datavorous/entrance-exam-dataset huggingface dataset:
Number of bromo derivatives obtained on treating ethane with excess of Br2 in diffused sunlight is _________

Options:

The correct answer is:
9

Solution:
A substitution reaction is a chemical reaction during which one functional group in a chemical compound is replaced by another functional group.The number of different substitution products possible when ethane is allowed to react with bromine in sunlight are 9.

CH3-CH3+Br2→sunlightCH3-CH2Br+CH2Br-CH2Br+CH2Br-CHBr2+CHBr2-CHBr2+CBr3-CBr3+CH3-CHBr2+CH3-CBr3+CH2Br-CBr3+CHBr2-CBr3

6 Bromine atoms: 1 product possible

5 Bromine atoms : 1 product possible

4 Bromine atoms : 2 products possible

3 Bromine atoms : 2 products possible

2 Bromine atoms : 2 products possible

1 Bromine atom : 1 product possible

Total 9 products possible

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Number of carbocations from the following that are not stabilized by hyperconjugation is _______

Options:

The correct answer is:
5

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Number of Cl=O bonds in chlorous acid, chloric acid and perchloric acid respectively are:

Options:
A 3,1 and 1
B 4,1 and 0
C 1,1 and 3
D 1,2 and 3

The correct answer is:
1,2 and 3

Solution:
Number of Cl=O bonds

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (27 Jul Shift 2)
Number of common points to the curves C1:-1+2cosâ¡Î±,2sinâ¡Î± and C2:4+3sinâ¡Î¸,3cosâ¡Î¸ is/are equal to

Options:

The correct answer is:
1

Solution:
C1 is x+12+y2=22
C2 is x-42+y2=32
1,0 is the only common point

Topic: Mathematics, Circle, JEE Main
Number of common tangents to the parabola \( y^{2}=4 a x \) and \( x^{2}=4 b y \) is

Options:
A \( 4 \)
B \( 3 \)
C \( 2 \)
D \( 1 \)

The correct answer is:
\( 1 \)

Solution:
It is clear from the figure, that only one common tangent is possible

Alternate Solution

Tangent of y2=4ax

is y=mx+am

tangent of x2=4by

is y=mx-bm2

∴ am=-bm2

m3=-ab

m=-ab13

Only one common tangent is possible.

Topic: Mathematics, Parabola, JEE Main
Number of complex numbers satisfying \( |z|=\max \{|z-1|,|z+1|\} \).

Options:

The correct answer is:
0

Solution:
Let, z=x+iy.

So, z-1=x-12+y2=x2+y2+1-2x

and,

z+1=x+12+y2=x2+2x+1+y2

Now

Case 1: For x≥0,

maxz-1,z+1=maxx2-2x+1+y2,x2+2x+1+y2=x2+2x+1+y2

⇒x2+y2=x2+2x+1+y2⇒2x+1=0⇒x=-12

and

Case 2: For x<0,

maxz-1,z+1=maxx2-2x+1+y2,x2+2x+1+y2=x2-2x+1+y2

⇒x2+y2=x2-2x+1+y2⇒-2x+1=0⇒x=12

Examine both cases, you can conclude that x has no value that can satisfy any of the above conditions.

Hence, number of such complex numbers =0.

Topic: Mathematics, Complex Number, JEE Main
Number of complexes which show optical isomerism among the following is _________.

cis-Cr(ox)2Cl23-,Co(en)33+, cis -Pt(en)2Cl22+, cis -Co(en)2Cl2+

​​​​​​​,trans-Pt(en)2Cl22+, trans -Cr(ox)2Cl23-

Options:

The correct answer is:
4

Solution:
Pt(en)2Cl22+ is a square planar complex forms geometrical isomers (cis and trans). Trans isomer does not show optical isomerism since it is symmetrical, while cis isomer shows optical isomerism as it is unsymmetrical.

