datavorous/entrance-exam-dataset huggingface dataset:
The electronic configuration of $\mathrm{Cr}^{3+}$ is

Options:
A $[\mathrm{Ar}] 3 d^{4} 4 s^{2}$
B $[\mathrm{Ar}] 3 d^{3} 4 s^{0}$
C $[\mathrm{Ar}] 3 d^{2} 4 s^{1}$
D $[\mathrm{Ar}] 3 d^{5} 4 s^{1}$

The correct answer is:
$[\mathrm{Ar}] 3 d^{3} 4 s^{0}$

Solution:
$\mathrm{Cr}(24):[\mathrm{Ar}] 3 d^{5} 4 s^{1}$
$$
\mathrm{Cr}^{3+}:[\mathrm{Ar}] 3 d^{3} 4 s^{0}
$$

Topic: Chemistry, d and f Block Elements, JEE Main
The electronic configuration of element with atomic number 24 is

Options:
A $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^4, 4 s^2$
B $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^{10}$
C $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^6$
D $1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1$

The correct answer is:
$1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^5 4 s^1$

Solution:
The electronic configuration of atomic number $24=1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^5, 4 s^1$

Topic: Chemistry, Structure of Atom, JEE Main
The electronic configuration of elements $A, B$ and $C$ are $[\mathrm{He}] 2 s^1,[\mathrm{Ne}] 3 s^1$ and [Ar] $4 s^1$ respectively, which one of the following order is correct for the first ionization potentials (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of $A, B$ and $C$ ?

Options:
A $A>B>C$
B $C>B>A$
C $B>C>A$
D $C>A>B$

The correct answer is:
$A>B>C$

Solution:
Ionisation potential of $A, B$ and $C$ elements decreases with the ease of removal of electron from $A$ to $C$. Atomic size increase from $A$ to $C$, hence ionisation potential decrease in the same order.
So, order of ionisation potential is $A>B>C$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The electronic configuration of few elements is given below. Mark the statement which is not correct about these elements.
(i) 1s22s22p63s1
(ii) 1s22s22p5
(iii) 1s22s22p6
(iv) 1s22s22p3

Options:
A (i) is an alkali metal
B (iii) is a noble metal
C (i) and (ii) form ionic compounds
D (iv) has high ionisation enthalpy than accepted

The correct answer is:
(iii) is a noble metal

Solution:
As (iii) has 1s22s22p6 type of noble gas configuration so it is an inert gas Neon not a metal.

Topic: Chemistry, Structure of Atom, JEE Main
The electronic configuration of Pt (atomic number 78) is

Options:
A Kr4f145d10
B Xe4f145d10
C Xe4f145d86s2
D [Xe]4f145d96s1

The correct answer is:
[Xe]4f145d96s1

Solution:
The atomic number of Platinum is 78. Platinum belongs Nickel group, and it is 5d series element. Electronic configuration of Platinum is1s22s2p63s2p6d104s2p6d10f145s2p6d96s1

Pt78=Xe4f145d96s1 (Exceptional electronic configuration)

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The electronic configuration with the highest ionization enthalpy is:

Options:
A Ar 3d10 4s2 4p3
B Ne3s2Â 3p1
C Ne 3s2 3p3
D NeÂ 3s2Â 3p2

The correct answer is:
Ne 3s2 3p3

Solution:
(i) Ionisation energy decrease with increasing atomic radius.
(ii) Half filled and completely filled electronic configuration are extra stable.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2017 (09 Apr Online)
The electronic configurations of bivalent europium and trivalent cerium are:
(atomic number: Xe=54 , Ce=58 , Eu=63)

Options:
A Xe4f2 and Xe4f7
B Xe4f7 and Xe4f1
C Xe4f76s2 and Xe4f26s2
D Xe4f4 and Xe4f9

The correct answer is:
Xe4f7 and Xe4f1

Solution:
Electronic configuration of Ce and Eu in ground state are following:

Ce=Xe 4f1 5d1 6s2Eu=Xe 4f7 6s2

Trivalent cerium means cerium loss three electrons and obtain +3 charge and bivalent Europium means Europium loss two electrons and obtain +2 charge, so their electronic configurations are following

Eu2+:Xe4f7

Ce3+:Xe4f1

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The electronic energy levels of the hydrogen atom in the Bohr's theory are called

Options:
A Rydberg levels
B Orbits
C Ground states
D Orbitals

The correct answer is:
Orbits

Solution:
Generally electron moving in orbits according to Bohtss principle.

