datavorous/entrance-exam-dataset huggingface dataset:
Which is the correct order of decreasing boiling point of azeotropic mixtures X-H2O ?

Options:
A HI>HBr>HF>HCl
B HI>HCl>HBr>HF
C HF>HCl>HBr>HI
D HCI>HF>HBr>HI

The correct answer is:
HI>HCl>HBr>HF

Solution:
The decreasing order of boiling point of azeotropic mixture, HX-H2O is HI127oC>HCl1100C>HBr1060C>HF(20oC)

Topic: Chemistry, Solutions, JEE Main
Which is the correct order of increasing basicity?

Options:
A CH 3 CH 2 CH 3 < CH 3 CH 2 SH < CH 3 CH 2 OH < CH 3 CH 2 NH 2
B CH 3 CH 2 CH 3 < CH 3 CH 2 OH < CH 3 CH 2 SH < CH 3 CH 2 NH 2
C CH 3 CH 2 NH 2 < CH 3 CH 2 SH < CH 3 CH 2 OH < CH 3 CH 2 CH 3
D CH 3 CH 2 CH 3 < CH 3 CH 2 OH < CH 3 CH 2 NH 2 < CH 3 CH 2 SH

The correct answer is:
CH 3 CH 2 CH 3 < CH 3 CH 2 SH < CH 3 CH 2 OH < CH 3 CH 2 NH 2

Solution:
CH 3 - CH 2 - CH 3 } No lone pair, least basic
CH 3 - CH 2 - OH CH 3 - CH 2 - NH 2 } E.N. â B.S. â Size same
CH3CH2SH size increases, basicity decreases

Topic: Chemistry, General Organic Chemistry, JEE Main
Which is the correct order of ionic sizes ?
(Atomic Number : Ce = 58, Sn = 50, Yb = 70 and Lu = 71)

Options:
A
Ce > Sn > Yb > Lu

B
Sn > Ce > Lu > Yb

C
Lu > Yb > Sn > Ce

D
Sn > Yb > Ce > Lu

The correct answer is:
Ce > Sn > Yb > Lu

Solution:
No Solution Available

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2002
Which is the correct order of second ionization potential of $\mathrm{C}, \mathrm{N}, \mathrm{O}$ and $\mathrm{F}$ in the following ?

Options:
A
$\mathrm{O}>\mathrm{N}>\mathrm{F}>\mathrm{C}$

B
$\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

C
F $>$ O $>\mathrm{N}>\mathrm{C}$

D
$\quad$ C $>$ N $>\mathrm{O}>\mathrm{F}$

The correct answer is:
$\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

Solution:
The second ionization potential means removal of electron from cation $\mathrm{C}^{+}=1 s^2 2 s^2 2 p^1, \mathrm{~N}^{+}=1 s^2 2 s^2 2 p^2$ $\mathrm{O}^{+}=1 s^2 2 s^2 2 p^3, \mathrm{~F}^{+}=1 s^2 2 s^2 2 p^4$ Therefore $\mathrm{O}>\mathrm{F}>\mathrm{N}>\mathrm{C}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2013 (23 Apr Online)
Which is the correct relation between glucose and mannose?

Options:
A Structural isomers
B Disaccharides
C Epimers
D Anomers

The correct answer is:
Epimers

Solution:
Glucose and mannose are epimers because they both differ in configuration at C-2 and the Pair of diastereomer which differ at in configuration at only one chiral atom is known as epimers of each other.

Topic: Chemistry, Biomolecules, JEE Main
Which is the decreasing order of stability
(i) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3$
(ii) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{O}-\mathrm{CH}_3$
(iii) $\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{CO}-\mathrm{CH}_3$

Options:
A $(i) \lt (i i) \lt (i i i)$
B (i) $\gt($ ii $)\gt($ iii $)$
C $(i i i)\gt(i i)\gt(i)$
D $(i i)\gt(i i i)\gt(i)$

The correct answer is:
(i) $\gt($ ii $)\gt($ iii $)$

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Which is the example of hexadentate ligand

Options:
A 2, 2—dipyridyl
B Dimethyl glyoxime
C Aminodiacetate ion
D Ethylene diammine tetra acetate ion [EDTA]

The correct answer is:
Ethylene diammine tetra acetate ion [EDTA]

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main
Which is the incorrect statement for 1 L sample of N2(g) and CO2(g) at 298 K and 1 atm pressure

Options:
A The average translational KE per molecule is the same in N2 and CO2
B The rms speed remains same for both N2 and CO2
C The density of N2 is less than that of CO2
D The most probable velocity is different for the two gases.

