Chemical Bonding and Molecular Structure — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Chemical Bonding and Molecular Structure" — 5 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Chemical Bonding and Molecular Structure" — 5 important questions with detailed answ…
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NCERT Solutions for Class 11 Chemistry — Chemical Bonding and Molecular Structure. Step-by-step answers to all 5 textbook questions from this chapter, written for CBSE board preparation and free to use.
Chemical Bonding and Molecular Structure — All 5 Questions Solved
Q1. What is a covalent bond? Explain the difference between nonpolar and polar covalent bonds with examples.
Covalent Bond Definition:
A covalent bond is a chemical bond formed by the sharing of electrons between two atoms.
In a covalent bond:
→ Valence electrons of atoms are shared in pairs
→ Both atoms are attracted to the shared electron pair
→ Forms between nonmetallic elements (and sometimes metals)
→ Denoted by a line connecting the atoms (e.g., H-H, C=O)
Nonpolar Covalent Bond:
Definition: A covalent bond where the electron pair is shared equally between the two atoms.
Characteristics:
→ Electronegativity difference (Δ EN) ≤ 0.4
→ Electron density is equally distributed
→ Neither atom has a net charge
→ No dipole moment
Examples:
1. H-H (Hydrogen molecule)
Δ EN = 2.20 - 2.20 = 0
Both atoms are identical; electrons shared equally
2. Cl-Cl (Chlorine molecule)
Δ EN = 3.16 - 3.16 = 0
Both atoms are identical; electrons shared equally
3. C-C (Carbon-carbon bond in alkanes)
Δ EN = 2.55 - 2.55 = 0
Both atoms have same electronegativity
4. C-H (Carbon-hydrogen bond)
Δ EN = 2.55 - 2.20 = 0.35 (≤ 0.4)
Nearly equal sharing; only slight difference
Polar Covalent Bond:
Definition: A covalent bond where the electron pair is unequally shared between the two atoms.
Characteristics:
→ Electronegativity difference (Δ EN): 0.4 < Δ EN < 1.7
→ Electron density is shifted toward the more electronegative atom
→ Atoms develop partial charges (δ+ and δ-)
→ Has a dipole moment (μ ≠ 0)
→ More electronegative atom: δ- (partial negative)
→ Less electronegative atom: δ+ (partial positive)
Examples:
1. H-F (Hydrogen fluoride)
Δ EN = 4.0 - 2.20 = 1.80
F is much more electronegative; electron density shifts toward F
H-F becomes Hδ+-Fδ-
2. H-Cl (Hydrogen chloride)
Δ EN = 3.16 - 2.20 = 0.96
Cl is more electronegative; electron density shifts toward Cl
H-Cl becomes Hδ+-Clδ-
3. H-O (Oxygen-hydrogen bond in water)
Δ EN = 3.44 - 2.20 = 1.24
O is more electronegative; electron density shifts toward O
H-O becomes Hδ+-Oδ-
4. C-O (Carbon-oxygen bond)
Δ EN = 3.44 - 2.55 = 0.89
O is more electronegative
Comparison:
Property Nonpolar Covalent Polar Covalent
electronegativity diff Δ EN ≤ 0.4 0.4 < Δ EN < 1.7
Electron sharing Equal Unequal
Partial charges Absent Present (δ+ and δ-)
Dipole moment Zero Nonzero
Examples H₂, Cl₂, C-C, C-H H-F, H-Cl, H-O, C=O
Conclusion: Polar and nonpolar covalent bonds differ in the equality of electron sharing, determined by electronegativity difference between bonded atoms.
Q2. Explain Lewis dot structure. Draw the Lewis dot structure of CO₂ (carbon dioxide).