Topic: Chemistry, Structure of Atom, JEE Main
The electronic spectrum of TiH2O63+ shows a single broad peak with a maximum at 20,300cm-1.The crystal field stabilization energy (CFSE) of the complex ion, in kJmol-1, is : 1kJmol-1=83.7cm-1

Options:
A 145.5
B 242.5
C 83.7
D 97

The correct answer is:
97

Solution:
Ti(H2O)63+ ⇒Ti3+=3d1 4s0

⇒t2g1,0,0,eg0,0

CFSE =[-0.4nt2g+0.6neg] ∆0+n(p)

=[-0.4×1 ]20300

=-812083.7kJ/mole

=-97 kJ/mole

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The electronic structure $(n-1) d^{1-10} n s^{-2}$ is characteristic of

Options:
A Transition elements
B Lanthanides
C Actinides
D Rare gases

The correct answer is:
Transition elements

Solution:
The electronic structure $(n-1)d^{1-10} ns^{0-2}$ is characteristic of transition metals. This is because the last electron is filled in the $(n-1)d$ orbital, according to the $(N+1)$ rule. Transition elements have a characteristic electronic configuration that can be represented as $(n-1)s^2p^6d^{1-10}ns^1$ or $ns^2$. As the transition series progresses, the d-orbitals are gradually filled.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The electronic structure of molecule $O F_2$ is a hybrid of

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

The correct answer is:
$s p^3$

Solution:
ANSWER (3) $s p^3$
$\mathrm{OF}_2$ has a hybridization of $s p^3$. Hyb is by atom, not molecule. The $O$ here is $s p^3$. The $F$ can be taken as either $s p^3$ or the original p.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The electrons are more likely to be found:

Options:
A Only in the region a
B In the region a and b
C In the region a and c
D Only in the region c

The correct answer is:
In the region a and c

Solution:
ψ2(r) is probability of finding an electron and it will be maximum at a and c.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The electrons identified by quantum numbers $\mathrm{n}$ and $\mathrm{I}$ :
(a) $n=4, I=1$
(b) $n=4, l=0$
(c) $n=3, I=2$
(d) $n=3, I=1$
Can be placed in order of increasing energy as:

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

The correct answer is:
(d) $ < $ (b) $ < $ (c) $ < $ (a)

Solution:
(a) $(n+l)=4+1=5$
(b) $(n+l)=4+0=4$
(c) $(n+1)=3+2=5$
(d) $(n+1)=3+1=4$

Topic: Chemistry, Structure of Atom, JEE Main
The electrons identified by quantum numbers $\mathrm{n}$ and $\mathrm{I}$ :
(a) $n=4, I=1$
(b) $n=4, l=0$
(c) $n=3, I=2$
(d) $n=3, I=1$
Can be placed in order of increasing energy as:

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

The correct answer is:
(d) $ < $ (b) $ < $ (c) $ < $ (a)

Solution:
(a) $(n+l)=4+1=5$
(b) $(n+l)=4+0=4$
(c) $(n+1)=3+2=5$
(d) $(n+1)=3+1=4$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2012 (Offline)
The electrons of Rutherford's model would be expected to lose energy because, they

Options:
A move randomly.
B jump on nucleus.
C radiate electromagnetic waves.
D escape from the atom.

The correct answer is:
radiate electromagnetic waves.

Solution:
In classical mechanics, when any charged particle will be in acceleration, it will radiate energy in form of electromagnetic waves and will lose its energy.

