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Chemistryp-Block Elements
Identify the correct statement about ClF₃ from the following options :
Options
1
It has a trigonal pyramidal geometry with two lone pairs on Cl atom.
2
It has T-shaped geometry with two lone pairs on Cl atom.
3
It has a planar trigonal geometry with two lone pairs on Cl atom.
4
It has T-shaped geometry with three lone pairs on Cl atom.
Correct Answer
Option 2 : T-shaped, 2 lone pairs on Cl
Step-by-Step Solution
1

Valence electrons on Cl: 7 (Group 17)

Steric number = ½(valence electrons + monovalent atoms attached) = ½(7 + 3) = 5

2

Electron pairs around Cl: 5 total

Bond pairs = 3 (three Cl−F bonds)

Lone pairs = 5 − 3 = 2 lone pairs

3

VSEPR geometry: 5 electron pairs = trigonal bipyramidal arrangement.

2 lone pairs occupy equatorial positions (more space) → 3 bond pairs: 2 equatorial + 1 axial... wait — lone pairs prefer equatorial in trigonal bipyramidal.

Actual arrangement: 2 lone pairs in equatorial positions, 3 F atoms: 2 axial + 1 equatorial → gives T-shape

ClF₃: Steric number = 5

3 bond pairs + 2 lone pairs

Geometry: T-shaped (from trigonal bipyramidal)

Lone pairs on Cl = 2

Theory: VSEPR Theory & Molecular Geometry
1. VSEPR Theory — Key Principles

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry based on electron pair repulsions. Electron pairs (bonding and lone pairs) arrange themselves to minimize repulsion. Repulsion order: Lone pair–Lone pair (lp-lp) > Lone pair–Bond pair (lp-bp) > Bond pair–Bond pair (bp-bp). Lone pairs occupy more angular space than bond pairs (they are closer to central atom, more diffuse). Multiple bonds count as single region of electron density. Steric number = (number of atoms bonded to central atom) + (number of lone pairs on central atom).

2. VSEPR Geometries — Complete Table
Steric No.BPLPShapeExample
220LinearBeCl₂, CO₂
330Trigonal planarBF₃, SO₃
321Bent/V-shapeSO₂, SnCl₂
440TetrahedralCH₄, NH₄⁺
431Trigonal pyramidalNH₃, PCl₃
422BentH₂O, H₂S
550Trigonal bipyramidalPCl₅
541See-sawSF₄, IF₄⁺
532T-shapedClF₃, BrF₃
523LinearXeF₂, I₃⁻
660OctahedralSF₆, [Co(NH₃)₆]³⁺
651Square pyramidalIF₅, BrF₅
642Square planarXeF₄, [Pt(NH₃)₂Cl₂]
3. Why T-Shape and Not Trigonal Planar for ClF₃?

ClF₃ has steric number 5 (trigonal bipyramidal parent). In trigonal bipyramidal geometry, two positions are axial (90° to equatorial) and three are equatorial (120° to each other). Lone pairs prefer equatorial positions because equatorial positions have less repulsion — they only have 2 bonds at 90° (the axial bonds), while axial positions have 3 bonds at 90° (the equatorial bonds). With 2 lone pairs: both go to equatorial positions, leaving 3 F atoms (2 axial + 1 equatorial). This gives T-shaped molecular geometry (the 2 axial F + 1 equatorial F form a T).

4. Important Interhalogen Compounds

📌 ClF (AB): Linear, no lone pairs

📌 ClF₃ (AB₃): T-shaped, 2 lp on Cl — this question!

📌 ClF₅ (AB₅): Square pyramidal, 1 lp on Cl

📌 BrF₃ (AB₃): T-shaped, 2 lp on Br (same as ClF₃)

📌 BrF₅ (AB₅): Square pyramidal, 1 lp on Br

📌 IF₇ (AB₇): Pentagonal bipyramidal, 0 lp on I

📌 ICl (AB): Linear

📌 ICl₃ (AB₃): T-shaped, 2 lp on I

📌 Interhalogen compounds: between two different halogens, more electronegative halogen is terminal

5. Bond Angles in ClF₃

In ideal T-shape (from trigonal bipyramidal): F(axial)−Cl−F(axial) = 180°, F(axial)−Cl−F(equatorial) = 90°. However, due to lone pair repulsions (lp-lp and lp-bp), the actual bond angles are slightly less than ideal: F(axial)−Cl−F(equatorial) ≈ 87·5° (slightly less than 90° due to 2 lone pairs pushing the bond pairs inward). This distortion from ideal geometry is a key feature explained by VSEPR theory.

6. Hybridisation of ClF₃

Steric number = 5 → sp³d hybridisation. Cl uses one 3s, three 3p, and one 3d orbital to form 5 hybrid orbitals. Three hybrid orbitals form bonds with F atoms. Two hybrid orbitals hold lone pairs. The d orbital participation allows Cl to expand its octet (have more than 8 electrons around it) — this is possible for elements in Period 3 and beyond (they have available d orbitals). Period 2 elements (C, N, O, F) cannot expand octet — no available d orbitals.