Lewis Dot Structure Definition:
A Lewis dot structure (Lewis structure or electron dot structure) is a representation of a molecule showing:
→ All valence electrons of the atoms
→ The arrangement of atoms in the molecule
→ Covalent bonds as lines or pairs of dots
→ Lone pairs of electrons as pairs of dots
Purpose:
→ Shows how valence electrons are distributed in a molecule
→ Helps determine bonding and lone pairs
→ Used to predict molecular geometry and polarity
Rules for Drawing Lewis Structures:
1. Count total valence electrons (sum of all atoms' valence electrons)
2. Arrange atoms with the least electronegative as central (except H)
3. Connect atoms with single bonds (1 pair of electrons)
4. Place remaining electrons as lone pairs on outer atoms
5. Form multiple bonds if needed to satisfy octet rule
6. Each atom should achieve octet (2 for H)
Example: Lewis Dot Structure of CO₂
Step 1: Count valence electrons
C has 4 valence electrons
O has 6 valence electrons each, and there are 2 O atoms
Total valence electrons = 4 + 6 + 6 = 16 electrons
Step 2: Arrange atoms
Carbon is the central atom (least electronegative, except H)
Oxygen atoms are on the sides
O - C - O
Step 3: Connect with single bonds
Each C-O bond uses 2 electrons
O - C - O (uses 4 electrons, leaving 16 - 4 = 12 electrons)
Step 4: Distribute remaining electrons on outer atoms
Each O needs 6 valence electrons (octet = 8, but one pair is in bond)
First O: give 3 lone pairs (6 electrons)
Second O: give 3 lone pairs (6 electrons)
Used: 6 + 6 = 12 electrons ✓
Structure now:
··
O = C = O (with lone pairs shown)
Or written more clearly:
:Ö:::C:::Ö:
Step 5: Check octet rule
C has 4 bonds (counts as 8 electrons around C) — octet satisfied ✓
Each O has 2 lone pairs (4 electrons) + 2 from one double bond (4 electrons) = 8 ✓
Final Lewis Structure of CO₂:
••
O = C = O
•• ••
(dots represent lone pairs)
Or with lines:
O = C = O
where: = represents a double bond (2 pairs of electrons)
Key Points:
→ CO₂ has two C=O double bonds (polar bonds)
→ Each O has 2 lone pairs
→ C has no lone pairs
→ Linear molecular geometry (180° bond angle)
→ Although individual C=O bonds are polar, CO₂ molecule is nonpolar (dipoles cancel due to linear geometry)
Q3. What is the octet rule? Explain exceptions to the octet rule with examples.
Octet Rule Definition:
The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable electron configuration with 8 valence electrons (like noble gases).
For hydrogen, the duet rule applies: H tends to achieve 2 electrons (like He).
Explanation:
→ Noble gases (He, Ne, Ar, Kr, etc.) are very stable and unreactive
→ Noble gases have filled valence shells:
- He: 1s² (2 electrons)
- Ne: 1s² 2s² 2p⁶ (8 electrons)
- Ar: [Ne] 3s² 3p⁶ (8 electrons)
→ Other atoms "want" to achieve this stability by bonding
Applications of Octet Rule:
→ Predicts covalent bonding patterns
→ Explains formation of stable molecules (H₂, H₂O, CO₂, NH₃, CH₄)
→ Predicts ionic charges
Examples Satisfying Octet Rule:
1. H₂: Each H needs 2 electrons (duet) — satisfied ✓
2. H₂O: O has 8 electrons (2 lone pairs + 2 bonds); H has 2 each
3. NH₃: N has 8 electrons (1 lone pair + 3 bonds); H has 2 each
