datavorous/entrance-exam-dataset huggingface dataset:
The number of antibonding electron pairs in $\mathrm{O}_{2}^{2-}$ molecular ion on the basis of molecular orbital theory is (Atomic number of $\mathrm{O}$ is 18 )

Options:
A 5
B 4
C 3
D 2

The correct answer is:
4

Solution:
The electronic configuration of $\mathrm{O}_{2}^{2-}$ ion is
$$
\begin{aligned}
&\mathrm{O}_{2}^{2-}-\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \dot{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}, \sigma 2 p_{x}^{2}, \\
&\pi 2 p_{y}^{2}, \pi 2 p_{x}^{2}, \pi 2 p_{y}^{2}
\end{aligned}
$$
Hence, number of antibonding electron pairs in $\mathrm{O}_{2}^{2-}$ molecular ion are four.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is

Options:
A 16800
B 33600
C 18000
D 14800

The correct answer is:
16800

Solution:
There are 5 vowels in the given word which are 4E's & 1I.

Since they have to always occur together we take them as a single object E E E E I for the time being.

This single object together with 7 remaining object will account for 8 objects.

There 8 objects in which there are 3N's & 2D's can be arrangement in 8!3!2! ways.

Corresponding to each of their arrangements the 5 vowels E,E,E,E & I which can be arranged in 5!4!

Hence, required number of arrangements.

=8!3!2!×5!4!=16800

Hence this is the correct option.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The number of arrangements that can be formed from the letters $a, b, c, d, e, f$ taken 3 at a time without repetition and each arrangement containing at least one vowel, is

Options:
A
96

B
128

C
24

D
72

The correct answer is:
96

Solution:
There are 2 vowels and 4 consonants in the letters $a, b, c, d, e, f$.
If we select one vowel, then number of arrangements
$$
={ }^2 C_1 \times{ }^4 C_2 \times 3 !=2 \times \frac{4 \times 3}{2} \times 3 \times 2=72
$$
If we select two vowels, then number of arrangements
$$
={ }^2 C_2 \times{ }^4 C_1 \times 3 !=1 \times 4 \times 6=24
$$
Hence, total number of arrangements
$$
=72+24=96
$$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2012 (19 May Online)
The number of atomic orbitals from the following having 5 radial nodes is
7s, 7p, 6s, 8p, 8d

Options:

The correct answer is:
3

Solution:
The number of radial nodes for an orbital with principal quantum number n and azimuthal quantum number l is given by the expression:

Number of radial nodes = n - l - 1

Using this expression, we can determine the number of radial nodes for each of the given atomic orbitals:

7s:n=7,l=0 7-0-1=6
7p:n=7,l=1 7-1-1=5
6s:n=6,l=0 6-0-1=5
8p:n=8,l=1 8-1-1=6
8d:n=8,l=2 8-2-1=5

Therefore, the atomic orbitals with 5 radial nodes are 7p, 6s, and 8d.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The number of atoms in 8 g of sodium is x×1023. The value of x is _____ . (Nearest integer)

Given : NA=6.02×1023 mol-1 Atomic mass of Na=23.0u]

Options:

The correct answer is:
2

Solution:
Number of mole of sodium =823

Number of sodium atom =823×6×1023=2.08×1023

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The number of atoms in the cyclic structure of $D$-fructose is

Options:
A $5$
B $6$
C $4$
D $7$

The correct answer is:
$5$

Solution:
5 atoms in the ring

Topic: Chemistry, Biomolecules, JEE Main
The number of bent-shaped molecule/s from the following is ______ N3-, NO2-, I3-, O3, SO2

Options:

The correct answer is:
3

Solution:
N3-Azide: N¯=N+=N¯↔N¯-N+≡N

It has sp hybridised central atom. Hence it is linear in shape.

I3-(triiodide):

It has linear geometry, sp3d hybridisation with three lone pairs at central atom.

NO2- (nitrite) :

It is (nonlinear) bent shaped as it has sp2 hybridisation with one lone pair at central atom.

O3 (ozone) :

It Is bent in shape with sp2 hybridisation and one lone pair at central atom.

