This MCQ module is based on: NCERT Exercises and Solutions: Chemical Bonding and Molecular Structure
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NCERT Exercises and Solutions: Chemical Bonding and Molecular Structure
🎓 Class 11
Chemistry
CBSE
Theory
Ch 4 – Chemical Bonding and Molecular Structure
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NCERT Exercises and Solutions: Chemical Bonding and Molecular Structure
Chapter 4 — Summary
Chemical bonding holds atoms together to form molecules and crystals. Three central theories — Kössel–Lewis, Valence Bond Theory and Molecular Orbital Theory — together with VSEPR for shape and the concept of hydrogen bonding, give us the modern picture of molecular structure. Below is a quick tour of the entire chapter at a glance.
Kössel–Lewis
Atoms transfer or share electrons to attain noble-gas octets. Lewis dot symbols summarise valence electrons.Ionic Bond
Electron transfer between low-IE metals and high-EA non-metals → cation + anion → 3-D crystal lattice. Stability measured by lattice enthalpy.Covalent Bond
Electron sharing. Bond parameters: length, angle, enthalpy, order, polarity.Formal Charge
FC = V − N − ½B; helps choose the most plausible Lewis structure.Resonance
Single Lewis structures inadequate (O₃, CO₃²⁻, NO₃⁻, benzene). Real molecule is the resonance hybrid.Polarity
Δ(EN) → polar bond; vector sum gives net molecular μ. Symmetry can cancel μ (CO₂, BF₃, CH₄).VSEPR
Geometry minimises electron-pair repulsion. Order: lp–lp > lp–bp > bp–bp.Valence Bond Theory
Bond forms by overlap of half-filled AOs with opposite spins. σ (axial) > π (lateral) in strength.Hybridisation
sp (linear), sp² (planar), sp³ (tetrahedral), sp³d (TBP), sp³d² (octahedral).Molecular Orbital Theory
LCAO → bonding & antibonding MOs. BO = (Nb − Na)/2. Explains O₂ paramagnetism.Hydrogen Bond
X–H···Y where X = F/O/N. Intra- or intermolecular. Critical for water, ice, DNA, proteins.Bond Order Trends
Removing antibonding e⁻ → stronger bond (O₂⁺ > O₂); adding antibonding e⁻ → weaker (O₂⁻).Key Terms
Lattice Enthalpy, Bond Order, Dipole Moment, Resonance Hybrid, Hybridisation, Molecular Orbital, Hydrogen Bond.
Interactive: Chapter Concept Lookup
Pick a topic and reveal the key relationship/formula instantly.
Formula / Rule: BO = (Nb − Na)/2
Higher BO → shorter, stronger bond.
Activity 4.5 — Chapter Recall Sprint
Setup: Without looking back, list (a) one example each of sp, sp², sp³, sp³d, sp³d² hybridisation; (b) one molecule with zero μ despite polar bonds; (c) one paramagnetic homonuclear diatomic; (d) one species where resonance gives equal bonds.
Predict yourself first: write down your answers, then reveal.
(a) sp — BeCl₂; sp² — BCl₃ or C in C₂H₄; sp³ — CH₄; sp³d — PCl₅; sp³d² — SF₆.
(b) CO₂ (linear), BF₃ (trigonal planar) or CH₄ (tetrahedral) — bond moments cancel.
(c) O₂ (or B₂, NO).
(d) CO₃²⁻ (three equal C–O bonds), NO₃⁻ (three equal N–O), benzene (six equal C–C).
NCERT Exercises (Solved)
4.1 Explain the formation of a chemical bond.
A chemical bond is the attractive force that holds atoms together in a molecule. Atoms combine because the resulting aggregate has lower energy than the isolated atoms; bond formation releases energy. Two main routes: electron transfer → ionic bond (e.g., NaCl) and electron sharing → covalent bond (e.g., H₂, CH₄). Both routes typically lead the atoms to a noble-gas octet (octet rule).
4.2 Write Lewis dot symbols for atoms of: Li, Na, Mg, K, Al, F, S.
Number of dots = number of valence electrons.
Li· , Na· , ·Mg· , K· , Al with 3 dots, F with 7 dots (3 lp + 1 unpaired), S with 6 dots (2 lp + 2 unpaired).
Li· , Na· , ·Mg· , K· , Al with 3 dots, F with 7 dots (3 lp + 1 unpaired), S with 6 dots (2 lp + 2 unpaired).
