This MCQ module is based on: Resonance Vsepr
TOPIC 15 OF 28
Resonance Vsepr
🎓 Class 11
Chemistry
CBSE
Theory
Ch 4 – Chemical Bonding and Molecular Structure
⏱ ~14 min
🌐 Language: [gtranslate]
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Resonance and VSEPR Theory
4.5 Resonance Structures
It is often observed that a single Lewis structure is inadequate for the representation of a molecule. For example, the O₃ molecule is best represented by two equivalent structures (I and II), each having one O–O single and one O=O double bond. Yet experiment shows that both O–O bond lengths in O₃ are equal (128 pm) — intermediate between a single (148 pm) and a double (121 pm) O–O bond.
Resonance: When the actual structure of a molecule cannot be represented adequately by a single Lewis structure, several similar structures (canonical forms) are written. The actual structure (resonance hybrid) is an average of these canonical forms. The hybrid is more stable than any single canonical form by an amount called the resonance energy.
Resonance in CO₃²⁻ and NO₃⁻
Important rules: (i) Resonance structures differ only in the position of electrons, never of atoms. (ii) The greater the number of equivalent structures, the greater the resonance stabilisation. (iii) The hybrid is the only real structure — the canonical forms are imaginary.
4.6 Polarity of Bonds
In a purely covalent bond between identical atoms (e.g., H–H, Cl–Cl) the shared electron pair is equally attracted by both nuclei → non-polar covalent bond. When the atoms differ in electronegativity (e.g., H–Cl), the pair is pulled towards the more electronegative atom → polar covalent bond. Partial charges δ⁺ and δ⁻ develop.
4.6.1 Dipole Moment
Dipole moment (μ): The product of the magnitude of the partial charge (Q) and the distance (r) between the centres of the positive and negative charges.
\(\mu = Q \times r\) — a vector quantity, pointing from + to −. Unit: Debye (D); 1 D = 3.33564 × 10⁻³⁰ C·m.
\(\mu = Q \times r\) — a vector quantity, pointing from + to −. Unit: Debye (D); 1 D = 3.33564 × 10⁻³⁰ C·m.
| Molecule | μ (D) | Molecule | μ (D) |
|---|---|---|---|
| HF | 1.78 | BF₃ | 0 (planar, vectors cancel) |
| HCl | 1.07 | CO₂ | 0 (linear) |
| HBr | 0.79 | CH₄ | 0 (tetrahedral) |
| HI | 0.38 | CHCl₃ | 1.04 |
| H₂O | 1.85 | NH₃ | 1.47 |
| NF₃ | 0.23 | H₂S | 0.95 |
Why is μ(NH₃) > μ(NF₃)? Both molecules are pyramidal. In NH₃ the lone-pair moment and the sum of N–H bond moments point in the same direction → larger μ. In NF₃ the bond moments point towards the highly electronegative F atoms (opposite to the lone-pair moment), giving partial cancellation → smaller μ (only 0.23 D).
Percentage Ionic Character
For the H–Cl bond, μ(observed) = 1.03 D. If the bond were 100% ionic, μ(theor) = 4.8 × 1.275 = 6.12 D (for r = 127.5 pm).
Percentage ionic character = (μ(obs)/μ(ionic)) × 100 = (1.03/6.12) × 100 ≈ 17%. The HCl bond is therefore mainly covalent.
4.7 The Valence Shell Electron Pair Repulsion (VSEPR) Theory
The Lewis approach explains the formation of a bond but is silent about the shape of a molecule. VSEPR theory (Sidgwick & Powell, 1940; Nyholm & Gillespie, 1957) fills this gap.
Main postulates of VSEPR:
- Geometry is determined by the repulsion between all electron pairs (bonded + lone) in the valence shell of the central atom.
- Electron pairs orient themselves so as to minimise repulsion and maximise their separation.
- Order of repulsion: lone pair–lone pair (lp–lp) > lone pair–bond pair (lp–bp) > bond pair–bond pair (bp–bp).
- A multiple bond is treated as a single super-pair occupying one position.