SO2(sulphur dioxide) :

It is bent in shape with sp2 hybridisation and one lone pair at central atom.

So 3 among the given molecules is bent in shape.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an :

Options:
A Isotope of parent
B Isobar of parent
C Isomer of parent
D Isotone of parent

The correct answer is:
Isotope of parent

Solution:
$\quad \mathrm{Z}^{\mathrm{A}} \stackrel{\alpha}{\longrightarrow} \mathrm{Z}_{\mathrm{Z}-2} \mathrm{Y}^{\mathrm{A}-4} \stackrel{2 \beta}{\longrightarrow} \mathrm{Z}^{\mathrm{A}-4}$
As the resulting daughter and parent nucleus has same atomic number. So they are isotope.

Topic: Physics, Nuclear Physics, NEET, NEET 2009 (Mains)
The number of bijective functions $f: \mathbf{Z} \rightarrow \mathbf{Z}$ such that $f(x+y)=f(x)+f(y) \forall x, y \in \mathbf{Z}$, is

Options:
A two
B four
C zero
D infinitely many

The correct answer is:
two

Solution:
Let $x$ and $y$ be any two elements in the domain $(Z)$, such that
$f(x+y)=f(x)+f(y)$ $\ldots$ (i)
Differentiating above expression w.r.t ' $y$ ', keeping $x$ constant, we get
$f^{\prime}(x+y)=f^{\prime}(y)$
Let $y=0 \Rightarrow f^{\prime}(x+0)=f^{\prime}(0)$
and Assume $f^{\prime}(0)=K$
$\therefore \quad f^{\prime}(x)=K$
Integrating on both sides, we get
$f(x)=K x+C\{C \rightarrow \text { integration constant }\}$ $\ldots$ (ii)
Now putting $x=y=0$ in Eq. (i), we get $f(0+0)=f(0)+f(0) \Rightarrow f(0)=2 f(0)$ or $f(0)=0$
Let $x=0$, from Eq. (ii)
$f(0)=K \times 0+C \text { or } 0=0+C \text { or } C=0$
$\therefore \quad f(x)=k x$
Case (i) If $k>0$, then $f(x)$ is strictly increasing.
Case (ii) If $k < 0$, then $f(x)$ is strictly decreasing So, function is injective Also $f(x)$ where every element in the codomain is a valid output of the function i.e., range is equal to codomain.
So, function is surjective also.
Therefore, there are two bijective functions.

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
The number of bond pairs is X and lone pairs is Y on Xe (central atom) in XeO3F2 . What is the sum of X+Y?

Options:

The correct answer is:
8

Solution:
8bpX+0lpY
X+Y=8+0=8

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of bonds between sulphur and oxygen atoms in S2O82- and the number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively are:

Options:
A 4 and 6
B 8 and 8
C 8 and 6
D 4 and 8

The correct answer is:
8 and 8

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (08 Jan Shift 1)
The number of π bonds in the major product will be _____.

Options:

The correct answer is:
1

Solution:
LiAlH 4 reduces aldehydes, ketones and carboxylic acids to corresponding alcohols and nitro -NO2 group to amino -NO2 group. But LiAlH4 cannot reduce carbon-carbon double bonds.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The number of $\pi$-bonds in the product formed by passing acetylene through dilute sulphuric acid containing mercuric sulphate is

Options:
A Zero
B One
C Two
D Three

The correct answer is:
One

Solution:
$\mathrm{CH} \equiv \mathrm{CH}+\mathrm{H}_2 \mathrm{O} \xrightarrow[\mathrm{HgSO}_4]{\text { dil. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{CH}_2=\mathrm{C}-\mathrm{OH} \rightarrow \underset{\large || \atop \stackrel{ \atop }{\huge \text{O}}}{\mathrm{CH}_3-\mathrm{C}-\mathrm{H}}$

Topic: Chemistry, Hydrocarbons, JEE Main
The number of bridging CO ligand(s) and Co-Co bond(s) in $\mathrm{Co}_{2}(\mathrm{CO})_{8},$ respectively are:

Options:
A 2 and 1
B 2 and 0
C 0 and 2
D 4 and 0

The correct answer is:
2 and 1

Solution:
The structure of $\mathrm{Co}_{2}(\mathrm{CO})_{8}$ is represented as

It contains two bridging CO ligands and one metal - metal $(\mathrm{Co}-\mathrm{Co})$ bond

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The number of carbonate ore among the following.

