Triiodide Polarity (I3- Molecular Geometry Guide)
I₃⁻ is a linear, symmetric ion with an I–I–I bond angle of 180°. Its central iodine has two bonding pairs and three lone pairs, giving an AX₂E₃ VSEPR classification. The I–I bonds are not polar because they join identical atoms. The ion has zero net dipole moment, but it still carries an overall −1 charge.
I₃⁻ Lewis Structure and VSEPR Overview
A Lewis structure shows how valence electrons are arranged as bonds and lone pairs. For I₃⁻, counting 22 valence electrons and drawing an I–I–I chain gives a structure with two bonds, three lone pairs on the central iodine, and a total formal charge of −1.
How to count the valence electrons
Valence electrons are the electrons in an atom’s outer shell that take part in bonding. Each iodine atom contributes seven, and the −1 charge means the ion has one extra electron. The total is therefore 3 × 7 + 1 = 22 valence electrons.
The extra electron matters: without it, the electron count would describe a different species. As a reliable first step, write the overall charge beside the formula, then include that charge when counting electrons.
How to draw the Lewis structure
Start with three iodine symbols in a row, connected by two single bonds: I–I–I. The two bonds use four electrons. Give each terminal iodine three lone pairs so that each has an octet; that uses another 12 electrons. The remaining six electrons form three lone pairs on the central iodine.
This arrangement gives the central iodine two bonding pairs and three lone pairs. It also gives each terminal iodine one bond and three lone pairs. When checking your drawing, count the electrons shown: four in bonds plus 18 in lone pairs equals 22.
Formal charge check
Formal charge is an electron-counting tool used to check whether a Lewis structure matches the molecule or ion’s total charge. The standard calculation is:
Formal charge = valence electrons − nonbonding electrons − ½(bonding electrons)
| Atom in the Lewis structure | Valence electrons | Nonbonding electrons | Bonding electrons | Formal charge |
|---|---|---|---|---|
| Central iodine | 7 | 6 | 4 | −1 |
| Each terminal iodine | 7 | 6 | 2 | 0 |
The formal charges add to −1, matching the ion’s charge. This is a bookkeeping result for the Lewis structure. It does not mean that the actual negative charge must sit entirely on one iodine in every description of the ion.
Electron-Domain Arrangement, Molecular Geometry, and Bond Angle
Electron-domain arrangement describes the positions of bonding regions and lone pairs around a central atom. Molecular geometry describes the positions of the atoms alone. In I₃⁻, the five electron domains form a trigonal-bipyramidal arrangement, while the three iodine atoms form a straight line.
Why I₃⁻ is AX₂E₃
VSEPR is a model that predicts molecular shape by treating regions of electron density around a central atom as domains. In the label AX₂E₃, A is the central iodine, X₂ represents two bonded iodine atoms, and E₃ represents three lone pairs.
Count each bond to a neighboring atom as one domain, even when the bond contains more than one shared pair. Then count each lone pair as one domain. The central iodine has two bonding domains and three lone-pair domains, for five electron domains in total.
Electron-domain arrangement versus molecular geometry
Five electron domains are arranged in a trigonal-bipyramidal pattern in the VSEPR model. This arrangement has two axial positions and three equatorial positions. The three lone pairs occupy the equatorial positions; the two bonded iodine atoms occupy the axial positions.
Molecular geometry considers only the atom positions, not the lone pairs. Since the terminal iodine atoms lie on opposite sides of the central iodine, the molecular geometry is linear. This distinction explains why three lone pairs do not make I₃⁻ bent.
| Feature | I₃⁻ result | What it describes |
|---|---|---|
| VSEPR class | AX₂E₃ | Two bonded atoms and three lone pairs around the center |
| Number of electron domains | 5 | Bonding and lone-pair regions |
| Electron-domain arrangement | Trigonal bipyramidal | Positions of all five domains |
| Molecular geometry | Linear | Positions of the three iodine atoms |
| I–I–I bond angle | 180° | Angle between the two bonds |
Why the bond angle is 180°
The linear shape puts the terminal iodine atoms in opposite directions from the central iodine. The angle between the two I–I bonds is therefore 180°. When reporting the angle, use the atom sequence I–I–I so it is clear that you mean the angle at the central iodine.
A common mistake is to see three lone pairs and assume the ion must be bent. VSEPR predicts the geometry from the arrangement of all five domains, then describes molecular shape using only the atoms. For AX₂E₃, those steps lead to a linear shape.
Polarity, Net Dipole, and Common Conceptual Errors
Polarity can describe uneven charge distribution in a bond or the overall dipole of a molecule. In I₃⁻, each I–I bond joins identical atoms, and the ion’s linear, symmetric shape gives it zero net dipole moment. That result does not remove its −1 charge.
Are the I–I bonds polar?
A polar bond has unequal sharing of electrons between its bonded atoms, often because they have different electronegativities. The two atoms in an I–I bond are both iodine, so the bond is treated as nonpolar. The Lewis structure contains two I–I bonds, but neither is polar in the usual bond-polarity sense.
Do not confuse the formal charge assigned in a Lewis structure with bond polarity. The central iodine’s formal charge of −1 comes from formal-charge bookkeeping. It does not turn either I–I bond into a polar bond between different elements.
