Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q1 - 24 January - Shift 2
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Q2 - 25 January - Shift 1
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Q3 - 25 January - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q4 - 25 January - Shift 2
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Q5 - 29 January - Shift 2
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Q6 - 30 January - Shift 1
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Q7 - 30 January - Shift 1
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
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Q8 - 30 January - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q9 - 31 January - Shift 1
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Q10 - 31 January - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Answer Key
(As per Official NTA Key released on 2 Feb)
Q1 (1)
Q2 (6)
Q3 (4)
Q4 (5)
Q5 (1)
Q6 (4)
Q7 (2)
Q8 (4)
Q9 (4)
Q10 (5)
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q1 (1)
Q2 (6)
Q3 (4)
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q4 (5)
Q5 (1)
Q6 (4)
Q7 (2)
Q8 (4)
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q9 (4)
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
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Chemical Bonding and Molecular Structure
JEE Main 2023 (January) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
Q10 (5)
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Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
1.
MathonGo
The number of species from the following which have square pyramidal structure is _______
PF5 , BrF
−
4
, IF5 , BrF5 , XeOF4 , ICl
−
4
[2023 (06 Apr Shift 1)]
2.
The number of species having a square planar shape from the following is
XeF4 ,
SF4 ,
SiF4 , BF4
−
, BrF4
−
2+
, [Cu (NH3 ) ]
4
, [FeCl4 ]
2−
, [PtCl4 ]
2−
[2023 (06 Apr Shift 2)]
3.
In an ice crystal, each water molecule is hydrogen bonded to _____ neighbouring molecules.
[2023 (06 Apr Shift 2)]
4.
Given below are two statements : One is labelled as Assertion
Assertion A
Reason R
:
:
A
and the other is labelled as Reason
R
.
Butan–1–ol has higher boiling point than ethoxyethane.
Extensive hydrogen bonding leads to stronger association of molecules.
In the light of the above statements, choose the correct answer from the options given below :
[2023 (08 Apr Shift 1)]
(1) A is true but R is false
(2) Both A and R are true and R is the correct explanation of A
(3) Both A and R are true but R is not the correct explanation of A
(4) A is false but R is true
5.
The number of following factors which affect the percent covalent character of the ionic bond is______
A) Polarising power of cation
B) Extent of distortion of anion
C) Polarisability of the anion
D) Polarising power of anion
[2023 (08 Apr Shift 1)]
6.
The number of species trom the following carrying a single lone pair on central atom Xenon is
XeF
+
5
, XeO3 , XeO2 F2 , XeF
−
5
, XeO3 F2 , XeOF4 , XeF4
[2023 (08 Apr Shift 2)]
7.
The pair from the following pairs having both compounds with net non-zero dipole moment is
[2023 (10 Apr Shift 1)]
(1)
1, 4
-Dichlorobenzene, 1, 3-Dichlorobenzene
(2) cis-butene, trans-butene
(3)
CH2 Cl2 ,
CHCl3
(4) Benzene, anisidine
8.
The compound which does not exist is
[2023 (10 Apr Shift 1)]
9.
(1)
NaO2
(2)
BeH2
(3)
PbEt4
(4)
(NH4 )
2
BeF4
The number of bent-shaped molecule/s from the following is ______ N
−
3
,
NO
−
2
, I
−
3
, O3 ,
SO2
[2023 (10 Apr Shift 1)]
10. The sum of lone pairs present on the central atom of the interhalogen IF and IF is ______
5
7
[2023 (10 Apr Shift 1)]
11. The number of molecules from the following which contain only two lone pair of electrons is _____________
H2 O, N2 , CO, XeF4 , NH3 , NO, CO2 , F2
[2023 (10 Apr Shift 2)]
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Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
12. Match List-I with List-II:
List-I Species
List-II Geometry/Shape
A.
H3 O
B.
Acetylide anion
C.
NH
D.
+
+
4
ClO
−
2
I.
Tetrahedral
II.
Linear
III.
Pyramidal
IV.
Bent
Choose the correct answer from the options given below:
[2023 (11 Apr Shift 1)]
(1) A(III), B(IV), C(I), D(II)
(2) A(III), B(I), C(II), D(IV)
(3) A(III), B(II), C(I), D(IV)
(4) A(III), B(IV), C(II), D(I)
13. Which one of the following pairs is an example of polar molecular solids?
[2023 (11 Apr Shift 2)]
(1)
SO2 ( s), CO2 ( s)
(2)
SO2 ( s), NH3 ( s)
(3)
MgO(s), SO2 (s)
(4)
HCl(s), AlN(s)
14. The maximum number of lone pairs of electron on the central atom from the following species is ......
ClO
3
−
,
XeF4 ,
SF4
and l
3
−
[2023 (11 Apr Shift 2)]
15. The bond order and magnetic property of acetylide ion are same as that of
[2023 (12 Apr Shift 1)]
(1)
O2
+
(2)
N2
(3)
NO
(4)
O2
+
+
−
16. Given below are two statements:
Statement I: SbCl is more covalent than SbCl
5
3
Statement II: The higher oxides of halogens also tend to be more stable than the lower ones.
