Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is textCl_2 > textBr_2 > textF_2 > textI_2. Statement II: The correct trend in the covalent character of the metal halides is [textSnCl_4 > textSnCl_2], [textPbCl_4 > textPbCl_2] and [textUF_4 > textUF_6] (or similar Fajan's rule trend). In the light of the above statements, choose the correct answer from the options given below:

Solution & Explanation

### Core Logic - Statement I: Bond dissociation energy order for halogens is textCl_2 > textBr_2 > textF_2 > textI_2 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 textUF_6 > textUF_4 (higher oxidation state has greater covalent character), making the statement II claim regarding textUF_4 > textUF_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

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More Chemical Bonding and Molecular Structure Previous-Year Questions

Q64 jee_main_2026_21_jan_morning VSEPR Theory
Given below are two statements: Statement I: The number of species among mathrmSF_4, mathrmNH_4^+, [mathrmNiCl_4]^2-, mathrmXeF_4, [mathrmPtCl_4]^2-, mathrmSeF_4 and [mathrmNi(CN)_4]^2-, that have tetrahedral geometry is 3. Statement II: In the set [NO_2, BeH_2, BF_3, AlCl_3], all the molecules have incomplete octet around central atom. In the light of the above statements, choose the correct answer from the options given below:
  • A. textStatement I is true but Statement II is false
  • B. textBoth Statement I and Statement II are false
  • C. textStatement I is false but Statement II is true
  • D. textBoth Statement I and Statement II are true

Solution

### Core Logic Evaluating Statement I: - mathrmSF_4: sp^3d (1 lone pair) rightarrow See-saw - mathrmXeF_4: sp^3d^2 (2 lone pairs) rightarrow Square planar - [mathrmPtCl_4]^2-: dsp^2 rightarrow Square planar - [mathrmNiCl_4]^2-: sp^3 rightarrow Tetrahedral - [mathrmNi(CN)_4]^2-: dsp^2 rightarrow Square planar - mathrmSeF_4: sp^3d (1 lone pair) rightarrow See-saw - mathrmNH_4^+: sp^3 (0 lone pairs) rightarrow Tetrahedral Total tetrahedral species = 2 ([mathrmNiCl_4]^2- and mathrmNH_4^+). Statement I says 3, so it is false. Evaluating Statement II: - NO_2: Central N has 7 valence electrons (odd-electron molecule, incomplete octet). - BeH_2: Central Be has 4 electrons (incomplete octet). - BF_3: Central B has 6 electrons (incomplete octet). - AlCl_3 (monomer): Central Al has 6 electrons (incomplete octet). Therefore, all molecules have incomplete octets. Statement II is true. ### Step 1: Final Conclusion Statement I is false, Statement II is true. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: Coordination Compounds
Q64 jee_main_2026_21_jan_evening Bond Length Trends
The correct increasing order of textC-H (A), textC-O (B), textC=O (C) and textCequivtextN (D) bonds in terms of covalent bond length is: (1) A < B < C < D (2) A < D < C < B (3) D < C < B < A (4) D < C < A < B
  • A. (1) \ A < B < C < D
  • B. (2) \ A < D < C < B
  • C. (3) \ D < C < B < A
  • D. (4) \ D < C < A < B

Solution

### Core Logic Comparing bond lengths: - C–H (A): sim 107 text pm - C≡N (D): sim 116 text pm - C=O (C): sim 121 text pm - C–O (B): sim 143 text pm ### Step 1: Final Conclusion Thus, the increasing order is A < D < C < B, corresponding to option (2). ### Pattern Recognition Sees: covalent bond length comparison across bond orders and atomic radii. Trap: Assuming triple bonds are always longer or shorter without accounting for smaller atomic radii like hydrogen. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q69 jee_main_2026_21_jan_evening Bond Dissociation Enthalpy and Fajan's Rules
Given below are two statements: Statement I: The correct order in terms of bond dissociation enthalpy is textCl_2 > textBr_2 > textF_2 > textI_2. Statement II: The correct trend in the covalent character of the metal halides is [textSnCl_4 > textSnCl_2], [textPbCl_4 > textPbCl_2] and [textUF_4 > textUF_6] (or similar Fajan's rule trend). In the light of the above statements, choose the correct answer from the options given below:
  • A. (1) text Statement I is true but Statement II is false
  • B. (2) text Both Statement I and Statement II are true
  • C. (3) text Statement I is false but Statement II is true
  • D. (4) text Both Statement I and Statement II are false

Solution

### Core Logic - Statement I: Bond dissociation energy order for halogens is textCl_2 > textBr_2 > textF_2 > textI_2 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 textUF_6 > textUF_4 (higher oxidation state has greater covalent character), making the statement II claim regarding textUF_4 > textUF_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
Q64 jee_main_2026_22_january_morning Bond Length Trends
The correct increasing order of textC-H (A), textC-O (B), textC=O (C) and textCequivtextN (D) bonds in terms of covalent bond length is: (1) A < B < C < D (2) A < D < C < B (3) D < C < B < A (4) D < C < A < B
  • A. (1) \ A < B < C < D
  • B. (2) \ A < D < C < B
  • C. (3) \ D < C < B < A
  • D. (4) \ D < C < A < B

Solution

### Core Logic Comparing bond lengths: - C–H (A): sim 107 text pm - C≡N (D): sim 116 text pm - C=O (C): sim 121 text pm - C–O (B): sim 143 text pm ### Step 1: Final Conclusion Thus, the increasing order is A < D < C < B, corresponding to option (2). ### Pattern Recognition Sees: covalent bond length comparison across bond orders and atomic radii. Trap: Assuming triple bonds are always longer or shorter without accounting for smaller atomic radii like hydrogen. ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)