Geometry when the central atom has NO lone pair (Table 4.6)
| Type | Electron pairs | Geometry | Bond angle | Examples |
|---|---|---|---|---|
| AB₂ | 2 | Linear | 180° | BeCl₂, HgCl₂, CO₂ |
| AB₃ | 3 | Trigonal planar | 120° | BF₃, BCl₃, NO₃⁻ |
| AB₄ | 4 | Tetrahedral | 109.5° | CH₄, NH₄⁺, SO₄²⁻ |
| AB₅ | 5 | Trigonal bipyramidal | 120° (eq) & 90° (ax) | PCl₅, PF₅ |
| AB₆ | 6 | Octahedral | 90° | SF₆ |
Geometry when the central atom has lone pairs (Table 4.7)
| Type (ABnEm) | Bond pairs | Lone pairs | Shape | Example |
|---|---|---|---|---|
| AB₂E | 2 | 1 | Bent (angular) | SO₂, O₃ |
| AB₃E | 3 | 1 | Trigonal pyramidal | NH₃ |
| AB₂E₂ | 2 | 2 | Bent | H₂O |
| AB₄E | 4 | 1 | See-saw | SF₄ |
| AB₃E₂ | 3 | 2 | T-shape | ClF₃ |
| AB₅E | 5 | 1 | Square pyramidal | BrF₅ |
| AB₄E₂ | 4 | 2 | Square planar | XeF₄ |
Interactive: VSEPR Shape Predictor
Choose the number of bond pairs (BP) and lone pairs (LP) on the central atom — the simulator returns the electron-pair geometry, molecular shape, ideal bond angle and an example.
Electron geometry: Tetrahedral
Molecular shape: Tetrahedral
Ideal bond angle: 109.5°
Example: CH₄
Activity 4.2 — VSEPR challenge
Setup: Apply VSEPR to predict the shape of (i) SF₄ (ii) ClF₃ (iii) XeF₂ (iv) IF₇.
Predict: Count the bond pairs and lone pairs around the central atom for each molecule, then assign electron-pair geometry and final shape.
SF₄ — 4 bp + 1 lp → trigonal bipyramidal e-geometry; lp in equatorial position → see-saw shape.
ClF₃ — 3 bp + 2 lp → trigonal bipyramidal e-geometry; 2 lp equatorial → T-shape.
XeF₂ — 2 bp + 3 lp → trigonal bipyramidal e-geometry; 3 lp equatorial → linear F–Xe–F.
IF₇ — 7 bp + 0 lp → pentagonal bipyramidal.
Worked Example 4.4 — Predict the shape of NH₄⁺
N has 5 valence e⁻; subtract 1 for + charge → 4. All 4 are used to bond 4 H atoms. Bond pairs = 4, lone pairs = 0. Type AB₄ → regular tetrahedral, all H–N–H = 109.5°.
Worked Example 4.5 — Why is the dipole moment of CO₂ zero but of H₂O ≠ 0?
CO₂ is linear (O=C=O, 180°). The two C=O bond moments are equal in magnitude but opposite in direction — they cancel exactly, so net μ = 0.
H₂O is bent (104.5°). The two O–H bond moments do not lie along the same line; they add as vectors to give a resultant ≈ 1.85 D pointing along the lone-pair axis.
H₂O is bent (104.5°). The two O–H bond moments do not lie along the same line; they add as vectors to give a resultant ≈ 1.85 D pointing along the lone-pair axis.
Competency-Based Questions
Q1. The shape of a molecule with 5 bond pairs and 1 lone pair around the central atom is:L1 Remember
Answer: (b) Square pyramidal (e.g., BrF₅).
Q2. Among CO₂, H₂O, BF₃ and NH₃, which two have a net dipole moment of zero?L2 Understand
Answer: CO₂ (linear) and BF₃ (trigonal planar). Their symmetric geometries cancel all bond moments. H₂O (bent, 1.85 D) and NH₃ (pyramidal, 1.47 D) have non-zero μ.
Q3. Apply VSEPR: predict the shape of XeF₄.L3 Apply
Xe = 8 valence e⁻; 4 bonding pairs to F + 2 lone pairs → 6 electron pairs (octahedral e-geometry). The two lp occupy opposite axial positions to minimise lp-lp repulsion → square planar shape.
Q4. Why is the H–N–H angle in NH₃ (107°) smaller than the tetrahedral angle (109.5°)?L4 Analyse
N has 4 electron pairs (3 bp + 1 lp). The lone pair occupies more space than a bond pair (lp–bp repulsion > bp–bp). It pushes the three N–H bonds closer together, compressing the H–N–H angle below 109.5°.
Q5. Compare the dipole moments of NH₃ (1.47 D) and NF₃ (0.23 D). Both are pyramidal — explain.L5 Evaluate
In NH₃ the resultant of the three N–H bond moments and the lone-pair moment point in the same direction (lp on top, H below) → vectors add, large μ. In NF₃ the bond moments point towards the highly electronegative F (lp opposes them) → partial cancellation, very small μ.
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: The three C–O bond lengths in CO₃²⁻ are equal.