The correct answer is:
The rms speed remains same for both N2 and CO2

Solution:
Kinetic energies per molecule will be same because it is proportional to absolute temperature only.
dN2dCO2=MN2MCO2=2844 i.e., d N 2 < d C O 2 .
Total translational kinetic energy will also be same because at same temperature & pressure number of molecules present in same volume would be same (according to Avogadro's Law)
Hint : rms speed has relation, USO2UO2=MO2TSO2MSO2TO2

Topic: Chemistry, States of Matter, NEET
Which is the main postulate of Bohr’s model of atom in the following

Options:
A Protons are present in the nucleus
B Electrons are revolving around the nucleus
C Centrifugal force produced due to the revolving electrons balances the force of attraction between the electron and the protons
D Angular momentum of electron is an integral multiple of h2π

The correct answer is:
Angular momentum of electron is an integral multiple of h2π

Solution:
The electron can move only in those circular orbits where the angular momentum is a whole number multiple of h2π or it is quantised.

Topic: Chemistry, Structure of Atom, NEET
Which is the major product formed when acetone is heated with iodine and potassium hydroxide?

Options:
A Acetophenone
B Iodoform
C Iodoacetone
D Acetic acid

The correct answer is:
Iodoform

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2014 (09 Apr Online)
Which is the monomer of teflon from the following?

Options:
A \( \mathrm{F}_{2} \mathrm{C}=\mathrm{CF}_{2} \)
B
C \( \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2} \)
D

The correct answer is:
\( \mathrm{F}_{2} \mathrm{C}=\mathrm{CF}_{2} \)

Solution:
F2C=CF2 is monomer of teflon

Topic: Chemistry, Polymers, JEE Main
Which is the most hazardous metal pollutant of automobile exhaust

Options:
A Mercury
B 2 Tin
C 3 Cadmium
D 4 Lead

The correct answer is:
4 Lead

Solution:
Lead acts as a toxic substance.

Topic: Chemistry, Environmental Chemistry, JEE Main
Which is the most stable carbocation among the following?

Options:
A Iso-butyl cation
B Triphenyl methyl cation
C Cyclopropyl methyl cation
D t-Butyl cation

The correct answer is:
Triphenyl methyl cation

Solution:
Here, triphenylmethyl cation is most stable as it is resonance stabilised by three benzene rings

Topic: Chemistry, General Organic Chemistry, JEE Main
Which is the most suitable reagent among the following to distinguish compound (iii) from the rest of the compounds

i. CH3-C≡C-CH3

ii. CH2-CH2-CH2-CH3

iii. CH3-CH2-C≡CH

iv. CH3-CH=CH2

Options:
A Br2 in CCl4
B Cold aqueous alkaline KMnO4
C Ammonical AgNO3 solution
D Os O 4

The correct answer is:
Ammonical AgNO3 solution

Solution:
Terminal alkyne reacts with ammonical AgNO3 solution and gives white precipitate because it has acidic hydrogen on sp hybridized carbon.

Topic: Chemistry, Hydrocarbons, JEE Main
Which is the most suitable reagent for the following transformation?

Options:
A Tollen's reagent
B I2/NaOH
C Alkaline KMnO4
D CrO2Cl2/CS2

The correct answer is:
I2/NaOH

Solution:
It is an iodoform reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Which is the most thermodynamically stable allotropic form of phosphorus

Options:
A Red
B White
C Black
D Yellow

The correct answer is:
Black

Solution:
The ignition temperature of black phosphorus is highest among all allottropes.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which is the most thermodynamically stable allotropic form of phosphorus

Options:
A Red
B White
C Black
D Yellow

The correct answer is:
Black

Solution:
The ignition temperature of black phosphorus is highest among all allottropes

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which is the most volatile?

Options:
A CH3CH2CH2NH2
B CH33N
C
D CH3–CH2–OH

The correct answer is:
CH33N

Solution:
Volatile nature Volatile nature∝ 1 Boiling point
CH33N→ No H-bond (less Boiling point)
Hence it is most volatile.

