Chemical Bonding & Molecular Structure: JEE Main Previous Year Questions (2021-2024) | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Chemical Bonding and 
Molecular Structure 
2024 
Q1: Given below are two statements: 
Statement (I) : A ?? bonding MO has lower electron density above and below the inter-nuclear axis. 
Statement (II) : The ?? *
 antibonding MO has a node between the nuclei. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true  [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (d) 
Let's analyze both statements: 
Statement (I): "A ?? bonding MO has lower electron density above and below the inter-nuclear axis." 
This statement is false. In molecular orbital (MO) theory, a pi bond (  bond) is formed by the sideways 
overlap of porbitals from two adjacent atoms. The characteristic electron density of a ?? bonding 
molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of 
electron density are where the p-orbitals overlap and electrons are likely to be found. This 
concentration of electron density above and below the inter-nuclear axis is what allows the ?? bond to 
provide additional stability to the molecule, alongside any sigma (?? ) bonds that may be present. 
Statement (II): "The ?? *
 antibonding MO has a node between the nuclei." 
This statement is true. In a ?? *
 (pi-star) antibonding molecular orbital, there is indeed a nodal plane 
between the nuclei where the probability of finding electrons is essentially zero. This node arises from 
the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive 
interference and a region of zero electron density in the bonding region. The presence of this nodal 
plane is what makes the ?? *
 orbital antibonding-the electrons in this orbital actually serve to destabilize 
the bond between the two atoms. 
Therefore, the correct answer is: 
Option D: Statement I is false but Statement II is true. 
Q2: Select the compound from the following that will show intramolecular hydrogen bonding. 
(A)  
 
(B) ?? ?? ?? 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Chemical Bonding and 
Molecular Structure 
2024 
Q1: Given below are two statements: 
Statement (I) : A ?? bonding MO has lower electron density above and below the inter-nuclear axis. 
Statement (II) : The ?? *
 antibonding MO has a node between the nuclei. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true  [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (d) 
Let's analyze both statements: 
Statement (I): "A ?? bonding MO has lower electron density above and below the inter-nuclear axis." 
This statement is false. In molecular orbital (MO) theory, a pi bond (  bond) is formed by the sideways 
overlap of porbitals from two adjacent atoms. The characteristic electron density of a ?? bonding 
molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of 
electron density are where the p-orbitals overlap and electrons are likely to be found. This 
concentration of electron density above and below the inter-nuclear axis is what allows the ?? bond to 
provide additional stability to the molecule, alongside any sigma (?? ) bonds that may be present. 
Statement (II): "The ?? *
 antibonding MO has a node between the nuclei." 
This statement is true. In a ?? *
 (pi-star) antibonding molecular orbital, there is indeed a nodal plane 
between the nuclei where the probability of finding electrons is essentially zero. This node arises from 
the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive 
interference and a region of zero electron density in the bonding region. The presence of this nodal 
plane is what makes the ?? *
 orbital antibonding-the electrons in this orbital actually serve to destabilize 
the bond between the two atoms. 
Therefore, the correct answer is: 
Option D: Statement I is false but Statement II is true. 
Q2: Select the compound from the following that will show intramolecular hydrogen bonding. 
(A)  
 