In the light of the above statements, choose the most appropriate answer from the options given below.
[2023 (12 Apr Shift 1)]
(1) Statement I is incorrect but statement II is correct
(2) Both Statement I and Statement II are incorrect
(3) Both Statement I and Statement II are correct
(4) Statement I is correct but statement II is incorrect
17. In which of the following processes, the bond order increases and paramagnetic character changes to diamagnetic one?
[2023 (13 Apr Shift 1)]
18.
(1)
O2 → O
(2)
O2 → O
(3)
NO → NO
(4)
N2 → N
ClF5
+
2
2−
2
+
+
2
at room temperature is a
[2023 (13 Apr Shift 1)]
(1) Colourless liquid with trigonal bipyramidal geometry
(2) Colourless gas with square pyramidal geometry
(3) Colourless gas with trigonal bipyramidal geometry
(4) Colourless liquid with square pyramidal geometry
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Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
19. Among the following compounds, the one which shows highest dipole moment is
[2023 (13 Apr Shift 1)]
(1)
(2)
(3)
(4)
20. Match Lis-I with List-II.
List-1
List-II
A. Weak intermolecular forces of attraction I.
Hexamethylenediamine
+ adipic acid
B. Hydrogen bonding
II.
AlEt3 + TiCl4
C. Heavily branched polymer
III.
2 − chloro − 1, 3 − butadiene
D. High density polymer
IV.
Phenol
+
formaldehyde
Choose the correct answer from the options given below
[2023 (13 Apr Shift 2)]
(1) A-IV, B-II, C-III, D-I
(2) A-IV, B-I, C-III, D-II
(3) A-II, B-IV, C-I, D-III
(4) A-III, B-I, C-IV, D-II
21. Given below are two statements :
Statement I : SO and H
2
2
both possess V-shaped structure
O
Statement II : The bond angle of SO is less than that of H
2
2O
.
In the light of the above statements, choose the most appropriate answer from the options given below:
[2023 (13 Apr Shift 2)]
(1) Both Statement I and Statement II are incorrect
(2) Both Statement I and Statement II are correct
(3) Statement I is incorrect but Statement II is correct
(4) Statement I is correct but Statement II is incorrect
22. Consider the following statement
(A) NF molecules has a trigonal planar structure.
3
(B) Bond Length of N is shorter than O .
2
2
(C) Isoelectronic molecules or ions have identical bond order.
(D) Dipole moment of H
2S
is higher than that of water molecule.
Choose the correct answer from the options given below:
[2023 (15 Apr Shift 1)]
(1)
(A)
(2)
(A)
and (B) are correct
and (D) are correct
(3)
(C)
and (D) are correct
(4)
(B)
and (C) are correct
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Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
ANSWER KEYS
1. (3)
2. (4)
3. (4)
4. (2)
5. (3)
6. (4)
7. (3)
8. (1)
9. (3)
10. (1)
11. (3)
12. (3)
13. (2)
14. (3)
15. (3)
16. (3)
17. (3)
18. (4)
19. (4)
20. (4)
21. (4)
22. (4)
1.
(3)
Among the given, IF
5;
BrF5
and XeOF have squre pyramidal structure. VSEPR theory helps us to predict the shape of molecules from the
4
number of electron pairs and lone pairs of electrons that surrounds the central atoms.
2.
(4)
XeF4
is sp
SF4
SiF4
BF4
is sp
3
3
d
d
2
hybridised with two lone pairs of electrons, hence, it is square planar.
hybridised with one lone pairs of electrons, hence, it is See saw.
is sp hybridised with zero lone pairs of electrons, hence, it is Tetrahedral
3
−
is sp hybridised with zero lone pairs of electrons, hence, it is Tetrahedral
3
[Cu(NH3 ) ]
2+
4
[FeCl4 ]
[PtCl4 ]
BrF4
−
is dsp hybridised and is Square planar
2
2–
is sp hybridised and is Tetrahedral
2–
is dsp hybridised and is Square planar
3
is sp
2
3
d
2
with Square planar geometry.
So, 4 square planer shape compounds are present.
3.
(4)
Hydrogen bonding occurs between hydrogen and strongly electronegative atoms like F,
N, O
. In water or ice hydrogen bonding occurs as there is
hydrogen as well as oxygen both are present. Water has 2 hydrogen atoms and 1 oxygen atom. The oxygen of one water molecule has two lone
pairs of electrons, each of which can form hydrogen bonds with hydrogen on other two water molecules.
Each water molecule is H-bonded to 4 neighbouring molecules.
