adsorption isotherm at very high pressure?

Options:
A It is valid for chemisorption
B In Xm=-ΔHoRT+ΔSoR
C Xm=ab
D At very high pressure rate of adsorption > rate of desorption

The correct answer is:
At very high pressure rate of adsorption > rate of desorption

Solution:
x m  =  ap 1 + bp
at high pressure
1 + bp≃bp
x m  =  ap bp ⇒ x m  =  a b  = constant
At very high P, rate of adsorption = rate or desorption
â´â X m â=â a b â=âconst.(K)
â´â ÎG o â=ââ2.303âlogâK
or ÎH o â TÎS o â=ââ2.303âlogâK

Topic: Chemistry, Surface Chemistry, JEE Main
Which one is not equal to zero for an ideal solution?

Options:
A ∆Hmixture
B ∆Smixture
C ∆Vmixture
D ∆P=PObserved-PRaoult

The correct answer is:
∆Smixture

Solution:
For the ideal solution
Δ H mixture =0, because new bond formed and break so no enthalpy gain or lose.
Δ S mixture >0, because randomness increasing on mixing on a solution.
Δ V mixture =0, e − because no new bond for or break so, no force of attraction change.
ΔP= P observed −Δ P Raoult =0; vapour pressure remains the same because of no force of attraction increase or decrease.

Topic: Chemistry, Solutions, NEET, NEET 2015 (Phase 1)
Which one is not equal to zero for an ideal solution

Options:
A ΔSmix
B ΔVmix
C ΔP=Pobserved-PRaoult
D ΔHmix

The correct answer is:
ΔSmix

Solution:
For an ideal solution ΔSmix>0 as in order to maintain spontaneity ΔSmix must be positive. for ideal solution ΔVmix , ΔHmix and ΔP=Pobserved-PRaoult are zero.

Topic: Chemistry, Solutions, NEET
Which one is not periodic

Options:
A
$|\sin 3 x|+\sin ^2 x$

B
$\cos \sqrt{x}+\cos ^2 x$

C
$\cos 4 x+\tan ^2 x$

D
$\cos 2 x+\sin x$

The correct answer is:
$\cos \sqrt{x}+\cos ^2 x$

Solution:
No Solution Available

Topic: Mathematics, Functions, JEE Main, JEE Main 2002
Which one is not the correct relation in the following

Options:
A $h=\frac{E}{V}$
B $E=m c^2$
C $\Delta x \times \Delta p=\frac{h}{4 \pi}$
D $\lambda=\frac{h}{m v}$

The correct answer is:
$\Delta x \times \Delta p=\frac{h}{4 \pi}$

Solution:
$\Delta x \times \Delta p=\frac{h}{4 \pi}$ is not the correct relation. But correct Heisenberg's uncertainty equation is $\Delta x \times \Delta p \geq \frac{h}{4 \pi}$

Topic: Chemistry, Structure of Atom, JEE Main
Which one is organometallic compound

Options:
A Lithium methoxide
B Lithium dimethyl amide
C Lithium acetate
D Methyl lithium

The correct answer is:
Methyl lithium

Solution:
Methyl lithium is the organometallic compound in which lithium bonded with carbon and organometallic are those in which metal-carbon bond found.

Topic: Chemistry, Coordination Compounds, JEE Main
Which one is polymer?

Options:
A Sucrose
B Glycogen
C Fructose
D Lactose

The correct answer is:
Glycogen

Solution:
Glycogen is the main storage polysaccharide in animals. Hence, it is commonly called 'animal starch'. Blue-green algae and fungi also store glycogen as 'fuel reserve'. Like those of amylopectin, glycogen molecules are also large and highly branched (branch points are Î± 1-6 linkages) polymers of thousands of D-glucose residues linked by Î± 1-4 glycosidic bonds. In comparison to amylopectin molecules, however, glycogen molecules are much larger, more extensively branched, and more compactly coiled concentrically.

Topic: Chemistry, Biomolecules, NEET
Which one is the most likely structure of CrCl3 .6H2O, if 1/3 of total chlorine of the compound is precipitated by adding AgNO3 to its aqueous solution?

Options:
A CrCl 3 &CenterDot; 6 H 2 O
B Cr H 2 O 3 Cl 3 &CenterDot; 3 H 2 O
C Cr Cl 2 H 2 O 4 Cl · 2 H 2 O
D Cr Cl H 2 O 5 Cl 2 &CenterDot; H 2 O

The correct answer is:
Cr Cl 2 H 2 O 4 Cl · 2 H 2 O

Solution:
CrCl 3 Â· 6 H 2 O has 3âCl atoms but 1 3 Cl atoms precipitated so one 1âCl atom get precipitated.
According to Werner's theory, only those ions are precipitated which are attached to the metal atoms with ionic bonds and are present outside the co-ordination sphere.
Cr H 2 O 4 Cl 2 Clâ(aq)â Cr H 2 O 4 Cl 2 + aq + Cl â aq
Ag++Cl- â¶AgCl âwhite

Topic: Chemistry, Coordination Compounds, JEE Main
Which one is the wrong pairing for the disease and its causal organism ?

