To determine the total number of lone pairs in the I₃^- ion, we need to understand its structure and bonding. The I₃^- ion is linear and symmetrical, with a central iodine atom bonded to two terminal iodine atoms.

Step 1: Lewis Structure
Iodine (I) has 7 valence electrons. In I₃^-, there is an extra electron due to the negative charge, so total valence electrons = 3*7 + 1 = 22.

Step 2: Bonding
The central I atom forms two single bonds with terminal I atoms, using 4 electrons (2 bonds * 2 electrons each).

Step 3: Lone Pairs
After bonding, remaining electrons = 22 - 4 = 18. These are distributed as lone pairs (each lone pair has 2 electrons). So, number of lone pairs = 18 / 2 = 9.

Distribution:
- Central I atom has 3 lone pairs (6 electrons).
- Each terminal I atom has 3 lone pairs (6 electrons each).
Total lone pairs = 3 (central) + 3 (terminal1) + 3 (terminal2) = 9.

Final Answer: The total number of lone pairs is 9.

Related Topics

Lewis Structures: Diagrams that show the bonding between atoms and lone pairs of electrons in a molecule.

Valence Shell Electron Pair Repulsion (VSEPR) Theory: Predicts molecular geometry based on electron pair repulsion. For I₃^-, it is linear.

Formulae

Total valence electrons = sum of valence electrons of all atoms + charge (if negative, add; if positive, subtract).

Number of bonding electrons = 2 * number of bonds.

Number of lone pair electrons = total valence electrons - bonding electrons.

Number of lone pairs = lone pair electrons / 2.