I3− Lewis Structure
Triiodide ion · Iodine dissolved in iodide solution, and the reason iodine turns starch blue-black.
The I3- Lewis structure has 22 valence electrons - the 1− charge adds one, drawn as two I–I bonds, with nine lone pairs in total, all inside square brackets. The central I is sp3d hybridized and the ion is linear, with a bond angle of 180 degrees. The I carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.
- 22 valence e−
- linear
- 180° bond angle
- sp3d
- 1− charge
I3− molecular geometry
Linear, from 5 electron domains on the central I: 2 bonded atoms and 3 lone pairs (AX2E3). The electron geometry is trigonal bipyramidal; the shape is what is left when the lone pairs are not counted as atoms. Wedges point toward you, hashed wedges away. Every shape in the VSEPR chart →
Calculated properties
| Total valence electrons | 22 |
|---|---|
| Bonding electrons | 4 (2 shared pairs) |
| Nonbonding electrons | 18 (9 lone pairs) |
| Lone pairs on I | 3 |
| Electron domains (steric number) | 5 |
| VSEPR notation | AX2E3 |
| Electron geometry | trigonal bipyramidal |
| Molecular geometry | linear |
| Bond angle | 180° |
| Hybridization | sp3d |
| Charge | 1− |
| Charge distribution | localised |
| Formal charges | I: -1 |
| Resonance structures | none |
| Molar mass | 380.7 g/mol |
I Iodine 3 atoms, 7 valence e− each
How to draw the I3− Lewis structure
Step through it. The bar shows how much of the 22-electron budget is spent at each stage - it never grows, which is the whole constraint.
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Skeleton. Join every connected pair with one bond. 2 bonds spends 4 electrons.
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Outer octets. Spend electrons as lone pairs on the outer atoms until each has eight. 6 are still unplaced.
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Centre. Everything still unplaced goes on I, which now has 10 electrons around it.
The reasoning at each step
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Count the valence electrons
Add up the outer-shell electrons every atom brings, then account for the charge. A 1- ion has gained one electron, so it is added to the budget. Skip this and every count that follows will be wrong.
3 x 7 (I) + 1 (1- charge) = 22 valence electrons -
Work out what is bonded to what
Three identical atoms form a chain: one I sits in the middle and bonds to the other two.
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Join everything with single bonds first
Every connection starts as one shared pair. Draw them all before worrying about double bonds - the arithmetic in the next step is what tells you where the double bonds have to go.
2 bonds x 2 = 4 electrons used, 18 left -
Work out how far short you are
Add up what every atom still needs to fill its shell - eight electrons for most atoms, two for hydrogen - minus what its single bonds already give it. Compare that with the electrons you have left. The difference decides everything that follows.
needs 16, has 18, 2 to spare -
Park the spare electrons on the central atom
There are 2 electrons more than the outer atoms can use. They become 1 extra lone pair on I, which is in period 5 and can hold more than eight electrons. A period-2 atom could not do this.
10 electrons around I -
Fill in the lone pairs
Every electron not in a bond sits as a lone pair on an atom. Each atom takes exactly what it needs to finish its shell - there is no choice left at this point.
I1: 3, I2: 3, I3: 3 (18 electrons) -
Check the central atom
I ends up with 10 electrons, an expanded octet. Only period 3 and below can do this.
I: 4 bonding + 6 nonbonding = 10 -
Check the formal charges
Formal charge is valence electrons, minus lone-pair electrons, minus the number of bonds. They are not real charges, and they are not the ion's charge either - but they must add up to it. Adding these up gives 1-, which is the check that the structure is complete.
I2: 7 - 6 - 2 = -1 -
Work out the shape
Count the electron domains on I: 2 bonded groups plus 3 lone pairs. A double or triple bond still counts as one domain, because it points in one direction. That gives trigonal bipyramidal electron geometry; ignore the lone pairs and the atoms themselves sit in a linear arrangement. Three equatorial lone pairs leave the two bonds opposite each other.
steric number 5 -> sp3d -> linear, bond angle 180 -
Decide whether it is polar
The bond dipoles and lone pairs are arranged symmetrically, so the vectors cancel and the molecule has no net dipole.
dipoles cancel -> nonpolar
Where the 1− charge sits
The I carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.
People also ask whether I3− is polar. Strictly it is neither: polar and nonpolar sort neutral molecules by whether their bond dipoles cancel. A 1− ion is pulled towards any nearby charge whatever its shape, which is why its salts dissolve in water regardless of geometry. The table below is still worth reading - it shows how unevenly each bond holds its electrons, which is what decides whether the charge ends up spread or concentrated.
| Bond | Electronegativity difference | Character |
|---|---|---|
| I–I | 0.00 | nonpolar covalent |
Common questions
How many valence electrons does I3- have?
I3- has 22 valence electrons. The atoms bring 21 between them, and the 1− charge adds one. 4 of them are in bonds and 18 sit in lone pairs.
Why does I3- have a 1− charge?
The I carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean. The brackets around the drawing are part of the answer: they say the charge belongs to the whole ion rather than to whichever atom happens to be nearest.
What is the molecular geometry of I3-?
I3- is a linear ion. The central I has 2 bonded groups and three lone pairs, a steric number of 5, which gives trigonal bipyramidal electron geometry and a linear ion.
What is the bond angle in I3-?
The bond angle in I3- is 180 degrees. Three equatorial lone pairs leave the two bonds opposite each other.
What is the hybridization of I3-?
The central I in I3- is sp3d hybridized. Steric number 5 means 5 orbitals have to be mixed, which is exactly what sp3d gives you.
Is I3- polar or nonpolar?
Neither, strictly. Polar and nonpolar describe neutral molecules, by asking whether their bond dipoles cancel out. I3- carries a net 1− charge, so it is pulled towards any nearby charge whatever its shape - which is why ionic compounds dissolve in water regardless of geometry. The question the shape does answer is whether the charge is spread evenly: it is, because the linear arrangement puts every identical bond at the same angle to the others.