datavorous/entrance-exam-dataset huggingface dataset:
$$
\text { If } A=\left[\begin{array}{ccc}
0 & c & -b \\
-c & 0 & a \\
b & -a & 0
\end{array}\right] \text { and } B=\left[\begin{array}{ccc}
a^2 & a b & a c \\
a b & b^2 & b c \\
a c & b c & c^2
\end{array}\right] \text {, }
$$
then $A B$ is equal to

Options:
A
B
B
A
C
O
D
I

The correct answer is:
O

Solution:
$\begin{aligned} &\quad A B=\left[\begin{array}{ccc}0 & c & -b \\ -c & 0 & a \\ b & -a & 0\end{array}\right]\left[\begin{array}{ccc}a^2 & a b & a c \\ a b & b^2 & b c \\ a c & b c & c^2\end{array}\right] \\ & A B=\left[\begin{array}{ccc}a b c-a b c & b^2 c-b^2 c & b c^2-b c^2 \\ -a^2 c+a^2 c & -a b c+a b c & -a c+a c \\ a^2 b-a^2 b & a b^2-a b^2 & a b c-a b c\end{array}\right] \\ & =\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]=\mathrm{O} \\ & \end{aligned}$

Topic: Mathematics, Matrices, JEE Main
BX3+NH3→BX3.NH3+ Heat of adduct formation ΔH
The numerical value of ΔH is found to be maximum for

Options:
A BF3
B BCl3
C BBr3
D BI3

The correct answer is:
BI3

Solution:
Lewis acidic strength
BF3<BCl3<BBr3<BI3
As BI3 is strongest Lewis acid among all boron halides due to back bonding therefore, heat of adduct formation will be maximum numerically, for BI3 .

Topic: Chemistry, Ionic Equilibrium, JEE Main
BX and BY are the magnetic field at the centre of two coils of two coils X and Y respectively, each carrying equal current. If coil X has 200 turns and 20 cm radius and coil Y has 400 turns and 20 cm radius, the ratio of BX and BY is

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

The correct answer is:
1:2

Solution:
Magnetic field at the centre of a single coil is B=μ0i2R

If N number of loops are kept closely B=Nμ0i2R

Therefore,

BxBy=NxNy

⇒BxBy=12

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Ba-122 has half life of 2 min. Experiment has to be done using Ba-122 and it takes 10 min to set up the experiment. If initially 80 g of Ba-122 was taken, how much Ba was left when experiment was started?

Options:
A 2.5 g
B 5 g
C 10 g
D 20 g

The correct answer is:
2.5 g

Solution:
As $\mathrm{T}_{\frac{1}{2}}=2 \mathrm{~min} \Rightarrow 10 \mathrm{~min}=5$ half-lives
We know, $\frac{\mathrm{N}}{\mathrm{N}_0}=\left(\frac{1}{2}\right)^{\mathrm{n}}$
where, $\quad \mathrm{N}=$ remaining of nuclei
$\mathrm{N}_0=$ initial number of nuclei
$\mathrm{n}=$ number of half lives
$\Rightarrow \quad \mathrm{N}=\frac{\mathrm{N}_0}{2^n}=\frac{80}{2^5}=\frac{80}{32}=2.5 \mathrm{~g}$

Topic: Physics, Nuclear Physics, JIPMER, JIPMER 2019
Bacillus thuringiensis forms protein crystals, which contain insecticidal protein

Options:
A binds with epithelial cells of midgut of the insect pest ultimately killing it
B is coded by several genes including the gene cry
C is activated by acid of $\mathrm{pH}$ of the foregut of the insect pest
D does not kill the carrier bacterium which is itself resistant to this toxin

The correct answer is:
binds with epithelial cells of midgut of the insect pest ultimately killing it

