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

$$\ce{\overset{-}{I}\bond{...}I-I &lt;-&gt; I-I\bond{...}\overset{-}{I}}$$

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:


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