(B) ?? ?? ?? 
(C) ?? ?? ?? ?? ???? 
(D) ????
??       [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Intramolecular hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly 
electronegative atom, like oxygen or nitrogen, and this hydrogen atom is also attracted to another 
electronegative atom within the same molecule. This phenomenon tends to happen when the molecule 
can form a six-membered or five-membered ring as a result of this bonding, which offers a stable 
configuration. 
Let's consider each of the given options: 
Option A: The structure in this option contains an ortho-nitrophenol molecule, in which the hydroxyl 
group (-OH) and the nitro group (-NO2) are positioned on the benzene ring such that they are adjacent 
to each other. This allows for the formation of an intramolecular hydrogen bond between the hydrogen 
of the hydroxyl group and the oxygen of the nitro group. Hence, this compound can exhibit 
intramolecular hydrogen bonding. 
Option B : Water (H
2
O) is capable of intermolecular hydrogen bonding, but it cannot form 
intramolecular hydrogen bonds as it is a simple molecule with just one oxygen atom and no possibility to 
form a stable ring structure within a single molecule. 
Option C : Ethanol (C
2
H
5
OH) can form intermolecular hydrogen bonds through its hydroxyl group with 
other ethanol molecules or water molecules, but like water, it cannot form intramolecular hydrogen 
bonds because it does not have the correct geometry for the hydroxyl group to bond within the same 
molecule. 
Option D : Ammonia (NH
3
) is known for intermolecular hydrogen bonding with other ammonia 
molecules or with water molecules, but there are no hydrogen bond acceptors within the molecule that 
would allow for intramolecular hydrogen bonding. 
Therefore, the compound that can show intramolecular hydrogen bonding is in Option A, the ortho-
nitrophenol molecule. 
Q3: Given below are two statements: one is labelled as Assertion (A) and the other is labelled as 
Reason (R). 
Assertion (A): ????
?? has lower boiling point than ????
?? . 
Reason (R) : In liquid state ????
?? molecules are associated through vander Waal's forces, but ????
?? 
molecules are associated through hydrogen bonding. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both (A) and (R) are correct and (R) is the correct explanation of (A) 
B. (A) is not correct but (R) is correct 
C. (A) is correct but (R) is not correct 
D. Both (A) and (R) are correct but (R) is not the correct explanation of (A)   [JEE Main 2024 (Online) 
1st February Morning Shift] 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Chemical Bonding and 
Molecular Structure 
2024 
Q1: Given below are two statements: 
Statement (I) : A ?? bonding MO has lower electron density above and below the inter-nuclear axis. 
Statement (II) : The ?? *
 antibonding MO has a node between the nuclei. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true  [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (d) 
Let's analyze both statements: 
Statement (I): "A ?? bonding MO has lower electron density above and below the inter-nuclear axis." 
This statement is false. In molecular orbital (MO) theory, a pi bond (  bond) is formed by the sideways 
overlap of porbitals from two adjacent atoms. The characteristic electron density of a ?? bonding 
molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of 
electron density are where the p-orbitals overlap and electrons are likely to be found. This 
concentration of electron density above and below the inter-nuclear axis is what allows the ?? bond to 
provide additional stability to the molecule, alongside any sigma (?? ) bonds that may be present. 
Statement (II): "The ?? *
 antibonding MO has a node between the nuclei." 
This statement is true. In a ?? *
 (pi-star) antibonding molecular orbital, there is indeed a nodal plane 
between the nuclei where the probability of finding electrons is essentially zero. This node arises from 
the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive 
interference and a region of zero electron density in the bonding region. The presence of this nodal 
plane is what makes the ?? *
 orbital antibonding-the electrons in this orbital actually serve to destabilize 
the bond between the two atoms. 
Therefore, the correct answer is: 
Option D: Statement I is false but Statement II is true. 
Q2: Select the compound from the following that will show intramolecular hydrogen bonding. 
(A)  
 
(B) ?? ?? ?? 
(C) ?? ?? ?? ?? ???? 
(D) ????
??       [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Intramolecular hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly 
electronegative atom, like oxygen or nitrogen, and this hydrogen atom is also attracted to another 
electronegative atom within the same molecule. This phenomenon tends to happen when the molecule 
can form a six-membered or five-membered ring as a result of this bonding, which offers a stable 
configuration. 
Let's consider each of the given options: 
Option A: The structure in this option contains an ortho-nitrophenol molecule, in which the hydroxyl 
group (-OH) and the nitro group (-NO2) are positioned on the benzene ring such that they are adjacent 
to each other. This allows for the formation of an intramolecular hydrogen bond between the hydrogen 
of the hydroxyl group and the oxygen of the nitro group. Hence, this compound can exhibit 
intramolecular hydrogen bonding. 
Option B : Water (H
2
O) is capable of intermolecular hydrogen bonding, but it cannot form 
intramolecular hydrogen bonds as it is a simple molecule with just one oxygen atom and no possibility to 
form a stable ring structure within a single molecule. 
Option C : Ethanol (C
2
H
5
OH) can form intermolecular hydrogen bonds through its hydroxyl group with 
other ethanol molecules or water molecules, but like water, it cannot form intramolecular hydrogen 
bonds because it does not have the correct geometry for the hydroxyl group to bond within the same 
molecule. 
Option D : Ammonia (NH
3
) is known for intermolecular hydrogen bonding with other ammonia 
molecules or with water molecules, but there are no hydrogen bond acceptors within the molecule that 
would allow for intramolecular hydrogen bonding. 
Therefore, the compound that can show intramolecular hydrogen bonding is in Option A, the ortho-
nitrophenol molecule. 
Q3: Given below are two statements: one is labelled as Assertion (A) and the other is labelled as 
Reason (R). 
Assertion (A): ????
?? has lower boiling point than ????
?? . 
Reason (R) : In liquid state ????
?? molecules are associated through vander Waal's forces, but ????
?? 
molecules are associated through hydrogen bonding. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both (A) and (R) are correct and (R) is the correct explanation of (A) 
B. (A) is not correct but (R) is correct 
C. (A) is correct but (R) is not correct 
D. Both (A) and (R) are correct but (R) is not the correct explanation of (A)   [JEE Main 2024 (Online) 
1st February Morning Shift] 
Ans: (c) 
The correct answer is Option C: (A) is correct but (R) is not correct. 
Assertion (A): PH
3
 has lower boiling point than NH
3
. 
This assertion is true. The boiling point of ammonia (NH
3
) is higher than that of phosphine (PH
3
). 
Ammonia has a boiling point of about -33.34
°
C, whereas phosphine has a boiling point of about 
-87.7
°
C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the 
strength and type of intermolecular forces present in the substances. 
Reason (R) : In liquid state NH
3
 molecules are associated through Vander Waals' forces, but PH
3
 