4.
(2)
Butan-1-ol can undergo hydrogen bonding due to the presence of an − OH group, ethoxyethane (also known as diethyl ether) can also participate
in hydrogen bonding.
Even though there are no hydrogen atoms directly bonded to a highly electronegative atom (like F,
O,
or N
) in ethoxyethane, the molecule still
has polar C − O bonds that can lead to dipole-dipole interactions with other polar molecules.
Owing to intermolecular hydrogen bonding in butanol, it has higher boiling point than ethoxyethane.
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Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
5.
MathonGo
(3)
The percentage of covalent character in an ionic bond is determined by the polarization power of the cation, the polarizing ability of the anion, and
the degree of distortion of the anion's electron cloud that occurs due to the electric field of the cation. The polarization power of a cation refers to
its ability to distort the electron cloud of the anion in the ionic bond. So the options A, B, C are correct.
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Chemical Bonding and Molecular Structure
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6.
MathonGo
(4)
The central Xe atom in XeO has three bonding domains and one lone pair of electrons. Hence, the electron geometry is tetrahedral and molecular
3
geometry is pyramidal.
XeO2 F2
XeOF4
is trigonal bipyramidal and the shape is a see-saw.
is in square pyramidal shape.
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Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
So, XeF
7.
+
5
MathonGo
, XeO3 , XeO2 F2
and XeOF have single lone pair on central atom.
4
(3)
The dipole moment is the vector sum of all the bond moments. The CH
2
Cl2
and CHCl both have net non-zero dipole moment because the bond
3
moments are not cancelled each other due to their tetrahedral structure. The 1, 4-dichlorobenzene, trans-butene, benzene have zero dipole moment
due to cancellation of bond moments. The 1, 3-dichlorobenzene, cis-butene, anisidine has non-zero dipole moment.
8.
(1)
Oxide and peroxides of sodium are stable (Na
BeH2
O
and Na
2
O2
) K,
Rb,
Cs
form superoxides. Beryllium forms a hydride with molecular formula
Organic lead (tetraethyl lead; TEL) is used as an antiknock agent in gasoline and jet fuels. The compound (NH
decomposition produces BeF and NH
2
9.
2
4
F
4)
2
BeF4
.
(3)
+
N
−
3
+
¯
¯ = N = ¯
¯
¯ ↔ ¯
¯
¯− N ≡ N
(Azide): ¯
N
N
N
It has sp hybridised central atom. Hence it is linear in shape.
I
−
3
(triiodide):
It has linear geometry, sp
NO
−
2
3
d
hybridisation with three lone pairs at central atom.
(nitrite) :
It is (nonlinear) bent shaped as it has sp hybridisation with one lone pair at central atom.
2
O3
(ozone) :
It Is bent in shape with sp hybridisation and one lone pair at central atom.
2
SO2
(sulphur dioxide) :
It is bent in shape with sp hybridisation and one lone pair at central atom.
2
So 3 among the given molecules is bent in shape.
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on thermal
Chemical Bonding and Molecular Structure
JEE Main 2023 (April) Chapter-wise Qs Bank
Questions with Solutions
MathonGo
10. (1)
In IF , the central atom is iodine, which has 7 valence electrons. Iodine in this molecule has 5 bonding pairs and 1 lone pair. Therefore, the number
5
of lone pairs on the central atom of IF is 1.
5
In IF , the central atom is also iodine, which has 7 valence electrons. Iodine in this molecule has 7 bond pairs and 0 lone pairs. Therefore, the
7
number of lone pairs on the central atom of IF is 0.
7
Therefore, the sum of the number of lone pairs in the central atom of IF and IF is 1 + 0 = 1..
5
7
11. (3)
The lone pair of electrons can be identified by drawing structure of the given moleules as follows,
The number of molecules having only 2 lone pair of electrons =
Which are H
2
O, N2
3
and CO. XeF have 2 lone pairs on central atom, but we are asked about lone pair in molecule
4
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MathonGo
12. (3)
The molecular geometry of H
3
O
+
is pyramidal. It consists of three hydrogen atoms bonded to the central oxygen atom, with one lone pair of electrons on the
oxygen atom.
The acetylide ion consists of two carbon atoms bonded together by a triple bond (C ≡ C) with a lone pair of electrons on the terminal carbon atom. So it is
linear.
The NH
+
4
ion consists of a central nitrogen atom bonded to four hydrogen atoms. It is tetrahedral.
The ClO ion is bent in shape due to its molecular geometry. It consists of one central chlorine atom bonded to two oxygen atoms, with an additional lone pair
−
2
of electrons on the central chlorine atom.