R: The carbonate ion is a resonance hybrid of three equivalent canonical forms.
Answer: (A). Both true; R explains A. Resonance averages the bond order to 1.33 for each C–O.
A: Methane has zero dipole moment although the C–H bond is polar.
R: The four C–H bond moments cancel in the regular tetrahedral geometry.
Answer: (A). Both true; R correctly explains A.
A: H₂O is bent and not linear.
R: Lone pairs on oxygen experience greater repulsion than bond pairs.
Answer: (A). Both true; R correctly explains A. The 2 lp + 2 bp arrangement is tetrahedral electron-geometry; the molecular shape (positions of atoms only) is bent, with H–O–H = 104.5°.
Frequently Asked Questions — Resonance and VSEPR Theory
What is resonance and why is it important?
Resonance is the phenomenon in which a single Lewis structure cannot adequately represent a molecule, and the true structure is described as a hybrid of two or more contributing canonical structures. The canonical structures differ only in the arrangement of electrons (not nuclei). For example, the carbonate ion CO₃²⁻ is best described as three equivalent C-O bonds, each with bond order 1.33 — a resonance hybrid of three Lewis structures. NCERT Class 11 Chemistry Chapter 4 uses resonance to explain stability, equal bond lengths in benzene (C₆H₆) and the structure of nitrate, nitrite and ozone.
What are the postulates of VSEPR theory?
VSEPR (Valence Shell Electron Pair Repulsion) theory was proposed by Sidgwick and Powell and refined by Gillespie. Its postulates are: (1) the shape of a molecule depends on the number of electron pairs around the central atom; (2) electron pairs repel each other and arrange themselves to maximise distance; (3) repulsion order is LP-LP > LP-BP > BP-BP; (4) multiple bonds count as a single electron domain but exert greater repulsion than single bonds. NCERT Class 11 Chemistry uses VSEPR to predict shapes of CH₄ (tetrahedral), NH₃ (pyramidal), H₂O (bent), PCl₅ (trigonal bipyramidal) and SF₆ (octahedral).
How do you predict molecular geometry using VSEPR?
To predict molecular geometry using VSEPR in NCERT Class 11 Chemistry: (1) draw the Lewis structure of the molecule, (2) count electron pairs (bonding + lone) around the central atom, (3) determine the electron pair geometry from the steric number — 2 (linear), 3 (trigonal planar), 4 (tetrahedral), 5 (trigonal bipyramidal), 6 (octahedral), (4) consider lone pairs — molecular geometry differs from electron pair geometry. Example: NH₃ has 4 electron pairs (3 BP + 1 LP) so electron geometry is tetrahedral but molecular geometry is trigonal pyramidal with bond angle 107° (less than 109.5° due to LP repulsion).
What is the order of repulsion in VSEPR theory?
The order of repulsion in VSEPR theory is: lone pair-lone pair (LP-LP) > lone pair-bond pair (LP-BP) > bond pair-bond pair (BP-BP). Lone pairs occupy more space than bond pairs because they are attracted to only one nucleus, while bond pairs are attracted to two nuclei. This explains why bond angles decrease as lone pairs increase: CH₄ has 109.5° (0 LP), NH₃ has 107° (1 LP), H₂O has 104.5° (2 LP). NCERT Class 11 Chemistry Chapter 4 uses this rule to explain bond-angle distortions in various molecules.
Why is the bond angle in water 104.5° and not 109.5°?
Although oxygen in water has four electron pairs (two bond pairs and two lone pairs) arranged tetrahedrally, the bond angle is 104.5° instead of the ideal tetrahedral angle 109.5°. This is because the two lone pairs on oxygen repel each other and the bond pairs more strongly than bond pairs repel each other. The greater lone pair-lone pair and lone pair-bond pair repulsion pushes the two O-H bond pairs closer together, decreasing the H-O-H angle from 109.5° to 104.5°. NCERT Class 11 Chemistry presents this as a key example of VSEPR predictions.
How does VSEPR explain shapes of SF₆ and PCl₅?
VSEPR predicts SF₆ has six bond pairs around S with no lone pairs, giving an octahedral geometry with all F-S-F angles = 90°. PCl₅ has five bond pairs and no lone pairs, giving trigonal bipyramidal geometry with three equatorial Cl-P-Cl angles = 120° and two axial Cl-P-Cl angles = 90°. The axial bonds are slightly longer than equatorial bonds in PCl₅ because of greater repulsion. NCERT Class 11 Chemistry Chapter 4 uses these as examples of expanded octet molecules where d-orbitals participate in bonding, requiring sp³d and sp³d² hybridisation respectively.
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