7. Xenon Fluoride Geometries

📌 XeF₂: Steric No. 5, 3 lp, 2 bp → Linear (lp in equatorial)

📌 XeF₄: Steric No. 6, 2 lp, 4 bp → Square planar (lp on opposite sides)

📌 XeF₆: Steric No. 7, 1 lp, 6 bp → Distorted octahedral

📌 XeO₃: Steric No. 4, 1 lp, 3 bp → Trigonal pyramidal

📌 XeO₄: Steric No. 4, 0 lp, 4 bp → Tetrahedral

📌 XeOF₄: Steric No. 6, 1 lp, 5 bp → Square pyramidal

8. Shapes of Common Molecules — Quick Reference

Linear (180°): CO₂, BeCl₂, CS₂, HCN, C₂H₂. Trigonal planar (120°): BF₃, BCl₃, SO₃, NO₃⁻, CO₃²⁻. Bent (~105° or ~120°): H₂O (104·5°), H₂S (92°), SO₂ (119·5°), NO₂ (134°). Tetrahedral (109·5°): CH₄, CCl₄, SiCl₄, NH₄⁺, PO₄³⁻. Trigonal pyramidal (~107°): NH₃ (107°), PCl₃, NF₃, H₃O⁺. T-shaped (~87·5°): ClF₃, BrF₃, ICl₃. Square planar (90°): XeF₄, [Pt(CN)₄]²⁻, [NiCl₄]²⁻ (d²sp³). Octahedral (90°): SF₆, [Co(NH₃)₆]³⁺.

Frequently Asked Questions
1. How to calculate steric number for ClF₃?
Steric number = ½(V + M) where V = valence electrons on central atom, M = number of monovalent atoms attached. For ClF₃: V = 7 (Cl is Group 17), M = 3 (three F atoms). Steric No. = ½(7 + 3) = 5. This means 5 electron pairs total around Cl: 3 bond pairs (Cl−F bonds) + 2 lone pairs. Parent geometry = trigonal bipyramidal. With 2 lone pairs in equatorial positions: molecular shape = T-shaped.
2. Why do lone pairs prefer equatorial positions in trigonal bipyramidal?
In trigonal bipyramidal: axial positions have 3 neighbours at 90° (the 3 equatorial bonds) + 1 at 180°. Equatorial positions have 2 neighbours at 90° (the 2 axial bonds) + 2 at 120°. 90° interactions cause much stronger repulsion than 120° interactions. Lone pairs (which cause more repulsion than bond pairs) prefer equatorial positions because they have fewer 90° interactions (2 axial × 90° vs 3 equatorial × 90° for axial positions). So lone pairs go equatorial to minimise total repulsion energy.
3. What is the hybridisation of ClF₃?
Steric number = 5 → sp³d hybridisation. Cl uses: 1 × 3s orbital + 3 × 3p orbitals + 1 × 3d orbital = 5 sp³d hybrid orbitals. Three sp³d orbitals form sigma bonds with 3 F atoms. Two sp³d orbitals hold lone pairs. Note: Only Period 3+ elements can use d orbitals (Cl, P, S, etc.). Period 2 elements (C, N, O) cannot exceed octet (no d orbitals available).
4. What is the difference between T-shaped and trigonal planar?
Trigonal planar: 3 bond pairs, 0 lone pairs, steric no. = 3. All 3 atoms in the same plane, 120° bond angles. Example: BF₃. T-shaped: 3 bond pairs, 2 lone pairs, steric no. = 5. Two atoms in axial positions, one in equatorial. The three atoms form a T pattern. Bond angles ~90° (axial-equatorial) and ~180° (axial-axial). The lone pairs are NOT visible in molecular geometry — only the positions of atoms matter. Example: ClF₃.
5. Why is XeF₂ linear despite Xe having 3 lone pairs?
XeF₂: V = 8 (Xe, Group 18), M = 2 → Steric No. = ½(8+2) = 5. Bond pairs = 2, lone pairs = 3. Parent: trigonal bipyramidal. All 3 lone pairs go to equatorial positions (minimise repulsion). Two F atoms occupy axial positions (180° apart). Three equatorial lone pairs arrange at 120° to each other. Molecular geometry = LINEAR (only F atoms considered). This is a beautiful example of how a molecule with 3 lone pairs can be linear!
6. What is the shape of SF₄ and how many lone pairs does S have?
SF₄: V = 6 (S, Group 16), M = 4 → Steric No. = ½(6+4) = 5. Bond pairs = 4, lone pairs = 1. Parent: trigonal bipyramidal. 1 lone pair goes to equatorial position. 4 F atoms: 2 axial + 2 equatorial. Molecular geometry = See-saw (or disphenoidal). Bond angles: axial-axial ~180° (slightly less), equatorial-equatorial ~120° (slightly less), axial-equatorial ~90° (slightly less). The lone pair pushes all bond angles inward.
7. Which molecules are exceptions to octet rule?
Three types of exceptions: (1) Incomplete octet (< 8 electrons): BeCl₂ (4e), BCl₃ (6e), AlCl₃ (6e) — Period 2 elements. (2) Odd electron species (free radicals): NO (11e), NO₂ (17e), ClO₂ (19e). (3) Expanded octet (> 8 electrons): Period 3+ elements with d orbitals: PCl₅ (10e), SF₆ (12e), ClF₃ (10e), XeF₄ (12e). Remember: only Period 3+ elements can have expanded octet. Period 2 (Li to Ne) never exceeds 8 electrons.
8. What are interhalogen compounds and why are they more reactive than halogens?
Interhalogen compounds form between two different halogens (XYₙ where X is larger halogen). Examples: ClF, ClF₃, ClF₅, BrF, BrF₃, BrF₅, IF, IF₃, IF₅, IF₇. More reactive than parent halogens because: (1) The X−Y bond is weaker than X−X or Y−Y bonds (polarisation makes it easier to break). (2) The X−Y bond is polarised (more electronegative halogen is δ⁻), making both atoms reactive. (3) Can act as both oxidising and fluorinating agents. Used in: nuclear fuel processing (ClF₃ + UF₄ → UF₆), fluorinating agents in organic synthesis.
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