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JEE Main Previous Year Questions (2026): Chemical Bonding and Molecular Structure
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Page 1 JEE Main Previous Year Questions (2025): Chemical Bonding and Molecular Structure Q1: The number of molecules/ions that show linear geometry among the following is ____. ???? ?? , ???????? ?? , ???? ?? , ?? ?? - , ???? ?? , ?? ?? ?? , ?????? ?? , ???? ?? + , ?? ?? - , ?? ?? JEE Main 2025 (Online) 22nd January Morning Shift Ans: 6 Solution: Linear species are Cl - Bc - Cl, O = C = O, N = N = N (sp) (sp) (sp) Q2: Total number of non-bonded electrons present in ???? ?? - ion based on Lewis theory is ____ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: NO 2 - ion Page 2 JEE Main Previous Year Questions (2025): Chemical Bonding and Molecular Structure Q1: The number of molecules/ions that show linear geometry among the following is ____. ???? ?? , ???????? ?? , ???? ?? , ?? ?? - , ???? ?? , ?? ?? ?? , ?????? ?? , ???? ?? + , ?? ?? - , ?? ?? JEE Main 2025 (Online) 22nd January Morning Shift Ans: 6 Solution: Linear species are Cl - Bc - Cl, O = C = O, N = N = N (sp) (sp) (sp) Q2: Total number of non-bonded electrons present in ???? ?? - ion based on Lewis theory is ____ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: NO 2 - ion Structure (Lewis Structure) Nitrogen ( N ) has 2 non-bonding electrons. The total valence electrons of N is 5 . Out of 5,3 electrons are bonding electrons and 2 electrons are non-bonding electrons. Double bonded oxygen has 4 non-bonding electrons. Oxygen valence electrons : 6 Out of 6,2 are bonding electrons are 4 are non-bonding electrons. Negatively charged oxygen has 6 non-bonding electrons. The negative charge is due to the extra electron present. So, the total number of non-bonded electrons present in NO 2 - ion based on Lewis theory is 2(???? N) + 4(???? O) + 6(???? O - ) = 2 + 4 + 6 = 12 Q3: Arrange the following compounds in increasing order of their dipole moment: ?????? , ?? ?? ?? , ???? ?? and ???????? ?? JEE Main 2025 (Online) 22nd January Evening Shift Options: A. HBr < H 2 S < NF 3 < CHCl 3 B. H 2 S < HBr < NF 3 < CHCl 3 C. NF 3 < HBr < H 2 S < CHCl 3 D. CHCl 3 < NF 3 < HBr < H 2 S Ans: C Solution: We need to compare the experimental (or well-established) dipole moments of the given molecules: HBr, H 2 S, NF 3 , and CHCl 3 . 1. NF 3 Structure: Trigonal pyramidal (like NH 3 ), but each bond is more polar toward fluorine. Net dipole moment: Quite small, about 0.23 - 0.24?? (the bond dipoles partly oppose the lone-pair contribution). 2. HBr Structure: Simple diatomic. Dipole moment: About 0.78?? . Page 3 JEE Main Previous Year Questions (2025): Chemical Bonding and Molecular Structure Q1: The number of molecules/ions that show linear geometry among the following is ____. ???? ?? , ???????? ?? , ???? ?? , ?? ?? - , ???? ?? , ?? ?? ?? , ?????? ?? , ???? ?? + , ?? ?? - , ?? ?? JEE Main 2025 (Online) 22nd January Morning Shift Ans: 6 Solution: Linear species are Cl - Bc - Cl, O = C = O, N = N = N (sp) (sp) (sp) Q2: Total number of non-bonded electrons present in ???? ?? - ion based on Lewis theory is ____ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: NO 2 - ion Structure (Lewis Structure) Nitrogen ( N ) has 2 non-bonding electrons. The total valence electrons of N is 5 . Out of 5,3 electrons are bonding electrons