datavorous/entrance-exam-dataset huggingface dataset:
The one that does not stabilize 2∘ and 3∘ structures of proteins is

Options:
A –S–S–linkage
B H-bonding
C –O–O–linkage
D van der Waals forces

The correct answer is:
–O–O–linkage

Solution:
Secondary structure of protein refers to local folded structures that form within a polypeptide due to interactions between atoms of the backbone. They are found to exist in two different types of structures α – helix and β – pleated sheet structures. This structure arises due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between -CO group and -NH groups of the peptide bond. Tertiary Structure of Protein arises from further folding of the secondary structure of the protein. H-bonds, electrostatic forces, disulphide linkages, and Vander Waals forces stabilise this structure. The tertiary structure of proteins represents overall folding of the polypeptide chains, further folding of the secondary structure.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The one that is extensively used as a piezoelectric material is:

Options:
A Tridymite
B Quartz
C Amorphous silica
D Mica

The correct answer is:
Quartz

Solution:
Piezoelectric materials are those that produce an electric current when placed under mechanical stress.

Quartz is used as a piezoelectric material in electronic devices.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The one that is not a carbonate are is:

Options:
A malachite
B bauxite
C calamine
D siderite

The correct answer is:
bauxite

Solution:
Bauxite is an oxide ore of Al. Its formula is Al2O3.

Formula of malachite is (Cu2(CO3)(OH)2)

Formula of calamine is ZnCO3

Formula of calamine is FeCO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The one that is not expected to show isomerism is:

Options:
A NiNH34H2O22+
B Ni(en)32+
C NiNH32Cl2
D PtNH32Cl2

The correct answer is:
NiNH32Cl2

Solution:
(1) NiNH32Cl2 does not show isomerism because it is tetrahedral and has a plane of symmetry.

(2)NiNH34H2O22+shows geometrical isomerism.

(3)Ni(en)32+ , shows optical isomerism.
(4)PtNH32C2, shows geometrical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The one that is NOT suitable for the removal of permanent hardness of water is :

Options:
A Clark's method
B Ion–exchange method
C Calgon's method
D Treatment with sodium carbonate

The correct answer is:
Clark's method

Solution:
Clark's Method is used to remove temporary hardness by using lime water (or) Ca(OH)2

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The one that will show optical activity is
(en=ethane-1, 2-diamine)

Options:
A
B
C

D

The correct answer is:

Solution:
Plane of symmetry exists.
Hence, does not show optical isomerism.
This structure has a plane of symmetry.
Hence, no optical isomerism.
No plane of symmetry it shows optical isomerism.
Plane of symmetry, so it does not show optical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The one which does not represent a hyperbola is

Options:
A $x y=1$
B $x^2-y^2=5$
C $(x-1)(y-3)=3$
D $x^2-y^2=0$

The correct answer is:
$x^2-y^2=0$

Solution:
For hyperbola $\Delta \neq 0$ and $h^2 \gt a b$. Here $\Delta=0$. So (4) \(x^2-y^2=0\) is not satisfying the condition.

Topic: Mathematics, Hyperbola, JEE Main
The only force acting on a block is along \( x \)-axis is given by \( F=-\left(\frac{4}{x^{2}+2}\right) N \) When the block moves from \( x=-2 m \) to \( \mathrm{x}=4 \) mthe change in kinetic energy of block is -

Options:
A Positive
B Negative
C Zero
D May be positive or negative

The correct answer is:
Negative

Solution:
As W =ΔK
Force is along negative x-axis and displacement is along + x-axis
∴W=negative
Hence
ΔK=negative

Topic: Physics, Work Power Energy, JEE Main
The open-chain glucose, (an aldohexose) and fructose (a 2-oxohexose) have X and Y chiral carbons respectively. Find the sum of X + Y.

Options:

The correct answer is:
7

Solution:
Open-chain glucose,
and open-chain fructose,
has three chiral carbons.
Hence sum of X + Y = 4 + 3 = 7.

Topic: Chemistry, Biomolecules, JEE Main
The open-chain glucose on oxidation with HIO4 gives

Options:
A 5HCOOH+H2C=O
B 4HCOOH+2H2C=O
C 3HCOOH+3H2C=O
D 2HCOOH+4H2C=O

The correct answer is:
5HCOOH+H2C=O

Solution:
Both CHO and CHOH form HCOOH; while HOCH2 forms H2C=O giving 5HCOOH+H2C=O .