(A) H
3O
+
→
(III) Pyramidal
(B) Acetylide ion → (II) Linear
(C) NH ion →
+
4
(D) ClO ion →
−
2
(I) Tetrahedral
(IV) Bent
13. (2)
SO2
is a bent molecule with a central sulfur atom bonded to two oxygen atoms. The oxygen atoms are more electronegative than sulfur, causing a separation of
charge. This results in a bent molecular geometry and a net dipole moment, making SO a polar molecule.
2
NH3
is a trigonal pyramidal molecule with a central nitrogen atom bonded to three hydrogen atoms. Again, the electronegativity difference between nitrogen and
hydrogen leads to a separation of charge. The arrangement of the atoms in NH gives it a net dipole moment, making it a polar molecule.
3
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MathonGo
14. (3)
ClO3
−
XeF4
SF4
I3
−
: The central atom is Cl, there is one lone pair of electrons on the central atom.
: The central atom is Xe. It has two lone pair on central atom.
: The central atom is S. It contains one lone pair on central atom.
: It contains three lone pair on central atom.
15. (3)
The bond order of the acetylide ion HC ≡ C is 3, indicating a triple bond between the carbon atoms.
−
The acetylide ion consists of two carbon atoms with a triple bond between them and a negative charge. In the triple bond, there are six bonding electrons (two pi
bonds and one sigma bond) and zero antibonding electrons.
Among the given options NO will have the same bond order and magnetic property like acetylide ion.
+
Species
HC ≡ C
O2
N2
+
+
NO
O2
Bond Order
−
3
0
2. 5
√3 B. M
2. 5
√3 B. M
3
0
1. 5
√3 B. M
+
−
Magnetic moment
16. (3)
The oxidation state of +5 in pentahalides is more covalent as compared to the +3 oxidation state in trihalides. Due to the higher positive oxidation state of the
central atom in pentahalide state, these atoms will have larger polarising power than the halogen atom attached to them since the polarising power is directly
proportional to the charge. More is the polarisation, larger will be the covalent character of the bond. Hence, due to larger polarisation of bond in pentahalide
state as compared to trihalide state, the SbCl is more covalent than SbCl .
5
3
Higher oxides of halogens are more stable than lower ones due to the formation of more number of bonds, which results in higher release of energy. So, both the
statements I and II are correct.
17. (3)
Process
N2 → N
+
2
NO → NO
O2 → O
+
−2
2
O2 → O
+
2
Change in
Bonad order
magnetic nature
change
Dia → Para
3 → 2. 5
Para →Dia
2. 5 → 3
Para→ Dia
2 → 1
Para → Para
2 → 2. 5
18. (4)
Chlorine pentafluoride is an inter halogen compound. ClF is colourless liquid. The central chlorine atom is having five bond pairs and one lone pair on it.
5
Hence, according to VSEPR theory it has square pyramidal geometry. It is toxic by inhalation and an irritant to skin, eyes and mucus membranes.
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19. (4)
The compound which possess aromatic character will show highest dipole moment. From the given compounds the following compound will have the highest
dipole moment. Because the Positive and negative ends acquire aromaticity.
20. (4)
(A) Weak Intermolecular Forces are present in polymer of 2−chloro, 1,
3−
butadiene.It is a monomer of neoprene which is a rubber(elastomer.).
(B). Hydrogen Bonding is present in NYLON−6, 6 which is a polymer of Hexamethylenediamine and adipic acid. This hydrogen bonding is due to the presence
of amide group.
(C) Heavily branched polymer is Bakelite which is polymer of phenol and formaldehyde. It is a crosslinked polymer.
(D) High density polymer(polyethylene) preparation requires Al(Et) and TiCl as a catalyst (Ziegler Natta Catalyst)
3
4
21. (4)
The molecule SO is having two sigma bond pairs and lone pair, hence, it exhibits angular or V-shape structure. The molecule H
2
2O
is having two sigma bond
pairs and a lone pair hence, it exhibits angular or V-shape structure. So, both molecules are having same shape. But double bond-double bond repulsions are
greater than single bond repulsions, hence, the bond angle in sulphur dioxide is greater than water.
SO2
and H
2O
both have V-Shape
Bond angle: SO
2
>
H2 O
22. (4)
In NF , nitrogen is having three bond pairs and a lone pair, hence, it has pyramidal shape.
3
The bond order of N
2
= 3,
and The bond order of O
2
= 2
. Bond length is inversely related to bond order. Hence, bond length of N is shorter than O
2
Isoelectronic species have identical bond order
Dipole moment of H
2O
is more than that of H
2S
due to higher electronegativity of O.