Options:
A Root-knot of vegetables: Meloidogyne sp
B Late blight of potato: Alternaria solani
C Black rust of wheat: Puccinia graminis
D Loose smut of wheat: Ustilago nuda

The correct answer is:
Late blight of potato: Alternaria solani

Solution:
No Solution Available

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2009 (Mains)
Which one is the wrong statement?

Options:
A Anhydrous AlCl3 exists as Al2Cl6 (dimer)
B Al 2 Cl 6 contains 3c− 4e − bonds
C Anhydrous AlCl3 fumes in moist air
D Anhydrous AlCl3 is ionic

The correct answer is:
Anhydrous AlCl3 is ionic

Solution:
Anhydrous AlCl3 exists as Al2Cl6 (dimer)
White fumes are due to the presence of HCl gas. Anhydrous aluminium chloride is hydrolysed partially with the moisture in the atmosphere to give HCl gas. This HCl combines with the moisture in the air and appears white in colour.
AlCl3â+3H2âOâââAlOH3â+3HClwhite fumes
Thus AlCl 3 is a covalent compound rather than ionic. Fajan's rule states that, small size of cation, higher the charge on cation and large size of anion, greater is the covalent character of the ionic bond.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Which one of following is a disproportionation reaction?

Options:
A $2 \mathrm{AgNO}_3(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
B $3 \mathrm{AgNO}_3(\mathrm{aq})+\mathrm{K}_3 \mathrm{PO}_4(\mathrm{aq}) \longrightarrow \mathrm{Ag}_3 \mathrm{PO}_4(\mathrm{~s}) +3 \mathrm{KNO}_3(\mathrm{aq})$
C $4 \mathrm{KClO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{KCl}(\mathrm{s})+3 \mathrm{KClO}_4(\mathrm{~s})$
D $4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3$

The correct answer is:
$4 \mathrm{KClO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{KCl}(\mathrm{s})+3 \mathrm{KClO}_4(\mathrm{~s})$

Solution:
Reaction (c) gives oxidation as well as reduction of same (one) species.

Topic: Chemistry, Redox Reactions, JEE Main
Which one of following is employed as a tranquilizer ?

Options:
A Chlorpheninamine
B Equanil
C Naproxen
D Tetracycline

The correct answer is:
Equanil

Solution:
Equanil is a diasaccharide [Everday life]

Topic: Chemistry, Polymers, NEET, NEET 2009 (Mains)
Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states ?

Options:
A $3 d^2 4 s^2$
B $3 \mathrm{~d}^3 4 \mathrm{~s}^2$
C $3 d^5 4 s^1$
D $3 d^5 4 s^2$

The correct answer is:
$3 d^5 4 s^2$

Solution:
$7^{\text {th }}$ group $\rightarrow$ largest number of oxidation state.

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2009 (Mains)
Which one of the factors does not affect rate of the reaction?

Options:
A Molecularity
B Concentration
C Nature of reactants
D Temperature

The correct answer is:
Molecularity

Solution:
The molecularity of a reaction is defined as the number of reacting molecules which collide simultaneously to bring about a chemical reaction. In other words, the molecularity of an elementary reaction is defined as the number of reactant molecules taking part in the reaction.

Molecularity does not affect the rate of the reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
Which one of the following 0.06 M aqueous solutions has lowest freezing point ?

Options:
A Al2SO43
B C6H12O6
C KI
D K2SO4

The correct answer is:
Al2SO43

Solution:
Tf∘-Tf=iKf·m

Tf = Freezing point of the solution.

i=Van 't Hoff factor

Kf = molal freezing point depression constant

m=molality
For minimum Tf, i should have the maximum value.

Al2SO43i=5C6H12O6i=1KIi=2K2SO4i=3

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?

Options:
A glycine
B hydrazine
C KHSO4
D glucose

The correct answer is:
KHSO4

Solution:
ΔTf=ikfm
As m is same for all electrolyte so
ΔTf α i m
Greater the value of i greater is depression in freezing point
For KHSO4i=3
And for Glucose, Ethylele Glycol & glycerol i=1

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Which one of the following acids does not exhibit optical isomerism?

Options:
A
Lactic acid

B
Tartaric acid

C
Maleic acid

D
$\alpha$-amino acids

The correct answer is:
Maleic acid

Solution:
Optical isomers contain an asymmetric (chiral) carbon atom (a carbon atom attached to four different atoms or groups). Therefore all except maleic acid exhibit optical isomerism.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2014 (12 Apr Online)
Which one of the following acids is a derivative of carotenoids ?