4.3 Write Lewis symbols for the following atoms and ions: S and S²⁻; Al and Al³⁺; H and H⁻.
S has 6 dots; S²⁻ has 8 dots inside [ ]²⁻. Al has 3 dots; Al³⁺ has none, [Al]³⁺. H has 1 dot; H⁻ has 2 dots inside [ ]⁻.
4.4 Draw Lewis structures for (a) F₂ (b) CO₂ (c) H₂SO₄ (d) HNO₃.
(a) F–F with three lone pairs on each F.
(b) Ö=C=Ö (each O has 2 lone pairs).
(c) H₂SO₄: central S double-bonded to two O atoms, single-bonded to two OH groups; each terminal O has 2 lone pairs, each O of OH has 2 lone pairs.
(d) HNO₃: N at centre; one N=O, one N–O⁻, one N–OH; ionic resonance allows the negative charge to switch between the two non-OH oxygens.
(b) Ö=C=Ö (each O has 2 lone pairs).
(c) H₂SO₄: central S double-bonded to two O atoms, single-bonded to two OH groups; each terminal O has 2 lone pairs, each O of OH has 2 lone pairs.
(d) HNO₃: N at centre; one N=O, one N–O⁻, one N–OH; ionic resonance allows the negative charge to switch between the two non-OH oxygens.
4.5 Define electronegativity. How does it differ from electron gain enthalpy?
Electronegativity is the relative tendency of an atom in a molecule to attract the shared bonding pair towards itself (Pauling scale: F = 3.98). It is a property of an atom in a chemical bond.
Electron gain enthalpy is the actual energy released when a free gaseous atom accepts an electron to form a gaseous anion. Electronegativity is a relative number; electron gain enthalpy is a measurable energy.
Electron gain enthalpy is the actual energy released when a free gaseous atom accepts an electron to form a gaseous anion. Electronegativity is a relative number; electron gain enthalpy is a measurable energy.
4.6 Explain with one example each: (i) Electrovalent (ionic) bond (ii) Covalent bond (iii) Coordinate bond.
(i) Ionic — Na transfers 1 e⁻ to Cl forming Na⁺Cl⁻ (NaCl).
(ii) Covalent — H and H share a pair to form H–H (H₂).
(iii) Coordinate (dative) — both shared electrons come from one atom only, e.g., NH₃ + BF₃ → H₃N→BF₃; or H₃N + H⁺ → NH₄⁺ where the lone pair of N forms the bond.
(ii) Covalent — H and H share a pair to form H–H (H₂).
(iii) Coordinate (dative) — both shared electrons come from one atom only, e.g., NH₃ + BF₃ → H₃N→BF₃; or H₃N + H⁺ → NH₄⁺ where the lone pair of N forms the bond.
4.7 Write Lewis symbols for: Mg²⁺, K⁺, Al³⁺, Cl⁻, O²⁻, N³⁻.
Cations [Mg]²⁺, [K]⁺, [Al]³⁺ — no dots (noble-gas core only). Anions [Cl with 8 dots]⁻, [O with 8 dots]²⁻, [N with 8 dots]³⁻ (each has the Ne or Ar octet).
4.8 Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, the bond angle in water is less than in ammonia. Discuss.
Both N (in NH₃) and O (in H₂O) are sp³ hybridised. NH₃ has 3 bp + 1 lp; H₂O has 2 bp + 2 lp. Two lone pairs on O exert greater lp–lp repulsion than the single lp on N → bonded H atoms in water are pushed closer together. Result: ∠HNH = 107°, ∠HOH = 104.5°.
4.9 How do you express the bond strength in terms of bond order?
Higher bond order → shorter and stronger bond. Examples: H₂ (BO 1, 436 kJ/mol, 74 pm); O₂ (BO 2, 498 kJ/mol, 121 pm); N₂ (BO 3, 946 kJ/mol, 110 pm). For iso-electronic species (N₂, CO, NO⁺) bond orders and energies are comparable.
4.10 Define bond length.
The equilibrium internuclear distance between the centres of two bonded atoms in a molecule. Determined by X-ray diffraction or spectroscopy. Unit: pm or Å.
4.11 Explain the important aspects of resonance with reference to CO₃²⁻ ion.
No single Lewis structure can show all three C–O bonds equal. We draw three equivalent canonical structures (each with one C=O and two C–O⁻). The actual carbonate ion is the resonance hybrid — bond order 1.33 for each C–O, all bond lengths equal (~129 pm), charge spread equally over the three O atoms. The hybrid is more stable than any canonical form by the resonance energy.
