Which way do the dipole moments of the I-I bonds in the triiodide anion point?

Question

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1. What the problem is asking

The triiodide ion, I₃⁻, is linear

I – I – I

We are asked to decide which way the dipole‑moment vectors of the two I–I bonds point (the “bond dipoles”).
The convention to use is the usual chemistry one: the arrow points from the positive side of the bond toward the negative side.
The overall dipole moment of the whole ion is zero because the two bond dipoles are equal in magnitude and opposite in direction, but the question wants the individual directions.

2. Step‑by‑step reasoning

2.1 Count electrons and assign formal charges

Valence electrons: 7 (e⁻) × 3 I + 1 (e⁻) (the extra electron of the anion) = 22 e⁻

Take a Lewis structure with two single I–I bonds:

atom	non‑bonding e⁻	bonding e⁻	total e⁻ counted	valence e⁻ (7)	formal charge
terminal I (each)	6 (three lone pairs)	2 (one bond)	8	7	0
central I	2 (one lone pair)	4 (two bonds)	6	7	–1

So the formal charge is –1 on the central iodine, 0 on the two terminals.

2.2 What does a formal charge tell us about electron density?

A formal charge is a bookkeeping tool; it does not equal the real charge distribution.
In a highly polarizable system such as I₃⁻ the extra electron is delocalised over the three atoms. Quantum‑chemical (MO) calculations give roughly

[ \begin{aligned} \text{central I} &\approx -0.30\,e
\text{each terminal I} &\approx -0.35\,e \end{aligned} ]

so the central atom is still the most electron‑rich region of each individual I–I bond.

2.3 Electronegativity argument

All three atoms are iodine; their Pauling electronegativity is the same (2.66).
Therefore, the only factor that can create a dipole across a given I–I bond is the unequal electron density that results from the extra negative charge residing more on the central atom than on the terminal atom of that bond.

2.4 Direction of a bond dipole

By the chemistry convention, the dipole‑moment arrow points from the less negative (more positive) side toward the more negative side.

For the left I–I bond: the left terminal iodine carries slightly less negative charge than the central iodine → the arrow points rightward, i.e. toward the central atom.
For the right I–I bond: the right terminal iodine is the less negative side, the central iodine is the more negative side → the arrow points leftward, i.e. toward the central atom.

Thus both bond dipoles point inward, toward the central iodine atom.

2.5 Verify that the vector sum is zero

If we assign the same magnitude ( \mu ) to each bond dipole (the two I–I bonds have the same length and the same charge separation), the two arrows are opposite vectors:

[ \boldsymbol{\mu}\text{left}=+\mu\;\hat{x},\qquad \boldsymbol{\mu}\text{right}=-\mu\;\hat{x} ]

[ \boldsymbol{\mu}\text{total}= \boldsymbol{\mu}\text{left}+ \boldsymbol{\mu}_\text{right}=0 ]

which matches the known fact that the overall dipole moment of I₃⁻ is zero.

3. Final answer

The bond dipole for the left I–I bond points from the left terminal iodine toward the central iodine.
The bond dipole for the right I–I bond points from the right terminal iodine toward the central iodine.

Both bond dipoles are equal in magnitude, opposite in direction, and therefore cancel, giving a net dipole moment of zero for the triiodide ion.

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Assuming the formal charge equals the actual charge distribution	Formal charges are a bookkeeping device; real electron density can be delocalised.	Look at resonance or MO results; remember that highly polarizable atoms (I) spread charge.
Thinking “negative charge moves outward → dipoles point outward”	The dipole arrow points toward the negative side, not away from it.	Remember the convention: positive → negative.
Ignoring that the three atoms are identical	If one assumes different electronegativities for the termini vs. the centre, the direction is mis‑assigned.	Use the fact that all three atoms are iodine; only charge asymmetry creates a dipole.
Concluding that the bond dipoles must be zero because the overall dipole is zero	Zero net dipole does not mean each bond dipole is zero; they can cancel.	Treat each bond separately, then sum vectors to check the total.
Using the vector arrow convention opposite to the chemistry one	Some physics textbooks draw arrows from negative to positive.	Explicitly state the chemistry convention at the start of the solution.

By keeping these points in mind, the direction of the individual I–I bond dipoles in I₃⁻ can be determined correctly.

Original question: Which way do the dipole moments of the I-I bonds in the triiodide anion point? on Chemistry Stack Exchange, licensed CC BY-SA.