POS = plane of symmetry

COS = centre of symmetry

The compounds, cis−[Co(en)2Cl2]+ and [Co(en)3]3+ show optical isomerism as they do not possess any element of symmetry.

cis-Cr(ox)2Cl23-→can show optical isomerism (no POS & COS)

trans-Pt(en)2Cl22+→can't show optical isomerism (contain POS & COS)

trans-Cr(ox)2Cl23-→can't show optical isomerism(contain POS & COS)

Hence, among the given complexes 4 complexes can show optical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Number of complexes which will exhibit synergic bonding amongst, CrCO6,MnCO5 and Mn2CO10 is

Options:

The correct answer is:
3

Solution:
Synergic bonding involves transference of electrons from ligands to metal. The transference of electrons takes place from filled metal orbitals to anti-bonding orbitals of ligands. It is a bond between a carbonyl group acting as a ligand and a metal. Synergic bonding means self strengthening bond. More the number of d electrons on the central metal ion more the synergic bonding. In metal carbonyls' oxidation state of the metal is zero.

In the complex MnCO5, the electronic configuration of Mn is 3d54s2. As carbon monoxide is a strong field, pairing of electrons occurs. Hence, the new electronic configuration of Mn is 3d7.

In the case of Mn2CO10 also electronic configuration after pairing of electrons is 3d7.

In the complex CrCO6 , the electronic configuration of Cr is 3d54s1 . As carbon monoxide is a strong field, pairing of electrons occurs. Hence, the new electronic configuration of Cr is 3d6

All of these compounds can show synergic bonding.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (28 Jun Shift 1)
Number of Complexes with even number of electrons in \(t_{2 g}\) orbitals is -
\(\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\)

Options:
A 2
B 3
C 1
D 5

The correct answer is:
3

Solution:

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Number of compounds among the following which contain sulphur as heteroatom is____.

Furan,Thiophene, Pyridine, Pyrrole, Cysteine, Tyrosine

Options:

The correct answer is:
2

Solution:
The structures of the compounds pyrrole, Furan, thiophene, pyridine, cysteine and tyrosine respectively are follows,

Hence from the given compounds, Thiophene and Cysteine are the compounds containing Sulphur.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Number of compounds from the following which cannot undergo Friedel-Crafts reactions is: ________.
toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, \(m\)-nitroaniline, \(m\)-dinitrobenzene

Options:

The correct answer is:
4

Solution:
Compounds which can not undergo Friedel Crafts reaction are

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Number of compounds giving (i) red colouration with ceric ammonium nitrate and also (ii) positive iodoform test from the following is _________

Options:

The correct answer is:
3

Solution:
Red colouration with ceric ammonium nitrate indicates presence of alcholic group, positive iodoform test indicates presence of a secondary alcohol which can oxidise to a methyl ketone. In total 3 compounds from the given will give positive test for both the reactions:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Number of compounds / species from the following with non-zero dipole moment is ______
\(\mathrm{BeCl}_2, \mathrm{BCl}_3, \mathrm{NF}_3, \mathrm{XeF}_4, \mathrm{CCl}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{~S}, \mathrm{HBr}, \mathrm{CO}_2, \mathrm{H}_2, \mathrm{HCl}\)

Options:

The correct answer is:
5

Solution:
Polar molecule: $\mathrm{NF}_3, \mathrm{H}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{~S}, \mathrm{HBr}, \mathrm{HCl}$
$(\mu \neq 0)$
Non Polar molecule: $\mathrm{BeCl}_2, \mathrm{BCl}_3, \mathrm{XeF}_4, \mathrm{CCl}_4, \mathrm{CO}_2, \mathrm{H}_2$ $(\mu-0)$
So answer is 5 .

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Number of compounds which give reaction with Hinsberg's reagent is _________.

Options:

The correct answer is:
5

Solution:
Hinsberg reagent is an alternative name for benzene sulfonyl chloride. This name is given for its use in the Hinsberg test for the detection and distinction of primary, secondary, and tertiary amines in a given sample.

This reagent is an organosulfur compound. Its chemical formula can be written as C6H5SO2Cl.

In the Hinsberg Test, the amines act as nucleophiles and attack the electrophile (sulfonyl chloride). This leads to the displacement of the chloride and the generation of the sulfonamides. When primary and secondary amines form sulfonamides, this sulfonamide product is not soluble and precipitates from the solution as a solid.

1° and 2° amines will give reaction with Hinsberg reagent. Compounds which gives reaction with Hinsberg reagent are as follows,

Topic: Chemistry, Amines, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Number of compounds with one lone pair of electrons on central atom amongst following is _
O3,H2O,SF4,ClF3,NH3,BrF5,XeF4

Options:

The correct answer is:
4

Solution:
A pair of electrons which is not shared by any of the reacting atoms is called a lone pair of electrons. Lone pairs are found in the outermost electron shell of atoms. They can be identified by using a Lewis structure.