Topic: Chemistry, Biomolecules, JEE Main
The optical rotation of the α -form of a pyranose is +150.7° , that of the β -form is +52.8° . In solution an equilibrium mixture of these anomers has an optical rotation of +80.2° . The percentage of the α -form in equilibrium mixture is

Options:
A 28%
B 32%
C 68%
D 72%

The correct answer is:
28%

Solution:
Let the percentage of α -form be X
150.7(X)+52.8(1-X)=+80.2
150.7X+52.8-52.8X=+80.2
97.9X=27.4
X=0.279=0.28
≈28%

Topic: Chemistry, General Organic Chemistry, JEE Main
The optically active tartaric acid is named as D - $(+)$ - tartaric acid because it has a positive

Options:
A optical rotation and is drived from D - glucose
B $\mathrm{pH}$ in organic solvent
C optical rotation and is derived from D - (+) glyceraldehyde
D optical rotation when substituted by deuterium

The correct answer is:
optical rotation and is derived from D - (+) glyceraldehyde

Solution:
Positive sign is for optical rotation (dextro rotatory) and D - is for configuration. It is derived from

Topic: Chemistry, General Organic Chemistry, JEE Main
The optically inactive amino acid is

Options:
A Lysine
B Glycine
C Arginine
D Alanine

The correct answer is:
Glycine

Solution:
Glycine has no chiral center. So, it is optically inactive

Topic: Chemistry, Biomolecules, JEE Main
The optically inactive compound from the following is:

Options:
A 2-chloropropanal
B 2-chloro-2-methylbutane
C 2-chlorobutane
D 2-chloropentane

The correct answer is:
2-chloro-2-methylbutane

Solution:
(Optically active)
(Optically inactive because of 2-CH3 groups present on same C atom)
(Optically active)

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2015 (10 Apr Online)
The orange colour of K2Cr2O7 and purple colour of KMnO4 is due to

Options:
A Charge transfer transition in both.
B d→d transition in KMnO4 and charge transfer transitions in K2Cr2O7.
C d→d transition in K2Cr2O7 and charge transfer transitions in KMnO4.
D d→d transition in both.

The correct answer is:
Charge transfer transition in both.

Solution:
The deep blue colour of the solution K2Cr2O7 is because an electron from the Oxygen lone pair is transferred to low lying Chromate ion orbital.

Oxidation number of Chromium in Cr2O72-⇒2x+7×-2 x=+6. The electronic configuration of Chromium is 1s22s22p63s23p64s13d5so it will lose six electrons to form Cr+6.

Electronic configuration of Cr6+=1s22s22p63s23p6since there are no electrons present in the d-orbital, the d-d transition will not occur.

The oxygen atom in Cr2O72-that act as a ligand and due to charge transfer transfers its one electron to chromium and due to which the coloured compound is formed.

Similarly, in MnO4-the oxidation number of Manganese⇒x+4(-2)=-1 x=+7. The electronic configuration of Manganese is 1s22s22p63s23p64s23d5so it will lose seven electrons to form Mn+7

Electronic configuration of Mn+7=1s22s22p63s23p6, here also no d-orbital is present so no d-d transition will take place.

The oxygen atom in MnO4- acts as a ligand and due to charge transfer transfers its one electron to manganese due to which a coloured compound is formed.

Therefore, K2Cr2O7 and KMnO4 are coloured compounds due to charge transfer or ligand to metal transfer.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (30 Jan Shift 2)
The orange solid on heating gives a colourless gas and a green solid which can be reduced to metal by aluminium powder. The orange and the green solids are, respectively

Options:
A $\mathrm{NH}_{4} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$
B $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$
C $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{CrO}_{3}$
D $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CrO}_{4}$ and $\mathrm{CrO}_{3}$

The correct answer is:
$\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$

Solution:
Orange solid is ammonium dichromate.
$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2} \uparrow+\mathrm{Cr}_{2} \mathrm{O}_{3}+4 \mathrm{H}_{2} \mathrm{O}$
$$
\mathrm{Cr}_{2} \mathrm{O}_{3}+2 \mathrm{Al} \longrightarrow 2 \mathrm{Cr}+\mathrm{Al}_{2} \mathrm{O}_{3}
$$

