datavorous/entrance-exam-dataset huggingface dataset:
Point mutation involves :

Options:
A Deletion
B Insertion
C Change in single base pair
D Duplication

The correct answer is:
Change in single base pair

Solution:
No Solution Available

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2009 (Mains)
Point out in which of the following properties oxygen differs from the rest of the members of its family (Group-VIA)

Options:
A High value of ionisation energies
B Oxidation states $(2,4,6)$
C Polymorphism
D Formation of hydrides

The correct answer is:
Oxidation states $(2,4,6)$

Solution:
Oxidation states are $2,4,6$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Point out the incorrect statement among the following?

Options:
A The oxidation state of oxygen is +2 in OF2 .
B Acidic character follows the order H2O<H2S<H2Se<H2Te.
C The tendency to form multiple bonds increases in moving down the group from sulphur to tellurium (towards C and N)
D Sulphur has a strong tendency to catenate while oxygen shows this tendency to a limited extent.

The correct answer is:
The tendency to form multiple bonds increases in moving down the group from sulphur to tellurium (towards C and N)

Solution:
Multiple bonds formation tendency with carbon and nitrogen decreases from sulphur to tellurium.
CS2(S=C=S) is moderately stable.
CSe 2 (Se = C = Se) decomposes readily whereas,
CTe2(Te=C=Te) does no exist.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Point P-1,7 lies on the line 4x+3y=17. Then the coordinates of the points farthest from the line which are at a distance of 10 units from the point P are

Options:
A 7,13 and -9,1
B 5,15 and -1,-7
C -1,5 and 15,-7
D 15,5 and -7,-1

The correct answer is:
7,13 and -9,1

Solution:
There are two points A & B which are at a distance of 10 units from P and farthest ( 10 unit distance) from the line 4x+3y=17.

The slope of AB=34 {Because the slope of 4x+3y=17 is -43 }
Let line AB makes an angle θ with x-axis then tan⁡θ=34⇒cos⁡θ,sin⁡θ=45,35
⇒A or B=10cos⁡θ-1,10sin⁡θ+7 or ⇒A or B=-10cos⁡θ-1,-10sin⁡θ+7
⇒A or B=7,13 or -9,1

Topic: Mathematics, Straight Lines, JEE Main
Points I, II and III in the following plot respectively correspond to (Vmp: most probable velocity)

Options:
A Vmp of N2300 K; Vmp of H2300 K; Vmp of O2400 K;
B Vmp of H2300Â K; Vmp of N2300Â K;Â Vmp of O2400Â K;
C Vmp of N2300Â K; Vmp of O2400Â K;Â Vmp of H2300Â K;
D Vmp of O2400Â K; Vmp of N2300Â K;Â Vmp of H2300Â K;

The correct answer is:
Vmp of N2300 K; Vmp of O2400 K; Vmp of H2300 K;

Solution:
Vmp=2RTM ∝TM
For N2 at 300 K, Vmp∝TM ∝30028=3.2
For O2 at 300 K,Vmp∝TM ∝40032=3.53
For H2 at 300 K, Vmp∝TM ∝3002=12.24
Vmp of N2<Ump of O2<Ump of H2

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Points P-3,2,Q9,10 and Rα,4 lie on a circle C with PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line 2x-ky=1, then k is equal to _____ .

Options:

The correct answer is:
3

Solution:
Let S≡x1,y1

Given:

P≡-3,2Q≡9,10R≡α,4

Since, PQ⊥RQ, so

mPQ×mQR=-1

⇒10-29+3×10-49-α=-1

⇒19-α=-14

⇒α=13

So, R≡13,4

So, centre of circle is O≡13-32,4+22≡5,3

Also,

mOQ·mQS=-1

⇒3-105-9mQS=-1

⇒mQS=-47

Equation of QS is

y-10=-47x-9

⇒4x+7y=106 …1

Also,

mOR·mRS=-1

⇒4-313-5mRS=-1

⇒mRS=-8

Equation of RS

y-4=-8x-13

⇒8x+y=108 …2

Solving equations 1 & 2, we get

x1=252; y1=8

Since, Sx1,y1 lies on 2x-ky=1

25-8k=1

⇒8k=24

⇒ k=3

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Polar of origin $(0,0)$ with respect to the circle $x^2+y^2+2 \lambda x+2 \mu y+c=0$ touches the circle $x^2+y^2=r^2$, if