In Rutherford's model, electron revolves around the nucleus and will be in accelerated motion. So it will lose kinetic energy in form of electromagnetic waves and will cause decrease in radius.

Topic: Physics, Atomic Physics, JEE Main
The electrophile, $\mathrm{E}^{\oplus}$ attacks the benzene ring to generate the intermediate $\sigma$-complex. Of the following, which $\sigma$-complex is of lowest energy?

Options:
A

B

C

D

The correct answer is:

Solution:
$-\mathrm{NO}_2$ is electron withdrawing which will destabilize $\sigma$ - complex.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2008
The electrophile involved in above reaction has ______ lone pair of electrons on central carbon
atom.

Options:

The correct answer is:
1

Solution:
The electrophile
Eâ is singlet carbene so it has on lone pair.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The electrophile involved in the above reaction is

Options:
A

B
dichlorocarbene $\left(: \mathrm{CCl}_2\right)$

C

D

The correct answer is:
dichlorocarbene $\left(: \mathrm{CCl}_2\right)$

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2006
The electrophile involved in the sulphonation of benzene is

Options:
A $\mathrm{SO}_{3}^{+}$
B $\mathrm{SO}_{3}^{2-}$
C $\mathrm{H}_{3}^{+} \mathrm{O}$
D $\mathrm{SO}_{3}$

The correct answer is:
$\mathrm{SO}_{3}$

Solution:
The electrophile involved in the sulphonation of benzene is $\mathrm{SO}_{3}$.
$$
2 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{SO}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HSO}_{4}^{-}
$$

Topic: Chemistry, Hydrocarbons, JEE Main
The electrophile, \( \mathrm{E}^{\oplus} \) attacks the benzene ring to generate the intermediate \( \sigma \)-complex.Of the following. which \( \sigma \) - complex
is of lowest energy ?

Options:
A
B
C
D

The correct answer is:

Solution:
This structure will be of lowest energy due to resonance stabilisation of +ve charge. In all other three structures, the presence of electron-withdrawing NO2 group will destabilize the +Ve charge and hence they will have greater energy.

Topic: Chemistry, Hydrocarbons, JEE Main
The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

Options:
A proportional to the square to the root of the distance between the plates.
B linearly proportional to be distance between the plates.
C independent of the distance between the plates.
D inversely proportional to the distance between the plates.

The correct answer is:
independent of the distance between the plates.

Solution:
For isolated capacitor Q= constant.

Now electric field because of one plate of the capacitor is E=σ2∈0=Q2A∈0.
∴ Force on 1 metal plate is
F=QE
F=Qσ2∈0=Q22A∈0
∴ The electrostatic force is independent of the distance between plates.

Topic: Physics, Capacitance, JEE Main
The electrostatic force $\left(\overrightarrow{F_1}\right)$ and magnetic force $\left(\vec{F}_2\right)$ acting on a charge $q$ moving with velocity $v$ can be written :

Options:
A $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B})$
B $\vec{F}_1=q \vec{B}, \vec{F}_2=q(\vec{B} \times \vec{V})$
C $\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{B} \times \vec{V})$
D $\vec{F}_1=q \vec{V} \cdot \vec{E}, \vec{F}_2=q(\vec{B} \cdot \vec{V})$

The correct answer is:
$\vec{F}_1=q \vec{E}, \vec{F}_2=q(\vec{V} \times \vec{B})$

Solution:
$\begin{aligned}
& \overrightarrow{\mathrm{F}}_1=\mathrm{q} \overrightarrow{\mathrm{E}} \\
& \overrightarrow{\mathrm{F}}_2=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})
\end{aligned}$
(Theory)

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The electrostatic potential due to an electric dipole at a distance r varies as :

Options:
A r
B 1r2
C 1r3
D 1r

The correct answer is:
1r2

Solution:
The potential at a distance r due to a short dipole is Vdipr=14πε∘ p→·r^r2, where p→=dipole moment.

