datavorous/entrance-exam-dataset huggingface dataset:
In blast furnace iron oxide is reduced by

Options:
A Silica
B CO
C NaOH
D Lime stone

The correct answer is:
CO

Solution:
Fe2O3+3CO→2Fe+3CO2
In blast furnace, iron oxide is reduced by carbon monoxide to give iron and carbon dioxide.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen

Options:
A
$5 \rightarrow 2$

B
$4 \rightarrow 1$

C
$2 \rightarrow 5$

D
$3 \rightarrow 2$

The correct answer is:
$5 \rightarrow 2$

Solution:
The lines falling in the visible region comprise Balmer series. Hence the third line would be $n_1=2, n_2=5$ i.e. $5 \rightarrow 2$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2003
In Bohr's model, atomic radius of the first orbit is $\gamma$, the radius of the $3^{\text {rd }}$ orbit, is

Options:
A $\gamma / 3$
B $\gamma$
C $3 \gamma$
D $9 \gamma$

The correct answer is:
$9 \gamma$

Solution:
$\quad r \propto n^2$
For $I^{\text {st }}$ orbit $\gamma=1$
For $III^{\text{rd}}$ orbit $=\gamma \propto 3^2=9$
So it will $9^\gamma$.

Topic: Chemistry, Structure of Atom, JEE Main
In borax bead test, which of the following compound is formed

Options:
A Meta borate
B Tetra borate
C Double oxide
D Ortho borate

The correct answer is:
Meta borate

Solution:
$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 .10 \mathrm{H}_2 \mathrm{O} \xrightarrow[-10 \mathrm{H}_2 \mathrm{O}]{ } \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$$\xrightarrow{\Delta} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3$
$\mathrm{CuO}+\mathrm{B}_2 \mathrm{O}_3 \rightarrow \mathrm{Cu}\left(\mathrm{BO}_2\right)_2$
(Copper meta borate blue)

Topic: Chemistry, Practical Chemistry, JEE Main
In borax if m is number of hydroxy groups and 'n' is number of water molecules. What is the sum of m + n?

Options:

The correct answer is:
12

Solution:
Borax is Na2B4O5OH48H2O
m = 4, n = 8
m+n=4+8=12

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In borax if \( \mathrm{m} \) is number of hydroxy groups and ' \( \mathrm{n} \) ' is number of water molecules. What is the sum of \( \mathrm{m}+\mathrm{n} \) ?

Options:

The correct answer is:
12

Solution:
Borax is Na2B4O5OH48H2O
m = 4, n = 8
m+n=4+8=12

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In Borax Na2B4O7.10H2O if number of sp2 hybridised B-atoms are X and number of sp3 hybridised B-atom are Y. What is the value of sum X+Y ?

Options:

The correct answer is:
4

Solution:
In Borax 2-B-atoms are sp2 and 2-B-atoms are sp3 hybridised
X=2,Y=2
X+Y=4

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In borax number of B−O−B bonds are X and number of boron atoms are Y. Then X+Y is

Options:

The correct answer is:
9

Solution:
X=4
Y=4
X+Y=9

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In both DNA and RNA, heterocyclic base and phosphate ester linkages are at-

Options:
A
$\mathrm{C}_5^{\prime}$ and $\mathrm{C}_2^{\prime}$ respectively of the sugar molecule

B
$\mathrm{C}_2^{\prime}$ and $\mathrm{C}_5^{\prime}$ respectively of the sugar molecule

C
$\mathrm{C}_1^{\prime}$ and $\mathrm{C}_5^{\prime}$ respectively of the sugar molecule

D
$\mathrm{C}_5^{\prime}$ and $\mathrm{C}_1^{\prime}$ respectively of the sugar molecule

The correct answer is:
$\mathrm{C}_1^{\prime}$ and $\mathrm{C}_5^{\prime}$ respectively of the sugar molecule

Solution:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2005
In Bredig's arc method some alkali is added because

Options:
A It increases electrical conductance
B To obtain molecular colloid
C To obtain colloidal particles of same size
D To stabilise the sol

The correct answer is:
To stabilise the sol

Solution:
Correct option is 4. To stabilise the sol
The purpose of alkali is to act as a stabilizer of the sols.