Pyrolusite, Malachite, Diaspore, Calamine, Cassiterite, Siderite, Cerussite, Pyrite, Dolomite, Magnesite.

Options:

The correct answer is:
6

Solution:
Pyrolusite - MnO2 , Malachite - CuCO3.CuOH2 , Diaspore - Al2O3.H2O , Calamine- ZnCO3 , Cassiterite- SnO2 Siderite- FeCO3 , Cerussite- PbCO3 , Pyrite- FeS2 , Dolomite- MgCO3.CaCO3 , Magnesite- MgCO3

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
The number of carbonyl carbon in the product X is
Propyne → Zn/H 2 O O 3 /CH 2 Cl 2 ,Δ X

Options:

The correct answer is:
2

Solution:
Ozonolysis of propyne results in the formation of 2 -oxopropanal which contains keto group and aldehyde (-CHO) group. Hence, 2 -oxopropanal contains two carbon atoms.

Topic: Chemistry, Hydrocarbons, JEE Main
The number of chiral carbons atoms in chloramphenicol are

Options:

The correct answer is:
2

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of chiral carbons in β -D(+)-glucose is

Options:

The correct answer is:
5.00

Solution:
The number of chiral carbons in β -D(+)-glucose is 5.

Topic: Chemistry, Biomolecules, JEE Main
The number of chiral carbon(s) present in peptide, lle-Arg-Pro, is ................

Options:

The correct answer is:
4

Solution:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The number of chiral carbons present in sucrose is ……….

Options:

The correct answer is:
9

Solution:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The number of chiral carbons present in the molecule given below is..............

Options:

The correct answer is:
5

Solution:
sp3 carbon attached to four different group is known as chiral carbon and total nimber of chiral carbons in given molecules is 5.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The number of chiral centres in penicillin is _________.

Options:

The correct answer is:
3

Solution:
Star marked atoms are chiral centers.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The number of chiral centres in penicillin is _________.

Options:

The correct answer is:
3

Solution:
Star marked atoms are chiral centers.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The number of chiral centres in the given compound is

Options:

The correct answer is:
5

Solution:
It has 5 chiral centres

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of chiral centres present in B is _____

Options:

The correct answer is:
4

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The number of chlorine atoms in 20 mL of chlorine gas at STP is _______ 1021. (Round off to the Nearest Integer). [Assume chlorine is an ideal gas at STP R=0.083 L bar mol-1K-1, NA=6.023×1023]

Options:

The correct answer is:
1

Solution:
PV=nRT

1.0×201000=N6.023×1023×0.083×273

∴ Number of Cl2 molecules, N=5.3×1020

Hence, Number of Cl -atoms =1.06×1021

≈1×1021

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (17 Mar Shift 2)
The number of chlorine atoms in bithionol is

Options:

The correct answer is:
4

Solution:
Bithionol is an antibacterial, anthelmintic, and algaecide. Bithionol (the compound is also called bithional) is added to soaps to impart antiseptic properties. It is aryl sulfide that is diphenyl sulfide in which each phenyl group is substituted at position 2 by hydroxy and at positions 3 and 5 by chlorine.

So, as we can see in the structure it has total of 4 chlorine atoms.

Chlorine atoms =4

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The number of chromosomes in the shoot tip cells of a maize plant is 20 . The number of chromosomes in the microspore mother cells of the same plant shall be

Options:
A
20

B
10

C
40

D
15

The correct answer is:
20

Solution:
The whole plant body of maize plant including shoot tip cells remains in diploid (2n) state. As the microspore mother cell is a part of reproductive organ, the chromosome number in these cells will remain same as the individual i.e., $2 \mathrm{n}=20$. These microspore mother cells are responsible for producing male gametes, i.e., haploid (n) by reduction division.

Topic: Biology, Reproduction in Organisms
The number of circular permutations of $n$ different objects is

Options:
A $n$ !
B $n$
C $(n-2)$ !
D $(n-1)$ !