Topic: Chemistry, d and f Block Elements, JEE Main
The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt{l(l+1)} \cdot \frac{\mathrm{h}}{2 \pi}$. This momentum for an s-electron will be given by

Options:
A
zero

B
$\frac{\mathrm{h}}{2 \pi}$

C
$\sqrt{2} \cdot \frac{\mathrm{h}}{2 \pi}$

D
$+\frac{1}{2} \cdot \frac{\mathrm{h}}{2 \pi}$

The correct answer is:
zero

Solution:
For s-electron, $l=0 \quad \therefore$ angular momentum $=$ zero

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2003
The orbital angular momentum of \( 3^{\text {rd }} \) orbital is equal to \( \sqrt{\mathrm{x}} \frac{\mathrm{h}}{2 \pi} \) the value of ' \( \mathrm{x} \) ' is

Options:

The correct answer is:
6

Solution:
Orbital angular momentum = l(l+1) h/2π,

Equate with the given orbital angular momentum value, calculate the value of x as:

2(3)h2π=6h2π.

So, x=6.

Topic: Chemistry, Structure of Atom, JEE Main
The orbital angular momentum of a p-electron is given as

Options:
A $\frac{h}{\sqrt{2} \pi}$
B $\sqrt{3} \frac{h}{2 \pi}$
C $\sqrt{\frac{3}{2}} \frac{h}{\pi}$
D $\sqrt{6} \cdot \frac{h}{2 \pi}$

The correct answer is:
$\frac{h}{\sqrt{2} \pi}$

Solution:
Orbital angular momentum $=\sqrt{l(l+1)} \times \frac{h}{2 \pi}$
$\because$ For $p$-electron, $l=1$
$\therefore$ Orbital angular momentum,
$$
\begin{aligned}
& =\sqrt{1(1+1)} \times \frac{h}{2 \pi} \\
& =\sqrt{2} \times \frac{h}{2 \pi} \\
& =\frac{h}{\sqrt{2} \pi}
\end{aligned}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2012 (Mains)
The orbital angular momentum of a satellite is L, when it is revolving in a circular orbit at height h from earth surface. If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be

Options:
A 8 L
B 9 L
C 4 L
D 3 L

The correct answer is:
3 L

Solution:
For a satellite revolving around the Earth in a circular orbit, it can be written that

mv2r=GMmr2 ...1 (r=R+h is the distance from the centre of the Earth)

Simplify equation (1) to obtain the linear speed of the satellite.

v=GMr

Thus, the angular momentum of the satellite is given by

L=mvr=mrGMr=mGMr ...2

Similarly, for distance r' from the centre of the Earth, the angular momentum is given by

L'=mGMr' ...3

Divide equation (3) by equation (2) to obtain the required angular momentum.

L'L=mGMr'mGMr= r'r ...4

Now, r'=r+8r=9r

Substitute the known values of the parameters into equation (4) to obtain the required angular momentum.

L'L=9⇒L'=3L

Topic: Physics, Gravitation, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The orbital angular momentum of an electron in 3 s orbital is xh2π. The value of x is (nearest integer)

Options:

The correct answer is:
0

Solution:
The orbital angular momentum
=h2πll+1
For 3s-electron, l=0.
∴ The orbital angular momentum
=h2π00+1
=0 zero

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The orbital angular momentum of an electron is 2s orbital is

Options:
A h4π
B zero
C h2π
D 2·h2π

The correct answer is:
zero

Solution:
For s-subshell l=0

Orbital angular momentum (mvr)=h2πl(l+1)

For 2s orbital, l (azimuthal quantum number) =0
∴ orbital angular momentum =h2π0(0+1)

=h2π0=0

Thus, the orbital angular momentum of an electron in 2s -orbital is Zero.

Topic: Chemistry, Structure of Atom, JEE Main
The orbital diagram in which the Aufbau principle is violated is

Options:
A
B
C
D

The correct answer is:

Solution:
According to Aufbau principle, the orbital of lower energy (2s) should be fully filled before the filling of orbital of higher energy starts.