Topic: Chemistry, Surface Chemistry, JEE Main
In bringing an electron towards another electron, the electrostatic potential energy of the system

Options:
A decreases
B increases
C remains same
D becomes zero

The correct answer is:
increases

Solution:
The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative Electrostatic potential energy of system of two electrons
$$
U=\frac{1}{4 \pi \varepsilon_{0}} \frac{(-e)(-e)}{r}=\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r}
$$
Thus, as $r$ decreases, potential energy $U$ increases.

Topic: Physics, Dual Nature of Matter, MHT CET, MHT CET 2007
In brown ring Fe and NO exist as Fe+ and NO+ rather than Fe2+ and NO. These forms can be differentiated by

Options:
A Measuring the conductivity of an aqueous solution of the complex
B By careful crystallization of the complex under special conditions and measuring the magnetic moment in solid state
C Estimating the concentration of iron
D Determining elevation in boiling point using the complex as solute

The correct answer is:
By careful crystallization of the complex under special conditions and measuring the magnetic moment in solid state

Solution:
When the combination is Fe2+ and NO, there will be 5 unpaired electrons in the complex ( 4 in Fe2+ and 1 in NO) and when the combination is Fe+ and NO+, there will be only 3 unpaired electrons which can be differentiated by measuring the magnetic moments.

Topic: Chemistry, Coordination Compounds, JEE Main
In Bryophyte, the main plant body will be

Options:
A Haploid and produces gamete
B Diploid and produces gamete
C Haploid and produces spores
D Diploid and produces spores

The correct answer is:
Haploid and produces gamete

Solution:
Gametophytic generation is the dominant phase in the life cycle of Bryophytes. The main plant body of bryophytes is found to be haploid and since they produce gametes are called gametophyte. They do produce diploid sporophyte which is attached and dependent on the gametophyte for nourishment.

Topic: Biology, Plant Kingdom, NEET
In C1H2=C2=C3H-C4H3 molecule, the hybridization of carbon 1, 2, 3 and 4 respectively, are :

Options:
A sp2, sp3, sp2, sp3
B sp2, sp, sp2, sp3
C sp2, sp2, sp2, sp3
D sp3, sp, sp3, sp3

The correct answer is:
sp2, sp, sp2, sp3

Solution:
C1H2=C2=C3H-C4H3

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (26 Feb Shift 2)
In callus culture, roots can be induced by the supply of

Options:
A auxin
B cytokinin
C gibberellin
D ethylene

The correct answer is:
auxin

Solution:
Tissue culture is the technique of maintaining and growing cells, tissue, etc. and the differentiation on artificial medium under aseptic condition.
Callus is a irregular, unorganised and undifferentiated mass of actively dividing cell. Differentiation of callus is under control of two hormones, ie, auxin and cytokinin.
$\begin{array}{l|l}
\hline \text{Development of} & \text{Condition} \\
\hline \text{Root} & \frac{\text { Auxin }}{\text { Cytokinin }}= \text{High} \\
\text{Shoot} & \frac{\text { Auxin }}{\text { Cytokinin }}= \text{Low} \\
\text{Both root and shoot} & \frac{\text { Auxin }}{\text { Cytokinin }}= \text{Intermediate} \\
\hline
\end{array}$

Topic: Biology, Strategies for Enhancement in Food Production, JIPMER, JIPMER 2010
In CAM plants

Options:
A stomata open in day and close at night
B stomata remain open in day and night both
C stomata remain closed in day and night both
D stomata open at night and close in day

The correct answer is:
stomata open at night and close in day

Solution:
CAM is found in members of the family Crassulaceae (e.g., Kalanchoe), family Cactaceae (e.g., Opuntia, Cereus), and some members of Liliaceae, Orchidaceae, Bromeliaceae, Euphorbia sp. All CAM plants are not succulent and stomata open at night.