and 2 electrons are non-bonding electrons. Double bonded oxygen has 4 non-bonding electrons. Oxygen valence electrons : 6 Out of 6,2 are bonding electrons are 4 are non-bonding electrons. Negatively charged oxygen has 6 non-bonding electrons. The negative charge is due to the extra electron present. So, the total number of non-bonded electrons present in NO 2 - ion based on Lewis theory is 2(???? N) + 4(???? O) + 6(???? O - ) = 2 + 4 + 6 = 12 Q3: Arrange the following compounds in increasing order of their dipole moment: ?????? , ?? ?? ?? , ???? ?? and ???????? ?? JEE Main 2025 (Online) 22nd January Evening Shift Options: A. HBr < H 2 S < NF 3 < CHCl 3 B. H 2 S < HBr < NF 3 < CHCl 3 C. NF 3 < HBr < H 2 S < CHCl 3 D. CHCl 3 < NF 3 < HBr < H 2 S Ans: C Solution: We need to compare the experimental (or well-established) dipole moments of the given molecules: HBr, H 2 S, NF 3 , and CHCl 3 . 1. NF 3 Structure: Trigonal pyramidal (like NH 3 ), but each bond is more polar toward fluorine. Net dipole moment: Quite small, about 0.23 - 0.24?? (the bond dipoles partly oppose the lone-pair contribution). 2. HBr Structure: Simple diatomic. Dipole moment: About 0.78?? . 3. ?? ?? ?? Structure: Bent (like H 2 O ) but S is less electronegative than O , and the S - H bond angle is wider than the O - H angle in water. Dipole moment: About 0.95?? . 4. ???????? ?? (chloroform) Structure: Tetrahedral around carbon with three Cl and one H . Dipole moment: About 1.01?? . Ordering from smallest to largest dipole moment NF 3 < HBr < H 2 S < CHCl 3 . Hence, the correct choice is: Option (C) NF 3 < HBr < H 2 S < CHCl 3 . Q4: Match the List - I with List - II List - I (Classification of molecules based on octet rule) List - II (Example) (A) Molecules obeying octet rule (I) ???? , ???? ?? (B) Molecules with incomplete octet (II) ?????? ?? , ???????? ?? (C) Molecules with incomplete octet with odd electron (III) ?? ?? ???? ?? , ?????? ?? (D) Molecules with expanded octet (IV) ?????? ?? , ???? ?? Choose the correct answer from the options given below: JEE Main 2025 (Online) 23rd January Morning Shift Options: A. A-IV, B-I, C-III, D-II B. A-II, B-IV, C-III, D-I C. A-IV, B-II, C-I, D-III D. A-III, B-II, C-I, D-IV Ans: C Solution: (A) A ? IV (B) B ? II (C) C ? I (D) D ? III Page 4 JEE Main Previous Year Questions (2025): Chemical Bonding and Molecular Structure Q1: The number of molecules/ions that show linear geometry among the following is ____. ???? ?? , ???????? ?? , ???? ?? , ?? ?? - , ???? ?? , ?? ?? ?? , ?????? ?? , ???? ?? + , ?? ?? - , ?? ?? JEE Main 2025 (Online) 22nd January Morning Shift Ans: 6 Solution: Linear species are Cl - Bc - Cl, O = C = O, N = N = N (sp) (sp) (sp) Q2: Total number of non-bonded electrons present in ???? ?? - ion based on Lewis theory is ____ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: NO 2 - ion Structure (Lewis Structure) Nitrogen ( N ) has 2 non-bonding electrons. The total valence electrons of N is 5 . Out of 5,3 electrons are bonding electrons and 2 electrons are non-bonding electrons. Double bonded oxygen has 4 non-bonding electrons. Oxygen valence electrons : 6 Out of 6,2 are bonding electrons are 4 are non-bonding electrons. Negatively charged oxygen has 6 non-bonding electrons. The negative charge is due to the extra electron present. So, the total number of non-bonded electrons present in NO 2 - ion based on Lewis theory is 2(???? N) + 4(???? O) + 6(???? O - ) = 2 + 4 + 6 = 12 Q3: Arrange the following compounds in increasing order of their dipole moment: ?????? , ?? ?? ?? , ???? ?? and ???????? ?? JEE Main 2025 (Online) 22nd January Evening Shift Options: A. HBr < H 2 S < NF 3 < CHCl 3 B. H 2 S < HBr < NF 3 < CHCl 3 C. NF 3 < HBr < H 2 S < CHCl 3 D. CHCl 3 < NF 3 < HBr < H 2 S Ans: C Solution: We need to compare the experimental (or well-established) dipole moments of the given molecules: HBr, H 2 S, NF 3 , and CHCl 3 . 