Options:
A $c=r\left(\lambda^2+\mu^2\right)$
B $r=c\left(\lambda^2+\mu^2\right)$
C $c^2=r^2\left(\lambda^2+\mu^2\right)$
D $r^2=c^2\left(\lambda^2+\mu^2\right)$

The correct answer is:
$c^2=r^2\left(\lambda^2+\mu^2\right)$

Solution:
Polar is $\lambda x+\mu y+c=0$. The condition of tangency $p=r$ gives the result (option 3).

Topic: Mathematics, Circle, JEE Main
Polar stratospheric clouds facilitate the formation of

Options:
A ClONO2
B HOCl
C ClO
D CH4

The correct answer is:
HOCl

Solution:
Polar stratospheric clouds surface act on catalyst which provided surface on which hydrolysis of ClONO2 takes place to form HOCl (Hypochlorous acid)

ClONO2g+H2Og→HOClg+HNO3g

Polar stratospheric clouds (PSCs) play a central role in the formation of the ozone hole in the Antarctic and Arctic.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (26 Jun Shift 1)
Polarization means "the distortion of the shape of an anion by an adjacently placed cation". Which of the following statements about polarization is correct

Options:
A Maximum polarization is brought about by a cation of high charge
B Minimum polarization is brought about by a cation of low radius
C A large cation is likely to bring about a large degree of polarization
D A small anion is likely to undergo a large degree of polarization

The correct answer is:
Maximum polarization is brought about by a cation of high charge

Solution:
According to Fajan's rule, polarisation of anion is influenced by charge and size of cation more is the charge and small size of cation causes more polarisation of anion having larger size and greater charge.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Polarization of electrons in acrolein may be written as:

Options:
A C - δ H 2 = CH - C + δ = O
B C - δ H 2 = CH - CH = O + δ
C C - δ H 2 = C + δ H - CH = O
D C + Î´ H 2 = CH - CH = O - Î´

The correct answer is:
C + δ H 2 = CH - CH = O - δ

Solution:
O atom is more electronegative than C atom, so it acquires -δ charge and C atom acquires +δ charge which is transferred to the last C atom.

Topic: Chemistry, General Organic Chemistry, JEE Main
Polarization of light is a phenomenon that supports

Options:
A Particle nature of light
B Dual nature of light
C Wave nature of light
D None of the above

The correct answer is:
Wave nature of light

Solution:
The direction of vibration of particles is a property associated with waves. Since light shows the vibration of the electric and magnetic field through polarization, the wave nature of light can be supported by polarization.

Topic: Physics, Dual Nature of Matter, JEE Main
Pollen grains remain preserved as fossils due to the presence of :

Options:
A Epidermal layer
B Tapetum
C Exine layer
D Intine layer

The correct answer is:
Exine layer

Solution:
Pollen grains are well-preserved as fossils because of the presence of sporopollenin, which is found in the outer layer of pollen, i.e., exine.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2024 (Re-NEET)
Pollination by water is

Options:
A rare in flowering plants.
B limited to about 30 genera.
C most of the genera are monocotyledons.
D all the above.

The correct answer is:
all the above.

Solution:
Transfer of pollen grains from anther lobe to stigma is called as pollination. Abiotic agents for pollination are air and water. Most plant uses air as agent of pollination. Pollination by water is rare in flowering plants and it is restricted to only 30 genera mainly monocots.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Poly-β-hydroxybutyrate-co-β-hydroxyvalerate PHBV is a copolymer of _____.

