JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Valence Bond Hybridisation
TOPIC 16 OF 28
Valence Bond Hybridisation
🎓 Class 11
Chemistry
CBSE
Theory
Ch 4 – Chemical Bonding and Molecular Structure
⏱ ~14 min
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Valence Bond Theory and Hybridisation
4.8 Valence Bond Theory (VBT)
The Lewis approach and VSEPR theory tell us about which atoms bond and the shape of the molecule but say nothing about how a covalent bond forms or about its energetics. Valence Bond Theory (VBT), developed by Heitler and London in 1927 and extended by Pauling and Slater, gives a quantum-mechanical picture of bond formation.
4.8.1 Formation of H₂ Molecule (Heitler–London Treatment)
Consider two H atoms approaching each other along the x-axis. Let nuclei be HA, HB and electrons eA, eB. As the atoms come closer, two new forces appear:
- Attractive forces: nucleus A – electron B, nucleus B – electron A.
- Repulsive forces: nucleus A – nucleus B, electron A – electron B.
At a particular internuclear distance the net attraction is maximum (system energy minimum). This distance is the equilibrium bond length (74 pm for H₂); the energy released equals the bond enthalpy (435.8 kJ mol⁻¹).
4.9 Orbital Overlap Concept
According to the orbital overlap concept, a covalent bond is formed by partial inter-penetration (overlap) of half-filled atomic orbitals containing electrons of opposite spin. The greater the overlap, the stronger the bond.
4.9.1 Sigma (σ) and Pi (π) Bonds
σ-bond — formed by head-on (axial) overlap along the inter-nuclear axis. Three sub-types:
- s–s: H₂
- s–p: HF, HCl
- p–p (axial): F₂, Cl₂
σ vs π — strength comparison: Axial overlap is more effective than lateral overlap, so σ-bonds are stronger than π-bonds. A double bond (1 σ + 1 π) is shorter and stronger than a single (σ) bond, but not exactly twice as strong.
4.9.2 Strength of σ and π Bonds
Bond enthalpies (kJ mol⁻¹): C–C (σ) ≈ 348; C=C (σ+π) ≈ 614; C≡C (σ+2π) ≈ 839. The π-bond contribution is roughly 264 kJ mol⁻¹ — clearly weaker per bond than the σ-bond.
4.10 Hybridisation
Pure atomic orbitals do not always explain the observed shape (e.g., 109.5° angles in CH₄ from 90° p-orbitals). To resolve this, Pauling proposed hybridisation.
Salient features of hybridisation:
- Number of hybrid orbitals = number of atomic orbitals mixed.
- Hybrid orbitals are equivalent in energy and shape.
- Hybrid orbitals are more effective in forming stable bonds than pure atomic orbitals.
- Hybrid orbitals are oriented in space to minimise repulsion (giving the molecular geometry).
4.10.1 Types of Hybridisation
| Type | Orbitals mixed | Geometry | Bond angle | Examples |
|---|---|---|---|---|
| sp | 1 s + 1 p | Linear | 180° | BeCl₂, BeF₂, C₂H₂, CO₂ |
| sp² | 1 s + 2 p | Trigonal planar | 120° | BCl₃, BF₃, C₂H₄, NO₃⁻ |
| sp³ | 1 s + 3 p | Tetrahedral | 109.5° | CH₄, NH₃, H₂O, NH₄⁺ |
| sp³d | 1 s + 3 p + 1 d | Trigonal bipyramidal | 120°/90° | PCl₅, PF₅ |
| sp³d² | 1 s + 3 p + 2 d | Octahedral | 90° | SF₆, [Co(NH₃)₆]³⁺ |
4.10.2 sp Hybridisation — BeCl₂
Be (1s² 2s²) excites one 2s electron to 2p (1s² 2s¹ 2p¹). The 2s and one 2p mix → 2 sp hybrids at 180°. Each sp overlaps with a Cl 3p orbital → linear Cl–Be–Cl.
4.10.3 sp² Hybridisation — BCl₃
B (1s² 2s² 2p¹) → excited (1s² 2s¹ 2p²) → 1 s + 2 p mix to give 3 sp² hybrids in a plane at 120°. Each forms a σ-bond with Cl → trigonal planar BCl₃.
4.10.4 sp³ Hybridisation — CH₄
C (1s² 2s² 2p²) → excited (1s² 2s¹ 2p³) → 1 s + 3 p form 4 sp³ hybrids at 109.5°. Each overlaps with H(1s) → tetrahedral CH₄.
4.10.5 NH₃ & H₂O via sp³
NH₃: N is sp³ hybridised. Three sp³ form N–H bonds, the fourth holds the lone pair → pyramidal, ∠HNH ≈ 107°.
H₂O: O is sp³ hybridised. Two sp³ make O–H bonds, two hold lone pairs → bent, ∠HOH ≈ 104.5°.