1. NF 3 Structure: Trigonal pyramidal (like NH 3 ), but each bond is more polar toward fluorine. Net dipole moment: Quite small, about 0.23 - 0.24?? (the bond dipoles partly oppose the lone-pair contribution). 2. HBr Structure: Simple diatomic. Dipole moment: About 0.78?? . 3. ?? ?? ?? Structure: Bent (like H 2 O ) but S is less electronegative than O , and the S - H bond angle is wider than the O - H angle in water. Dipole moment: About 0.95?? . 4. ???????? ?? (chloroform) Structure: Tetrahedral around carbon with three Cl and one H . Dipole moment: About 1.01?? . Ordering from smallest to largest dipole moment NF 3 < HBr < H 2 S < CHCl 3 . Hence, the correct choice is: Option (C) NF 3 < HBr < H 2 S < CHCl 3 . Q4: Match the List - I with List - II List - I (Classification of molecules based on octet rule) List - II (Example) (A) Molecules obeying octet rule (I) ???? , ???? ?? (B) Molecules with incomplete octet (II) ?????? ?? , ???????? ?? (C) Molecules with incomplete octet with odd electron (III) ?? ?? ???? ?? , ?????? ?? (D) Molecules with expanded octet (IV) ?????? ?? , ???? ?? Choose the correct answer from the options given below: JEE Main 2025 (Online) 23rd January Morning Shift Options: A. A-IV, B-I, C-III, D-II B. A-II, B-IV, C-III, D-I C. A-IV, B-II, C-I, D-III D. A-III, B-II, C-I, D-IV Ans: C Solution: (A) A ? IV (B) B ? II (C) C ? I (D) D ? III Q5: Which of the following statement is true with respect to ?? ?? ?? , ???? ?? and ???? ?? ? A. The central atoms of all the molecules are ???? ?? hybridized. B. The ?? - ?? - ?? , ?? - ?? - ?? and ?? - ?? - ?? angles in the above molecules are ?????? . ?? ° , ?????? . ?? ° and ?????? . ?? ° , respectively. C. The increasing order of dipole moment is ???? ?? < ???? ?? < ?? ?? ?? . D. Both ?? ?? ?? and ???? ?? are Lewis acids and ???? ?? is a Lewis base. E. A solution of ???? ?? in ?? ?? ?? is basic. In this solution ?? ?? ?? and ?? ?? ?? act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below: JEE Main 2025 (Online) 24th January Morning Shift Options: A. A, D and E Only B. C, D and E Only C. A, B and C Only D. A, B, C and E Only Solution: Dipole moment H 2 O > NH 3 > CH 4 H 2 O & NH 3 are Lewis Bases NH 3 act as Lowry- Bronsted base Hence, A, B&C are correct Q6: Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in ?? -direction] ? A. ?? ?? ?? and ?? ?? ?? B. ?? ?? and ?? ?? ?? C. ?? ?? ???? and ?? ?? ?? ?? -?? ?? Page 5 JEE Main Previous Year Questions (2025): Chemical Bonding and Molecular Structure Q1: The number of molecules/ions that show linear geometry among the following is ____. ???? ?? , ???????? ?? , ???? ?? , ?? ?? - , ???? ?? , ?? ?? ?? , ?????? ?? , ???? ?? + , ?? ?? - , ?? ?? JEE Main 2025 (Online) 22nd January Morning Shift Ans: 6 Solution: Linear species are Cl - Bc - Cl, O = C = O, N = N = N (sp) (sp) (sp) Q2: Total number of non-bonded electrons present in ???? ?? - ion based