Topic: Chemistry, Amines, JEE Main
Which is the right order for the solubility of sulphates of alkaline earth metals in water

Options:
A Be>Ca>Mg>Ba>Sr
B Be>Mg>Ca>Sr>Ba
C Mg>Be>Ba>Ca>Sr
D Mg>Ca>Ba>Be>Sr

The correct answer is:
Be>Mg>Ca>Sr>Ba

Solution:
Be>Mg>Ca>Sr>Ba
On moving down the group lattice energy remains almost constant as the sulphate is so big that small increase in the size of the cations from Be to Ba does not make any difference. However, the hydration energy decreases from Be+2 to Ba+2 . This causes decrease in the solubility of the sulphates as the ionic size increases.

Topic: Chemistry, s Block Elements, NEET
Which is the SI unit of radioactivity?

Options:
A Henery
B Becquerel
C Rutherford
D Farad

The correct answer is:
Becquerel

Solution:
Becquerel is the unit of radioactivity.

Topic: Physics, Nuclear Physics, JEE Main
Which is the strongest base among the following?

Options:
A $$
\text { } \mathrm{H}_3 \mathrm{CNH}_2
$$
B

C

D

The correct answer is:
$$
\text { } \mathrm{H}_3 \mathrm{CNH}_2
$$

Solution:
The species having more lone pair of electrons for donation are strong base.
In case of $\mathrm{CH}_3 \mathrm{NH}_2$, lone pair of electrons on $\mathrm{N}$-atom is available for donation. Hence, it is most basic. Order of basic nature of the given compounds is

Topic: Chemistry, Amines, JEE Main
Which is the strongest base among the following

Options:
A C6H5NH2
B p-NO2C6H4NH2
C m-NO2C6H4NH2
D C6H5CH2NH2

The correct answer is:
C6H5CH2NH2

Solution:
Lone pair is not taking part in resonance, most basic. In other cases, lone pair of nitrogen is part of delocalisation which decreases Lewis base strength.

Topic: Chemistry, General Organic Chemistry, JEE Main
Which is true about Buna-N?

Options:
A It is a linear polymer of 1,3-butadiene.
B It is obtained by copolymerization of 1,3 -butadiene and styrene.
C It is obtained by copolymerization of 1,3 -butadiene and acrylonitrile.
D The suffix N in Buna- N stands for its natural occurrence.

The correct answer is:
It is obtained by copolymerization of 1,3 -butadiene and acrylonitrile.

Solution:
Buna-N is formed by the co-polymerization of 1,3-butadiene and vinylcyanide(acrylonitrile) in the presence of a peroxide catalyst.

It is a synthetic rubber.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Which is true with regard to the properties of $\mathrm{PH}_3$

Options:
A $\mathrm{PH}_3$ is not much stable
B $\mathrm{PH}_3$ is neutral towards litmus
C $\mathrm{PH}_3$ has fishy smell
D $\mathrm{PH}_3$ is insoluble in water

The correct answer is:
$\mathrm{PH}_3$ is insoluble in water

Solution:
$\mathrm{PH}_3$ insoluble in water because does not consist of hydrogen bond.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which is used in the Haber process for the manufacture of $\mathrm{NH}_3$

Options:
A $P_t$
B $\mathrm{Fe}+\mathrm{Mo}$
C $\mathrm{CuO}$
D $\mathrm{Al}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{Fe}+\mathrm{Mo}$

Solution:
In the Haber process for the manufacture of $\mathrm{NH}_3, \mathrm{Fe}$ is used catalyst and $\mathrm{Mo}$ as a promoter

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which is wrong about oogenesis?

Options:
A Diploid oogonia
B Growth phase
C Formation of polar bodies
D Equal meiotic division

The correct answer is:
Equal meiotic division

Solution:
During oogenesis primary oocyte undergoes first meiotic division. This results in the formation of two daughter cells. These are secondary oocyte and a first polar body. These are unequal in size. Polar bodies are much smaller in diameter than the oocyte because the cytoplasm is retained in bulk in the secondary oocyte. This occurs because the secondary oocyte will get converted into ovum and if it gets fertilized will form zygote. Hence the retained cytoplasm will act as stored food for the developing zygote. This is the main difference between oogenesis and meiosis division in other cells.

Topic: Biology, Human Reproduction, NEET
Which is wrong with respect to our responsibility as a human being to protect our environment?