Options:
A 3- hydroxybutanoic acid and 4- hydroxypentanoic acid
B 3- hydroxybutanoic acid and 2- hydroxypentanoic acid
C 3- hydroxybutanoic acid and 3- hydroxypentanoic acid
D 2- hydroxybutanoic acid and 3- hydroxypentanoic acid

The correct answer is:
3- hydroxybutanoic acid and 3- hydroxypentanoic acid

Solution:
PHBV is obtained by copolymerization of 3- hydroxybutanoic acid & 3- hydroxypentanoic acid.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2019 (12 Jan Shift 1)
Polyethylene glycol method is used for :

Options:
A Energy production from sewage
B Gene transfer without a vector
C Biodiesel production
D Seedless fruit production

The correct answer is:
Gene transfer without a vector

Solution:
No Solution Available

Topic: Biology, Biotechnology - Principles and Processes, NEET, NEET 2009 (Mains)
Polyethylene is obtained from calcium carbide.

CaC2+2H2O→CaOH2+C2H2

C2H2+H2→C2H4

nC2H4→-CH2-CH2-n

Therefore, the amount of polyethylene (in kg) obtained for 64 kg CaC2 is

Options:

The correct answer is:
28

Solution:
Moles of CaC2=64×10364≅1×103
So from the balanced chemical equation, moles of C2H2= moles of C2H4
= moles of CaC2=1×103
So moles of polythene =1n×1×103
So weight of polythene =1n×1×1×28n kg
= 28 kg

Topic: Chemistry, Polymers, JEE Main
Polymer formation from monomers starts by

Options:
A
condensation reaction between monomers

B
coordinate reaction between monomers

C
conversion of monomer to monomer ions by protons

D
hydrolysis of monomers

The correct answer is:
condensation reaction between monomers

Solution:
No Solution Available

Topic: Chemistry, Polymers, JEE Main, JEE Main 2002
Polymer used in orlon is:

Options:
A

Polyethene

B

Polycarbonate

C

Polyamide

D

Polyacrylonitrile

The correct answer is:

Polyacrylonitrile

Solution:
The addition polymerisation of acrylonitrile in presence of a peroxide catalyst leads to the formation of polyacrylonitrile. The commercial name of polyacrylonitrile is orlon (acrilan). That is, it is used as a substitute for wool in making commercial fibres as orlon or acrilan.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Polymoniales are mainly based on the floral characters.
What is the animal order that includes families like Felidae and Canidae?

Options:
A Carnivora
B Primates
C Rodentia
D Marsupialia

The correct answer is:
Carnivora

Solution:
The animal order, Carnivora, includes families like Felidae and Canidae.

Topic: Biology, The Living World, NEET
Polysome is formed by

Options:
A Ribosomes attached to each other in a linear arrangement
B Several ribosomes attached to a single mRNA
C Many ribosomes attached to a strand of endoplasmic reticulum
D A ribosome with several subunits

The correct answer is:
Several ribosomes attached to a single mRNA

Solution:
This structure appears during protein synthesis in cytoplasm to translate different parts of a polypeptide.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Mains)
Polystyrene, Dacron and Orlon are classified respectively as

Options:
A Chain growth; step growth; step growth
B Chain growth; chain growth; step growth
C Chain growth; step-growth; chain growth
D Step growth; step growth; chain growth

The correct answer is:
Chain growth; step-growth; chain growth

Solution:
Polystyrene and orlon are vinyl derivative and they are chain growth polymers while Dacron is a step growth polymer

Topic: Chemistry, Biomolecules, JEE Main
Polysubstitution is a major drawback in:

Options:
A Reimer Tiemann reaction
B Friedel Craft's alkylation
C Acetylation of aniline
D Friedel Craft's acylation