4.12 H₃PO₃ can be represented as having two equivalent and one different P–O bond. Discuss its resonance structures.
In H₃PO₃ (phosphorous acid): two structures place the P=O double bond at either of the two equivalent OH oxygens, giving resonance. The third H is bonded directly to P (P–H) and is non-acidic. Two of the OH protons are acidic; H₃PO₃ is dibasic, not tribasic, because of this structural feature.
4.13 Write the resonance structures for SO₃, NO₂ and NO₃⁻.
SO₃: three structures with the S=O double bond at each of the three O atoms in turn → all three S–O bonds equal (143 pm), bond order 1.33. NO₂: two structures interchanging the position of N=O and N–O. NO₃⁻: three equivalent structures (analogous to CO₃²⁻).
4.14 Use Lewis symbols to show electron transfer between the following atoms to form ions: K and S; Ca and O; Al and N.
K loses 1 e⁻ × 2; S gains 2 e⁻ → 2K⁺ + S²⁻ → K₂S.
Ca loses 2 e⁻; O gains 2 e⁻ → Ca²⁺ + O²⁻ → CaO.
Al loses 3 e⁻ × 1; N gains 3 e⁻ × 1 → Al³⁺ + N³⁻ → AlN.
Ca loses 2 e⁻; O gains 2 e⁻ → Ca²⁺ + O²⁻ → CaO.
Al loses 3 e⁻ × 1; N gains 3 e⁻ × 1 → Al³⁺ + N³⁻ → AlN.
4.15 Although both CO₂ and H₂O are triatomic, the shape of H₂O is bent while that of CO₂ is linear. Explain.
CO₂: central C is sp hybridised (no lone pair) → linear (180°). H₂O: central O is sp³ hybridised with 2 bond pairs and 2 lone pairs → bent (104.5°). The lone pairs on oxygen distort the geometry from linear/tetrahedral to bent.
4.16 Write the favourable factors for the formation of an ionic bond.
(i) Low ionisation enthalpy of metal (easy to lose e⁻).
(ii) High (negative) electron gain enthalpy of non-metal.
(iii) Large lattice enthalpy of the resulting solid (small, highly charged ions favour large lattice energy).
(iv) Large electronegativity difference between the two atoms.
(ii) High (negative) electron gain enthalpy of non-metal.
(iii) Large lattice enthalpy of the resulting solid (small, highly charged ions favour large lattice energy).
(iv) Large electronegativity difference between the two atoms.
4.17 Discuss the shape of the following molecules using VSEPR: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.
BeCl₂ — 2 bp, 0 lp → linear (180°). BCl₃ — 3 bp → trigonal planar (120°). SiCl₄ — 4 bp → tetrahedral (109.5°). AsF₅ — 5 bp → trigonal bipyramidal (90°/120°). H₂S — 2 bp + 2 lp → bent (~92°). PH₃ — 3 bp + 1 lp → pyramidal (~93.5°).
4.18 Define the dipole moment. List its applications.
Dipole moment μ = Q × r; vector pointing from + to −; unit Debye (3.336 × 10⁻³⁰ C·m).
Applications: (i) decide whether a molecule is polar or non-polar (μ ≠ 0 or μ = 0); (ii) predict molecular geometry (CO₂ vs H₂O); (iii) calculate percentage ionic character; (iv) distinguish between cis–trans isomers (cis usually has μ ≠ 0, trans has μ = 0); (v) compare extent of polarity (HF > HCl > HBr > HI).
Applications: (i) decide whether a molecule is polar or non-polar (μ ≠ 0 or μ = 0); (ii) predict molecular geometry (CO₂ vs H₂O); (iii) calculate percentage ionic character; (iv) distinguish between cis–trans isomers (cis usually has μ ≠ 0, trans has μ = 0); (v) compare extent of polarity (HF > HCl > HBr > HI).
4.19 Arrange the bonds in order of increasing ionic character: N–H, F–H, C–H and O–H.
Ionic character ∝ Δ(EN). Δ(EN): C–H 0.4, N–H 0.9, O–H 1.4, F–H 1.9. Increasing order: C–H < N–H < O–H < F–H.