The correct answer is:
$(n-1)$ !

Solution:
Required number $=(n-1)!$.

Topic: Mathematics, Permutation Combination, JEE Main
The number of colloidal systems from the following, which will have 'liquid' as the dispersion medium, is _____.

Gem stones, paints, smoke, cheese, milk, hair cream, insecticide sprays, froth, soap lather

Options:

The correct answer is:
5

Solution:
The phase that is scattered or present in the form of colloidal particles is called dispersed phase and the medium in which the colloidal particles are dispersed is called the dispersion medium.

Coloured gemstones are solid sols, colloids in which solid particles are dispersed in solid medium. In paints, small amount of solid pigments are dispersed in a large amount of liquid solvent. So, it is an example of a solid in liquid mixture. Smoke is a solid aerosol-type colloid consisting of the solid dispersed phase in the gaseous dispersed medium. Cheese is an example of gel in the colloidal systems, in which the dispersed phase is a liquid while the dispersion medium is a solid. Milk is a type of 'liquid in liquid' colloid, also known as emulsion. Hair cream is an example of a colloidal system where dispersed Phase is Liquid and dispersion Medium is also Liquid. Insecticide sprays is an example of a colloidal solution in which liquid particles are dispersed in the gaseous phase. Froth is a colloidal solution of gas in liquid. Soap lather is an example of colloidal system foam in which the dispersed phase is gas and the dispersion medium is liquid.

Paints, milk, froth, soap lather and hair cream have liquid as dispersion medium.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The number of common tangents of the circles given by $x^2+y^2-8 x-2 y+1=0$ and $x^2+y^2+6 x+8 y=0$ is

Options:
A
one

B
four

C
two

D
three

The correct answer is:
two

Solution:
Given circles are $x^2+y^2-8 x-2 y+1=0$ and $x^2+y^2+6 x+8 y=0$ Their centres and radius are
$$
C_1(4,1), r_1=\sqrt{16}=4
$$
$$
C_2(-3,-4), r_2=\sqrt{25}=5
$$
Now, $C_1 C_2=\sqrt{49+25}=\sqrt{74}$
$$
r_1-r_2=-1, r_1+r_2=9
$$
Since, $r_1-r_2 < C_1 C_2 < r_1+r_2$
$\therefore$ Number of common tangents $=2$

Topic: Mathematics, Circle, JEE Main, JEE Main 2012 (26 May Online)
The number of common tangents to the circles $x^2+y^2=1$ and $x^2+y^2-4 x+3=0$ is

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

The correct answer is:
$3$

Solution:
Here $C_1=(0,0)~ r_1=1$ unit $C_2=(2,0)~ r_2=1$
Clearly $C_1 C_2=r_1+r_2=1+1=2$
Thus two circles touch externally, Hence the number of common tangents is 3.

Topic: Mathematics, Circle, JEE Main
The number of common tangents to the circles x2+y2+2x+8y-23=0 and x2+y2-4x-10y+9=0, are

Options:

The correct answer is:
2

Solution:
The centres and radii of given circles are C1-1, -4, C2(2, 5)
and r1=1+16+23=40,
r2=4+25-9=20
Now, C1C2=2+12+5+42=90
And r1+r2=40+20
Here, C1C2<r1+r2
∴ Two common tangents can be drawn

Topic: Mathematics, Circle, JEE Main
The number of common tangents to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6 x-8 y=24$ is

Options:
A 0
B 1
C 3
D 4

The correct answer is:
1

Solution:
The centres of the given circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6 x-8 y=24$ are $C_{1}(0,0)$ and $\mathrm{C}_{2}(3,4)$ respectively. Their radii are $\mathrm{r}_{1}=2$ and $r_{2}=7$ respectively.
$\mathrm{C}_{1} \mathrm{C}_{2}=5 < $ sum of radii
But $C_{1} C_{2}=$ difference of radii
Thus, the given circles touch each other internally.
Hence, number of common tangent is only one.