on Lewis theory is ____ . JEE Main 2025 (Online) 29th January Evening Shift Ans: 12 Solution: NO 2 - ion Structure (Lewis Structure) Nitrogen ( N ) has 2 non-bonding electrons. The total valence electrons of N is 5 . Out of 5,3 electrons are bonding electrons and 2 electrons are non-bonding electrons. Double bonded oxygen has 4 non-bonding electrons. Oxygen valence electrons : 6 Out of 6,2 are bonding electrons are 4 are non-bonding electrons. Negatively charged oxygen has 6 non-bonding electrons. The negative charge is due to the extra electron present. So, the total number of non-bonded electrons present in NO 2 - ion based on Lewis theory is 2(???? N) + 4(???? O) + 6(???? O - ) = 2 + 4 + 6 = 12 Q3: Arrange the following compounds in increasing order of their dipole moment: ?????? , ?? ?? ?? , ???? ?? and ???????? ?? JEE Main 2025 (Online) 22nd January Evening Shift Options: A. HBr < H 2 S < NF 3 < CHCl 3 B. H 2 S < HBr < NF 3 < CHCl 3 C. NF 3 < HBr < H 2 S < CHCl 3 D. CHCl 3 < NF 3 < HBr < H 2 S Ans: C Solution: We need to compare the experimental (or well-established) dipole moments of the given molecules: HBr, H 2 S, NF 3 , and CHCl 3 . 1. NF 3 Structure: Trigonal pyramidal (like NH 3 ), but each bond is more polar toward fluorine. Net dipole moment: Quite small, about 0.23 - 0.24?? (the bond dipoles partly oppose the lone-pair contribution). 2. HBr Structure: Simple diatomic. Dipole moment: About 0.78?? . 3. ?? ?? ?? Structure: Bent (like H 2 O ) but S is less electronegative than O , and the S - H bond angle is wider than the O - H angle in water. Dipole moment: About 0.95?? . 4. ???????? ?? (chloroform) Structure: Tetrahedral around carbon with three Cl and one H . Dipole moment: About 1.01?? . Ordering from smallest to largest dipole moment NF 3 < HBr < H 2 S < CHCl 3 . Hence, the correct choice is: Option (C) NF 3 < HBr < H 2 S < CHCl 3 . Q4: Match the List - I with List - II List - I (Classification of molecules based on octet rule) List - II (Example) (A) Molecules obeying octet rule (I) ???? , ???? ?? (B) Molecules with incomplete octet (II) ?????? ?? , ???????? ?? (C) Molecules with incomplete octet with odd electron (III) ?? ?? ???? ?? , ?????? ?? (D) Molecules with expanded octet (IV) ?????? ?? , ???? ?? Choose the correct answer from the options given below: JEE Main 2025 (Online) 23rd January Morning Shift Options: A. A-IV, B-I, C-III, D-II B. A-II, B-IV, C-III, D-I C. A-IV, B-II, C-I, D-III D. A-III, B-II, C-I, D-IV Ans: C Solution: (A) A ? IV (B) B ? II (C) C ? I (D) D ? III Q5: Which of the following statement is true with respect to ?? ?? ?? , ???? ?? and ???? ?? ? A. The central atoms of all the molecules are ???? ?? hybridized. B. The ?? - ?? - ?? , ?? - ?? - ?? and ?? - ?? - ?? angles in the above molecules are ?????? . ?? ° , ?????? . ?? ° and ?????? . ?? ° , respectively. C. The increasing order of dipole moment is ???? ?? < ???? ?? < ?? ?? ?? . D. Both ?? ?? ?? and ???? ?? are Lewis acids and ???? ?? is a Lewis base. E. A solution of ???? ?? in ?? ?? ?? is basic. In this solution ?? ?? ?? and ?? ?? ?? act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below: JEE Main 2025 (Online) 24th January Morning Shift Options: A. A, D and E Only B. C, D and E Only C. A, B and C Only D. A, B, C and E Only Solution: Dipole moment H 2 O > NH 3 > CH 4 H 2 O & NH 3 are Lewis Bases NH 3 act as Lowry- Bronsted base Hence, A, B&C are correct Q6: Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in ?? -direction] ? A. ?? ?? ?? and ?? ?? ?? B. ?? ?? and ?? ?? ?? C. ?? ?? ???? and ?? ?? ?? ?? -?? ?? D. 2 s and ?? ?? ?? E. ?? ?? ?? and ?? ?? ?? ?? -?? ?? Choose the correct answer from the options given below: JEE Main 2025 (Online) 24th January Morning Shift Options: A. D Only B. E Only C. C and D Only D. A and B Only Ans: A Solution:Read More