Solution:
Soil bacterium Bacillus thuringiensis (Bt) possesses is a gene family cry gene (1-40) over its plasmid, which forms an endotoxic proteins, cry protein (protein crystals) during a particular phase of its growth. These crystals contain a toxic insecticidal protein. This toxin, however does not kill the Bacillus. Actually, the Bt toxin protein exists as inactive protoxins but once an insect ingests the inactive toxin, it is converted into an active form of toxin due to the alkaline $\mathrm{pH}$ of the gut which solubilise the crystals. The activated toxin binds to the surface of midgut epithelial cells and creates pores that cause cell swelling and lysis, and eventually cause death of the insect.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2011 (Mains)
Back bonding always changes

Options:
A bond length
B bond angle
C hybridisation of central atom
D planarity

The correct answer is:
hybridisation of central atom

Solution:
Back bonding is the type of bonding occurs between the atoms in compound in which one atom has lone pair of $e^{-}$and the other has vacant orbital placed adjacent to each other.
planarity changes with back bonding due to the orbital onercaping heads to the election shifting to the neighbouring atom.
$\therefore$ The correct option is (3).

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Back bonding is involved in which of the organometallic compounds

Options:
A $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$
B $\mathrm{Mg}^{2+}\left(\mathrm{C}_5 \mathrm{H}_5^{-}\right)_2$
C $R-M g-X$
D $\left[\left(\mathrm{C}_5 \mathrm{H}_5\right)_2 \mathrm{Fe}\right]$

The correct answer is:
$\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$

Solution:
$\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$

Topic: Chemistry, Coordination Compounds, JEE Main
Back bonding is involved in which of the organometallic compounds

Options:
A $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$
B $\mathrm{Mg}^{2+}\left(\mathrm{C}_5 \mathrm{H}_5^{-}\right)_2$
C $R-M g-X$
D $\left[\left(\mathrm{C}_5 \mathrm{H}_5\right)_2 \mathrm{Fe}\right]$

The correct answer is:
$\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$

Solution:
Back bonding occurs when there is a transfer of electron density from a filled orbital of one atom to an empty orbital of another atom within a molecule. In $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$, aluminum atoms form a dimer through a bridge involving back bonding. Each aluminum has an empty p-orbital that can accept electron density from the filled p-orbitals of the bridging methyl groups. Therefore, the correct answer is $\left[\left(\mathrm{CH}_3\right)_3 \mathrm{Al}\right]_2$.

Topic: Chemistry, Coordination Compounds, JEE Main
Bacteria are considered plant because they

Options:
A are green in colour
B have rigid cell wall
C have chlorophyll
D have stomata

The correct answer is:
have rigid cell wall

Solution:
Main character of plants is the presence of rigid cell wall around the cell. Bacteria are considered as plants because they have rigid cell wall. Presence of stomata, chlorophyll and green colour are not the features of all plants, eg, chlorophyll is absent in fungi, stomata is absent in submerged acquatic hydrophytes or algae.

Topic: Biology, Biological Classification, JIPMER, JIPMER 2010
Bacteria, fungi, lower plants survive in adverse conditions by

Options:
A diapause
B suspended growth
C migration
D formation of thick walled spores.

The correct answer is:
formation of thick walled spores.

Solution:
Bacteria, fungi and lower plants survive in adverse conditions through formation of thick walled spores. Algae form resting spore or akinetes which are thick walled spores and store food material. In fungi, thick walled perennating spores called chlamydospores are formed. Bacteria form endospores which is probably the most resistant living thing known. They may remain alive in boiling water for more than 2 hours.

Topic: Biology, Plant - Growth and Development, AIIMS, AIIMS 2011
Bacterial and fungal enzymes degrade the detritus into simple inorganic substances; this process is called

Options:
A leaching
B catabolism
C fragmentation
D mineralisation

The correct answer is:
catabolism

Solution:
The organic content is broken down into inorganic substances by the process called decomposition. It is brought about by decomposers. Decomposition is mainly an oxygen-dependent process and its rate is dependent on chemical composition and climatic factors. It is brought about by bacterial and fungal microbes called decomposers. Decomposition is a catabolic process.

