How to work it out
- Central atom: I, with 7 valence electrons.
- Add 1 for each H or halogen around it: 2. O and S add nothing.
- Steric number = ½ × (7 + 2 + 1 (negative charge)) = 5.
- 2 atoms are bonded, so 5 − 2 = 3 lone pairs.
- 5 electron pairs → sp³d, trigonal bipyramidal arrangement; the lone pairs make the shape linear.
Why I₃⁻ has this shape
The central I has three lone pairs in the equatorial plane, so the three I atoms line up: linear, like XeF₂.
I₃⁻ is nonpolar: its shape is symmetrical, so the bond dipoles cancel out.
Other molecules with the same linear shape: XeF₂.
Hybridization of common molecules
| Molecule or ion | Hybridization | Shape | Lone pairs |
|---|---|---|---|
| XeF₄ | sp³d² | Square planar | 2 |
| XeF₂ | sp³d | Linear | 3 |
| XeF₆ | sp³d³ | Distorted octahedral | 1 |
| XeO₃ | sp³ | Trigonal pyramidal | 1 |
| XeOF₄ | sp³d² | Square pyramidal | 1 |
| SF₄ | sp³d | See-saw | 1 |
| SF₆ | sp³d² | Octahedral | 0 |
| ClF₃ | sp³d | T-shaped | 2 |
| BrF₃ | sp³d | T-shaped | 2 |
| BrF₅ | sp³d² | Square pyramidal | 1 |
| IF₅ | sp³d² | Square pyramidal | 1 |
| IF₇ | sp³d³ | Pentagonal bipyramidal | 0 |
| PCl₅ | sp³d | Trigonal bipyramidal | 0 |
| PCl₃ | sp³ | Trigonal pyramidal | 1 |
| NH₃ | sp³ | Trigonal pyramidal | 1 |
| H₂O | sp³ | Bent (V-shaped) | 2 |
| CH₄ | sp³ | Tetrahedral | 0 |
| BF₃ | sp² | Trigonal planar | 0 |
| BeCl₂ | sp | Linear | 0 |
| CO₂ | sp | Linear | 0 |
| SO₂ | sp² | Bent (V-shaped) | 1 |
| SO₃ | sp² | Trigonal planar | 0 |
| H₂S | sp³ | Bent (V-shaped) | 2 |
| NH₄⁺ | sp³ | Tetrahedral | 0 |
| H₃O⁺ | sp³ | Trigonal pyramidal | 1 |
| NO₃⁻ | sp² | Trigonal planar | 0 |
| NO₂⁺ | sp | Linear | 0 |
| CO₃²⁻ | sp² | Trigonal planar | 0 |
| SO₄²⁻ | sp³ | Tetrahedral | 0 |
| ClO₄⁻ | sp³ | Tetrahedral | 0 |
| I₃⁻ | sp³d | Linear | 3 |
| ICl₄⁻ | sp³d² | Square planar | 2 |
| PF₆⁻ | sp³d² | Octahedral | 0 |
| SnCl₂ | sp² | Bent (V-shaped) | 1 |
Questions students ask
What is the hybridization of I₃⁻?
sp³d. The central I has 5 electron pairs around it: 2 bonds and 3 lone pairs.
What is the shape of I₃⁻?
Linear. The electron pairs are arranged trigonal bipyramidal, but the 3 lone pairs are not counted in the shape.
What is the bond angle in I₃⁻?
180°.
Is I₃⁻ polar or nonpolar?
I₃⁻ is nonpolar: its shape is symmetrical, so the bond dipoles cancel out.