FAQs on JEE Main Previous Year Questions (2026): Chemical Bonding and Molecular Structure
| 1. What is the significance of hybridization in chemical bonding? |  |
Ans.Hybridization is a concept used to explain the geometry of molecular structures formed by the combination of atomic orbitals. It helps in predicting the shapes of molecules based on the arrangement of electrons. For example, in methane (CH₄), the carbon atom undergoes sp³ hybridization, resulting in a tetrahedral shape. This concept also aids in understanding bond angles and the type of bonds (single, double, or triple) formed between atoms.
| 2. How does the VSEPR theory predict the shape of molecules? | |
Ans.The Valence Shell Electron Pair Repulsion (VSEPR) theory states that the shape of a molecule is determined by the repulsion between the pairs of electrons in the valence shell of the central atom. The theory assumes that electron pairs, whether bonding or lone pairs, will arrange themselves as far apart as possible to minimize repulsion. For instance, in water (H₂O), the two hydrogen atoms and the lone pairs on oxygen create a bent shape due to this repulsion.
| 3. What are the different types of chemical bonds, and how do they differ? | |
Ans.Chemical bonds primarily consist of ionic, covalent, and metallic bonds. Ionic bonds occur when electrons are transferred from one atom to another, resulting in the formation of charged ions, as seen in sodium chloride (NaCl). Covalent bonds involve the sharing of electrons between atoms, as demonstrated in molecules like H₂O and O₂. In contrast, metallic bonds involve a sea of delocalized electrons around positively charged metal ions, which allows for conductivity and malleability in metals.
| 4. What is the role of electronegativity in determining bond polarity? | |
Ans.Electronegativity is the ability of an atom to attract electrons in a chemical bond. The difference in electronegativity between two bonded atoms determines the bond's polarity. If the difference is significant (greater than 1.7), the bond is typically ionic; if it is moderate (between 0.4 and 1.7), the bond is polar covalent; and if it is small (less than 0.4), the bond is considered nonpolar covalent. For example, in HCl, chlorine is more electronegative than hydrogen, leading to a polar covalent bond.
| 5. How does resonance influence the stability of molecules? | |
Ans.Resonance refers to the phenomenon where a molecule can be represented by two or more valid Lewis structures that contribute to its overall structure. These structures, known as resonance structures, differ only in the placement of electrons, not in the arrangement of atoms. Resonance enhances the stability of a molecule by delocalizing electrons over multiple bonds, which lowers the energy of the molecule. For instance, benzene (C₆H₆) exhibits resonance between its different structures, resulting in equal bond lengths and increased stability.
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