The correct answer is:
Friedel Craft's alkylation

Solution:
In Friedel crafts alkylation product obtained is more activated due to the activating nature of the alkyl group and hence poly substitution will take place.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Polyvinyl alcohol is an important polymer. The structure is given below

It is prepared by polymerization of

Options:
A CH2=CH-OH
B CH2=CH-OCOCH3, followed by hydrolysis
C CH2=CH-CN, followed by hydrolysis
D CH2=CH-COOCH3, followed by hydrolysis

The correct answer is:
CH2=CH-OCOCH3, followed by hydrolysis

Solution:
PVA is prepared by hydrolysis of polyvinyl acetate. Other vinyl ester-derived polymers are sometimes used, with formate, chloroacetate groups instead of acetate. The conversion of the polyesters is usually conducted by base-catalysed transesterification with ethanol:
CH2CHOAcn+C2H5OH→CH2CHOHn+C2H5OAc

Topic: Chemistry, Polymers, JEE Main
Portland cement contains 'X' to enhance the setting time. What is 'X'?

Options:
A CaSO4·12H2O
B CaSO4·2H2O
C CaSO4
D CaCO3

The correct answer is:
CaSO4·2H2O

Solution:
Cement is an important building material. It was first introduced in England in 1824 by Joseph Aspdin. It is also called Portland cement because it resembles with the natural limestone quarried in the Isle of Portland, England.

The raw materials for the manufacture of cement are limestone and clay.

When mixed with water, the setting of cement takes place to give a hard mass. This is due to the hydration of the molecules of the constituents and their rearrangement. The purpose of adding gypsum is only to slow down the process of setting of the cement so that it gets sufficiently hardened.

Hence,GypsumCaSO4·2H2O is used to enhance setting time in portland cement.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Position of a body with acceleration a is given by x=kam tn. Here t is time. Find the dimensions of m and n.

Options:
A m=1, n=1
B m=1, n=2
C m=2, n=1
D m=2, n=2

The correct answer is:
m=1, n=2

Solution:
Positions x=kamtn
M0LT0=LT-2mTn
=M0LmT-2m+n
On comparing both sides
m=1
-2m+n=0
n=2m
n=2×1=2

Topic: Physics, Units and Dimensions, JEE Main
Position of a particle in a rectangular-coordinate system is $(3,2,5)$. Then its position vector will be

Options:
A $3 \hat{i}+5 \hat{j}+2 \hat{k}$
B $3 \hat{i}+2 \hat{j}+5 \hat{k}$
C $5 \hat{i}+3 \hat{j}+2 \hat{k}$
D None of these

The correct answer is:
$3 \hat{i}+2 \hat{j}+5 \hat{k}$

Solution:
If a point have coordinate $(x, y, z)$ then its position vector $=x \hat{i}+y \hat{j}+z \hat{k}$.

Topic: Physics, Mathematics in Physics, JEE Main
Position of an ant ( S in metres) moving in Y-Z plane is given by S=2t2j^+5k^ (where t is in second). The magnitude and direction of velocity of the ant at t=1 s will be :

Options:
A 16 m s-1 in y-direction
B 4 m s-1 in x-direction
C 9 m s-1 in z-direction
D 4 m s-1 in y-direction

The correct answer is:
4 m s-1 in y-direction

Solution:
The velocity of the ant can be found as follows:

v→=dS→dt=ddt2t2j^+5k^=4tj^ ...1

At t=1 s, from equation (1), it follows that

v→=4j^

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Position of the point $(1,1)$ with respect to the circle $x^2+y^2-x+y-1=0$ is

Options:
A Outside the circle
B Upon the circle
C Inside the circle
D None of these

The correct answer is:
Outside the circle

Solution:
$S_1\gt0$, hence point lies outside.