Options:
A Using plastic bags.
B Restricting the use of vehicles
C Avoiding the use of floodlighted facilities
D Setting up compost tin in gardens

The correct answer is:
Using plastic bags.

Solution:
Plastic is non-biodegradable. Hence, to protect our environment, we should not use plastic bags.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Which is/are correct statement(s) for XeF2 ?

Options:
A It has bent structure
B It is hydrolysed rapidly in aqueous solution of a base
C It oxidizes Cl- and I- to Cl2 and I2 respectively
D Both B and C

The correct answer is:
Both B and C

Solution:
Here statements in option B and C are correct while A is incorrect as the five electrons pairs form trigonal bipyramid with three lone pairs in equatorial positions. Hence, the structure of XeF2 becomes linear in accordance with VSEPR theory.
(i) XeF2+2HCl→2HF+Xe+Cl2
(ii) XeF2+2KI→Xe+2KF+I2

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Which is/are correct statement/s?

Options:
A In flowering plants, tracheids and vessels are the main water transporting elements.
B In gymnosperms, tracheids are the chief water transporting elements
C The companion cells play an important role in the maintenance of a pressure gradient in the sieve tubes
D All of these

The correct answer is:
All of these

Solution:
Gymnosperms lack vessels in their xylem. Tracheids are elongated or tube-like cells with thick and lignified walls and tapering ends. These are dead and are without protoplasm. The inner layers of the cell walls have thickenings which vary in form. In flowering plants, tracheids and vessels are the main water transporting elements. The vessel is a long cylindrical tube-like structure made up of many cells called vessel members, each with lignified walls and a large central cavity.
Companion cells, along with phloem parenchyma play an important role in the maintenance of a pressure gradient in the sieve tubes.

Topic: Biology, Anatomy of Flowering Plants, NEET
Which is/are incorrect statements regarding photoelectric effect?

Options:
A Photoelectric effect was discovered by Hertz and investigated by Hallwach's and Lenard.
B For a given frequency of incident radiation stopping potential is independent of its intensity.
C For a given intensity of radiation, saturation current is independent of frequency.
D For a given frequency of incident radiation stopping potential depends on its intensity.

The correct answer is:
For a given frequency of incident radiation stopping potential depends on its intensity.

Solution:
The amount of potential that is required to stop the electron which has the maximum kinetic energy from moving is known as stopping potential.

The stopping potential in the photoelectric effect only depends on the frequency of the incident light and the nature of the material.

It is independent of the intensity of incident light for a given frequency.

Topic: Physics, Dual Nature of Matter, JEE Main
Which kind of population interaction is found between organism showing pseudocopulation?

Options:
A Ammensalism
B Commensalism
C Proto-cooperation
D Co-evolution

The correct answer is:
Co-evolution

Solution:
The Mediterranean orchid Ophrys employs ‘sexual deceit’ to get pollination done by a species of bee. One petal of its flower bears an uncanny resemblance to the female of the bee in size, colour and markings. The male bee perceives it as female and dusted with pollen from the flower during the process of pseudocopulation. When same bee pseudo-copulates with another flower, it transfers pollen to it and thus pollinates the flower. Thus, best cheater plant is Ophrys.
The significance of co-evolution process can be understood here. If female bee changes its colour pattern ever slightly, the success of pollination will be reduced unless orchid flower co-evolves to maintain the resemblance of its petal to female bee.

Topic: Biology, Organisms and Populations, NEET
Which law of the thermodynamics helps in calculating the absolute entropies of various substances at different temperatures?

Options:
A First law
B Second law
C Third law
D Zeroth law

The correct answer is:
Third law

Solution:
The third law helps to calculate the absolute entropies of pure substances at different temperature. The entropy of the substance at different temperature. $\mathrm{T}$ may be calculated by the measurement of heat capacity change

$$

\begin{array}{l}

\mathrm{S}_{\mathrm{T}}-\mathrm{S}_{0}=\Delta \mathrm{S}=\int_{0}^{\mathrm{T}} \frac{\mathrm{C}_{\mathrm{p}} \mathrm{dt}}{\mathrm{T}} \\

\text { Where } \mathrm{S}_{\mathrm{T}}=\text { Entropy at } \mathrm{T} \mathrm{K}

\end{array}

$$

$\mathrm{S}_{0}=$ Entropy at $0 \mathrm{~K}$

$$

\begin{array}{l}

=\mathrm{C}_{\mathrm{p}} \cdot \log _{e} \mathrm{~T} \\

=2.303 \mathrm{C}_{\mathrm{p}} \mathrm{log} \mathrm{T}

\end{array}

$$

Topic: Chemistry, Thermodynamics (C), JEE Main
Which leaves the liver and moves towards kidney?

