Draw and Explain the Lewis Structure for the I3 Ion

The I₃^– ion, also known as the triiodide ion, is a polyatomic ion composed of three iodine atoms. To draw its Lewis structure, follow these steps:

1. Count the Total Number of Valence Electrons:
   • Iodine (I) has 7 valence electrons.
   • Since there are three iodine atoms, the total number of valence electrons is 7 × 3 = 21.
   • Add one extra electron for the negative charge: 21 + 1 = 22 valence electrons.
2. Determine the Central Atom:
   • In the I₃^– ion, the central atom is the middle iodine atom.
3. Draw the Skeleton Structure:
   • Place the central iodine atom in the middle and connect it to the other two iodine atoms with single bonds.
4. Distribute the Remaining Electrons:
   • After forming the single bonds, you have used 4 electrons (2 for each bond).
   • You have 18 electrons left to distribute.
   • Place three lone pairs (6 electrons) on each of the terminal iodine atoms.
   • Place two lone pairs (4 electrons) on the central iodine atom.
5. Check the Octet Rule:
   • Each iodine atom should have 8 electrons around it.
   • The central iodine atom has 2 bonding pairs and 2 lone pairs, totaling 8 electrons.
   • The terminal iodine atoms each have 1 bonding pair and 3 lone pairs, totaling 8 electrons.

The final Lewis structure for the I₃^– ion looks like this:

    [I] - [I] - [I]
    :     :     :
    :     :     :

In this structure, the central iodine atom is bonded to two terminal iodine atoms, and each terminal iodine atom has three lone pairs of electrons. The central iodine atom also has two lone pairs of electrons.

This structure satisfies the octet rule for all atoms and accounts for the total number of valence electrons in the ion.