4. CH₄: C has 8 electrons (4 bonds); H has 2 each
Exceptions to the Octet Rule:
Exception 1: Atoms with Fewer than 8 Valence Electrons
Example: Boron (B) compounds
→ BF₃ (Boron trifluoride)
→ B forms 3 bonds with F atoms
→ B has only 6 valence electrons (not 8)
→ B remains electron-deficient
→ This is an exception, but BF₃ forms because B can expand its bonding
→ H₂: Each H has only 2 electrons
→ He: Noble gas with only 2 electrons (not 8)
Example: Beryllium (Be) compounds
→ BeH₂ (Beryllium hydride)
→ Be forms 2 bonds with H atoms
→ Be has only 4 valence electrons (not 8)
→ Be is electron-deficient
Exception 2: Atoms with More than 8 Valence Electrons (Expanded Octet)
Example: Phosphorus (P) compounds
→ PCl₅ (Phosphorus pentachloride)
→ P forms 5 bonds (10 electrons around P)
→ P "expands" its octet to 10 electrons
→ Possible because P is in the 3rd period and can use d orbitals
→ PF₆⁻ (Phosphorus hexafluoride ion)
→ P has 6 bonds + 1 lone pair = 12 electrons (expanded octet)
Example: Sulfur (S) compounds
→ SF₆ (Sulfur hexafluoride)
→ S forms 6 bonds (12 electrons around S)
→ Expanded octet to 12 electrons
→ S is in 3rd period, can use d orbitals
→ H₂SO₄ (Sulfuric acid)
→ S has expanded octet with 12 electrons
Example: Chlorine (Cl) compounds
→ ClF₃ (Chlorine trifluoride)
→ Cl has 3 bonds + 2 lone pairs = 10 electrons (expanded octet)
→ ClF₅ (Chlorine pentafluoride)
→ Cl has 5 bonds + 1 lone pair = 12 electrons (expanded octet)
Exception 3: Free Radicals (Odd Electrons)
Example: NO (Nitric oxide)
→ N: 5 valence electrons, O: 6 valence electrons
→ Total: 11 valence electrons (odd number)
→ Cannot form pairs perfectly
→ NO has an unpaired electron (free radical)
→ Does not satisfy octet rule
Example: O₂⁻ (Superoxide ion)
→ Oxygen-containing species with unpaired electrons
Exception 4: Molecules with Incomplete Octets (Electron Deficiency)
Example: BH₃ (Borane)
→ B has only 6 electrons
→ Actual form: B₂H₆ (diborane), where B still has incomplete octet
Why Exceptions Occur:
1. Expanded Octet (Elements in 3rd period and beyond):
→ Elements in period 3+ can use d orbitals
→ d orbitals allow accommodation of >8 electrons
→ P, S, Cl, etc., can exceed octet
2. Electron Deficiency (Light elements):
→ B, Be cannot achieve octet (valence orbitals too small)
→ Form compounds with <8 electrons
3. Odd Electrons:
→ Odd total valence electrons prevent pairing for all atoms
→ Free radicals form
Summary of Exceptions:
1. Expanded Octet: PCl₅, SF₆, ClF₅, H₂SO₄ (>8 electrons)
2. Incomplete Octet: BF₃, BeH₂, BH₃ (<8 electrons)
3. Odd Electrons: NO, NO₂ (free radicals)
4. Duet Rule: H₂, HF (2 electrons for H is sufficient)
Conclusion: While the octet rule is very useful for predicting bonding in main group elements, exceptions exist for electron-deficient compounds and elements that can expand their octets.
Q4. Define electronegativity. Explain how electronegativity differences determine bond type (ionic, polar covalent, nonpolar covalent).
Electronegativity Definition:
Electronegativity is the relative tendency of an atom to attract the electron pair towards itself in a covalent bond.
It is a measure of the "pulling power" of an atom for electrons in a bond.