The structures of the compounds are as follows,

Hence the compounds NH3,O3,SF4 and BrF5 have one lone pair on central metal atom.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Number of constitutional isomers with formula C5H11Br is

Options:

The correct answer is:
8

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Number of crystal systems from the following where body centred unit cell can be found, is _____.

Cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic, triclinic

Options:

The correct answer is:
3

Solution:
Bravais Lattice refers to the 14 different 3-dimensional configurations into which atoms can be arranged in crystals. The smallest group of symmetrically aligned atoms which can be repeated in an array to make up the entire crystal is called a unit cell. Crystal systems where body centred unit cell can be found are Cubic, orthorhombic and tetragonal

Hence, correct answer is 3

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Number of crystal systems having only 2 types of bravais lattices =x , number of crystal system having at least two interfacial angles equal =y and number of crystal systems having all the three edge lengths equal =z . Then find the value of x×y×z.

Options:

The correct answer is:
24

Solution:
x=2 (Tetragonal, Monoclinic)
y=6 (except triclinic all the bravis lattices)
z=2 (Cubic, Rhombohedral)
xyz=262=24.00

Topic: Chemistry, Solid State, JEE Main
Number of cyclic tripeptides formed with 2 amino acids A and B is:

Options:
A 2
B 3
C 5
D 4

The correct answer is:
4

Solution:
Let two amino acid be

H2N-CHR1-COOH(A), H2N-CHR2-COOH(B)

Tripeptide are formed from three amino acids in a combination, thus 4 tripeptides can be formed from two amino acids in the following combination:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Number of Deuterium exchange in the given tautomer when the compound is kept in NaOD/D2O for a long time is:

Options:
A 2
B 4
C 6
D 8

The correct answer is:
2

Solution:
Deuterium exchange takes place with Hpresent at α-position.
Total α−H = 2

Topic: Chemistry, General Organic Chemistry, JEE Main
Number of electron-rich hydrides among the following are: \( \mathrm{CH}_{4}, \mathrm{NH}_{3}, \mathrm{PH}_{3}, \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{BH}_{3}, \mathrm{HF}, \mathrm{AIH}_{3}, \mathrm{AsH}_{3} \).

Options:

The correct answer is:
6

Solution:
Hydrides of group 15, 16 and 17 ( i.e., NH3, PH3, H2O, H2S,HF, AsH3)have more electrons than required to form normal covalent bonds and hence are called electron-rich hydrides. These extra electrons are present as lone pair electron on the central metal. These lone pair electron makes atom electron rich atom.

Topic: Chemistry, Hydrogen, JEE Main
Number of electrons having l + m = 0 in Mn (z = 25) is

Options:

The correct answer is:
13

Solution:
Mn = 1s2 2s2 2p6 3s2 3p6 4s2 3d5
Total 13 e-

Topic: Chemistry, Structure of Atom, JEE Main
Number of electrons having \( \mathrm{l}+\mathrm{m}=0 \) in \( \mathrm{Mn}(\mathrm{z}=25) \) is

Options:

The correct answer is:
13

Solution:
Mn = 1s2 2s2 2p6 3s2 3p6 4s2 3d5
Total 13 e-

Topic: Chemistry, Structure of Atom, JEE Main
Number of electrons present in 4f orbital of Ho3+ ion is ________ (Given Atomic No. of Ho =67)

Options:

The correct answer is:
10

Solution:
Ho=[Xe]4f116 s2

Ho3+=[Xe]4f10

so number of e- present in 4f is 10 .

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Number of electrons that Vanadium (Z=23) has in p-orbitals is equal to:

Options:

The correct answer is:
12

Solution:
V 23: 1s2 2s2 2p6 3s2 3p6 3d3 4s2
Number of electrons in p-orbitals is equal to 12.00

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Number of electrophillic centres in the given compound is___

Options:

The correct answer is:
3

Solution:
Electrophilic centre are areas of low electron density. Most often they are atoms which contain a full of partial positive charge. In the given structure the -CN group, the Keto group & the double bond in conjugation with Keto group are the 3 electrophilic centres.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Number of elements from the following that CANNOT form compounds with valencies which match with their respective group valencies is ______.
B, C, N, S, O, F, P, Al, Si

Options:
A 7
B 3
C 5
D 6

The correct answer is:
3

Solution:
$\mathrm{N}, \mathrm{O}, \mathrm{F}$ can't extend their valencies upto their group number due to the non-availability of vacant $2 \mathrm{~d}$ like orbital.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Number of geometrical isomers possible for the given structure is/are _________.