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2
Chemical Bonding and Molecular Structure
JEE Main 2022 (June) Chapter-wise Qs Bank
Questions
MathonGo
Q1 - 24 June - Shift 2
Space for your notes:
Q2 - 25 June - Shift 1
Space for your notes:
Q3 - 25 June - Shift 2
Space for your notes:
Q4 - 26 June - Shift 1
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MathonGo
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Q5 - 26 June - Shift 2
Space for your notes:
Q6 - 26 June - Shift 2
Space for your notes:
Q7 - 27 June - Shift 1
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Q8 - 27 June - Shift 2
Space for your notes:
Q9 - 27 June - Shift 2
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Q10 - 28 June - Shift 1
Space for your notes:
Q11 - 28 June - Shift 2
Space for your notes:
Q12 - 29 June - Shift 1
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Q13 - 29 June - Shift 2
Space for your notes:
Q14 - 29 June - Shift 2
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Questions
MathonGo
Answer Key
Q1 (B)
Q2 (C)
Q3 (3)
Q4 (A)
Q5 (B)
Q6 (3)
Q7 (B)
Q8 (C)
Q9 (C)
Q10 (1)
Q11 (A)
Q12 (C)
Q13 (B)
Q14 (C)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (June) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q1 (B)
Q2 (C)
Q3 (3)
Q4 (A)
Q5 (B)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (June) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q6 (3)
Q7 (B)
Q8 (C)
Q9 (C)
Q10 (1)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (June) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q11 (A)
Q12 (C)
Q13 (B)
Q14 (C)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (June) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Q1 - 25 July - Shift 1
Space for your notes:
Q2 - 25 July - Shift 2
Space for your notes:
Q3 - 25 July - Shift 2
Space for your notes:
Q4 - 25 July - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Space for your notes:
Q5 - 26 July - Shift 1
Space for your notes:
Q6 - 26 July - Shift 2
Space for your notes:
Q7 - 27 July - Shift 1
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Space for your notes:
Q8 - 27 July - Shift 1
Space for your notes:
Q9 - 27 July - Shift 1
Space for your notes:
Q10 - 27 July - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Space for your notes:
Q11 - 27 July - Shift 2
Space for your notes:
Q12 - 28 July - Shift 1
Space for your notes:
Q13 - 28 July - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Space for your notes:
Q14 - 29 July - Shift 1
Space for your notes:
Q15 - 29 July - Shift 2
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Space for your notes:
#MathBoleTohMathonGo
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Questions
MathonGo
Answer Key
Q1 (2)
Q2 (A)
Q3 (B)
Q4 (3)
Q5 (C)
Q6 (B)
Q7 (A)
Q8 (4)
Q9 (3)
Q10 (C)
Q11 (6)
Q12 (4)
Q13 (A)
Q14 (B)
Q15 (4)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q1 (2)
Q2 (A)
Q3 (B)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q4 (3)
Q5 (C)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q6 (B)
Q7 (A)
Q8 (4)
Q9 (3)
Q10 (C)
Q11 (6)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q12 (4)
Q13 (A)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
Q14 (B)
Q15 (4)
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Chemical Bonding and Molecular Structure
JEE Main 2022 (July) Chapter-wise Qs Bank
Hints and Solutions
MathonGo
#MathBoleTohMathonGo
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Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Questions with Answer Keys
MathonGo
Q1 2021 (01 Sep Shift 2)
The spin-only magnetic moment value of B
species is ____ ×10
−2
BM
+
2
. (Nearest integer)
[Given : √3 = 1.73 ]
Q2 2021 (01 Sep Shift 2)
Number of paramagnetic oxides among the following given oxides is ____.
Li2 O, CaO, Na2 O2 , KO2 , MgO
and K
2O
(1) 1
(2) 2
(3) 3
(4) 0
Q3 2021 (31 Aug Shift 2)
According to molecular orbital theory, the number of unpaired electron(s) in O
2−
2
Q4 2021 (27 Aug Shift 2)
The number of species having non-pyramidal shape among the following is ____.