4.20 The skeletal structure of CH₃COOH is given. Draw the Lewis structure and find the bond orders of C–O and C=O.
Total v.e. = 24. Lewis structure: H₃C–C(=O)–O–H. Carboxyl C is sp²; the C=O bond order is 2; the C–O(H) bond order is 1. (In acetate ion CH₃COO⁻ the two C–O bonds become equivalent due to resonance, BO = 1.5 each.)
4.21 Apart from tetrahedral geometry, another arrangement of XeF₄ may be square planar with four F atoms at the corners. On the basis of VSEPR, predict the actual shape.
XeF₄: Xe has 4 σ bonds + 2 lp → 6 electron pairs → octahedral electron geometry. The 2 lp occupy opposite axial positions (180°) to minimise lp–lp repulsion → square planar arrangement of 4 F atoms. (Tetrahedral with 4 F + 2 lp would force lp at 109.5° which is energetically unfavourable.)
4.22 Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than ammonia. Discuss. (Same as 4.8 — alternate phrasing for completeness.)
See solution to Q 4.8: H₂O has 2 lp on O ⇒ greater lp–lp + lp–bp repulsion ⇒ smaller H–O–H angle (104.5°) than the H–N–H angle in NH₃ (107°).
4.23 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp² and sp³ hybrid orbitals.
Hybridisation = mixing of two or more atomic orbitals of comparable energy on the same atom to give the same number of equivalent hybrid orbitals.
sp — two hybrids, linear, 180° (50% s, 50% p).
sp² — three hybrids, trigonal planar, 120° (33% s, 67% p).
sp³ — four hybrids, tetrahedral, 109.5° (25% s, 75% p).
Each hybrid is unsymmetrical, with a large lobe (used for bonding) and a small lobe.
sp — two hybrids, linear, 180° (50% s, 50% p).
sp² — three hybrids, trigonal planar, 120° (33% s, 67% p).
sp³ — four hybrids, tetrahedral, 109.5° (25% s, 75% p).
Each hybrid is unsymmetrical, with a large lobe (used for bonding) and a small lobe.
4.24 Describe the change in hybridisation (if any) of the Al atom in: AlCl₃ + Cl⁻ → AlCl₄⁻.
In AlCl₃ Al uses sp² hybrids (trigonal planar, vacant p). On accepting Cl⁻ via coordinate bond, the vacant p mixes with the sp² set → 4 sp³ hybrids → AlCl₄⁻ is tetrahedral. Hybridisation: sp² → sp³.
4.25 Is there any change in the hybridisation of B and N atoms when BF₃ combines with NH₃ to form F₃B·NH₃?
In BF₃ boron is sp² (trigonal planar, vacant p). In NH₃ nitrogen is sp³ (pyramidal, with one lone pair). On bond formation N donates the lone pair to the empty B p-orbital. B re-hybridises to sp³ (tetrahedral around B); N remains sp³. Adduct is F₃B–NH₃ with both atoms sp³.
4.26 Draw diagrams showing the formation of double and triple bonds between carbon atoms in C₂H₄ and C₂H₂.
C₂H₄: each C is sp²; C–C σ bond by sp²–sp² overlap; one π bond by sideways overlap of un-hybridised 2p orbitals; planar molecule with ∠HCH = 117°.
C₂H₂: each C is sp; C–C σ bond by sp–sp overlap; two π bonds by sideways overlap of two pairs of mutually perpendicular un-hybridised 2p orbitals; linear, ∠HCC = 180°.
C₂H₂: each C is sp; C–C σ bond by sp–sp overlap; two π bonds by sideways overlap of two pairs of mutually perpendicular un-hybridised 2p orbitals; linear, ∠HCC = 180°.
4.27 What is the total number of σ and π bonds in C₂H₄ and C₂H₂?
C₂H₄: 5 σ (4 C–H + 1 C–C) and 1 π → total 5σ + 1π.
C₂H₂: 3 σ (2 C–H + 1 C–C) and 2 π → total 3σ + 2π.
C₂H₂: 3 σ (2 C–H + 1 C–C) and 2 π → total 3σ + 2π.
4.28 Considering the X-axis as the internuclear axis, which of the following overlaps will not form a bond: 1s & 1s; 1s & 2px; 2py & 2py; 1s & 2s.