Options:
A Ammonia
B Urea
C Glycogen
D Bile

The correct answer is:
Urea

Solution:
Urea is the main nitrogenous excretory product of ureotelic animals. It is produced by liver cells from deaminated excess amino acid via urea cycle or Ornithine cycle. Urea is released outside the body in the form of urine through the kidneys.

Topic: Biology, Excretory Products and their Elimination, JIPMER, JIPMER 2006
Which level of the single ionized carbon has the same energy as the ground state energy of hydrogen atom?

Options:
A 1
B 6
C 4
D 8

The correct answer is:
6

Solution:
Energy of H -atom is E=-13.6Z2/n2 for H-atom Z=1 and for ground state, n=1

⇒E=-13.6×1212=-13.6 eV

Now for carbon atom (single ionised), Z=6

E=-13.6Z2n2=-13.6 (given)

⇒n2=62⇒n=6

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Which logic gate is represented by the following combination of logic gates

Options:
A OR
B NAND
C AND
D NOR

The correct answer is:
AND

Solution:
$Y=\overline{\bar{A}+\bar{B}}$
According to De morgan's theorem
$Y=\overline{\bar{A}}+\overline{\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}=A \cdot B$
This is the output equation of 'AND' gate.

Topic: Physics, Semiconductors, JEE Main
Which logic gate with inputs $\mathrm{A}$ and $\mathrm{B}$ performs the same operation as that performed by the following circuit?

Options:
A
NAND gate

B
OR gate

C
NOR gate

D
AND gate

The correct answer is:
OR gate

Solution:
When either of $A$ or $B$ is 1 i.e. closed then lamp will glow.
In this case, Truth table

Topic: Physics, Semiconductors, JEE Main, JEE Main 2012 (07 May Online)
Which lupine was excavated from the tundra that germinated and flowered after a record 10,000 years?

Options:
A Lupinus polyphyllus
B Lupinus Lepidus
C Lupinus littoralis
D Lupinus arcticus

The correct answer is:
Lupinus arcticus

Solution:
After their dispersal, seeds remain viable for different periods. Some seeds can be very old yet viable. Lupinus arcticus excavated from Arctic Tundra has a seed that germinated after 10,000 years of dormancy. Date palm, Phoenix dactylifera, has a 2000-year-old viable seed. It was discovered during the archeological excavation at King Herod's palace near the Dead Sea.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Which mechanism best describes the formation of oxiranes from alkenes using peracids?

Options:
A Radical mechanism
B Electrophilic addition
C Nucleophilic substitution
D Concerted mechanism

The correct answer is:
Concerted mechanism

Solution:
Concerted mechanism
Explanation: The formation of oxiranes from alkenes using peracids proceeds via a concerted mechanism where the peracid adds across the double bond in a single, synchronous step.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Which member of group 13 does not exhibit the group valency in its compounds?

Options:
A Boron
B Aluminium
C Gallium
D Thallium

The correct answer is:
Thallium

Solution:
The usual oxidation state of group 13 is +3, but due to inert pair effect, both the s-electrons of the outer shell remains paired, due to energy considerations in case of thallium. Consequently, Tl show +1 oxidation state.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Which metal has a greater tendency to form metal oxide?

Options:
A Cr
B $\mathrm{Fe}$
C Al
D Ca

The correct answer is:
Ca

Solution:
Greater the stability of oxide, greater is the ease of its formation. Generally ionic oxides are more stable than covalent oxides and among the given metals only Ca form ionic oxide. Hence, Ca has greater tendency to form oxide.