(A) SO
3
(B) NO
−
3
(C) PCl
(D) CO
3
2−
3
Q5 2021 (27 Aug Shift 2)
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is :
Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Questions with Answer Keys
MathonGo
The compound/s which will show significant intermolecular H-bonding is/are :
(1) (b) only
(2) (c) only
(3) (a) and (b) only
(4) (a), (b) and (c)
Q6 2021 (26 Aug Shift 2)
The bond order and magnetic behaviour of O ion are, respectively :
−
2
(1) 1.5 and paramagnetic
(2) 1.5 and diamagnetic
(3) 2 and diamagnetic
(4) 1 and paramagnetic
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Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Questions with Answer Keys
MathonGo
Q7 2021 (26 Aug Shift 2)
The interaction energy of London forces between two particles is proportional to r , where r is the distance
x
between the particles. The value of x is :
(1) 3
(2) −3
(3) −6
(4) 6
Q8 2021 (26 Aug Shift 1)
AB3
is an interhalogen T-shaped molecule. The number of lone pairs of electrons on A is ____. (Integer
answer)
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Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Questions with Answer Keys
MathonGo
Answer Key
Q1 (173)
Q2 (1)
Q3 (0)
Q4 (3)
Q5 (1)
Q6 (1)
Q7 (3)
Q8 (2)
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Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Hints and Solutions
MathonGo
Q1 (173)
B
+
2
⇒ σ
⇒ 9e
2
1 s
σ
∗2
1s
σ
2
2s
σ
∗2
2 s
π
1
≃ π
2py
0
2pz
−
μ = √1(1 + 2) = √3 BM
= 1.73 BM
= 1.73 × 10
−2
BM
Q2 (1)
+
Li2 O
⇒ 2Li
CaO
⇒ Ca
Na2 O2
⇒ 2Na
KO2
⇒ K
2+
K2 O
⇒ 2 K
O
O
2
2
−
2
⇒
+
O
O
2+
⇒ Mg
−
O
+
MgO
O
O
O
+
2−
2−
2−
2
−
2
2−
O
2
Complete octet, diamagnetic
⇒ σ
2
1s
⇒ σ
σ
2
1 s
∗2
1s
σ
σ
∗2
1s
2
2 s
σ
σ
2
2 s
∗2
2 s
σ
σ
∗2
2 s
2
2px
σ
π
2
2px
2
2py
π
≃ π
2
2py
2
2pz
≃ π
π
2
2pz
∗2
2py
π
≃ π
∗2
2py
∗2
2pz
≃ π
(dia)
∗1
2pz
(
para )
Q3 (0)
Molecular orbital configuration of O
σ
2
1 s
σ
∗2
1 s
σ
2
2 s
σ
∗2
2 s
2
2
(π2px = π2py ) (π
2−
2
∗2
2px
is
= π
∗2
2py
)
Zero unpaired electron
Q4 (3)
Trigonal planar
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Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Hints and Solutions
MathonGo
Trigonal planar
Trigonal planar
Pyramidal
Hence non-pyramidal species are SO
CO
2−
3
3,
NO
−
3
and
.
Q5 (1)
(a) Shows intra molecular H−bonding
(b) Shows significant intermolecular H−bonding
(c) It do not show intermolecular H−bonding due to steric hindrance.
Q6 (1)
O
−
2
(π
= (σ1
2
2px
= π
s)
2
2
2py
(σ
2
∗
1 s
) (π
) (σ2
∗2
2px
s)
= π
2
(σ
∗1
2py
∗
2 s
2
2
) (σ2p )
z
)
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Chemical Bonding and Molecular Structure
JEE Main 2021 (August) Chapter-wise Questions
Hints and Solutions
Bond order =
MathonGo
10−7
2
= 1.5
and paramagnetic.
Q7 (3)
For london dispersion forces.
E ∝
1
r
6
Hence x = −6
Q8 (2)
T-shaped molecule means 3 sigma bond and 2 lone pairs of electron on central atom.
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Chemical Bonding and Molecular Structure
JEE Main 2021 (July) Chapter-wise Questions
Questions with Answer Keys
MathonGo
Q1 (20 July 2021 Shift 1)
The number of lone pairs of electrons on the central I atom in I is ___
−
3
Q2 (20 July 2021 Shift 2)
The hybridisations of the atomic orbitals of nitrogen in NO
(1) sp
3
, sp
2
−
2
, NO
+
2
and sp
(2) sp, sp and sp
3
(3) sp
3
(4) sp
2
2
, sp
and sp
2
, sp
and sp
3
Q3 (22 July 2021 Shift 1)
Match List-I with List-II :
Choose the correct answer from the options given below:
(1) (a)-( i), (b)-( ii), (c)-(v) and (d)-(iii)
(2) (a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)
(3) (a)-(iii), (b)-(i), (c)-( v) and (d)-(iv)
(4) (a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)
Q4 (25 July 2021 Shift 2)
In the following the correct bond order sequence is:
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and NH respectively are.
+
4
Chemical Bonding and Molecular Structure
JEE Main 2021 (July) Chapter-wise Questions
Questions with Answer Keys
(1) O
2−
(2) O
+
(3) O
+
(4) O
2
> O
2
+
2
−
2
> O
> O
> O2 > O
2
2
> O
> O
−
2
> O
−
2
2−
2
−
2
2
> O2
> O2
> O
2−
MathonGo
2−
2
> O
+
2
Q5 (25 July 2021 Shift 2)
Identify the species having one π -bond and maximum number of canonical forms from the
following :
(1) SO
(2) O
3
2
(3) SO
2
(4) CO
2−
3
Q6 (27 July 2021 Shift 1)
Given below are two statements: One is labelled as Assertion A and the other labelled as Reason R.
Assertion A : Lithium halides are some what
covalent in nature.
Reason R : Lithium possess high polarisation capability.