Topic: Biology, Ecosystem, NEET
Bacterial leaf blight of rice is caused by a species of

Options:
A Erwinia
B Xanthomonas
C Pseudomonas
D Alternaria

The correct answer is:
Xanthomonas

Solution:
Xanthomonas oryzae.

Topic: Biology, Strategies for Enhancement in Food Production, NEET, NEET 2008 (Mains)
Baeyer's reagent is

Options:
A alkaline potassium permanganate
B acidified potassium permanganate
C neutral potassium permanganate
D alkaline potassium manganate

The correct answer is:
alkaline potassium permanganate

Solution:
Baeyer's reagent is $1 \%$ cold dilute alkaline potassium permanganate. It is used to identify unsaturation. All unsaturated compounds lose its purple colour.

Topic: Chemistry, d and f Block Elements, JEE Main
Bag A contains 3 white, 7 red balls and bag B contains 3 white, 2 red balls. One bag is selected at random and a ball is drawn from it. The probability of drawing the ball from the bag A, if the ball drawn in white, is :

Options:
A 14
B 19
C 13
D 310

The correct answer is:
13

Solution:
Given: Given balls in bags are A→3W, 7R, B→3W, 2R

Let, E1:A is selected

, E2:B is selected

and E: white ball is drawn

Using Baye's theorem,
⇒PE1E=P(E1)·PEE1PE1·PEE1+PE2·PEE2

⇒PE1E=12×31012×310+12×35

⇒PE1E=320320+620

⇒PE1E=33+6

⇒PE1E=13

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Bag $A$ contains 4 green and 3 red balls and bag $B$ contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted it is green. The probability that it comes bag $B$

Options:
A ${ }^{\frac{2}{7}}$
B $\frac{2}{3}$
C ${ }^{\frac{3}{7}}$
D $)^{\frac{1}{3}}$

The correct answer is:
${ }^{\frac{3}{7}}$

Solution:
It is based on Baye's theorem.
Probability of picked bag $A$
$P(A)=\frac{1}{2}$
Probability of picked bag $B$
$P(B)=\frac{1}{2}$
Probability of green ball picked from bag $A$
$=P(A) \cdot P\left(\frac{G}{A}\right)=\frac{1}{2} \times \frac{4}{7}=\frac{2}{7}$
Probability of green ball picked from bag $B$
$=P(B) \cdot P\left(\frac{G}{B}\right)=\frac{1}{2} \times \frac{3}{7}=\frac{3}{14}$
Total probability of green ball $=\frac{2}{7}+\frac{3}{14}=\frac{1}{2}$
Probability of fact that green ball is drawn from bag $B$
$P\left(\frac{G}{B}\right)=\frac{P(B) P\left(\frac{G}{B}\right)}{P(A) P\left(\frac{G}{A}\right)+P(B) P\left(\frac{G}{B}\right)}$
$=\frac{\frac{1}{2} \times \frac{3}{7}}{\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{3}{7}}=\frac{3}{7}$

Topic: Mathematics, Probability, JEE Main
Bag A contains 4 green and 3 red balls and bag B contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted to be green. The probability that it comes from bag B, is

Options:
A 27
B 23
C 37
D 13

The correct answer is:
37

Solution:
Let A is the event that ball is drawn from Bag A, B is the event that ball is drawn from Bag B and G be the event that a green ball is drawn
⇒PGA= 4C1 7C1=47 and PGB= 3C1 7C1=37
Now, PBG=P(B).PGBP(A)PGA+P(B)PGB
PBG=12.3712.47+12.37=37

Topic: Mathematics, Probability, JEE Main
Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is

Options:
A 49
B 518
C 16
D 310

The correct answer is:
518

Solution:
Given,

In bag 1→3R, 4B & 3W

And in bag 2→2R, 5B & 2W

Let A: Drawn ball from bag II is black

And let B: Red ball transferred

Now by total probability theorem we have PA=39×510+49×610+39×510=5490

And PA∩B=39×510

So, by bayes theorem we have, PBA=PA∩BPA

=39×51039×510+49×610+39×510

=1515+24+15=1554=518

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Bakelite is a cross-linked polymer of formaldehyde and:

Options:
A PHBV
B Buna-S
C Novolac
D Dacron

The correct answer is:
Novolac

Solution:
Novolac (phenol formaldehyde Resin) + Formaldehyde → Bakelite

Topic: Chemistry, Polymers, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Bakelite is obtained from phenol by reacting it with

Options:
A
Acetaldehyde

B
Chlorobenzene

C
Formaldehyde

D
Acetamide

The correct answer is:
Formaldehyde

Solution:
Phenol and formaldehyde undergo condensation polymerisation under two different conditions to give a cross linked polymer called bakelite.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2012 (26 May Online)
Bakelite is obtained from phenol by reacting with:

Options:
A
Acetal

B
$\mathrm{CH}_3 \mathrm{CHO}$

C
$\mathrm{HCHO}$

D
Chlorobenzene

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

Solution:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2013 (25 Apr Online)
Bakelite is obtained from phenol by reacting with

Options:
A
$\left(\mathrm{CH}_2 \mathrm{OH}\right)_2$

B
$\mathrm{CH}_3 \mathrm{CHO}$

C
$\mathrm{CH}_3 \mathrm{COCH}_3$

D
$\mathrm{HCHO}$

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

Solution:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2008
Balance the following equation by choosing the correct option.
$x \mathrm{KNO}_3+y \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow p \mathrm{~N}_2+q \mathrm{CO}_2+r \mathrm{H}_2 \mathrm{O}+s \mathrm{~K}_2 \mathrm{CO}_3$

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

The correct answer is:
(b)

Solution:
$48 \mathrm{KNO}_3+5 \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \longrightarrow 24 \mathrm{~N}_2+36 \mathrm{CO}_2+55 \mathrm{H}_2 \mathrm{O}+24 \mathrm{~K}_2 \mathrm{CO}_3$

Topic: Chemistry, Redox Reactions, JEE Main
Balance the following redox reactions by ion - electron method.
\( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+\mathrm{SO}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \quad \) (in acidic solution) by selecting any of the choices correctly

Options:
A One moles of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) oxidises \( 3 \) moles of \( \mathrm{SO}_{2} \)
B One mole of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) produces \( 3 \) moles of \( \mathrm{SO}_{4}^{2-} \)
C One moles of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) oxidises \( 3 \) moles of \( \mathrm{SO}_{2} \)
e mole of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) produces \( 3 \) moles of \( \mathrm{SO}_{4}^{2-} \)
D One mole of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) ion oxidises \( 2 \) moles of \( \mathrm{SO}_{2} \)

The correct answer is:
One moles of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) oxidises \( 3 \) moles of \( \mathrm{SO}_{2} \)
e mole of \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \) produces \( 3 \) moles of \( \mathrm{SO}_{4}^{2-} \)

Solution:
Cr 2 O 7 2 - + 1 4 H aq + + 6 e - â¶ 2 Cr aq 3 + + 7 H 2 O RHR
SO 2 + 2 H 2 O â¶ SO 4 2 - + 4 H + + 2 e - OHR } × 3
-------------------------------------------------------------------
Cr 2 O 7 aq 2 - + 2 H aq + + 3 SO 2 â¶ 2 Cr aq 3 + + 3 SO 4 aq 2 - + H 2 O e

Topic: Chemistry, Redox Reactions, JEE Main
Band theory predicts that magnesium is an insulator. However, in practice it acts as a conductor due to

Options:
A presence of filled 3s – orbital
B overlap of filled 2p and filed 3s- orbital
C overlap of filled 3s and empty 3p-orbital
D presence of unfilled 3p –orbital

The correct answer is:
overlap of filled 3s and empty 3p-orbital

Solution:
The electronic configuration of Mg is
1s2,2s2,2p6,3s2,3p0
In it 3s – orbital is completely filed and 3p-orbital is empty. There is very small energy difference between 3s and 3p – orbitals. Thus, they overlap and make the Mg a conductor.
Band model of Mg