Thus, Vdip(r) is proportional to 1r2.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (30 Jan Shift 1)
The electrostatic potential inside a charged sphere is given as $V=A r^2+B$, where $r$ is the distance from the centre of the sphere, $A$ and $B$ are constants. Then, the charge density in the sphere is

Options:
A $16 A \varepsilon_0$
B $-6 A \varepsilon_0$
C $20 A \varepsilon_0$
D $-15 A \varepsilon_0$

The correct answer is:
$-6 A \varepsilon_0$

Solution:
Let the volume charge density be $\rho$.
The electric field inside the charged sphere,
$$
E=\frac{K q r}{R^3}
$$
where, $R=$ Radius of charged sphere $r=$ Distance of the point $(P)$ inside the sphere
As volume charge density is $\rho$, then in terms of charge density,
$$
\begin{aligned}
E & =\frac{\rho r}{3 \varepsilon_0} \\
\quad \quad \quad \rho & =\frac{3 E \varepsilon_0}{r}
\end{aligned}
$$
Now, electric field, $E=-\frac{d V}{d r}$
Here, $V=A r^2+B$
$$
\begin{array}{ll}
\therefore & E=-\frac{d V}{d r}=-2 A r+0=-2 A r \\
\therefore & \rho=\frac{3 E \varepsilon_0}{r}
\end{array}
$$
$$
=\frac{3\left(-2 A r \cdot \varepsilon_0\right)}{r}=-6 A \varepsilon_0
$$

Topic: Physics, Electrostatics, JEE Main
The electrostatic potential inside a charged spherical ball is given by $\phi=\alpha \rho^2+b$ where $r$ is the distance from the centre; $a, b$ are constants. Then the charge density inside ball is

Options:
A
$-6 a \varepsilon_0 r$

B
$-24 \pi a \varepsilon_0 r$

C
$-6 a \varepsilon_0$

D
$-24 \pi a \varepsilon_0 r$

The correct answer is:
$-6 a \varepsilon_0$

Solution:
Potential inside $(\phi)=a r^2+b$
$$
\therefore \mathrm{E}_{\mathrm{r}}=-\frac{\delta \mathrm{v}}{\delta \mathrm{r}}=-2 \mathrm{ar}
$$
Electric field inside uniformly charged solid volume varies with ' $r$ '. So charge density is constant
$$
\begin{aligned}
& \phi_{\text {net }}=(-2 \mathrm{ar}) 4 \pi \mathrm{r}^2=-8 \pi a r^3 \\
& -8 \pi \mathrm{ar}^3=\frac{\sigma \times \frac{4}{3} \pi \mathrm{r}^3}{\varepsilon_0} \\
& \therefore \sigma=-6 \mathrm{a} \varepsilon_0
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2011
The element A burns in nitrogen to give an ionic compound B. The compound B reacts with water to give C and D. A solution of C becomes milky on bubbling carbon dioxide. What is the nature of compound D?

Options:
A Acidic
B Basic
C Amphoteric
D Neutral

The correct answer is:
Basic

Solution:
A → burns, nitrogen B ionic compound → H 2 O C milky solution  + D → CO 2
Since the solution of C becomes milky on bubbling CO2 into it, C is lime water (CaOH)2. Thus, the element A is calcium and B is the compound of calcium with nitrogen. Nitrides on reacting with water give the corresponding hydroxide and ammonia. Therefore,
A is calcium (Ca)
B is calcium nitride Ca3N2 .
C is lime water CaOH2
D is ammonia NH3 .
Ca → burns, nitrogen Ca 3 N 2 ionic compound → H 2 O Ca (OH) 2 milky solution  +  NH 3 → CO 2 CaCO 3

Topic: Chemistry, s Block Elements, JEE Main
The element having the greatest difference between its first and second ionization energies, is