Key Points:
→ Electronegativity is a property of atoms in bonds (not isolated atoms)
→ Higher electronegativity = stronger attraction for bonding electrons
→ Measured on the Pauling scale (0 to 4)
→ Fluorine has the highest electronegativity (4.0)
Pauling Scale Values (Common Elements):
F = 4.0 (highest)
O = 3.44
N = 3.0
Cl = 3.16
Br = 2.96
I = 2.66
C = 2.55
H = 2.20
B = 2.04
Be = 1.57
Mg = 1.31
Na = 0.93
Cs = 0.79 (lowest for reactive elements)
How Electronegativity Differences Determine Bond Type:
Calculate: Δ EN = |EN(more electronegative atom) - EN(less electronegative atom)|
Then classify based on Δ EN value:
Bond Type 1: IONIC BONDS
Criterion: Δ EN ≥ 1.7
Characteristics:
→ Large electronegativity difference
→ More electronegative atom pulls electrons completely
→ Electron transfer occurs (not sharing)
→ Forms cation (positive ion) and anion (negative ion)
→ Electrostatic attraction between ions
→ Found between metals and nonmetals
→ Solid form at room temperature
→ Conduct electricity when molten or dissolved
→ High melting and boiling points
→ Usually soluble in polar solvents (water)
Examples:
1. NaCl: Δ EN = |0.93 - 3.16| = 2.23 ≥ 1.7 → IONIC ✓
Na loses electron completely to Cl⁻
2. MgO: Δ EN = |1.31 - 3.44| = 2.13 ≥ 1.7 → IONIC ✓
Mg²⁺ and O²⁻ ions
3. CaCl₂: Δ EN = |1.00 - 3.16| = 2.16 ≥ 1.7 → IONIC ✓
4. NaF: Δ EN = |0.93 - 4.0| = 3.07 ≥ 1.7 → IONIC ✓
Very large difference, very ionic
5. KBr: Δ EN = |0.82 - 2.96| = 2.14 ≥ 1.7 → IONIC ✓
Bond Type 2: POLAR COVALENT BONDS
Criterion: 0.4 < Δ EN < 1.7
Characteristics:
→ Moderate electronegativity difference
→ Electrons are shared, but unequally
→ More electronegative atom attracts electron density more
→ Develops partial charges: δ+ and δ-
→ Has a dipole moment (μ ≠ 0)
→ Found between nonmetals of different electronegativities
→ Often gases or liquids at room temperature
→ Polar molecules have higher boiling points than nonpolar
→ Many are soluble in polar solvents (water)
Examples:
1. HCl: Δ EN = |2.20 - 3.16| = 0.96, where 0.4 < 0.96 < 1.7 → POLAR COVALENT ✓
Hδ+-Clδ- (H is slightly positive, Cl is slightly negative)
2. HF: Δ EN = |2.20 - 4.0| = 1.80 ≥ 1.7 → borderline IONIC/POLAR
Actually considered polar covalent (some texts say borderline)
H is significantly positive, F is significantly negative
3. H₂O: Δ EN = |2.20 - 3.44| = 1.24, where 0.4 < 1.24 < 1.7 → POLAR COVALENT ✓
Hδ+-Oδ- (both O-H bonds are polar)
Molecule is polar overall
4. NH₃: Δ EN = |2.20 - 3.0| = 0.80, where 0.4 < 0.80 < 1.7 → POLAR COVALENT ✓
N-H bonds are polar
Molecule is polar (due to lone pair on N)
5. CO: Δ EN = |2.55 - 3.44| = 0.89, where 0.4 < 0.89 < 1.7 → POLAR COVALENT ✓
6. C-O (in C=O): Δ EN = 0.89 → POLAR COVALENT ✓
Bond Type 3: NONPOLAR COVALENT BONDS
Criterion: Δ EN ≤ 0.4
Characteristics:
→ Small or zero electronegativity difference
→ Electrons are shared equally (or nearly equally)
→ No partial charges (or negligible)
→ No dipole moment (μ = 0)
→ Found between identical atoms or very similar atoms
→ Often gases at room temperature
→ Generally insoluble in polar solvents (water)
→ Often soluble in nonpolar solvents (benzene, CCl₄)
Examples:
1. H₂: Δ EN = |2.20 - 2.20| = 0 ≤ 0.4 → NONPOLAR COVALENT ✓
Identical atoms, electrons shared equally