Topic: Chemistry, Structure of Atom, JEE Main
The orbital frequency of an electron in the hydrogen atom is proportional to

Options:
A \( n^{3} \)
B \( n^{-3} \)
C \( n \)
D \( n^{0} \)

The correct answer is:
\( n^{-3} \)

Solution:
By Bohr's model of hydrogen atom,

kZe2r2=mv2r⇒Ze24πε0=mv2r ...1
From angular momentum of electron in the nth orbit is
mvr=nh2π ...2
from 1, and 2, we get
v=Ze22ε0nh, and r=n2h2ε0πme2Z.

The time period of revolution of electron,

T=2πrv

T=4ε02n3h3mZ2e4

T∝n3

∴ 1 frequency (f)∝n3 or f∝n-3

Thus, the frequency of revolving electron is

f∝n-3

Topic: Physics, Atomic Physics, JEE Main
The orbital having two radial as well as two angular nodes is:

Options:
A 5d
B 4d
C 4f
D 3p

The correct answer is:
5d

Solution:
Nodes are the spaces where the probability of finding an electron is zero.

Radial nodes are the nodes that appear along the radius of an atom while angular nodes are the nodes that appear along the plane of an angle.

n-l-1=2

l=2

n-2-1=2

n=5

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is $v$. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is

Options:
A $\frac{3}{2} v$
B $\sqrt{\frac{3}{2}} v$
C $\sqrt{\frac{2}{3}} v$
D $\frac{2}{3} v$

The correct answer is:
$\sqrt{\frac{2}{3}} v$

Solution:
$v=\sqrt{\frac{G M}{R+h}}$
For first satellite $h=0$,
$v_1=\sqrt{\frac{G M}{R}}$
$\begin{aligned}& \text { For second satellite } h=\frac{R}{2}, \quad v_2=\sqrt{\frac{2 G M}{3 R}} \\& v_2=\sqrt{\frac{2}{3}} v_1=\sqrt{\frac{2}{3}} v\end{aligned}$

Topic: Physics, Gravitation, JEE Main
The order and degree of the differential equation $\left(1+3 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^3 y}{d x^3}$ are

Options:
A
$\left(1, \frac{2}{3}\right)$

B
(3, 1)

C
(3, 3)

D
(1, 2)

The correct answer is:
(3, 3)

Solution:
Order = 3, degree = 3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2002
The order and degree of the differential equation dâ¡ 2 y dx 2 = y + dâ¡ y dâ¡ x 6 1 / 4 are

Options:
A 2,4
B 3,4
C 2,5
D 2,6

The correct answer is:
2,4

Solution:
d⁡ 2 y dx 2 = y + d⁡ y d⁡ x 6 1 / 4
⇒ ( d⁡ 2 y dx 2 ) 4 = y + d⁡ y d⁡ x 6
Hence, order is 2 and degree is 4.

Topic: Mathematics, Differential Equations, JEE Main
The order and degree of the differential equation of all the parabolas which have a fixed length of latus rectum and their axes are parallel to the x-axis, are respectively

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

The correct answer is:
2,1

Solution:
Let, the length of the latus rectum is 4a, then the equation of such family of parabolas is
y-k2=4ax-h, where, h,k are arbitrary constants

On differentiating, we get,

2y-kdydx=4a â¦ 1
y-kd2ydx2+dydx2=0 â¦ 2
From 1 putting the value of y-k, we get,
2adydxd2ydx2+dydx2=0
2ad2ydx2+dydx3=0
Hence, order =2, degree =1

Topic: Mathematics, Differential Equations, JEE Main
The order and degree of the differential equation $x \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+y^2=0$ are respectively

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

The correct answer is:
2 and 1

Solution:
Order and degree of given differential equation will be 2 and 1 respectively.

Topic: Mathematics, Differential Equations, JEE Main
The order and degree of the differential equation$x\left(\frac{d y}{d x}\right)^3+2\left(\frac{d^2 y}{d x^2}\right)^2+3 y+x=0$ are respectively

Options:
A 3,2
B 2,1
C 2,2
D 2,3

The correct answer is:
2,2

Solution:
Clearly, order of highest derivative $=2$ and degree of the highest order derivative $=2$
$\therefore$ order $=2$, degree $=2$.