Options:
A Ba
B K
C Ca
D Sc

The correct answer is:
K

Solution:
According to given options, except for potassium and Sc rest, others are alkaline earth metals with an outer configuration ns2.
In the case of potassium, the outer configuration is 1s2 2s2 2p6 3s2 3p6 4s1.
First ionization enthalpy is low, but once an electron is removed, it will attain noble gas configuration and will become more stable. Hence, removing the second electron is difficult and requires a large amount of energy.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The element playing significant role in neuromuscular function and interneuronal transmission is :

Options:
A Be
B Mg
C Ca
D Li

The correct answer is:
Ca

Solution:
About 99 % of body calcium is present in bones and teeth. Calcium plays important role in neuromuscular function, interneuronal transmission. Cell membrane integrity and blood coagulation. An adult body contains about 1200 g of Ca.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (31 Jan Shift 2)
The element that can be refined by distillation is :

Options:
A nickel
B zinc
C tin
D gallium

The correct answer is:
zinc

Solution:
Zn, Cd & Hg are purified by fractrional distillation process.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The element that does not show catenation is

Options:
A Sn
B Si
C Ge
D Pb

The correct answer is:
Pb

Solution:
Catenation is the tendency to link with one another through covalent bonds to form chain and rings, i.e., catenation. This is because bonds are very strong in the carbon family. On moving down the group, the size increases and the tendency to show catenation decreases because bond enthalpy decreases. The order of catenation is C>Si>Ge>Pb. Pb does not show catenation.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The element that shows greater ability to form pπ-pπ multiple bonds is:

Options:
A Sn
B C
C Si
D Ge

The correct answer is:
C

Solution:
C forms most stable pπ-pπ bonds because it has high bond dissociation energy, more electronegativity, high ionization energy and vacant d- orbital is unavailable. While others have vacant orbital and low bond dissociation enthalpy and low ionisation enthalpy

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The element that usually does NOT show variable oxidation states is:

Options:
A $\mathrm{Cu}$
B Ti
C Sc
D $V$

The correct answer is:
Sc

Solution:
Sc shows oxidation state of +3 only.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The element used by Rutherford in his famous scattering experiment was

Options:
A Gold
B Tin
C Silver
D Lead

The correct answer is:
Gold

Solution:
Gold used by Rutherford in scatting experiment.

Topic: Chemistry, Structure of Atom, JEE Main
The element which shows only one oxidation state other than its elemental form is :

Options:
A Cobalt
B Titanium
C Nickel
D Scandium

The correct answer is:
Scandium

Solution:
Co, $\mathrm{Ti}, \mathrm{Ni}$ can show $+2,+3$ and +4 oxidation state, But ' $\mathrm{Sc}$ ' only shows +3 stable oxidation state.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (04 Apr Shift 1)
The element with atomic number \( 25 \) will be found in group

Options:

The correct answer is:
7

Solution:
For d block elements:

Group number = number of electrons in (n -1)d subshell + number of electrons in nS subshell .

The E.C. of element = Ar 3 d 5 4 s 2 ; Group = 7th.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The element with which of the following outer electron configuration may exhibit the largest number of oxidation states in its compounds :

Options:
A
$3 d^5 4 s^2$

B
$3 d^8 4 s^2$

C
$3 d^7 4 s^2$

D
$3 d^6 4 s^2$

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

Solution:
The element with outer electron configuration $3 d^5 4 s^2$ is $\mathrm{Mn}$ which exhibits oxidation states from $+2$ to $+7$.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2013 (09 Apr Online)
The elements having the electronic configuration, $[K r] 4 d^{10} f^{14}, 5 s^2 p^6 d^2, 6 s^2$ belongs to

Options:
A $s$-block
B $p$-block
C $d$-block
D $f$-block

The correct answer is:
$d$-block

Solution:
$d$-Block because the last electron enters $d$-subshell.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The elements of group IA provide a colour to the flame of Bunson burner due to

Options:
A Low ionization potential
B Low melting point
C Softness
D Presence of one electron in the outermost orbit

The correct answer is:
Low ionization potential

Solution:
Flame test is given by the alkali metals by a specific color since they have minimum ionization potential so when they are heated electrons jump from lower to higher level and then fall back releasing light energy in the form of color of flame.