2. Cl₂: Δ EN = |3.16 - 3.16| = 0 ≤ 0.4 → NONPOLAR COVALENT ✓
Identical atoms
3. C-C: Δ EN = |2.55 - 2.55| = 0 ≤ 0.4 → NONPOLAR COVALENT ✓
Identical atoms
4. C-H: Δ EN = |2.55 - 2.20| = 0.35 ≤ 0.4 → NONPOLAR COVALENT ✓
Very small difference
5. B-C: Δ EN = |2.04 - 2.55| = 0.51 > 0.4 → POLAR COVALENT
(Actually slightly polar, but often treated as nearly nonpolar)
6. CO₂: Contains C=O bonds (polar), but overall molecule is NONPOLAR
(due to linear geometry, dipoles cancel)
Linear: O=C=O — dipoles point in opposite directions and cancel
Comparison Table:
Property Ionic Bond Polar Covalent Nonpolar Covalent
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Δ EN Range ≥ 1.7 0.4 < Δ EN < 1.7 Δ EN ≤ 0.4
Electron interaction Transfer Unequal sharing Equal sharing
Partial charges Complete charges δ+ and δ- No charges
Dipole moment Yes (ionic) Yes (μ ≠ 0) No (μ = 0)
State (25°C) Usually solid Gas/liquid Gas/liquid
Melting point High Low to moderate Low
Solubility in water Usually high Variable Usually low
Electrical property Conducts when Nonconductor Nonconductor
molten/dissolved (in pure form) (in pure form)
Bond strength Electrostatic Covalent Covalent
Atoms involved Metal + nonmetal Nonmetal + nonmetal Identical/similar nonmetals
Conclusion:
Electronegativity difference is the key factor determining bond type:
→ Large Δ EN (≥1.7) → electron transfer → ionic bond
→ Moderate Δ EN (0.4-1.7) → unequal sharing → polar covalent bond
→ Small Δ EN (≤0.4) → equal sharing → nonpolar covalent bond
Q5. Explain molecular geometry and hybridization. Use CH₄ as an example.
Molecular Geometry Definition:
Molecular geometry is the three-dimensional spatial arrangement of atoms in a molecule.
It is determined by:
→ The positions of bonded atoms
→ The positions of lone pairs on the central atom
→ Bond angles between bonds
Note: Molecular geometry differs from electron geometry
→ Electron geometry considers both bonding and lone pairs
→ Molecular geometry considers only the positions of atoms (not lone pairs)
Hybridization Definition:
Hybridization is the mixing of atomic orbitals of similar energy to form new hybrid orbitals suitable for bonding.
Purpose of Hybridization:
→ Explains molecular geometry
→ Explains bond angles
→ Explains bonding in molecules
→ Accounts for equivalent bonds
Types of Hybridization:
1. sp hybridization:
→ One s orbital + one p orbital mix
→ Forms 2 hybrid orbitals (linear)
→ Bond angle = 180°
→ Example: C₂H₂ (acetylene), BeCl₂
2. sp² hybridization:
→ One s orbital + two p orbitals mix
→ Forms 3 hybrid orbitals (trigonal planar)
→ Bond angle ≈ 120°
→ Example: C₂H₄ (ethene), BF₃
3. sp³ hybridization:
→ One s orbital + three p orbitals mix
→ Forms 4 hybrid orbitals (tetrahedral)
→ Bond angle ≈ 109.5°
→ Example: CH₄, NH₃, H₂O, CCl₄
Example: CH₄ (Methane)
Step 1: Determine electron configuration of carbon.
C (atomic number 6): 1s² 2s² 2p²
Valence shell (n=2): 2s² 2p²
Step 2: In ground state (unexcited).
C has 4 valence electrons in s and p orbitals
2s orbital: ↑↓ (2 paired electrons)
2p orbitals: ↑ ↑ _ (2 unpaired electrons)
Orbital notation: ↑↓ | ↑ ↑ _
C can form only 2 bonds in ground state (limited by unpaired electrons)
But CH₄ has 4 bonds → Need to explain this!