Topic: Biology, Photosynthesis in Higher Plants, JIPMER, JIPMER 2004
In Cannizzaro reaction qiven below

Options:
A
the attack of : $\mathrm{OH}$ at the carboxyl group

B
the transfer of hydride to the carbonyl group

C
the abstraction of proton from the carboxylic group

D
the deprotonation of $\mathrm{Ph} \mathrm{CH}_2 \mathrm{OH}$

The correct answer is:
the transfer of hydride to the carbonyl group

Solution:
Hydride transfer is the slowest step.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2009
In Carius method for estimation of halogens, 0.2 g of an organic compound gave 0.188 g of AgBr. The percentage of bromine in the compound is __________. (Nearest integer)

[Atomic mass: Ag=108, Br=80]

Options:

The correct answer is:
40

Solution:
= Atomic mass of X Molecular mass of AgX× wt. of AgX wt. of organic helides

=80188×0.1880.2×100=40%

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (27 Aug Shift 1)
In Carius method of estimation of halogen, 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which of these is the correct structure of the compound ?

Options:
A CH3CHBr2
B
C
D H3C−CH2−Br

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg AgBr. What is the percentage of bromine in the compound (atomic mass of Ag=108 and atomic number of Br=80)?

Options:
A 60
B 24
C 36
D 48

The correct answer is:
24

Solution:
R - Br → Carius method AgBr
250 mg organic compound is RBr.
141 mgAgBr⇒141×80188mgBr
From principle of atom coservation(POAC), mass of Bromine is conserved in reactant and product

So, mass of Bromine in organic compound = 141×80188 mg

% of Br in organic compound = mass of Brmass of compound×100
=141×80188×1250×100=24%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2015 (04 Apr)
In carprolactum if 'a' is number of lone pairs of e- and 'b' is number of sp3 hybridised atoms. What is the value of a + b?

Options:

The correct answer is:
9

Solution:
number of lone pairs =a=3
number of sp3 hybridised atoms =b=5'C'+1'N'=6
sum of a+b=3+6=9

Topic: Chemistry, Amines, JEE Main
In case of dimensions of electric field and electric dipole moment the power of mass is respectively.

Options:
A 1, 1
B 1, 0
C 0, 1
D 0, 0

The correct answer is:
1, 0

Solution:
Electric field, E=Fq
Where, F = force and q = electric charge
Dimension of E or E=MLT-2AT=MLT-3A-1
Power of mass = 1
Electric moment, m=l×A
where, l = current and A = area
Dimension of m = [M] = M0L2T0A
Hence, power of mass =0

Topic: Physics, Units and Dimensions, JEE Main
In case of isoelectronic species the size of F–, Ne and Na+ is affected by:

Options:
A Principal quantum number n
B None of the factors because their size is the same
C Electron-electron interaction in the outer orbitals
D Nuclear charge z

The correct answer is:
Nuclear charge z

Solution:
In F-,Ne,Na+all have 1 s2,2 s2,2p6configuration. They have different size due to the difference in nuclear charge.

For isoelectronic species, the more nuclear charges, the less will be the size of the species. The valence electrons will experience a greater attractive force due to the increase in nuclear charge.

Hence, the size of isoelectronic species - F-, Ne and Na+ is affected by nuclear charge Z.

Thus, the ionic size of Na+ is the lowest, and the ionic size of F- ion is the highest.

Hence, the answer is D.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2024 (01 Feb Shift 1)
In case of mammals, testis are found in scrotal sacs outside the viscera because

Options:
A spermatogenesis requires a lower temperature.
B mammals are highly evolved animals.
C of the presence of a long vas deferens.
D of limited space due to the presence of urinary bladder.

The correct answer is:
spermatogenesis requires a lower temperature.

Solution:
In man, both testes normally remain suspended in a pouch called Scrotum outside the abdominal cavity. This keeps the testes at a cooler temperature than the body temperature (about 2°C below) this is essential for the maintenance and function of the spermatogenic tissue of the testes. Testes descend in the scrotal sac when fetus is about 7 month old and occurs under the influence of FSH and testosterone.