In the light of the above statements, choose the most appropriate answer from the options given below:
(1) A is true but R is false
(2) A is false but R is true
(3) Both A and R are true but R is NOT the
correct explanation of A
(4) Both A and R are true and R is the correct explanation of A
Q7 (27 July 2021 Shift 1)
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Chemical Bonding and Molecular Structure
JEE Main 2021 (July) Chapter-wise Questions
Questions with Answer Keys
MathonGo
In gaseous triethyl amine the "-C-N-C-" bond
angle is ______ degree.
Q8 (27 July 2021 Shift 1)
The difference between bond orders of CO and
NO
⊕
is
x
2
where x =
(Round off to the Nearest Integer)
Q9 (27 July 2021 Shift 2)
The total number of electrons in all bonding
molecular orbitals of O
2−
2
is ........
(Round off to the nearest integer)
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Chemical Bonding and Molecular Structure
JEE Main 2021 (July) Chapter-wise Questions
Questions with Answer Keys
MathonGo
Answer Key
Q1 (3)
Q2 (4)
Q3 (3)
Q4 (3)
Q5 (4)
Q6 (4)
Q7 (108)
Q8 (0)
Q9 (10)
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Chemical Bonding and Molecular Structure
JEE Main 2021 (July) Chapter-wise Questions
Hints and Solutions
MathonGo
Q1
I
−
3
:
The number of lone pairs of electron on the central atom is 3.
Q2
Q3
(a) SF
(b) IF
4−
− sp
5
(c) NO
sp
+
2
(d) NH
−
+
4
3
3
hybridisation
d
d
hybridisation
2
sp hybridisation
− sp
3
hybridisation
Q4
O2
π
(16 electrons) σ
2
2px
= π
2
2py
,π
y
∗1
2p
Bond order of O
2
1s
,σ
= π
x
2
−
Bond order of O
2−
Bond order of O
+
2
2
,σ
∗1
2py
2
2s
,σ
y
,σ
∗2
2s
,σ
2
2p2
∗
2p
p
⇒ 2
Bond order of O
2
∗2
1s
⇒ 1.5
⇒ 1
⇒ 2.5
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Chemical Bonding and Molecular Structure
JEE Main 2021 (July) Chapter-wise Questions
Hints and Solutions
MathonGo
Q5
Among SO
CO
2−
3
3,
O2 , SO2
and CO , only O and
2−
2
3
has only one π -bond
Q6
Lithium due to small size has very high
polarization capability and thus increases covalent nature in Halides.
Q7
In gaseous triethyl amine the "-C-N-C-" bond
angle is 108 degree.
Q8
Bond order of CO = 3
Bond order of NO
Difference = 0 =
+
= 3
x
2
x = 0
Q9
M. O. Configuration of O
2
2
2
2
2−
2
2
( (18e )
¯
¯
¯
2
2
σ1 s σ1 s σ2 s σ2 s σ2pz π2px = π2py
2
2
π2px = π2py
Total B.M.O electrons = 10
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Chemical Bonding and Molecular Structure
JEE Main 2021 (March)
Questions with Answer Keys
MathonGo
Q1: 16 March (Shift 1) - Single Correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as
Reason R :
Assertion A : The H − O − H bond angle in water molecule is 104.5
∘
Reason R : The lone pair − lone pair repulsion of electrons is higher than the bond pair - bond pair repulsion.
(1) A is false but R is true
(2) Both A and R are true, but R is not the correct correct explanation of A
(3) A is true but R is false
(4) Both A and R are true, and R is the correct explanation of A
Q2: 17 March (Shift 1) - Single Correct
A central atom in a molecule has two lone pairs of electrons and forms three single bonds. The shape of this
molecule is:
(1) see-saw
(2) planar triangular
(3) T-shaped
(4) trigonal pyramidal
Q3: 17 March (Shift 2) - Single Correct
Amongst the following, the linear species is:
(1) NO
(2) Cl
(3) O
(4) N
2
2O
3
−
3
Q4: 18 March (Shift 1) - Numerical
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Chemical Bonding and Molecular Structure
Questions with Answer Keys
AX
JEE Main 2021 (March)
MathonGo
is a covalent diatomic molecule where A
and X are second row elements of periodic table. Based on Molecular orbital theory, the bond order of AX is
25 . The total number of electrons in AX is____( Round off to the
Nearest Integer).