Topic: Mathematics, Circle, JEE Main
The number of common tangents to the circles x 2 + y 2 - 4 x - 6 y - 1 2 = 0 and x2+y2+6x+18y+26=0, is

Options:
A 4
B 1
C 2
D 3

The correct answer is:
3

Solution:
Given x 2 + y 2 - 4 x - 6 y - 1 2 = 0
C12, 3, r1=22+32+12=25=5
And x2+y2+6x+18y+26=0
C2-3,-9, r2=32+92-26
=90-26=8
Then, C1C2=52+122=13
C1C2=r1+r2
⇒Externally touching circles.
⇒3 common tangents.

Topic: Mathematics, Circle, JEE Main, JEE Main 2015 (04 Apr)
The number of common tangents, to the circles x2+y2-18x-15y+131=0 and x2+y2-6x-6y-7=0, is

Options:
A 3
B 1
C 4
D 2

The correct answer is:
3

Solution:
If S≡x2+y2+2gx+2fy+c=0.

C≡-g,-f and r=g2+f2-c.

The given circles are x2+y2-18x-15y+131=0

⇒C19, 152, r1=81+2254-131=52

and x2+y2-6x-6y-7=0

⇒C23, 3, r2=9+9+7=5

⇒d=C1C2=9-32+152-32=36+814=152

⇒r1+r2=52+5=152

⇒C1C2=r1+r2

∴ Circles touch each other externally, 3 common tangents.

Hence this is the correct option.

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The number of common tangents to the circles $x^2+y^2-x=0, x^2+y^2+x=0$ is

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

The correct answer is:
$3$

Solution:
The centres and radii of the circles are Centres: $C_1\left(\frac{1}{2}, 0\right), C_2\left(-\frac{1}{2}, 0\right)$
Radii: $r_1=\frac{1}{2}, r_2=\frac{1}{2}$
Clearly, $C_1 C_2=r_1+r_2$. Therefore the circles touch each other externally. Hence there are $3$ common tangents.

Topic: Mathematics, Circle, JEE Main
The number of common tangents to the circles \( x^{2}+y^{2}+2 x+8 y-23=0 \) and \( x^{2}+y^{2}-4 x-10 y+9=0 \), are

Options:

The correct answer is:
2

Solution:
The centres and radii of given circles are C1-1, -4, C2(2, 5)
and r1=1+16+23=40,
r2=4+25-9=20
Now, C1C2=2+12+5+42=90
And r1+r2=40+20
Here, C1C2<r1+r2
∴ Two common tangents can be drawn

Topic: Mathematics, Circle, JEE Main
The number of common tangents to two circles $x^2+y^2=4$ and $x^2-y^2-8 x+12=0$ is

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

The correct answer is:
$3$

Solution:
Here $C_1=(0,0), r_1=2, C_2=(4,0), r_2=2$
Here $C_1 C_2=4=r_1+r_2$.
Thus two circles touch externally, Hence, the number of common tangents is $3$.

Topic: Mathematics, Circle, JEE Main
The number of common terms in the progressions 4,9,14,19,……, up to 25th term and 3,6,9,12,.... up to 37th term is :

Options:
A 9
B 5
C 7
D 8

The correct answer is:
7

Solution:
4,9,14,19,…, up to 25th term

⇒T25=4+(25-1)5

⇒T25=4+120

⇒T25=124

3,6,9,12,…, up to 37th term

⇒T37=3+(37-1)3

⇒T37=3+108

⇒T37=111

Common difference of Ist series is d1=5

Common difference of IInd series is d2=3

First common term =9, and their common difference =15(LCM of d1 and d2.

So, the common terms are 9, 24, 39, 54, 69, 84, 99.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (27 Jan Shift 1)
The number of complex numbers $z$ such that $|z-1|=|z+1|=|z-i|$ equals

Options:
A
1

B
2

C
$\infty$

D
0

The correct answer is:
1

Solution:
Let $z=x+i y$
$$
\begin{array}{lll}
|z-1|=|z+1| & \Rightarrow \operatorname{Re} z=0 & \Rightarrow x=0 \\
|z-1|=|z-i| & \Rightarrow x=y & \\
|z+1|=|z-i| & \Rightarrow y=-x &
\end{array}
$$
Only $(0,0)$ will satisfy all conditions.
$\Rightarrow$ Number of complex number $z=1$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2010
The number of complexes from the following with no electrons in the \(t_2\) orbital is \(\qquad\) \(\mathrm{TiCl}_4,\left[\mathrm{MnO}_4\right]^{-},\left[\mathrm{FeO}_4\right]^{2-},\left[\mathrm{FeCl}_4\right]^{-},\left[\mathrm{CoCl}_4\right]^{2-}\)