Why the ion has zero net dipole moment
A dipole moment describes separation of positive and negative charge. For a molecule with polar bonds, shape determines whether the bond dipoles cancel. In I₃⁻, the individual I–I bonds are nonpolar, and the linear structure is symmetric. The ion therefore has zero net dipole moment.
The key distinction is between dipole and net charge:
| Question | Answer for I₃⁻ | Why |
|---|---|---|
| Does the ion have polar I–I bonds? | No | Each bond joins identical iodine atoms |
| Is its molecular geometry linear? | Yes | The iodine atoms lie in a straight line |
| Is its net dipole moment zero? | Yes | The structure is symmetric |
| Is the ion neutral? | No | Its overall charge is −1 |
| Is “nonpolar molecule” the best label? | No | I₃⁻ is an ion, not a neutral molecule |
For accuracy, describe I₃⁻ as a charged molecular ion with zero net dipole moment. “Nonpolar ion” may be used informally to refer to its dipole behavior, but it should not suggest that the ion has no charge.
Common errors and how to correct them
-
Error: “I₃⁻ is bent because it has lone pairs.”
Correction: It has five electron domains. In AX₂E₃, the lone pairs occupy equatorial positions, leaving the bonded atoms opposite each other. -
Error: “Zero dipole means I₃⁻ is neutral.”
Correction: Dipole moment and net charge are different properties. I₃⁻ still has a −1 charge. -
Error: “The central iodine has −1 formal charge, so the bonds are polar.”
Correction: Formal charge is calculated for a Lewis structure; bond polarity concerns electron sharing between the two atoms in a bond. -
Error: “Trigonal bipyramidal is the molecular geometry.”
Correction: That term describes the arrangement of all five electron domains. The molecular geometry, which counts only atoms, is linear.
Polyiodide Chemistry: Where the Ion Fits
Polyiodide chemistry concerns species that contain more than two iodine atoms and have a net charge. I₃⁻ is one such ion. Its Lewis structure and shape are useful for predicting its basic bonding and dipole behavior, while its charged nature must remain part of any chemical description.
A practical context for using the structure
In chemistry problems, I₃⁻ may appear in a formula or reaction as a polyiodide ion. When you see it, start with the species itself rather than trying to infer polarity from its surroundings: count its electrons, check its formal charge, and use VSEPR to identify its shape.
The same structural analysis helps when comparing I₃⁻ with other species. A formula with three atoms is not automatically bent, and a species with zero dipole is not automatically neutral. Here, electron-domain counting, symmetry, and charge answer three separate questions.
A four-step check for homework and exams
I use this order because each step checks a different part of the answer:
- Count: Three iodine atoms contribute 21 valence electrons; the −1 charge adds one, giving 22.
- Draw: Connect the atoms in a chain, complete the terminal octets, and place the six remaining electrons as three lone pairs on the center.
- Verify and shape: Check that formal charges sum to −1. Then count five domains and classify the center as AX₂E₃, giving a linear shape.
- Assess polarity: Note that identical atoms form nonpolar I–I bonds. The symmetric linear ion has zero net dipole but remains charged.
If one result seems to conflict with another, identify which property you are checking. Formal charge, molecular geometry, bond polarity, dipole moment, and overall charge are related, but they are not interchangeable.
Frequently Asked Questions
These short answers distinguish the ion’s Lewis structure, geometry, polarity, and charge. Keep each property separate when writing a complete chemistry response: that makes it easier to avoid calling I₃⁻ neutral or predicting the wrong shape.
Is I₃⁻ polar or nonpolar?
It has zero net dipole moment. Since it is charged, a precise answer is that I₃⁻ is a molecular ion with zero dipole, not a neutral nonpolar molecule.
What is the molecular geometry of I₃⁻?
Its molecular geometry is linear. The iodine atoms form a straight line, with the central iodine between the two terminal iodines.
What is the I–I–I bond angle?
The bond angle is 180°. The two terminal iodine atoms lie in opposite directions from the central iodine.
What is the VSEPR notation for I₃⁻?
The notation is AX₂E₃. The central iodine is bonded to two iodine atoms and has three lone pairs.
How many valence electrons does I₃⁻ have?
It has 22 valence electrons: three iodine atoms contribute 21, and the negative charge adds one more.
How many lone pairs are on the central iodine?
The central iodine has three lone pairs in the standard Lewis structure, along with two bonding pairs.
Which atom has a formal charge of −1 in the Lewis structure?
The central iodine has a formal charge of −1; each terminal iodine has a formal charge of zero. The formal charges total −1.
Are the I–I bonds polar?
No. Each I–I bond joins two atoms of the same element, so it is treated as nonpolar.
Why does I₃⁻ have zero dipole but still have a charge?
Dipole moment measures charge separation, while net charge measures the total charge of the species. I₃⁻ is symmetric and has zero net dipole, but its total charge remains −1.
Is the electron-domain arrangement also linear?
No. The five electron domains around the central iodine have a trigonal-bipyramidal arrangement. The atom-only molecular geometry is linear.
(This article was written by one of our staff writers, Alan Westbrook. Visit our Meet the Team page.)