Options:
A Abscisic acid
B Indole butyric acid
C Indole-3-acetic acid
D Gibberellic acid

The correct answer is:
Abscisic acid

Solution:
No Solution Available

Topic: Biology, Plant - Growth and Development, NEET, NEET 2009 (Mains)
Which one of the following acids is the weakest

Options:
A $\mathrm{HClO}$
B $\mathrm{HBr}$
C $\mathrm{HClO}_3$
D $\mathrm{HCl}$

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

Solution:
$\underset{\text { Weak acid }}{\mathrm{HClO}} \rightleftharpoons \mathrm{H}^{+}+\underset{\substack{\text { Strong conjugate } \\ \text { base }}}{\mathrm{ClO}^{-}}$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which one of the following alkaline earth metal ions has the highest ionic mobility in its aqueous solution?

Options:
A Sr2+
B Mg2+
C Ca2+
D Be2+

The correct answer is:
Sr2+

Solution:
Larger the size of cation, lesser it gets hydrated. Hence, show higher mobility in aqueous solution. Among the given species Sr2+ ion has the largest size, so hydrate to smaller extent and show the highest mobility.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?

Options:
A SrSO4
B CaSO4
C BeSO4
D BaSO4

The correct answer is:
BeSO4

Solution:
Δ H Hydration > Δ H Lattice
Salt is soluble. BeSO4 is soluble due to high hydration energy of small Be2+ ion. Ksp for BeSO4 is very high.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2015 (04 Apr)
Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?

Options:
A CH3O-CH=CH2
B Cl-CH=CH2
C H2N-CH=CH2
D F3C-CH=CH2

The correct answer is:
F3C-CH=CH2

Solution:
Markovnikov’s product is formed when a stable carbocation is formed on the more substituted carbon.

In CF3-CH=CH2, the carbocation formed will be destabilized by strong -I group of CF3. Hence, the Anti-Markovnikov’s product will be formed.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Which one of the following animals may occupy more than one trophic levels in the same ecosystem at the same time?

Options:
A Sparrow
B Lion
C Goat
D Frog

The correct answer is:
Sparrow

Solution:
Sparrow feeds upon grains hence, is called primary consumer or herbivore and belongs to second trophic level $\left(\mathrm{T}_2\right)$. It also feeds upon insects in the same ecosystem at the same time, hence, called secondary consumer or primary carnivores, and now belongs to third trophic level $\left(\mathrm{T}_3\right)$.

Topic: Biology, Ecosystem, NEET, NEET 2011 (Mains)
Which one of the following aqueous solutions will exhibit highest boiling point?

Options:
A
$0.01 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4$

B
$0.015 \mathrm{M}$ glucose

C
$0.015 \mathrm{M}$ urea

D
$0.01 \mathrm{M} \mathrm{KNO}_3$

The correct answer is:
$0.01 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4$

Solution:
Von't Hoffs factor (i) for $\mathrm{Na}_2 \mathrm{SO}_4$ is maximum i.e. 3( maximum no of particles)
$$
\mathrm{Na}_2 \mathrm{SO}_4 \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{SO}_4^{-}
$$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2004
Which one of the following are examples of Sandmeyer reaction?

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

The correct answer is:
1, 3, 4

Solution:
Here reactions in 1, 3 and 4 are example of Sandmeyer reaction but not given reaction 2.
The reactions of cuprous salts with is called Sandmeyer reaction.

Topic: Chemistry, Amines, JEE Main
Which one of the following arrangements does not give the correct picture of the trends indicated against it?

Options:
A $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Electronegativity
B $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Oxidizing power
C $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Electron gain enthalpy
D $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Bond dissociation energy

The correct answer is:
$\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$ : Bond dissociation energy

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET, NEET 2008 (Mains)
Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $\mathrm{Hg}_2 \mathrm{Cl}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$, $\mathrm{BaSO}_4$ and $\mathrm{CrCl}_3$ respectively?