1s + 1s → σ (bond). 1s + 2px → σ (axial overlap, bond). 2py + 2py → π (lateral overlap, bond). 1s + 2s → σ (bond). All four can form bonds! (If the question asks 1s + 2py with X-axis as internuclear axis, 1s is symmetric about X but 2py changes sign on going through the YZ plane — net overlap = 0 → no bond.)
4.29 Which hybrid orbitals are used by C atoms in: (i) CH₃–CH₃, (ii) CH₃–CH=CH₂, (iii) CH₃–CH₂–OH, (iv) CH₃–CHO, (v) CH₃COOH?
(i) Both C — sp³.
(ii) C¹(sp³)–C²(sp²)=C³(sp²).
(iii) C¹(sp³)–C²(sp³)–OH.
(iv) C¹(sp³)–C²(sp²)=O.
(v) C¹(sp³)–C²(sp²)(=O)(OH).
(ii) C¹(sp³)–C²(sp²)=C³(sp²).
(iii) C¹(sp³)–C²(sp³)–OH.
(iv) C¹(sp³)–C²(sp²)=O.
(v) C¹(sp³)–C²(sp²)(=O)(OH).
4.30 What do you understand by bond pairs and lone pairs of electrons? Illustrate with one example.
Bond pair — pair of electrons shared between two atoms forming a covalent bond.
Lone pair — pair of valence electrons not involved in bonding; localised on one atom.
In H₂O, oxygen has 2 bond pairs (one in each O–H) and 2 lone pairs.
Lone pair — pair of valence electrons not involved in bonding; localised on one atom.
In H₂O, oxygen has 2 bond pairs (one in each O–H) and 2 lone pairs.
4.31 Distinguish between σ and π bonds.
σ-bond: axial (head-on) overlap, electron density max along inter-nuclear axis, stronger, can exist alone, free rotation possible. π-bond: lateral (sideways) overlap, density above & below the inter-nuclear axis, weaker, exists only with a σ-bond, restricts rotation.
4.32 Explain the formation of H₂ molecule on the basis of valence bond theory.
Two H atoms with electrons of opposite spin approach each other along the X-axis. Attractive forces (nuclear–electron) increase faster than repulsions (nuclear–nuclear, electron–electron). At 74 pm internuclear separation potential energy is minimum: a covalent σ-bond (overlap of 1s + 1s) forms. Energy released = 435.8 kJ mol⁻¹ = bond enthalpy.
4.33 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
(i) Combining AOs must have comparable energies. (ii) They must overlap to a significant extent. (iii) They must have the same symmetry about the inter-nuclear axis.
4.34 Use molecular orbital theory to explain why the Be₂ molecule does not exist.
Be₂ has 8 electrons. MO config: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)². Nb = 4, Na = 4 → BO = 0. No net bond → Be₂ cannot exist as a stable molecule.
4.35 Compare the relative stability of O₂, O₂⁺, O₂⁻ and O₂²⁻ and comment on their magnetic properties.
| Species | e⁻ | BO | Magnetic |
|---|---|---|---|
| O₂⁺ | 15 | 2.5 | Paramagnetic (1 unpaired) |
| O₂ | 16 | 2.0 | Paramagnetic (2 unpaired) |
| O₂⁻ | 17 | 1.5 | Paramagnetic (1 unpaired) |
| O₂²⁻ | 18 | 1.0 | Diamagnetic |
4.36 Write the significance of a plus and a minus sign shown in representing the orbitals.
+ and − represent the algebraic sign (phase) of the wave function ψ, not electric charge. Same-phase (++) overlap → constructive interference → bonding MO. Opposite-phase (+−) overlap → destructive interference → antibonding MO with a node.
4.37 Describe the hybridisation in case of PCl₅. Why are the axial bonds longer compared to equatorial?
P uses one s + three p + one d → 5 sp³d hybrid orbitals → trigonal bipyramidal. Each axial Cl experiences three perpendicular (90°) repulsions from equatorial bonds; each equatorial Cl experiences only two (axial). Greater repulsion on axial bonds → axial bonds longer (240 pm) and weaker than equatorial (202 pm).
4.38 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Hydrogen bond: electrostatic attraction between an H atom covalently bonded to F/O/N and a lone pair on another electronegative atom. Energy 10–40 kJ/mol. Stronger than van der Waals forces (1–10 kJ/mol) but much weaker than covalent bonds (~150–500 kJ/mol).
4.39 What is meant by the term bond order? Calculate the bond orders of N₂, O₂, O₂⁺ and O₂⁻.
Bond order = (Nb − Na)/2.
