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NEET UG Unit 2A: Chemical Bonding, Lewis Diagrams & Molecular Geometry — practice questions

Explore 32 free Unit 2A: Chemical Bonding, Lewis Diagrams & Molecular Geometry questions in NEET UG Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 2A: Chemical Bonding, Lewis Diagrams & Molecular Geometry questions and exercises with answers

Showing questions 1–30 of 32.

The number of valence electrons in the Lewis dot symbol of nitrogen is:Read the answer and worked solution →In the Lewis structure of CO₃²⁻, the total number of valence electrons to be accounted for is:Read the answer and worked solution →For the NH₄⁺ ion, the total valence electrons available for writing the Lewis structure is:Read the answer and worked solution →In a Lewis structure, the formal charge on an atom is calculated as:Read the answer and worked solution →Which of the following compounds is an example of *incomplete* octet of the central atom?Read the answer and worked solution →Which of the following is most likely to have the greatest covalent character (Fajans' rules)?Read the answer and worked solution →Among H–H, F–F, Cl–Cl, Br–Br, I–I, the bond length is shortest for:Read the answer and worked solution →The average O–H bond enthalpy in water is calculated from the two stepwise dissociation values (502 and 427 kJ mol⁻¹) as:Read the answer and worked solution →Among the following, which molecule has zero net dipole moment?Read the answer and worked solution →One Debye unit (1 D) is equivalent to:Read the answer and worked solution →According to VSEPR theory, the order of repulsion between electron pairs is:Read the answer and worked solution →The shape of a molecule of type AB₄ (no lone pair on A), e.g., CH₄, is:Read the answer and worked solution →The shape of SF₄ is:Read the answer and worked solution →The shape of ClF₃ is:Read the answer and worked solution →XeF₄ adopts which geometry?Read the answer and worked solution →Which of the following overlaps is positive (gives bonding)?Read the answer and worked solution →A sigma (σ) bond is formed by:Read the answer and worked solution →In sp hybridisation, each hybrid orbital has the following s and p character:Read the answer and worked solution →In BeCl₂ (gaseous), the central Be is sp-hybridised. The molecular geometry and Cl–Be–Cl bond angle are:Read the answer and worked solution →The hybridisation of the central atom and shape of NH₃ are:Read the answer and worked solution →Hybridisation of the central atom in PCl₅ is:Read the answer and worked solution →The hybridisation of S in SF₆ is:Read the answer and worked solution →The hybridisation of each carbon in ethyne (C₂H₂) is:Read the answer and worked solution →According to MO theory, the bond order in a molecule is:Read the answer and worked solution →The paramagnetism of O₂ is explained by MO theory because the molecule has:Read the answer and worked solution →The bond order in N₂, O₂ and O₂²⁻ (peroxide) are respectively:Read the answer and worked solution →A hydrogen bond is formed when an H atom is covalently bonded to:Read the answer and worked solution →Which of the following is an example of *intramolecular* hydrogen bonding?Read the answer and worked solution →Isoelectronic species N₂, CO and NO⁺ all have a bond order of:Read the answer and worked solution →The order of increasing polarity in HCl, $\mathrm{CO}_{2}$ , $\mathrm{H}_{2}\mathrm{O}$ and HF molecules isRead the answer and worked solution →

Worked practice questions

The number of valence electrons in the Lewis dot symbol of nitrogen is:

  1. 4
  2. 5
  3. 3
  4. 7
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Correct answer: B. 5

Nitrogen (group 15) has electronic configuration 2s² 2p³ — **5 valence electrons** — represented as five dots around the symbol N.

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In the Lewis structure of CO₃²⁻, the total number of valence electrons to be accounted for is:

  1. 26
  2. 24
  3. 22
  4. 28
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Correct answer: B. 24

C contributes 4 + 3 × O contributes 6 = 22 from neutral atoms. Add 2 for the −2 charge: **22 + 2 = 24**.

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For the NH₄⁺ ion, the total valence electrons available for writing the Lewis structure is:

  1. 10
  2. 12
  3. 8
  4. 9
Show answer and solution

Correct answer: C. 8

N (5) + 4 × H (1) = 9 from neutral atoms. **Subtract 1 for the +1 charge:** **9 − 1 = 8** electrons (i.e., 4 bonding pairs in the N–H bonds).

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More Chemistry topics

Amines and BiomoleculesChemical KineticsAldehydes, Ketones and Carboxylic AcidsEquilibriumRedox ReactionsHydrocarbonsChemical Bonding and Molecular StructureSolutions