Topic: Mathematics, Circle, JEE Main
Positive and negative point charges of equal magnitude are kept at \( \left(0,0, \frac{a}{2}\right) \) and \( \left(0,0, \frac{-a}{2}\right) \), respectively. The work done by the electric field when another positive point charge is moved from \( (-a, 0,0) \) to \( (0, a, 0) \) is

Options:
A positive
B negative
C zero
D depends on the path connecting the initial and final positions.

The correct answer is:
zero

Solution:
It is clear from the figure, that the dipole is present along the Z-axis and another point positive charge is moved on X-Y plane from A to B, both are axial points of the dipole. So,
VA=VB=0⇒W=QΔV=+1VA-VB=0
Hence, work done is zero.

Topic: Physics, Electrostatics, JEE Main
Positive deviation from ideal behaviour takes place because of

Options:
A molecular interaction between atoms and PVnR>1
B molecular interaction between atoms and PVnRT<1
C finite size of atoms and PVnRT>1
D finite size of atoms and PVnRT<1

The correct answer is:
finite size of atoms and PVnRT>1

Solution:
For positive deviation
PVnRT>1
PV=nRT+nPb
PV/nRT=1+Pb/RT
PV/nRT>1.

Topic: Chemistry, States of Matter, JEE Main
Positive dipole moment is present in

Options:
A $\mathrm{CCl}_4$
B $\mathrm{C}_6 \mathrm{H}_6$
C ${ }^{B F_3}$
D $H F$

The correct answer is:
$H F$

Solution:
$H-F$ is polar due to difference of electronegativity of hydrogen and fluorine so it shows positive dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Positive rays was discovered by:

Options:
A Thomson
B Goldstein
C W. Crookes
D Rutherford

The correct answer is:
Goldstein

Solution:
German scientist Eugen Goldstein is the first to identify the presence of Positive rays in 1886 , properties of positive rays were investigated by J J Thomson in 1906. Hence, option (2) is the answer

Topic: Physics, Dual Nature of Matter, JEE Main
Positive rays was discovered by

Options:
A Thomson
B Goldstein
C W. Crookes
D Rutherford

The correct answer is:
Goldstein

Solution:
Positive rays was discovered by Goldstein.

Topic: Physics, Atomic Physics, JEE Main
Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.

6OH-+Cl-→ClO3-+3H2O+6e-

A current of xA has to be passed for 10 h to produce 10.0 g of potassium chlorate. the value of x is __________. (Nearest integer)

(Molar mass of KClO3=122.6 g mol-1 F=96500 C)

Options:

The correct answer is:
1

Solution:
Given balanced equation is

60H⊝+C⊝l⟶ClO3-+3H2O+6e-

→10 gKClO3⇒10122.6 molKCO3 in obtained

→ from the above reaction, it is concluded that by
6 F charge 1 molKClO3 is obtained.

→ By the passage of 6 F charge =1 molKClO3

∴ By the passage of x×10×60×6096500F charge

=16×x×10×60×6096500

Now x×10×60×606×96500=10122.6

⇒x=10×96560×122.6=965735.6=1.311≃1

OR

W=EF×I×t

10=122.696500×6×x×10×3600

X=1.311

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Potassium dichromate acts as a strong oxidizing agent in acidic solution. During this process, the oxidation state changes from

Options:
A +3 to +1
B +6 to +3
C +2 to +1
D +6 to +2

The correct answer is:
+6 to +3

Solution:
Dichromate oxidises to Cr3+ from Cr2O72- in acidic medium according to the reaction.