Options:
A
$\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3>\mathrm{CrCl}_3$

B
$\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

C
$\mathrm{BaSO}_4>\mathrm{CrCl}_3>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

D
$\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{BaSO}_4>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

The correct answer is:
$\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

Solution:
$$
\text { } \begin{aligned}
& \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3 \leftrightharpoons 2 \mathrm{Cr}^{+++}+3 \mathrm{SO}_4^{--} \\
& \mathrm{K}_{\mathrm{sp}}=(2 s)^2(3 s)^3=4 \mathrm{~s}^2 \times 27 s^3=108 s^5 \\
& \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{108}\right)^{1 / 5} \\
& \mathrm{Hg}_2 \mathrm{Cl}_2 \leftrightharpoons 2 \mathrm{Hg}^{2+}+2 \mathrm{Cl}^{-} \\
& \mathrm{K}_{\mathrm{sp}}=(2 s)^2 \times(2 s)^2=16 s^4 \\
& \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{16}\right)^{1 / 4} \\
& \mathrm{BaSO}_4 \leftrightharpoons \mathrm{Ba}^{++}+\mathrm{SO}_4^{-} \\
& \mathrm{K}_{\mathrm{sp}}=s^2 \\
& \mathrm{~s}=\sqrt{\mathrm{K}_{\mathrm{sp}}} \\
& \mathrm{CrCl}_3 \leftrightharpoons \mathrm{Cr}^{3+}+3 \mathrm{Cl}^{-} \\
& s
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=s \times(3 s)^3=27 s^4 \\
& s=\left(\frac{\mathrm{K}_{s p}}{27}\right)^{1 / 4}
\end{aligned}
$$
Hence the correct order of solubilities of salts is
$$
\sqrt{\mathrm{K}_{\mathrm{sp}}}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{16}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main
Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts $\mathrm{Hg}_2 \mathrm{Cl}_2, \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$, $\mathrm{BaSO}_4$ and $\mathrm{CrCl}_3$ respectively?

Options:
A
$\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3>\mathrm{CrCl}_3$

B
$\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

C
$\mathrm{BaSO}_4>\mathrm{CrCl}_3>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

D
$\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{BaSO}_4>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

The correct answer is:
$\mathrm{BaSO}_4>\mathrm{Hg}_2 \mathrm{Cl}_2>\mathrm{CrCl}_3>\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3$

Solution:
$$
\text { } \begin{aligned}
& \mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3 \leftrightharpoons 2 \mathrm{Cr}^{+++}+3 \mathrm{SO}_4^{--} \\
& \mathrm{K}_{\mathrm{sp}}=(2 s)^2(3 s)^3=4 \mathrm{~s}^2 \times 27 s^3=108 s^5 \\
& \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{108}\right)^{1 / 5} \\
& \mathrm{Hg}_2 \mathrm{Cl}_2 \leftrightharpoons 2 \mathrm{Hg}^{2+}+2 \mathrm{Cl}^{-} \\
& \mathrm{K}_{\mathrm{sp}}=(2 s)^2 \times(2 s)^2=16 s^4 \\
& \mathrm{~s}=\left(\frac{\mathrm{K}_{s p}}{16}\right)^{1 / 4} \\
& \mathrm{BaSO}_4 \leftrightharpoons \mathrm{Ba}^{++}+\mathrm{SO}_4^{-} \\
& \mathrm{K}_{\mathrm{sp}}=s^2 \\
& \mathrm{~s}=\sqrt{\mathrm{K}_{\mathrm{sp}}} \\
& \mathrm{CrCl}_3 \leftrightharpoons \mathrm{Cr}^{3+}+3 \mathrm{Cl}^{-} \\
& s
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=s \times(3 s)^3=27 s^4 \\
& s=\left(\frac{\mathrm{K}_{s p}}{27}\right)^{1 / 4}
\end{aligned}
$$
Hence the correct order of solubilities of salts is
$$
\sqrt{\mathrm{K}_{\mathrm{sp}}}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{16}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 4}>\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5}
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (22 Apr Online)
Which one of the following arrangements represents the correct order of the proton affinity of the given species :

Options:
A
$\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-} < \mathrm{NH}_2^{-}$
B
$\mathrm{HS}^{-} < \mathrm{NH}_2^{-} < \mathrm{F}^{-} < \mathrm{I}^{-}$
C
$\mathrm{F}^{-} < \mathrm{I}^{-} < \mathrm{NH}_2^{-} < \mathrm{HS}^{-}$
D
$\mathrm{NH}_2^{-} < \mathrm{HS}^{-} < \mathrm{I}^{-} < \mathrm{F}^{-}$

The correct answer is:
$\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-} < \mathrm{NH}_2^{-}$

Solution:
The species with the greatest proton affinity will be the strongest base, and its conjugate acid will be the weakest acid. The weakest acid will have the smallest value of $\mathrm{K}_a$. Since $\mathrm{HI}$ is a stronger acid than $\mathrm{HF}$ which is a stronger acid than $\mathrm{H}_2 \mathrm{S}$, a partial order of proton affinity is
$\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-}$
Since $\mathrm{NH}_3$ is a very weak acid, $\mathrm{NH}_2^{-}$must be a very strong base. Therefore the correct order of proton affinity is
$\mathrm{I}^{-} < \mathrm{F}^{-} < \mathrm{HS}^{-} < \mathrm{NH}_2^{-}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2013 (25 Apr Online)