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Class 11 Chemistry (Part-1) Chapter 4: Chemical Bonding and Molecular Structure has introduced you to learn how atoms combine or bond, the different bonding theories, their types, and the underlying principles that govern the formation of chemical compounds.
This blog will provide you with exercises along with their solutions, which will help you understand the concepts more simply.
Contents
Explore Notes of Class 11 Chemistry
NCERT Solutions Class 11 Chemistry (Part-1) Chapter 4: Chemical Bonding and Molecular Structure
Below, we have provided you with exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 4: Chemical Bonding and Molecular Structure.
Exercises
- Explain the formation of a chemical bond.
- Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
- Write Lewis symbols for the following atoms and ions: S and S²⁻; Al and Al³⁺; H and H⁻
- Draw the Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₃²⁻, HCOOH
- Define the octet rule. Write its significance and limitations.
- Write the favourable factors for the formation of an ionic bond.
- Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃
- Although the geometries of NH₃ and H₂O molecules are distorted tetrahedral, the bond angle in water is less than that of ammonia. Discuss.
- How do you express the bond strength in terms of bond order?
- Define the bond length.
- Explain the important aspects of resonance with reference to the CO₃²⁻ ion.
- H₃PO₃ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.
- Write the resonance structures for SO₃, NO₂ and NO₃⁻.
- Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S, (b) Ca and O, (c) Al and N.
- Although both CO₂ and H₂O are triatomic molecules, the shape of the H₂O molecule is bent, while that of CO₂ is linear. Explain this on the basis of the dipole moment.
- Write the significance/applications of the dipole moment.
- Define electronegativity. How does it differ from electron gain enthalpy?
- Explain with the help of a suitable example polar covalent bond.
- Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂, and ClF₃.
- The skeletal structure of CH₃COOH, as shown below, is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
- Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH₄ is not square planar?
- Explain why the BeH₂ molecule has a zero dipole moment, although the Be–H bonds are polar.
- Which out of NH₃ and NF₃ has a higher dipole moment and why?
- What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
- Describe the change in hybridisation (if any) of the Al atom in the following reaction: AlCl₃ + Cl⁻ → AlCl₄⁻
- Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
- What is the total number of sigma and pi bonds in the following molecules?
- Considering x-axis as the internuclear axis, which of the following will not form a sigma bond and why?
(a) 1s and 1s (b) 1s and 2pₓ (c) 2pᵧ and 2pᵧ (d) 1s and 2s.
- Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) C₂H₂ (b) C₂H₄ (c) CH₃–CH₃ (d) CH₃–CH=CH₂ (e) CH₃–CH₂–OH (f) CH₃–CHO (g) CH₃COOH
- What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
- Distinguish between a sigma and a pi bond.
- Explain the formation of the H₂ molecule on the basis of valence bond theory.
- Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
- Use molecular orbital theory to explain why the Be₂ molecule does not exist.
- Compare the relative stability of the following species and indicate their magnetic properties: O₂²⁻ (peroxide), O₂⁻ (superoxide), O₂
- Write the significance of a plus and a minus sign shown in representing the orbitals.
- Describe the hybridisation in the case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
- Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
- What is meant by the term bond order? Calculate the bond order of: N₂, O₂, O₂⁺ , and O₂⁻.
Solutions
- Atoms combine to complete their octet and achieve a stable electronic configuration. This leads to formation of a chemical bond, either by transfer (ionic bond) or sharing (covalent bond) of electrons. [p. 99]
- Lewis dot symbols for: Mg, Na, B, O, N, Br
- Mg: •
- Na: •
- B: •••
- O: •• ••
- N: •• •• •
- Br: •• •• •• •• ••
- Lewis symbols for:
- S: •• •• •• ••
- S²⁻: •• •• •• •• (2 extra electrons shown in square brackets with 2⁻ charge)
- Al: •
- Al³⁺: [Al]³⁺
- H: •
- H⁻: [H]⁻
- Lewis structures for:
- H₂S: H–S–H (S has two lone pairs)
- SiCl₄: Cl–Si–Cl (tetrahedral, each Cl with 3 lone pairs)
- BeF₂: F–Be–F (linear)
- CO₃²⁻: Resonance structures with double bond rotating among 3 O atoms
- HCOOH: H–C(=O)–OH
- Definition: Atoms tend to gain, lose or share electrons to complete 8 electrons in their valence shell.
Significance: Explains stability of noble gases and bond formation.