Topic: Chemistry, d and f Block Elements, JEE Main
Which method of purification is represented by the equation
$\underset{\text { Impure }}{T i}+2 I_2 \xrightarrow{500 \mathrm{~K}} T i I_4 \xrightarrow{1675 \mathrm{~K}} T i+2 I_2$

Options:
A Cupellation
B Poling
C Van Arkel
D Zone refining

The correct answer is:
Van Arkel

Solution:
Van Arkel method $T i$ and $Z n$ are refined by this method. It is used for obtaining ultra pure metals.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Which method of purification is represented by the following equation:
$\mathrm{Ti}(\mathrm{s})+2 \mathrm{I}_2(\mathrm{g}) \stackrel{523 \mathrm{~K}}{\longrightarrow} \mathrm{TiI}_4(\mathrm{g}) \stackrel{1700 \mathrm{~K}}{\longrightarrow} \mathrm{Ti}(\mathrm{s})+2 \mathrm{I}_2(\mathrm{g})$

Options:
A
zone refining

B
cupellation

C
Poling

D
Van Arkel

The correct answer is:
Van Arkel

Solution:
Van Arkel method
$\begin{aligned}
& \mathrm{Ti}(\mathrm{s})+2 \mathrm{I}_2(\mathrm{g}) \stackrel{523 \mathrm{~K}}{\longrightarrow} \mathrm{TiI}_4(\mathrm{g}) \\
& \mathrm{Til}_4(\mathrm{g}) \stackrel{1700 \mathrm{~K}}{\longrightarrow} \mathrm{Ti}(\mathrm{s})+2 \mathrm{I}_2(\mathrm{g})
\end{aligned}$

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2012 (Offline)
Which mixture is separated by conc. aqueous solution of sodium hydroxide

Options:
A $A l^{3+}$ and $S n^{2+}$
B $A l^{3+}$and $F e^{3+}$
C $A l^{3+}$ and $Z n^{2+}$
D $\mathrm{Zn}^{2+}$ and $\mathrm{Pb}^{2+}$

The correct answer is:
$A l^{3+}$and $F e^{3+}$

Solution:
$\begin{aligned} & \mathrm{NaOH}+\mathrm{Al}^{+3} \rightarrow \mathrm{NaAlO}_2 \\ & \mathrm{NaOH}+\mathrm{Fe}^{+3} \rightarrow \text { Noreaction(Insoluble in } \mathrm{NaOH} \text { ) }\end{aligned}$

Topic: Chemistry, Practical Chemistry, JEE Main
Which mixture of the solutions will lead to the formation of negatively charged colloidal AgII- sol?

Options:
A 50 mL of 1 M AgNO3+50 mL of 0.5 M KI
B 50Â mLÂ ofÂ 1Â MÂ AgNO3+50Â mLÂ ofÂ 2Â MÂ KI
C 50Â mLÂ ofÂ 2Â MÂ AgNO3+50Â mLÂ ofÂ 1.5Â MÂ KI
D 50Â mLÂ ofÂ 0.1Â MÂ AgNO3+50Â mLÂ ofÂ 0.1Â MÂ KI

The correct answer is:
50 mL of 1 M AgNO3+50 mL of 2 M KI

Solution:
The common ion between the colloidal particles and the remaining solution gets adsorbed on the surface of the colloidal particles and thus the particles acquire charge.

Formation of negatively charged colloidal [AgI]I- sol:

50 ml of 1 M AgNO3+50 ml  2.0 M KI

Millimoles of Ag+=50×1 m=50 m moles

Millimoles of I-=50×2=100 m  moles

nI->nAg+

∴ it will form negatively charged solution.
[AgI/I-]

Topic: Chemistry, Surface Chemistry, NEET, NEET 2019 (All India)
Which molecule has the largest dipole moment

Options:
A $\mathrm{HCl}$
B $\mathrm{HI}$
C $\mathrm{HBr}$
D $H F$

The correct answer is:
$H F$

Solution:
$H F$ has largest dipole moment because electronegativity difference of both is high so it is highly polar.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Which molecule is exclusively responsible for maintaining the identity of different species of organisms over millions of years?

Options:
A RNA
B Carbohydrates
C DNA
D Proteins

The correct answer is:
DNA

Solution:
DNA is responsible for maintaining the genetic identity of different species over long periods of time through its ability to store and transmit genetic information.

Topic: Chemistry, Biomolecules, NEET
Which molecule/ion among the following cannot act as a ligand in complex compounds?

Options:
A CN-
B CH4
C CO
D Br-

The correct answer is:
CH4

Solution:
For a molecule to act as a ligand, it should contain lone pair or negative charge. CH4 does not have either a lone pair or π -electron pair it cannot act as ligand.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2015 (10 Apr Online)