Options:

The correct answer is:
4

Solution:
The isomerism that arises when atoms or groups having different spatial arrangements are restricted to rotate around a bond or bonds in a molecule is called geometric isomerism. This restricted rotation mainly occurs in a carbon-carbon double bond.

The conditions to form geometrical isomers are-

restricted rotation (often involving a carbon-carbon double bond);

Presence of two different groups on the left-hand end of the double bond and two different groups on the right-hand end. It is not necessary to have similar kinds of groups or atoms at both ends.

In the given compound, the central double bond is not give geometrical isomerism as it contains the same group on carbon, where the terminal double bonds (two double bonds) are exhibiting geometrical isomerism. Hence, the total number of isomers is 22=4 isomers.

Hence, the given compound shows four geometrical isomers.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Number of grams of bromine that will completely react with 5.0 g of pent-1-ene is____×10-2 g. (Atomic mass of Br=80 g/mol) [Nearest integer]

Options:

The correct answer is:
1143

Solution:
moles of $\mathrm{Br}_2=$ moles of $\mathrm{C}_5 \mathrm{H}_{10}$
$\Rightarrow \mathrm{w} / 160=5 / 70$
$\Rightarrow \mathrm{w}=5 \times 160 / 70 \mathrm{~g}$
$=11.428 \mathrm{~g}$
$=1142.8 \times 10^{-2} \mathrm{~g} \approx 1143 \times 10^{-2} \mathrm{~g}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Number of hydrocarbons produces, when mixture of sodium ethanoate and sodium propanoate undergoes kolbe
electrolysis.

Options:
A \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\)
B \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)
C \(\mathrm{CH}_{3} \mathrm{CH}_{3}\)
D all of these

The correct answer is:
\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Solution:
This problem is based on mixing of kolbe process of electrolysis and mechanism of kolbe process. To solve this process students are advised to undergo proper mechanistic approach.

Topic: Chemistry, Hydrocarbons, JEE Main
Number of hydrogen atoms per molecule of a hydrocarbon A having 85.8% carbon is (Given : Molar mass of A=84 g mol-1 )

Options:

The correct answer is:
12

Solution:
Using table of empirical formula for the compound:

Element	Percentage	Mole	ratio

C	85.8	85.812=7.15	1

H	14.2	14.21=14.2	2

Thus, empirical formula is CH2, empirical molecular mass = 14
Thus, Molecular MassEmpirical Mass = n

14×n=84

n=6

∴ Molecular formula C6H12

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Number of incorrect statements are‐
A The π bond between metal and carbonyl carbon reduces the bond order of C-O in carbon monoxide
(B)dz2 orbital of central metal atom/ ion is used in dsp2 hybridisation.
CCN- is a π‐ acid Ligand.
D All negative ligands are stronger than neutral ligands.

Options:

The correct answer is:
2

Solution:
A The π bond between metal and carbonyl carbon reduces the bond order of C-O in carbon monoxide
CCN- is a π‐ acid Ligand.
The above statements are correct.
d x 2 − y 2 orbital of central metal atom/ ion is used in dsp2 hybridisation.

Topic: Chemistry, Coordination Compounds, JEE Main
Number of integral solutions to the equation x+y+z=21, where x≥1,y≥3,z≥4, is equal to _____ .

Options:

The correct answer is:
105

Solution:
We have,

x≥1, y≥3, z≥4

⇒x-1≥0, y-3≥0, z-4≥0

So,

x+y+z=21

⇒x-1+y-3+z-4=13

So, number of non-negative integral solutions is

=C3-113+3-1=C215=15×142=105

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Number of integral terms in the expansion of 5+781024 is

Options:

The correct answer is:
129

Solution:
We have, Tr=1=1024Cr 51024-r27r8
This is an integer, if 1024-r is an even number and r is a multiple of 8

i.e. when r is a multiple of 8 since 0â¤râ¤1024

The number of multiples of 8 which lies between 0 and 1024 is
1024=0+n-18
ân-1=10248=128
ân=129

Topic: Mathematics, Binomial Theorem, JEE Main
Number of isomeric ethers with chiral carbon in molecular formula \( \mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O} \) is

Options:

The correct answer is:
2

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
Number of isomeric products formed by monochlorination of 2-methylbutane in presence of sunlight is _______.