Topic: Chemistry, s Block Elements, JEE Main
The elements on the right side of the periodic table are

Options:
A Metals
B Metalloids
C Non-metals
D Transition elements

The correct answer is:
Non-metals

Solution:
Elements on the right side of the periodic table are p-block. Mostly non-metals.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The elements with atomic number 10, 18, 36, 54 and 86 are all

Options:
A Light metals
B Inert gases
C Halogens
D Rare–earths

The correct answer is:
Inert gases

Solution:
Inert gases, these have $n s^2 n p^6$ configuration

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The elements with the highest and the lowest enthalpy of atomisation respectively, for first row transition elements are:

Options:
A Sc, Zn
B Ti, Ni
C V, Zn
D Cr, Mn

The correct answer is:
Sc, Zn

Solution:
The enthalpy of atomisation is the enthalpy change that accompanies the total separation of all atoms in a chemical substance.

The extent of metallic bonding an element undergoes decides the enthalpy of atomization. The more extensive the metallic bonding of an element, the more will be its enthalpy of atomisation.

Sc has the lowest enthalpy of atomisation, for first row of transition elements and Zn (4d10 5s2) has completely filled d-orbital and has no unpaired electron to take part in the formation of metallic bonds. So, the metallic bonding in zinc is the weakest and it has the least enthalpy of atomisation.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The elements X and Y form compound having molecular formula XY2 and XY4 (both are non-electrolyses). When dissolved in 20 g benzene, 1 g XY2 lowers the freezing point by 2 .3 o C whereas 1 g of XY4 lowers the freezing point by 1 .3 o C . Molal depression constant for benzene is 5.1 . Thus atomic masses of X and Y respectively are

Options:
A 42.64, 21.10
B 21.10, 42.64
C 25.59, 42.64
D 42.64, 25.69

The correct answer is:
25.59, 42.64

Solution:
ΔTf=KfwM.W×1000
2.3=5.1×1MXY2×100020
MXY2=110.87
and 1.3=5×1MXY4×100020
MXY4=196.15
Now, let atomic mass of X = a and atomic mass of Y=b
Thus, a+2b=110.87
and -a±4b=-196.15-2b=-85.28
b=42.64
and thus a=25.59

Topic: Chemistry, Solutions, JEE Main
The elevation in boiling point for 1 molal solution of glucose is 2 K. The depression in freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is:

Options:
A Kb=2 Kf
B Kb=1.5 Kf
C Kb=Kf
D Kb=0.5 Kf

The correct answer is:
Kb=2 Kf

Solution:
Freezing point depression refers to the lowering of the freezing point of solvent upon the addition of solute. Boiling point elevation refers to the increase in the boiling point of a solvent upon the addition of a solute.

∆Tb=iKbm⇒2=iKb×1............i
∆Tf=iKfm⇒2=iKf×2...............ii
Dividing iii,
1=Kb2Kf
Kb=2 Kf

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of Kb and Kf i.e., Kb/Kf is 1:X. The value of X is

Options:

The correct answer is:
1

Solution:
We know that

ΔTb=iKbm1ΔTf=iKfm2

And for non-volatile solute i=1.

ΔTbΔTf=Kb×1 Kf×2⇒36=12=KbKf×12

KbKf=11⇒X=1

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The elevation in boiling point of a solution of a solution dTb is related with molality of solution (m) by the relation
dTb=RTb2ΔΗvapM11+mM1, where M1 is molar mass of solvent and ∆Hvap is heat of vaporization of solvent. For a dilute solution, the relation ∂Tb∂mm→0 gives:

Options:
A Molal ebullioscopic constant
B Elevation in boiling point
C Boiling point of solvent
D Elevation in boiling point (ΔT b ) becomes more predominant

The correct answer is:
Molal ebullioscopic constant

Solution:
∂Tb∂mm→0=RTb2M1ΔΗvap=Kb.

Topic: Chemistry, Solutions, JEE Main