In dichromate, chromium is in +6 oxidation state and on the right hand side we have Cr in +3 state. So, the change is from +6 to +3 in acidic medium.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown-red vapours of :

Options:
A
$\mathrm{CrO}_3$

B
$\mathrm{CrCl}_3$

C
$\mathrm{CrO}_2 \mathrm{Cl}_2$

D
$\mathrm{Cr}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{CrO}_2 \mathrm{Cl}_2$

Solution:
Solid potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride gives orange red vapours of a volatile oily liquid $\mathrm{CrO}_2 \mathrm{Cl}_2$
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{NaCl}+6 \mathrm{H}_2 \mathrm{SO}_4\longrightarrow 2 \mathrm{KHSO}_4+4 \mathrm{NaHSO}_4+2 \mathrm{CrO}_2 \mathrm{Cl}_2$ chromyl chloride

Topic: Chemistry, d and f Block Elements, JEE Main
Potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride, gives brown-red vapours of :

Options:
A
$\mathrm{CrO}_3$

B
$\mathrm{CrCl}_3$

C
$\mathrm{CrO}_2 \mathrm{Cl}_2$

D
$\mathrm{Cr}_2 \mathrm{O}_3$

The correct answer is:
$\mathrm{CrO}_2 \mathrm{Cl}_2$

Solution:
Solid potassium dichromate when heated with concentrated sulphuric acid and a soluble chloride gives orange red vapours of a volatile oily liquid $\mathrm{CrO}_2 \mathrm{Cl}_2$
$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{NaCl}+6 \mathrm{H}_2 \mathrm{SO}_4\longrightarrow 2 \mathrm{KHSO}_4+4 \mathrm{NaHSO}_4+2 \mathrm{CrO}_2 \mathrm{Cl}_2$ chromyl chloride

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2013 (09 Apr Online)
Potassium ferrocyanide is used in the detection of

Options:
A Cu2+ ions
B Fe3+ ions
C Mn2+ ions
D Both A and B

The correct answer is:
Both A and B

Solution:
Both Cu2+ and Fe3+ are detected by potassium ferrocyanide
2CuSO4+K4FeCN6âCu2FeCN6Chocolate ppt. +2K2SO4
4FeCl 3  +  3K 4 Fe CN 6 → Fe 4 Fe CN 6 3 Ferri-ferrocyanide Prussian blue +12KCI

Topic: Chemistry, Coordination Compounds, JEE Main
Potassium fluoride has $\mathrm{NaCl}$ type structure. What is the distance between $\mathrm{K}^{+}$and $F^{-}$ions if cell edge is a $\mathrm{cm}$

Options:
A $2 \mathrm{a} \mathrm{cm}$
B $a / 2 \mathrm{~cm}$
C $4 a \mathrm{~cm}$
D $a / 4 \mathrm{~cm}$

The correct answer is:
$a / 2 \mathrm{~cm}$

Solution:
Distance between $K^{+}$and $F^{-}=\frac{1}{2} \times$ length of the edge

Topic: Chemistry, Solid State, JEE Main
Potassium is kept in

Options:
A Alcohol
B Water
C Kerosene
D Liquid ammonia

The correct answer is:
Kerosene

Solution:
Alkali metals are highly reactive metals. but they do not react with kerosene

Topic: Chemistry, s Block Elements, JEE Main
Potassium ozonide on decomposition gives

Options:
A K+O2
B K2O+O2
C KO2+O2
D KO2+O3

The correct answer is:
KO2+O2

Solution:
2KO3→2KO2+O2

Topic: Chemistry, s Block Elements, JEE Main
Potassium permanganate acts as an oxidant in neutral, alkaline as well as acidic media. The final products obtained from it in the three conditions are, respectively

Options:
A $\mathrm{MnO}_2, \mathrm{MnO}_2, \mathrm{Mn}^{2+}$
B $\mathrm{MnO}_4^{2-}, \mathrm{Mn}^{3+}, \mathrm{Mn}^{2+}$
C $\mathrm{MnO}_2, \mathrm{MnO}_4^{2-}, \mathrm{Mn}^{3+}$
D $\mathrm{MnO}, \mathrm{MnO}_4, \mathrm{Mn}^{2+}$