Topic: Chemistry, Solid State, JEE Main
Base strength of
a. $\mathrm{H}_3 \mathrm{CC°H}_2$
b. $\mathrm{H}_2 \mathrm{C}=\mathrm{CH}$
c. $\mathrm{H}-\mathrm{C}=\mathrm{C}^{\circ}$
is in the order of

Options:
A $a>b>c$
B $b>a>c$
C $c>b>a$
D $\mathrm{a}>\mathrm{c}>\mathrm{b}$

The correct answer is:
$a>b>c$

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2008 (Mains)
Based on a 10% law of energy transfer, identify values for A, B, and C in the figure given below.

Options:
A

A	B	C

10	0.1	0.01

B

A	B	C

100	10	1

C

A	B	C

100	10	0.1

D

A	B	C

1000	100	10

The correct answer is:

A	B	C

100	10	1

Solution:
The 10% law was proposed by Lindemann in 1942. The transfer of energy from one trophic level to another trophic level is accompanied by loss of energy at each level or step. When the plants are eaten by the herbivore, approximately 10% of energy in the food is fixed into animal flesh, whereas 90% is consumed in ingestion, respiration, maintenance of body heat and other activities. Similarly, when a carnivore consumes that herbivore, again approimately 10% of energy is fixed. Therefore, at each transfer, only 10% of the total energy is actually available to the next trophic level. It is called 10% law.

Topic: Biology, Ecosystem, NEET
Based on cellular mechanisms there are two major types of regeneration found in the animals. Which one of the following is the correct example of the type mentioned?

Options:
A Morphallaxis - Regeneration of two transversely cut equal pieces of a Hydra into two small Hydras
B Epimorphosis - Replacement of old and dead erythrocytes by the new ones
C Morphallaxis - Healing up of a wound in the skin
D Epimorphosis - Regeneration of crushed and filtered out pieces of a Planaria into as many new Planarians

The correct answer is:
Morphallaxis - Regeneration of two transversely cut equal pieces of a Hydra into two small Hydras

Solution:
Based on cellular mechanisms, regeneration can be divided into two broad categories : morphallaxis and epimorphosis.
Morphallaxis refers to the type of regeneration in which lost body parts are replaced by the remodelling of the remaining tissue. In this type of regeneration, little or no cellular proliferation takes place during the regeneration process. A classic example of an organism that regenerates using this mechanism is the Hydra. When a Hydra is cut into two pieces, two Hydra will be regenerated, both smaller than the parental Hydra. Once regeneration is completed, the two Hydra can continue to grow and reach the size of their original parent. In contrast to morphallaxis epimorphosis require active cellular proliferation prior to the replacement of the lost body part. For eg. Planarian, which are flatworms, regenerate using a mechanism in which pre-existing stem cells, known as neoblasts, begin to proliferate and migrate to the injured site in response to injury. These cells then form a mass of proliferating cells, known as regeneration blastema, that will later differentiate into the specialized cells that comprise the regenerated structure. Most tissue regeneration in mammals also belongs to the epimorphosis. For example, mammals can regenerate their muscle, bone epithelia of the skin and gut, blood, and some neurons by activating pre-existing stem cells or progenitor cells.

Topic: Biology, Reproduction in Organisms, AIIMS, AIIMS 2005
Based on cellular mechanisms there are two major types of regeneration found in the animals. Which one of the following is the correct example of the type mentioned?

Options:
A Epimorphosis - Regeneration of crushed and filtered out pieces of a Planaria into as many new Planarians.
B Morphallaxis - Regeneration of two transversely cut equal pieces of a Hydra into two small hydras
C Epimorphosis - Replacement of old and dead erythrocytes by the new ones.
D Morphallaxis - Healing up of a wound in the skin.