4.10.6 Hybridisation in Multiply-Bonded Carbons
Ethene (C₂H₄) — each C is sp² hybridised. Three sp² orbitals on each C form three σ-bonds (two C–H and one C–C). The remaining un-hybridised p-orbitals on the two C atoms overlap sideways → π-bond. C=C bond length 134 pm, ∠HCH = 117°, ∠HCC = 121°.
Ethyne (C₂H₂) — each C is sp hybridised. Two sp orbitals on each C form one C–H σ and one C–C σ-bond (linear). The two un-hybridised p-orbitals on each C give two mutually perpendicular π-bonds → triple bond. C≡C = 120 pm.
Benzene (C₆H₆) — each C is sp² hybridised; three sp² form σ-bonds with two adjacent C and one H. The six un-hybridised p-orbitals form a delocalised π-cloud above and below the planar hexagon → all C–C bonds equal (139 pm).
4.10.7 sp³d Hybridisation — PCl₅
P (1s² 2s² 2p⁶ 3s² 3p³) → excited (3s¹ 3p³ 3d¹) → 5 sp³d hybrids form a trigonal bipyramid. Three equatorial Cl at 120° (P–Cl = 202 pm); two axial Cl at 90° to the equatorial plane (P–Cl = 240 pm). The longer axial bonds make PCl₅ reactive.
4.10.8 sp³d² Hybridisation — SF₆
S excited to 3s¹ 3p³ 3d² → 6 sp³d² hybrids forming an octahedron. Six S–F bonds at 90° (S–F = 158 pm). Highly stable and inert.
Interactive: Hybridisation Identifier
Use the formula: Steric number = (no. of σ-bonds) + (no. of lone pairs on central atom). Enter the steric number to get the hybridisation, geometry and bond angle.
Hybridisation: sp³
Geometry of hybrid orbitals: Tetrahedral
Bond angle (no lp): 109.5°
Examples: CH₄, NH₄⁺, SO₄²⁻
Activity 4.3 — Identify the hybridisation
Setup: For each species, state the hybridisation of the central atom and the geometry: (i) BF₃ (ii) NH₃ (iii) H₂O (iv) PCl₅ (v) SF₆ (vi) CO₂ (vii) C₂H₄.
Predict: Find the steric number, then assign hybridisation.
(i) BF₃ — sp², trigonal planar.
(ii) NH₃ — sp³, pyramidal.
(iii) H₂O — sp³, bent.
(iv) PCl₅ — sp³d, trigonal bipyramidal.
(v) SF₆ — sp³d², octahedral.
(vi) CO₂ — sp on C, linear (with two π-bonds).
(vii) C₂H₄ — sp² on each C, trigonal planar at each C.
Worked Example 4.6 — Hybridisation in XeF₄
Xe: 5s² 5p⁶. Promote 2 e⁻ to 5d → 4 unpaired e⁻ form 4 σ-bonds with F. With 2 lone pairs remaining: steric number = 4 + 2 = 6 → sp³d² hybridisation. Six hybrid orbitals point to corners of an octahedron; lone pairs occupy opposite axial positions → square planar molecular geometry.
Worked Example 4.7 — Bond order and hybridisation in NO₃⁻
NO₃⁻ has 3 σ-bonds + 0 lp on N → steric number 3 → sp². Three resonance structures place the C=O double bond at each O in turn → bond order = 4/3 ≈ 1.33. Geometry: trigonal planar, ∠O–N–O = 120°.
Competency-Based Questions
Q1. Hybridisation of carbon in ethyne (C₂H₂) is:L1 Remember
Answer: (c) sp. Each C forms 1 σ + 1 σ + 2 π = 2 σ → steric number 2.
Q2. Why is a σ-bond stronger than a π-bond?L2 Understand
σ-bonds form by axial overlap, which gives maximum electron density between the nuclei. π-bonds form by sideways overlap, which is less effective and the electron density lies above and below (not between) the nuclei → smaller overlap integral, smaller bond enthalpy.
Q3. Identify the hybridisation and shape of XeF₂.L3 Apply
Xe forms 2 σ + 3 lp → steric number 5 → sp³d. The 3 lp occupy equatorial positions of the trigonal bipyramid → linear F–Xe–F (180°).
Q4. Compare the bond lengths in C–C, C=C and C≡C and explain in terms of hybridisation.L4 Analyse
C–C (sp³–sp³, 154 pm) > C=C (sp²–sp², 134 pm) > C≡C (sp–sp, 120 pm). Higher s-character → smaller orbital → shorter bond. Also, the additional π-overlaps in double and triple bonds pull the carbons closer.