Topic: Biology, Human Reproduction, NEET
In case of nitrogen, NCl3 is possible but not NCl5 while in case of phosphorous, PCl3 as well as PCl5 are possible. It is due to

Options:
A Availability of vacant d orbitals in P but not in N
B Lower electronegativity of P than N
C Lower tendency of H-bond formation in P than N
D Occurrence of P in solid while N in gaseous state at room temperature.

The correct answer is:
Availability of vacant d orbitals in P but not in N

Solution:
N7=1s22s22p3; P15=1s22s22p63s23p3
In phosphorous the 3d-orbitals are available. Hence phosphorus can form pentahalides also but nitrogen cannot form pentahalide due to absence of d-orbitals.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
In case of nitrogen, $\mathrm{NCl}_3$ is possible but not $\mathrm{NCl}_5$ while in case of phosphorous, $\mathrm{PCl}_3$ as well as $\mathrm{PCl}_5$ are possible. It is due to

Options:
A
availability of vacant d orbitals in P but not in N

B
lower electronegativity of P than N

C
lower tendency of H - bond formation in P than N

D
occurrence of P in solid while N in gaseous state at room temperature

The correct answer is:
availability of vacant d orbitals in P but not in N

Solution:
No Solution Available

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2002
In case of nitrogen, $\mathrm{NCl}_{3}$ is possible but not $\mathrm{NCl}_{5}$ while in case of phosphorus, $\mathrm{PCl}_{3}$ as well as $\mathrm{PCl}_{5}$. are possible. It is due to

Options:
A availability of vacant $d$ orbitals in P but not in $\mathrm{N}$
B lower electronegativity of P than $\mathrm{N}$
C lower tendency of $\mathrm{H}$-bond formation in $\mathrm{P}$ than $\mathrm{N}$
D occurrence of $\mathrm{P}$ in solid while $\mathrm{N}$ in gaseous state at room temperature.

The correct answer is:
availability of vacant $d$ orbitals in P but not in $\mathrm{N}$

Solution:
${ }_{7} \mathrm{~N}=1 s^{2} 2 s^{2} 2 p^{3} ; \quad{ }_{15} \mathrm{P}=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}$
In phosphorous the $3 d$-orbitals are available. Hence phosphorus can form pentahalides also but nitrogen cannot form pentahalide due to absence of $d$-orbitals.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In case of photoelectric effect, slope of V 0 vs ν curve is (where V 0 - stopping potential, ν = subjected frequency)

Options:
A e
B h e
C ϕ
D h

The correct answer is:
h e

Solution:
Ïâ=â Ï 0 â+â eV 0 ...(1)
Ï= work function
Ï 0 â= threshold energy
hÎ½â=â hÎ½ 0 â+â eV 0
V 0 â=â hÎ½ 0 e â hÎ½ 0 e ...(2)
Î½â= subjected frequency
Î½ 0 â= threshold frequency
Comparing equation (2) with straight line equation:
y = mx + c
Slope of Î½ 0 v/s Î½ curve = h e

Topic: Chemistry, Structure of Atom, JEE Main
In case of projectile motion, which one of the following figures represent variation of horizontal component of velocity $\left(\mathrm{u}_{\mathrm{x}}\right)$ with time $\mathrm{t} ?$ (assume that air resistance is negligible)

Options:
A
B
C
D

The correct answer is:

Solution:
Horizontal component of velocity remains constant.

Topic: Physics, Motion In Two Dimensions, WBJEE, WBJEE 2021
In case of reflection from curved surface

Options:
A the normal is considered to be tangent to the spherical surface at the point of incidence
B the normal is considered to be drawn at the pole of the spherical surface only.
C the normal is the principal axis only.
D the normal is not required.

The correct answer is:
the normal is considered to be tangent to the spherical surface at the point of incidence

Solution:
The normal to the curved surface along the centre or radius of the spherical surface, i.e., the normal is considered to be tangent to the spherical surface at the point of incidence.

The geometrical centre of spherical surface is called pole, while in lens it is called optical centre.