The correct answer is:
Morphallaxis - Regeneration of two transversely cut equal pieces of a Hydra into two small hydras

Solution:
Morphallaxis refers to the type of regeneration in which lost body parts are replaced by the remodelling of the remaining tissue. In this type of regeneration, little or no cellular proliferation takes place during the regeneration process. A classic example of an organism that regenerates using this mechanism is the Hydra. When a Hydra is cut into two pieces, two Hydra's will be regenerated, both smaller than the parental Hydra. Once regeneration is completed, the two Hydra can continue to grow and reach the size of their original parent. Growth requires cellular proliferation but during the regenerative process very little cellular proliferation takes place.

Topic: Biology, Reproduction in Organisms, NEET
Based on equation $E=-2.178 \times 10^{-18} \mathrm{~J}\left(\frac{Z^2}{n^2}\right)$ certain conclusions are written. Which of them is not correct?

Options:
A The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus
B Larger the value of $n$, the larger is the orbit radius
C Equation can be used to calculate the change in energy when the electron changes orbit
D For $n=1$ the electron has a more negative energy than it does for $n=6$ which means that the electron is more loosely bound in the smallest allowed orbit

The correct answer is:
For $n=1$ the electron has a more negative energy than it does for $n=6$ which means that the electron is more loosely bound in the smallest allowed orbit

Solution:
If $n=1$,
$$
E_1=-2.178 \times 10^{-18} Z^2 \mathrm{~J}
$$
If $n=6$
$$
\begin{aligned}
E_6 & =\frac{-2.178 \times 10^{-18} Z^2}{36} \mathrm{~J} \\
& =6.05 \times 10^{-20} Z^2 \mathrm{~J}
\end{aligned}
$$
From the above calculation, it is abvious that electron has a more negative energy than it does for $n=6$. It means that electron is more strongly bound in the smallest allowed orbit.

Topic: Chemistry, Structure of Atom, JEE Main
Based on Heisenberg's uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter \(10^{-15} \mathrm{~m}\) is ________. \(10^9 \mathrm{~ms}^{-1}\) (nearest integer)
[Given : mass of electron \(=9.1 \times 10^{-31} \mathrm{~kg}\), Plank's constant \((h)=6.626 \times 10^{-34} \mathrm{Js}\)] (Value of \(\pi=3.14\))

Options:

The correct answer is:
58

Solution:
$\begin{aligned} & \mathrm{m} \Delta \mathrm{V} . \Delta \mathrm{x}=\frac{\mathrm{h}}{4 \pi} \\ & \begin{aligned} \Delta \mathrm{V} & =\frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{-15} \times 4 \times 3.14} \\ & =57.97 \times 10^{+9} \mathrm{~m} / \mathrm{sec}\end{aligned}\end{aligned}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Based on lattice energy and other considerations, which one of the following alkali metal chloride is expected to have the highest melting point?

Options:
A
$\mathrm{NaCl}$

B
$\mathrm{KCl}$

C
$\mathrm{LiCl}$

D
$\mathrm{RbCl}$

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

Solution:
$\mathrm{NaCl}$ due it to its largest lattice energy has the highest melting point.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2012 (07 May Online)
Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point

Options:
A $\mathrm{LiCl}$
B $\mathrm{NaCl}$
C $\mathrm{KCl}$
D $\mathrm{RbCl}$

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

Solution:
Although lattice energy of $\mathrm{LiCl}$ higher than $\mathrm{NaCl}$ but $\mathrm{LiCl}$ is covalent in nature and $\mathrm{NaCl}$ ionic there after, the melting point decreases as we move $\mathrm{NaCl}$ because the lattice energy decreases as a size of alkali metal atom increases (lattice energy melting point of alkali metal halide)

Topic: Chemistry, s Block Elements, JEE Main
Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point.

Options:
A
LiCl

B
NaCl

C
KCl

D
RbCl

The correct answer is:
NaCl

Solution:
Although lattice energy of LiCl higher than NaCl but LiCl is covalent in nature and NaCl ionic there after , the melting point decreases as we move NaCl because the lattice energy decreases as a size of alkali metal atom increases (lattice energy ∝ to melting point of alkali metal halide)

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2005