Topic: Mathematics, Differential Equations, JEE Main
The order of a reaction which has the rate expression $\frac{d c}{d t}=K[E]^{3 / 2}[D]^{3 / 2}$ is

Options:
A 3/2
B 3
C 2
D 0

The correct answer is:
3

Solution:
Order of reaction $=\frac{3}{2}+\frac{3}{2}=\frac{3+3}{2}=\frac{6}{2}=3$

Topic: Chemistry, Chemical Kinetics, JEE Main
The order of basic strength of the following in aqueous solution is
I. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
II. $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$
III. $\mathrm{NH}_3$
IV. $\mathrm{CH}_3 \mathrm{NH}_2$
V. $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$

Options:
A $\mathrm{IV}>\mathrm{I}>\mathrm{V}>\mathrm{III}>\mathrm{II}$
B $\mathrm{II}>\mathrm{V}>\mathrm{IV}>\mathrm{III}>\mathrm{I}$
C $\mathrm{V}>\mathrm{IV}>\mathrm{II}>\mathrm{III}>\mathrm{I}$
D $\mathrm{IV}>\mathrm{III}>\mathrm{V}>\mathrm{II}>\mathrm{II}$

The correct answer is:
$\mathrm{V}>\mathrm{IV}>\mathrm{II}>\mathrm{III}>\mathrm{I}$

Solution:
$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}>\mathrm{CH}_3 \mathrm{NH}_2>\left(\mathrm{CH}_3\right)_3 \mathrm{~N}>\mathrm{NH}_3>\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$

Topic: Chemistry, Amines, JEE Main
The order of basicity among the following compounds is

Options:
A II > I > IV > III
B I > IV > III > II
C IV > II > III > I
D IV > I > II > III

The correct answer is:
IV > I > II > III

Solution:
The conjugate acid is stabilized by resonance with two different -NH2 group.
The conjugate acid is stabilized by resonance with one -NH2 group and by hyper conjugation of -CH3 group.
The conjugate acid is stabilized by resonance with only one NH2 group.
(III) Least basic, as the LP is used in aromaticity.

Topic: Chemistry, General Organic Chemistry, JEE Main
The order of basicity among the following nitrogen compounds is

Options:
A iv $>$ i $>$ iii $>$ ii
B iii $>$ i $>$ iv $>$ ii
C ii $>$ i $>$ iii $>$ iv
D i $>$ iii $>$ ii $>$ iv

The correct answer is:
i $>$ iii $>$ ii $>$ iv

Solution:
Aryl amines have amine group directly attached to aromatic ring. These are weaker bases than arylalkyl amines due to involvement of lone pair of nitrogen in resonance with benzene ring which decreases the availability of lone pair of $e^{-}$ on nitrogen.

Topic: Chemistry, Amines, JEE Main
The order of basicity of amines in gaseous state is :

Options:
A
$1^{\circ}>2^{\circ}>3^{\circ}>\mathrm{NH}_3$

B
$3^{\circ}>2^{\circ}>\mathrm{NH}_3>1^{\circ}$

C
$3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{NH}_3$

D
$\mathrm{NH}_3>1^{\circ}>2^{\circ}>3^{\circ}$

The correct answer is:
$3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{NH}_3$

Solution:
The correct order of relative basicity of amines in the gas phase is $3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{NH}_3$ The alkyl group releases electron and thus, tends to disperse the positive charge of the alkyl ammonium ion and therefore stabilises it. Since, $\mathrm{NH}_4{ }^{+}$(from $\mathrm{NH}_3$ ) has no such alkyl group, it is not stabilised to such an extent as alkylammonium ion.

Topic: Chemistry, Amines, JEE Main, JEE Main 2013 (23 Apr Online)
The order of basicity of the compounds
is

Options:
A
IV $>$ I $>$ III $>$ II

B
I $>$ III $>$ II $>$ IV

C
III $>$ I $>$ IV $>$ II

D
II $>$ I $>$ III $>$ IV

The correct answer is:
I $>$ III $>$ II $>$ IV

Solution:
The order of basicity is I $>$ III $>$ II $>$ IV The lone pair of electrons on $\mathrm{N}$ is more readily available for protonation in I and III then in II. III contains an oxygen atom which has - I effect due to which it is less basic than I. In compound IV lone pair of $\mathrm{e}^{-} \mathrm{s}$ on $\mathrm{N}$-atom is contributed towards the aromatic sextet formation and hence is not at all available for protonation. Hence option (b) is correct.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2012 (19 May Online)
The order of decreasing rate of reaction with ammonia is

