Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: The correct statement for the molecule ${\text{CsI}}_3$ is

When we encounter a molecule like ${\text{CsI}}_3$, our first instinct might be to assume it's a covalent molecule, much like ${\text{PCl}}_3$ or ${\text{NH}}_3$. However, we must always look at the constituent elements.

Cesium ($\text{Cs}$) is a heavy alkali metal located in Group 1 of the periodic table. It is highly electropositive. On the other hand, Iodine ($\text{I}$) is a Group 17 halogen, which is highly electronegative.

When an element from the far left of the periodic table meets an element from the far right, the massive difference in electronegativity dictates that they will form an ionic bond. Therefore, ${\text{CsI}}_3$ is fundamentally an ionic compound, not a covalent one.

Now that we know it's an ionic compound, we need to determine the charges on the ions. Let's start with the cation.

Cesium, being an alkali metal, has a valence electronic configuration of $6s^1$. It has exactly one electron in its outermost shell. By losing this single electron, Cesium achieves a highly stable noble gas configuration (isoelectronic with Xenon).

Because removing a second or third electron would require breaking into this incredibly stable, full inner shell, the ionization energy required is astronomically high. Therefore, Cesium strictly forms a $+1$ cation: ${\text{Cs}}^{+}$. A $+3$ state, as suggested in one of the options, is chemically impossible under normal conditions.

Since the Cesium cation carries a $+1$ charge, the entire anionic portion of the molecule must carry a net $-1$ charge to maintain electrical neutrality.

This means the three iodine atoms must collectively carry a $-1$ charge. They do not exist as three separate ${\text{I}}^{-}$ ions, nor do they exist as a loose mixture of ${\text{I}}^{-}$ and a neutral ${\text{I}}_2$ molecule in the lattice. Instead, they bond together to form a single, stable polyatomic anion known as the triiodide ion, denoted as ${\text{I}}_3^{-}$.

You might wonder how three iodine atoms can form a single ion. The formation of the triiodide ion is a classic example of coordinate bonding.

A standard iodide ion (${\text{I}}^{-}$) donates a lone pair of electrons to the empty antibonding orbital of a neutral iodine molecule (${\text{I}}_2$). This interaction forms a coordinate covalent bond, resulting in the ${\text{I}}_3^{-}$ complex.

Geometrically, the central iodine atom in the ${\text{I}}_3^{-}$ ion undergoes $s{p}^{3}d$ hybridization. It possesses three lone pairs and two bond pairs, which arrange themselves in a trigonal bipyramidal geometry. The three lone pairs occupy the equatorial positions to minimize repulsion, forcing the three iodine atoms into a perfectly linear shape.

Large, highly polarizable anions like ${\text{I}}_3^{-}$ are relatively unstable on their own, but they are beautifully stabilized in a solid crystal lattice by large cations like ${\text{Cs}}^{+}$.

Bringing all our logical deductions together, we can confidently state that Cesium triiodide (${\text{CsI}}_3$) is an ionic compound composed of a ${\text{Cs}}^{+}$ cation and an ${\text{I}}_3^{-}$ anion.

It is not covalent, it does not contain a ${\text{Cs}}^{3+}$ ion, and it is not a mere physical mixture of ${\text{I}}^{-}$ and ${\text{I}}_2$ in the lattice. The correct statement is that it contains ${\text{Cs}}^{+}$ and ${\text{I}}_3^{-}$ ions.