Q5: 18 March (Shift 2) - Single Correct
The oxide that shows magnetic property is :
(1) SiO
(2) Mn
(3) Na
2
3 O4
2O
(4) MgO
Q6: 18 March (Shift 2) - Single Correct
In the following molecules,
Hybridisation of carbon a, b and c respectively
are :
(1) sp
3
(2) sp
3
(3) sp
3
(4) sp
3
, sp, sp
2
, sp , sp
2
, sp , sp
, sp, sp
2
2
Q7: 18 March (Shift 2) - Numerical
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Chemical Bonding and Molecular Structure
Questions with Answer Keys
MathonGo
The number of species below that have two
lone pairs of electrons in their central atom is_______
(Round off to the Nearest integer)
SF4 , BF
−
4
JEE Main 2021 (March)
, CIF3 , AsF3 , PCl5 , BrF5 , XeF4 , SF6
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Chemical Bonding and Molecular Structure
JEE Main 2021 (March)
Questions with Answer Keys
MathonGo
Answer Key
Q1 (4)
Q2 (3)
Q3 (4)
Q5 (2)
Q6 (3)
Q7 (2)
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Q4 (15)
Chemical Bonding
JEE Main 2021 (February)
Questions with Answer Keys
MathonGo
Q1: 24 Feb (Shift 1) - Single Correct
Which of the following are isostructural pairs?
(A) SO
2−
4
and CrO
2−
4
(B) SiCl and TiCl
4
(c) NH and NO
3
4
−
3
(D) BCl and BrCl
3
3
(1) A and C only
(2) A and B only
(3) B and C only
(4) C and D only
Q2: 24 Feb (Shift 2) - Single Correct
The correct shape and I-I-I bond angles respectively in I ion are :
−
3
(1) Trigonal planar; 120
∘
(2) Distorted trigonal planar; 135 and 90
∘
(3) Linear; 180
∘
∘
(4) T-shaped; 180 and 90
∘
∘
Q3: 25 Feb (Shift 1) - Single Correct
According to molecular orbital theory, the species among the following that does not exist is:
(1) He
(2) He
(3) O
−
2
+
2
2
2−
2
(4) Be
2
Q4: 25 Feb (Shift 2) - Single Correct
Which among the following species has unequal bond lengths ?
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Chemical Bonding
JEE Main 2021 (February)
Questions with Answer Keys
(1) XeF
(2) SiF
(3) BF
(4) SF
MathonGo
4
4
−
4
4
Q5: 26 Feb (Shift 1) - Single Correct
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : Dipole-dipole interactions are the only non-covalent interactions, resulting in hydrogen bond
formation
Reason R : Fluorine is the most electronegative element and hydrogen bonds in HF are symmetrical In the
light of the above statements, choose the most appropriate answer from the options given below :
(1) A is false but R is true
(2) Both A and R are true and R is the correct explanation of A
(3) A is true but R is false
(4) Both A and R are true and R is not the correct explanation of A
Q6: 26 Feb (Shift 2) - Single Correct
Match List-I with List-II.
Choose the correct answer from the options given below:
(1) (a) − (iii), (b) − (iv), (c) − (i), (d) − (iii)
(2) (a) − (i), (b) − (ii), (c) − (iii), (d) − (iv)
(3) (a) − (ii), (b) − (i), (c) − (iv), (d) − (iii)
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Chemical Bonding
JEE Main 2021 (February)
Questions with Answer Keys
MathonGo
(4) (a) − (iv), (b) − (iii), (c) − (ii), (d) − (i)
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Chemical Bonding
JEE Main 2021 (February)
Questions with Answer Keys
MathonGo
Answer Key
Q1 (2)
Q2 (3)
Q5 (3)
Q6 (1)
Q3 (4)
#PaperPhodnaHai
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Q4 (4)
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Chemical Bonding
JEE Mains 2020 Jan Chapter wise Question Bank
JEE Mains 2020 Jan Chapter wise Question Bank
Chemical Bonding
Q1
7th Jan Morning
Sol
Q2
7th Jan Morning
Sol
Q3
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Chemical Bonding
JEE Mains 2020 Jan Chapter wise Question Bank
7th Jan Evening
Sol
Q4
8th Jan Morning
Sol
Q5
8th Jan Evening
Sol
Q6
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Chemical Bonding
JEE Mains 2020 Jan Chapter wise Question Bank
8th Jan Evening
Sol
Q7
9th Jan Morning
Sol
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Chemical Bonding
JEE Mains 2019 Chapter wise Question Bank
JEE Mains 2019 Chapter wise Question Bank
Chemical Bonding - Questions
Q1
9 April Morning
Q6
9 Jan Morning
Q2
9 April Evening
Q7
10 April Morning
9 Jan Evening
Q8
Q3
10 Jan Morning
12 April Morning
Q4
8 April Evening
Q5
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Chemical Bonding
JEE Mains 2019 Chapter wise Question Bank
JEE Mains 2019 Chapter wise Question Bank
Chemical Bonding - Solutions
Q1
Q4
9 Jan Morning
Q2
8 April Evening
Q5
9 Jan Evening
Q3
9 April Morning
Q6
10 Jan Morning
To download more free study materials, visit www.mathongo.com
Chemical Bonding
JEE Mains 2019 Chapter wise Question Bank
9 April Evening
Q7
10 April Morning
Q8
12 April Morning
To download more free study materials, visit www.mathongo.com