Match the LIST-I with LIST-II.
LIST-I (Molecule/ion)LIST-II (Bond pair : lone pair on the central atom)
(A) mathrmICl_2^-(I) 4 : 2
(B) mathrmH_2mathrmO(II) 4 : 1
(C) mathrmSO_2(III) 2 : 3
(D) mathrmXeF_4(IV) 2 : 2
Choose the correct answer from the options given below:

Solution & Explanation

### Core Logic Let's find the number of bond pairs (sigma-bonds or regions) and lone pairs on the central atom of each species: 1. **mathrmICl_2^-**: - Central atom Iodine has 7 valence electrons + 1 negative charge = 8 electrons. - Forms 2 single bonds (bond pairs = 2). - Remaining 6 electrons form 3 lone pairs. - Ratio is 2 : 3 (Matches LIST-II, III).
VSEPR linear shape of ICl2-
VSEPR linear shape of ICl2-
2. **mathrmH_2mathrmO**: - Oxygen has 6 valence electrons. - Forms 2 bond pairs with Hydrogens. - Remaining 4 electrons form 2 lone pairs. - Ratio is 2 : 2 (Matches LIST-II, IV).
VSEPR linear shape of ICl2-
VSEPR linear shape of ICl2-
3. **mathrmSO_2**: - Sulfur has 6 valence electrons. - Forms 2 double bonds (which are counted as 4 bonding pairs of electrons/bond pairs in typical VSEPR representations here). - Remaining 2 electrons form 1 lone pair. - Ratio is 4 : 1 (Matches LIST-II, II).
VSEPR linear shape of ICl2-
VSEPR linear shape of ICl2-
4. **mathrmXeF_4**: - Xenon has 8 valence electrons. - Forms 4 bond pairs with Fluorines. - Remaining 4 electrons form 2 lone pairs. - Ratio is 4 : 2 (Matches LIST-II, I).
VSEPR linear shape of ICl2-
VSEPR linear shape of ICl2-
Thus, the correct mapping is: A-III, B-IV, C-II, D-I. ### Pattern Recognition For match-the-column with VSEPR structures: - Always find steric number: textSteric Number = frac12(V + M - C + A). - Water is 2 bond pairs, 2 lone pairs (sp^3) rightarrow B-IV. This alone helps eliminate multiple incorrect options immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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
Q52 jee_main_2026_22_january_morning Lewis Structures and Formal Charge
The formal changes on the atoms marked as (1) to (4) in the Lewis representation of HNO_3 molecule respectively are
Lewis structure of HNO3 diagram for Q52 - JEE Main 2026 Morning
The image shows the Lewis structure of HNO3 with four atoms numbered 1 through 4.
  • A. text+1, 0, 0, -1
  • B. text0, -1, 0, +1
  • C. text0, +1, 0, -1
  • D. text0, 0, -1, +1

Solution

### Related Formula textFormal charge = (textValence e^-) - (textNon-bonding e^-) - fractextBonding e^-2 ### Core Logic Evaluate the structure of HNO_3 shown in the solution image:
Detailed Lewis structure of HNO3 diagram
The image shows the Lewis structure of HNO3 with four atoms numbered 1 through 4.
Atom 1 (Oxygen with H and N): F.C. = 6 - 4 - frac42 = 0 Atom 2 (Nitrogen): F.C. = 5 - 0 - frac82 = +1 Atom 3 (Double bonded Oxygen): F.C. = 6 - 4 - frac42 = 0 Atom 4 (Single bonded Oxygen): F.C. = 6 - 6 - frac22 = -1 ### Step 1: Final Conclusion The formal charges on atoms (1), (2), (3), and (4) are respectively 0, +1, 0, -1. ### Pattern Recognition In nitro groups (-NO_2), the central nitrogen is always +1, the single bonded oxygen is -1, and the double-bonded oxygen is 0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure
Q68 jee_main_2026_22_january_morning Hybridization and VSEPR Theory
Two p-block elements X and Y form fluroides of the type EF_3. The fluoride compound XF_3 is a Lewis acid and YF_3 is a Lewis base. The hybridization of the central atoms of XF_3 and YF_3 respectively are
  • A. textBoth sp^3
  • B. sp^2 text and sp^3
  • C. sp^3 text and sp^2
  • D. textBoth sp^2

Solution

### Core Logic XF_3 acts as a Lewis acid, meaning it is an electron-deficient species capable of accepting an electron pair. A common example from the p-block is BF_3. The central Boron atom has 3 bond pairs and 0 lone pairs. Therefore, its steric number is 3, corresponding to sp^2 hybridization. YF_3 acts as a Lewis base, meaning it has an available lone pair to donate. A common example is NF_3. The central Nitrogen atom has 3 bond pairs and 1 lone pair. Its steric number is 4, corresponding to sp^3 hybridization. ### Step 1: Final Conclusion The hybridization of X and Y respectively are sp^2 and sp^3. ### Pattern Recognition Electron deficient central atoms (Group 13) form sp^2 planar molecules. Atoms with a lone pair (Group 15) form sp^3 pyramidal molecules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 11 Chemistry: p-Block Elements
Rankbit System
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%)