Step 3: Promotion and Hybridization.
One electron from 2s orbital is promoted to empty 2p orbital (requires energy but compensated by bonding).
Promoted state:
2s orbital: ↑ (1 unpaired electron)
2p orbitals: ↑ ↑ ↑ (3 unpaired electrons)
Orbital notation: ↑ | ↑ ↑ ↑
Now C has 4 unpaired electrons → can form 4 bonds!
Step 4: Hybridization of orbitals.
The one 2s orbital and three 2p orbitals mix (undergo hybridization) to form 4 equivalent sp³ hybrid orbitals.
1 s orbital + 3 p orbitals → 4 sp³ hybrid orbitals (tetrahedral arrangement)
These 4 sp³ hybrid orbitals are equivalent in energy and shape.
Step 5: Bond formation.
Each of the 4 sp³ hybrid orbitals on C overlaps with the 1s orbital of a hydrogen atom.
Forms 4 C-H sigma (σ) bonds.
Bond description:
→ All C-H bonds are equivalent
→ All bond lengths are equal (1.09 Å)
→ All bond angles are equal (109.5°)
Step 6: Molecular geometry of CH₄.
Central atom (C) has:
→ 4 bonding pairs (4 C-H bonds)
→ 0 lone pairs
→ Electron geometry: tetrahedral (based on 4 electron pairs)
→ Molecular geometry: tetrahedral (based on 4 atoms)
Structure representation:
H
|
H-C-H (looks like a flat representation, but actually 3D)
|
H
3D representation (tetrahedral):
The H atoms are arranged at the vertices of a tetrahedron, with C at the center.
Bond angle: H-C-H = 109.5° (tetrahedral angle)
Step 7: Properties explained by hybridization.
1. All C-H bonds are identical:
→ Explained by sp³ hybridization (4 equivalent hybrid orbitals)
→ Each forms identical C-H bond
2. Tetrahedral geometry:
→ sp³ hybrid orbitals arrange tetrahedral in space (minimizes electron repulsion)
→ Maximizes distance between bonding pairs
→ VSEPR theory explains this
3. Bond angle = 109.5°:
→ Tetrahedral geometry
→ This angle minimizes repulsion between electron pairs
4. CH₄ is nonpolar:
→ Tetrahedral symmetry
→ Dipoles cancel out
→ μ = 0
5. Melting point is very low:
→ Nonpolar molecule
→ Only weak van der Waals forces
→ Boiling point = -161.5°C
Summary for CH₄:
Property Value/Description
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Central atom C (carbon)
Valence electrons 4
Hybridization type sp³
Number of hybrid orbitals 4
Number of bonding pairs 4 (C-H bonds)
Number of lone pairs 0
Electron geometry Tetrahedral
Molecular geometry Tetrahedral
Bond angle 109.5°
Bond length 1.09 Å
Molecular polarity Nonpolar
Structure Regular tetrahedron
Comparison with other hybridizations:
Compound Hybridization Bonding pairs Lone pairs Geometry Bond angle
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
C₂H₂ sp 2 0 Linear 180°
C₂H₄ sp² 3 0 Trigonal planar 120°
CH₄ sp³ 4 0 Tetrahedral 109.5°
NH₃ sp³ 3 1 Trigonal pyram. 107°
H₂O sp³ 2 2 Bent 104.5°
Conclusion: Hybridization explains the geometry and bonding properties of CH₄ as a tetrahedral molecule with 4 equivalent C-H bonds and 109.5° bond angles.
Chemical Bonding and Molecular Structure — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following molecules obeys the octet rule completely?
- A. BF3
- B. PCl5
- C. SF6
- D. CCl4
Answer: D. CCl4 — CCl4 has carbon forming four single bonds with chlorine atoms, resulting in 8 electrons around carbon, satisfying the octet rule. BF3 is electron deficient, while PCl5 and SF6 have expanded octets.