Options:
A 1
B 4
C 3
D 2

The correct answer is:
3

Solution:
$\begin{array}{ll}\mathrm{TiCl}_4 \Rightarrow \mathrm{Ti}^{+4} & \mathrm{e}^0 \mathrm{t}_2^0 \\ \mathrm{MnO}_4^{-} \Rightarrow \mathrm{Mn}^{+7} & \mathrm{e}^0 \mathrm{t}_2^0 \\ \mathrm{FeO}_4^{2-} \Rightarrow \mathrm{Fe}^{+6} & \mathrm{e}^2 \mathrm{t}_2^0 \\ \mathrm{FeCl}_4^{2-} \Rightarrow \mathrm{Fe}^{+2} & \mathrm{e}^3 \mathrm{t}_2^3 \\ \mathrm{CoCl}_4^{2-} \Rightarrow \mathrm{Co}^{+2} & \mathrm{e}^4 \mathrm{t}_2^3\end{array}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The number of compound/s given below which contain/s -COOH group is ___ . (Integer answer)
(A) Sulphanilic acid
(B) Picric acid
(C) Aspirin
(D) Ascorbic acid

Options:

The correct answer is:
1

Solution:
COOH group present in

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The number of compounds having all atoms in a single plane are

XeF4 , SF4 , Al2Cl6 , SnCl4 , I2Cl6 , H2C=C=C=CH2 , PhCl,C2(CN)4 , ICl3 , PCl5

Options:

The correct answer is:
6

Solution:
Sol. XeF4 - square planar, SF4 - seesaw, Al2Cl6 - tetrahedral at aluminium, SnCl4 - tetrahedral, I2Cl6 -planar dimer, H2C=C=C=CH2 -terminal hydrogens in one plane, PhCl-planar molecules, C2(CN)4 - trigonal planar at carbon, ICl3 - T-shaped, PCl5 - trigonal pyramidal.
Here, XeF4, I2Cl6, H2C=C=CH2, PhâCl, C2CN4, ICl3 are planar

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The number of compounds in which complete delocalisation of π-electron can takes place is.

Options:

The correct answer is:
6

Solution:
The following molecules will have complete delocalisation of π-electron

Topic: Chemistry, General Organic Chemistry, JEE Main
The number of correct statements from the following is...........
A. Ecell is an intensive parameter
B. A negative Eo means that the redox couple is a stronger reducing agent than the H+/H2 couple.
C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

Options:

The correct answer is:
4

Solution:
An intensive property is a bulk property, meaning that it is a physical property of a system that does not depend on the system size or the amount of material in the system.

ΔG=−nFEcell ​

The ∆G in the above equation is an extensive property, Ecell is an intensive property. A negative Eomeans that the redox couple is a stronger reducing agent than the H+/H2couple. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The number of correct statements from the following is ...........
A. For 1s orbital, the probability density is maximum at the nucleus
B. For 2s orbital, the probability density first increases to maximum and then decreases sharply to zero.
C. Boundary surface diagrams of the orbitals encloses a region of 100% probability of finding the electron.
D. p and d-orbitals have 1 and 2 angular nodes respectively
E. probability density of p-orbital is zero at the nucleus

Options:

The correct answer is:
3

Solution:
Probability density curves for 1s, 2s and 2p orbitals are given below.

For 1s orbital, the probability density is maximum at the nucleus. For 2s orbital, the probability density first decreases and then increases. At any distance from nucleus the probability density of finding electron is never zero, and it always have some finite value. The number of angular nodes are equal to azimuthal quantum number. For p-orbital it is equal to 1 and for d-orbital it is equal to 2. Probability density of p-orbital is zero at the nucleus

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (11 Apr Shift 2)