The correct answer is:
$\mathrm{MnO}_2, \mathrm{MnO}_2, \mathrm{Mn}^{2+}$

Solution:
$\mathrm{MnO}_2, \mathrm{MnO}_2, \mathrm{Mn}^{2+}$
In neutral medium :
$2 \mathrm{KMnO}_4+3 \mathrm{MnSO}_4+2 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{MnO}_2$
In alkaline medium :
$2 \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{KOH}+3 \mathrm{O}$
In acidic medium :
$2 \mathrm{KMnO}_4+3 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4+5 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}$

Topic: Chemistry, d and f Block Elements, JEE Main
Potassium permanganate on heating at \( 513 \mathrm{~K} \) gives a product which is :

Options:
A paramagnetic and colourless
B diamagnetic and colourless
C diamagnetic and green
D paramagnetic and green

The correct answer is:
paramagnetic and green

Solution:
2KMnO4⟶K2MnO4+MnO2( s)+O2 Purple Green (Diamagnetic) (Paramagnetic)

Topic: Chemistry, d and f Block Elements, JEE Main
Potassium permanganate works as oxidising agent both in acidic and basic medium. In both state, product obtained by \( \mathrm{KMnO}_{4} \) is respectively

Options:
A \( \mathrm{MnO}_{2}{ }^{-} \)and \( \mathrm{Mn}^{3+} \)
B \( \mathrm{Mn}^{3+} \) and \( \mathrm{Mn}^{3+} \)
C \( \mathrm{Mn}^{2+} \) and \( \mathrm{Mn}^{3+} \)
D \( \mathrm{Mn}^{2+} \) and \( \mathrm{MnO}_{2} \)

The correct answer is:
\( \mathrm{Mn}^{2+} \) and \( \mathrm{MnO}_{2} \)

Solution:
i In alkaline medium, KMnO4 first reduces to manganate and then insoluble manganese dioxide.

Basic medium 2MnO4-+H2O→MnO2+2O-H+3O

ii In acidic medium:

MnO4-+8H++5e-→Mn+2+4H2O

Mn give five electrons oxidizing agent.

Topic: Chemistry, d and f Block Elements, JEE Main
Potassium permangnate works as oxidising agent both in acidic and basic medium. In both state product obtained by $\mathrm{KMnO}_4$ is respestively

Options:
A $\mathrm{MnO}_2^{-}$and $\mathrm{Mn}^{3+}$
B $\mathrm{Mn}^{3+}$ and $\mathrm{Mn}^{2+}$
C $\mathrm{Mn}^{2+}$ and $\mathrm{Mn}^{3+}$
D $\mathrm{MnO}_2$ and $\mathrm{Mn}^{2+}$

The correct answer is:
$\mathrm{MnO}_2$ and $\mathrm{Mn}^{2+}$

Solution:
In alkaline medium, $\mathrm{KMnO}_4$ first reduced in manganate \& then in insolusla manganese dioxide.
$2 \mathrm{MnO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{MnO}_2+2 \mathrm{OH}^{-}+3[\mathrm{O}]$
In acidic medium, Manganous s্রulphate formed
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, d and f Block Elements, JEE Main
Potassium superoxide on hydrolysis gives

Options:
A K++OH¯+O2
B K++K2O+O2+OH¯
C K++H2O2+O2+OH
D -K++H2O2+OH

The correct answer is:
K++H2O2+O2+OH

Solution:
On combustion in excess of air, lithium forms mainly the oxide, Li2O (plus some peroxide Li2O2 ), sodium forms the peroxide, Na2O2 (and some superoxide NaO2 ) whilst potassium, rubidium and caesium form the superoxides, MO2. Under appropriate conditions pure compounds M2O, M2O2 and MO2 may be prepared. The increasing stability of the peroxide or superoxide, as the size of the metal ion increases, is due to the stabilisation of large anions by larger cations through lattice energy effects.

Potassium superoxide on hydrolysis gives potassium hydroxide,hydrogen peroxide and oxygen.

2KO2+2H2O→2KOH+H2O2+O2

Topic: Chemistry, s Block Elements, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 2)