molecules are associated through hydrogen bonding. 
This reason is incorrect. Ammonia (NH
3
) can form hydrogen bonds due to the presence of a highly 
electronegative nitrogen atom bonded to hydrogen atoms. This allows NH
3
 molecules to strongly 
associate with each other through hydrogen bonding, which is a much stronger intermolecular force 
than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of 
ammonia. 
On the other hand, phosphine (PH
3
) does not form hydrogen bonds because the electronegativity 
difference between phosphorus and hydrogen is not significant enough to enable the formation of 
hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals 
forces, leading to a lower boiling point when compared to ammonia. 
In conclusion, the assertion that PH
3
 has a lower boiling point than NH
3
 is correct, due to the hydrogen 
bonding present in NH
3
 and absent in PH
3
. However, the reason provided is incorrect because it 
incorrectly states that PH
3
 forms hydrogen bonds and that NH
3
 is associated through Van der Waals 
forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false. 
Q4: Arrange the bonds in order of increasing ionic character in the molecules. ?????? , ?? ?? ?? , ?? ?? , ????
?? and 
?????? ?? : 
A. ?? ?? < ????
?? < ?????? ?? < ?? ?? ?? < ?????? 
B. ?????? ?? < ?? ?? < ????
?? < ?? ?? ?? < ?????? 
C. ?????? < ?? ?? ?? < ?????? ?? < ????
?? < ?? ?? 
D. ?? ?? < ?????? ?? < ????
?? < ?? ?? ?? < ??????   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
To determine the order of increasing ionic character in the bonds of the given molecules, we should 
consider the concept of electronegativity, which is a measure of the tendency of an atom to attract a 
bonding pair of electrons. The larger the difference in electronegativity between two atoms, the more 
ionic the bond is. Covalent bonds, on the other hand, have smaller differences in electronegativity. 
We can qualitatively assess the ionic character of the bonds in each molecule as follows : 
? N
2
 : Both nitrogen atoms have the same electronegativity since it is a diatomic molecule of the 
same element, therefore the bond is purely covalent with no ionic character. 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Chemical Bonding and 
Molecular Structure 
2024 
Q1: Given below are two statements: 
Statement (I) : A ?? bonding MO has lower electron density above and below the inter-nuclear axis. 
Statement (II) : The ?? *
 antibonding MO has a node between the nuclei. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true  [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (d) 
Let's analyze both statements: 
Statement (I): "A ?? bonding MO has lower electron density above and below the inter-nuclear axis." 
This statement is false. In molecular orbital (MO) theory, a pi bond (  bond) is formed by the sideways 
overlap of porbitals from two adjacent atoms. The characteristic electron density of a ?? bonding 
molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of 
electron density are where the p-orbitals overlap and electrons are likely to be found. This 
concentration of electron density above and below the inter-nuclear axis is what allows the ?? bond to 
provide additional stability to the molecule, alongside any sigma (?? ) bonds that may be present. 
Statement (II): "The ?? *
 antibonding MO has a node between the nuclei." 
This statement is true. In a ?? *
 (pi-star) antibonding molecular orbital, there is indeed a nodal plane 
between the nuclei where the probability of finding electrons is essentially zero. This node arises from 
the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive 
interference and a region of zero electron density in the bonding region. The presence of this nodal 
plane is what makes the ?? *
 orbital antibonding-the electrons in this orbital actually serve to destabilize 
the bond between the two atoms. 
Therefore, the correct answer is: 
Option D: Statement I is false but Statement II is true. 
Q2: Select the compound from the following that will show intramolecular hydrogen bonding. 
(A)  
 