Topic: Physics, Ray Optics, JEE Main
In cathode ray oscillograph, the focusing of beam on the screen is achieved by

Options:
A Convex lenses
B Magnetic field
C Electric potential
D All of these

The correct answer is:
Electric potential

Solution:
In a cathode ray oscillograph the focusing of beam on the screen is achieved by electric potential. There are two plates X and Y. X plates consists two plates X1 and X2 in vertical plane while Y plates also consist two plates Y1 and Y2 in a horizontal plane. An electric is applied between the X and Y plates by an external source.

Topic: Physics, Dual Nature of Matter, JEE Main
In certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64 . How many telephones numbers have all six digits distinct?

Options:

None

Solution:
If first two digits are 41 , the remaining 4 digits can be arranged in
$$
\begin{aligned}
={ }^8 P_4=\frac{8 !}{8-4 !} &=\frac{8 !}{4 !}=\frac{8 \times 7 \times 6 \times 5 \times 4 !}{4 !} \\
&=8 \times 7 \times 6 \times 5=1680
\end{aligned}
$$
Similarly, iffirst two digits are $42,46,62$, or 64 , the remaining 4 digits can be arranged in same way i.e., 1680 ways.
$\therefore$ Total number of telephone numbers having all six digits
$$
\text { distinct }=5 \times 1680=8400 \text {. }
$$

Topic: Mathematics, Permutation Combination
In certain matters lithium differs from other alkali metals, the main reason for this is

Options:
A Small size of $L i$ atom and $L i^{+}$ion
B Extremely high electropositivity of $L i$
C Greater hardness of $L i$
D Hydration of $\mathrm{Li}^{+}$ion

The correct answer is:
Small size of $L i$ atom and $L i^{+}$ion

Solution:
Small atomic and ionic size leads to high electronegativity and hydration energy. Small atomic and ionic size leads to high electronegativity and hydration energy.

Topic: Chemistry, s Block Elements, JEE Main
In CGS system of units, the density of a material is 4 g cm-3. What will be the value of the density of the material in a system of units in which unit of length is 10​ cm and unit of mass is 100 g ?

Options:
A 0.04
B 0.4
C 40
D 400

The correct answer is:
40

Solution:
As n1u1=n2u2
4 g cm-3=n2100 g10 cm3

⇒n2=40

Topic: Physics, Units and Dimensions, JEE Main
In CGS system the magnitude of the force is 100 dynes. In another system where the fundamental physical. quantities are kilogram, meter and minute, the magnitude of the force is

Options:
A $0.036$
B $0.36$
C $3.6$
D $36$

The correct answer is:
$3.6$

Solution:
$n_2=n_1\left[\left(\frac{m_1}{m_2}\right)^a\left(\frac{L_1}{L_2}\right)^b\left(\frac{T_1}{T_2}\right)^c\right]$
Dimensions of force is [ $\mathrm{M} \mathrm{L} \mathrm{T}^{-2}$ ]
$a=1, b=1, c=-2$
$\begin{aligned} n_2 & =100\left[\left(\frac{\mathrm{g}}{\mathrm{kg}}\right)^1\left(\frac{\mathrm{cm}}{\text { metre }}\right)^1\left(\frac{\mathrm{sec}}{\mathrm{min}}\right)^{-3}\right] \\ & =100\left[\frac{\mathrm{g}}{100 \mathrm{~g}} \times \frac{\mathrm{cm}}{100 \mathrm{~cm}} \times\left(\frac{\mathrm{sec}}{60 \mathrm{sec}}\right)^{-2}\right] \\ & =100\left[\frac{1}{1000} \times \frac{1}{100} \times(60)^2\right] \\ & =100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600=3.6\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
In C.G.S. system the magnitutde of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is