MCQ 2. The correct order of bond angles in H2O, NH3, and CH4 is:
- A. CH4 < NH3 < H2O
- B. NH3 < H2O < CH4
- C. H2O < NH3 < CH4
- D. CH4 < H2O < NH3
Answer: C. H2O < NH3 < CH4 — CH4 has sp3 hybridization with 0 lone pairs (109.5°), NH3 has sp3 hybridization with 1 lone pair (107°), and H2O has sp3 hybridization with 2 lone pairs (104.5°). Lone pair-bond pair repulsion decreases the bond angle.
MCQ 3. Which of the following species has a bond order of 2.5 and is paramagnetic?
- A. N2
- B. O2+
- C. C2
- D. B2
Answer: B. O2+ — O2+ has 15 electrons. Its molecular orbital configuration gives a bond order of (10-5)/2 = 2.5. It has an unpaired electron in π* molecular orbital, making it paramagnetic.
MCQ 4. What is the hybridization of carbon in ethyne (C2H2)?
- A. sp
- B. sp2
- C. sp3
- D. d2sp3
Answer: A. sp — Each carbon atom in ethyne forms one sigma bond with hydrogen and one sigma bond and two pi bonds with the other carbon atom. This corresponds to sp hybridization.
MCQ 5. Which of the following molecules has a non-zero dipole moment?
- A. CO2
- B. CCl4
- C. BF3
- D. NH3
Answer: D. NH3 — CO2, CCl4, and BF3 have symmetrical geometries with individual bond dipoles canceling out, resulting in a zero net dipole moment. NH3 has a trigonal pyramidal geometry with a lone pair, leading to a net dipole moment.
MCQ 6. According to Fajan's rules, covalent character is favored by:
- A. Large cation, large anion
- B. Small cation, small anion
- C. Small cation, large anion
- D. Large cation, small anion
Answer: C. Small cation, large anion — A small cation has high polarizing power, and a large anion is easily polarizable. This combination leads to a greater covalent character in an ionic bond.
MCQ 7. The geometry of SF4 molecule is:
- A. Tetrahedral
- B. Square planar
- C. See-saw
- D. Trigonal bipyramidal
Answer: C. See-saw — SF4 has 4 bond pairs and 1 lone pair of electrons around the central sulfur atom. According to VSEPR theory, this corresponds to a trigonal bipyramidal electron geometry but a 'see-saw' molecular geometry.
MCQ 8. Which of the following species cannot exhibit resonance?
- A. CO3^2-
- B. SO3
- C. NO3-
- D. SiO2
Answer: D. SiO2 — Resonance involves delocalization of pi electrons or lone pairs. SiO2 is a network covalent solid with localized single bonds and does not possess delocalized pi systems to exhibit resonance.
MCQ 9. The number of sigma and pi bonds in C2H4 are:
- A. 3 sigma, 2 pi
- B. 5 sigma, 1 pi
- C. 4 sigma, 2 pi
- D. 2 sigma, 2 pi
Answer: B. 5 sigma, 1 pi — In C2H4 (ethene), there are four C-H sigma bonds and one C-C sigma bond, making a total of 5 sigma bonds. The C=C double bond consists of one sigma and one pi bond, so there is 1 pi bond.
MCQ 10. Which of the following bond types has the shortest bond length?
- A. C-C
- B. C=C
- C. C≡C
- D. C-H
Answer: C. C≡C — Bond length decreases with an increase in the number of bonds between two atoms. A triple bond (C≡C) is shorter than a double bond (C=C), which is shorter than a single bond (C-C).
MCQ 11. Which of the following diatomic molecules is paramagnetic?
- A. N2
- B. F2
- C. O2
- D. Li2
Answer: C. O2 — According to MOT, O2 has two unpaired electrons in its π* antibonding molecular orbitals, making it paramagnetic. N2, F2, and Li2 are all diamagnetic as they have all electrons paired.