(B) ?? ?? ?? 
(C) ?? ?? ?? ?? ???? 
(D) ????
??       [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Intramolecular hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly 
electronegative atom, like oxygen or nitrogen, and this hydrogen atom is also attracted to another 
electronegative atom within the same molecule. This phenomenon tends to happen when the molecule 
can form a six-membered or five-membered ring as a result of this bonding, which offers a stable 
configuration. 
Let's consider each of the given options: 
Option A: The structure in this option contains an ortho-nitrophenol molecule, in which the hydroxyl 
group (-OH) and the nitro group (-NO2) are positioned on the benzene ring such that they are adjacent 
to each other. This allows for the formation of an intramolecular hydrogen bond between the hydrogen 
of the hydroxyl group and the oxygen of the nitro group. Hence, this compound can exhibit 
intramolecular hydrogen bonding. 
Option B : Water (H
2
O) is capable of intermolecular hydrogen bonding, but it cannot form 
intramolecular hydrogen bonds as it is a simple molecule with just one oxygen atom and no possibility to 
form a stable ring structure within a single molecule. 
Option C : Ethanol (C
2
H
5
OH) can form intermolecular hydrogen bonds through its hydroxyl group with 
other ethanol molecules or water molecules, but like water, it cannot form intramolecular hydrogen 
bonds because it does not have the correct geometry for the hydroxyl group to bond within the same 
molecule. 
Option D : Ammonia (NH
3
) is known for intermolecular hydrogen bonding with other ammonia 
molecules or with water molecules, but there are no hydrogen bond acceptors within the molecule that 
would allow for intramolecular hydrogen bonding. 
Therefore, the compound that can show intramolecular hydrogen bonding is in Option A, the ortho-
nitrophenol molecule. 
Q3: Given below are two statements: one is labelled as Assertion (A) and the other is labelled as 
Reason (R). 
Assertion (A): ????
?? has lower boiling point than ????
?? . 
Reason (R) : In liquid state ????
?? molecules are associated through vander Waal's forces, but ????
?? 
molecules are associated through hydrogen bonding. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both (A) and (R) are correct and (R) is the correct explanation of (A) 
B. (A) is not correct but (R) is correct 
C. (A) is correct but (R) is not correct 
D. Both (A) and (R) are correct but (R) is not the correct explanation of (A)   [JEE Main 2024 (Online) 
1st February Morning Shift] 
Ans: (c) 
The correct answer is Option C: (A) is correct but (R) is not correct. 
Assertion (A): PH
3
 has lower boiling point than NH
3
. 
This assertion is true. The boiling point of ammonia (NH
3
) is higher than that of phosphine (PH
3
). 
Ammonia has a boiling point of about -33.34
°
C, whereas phosphine has a boiling point of about 
-87.7
°
C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the 
strength and type of intermolecular forces present in the substances. 
Reason (R) : In liquid state NH
3
 molecules are associated through Vander Waals' forces, but PH
3
 