N₂ (14 e⁻): Nb = 10, Na = 4 → BO = 3.
O₂ (16 e⁻): Nb = 10, Na = 6 → BO = 2.
O₂⁺ (15 e⁻): Nb = 10, Na = 5 → BO = 2.5.
O₂⁻ (17 e⁻): Nb = 10, Na = 7 → BO = 1.5.
N₂ (14 e⁻): Nb = 10, Na = 4 → BO = 3.
O₂ (16 e⁻): Nb = 10, Na = 6 → BO = 2.
O₂⁺ (15 e⁻): Nb = 10, Na = 5 → BO = 2.5.
O₂⁻ (17 e⁻): Nb = 10, Na = 7 → BO = 1.5.
4.40 Discuss the effect of hydrogen bonding on the physical properties of compounds (with examples).
H-bonding raises melting and boiling points (HF, H₂O, NH₃ are anomalously high vs HCl, H₂S, PH₃). It increases viscosity (glycerol, conc. H₂SO₄) and surface tension (water). It decreases volatility. Solubility in water is enhanced for H-bond donors/acceptors (alcohols, sugars). Ice has lower density than water (anomalous expansion). In biology, H-bonds stabilise the DNA double helix and protein α-helix/β-sheet structures.
Competency-Based Questions
Q1. The hybridisation of S in SF₆ is:L1 Remember
Answer: (c) sp³d²; six bond pairs, octahedral geometry.
Q2. Which species has the highest bond order?L2 Understand
Answer: (d) O₂⁺ (BO = 2.5). Removing an antibonding electron strengthens the bond.
Q3. Apply the formal charge formula to the central N in N₂O (N≡N⁺=O⁻ vs N⁻=N⁺=O).L3 Apply
In structure N≡N⁺=O⁻: central N FC = 5 − 0 − ½(8) = +1; terminal N FC = 5 − 2 − ½(6) = 0; O FC = 6 − 4 − ½(4) = 0. Sum = 0 + (+1) + (−1)... actually let's recount: terminal N (≡N): FC = 5 − 2 − 3 = 0; central N: 5 − 0 − 4 = +1; O: 6 − 4 − 2 = 0; need −1 somewhere. Correct Lewis: N=N⁺=O⁻ gives terminal N FC = 5 − 4 − 2 = −1; central N = +1; O = 0; sum 0 ✓. The structure with smaller magnitudes and negative FC on more EN atom (O) is more stable.
Q4. Why does HF, despite the highest electronegativity of F, have a lower boiling point than H₂O?L4 Analyse
Each H₂O molecule can form 4 H-bonds (2 as donor + 2 as acceptor) due to 2 H atoms and 2 lone pairs on O. Each HF can form only 2 H-bonds (1 H + 3 lone pairs but only 1 donor). The greater connectivity in water → much more extensive H-bond network → higher boiling point (100 °C vs 19.5 °C).
Q5. HOT (Create): Using MOT, design an interstellar molecule with bond order 3.5. Is it possible? Justify.L6 Create
BO = 3.5 requires (Nb − Na) = 7, an odd number — needs an odd electron count. Example design: NO⁺ (10 valence e⁻, BO = 3) plus removal of one antibonding electron → impossible without further ionisation. Better candidates: hypothetical CO⁺ would be BO = 2.5; C₂⁻ has 13 v.e., BO = 2.5. A BO of 3.5 is mathematically possible only if a species had 13 bonding minus 6 antibonding electrons — not realised in any known molecule. Possible resonance-averaged BO of 3.5 occurs in some metal–oxo complexes. Overall: while idealised BO = 3.5 is rare in main-group diatomics, transition-metal multiple bonds (e.g., in Cr₂(CH₃COO)₄ the Cr–Cr is sometimes assigned a quadruple bond) achieve very high orders.
Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: All five P–Cl bonds in PCl₅ are not equivalent.
R: The trigonal bipyramidal geometry has two distinct positions: axial and equatorial.
Answer: (A). Both true; R correctly explains A.
A: The molecule of carbon monoxide (CO) has the highest bond enthalpy among diatomic molecules (1071 kJ/mol).
R: CO is iso-electronic with N₂ and has a triple bond with additional ionic stabilisation.
Answer: (A). Both true; R correctly explains A.
A: NH₃ is more basic than NF₃.