MCQ 12. How many lone pairs are present on the central atom in ICl4- ?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: C. 2 — In ICl4-, the central iodine atom has 7 valence electrons plus 1 for the negative charge, making 8 electrons. It forms 4 bonds with chlorine atoms, leaving (8 - 4) = 4 non-bonding electrons, which means 2 lone pairs.
MCQ 13. Intermolecular hydrogen bonding is strongest in:
- A. HCl
- B. H2S
- C. HF
- D. CH4
Answer: C. HF — Hydrogen bonding occurs when hydrogen is bonded to a highly electronegative atom like F, O, or N. Fluorine is the most electronegative element, leading to the strongest intermolecular hydrogen bonding in HF.
MCQ 14. What is the formal charge on the central sulfur atom in SO4^2- ion?
- A. 0
- B. +1
- C. +2
- D. -2
Answer: A. 0 — In the most stable Lewis structure of SO4^2- (with an expanded octet for S), sulfur forms two double bonds and two single bonds with oxygen atoms. The formal charge for S = [6 (valence e-) - 0 (lone pair e-) - 6 (bonding e- / 2)] = 0.
MCQ 15. The ideal bond angle in a molecule with sp3 hybridization is:
- A. 180°
- B. 120°
- C. 109.5°
- D. 90°
Answer: C. 109.5° — Molecules with sp3 hybridization and no lone pairs, like methane (CH4), adopt a tetrahedral geometry with an ideal bond angle of 109.5°.
MCQ 16. Which of the following molecules is linear?
- A. H2O
- B. SO2
- C. CO2
- D. CH4
Answer: C. CO2 — CO2 has sp hybridization on the central carbon atom and no lone pairs, resulting in a linear geometry. H2O and SO2 are bent, and CH4 is tetrahedral.
MCQ 17. Arrange the following species in increasing order of their stability: O2+, O2, O2-, O2^2-
- A. O2^2- < O2- < O2 < O2+
- B. O2+ < O2 < O2- < O2^2-
- C. O2- < O2^2- < O2 < O2+
- D. O2^2- < O2 < O2- < O2+
Answer: A. O2^2- < O2- < O2 < O2+ — Stability is directly proportional to bond order. Bond orders are: O2+ (2.5), O2 (2.0), O2- (1.5), O2^2- (1.0). Therefore, the increasing order of stability is O2^2- < O2- < O2 < O2+.
MCQ 18. Which of the following bonds has the highest ionic character?
- A. H-F
- B. H-Cl
- C. H-Br
- D. H-I
Answer: A. H-F — Ionic character increases with the difference in electronegativity between the bonded atoms. Fluorine is the most electronegative element, so the H-F bond has the highest electronegativity difference and thus the highest ionic character.
MCQ 19. The number of sp2 hybridized carbon atoms in propene (CH3-CH=CH2) is:
- A. 0
- B. 1
- C. 2
- D. 3
Answer: C. 2 — In propene, the two carbon atoms involved in the double bond (CH=CH2) are sp2 hybridized. The carbon atom in the CH3 group is sp3 hybridized.
MCQ 20. The correct order of electron pair geometry around the central atom for XeF4, ClF3, and SF6 is:
- A. Octahedral, Trigonal Bipyramidal, Octahedral
- B. Octahedral, Trigonal Bipyramidal, Square Planar
- C. Square Planar, T-shape, Octahedral
- D. Octahedral, Trigonal Pyramidal, Octahedral
Answer: A. Octahedral, Trigonal Bipyramidal, Octahedral — XeF4 has 4 bond pairs and 2 lone pairs, totaling 6 electron pairs (octahedral electron geometry). ClF3 has 3 bond pairs and 2 lone pairs, totaling 5 electron pairs (trigonal bipyramidal electron geometry). SF6 has 6 bond pairs and 0 lone pairs, totaling 6 electron pairs (octahedral electron geometry).
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.