molecules are associated through hydrogen bonding. 
This reason is incorrect. Ammonia (NH
3
) can form hydrogen bonds due to the presence of a highly 
electronegative nitrogen atom bonded to hydrogen atoms. This allows NH
3
 molecules to strongly 
associate with each other through hydrogen bonding, which is a much stronger intermolecular force 
than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of 
ammonia. 
On the other hand, phosphine (PH
3
) does not form hydrogen bonds because the electronegativity 
difference between phosphorus and hydrogen is not significant enough to enable the formation of 
hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals 
forces, leading to a lower boiling point when compared to ammonia. 
In conclusion, the assertion that PH
3
 has a lower boiling point than NH
3
 is correct, due to the hydrogen 
bonding present in NH
3
 and absent in PH
3
. However, the reason provided is incorrect because it 
incorrectly states that PH
3
 forms hydrogen bonds and that NH
3
 is associated through Van der Waals 
forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false. 
Q4: Arrange the bonds in order of increasing ionic character in the molecules. ?????? , ?? ?? ?? , ?? ?? , ????
?? and 
?????? ?? : 
A. ?? ?? < ????
?? < ?????? ?? < ?? ?? ?? < ?????? 
B. ?????? ?? < ?? ?? < ????
?? < ?? ?? ?? < ?????? 
C. ?????? < ?? ?? ?? < ?????? ?? < ????
?? < ?? ?? 
D. ?? ?? < ?????? ?? < ????
?? < ?? ?? ?? < ??????   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
To determine the order of increasing ionic character in the bonds of the given molecules, we should 
consider the concept of electronegativity, which is a measure of the tendency of an atom to attract a 
bonding pair of electrons. The larger the difference in electronegativity between two atoms, the more 
ionic the bond is. Covalent bonds, on the other hand, have smaller differences in electronegativity. 
We can qualitatively assess the ionic character of the bonds in each molecule as follows : 
? N
2
 : Both nitrogen atoms have the same electronegativity since it is a diatomic molecule of the 
same element, therefore the bond is purely covalent with no ionic character. 
? SO
2
 : Sulfur and oxygen have different electronegativities, with oxygen being more 
electronegative than sulfur, but not as drastically different as in ionic compounds. Therefore, 
the bond has some ionic character but is primarily covalent. 
? ClF
3
 : Chlorine and fluorine have different electronegativities, with fluorine being more 
electronegative. However, the difference is not as large as between metals and nonmetals, so 
while this bond has ionic character, it is not as ionic as a bond between a metal and a nonmetal. 
? K
2
O : Potassium is a metal with low electronegativity, and oxygen is a nonmetal with high 
electronegativity. Therefore, bonds between potassium and oxygen will have a high degree of 
ionic character. 
? LiF : Lithium is a metal with low electronegativity, and fluorine is a nonmetal with very high 
electronegativity. The difference in electronegativity is very large, indicating a very high ionic 
character, even more so than in K
2
O because fluorine is more electronegative than oxygen. 
Given these considerations, the correct order from least to most ionic character is as follows : 
N
2
< SO
2
< ClF
3
< K
2
O < LiF 
Therefore, the correct option is : 
Option A: N
2
< SO
2
< ClF
3
< K
2
O < LiF 
Q5: Which of the following is least ionic? 
A. ???????? ?? 
B. ?????? 
C. ???????? ?? 
D. ????????   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (d) 
AgCl < CoCl
2
< BaCl
2
< KCl (ionic character) 
Reason: Ag
+
has pseudo inert gas configuration. 
Q6: The linear combination of atomic orbitals to form molecular orbitals takes place only when the 
combining atomic orbitals 
A. have the same energy 
B. have the minimum overlap 
C. have same symmetry about the molecular axis 
D. have different symmetry about the molecular axis 
Choose the most appropriate from the options given below: 
A. B, C, D only 
B. A, B, C only 
C. B and D only 
D. A and C only   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (d) 
Molecular orbital should have maximum overlap & Symmetry about the molecular axis should be similar 
 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Chemical Bonding and 
Molecular Structure 
2024 
Q1: Given below are two statements: 
Statement (I) : A ?? bonding MO has lower electron density above and below the inter-nuclear axis. 
Statement (II) : The ?? *
 antibonding MO has a node between the nuclei. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true  [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (d) 
Let's analyze both statements: 
Statement (I): "A ?? bonding MO has lower electron density above and below the inter-nuclear axis." 
This statement is false. In molecular orbital (MO) theory, a pi bond (  bond) is formed by the sideways 
overlap of porbitals from two adjacent atoms. The characteristic electron density of a ?? bonding 
molecular orbital is concentrated above and below the inter-nuclear axis, not lower. These regions of 
electron density are where the p-orbitals overlap and electrons are likely to be found. This 
concentration of electron density above and below the inter-nuclear axis is what allows the ?? bond to 
provide additional stability to the molecule, alongside any sigma (?? ) bonds that may be present. 
Statement (II): "The ?? *
 antibonding MO has a node between the nuclei." 
This statement is true. In a ?? *
 (pi-star) antibonding molecular orbital, there is indeed a nodal plane 
between the nuclei where the probability of finding electrons is essentially zero. This node arises from 
the out-of-phase combination of the p-orbitals from adjacent atoms, which results in destructive 
interference and a region of zero electron density in the bonding region. The presence of this nodal 
plane is what makes the ?? *
 orbital antibonding-the electrons in this orbital actually serve to destabilize 
the bond between the two atoms. 
Therefore, the correct answer is: 
Option D: Statement I is false but Statement II is true. 
Q2: Select the compound from the following that will show intramolecular hydrogen bonding. 
(A)  
 