Q5. PCl₅ has two distinct P–Cl bond lengths (202 pm equatorial and 240 pm axial). Explain.L5 Evaluate
In the trigonal bipyramidal geometry the equatorial bonds experience repulsion from only two perpendicular axial bonds (90°), whereas the axial bonds experience repulsion from three equatorial bonds (90° each). Greater repulsion → longer (and weaker) axial P–Cl bonds. The axial bonds break preferentially, explaining PCl₅ ⇌ PCl₃ + Cl₂ in solution.
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 four C–H bonds in CH₄ are equivalent in length and energy.
R: The four sp³ hybrid orbitals on carbon are identical in shape and energy.
Answer: (A). Both true; R correctly explains A.
A: The bond angle in H₂O (104.5°) is less than that in NH₃ (107°).
R: Oxygen has two lone pairs which exert greater lp–lp and lp–bp repulsion than the single lone pair on nitrogen.
Answer: (A). Both true; R correctly explains A.
A: Ethene is planar.
R: The π-bond locks the two CH₂ planes into the same plane; rotation about C=C is restricted.
Answer: (A). Both true; R correctly explains A. Free rotation would break the π-bond (~264 kJ mol⁻¹).
Frequently Asked Questions — Valence Bond Theory and Hybridisation
What is the valence bond theory?
Valence bond theory (VBT), developed by Heitler, London, Pauling and Slater, describes covalent bond formation as the overlap of half-filled atomic orbitals of two atoms with electrons of opposite spin. The greater the overlap, the stronger the bond. Sigma (σ) bonds form by head-on overlap (s-s, s-p, p-p axial); pi (π) bonds form by sideways overlap of p-orbitals. NCERT Class 11 Chemistry Chapter 4 uses VBT to explain H₂, HF, N₂ and CH₄ bonding, and introduces hybridisation as an extension to handle multi-bond molecules and their geometries.
What is hybridisation and why is it needed?
Hybridisation is the process of mixing two or more atomic orbitals of slightly different energies to form an equal number of new hybrid orbitals with identical energy and shape, oriented in specific directions in space. It is needed because pure atomic orbitals cannot explain the observed geometry of molecules like CH₄ (tetrahedral, 109.5°) — carbon's 2s and three 2p orbitals must mix to form four equivalent sp³ hybrid orbitals. NCERT Class 11 Chemistry uses hybridisation to explain the shapes of molecules including BeF₂ (sp), BF₃ (sp²), CH₄ (sp³), PCl₅ (sp³d), SF₆ (sp³d²).
What are sp, sp² and sp³ hybridisations?
Sp hybridisation involves one s + one p orbital → two sp hybrid orbitals at 180° (linear), as in BeCl₂, CO₂, C₂H₂. Sp² hybridisation involves one s + two p orbitals → three sp² hybrid orbitals at 120° in a plane (trigonal planar), as in BF₃, BCl₃, C₂H₄, benzene. Sp³ hybridisation involves one s + three p orbitals → four sp³ hybrid orbitals at 109.5° (tetrahedral), as in CH₄, NH₄⁺, SiCl₄. NCERT Class 11 Chemistry Chapter 4 details each type with energy diagrams and geometries, foundational for organic chemistry too.
What is the difference between sigma and pi bonds?
A sigma (σ) bond forms by head-on (axial) overlap of two atomic orbitals along the internuclear axis, with maximum electron density between the nuclei. A pi (π) bond forms by sideways (lateral) overlap of two parallel p-orbitals above and below the bond axis, with zero electron density on the axis. Sigma bonds are stronger than pi bonds due to greater overlap. A single bond is one σ; a double bond is one σ + one π; a triple bond is one σ + two π. Free rotation is possible about σ bonds but restricted about π bonds. NCERT Class 11 Chemistry uses this in alkene/alkyne chemistry.
What is sp³d and sp³d² hybridisation?
Sp³d hybridisation involves one s + three p + one d orbital → five sp³d hybrid orbitals oriented in trigonal bipyramidal geometry (three equatorial at 120°, two axial at 90°), as in PCl₅, PF₅. Sp³d² hybridisation involves one s + three p + two d orbitals → six sp³d² hybrid orbitals oriented octahedrally at 90°, as in SF₆, [PCl₆]⁻. Both types require participation of d-orbitals, only possible from Period 3 onwards. NCERT Class 11 Chemistry Chapter 4 explains these for expanded-octet molecules and introduces them as the basis for transition metal coordination chemistry.
How do you determine hybridisation of an atom in a molecule?
To determine hybridisation of a central atom in NCERT Class 11 Chemistry: (1) calculate steric number = number of σ-bonded atoms + number of lone pairs on the central atom; (2) match steric number to hybridisation — 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d²; (3) the formula H = ½ (V + X − C + A), where V = valence electrons of central atom, X = monovalent atoms attached, C = positive charge, A = negative charge. Example: in NH₃, N has 3 bond pairs + 1 lone pair → steric number 4 → sp³ hybridisation. Practice with PCl₅, SF₆ and ICl₂⁻.
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