Options:
A $0.036$
B $0.36$
C $3.6$
D $36$

The correct answer is:
$3.6$

Solution:
$\begin{aligned} & n_2=n_1\left(\frac{M_1}{M_2}\right)^1\left(\frac{L_1}{L_2}\right)^1\left(\frac{T}{T_2}\right)^{-2} \\ & =100\left(\frac{g m}{k g}\right)^1\left(\frac{\mathrm{cm}}{\mathrm{m}}\right)^1\left(\frac{\mathrm{sec}}{\mathrm{min}}\right)^{-2} \\ & =100\left(\frac{\mathrm{gm}}{10^3 \mathrm{gm}}\right)^1\left(\frac{\mathrm{cm}}{10^2 \mathrm{~cm}}\right)^1\left(\frac{\mathrm{sec}}{60 \mathrm{sec}}\right)^{-2} \\ & n_2=\frac{3600}{10^3}=3.6\end{aligned}$

Topic: Physics, Units and Dimensions, JEE Main
In $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$, the bond that undergoes heterolytic cleavage most readily is

Options:
A $C-C$
B $C-O$
C $\mathrm{C}-\mathrm{H}$
D $\mathrm{O}-\mathrm{H}$

The correct answer is:
$\mathrm{O}-\mathrm{H}$

Solution:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow{\text { Heterolytic cleavage }} \mathrm{CH}_3 \mathrm{CH}_2^{+}+\mathrm{OH}^{-}$

Topic: Chemistry, General Organic Chemistry, JEE Main
In chemical reaction $A \rightleftharpoons B$, the system will be known in equilibrium when

Options:
A A completely changes to $B$
B $\mathbf{5 0} \%$ of $\boldsymbol{A}$ changes to $\boldsymbol{B}$
C The rate of change of $A$ to $B$ and $B$ to $A$ on both the sides are same
D Only $10 \%$ of $A$ changes to $B$

The correct answer is:
The rate of change of $A$ to $B$ and $B$ to $A$ on both the sides are same

Solution:
When rate of forward reaction is equal to rate of backward reaction the reaction is said to be in equilibrium.

Topic: Chemistry, Chemical Equilibrium, JEE Main
In chromatography technique, the purification of a compound is independent of:

Options:
A Mobility or flow of solvent system
B Solubility of the compound
C Length of the column or TLC Plate
D Physical state of the pure compound

The correct answer is:
Physical state of the pure compound

Solution:
In chromatography technique, the purification of a compound is independent of the physical state of the pure compound.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2021 (16 Mar Shift 1)
In chromatography, which of the following statements is incorrect for Rf?

Options:
A Higher Rf value means higher adsorption.
B The value of Rf cannot be more than one.
C Rf value depends on the type of chromatography.
D Rf value of dependent on the mobile phase.

The correct answer is:
Higher Rf value means higher adsorption.

Solution:
Rf or Retention factor or Retardation factor =distance moved by spotdistance moved by solvent
For low polarity compounds, lower will be adsorption and the lower the value of adsorption higher will be Rf value

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
In chromite ore, the oxidation number of iron and chromium are respectively.

Options:
A +3,+2
B +3,+6
C +2,+6
D +2,+3

The correct answer is:
+2,+3

Solution:
Actual composition of chromite ore(FeCr2O4) is FeO.Cr2O3. In FeO, the oxidation state of Fe is +2 while in Cr2O3, the oxidation state of Cr is +3.

Topic: Chemistry, d and f Block Elements, JEE Main
In chromyl chloride test for confirmation of Cl– ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and 10% H2O2 turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

Options:
A +6
B +5
C +10
D +3

The correct answer is:
+6

Solution:
The chromyl chloride test is used to detect chloride ions in the qualitative analysis. If any chloride salt like sodium chloride is heated with acidified potassium dichromate it produces red colour fumes of chromyl chloride. It confirms the presence of chloride ions in that salt.

Cl-+K2Cr2O7+H2SO4→CrO2Cl2→ Basic medium CrO42- yellow solution +Cl-

Chromate reacts rapidly with hydrogen peroxide to exchange two oxo ligands and become diperoxochromate(VI)

CrO42-yellow solution→3. 10% H2O21. Acidification2. Amyl alcoholCrO5blue compound

Hence, the answer is option A.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (29 Jan Shift 1)