Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is Cl2 > Br2 > F2 > I2. Statement II: The correct trend in the covalent character of the metal halides is [SnCl4 > SnCl2], [PbCl4 > PbCl2] and [UF4 > UF6] (or similar Fajan's rule trend). In the light of the above statements, choose the correct answer from the options given below:

- Statement I: Bond dissociation energy order for halogens is Cl2 > Br2 > F2 > I2 due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true. - Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so UF6 > UF4 (higher oxidation state has greater covalent character), making the statement II claim regarding UF4 > UF6 false. Statement I is true but Statement II is false, corresponding to option (1). Sees: halogen bond dissociation energy anomalies and Fajan's rules for covalent character. Trap: Assuming fluorine has the highest bond dissociation energy among halogens. Class 11 Chemistry: Chemical Bonding and Molecular Structure

Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is Cl2 > Br2 > F2 > I2. Statement II: The correct trend in the covalent character of the metal hal

- Statement I: Bond dissociation energy order for halogens is Cl2 > Br2 > F2 > I2 due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true. - Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so UF6 > UF4 (higher oxidation state has greater covalent character), making the statement II claim regarding UF4 > UF6 false. Statement I is true but Statement II is false, corresponding to option (1). Sees: halogen bond di

Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is Cl2 > Br2 > F2 > I2. Statement II: The correct trend in the covalent character of the metal halides is [SnCl4 > SnCl2], [PbCl4 > PbCl2] and [UF4 > UF6] (or similar Fajan's rule trend). In the light of the above statements, choose the correct answer from the options given below:

- Statement I: Bond dissociation energy order for halogens is Cl2 > Br2 > F2 > I2 due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true. - Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so UF6 > UF4 (higher oxidation state has greater covalent character), making the statement II claim regarding UF4 > UF6 false. Statement I is true but Statement II is false, corresponding to option (1). Sees: halogen bond dissociation energy anomalies and Fajan's rules for covalent character. Trap: Assuming fluorine has the highest bond dissociation energy among halogens. Class 11 Chemistry: Chemical Bonding and Molecular Structure

Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is Cl2 > Br2 > F2 > I2. Statement II: The correct trend in the covalent character of the metal hal

- Statement I: Bond dissociation energy order for halogens is Cl2 > Br2 > F2 > I2 due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true. - Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so UF6 > UF4 (higher oxidation state has greater covalent character), making the statement II claim regarding UF4 > UF6 false. Statement I is true but Statement II is false, corresponding to option (1). Sees: halogen bond di

Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is Cl2 > Br2

Year	2026	2025	2024	Total
Questions	12	14	16	42

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Statement I: Bond dissociation energy order for halogens is $Cl₂ > Br₂ > F₂ > I₂$ due to small size and lone-pair repulsions in fluorine weakening its bond. Statement I is true.
Statement II: According to Fajan's rules, higher charge on cation increases covalent character, so $UF₆ > UF₄$ $\text{UF}_6 > \text{UF}_4$ (higher oxidation state has greater covalent character), making the statement II claim regarding $UF₄ > UF₆$ $\text{UF}_4 > \text{UF}_6$ false.

Step 1: Final Conclusion

Statement I is true but Statement II is false, corresponding to option (1).

Pattern Recognition

Sees: halogen bond dissociation energy anomalies and Fajan's rules for covalent character. Trap: Assuming fluorine has the highest bond dissociation energy among halogens.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

More Chemical Bonding and Molecular Structure Previous-Year Questions — Page 9

Q77 jee_main_2024_31_jan_morning Molecular Orbital Theory

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. $A, B, C only$ $\text{A, B, C only}$
B. $A and C only$ $\text{A and C only}$
C. $B, C, D only$ $\text{B, C, D only}$
D. $B and D only$ $\text{B and D only}$

Solution

Core Logic

Conditions for the linear combination of atomic orbitals (LCAO) to form molecular orbitals:

The combining atomic orbitals must have the same or nearly the same energy.
The combining atomic orbitals must have the same symmetry about the molecular axis.
The combining atomic orbitals must overlap to the maximum extent (not minimum).

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Class 11 Chemistry: Chemical Bonding and Molecular Structure

Q85 jee_main_2024_31_jan_morning Hybridization

The number of species from the following in which the central atom uses

sp³

hybrid orbitals in its bonding is

NH₃, SO₂, SiO₂, BeCl₂, CO₂, H₂O, CH₄, BF₃

Numerical Answer. Answer: 4 to 4

Solution

Core Logic

Analyzing the hybridization of the central atom in each species:

$NH₃$ $NH_3$: 3 bp + 1 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$SO₂$ $SO_2$: 2 bp + 1 lp = 3 electron domains $arrow sp²$ $\rightarrow sp^2$
$SiO₂$ $SiO_2$: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally $arrow sp³$ $\rightarrow sp^3$
$BeCl₂$ $BeCl_2$: 2 bp + 0 lp = 2 electron domains $arrow sp$ $\rightarrow sp$
$CO₂$ $CO_2$: 2 bp + 0 lp = 2 electron domains $arrow sp$ $\rightarrow sp$
$H₂O$ $H_2O$: 2 bp + 2 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$CH₄$ $CH_4$: 4 bp + 0 lp = 4 electron domains $arrow sp³$ $\rightarrow sp^3$
$BF₃$ $BF_3$: 3 bp + 0 lp = 3 electron domains $arrow sp²$ $\rightarrow sp^2$

Total species with $sp³$ $sp^3$ hybridization: $NH₃, SiO₂, H₂O, CH₄$ $NH_3, SiO_2, H_2O, CH_4$. Total count = 4.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure

Solution & Explanation

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

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