How is the structure of triiodide ion (I3-) possible?

Question

I have seen the structure of triiodide ion ($\ce{I3-}$) but I cannot understand why this structure is even possible. I have seen in my textbook that

$\ce{I_3^-}$ is formed by combination of $\ce{I2}$ and $\ce{I^-}$ ion in which $\ce{I^-}$ ion acts as a donor and $\ce{I2}$ molecule act as acceptor. It accommodates electrons in empty d orbitals:

But I think $\ce{I-}$ is stable due to its complete octet. Then why it should combine with $\ce{I2}$ molecule to form another new compound? I am confused because the Lewis structure of $\ce{I3-}$ would not make any sense if we consider the octet of central iodine atom.

Answer 1

First things first: the bonding orbitals of $\ce{I3-}$ do not contain any significant d-orbital contributions. In fact, the $\ce{I-I}$ bond lengths are significantly longer than in $\ce{I2}$, suggesting a lower bond order.

In fact, the triiodide anion should rather be imagined as a 4-electron-3-centre bond (bond order approximately 0.5) which might be easier to understand using this mesomeric depiction:

$$\ce{\overset{-}{I}\bond{...}I-I <-> I-I\bond{...}\overset{-}{I}}$$

What happens orbital-wise in this? Simplistically speaking, a 4-electron-3-centre bond is made up of the following molecular orbitals stemming from $\mathrm p_z$ orbitals of the individual iodine atoms:

1. an all-bonding lowest level orbital ($\ce{{+}I-\bond{...}-I{+}\bond{...}{+}I-{}}$, where $-$ and $+$ denote the different phases of a p orbital)

2. A far-reaching bonding orbital to which the central atom does not contribute ($\ce{{+}I-\bond{...}OIO\bond{...}-I{+}}$, where O denotes no orbital lobes; there is no p-type orbital on the central iodine that could contribute. s-type contribution is possible.)

3. an all-antibonding highest level orbital ($\ce{{+}I-\bond{...}{+}I-\bond{...}{+}I-{}}$) which remains unoccupied.

Why does this happen although the starting compounds iodide and $\ce{I2}$ are stable on their own? Obviously, most reactions happen although both starting materials are full-octet compounds and inherently stable in some way or another. And most happen because the products are more stable. So we just need to determine why it is favourable for $\ce{I- + I2}$ to react to form $\ce{I3-}$.

The simplest answer is distribution of charge. Rather than have a single negative charge on one atom you end up with a structure that has approximately half a negative charge on two different atoms if you take the resonance structures at face value (hint: don’t). Whatever the final picture is, the negative charge will be more distributed over a greater number of atoms than before. This is, very generally speaking, a great stabilisation and we can stop here.

Answer 2

Expanding on Jan's reference to three-center four-electron bonding:

When the three iodine atoms all come together in a line the overlap among all three atoms gives the combination of $p$ orbitals rendered thusly (source: Wikipedia by Arun Sridharan):

Only the lowest orbital gives a bond but, because it involves overlapping all three atoms instead of just two, it's a stronger bond than just the two-center bond in $\ce{I_2}$. However, the three-center bond is distributed over two linkages instead of one and so each individual linkage is less strongly bonded than an $\ce{I_2}$ molecule.

Which leads to the reverse problem. Why is this a property of iodine? Why don't other halogens do the same thing? In fact the other halogens do form three-atom complexes, as we shall see; but they do so with increasing difficulty as we move to lighter halogens.

What enters here is the fact that you don't just have the trihalide ion. It must be bound to counterions by ionic bonds or to polar solvent molecules by ion-dipole interactions. These additional, nonmolecular electrostatistic interactions, are more favorable with a compact spherical ion consisting of one atom instead of an extended three-atom chain. The nonmolecular interactions with surrounding ions ir dipole thus tend to break up the trihalide ions. With iodine the single atom is still a bit bulky and therefore the nonmolecular interactions have little power to break up the triiodide complex. With smaller-atom halogens the nonmolecular interactions become stronger and more important.

What is the lightest known trihalide ion, then, given the fact of these evil nonmolecular electrostatic interactions? The answer may surprise a few people.

Tribromide ion

This ion is actually fairly commonplace if you know where to look, which could be an organic chem lab. Pyridinium tribromide, $\ce{(C5H6N)^+(Br3)^-}$, is used as a brominating agent for reactive substrates such as phenols; it is easier to handle than liquid bromine. We therefore move on to a bigger, or smaller, challenge:

Trichloride ion

The $\ce{Cl_3^-}$ ion has been obtained from electrochemical studies in suitable solvents such as acetonitrile, for example see here.

Trichloride ion is also obtained as an ionic liquid by adding chlorine to an ordinary chloride salt. These species can be used as chlorinating/oxidizing agents in circumstances where organic solvents would be prone to side reactions with the chlorine. The reference given here includes a list of trichloride ion bearing materials, both solid and liquid.

Trifluoride ion

Surely we have reached our wit's end? This reference reported that trifluoride ion has been observed in a low-temperature noble gas matrix and at low pressure in the gas phase. The same reference also reports that the main contributing structures for this ion involves radical combinations rather than the ion-molecule model we might ordinarily expect for this type of complex.

So at least under suitable conditions we can go all the way to $\ce{F_3^-}$!