R: The lone pair on N in NH₃ is more available for donation than in NF₃ where it is pulled towards highly electronegative F atoms.
Answer: (A). Both true; R correctly explains A. Same reason explains μ(NH₃) ≫ μ(NF₃).
Frequently Asked Questions — NCERT Exercises and Solutions: Chemical Bonding and Molecular Structure
What types of questions appear in Class 11 Chemistry Chapter 4 NCERT exercises?
NCERT Class 11 Chemistry Chapter 4 exercises include: (1) writing Lewis structures of given molecules and ions, (2) predicting geometry/shape using VSEPR theory, (3) identifying hybridisation of central atoms, (4) drawing MO diagrams and calculating bond order, (5) explaining bond properties (length, energy, polarity), (6) describing hydrogen bonding effects on boiling point and solubility, (7) comparing ionic character using Fajans rules. The MyAiSchool exercise set covers each type with step-by-step model solutions aligned to CBSE board exam patterns.
How do you draw Lewis structures correctly in NCERT exercises?
To draw Lewis structures in NCERT Class 11 Chemistry Chapter 4 exercises: (1) count total valence electrons (add for anions, subtract for cations); (2) place the least electronegative atom at the centre (except H); (3) connect outer atoms with single bonds; (4) complete octets of outer atoms with lone pairs; (5) place remaining electrons on central atom; (6) form double or triple bonds if central atom octet is incomplete; (7) calculate formal charges and choose structure with lowest |formal charge|. Practice with NO₃⁻, CO₃²⁻, SO₄²⁻, ClO₄⁻, XeF₄ — all commonly asked.
How do you predict molecular shape from VSEPR in exercises?
To predict molecular shape from VSEPR in NCERT Class 11 Chemistry exercises: (1) draw Lewis structure; (2) count electron pairs around central atom (bonding pairs + lone pairs); (3) total pairs determine electron geometry; (4) consider lone pair effects on bond angles; (5) state molecular shape. Examples: SF₄ has 4 BP + 1 LP → see-saw shape; XeF₂ has 2 BP + 3 LP → linear; IF₇ has 7 BP → pentagonal bipyramidal. Apply repulsion order LP-LP > LP-BP > BP-BP to predict bond angle distortions. The MyAiSchool exercise set drills all major molecule shapes.
How do you find hybridisation in exam questions?
To find hybridisation in NCERT Class 11 Chemistry Chapter 4 exam questions, use the steric number: SN = number of σ bonds + lone pairs on central atom. Matching: SN 2 = sp, SN 3 = sp², SN 4 = sp³, SN 5 = sp³d, SN 6 = sp³d². Alternative formula: H = ½ (V + X − C + A) where V = valence electrons of central atom, X = monovalent atoms attached, C = cationic charge, A = anionic charge. Example: in PCl₃, SN = 3 + 1 = 4 → sp³ hybridisation. Practice with NH₃, SF₆, IF₇, BrF₅, XeF₄.
How is bond order calculated in NCERT MOT problems?
To calculate bond order in NCERT Class 11 Chemistry Chapter 4 MOT problems: (1) write MO configuration of the diatomic species in order σ1s, σ*1s, σ2s, σ*2s, σ2p_z, π2p_x = π2p_y, π*2p_x = π*2p_y, σ*2p_z; (2) count N_b (electrons in bonding MOs) and N_a (electrons in antibonding MOs); (3) bond order = ½ (N_b − N_a). Examples: H₂ = 1, He₂⁺ = 0.5, N₂ = 3, O₂ = 2, O₂⁻ = 1.5, O₂²⁻ = 1. Higher bond order = stronger, shorter bond. Always state paramagnetism if unpaired electrons exist.
What 5-mark questions are common in Chapter 4 CBSE board exams?
Common 5-mark CBSE Class 11 Chemistry Chapter 4 questions: (1) draw MO diagrams of N₂, O₂, F₂ and explain magnetic properties using bond order; (2) discuss VSEPR theory and predict shapes of SF₄, XeF₂, BrF₅, IF₇; (3) compare hybridisation, geometry and bond angles of CH₄, NH₃ and H₂O; (4) explain hydrogen bonding in ice, water boiling point anomaly and o/p-nitrophenol; (5) discuss Fajans rules with examples. The MyAiSchool exercise set provides model answers with diagrams, marking schemes and conceptual links.
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Chemistry Class 11 Part I – NCERT (2025-26)
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