Options:
A Anhydrides, esters, ethers
B Anhydrides, ethers, esters
C Ethers, anhydrides, esters
D Esters, ethers, anhydrides

The correct answer is:
Anhydrides, ethers, esters

Solution:
The correct answer according to the provided solution is (2) Anhydrides, ethers, esters.
This order suggests that anhydrides react most rapidly with ammonia, followed by ethers, and then esters.
Anhydrides have two carbonyl groups, which are highly electron-withdrawing, making them very reactive towards nucleophilic attack by ammonia.
Ethers, while less reactive than anhydrides, are still more reactive than esters. This is because ethers lack the highly electronegative oxygen directly attached to the carbonyl carbon, which makes them more susceptible to nucleophilic attack compared to esters.
So, the correct order of decreasing rate of reaction with ammonia among the options provided is indeed (2) Anhydrides, ethers, esters.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The order of decreasing stability of the carbanions
(1) \(\left(\mathrm{CH}_3\right)_3 \overline{\mathrm{C}}\)
(2) \(\left(\mathrm{CH}_3\right)_2 \overline{\mathrm{C}}\mathrm{H}\)
(3) \(\mathrm{CH}_3 \overline{\mathrm{C}}\mathrm{H}_2\)
(4) \(\mathrm{C}_6 \mathrm{H}_5 \overline{\mathrm{C}}\mathrm{H}_2\)
is

Options:
A 1 > 2 > 3 > 4
B 4 > 3 > 2 > 1
C 4 > 1 > 2 > 3
D 1 > 2 > 4 > 3

The correct answer is:
4 > 3 > 2 > 1

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The order of Homo sapiens is

Options:
A primata
B diptera
C hominidae
D poales

The correct answer is:
primata

Solution:
The classification of all living organism has been done by diving them into different levels based on their features and characteristics. There is a hierarchy in this classification and each step is representing a rank or category known as taxonomic category. There are around seven main taxonomic ranks or categories - species, genus, family, order, class, phylum / division and kingdom. Families which exhibit few similar characters are placed under the same Order. The man (Homo sapiens) belongs to the animal kingdom, division chordate, class mammalia, order primata, family hominidae and genus Homo.

Topic: Biology, The Living World, NEET
The order of increasing sizes of atomic radii among the elements O, $\mathrm{S}$, Se and As is :

Options:
A
As $ < \mathrm{S} < \mathrm{O} < \mathrm{Se}$
B
$\mathrm{Se} < \mathrm{S} < \mathrm{As} < \mathrm{O}$
C
O $ < $ S $ < $ As $ < \mathrm{Se}$
D
$\mathrm{O} < \mathrm{S} < \mathrm{Se} < \mathrm{As}$

The correct answer is:
O $ < $ S $ < $ As $ < \mathrm{Se}$

Solution:
On moving down in a group atomic radii increases due to successive addition of extra shell hence
$$
\mathrm{O} < \mathrm{S} < \text { Se }
$$
Further As is in group 15 having one less electron in its $p$ orbital hence have higher atomic radii than group 16 elements.
i.e., $\mathrm{O} < \mathrm{S} < \mathrm{Se} < \mathrm{As}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2013 (22 Apr Online)
The order of ka values of the following acids is :-

Options:
A (i) > (iv) > (iii) > (iv)
B (iv) > (i) > (iii) > (ii)
C (iii) > (iv) > (i) > (ii)
D (iv) > (i) > (ii) > (iii)

The correct answer is:
(iv) > (i) > (iii) > (ii)

Solution:
KaâââAcidity
(iv) is most acidic as its conjugate base is stabilized by the two −OH groups by intra-molecular H -bonding.
(ii) is least acidic as conjugate base is not stabiilized by H -bond.

Topic: Chemistry, General Organic Chemistry, JEE Main
The order of $\left[\begin{array}{lll}x y z \right]\end{array}\left[\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ is

Options:
A $3 \times 1$
B $1 \times 1$
C $1 \times 3$
D $3 \times 3$

The correct answer is:
$1 \times 1$

Solution:
Order will be $(1 \times 3)(3 \times 3)(3 \times 1)=(1 \times 1)$

Topic: Mathematics, Matrices, JEE Main