Options:

The correct answer is:
6

Solution:
Monochlorination means the addition of a chlorine atom. When chlorine atom is added to the below structure. Chlorine will substitute at one of the hydrogens

Mono chlorination of 2-methyl butane gives four structural isomers.

Here only two, i.e., I and III are optically active. Since each optically active compound has two enantiomers, therefore, total six isomeric compounds are possible.

∴ Number of isomeric products =6

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Number of lone pair s of electrons on central atom and the shape of BrF3 molecule respectively, are :

Options:
A 0 , triangular planar.

B 1 , pyramidal.

C 2 , bent T-shape.

D 1, bent T-shape

The correct answer is:
2 , bent T-shape.

Solution:
Bromine undergo sp3d hybridization and the shape of the molecule is T-shape. Each fluorine is having 3 lone pairs and bromine is having 2 lone pairs. Hence, total 11 lone pairs are there.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (29 Jun Shift 2)
Number of lone pairs of electrons in the central atom of SCl2,O3,ClF3 and SF6, respectively, are

Options:
A 0,1,2 & 2
B 2, 1, 2 & 0
C 1, 2, 2 & 0
D None of these

The correct answer is:
2, 1, 2 & 0

Solution:
The Lewis dot structure of the given molecules will be the following.

As we can see in above structures that SCl2 has 2 lone pair and 2 bond pair, O3 has one lone pair and two bond pair,ClF3 has two lone pair and 3 bond pair and SF6 has 0 lone pair and 6 bond pair

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Number of \( \mathrm{H}_{2} \mathrm{O} \) molecule which is bonded by hydrogen bonding in \( \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \) :

Options:
A \( 5 \)
B \( 4 \)
C \( 3 \)
D \( 1 \)

The correct answer is:
\( 1 \)

Solution:
When CuSO4.5H2O undergo dissociation, then following reaction takes place:

CuSO4.5H2O → CuSO4.4H2O + H2O

The one water molecule is released because it is bonded through hydrogen bonding. Which can also be represented in its structure as below:

Topic: Chemistry, d and f Block Elements, JEE Main
Number of metal ions characterized by flame test among the following is _________.
Sr2+,Ba2+,Ca2+,Cu2+,Zn2+,Co2+,Fe2+

Options:

The correct answer is:
4

Solution:
The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic colour the salt turns the flame of a Bunsen burner. The heat of the flame excites the electrons of the metal ions, causing them to emit visible light.

Sr2+-Crimson red colourCu2+-Blue green colourCa2+-Brick red colourBa2+- Pale green colour

Hence, the answer is 4.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Number of metal oxygen bonds in the orange-red coloured compound formed when NaCl reacts with K2Cr2O7 and H2SO4 .

Options:
A 2
B 4
C 6
D 0

The correct answer is:
4

Solution:
K 2 Cr 2 O 7  + 4 NaCl + 6  H 2 SO 4 →2  KHSO 4 +  2  CrO 2 Cl 2 Chromyl Chloride  + 4  NaHSO 4
Orange-red coloured chromyl chloride.

Topic: Chemistry, d and f Block Elements, JEE Main
Number of molecules among the following having non‐zero dipole moment is:
O3,SO3 , SF4, SF6 , H2S,CS2, SO2 , H2O and H2O2

Options:

The correct answer is:
6

Solution:
SO 3 ,  SF 6 ,  CS 2 are molecules with symmetrical shapes and have zero dipole moment.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
Number of molecules from the following which are exceptions to octet rule is ______
\(\mathrm{CO}_2, \mathrm{NO}_2, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{SiF}_4, \mathrm{ClO}_2, \mathrm{PCl}_5, \mathrm{BeF}_2, \mathrm{C}_2 \mathrm{H}_6, \mathrm{CHCl}_3, \mathrm{CBr}_4\)

Options:

The correct answer is:
6

Solution:

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Number of molecules from the following which can exhibit hydrogen bonding is
_______ (nearest integer)

Options:

The correct answer is:
5

Solution:
Can show H-bonding.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2024 (06 Apr Shift 1)