(B) ?? ?? ?? 
(C) ?? ?? ?? ?? ???? 
(D) ????
??       [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (a) 
Intramolecular hydrogen bonding occurs when a hydrogen atom is covalently bonded to a highly 
electronegative atom, like oxygen or nitrogen, and this hydrogen atom is also attracted to another 
electronegative atom within the same molecule. This phenomenon tends to happen when the molecule 
can form a six-membered or five-membered ring as a result of this bonding, which offers a stable 
configuration. 
Let's consider each of the given options: 
Option A: The structure in this option contains an ortho-nitrophenol molecule, in which the hydroxyl 
group (-OH) and the nitro group (-NO2) are positioned on the benzene ring such that they are adjacent 
to each other. This allows for the formation of an intramolecular hydrogen bond between the hydrogen 
of the hydroxyl group and the oxygen of the nitro group. Hence, this compound can exhibit 
intramolecular hydrogen bonding. 
Option B : Water (H
2
O) is capable of intermolecular hydrogen bonding, but it cannot form 
intramolecular hydrogen bonds as it is a simple molecule with just one oxygen atom and no possibility to 
form a stable ring structure within a single molecule. 
Option C : Ethanol (C
2
H
5
OH) can form intermolecular hydrogen bonds through its hydroxyl group with 
other ethanol molecules or water molecules, but like water, it cannot form intramolecular hydrogen 
bonds because it does not have the correct geometry for the hydroxyl group to bond within the same 
molecule. 
Option D : Ammonia (NH
3
) is known for intermolecular hydrogen bonding with other ammonia 
molecules or with water molecules, but there are no hydrogen bond acceptors within the molecule that 
would allow for intramolecular hydrogen bonding. 
Therefore, the compound that can show intramolecular hydrogen bonding is in Option A, the ortho-
nitrophenol molecule. 
Q3: Given below are two statements: one is labelled as Assertion (A) and the other is labelled as 
Reason (R). 
Assertion (A): ????
?? has lower boiling point than ????
?? . 
Reason (R) : In liquid state ????
?? molecules are associated through vander Waal's forces, but ????
?? 
molecules are associated through hydrogen bonding. 
In the light of the above statements, choose the most appropriate answer from the options given 
below: 
A. Both (A) and (R) are correct and (R) is the correct explanation of (A) 
B. (A) is not correct but (R) is correct 
C. (A) is correct but (R) is not correct 
D. Both (A) and (R) are correct but (R) is not the correct explanation of (A)   [JEE Main 2024 (Online) 
1st February Morning Shift] 
Ans: (c) 
The correct answer is Option C: (A) is correct but (R) is not correct. 
Assertion (A): PH
3
 has lower boiling point than NH
3
. 
This assertion is true. The boiling point of ammonia (NH
3
) is higher than that of phosphine (PH
3
). 
Ammonia has a boiling point of about -33.34
°
C, whereas phosphine has a boiling point of about 
-87.7
°
C. The reason for the higher boiling point of ammonia as compared to phosphine lies in the 
strength and type of intermolecular forces present in the substances. 
Reason (R) : In liquid state NH
3
 molecules are associated through Vander Waals' forces, but PH
3
 