Limitations: Incomplete octet (e.g. BeCl₂), expanded octet (e.g. PCl₅), odd-electron molecules (e.g. NO). - BeCl₂: Linear
BCl₃: Trigonal planar
SiCl₄: Tetrahedral
AsF₅: Trigonal bipyramidal
H₂S: Bent (angular)
PH₃: Trigonal pyramidal
- Due to more lone pairs in H₂O (2 lone pairs) than NH₃ (1 lone pair), which causes greater repulsion and reduces the bond angle.
- Bond order is directly proportional to bond strength. Higher bond order → stronger bond.
- Distance between the nuclei of two bonded atoms. Depends on the size of the atoms and the bond order.
- CO₃²⁻ shows resonance with 3 equivalent Lewis structures where the double bond is delocalised among the 3 oxygen atoms.
- No. Structures are not canonical forms because the position of atoms changes, not just electrons.
- SO₃: 3 equivalent structures with a double bond rotating
NO₂: 2 resonance forms
NO₃⁻: 3 equivalent structures
- (a) K and S: K → S → K⁺ + S²⁻
(b) Ca and O: Ca → O → Ca²⁺ + O²⁻
(c) Al and N: Al → N → Al³⁺ + N³⁻
- H₂O is bent due to lone pairs → net dipole moment
CO₂ is linear → dipoles cancel → zero dipole moment - Applications of dipole moment
- Predict molecular shape
- Distinguish polar and non-polar molecules
- Measure bond polarity
- Electronegativity: Atom’s ability to attract shared electrons (relative scale)
Electron gain enthalpy: Energy change when an electron is added to an atom (absolute value)
- A polar covalent bond is formed when atoms share electrons unequally, e.g., HCl
N₂ < SO₂ < ClF₃ < K₂O < LiF - H–C–C(=O)–OH (Each atom satisfies valency)
- CH₄ is not square planar
- Due to sp³ hybridisation, which leads to a tetrahedral geometry. Square planar has 90° angles → more repulsion → unstable.
- Linear shape causes bond dipoles to cancel out.
- H₃ has a higher dipole moment than NF₃
- In NH₃, lone pair and bond dipoles add up; in NF₃, lone pair and bond dipoles oppose each other.
- sp: Linear (e.g., BeCl₂)
sp²: Trigonal planar (e.g., BCl₃)
sp³: Tetrahedral (e.g., CH₄)
- AlCl₃ + Cl⁻ → AlCl₄⁻
Al changes from sp² to sp³ hybridisation
- Double/triple bond in:
- C₂H₄ (ethene): sp²–sp² sigma + π bond
- C₂H₂ (ethyne): sp–sp sigma + 2 π bonds
- Total sigma and pi bonds:
- Count all single bonds as σ
- Each double bond = 1 σ + 1 π
- Each triple bond = 1 σ + 2 π
- Orbitals that do NOT form σ bond: (c) 2pᵧ and 2pᵧ → form π bond, not σ
- Hybrid orbitals used:
(a) C₂H₂ → sp
(b) C₂H₄ → sp²
(c) CH₃–CH₃ → sp³
(d) CH₃–CH=CH₂ → mix of sp³ and sp²
(e) CH₃CH₂OH → C: sp³, O: sp³
(f) CH₃–CHO → C1: sp³, C2: sp²
(g) CH₃COOH → C1: sp³, C2: sp², O: sp²
- Bond pair: Shared electrons (e.g., H–O–H)
Lone pair: Non-bonding (e.g., two on O in H₂O
- Sigma vs Pi bond:
Sigma (σ): Head-on overlap
Pi (π): Sideways overlap
- Overlap of 1s orbitals from two H atoms; spin pairing occurs; bond formed due to lowered energy.
- Conditions for LCAO
- Same symmetry
- Comparable energy
- Significant overlap
- Be₂ does not exist; MO diagram shows bond order = 0 → unstable → doesn’t exist
- Stability and magnetic properties:
- O₂: b.o. = 2, paramagnetic
- O₂⁻: b.o. = 1.5, paramagnetic
- O₂²⁻: b.o. = 1, diamagnetic
- Indicate sign of the wave function (not charge) → important in bonding/antibonding combinations
- sp³d → trigonal bipyramidal
Axial bonds longer due to more repulsion - Electrostatic attraction involving H bonded to N, O, or F
Stronger than van der Waals, but weaker than covalent - Bond order:
Use formula: (Nb – Na)/2
- N₂: (10–4)/2 = 3
- O₂: (10–6)/2 = 2
- O₂⁺: (10–5)/2 = 2.5
- O₂⁻: (10–7)/2 = 1.5
Download NCERT Solutions Class 11 Chemistry (Part-1) Chapter 4: Chemical Bonding and Molecular Structure
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