molecules are associated through hydrogen bonding. 
This reason is incorrect. Ammonia (NH
3
) can form hydrogen bonds due to the presence of a highly 
electronegative nitrogen atom bonded to hydrogen atoms. This allows NH
3
 molecules to strongly 
associate with each other through hydrogen bonding, which is a much stronger intermolecular force 
than Van der Waals forces. This hydrogen bonding is responsible for the relatively high boiling point of 
ammonia. 
On the other hand, phosphine (PH
3
) does not form hydrogen bonds because the electronegativity 
difference between phosphorus and hydrogen is not significant enough to enable the formation of 
hydrogen bonds. Instead, phosphine molecules are associated mainly through weaker Van der Waals 
forces, leading to a lower boiling point when compared to ammonia. 
In conclusion, the assertion that PH
3
 has a lower boiling point than NH
3
 is correct, due to the hydrogen 
bonding present in NH
3
 and absent in PH
3
. However, the reason provided is incorrect because it 
incorrectly states that PH
3
 forms hydrogen bonds and that NH
3
 is associated through Van der Waals 
forces. It is actually the other way around, so the correct answer is that (A) is true but (R) is false. 
Q4: Arrange the bonds in order of increasing ionic character in the molecules. ?????? , ?? ?? ?? , ?? ?? , ????
?? and 
?????? ?? : 
A. ?? ?? < ????
?? < ?????? ?? < ?? ?? ?? < ?????? 
B. ?????? ?? < ?? ?? < ????
?? < ?? ?? ?? < ?????? 
C. ?????? < ?? ?? ?? < ?????? ?? < ????
?? < ?? ?? 
D. ?? ?? < ?????? ?? < ????
?? < ?? ?? ?? < ??????   [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans: (a) 
To determine the order of increasing ionic character in the bonds of the given molecules, we should 
consider the concept of electronegativity, which is a measure of the tendency of an atom to attract a 
bonding pair of electrons. The larger the difference in electronegativity between two atoms, the more 
ionic the bond is. Covalent bonds, on the other hand, have smaller differences in electronegativity. 
We can qualitatively assess the ionic character of the bonds in each molecule as follows : 
? N
2
 : Both nitrogen atoms have the same electronegativity since it is a diatomic molecule of the 
same element, therefore the bond is purely covalent with no ionic character. 
? SO
2
 : Sulfur and oxygen have different electronegativities, with oxygen being more 
electronegative than sulfur, but not as drastically different as in ionic compounds. Therefore, 
the bond has some ionic character but is primarily covalent. 
? ClF
3
 : Chlorine and fluorine have different electronegativities, with fluorine being more 
electronegative. However, the difference is not as large as between metals and nonmetals, so 
while this bond has ionic character, it is not as ionic as a bond between a metal and a nonmetal. 
? K
2
O : Potassium is a metal with low electronegativity, and oxygen is a nonmetal with high 
electronegativity. Therefore, bonds between potassium and oxygen will have a high degree of 
ionic character. 
? LiF : Lithium is a metal with low electronegativity, and fluorine is a nonmetal with very high 
electronegativity. The difference in electronegativity is very large, indicating a very high ionic 
character, even more so than in K
2
O because fluorine is more electronegative than oxygen. 
Given these considerations, the correct order from least to most ionic character is as follows : 
N
2
< SO
2
< ClF
3
< K
2
O < LiF 
Therefore, the correct option is : 
Option A: N
2
< SO
2
< ClF
3
< K
2
O < LiF 
Q5: Which of the following is least ionic? 
A. ???????? ?? 
B. ?????? 
C. ???????? ?? 
D. ????????   [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: (d) 
AgCl < CoCl
2
< BaCl
2
< KCl (ionic character) 
Reason: Ag
+
has pseudo inert gas configuration. 
Q6: The linear combination of atomic orbitals to form molecular orbitals takes place only when the 
combining atomic orbitals 
A. have the same energy 
B. have the minimum overlap 
C. have same symmetry about the molecular axis 
D. have different symmetry about the molecular axis 
Choose the most appropriate from the options given below: 
A. B, C, D only 
B. A, B, C only 
C. B and D only 
D. A and C only   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (d) 
Molecular orbital should have maximum overlap & Symmetry about the molecular axis should be similar 
 
Q7: Given below are two statements: 
Statement - I: Since Fluorine is more electronegative than nitrogen, the net dipole moment of ????
?? is 
greater than ????
?? . 
Statement - II: In ????
?? , the orbital dipole due to lone pair and the dipole moment of ???? bonds are in 
opposite direction, but in ????
?? the orbital dipole due to lone pair and dipole moments of ?? - ?? bonds 
are in same direction. 
In the light of the above statements, choose the most appropriate from the options given below: 
A Statement I is true but Statement II is false. 
B Both Statement I and Statement II are true. 
C Both Statement I and Statement II are false. 
D Statement I is false but Statement II is true.   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (c) 
 
Q8: The molecule / ion with square pyramidal shape is 
A. ?????? ?? 
B. [???? (???? )
?? ]
?? -
 
C. ????
?? 
D. ?????? ??   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (d) 
BrF
5
 
 
 
Q9: Aluminium chloride in acidified aqueous solution forms an ion having geometry 
A. Square planar