Question

Sol. Key concept: Count total valence electrons and then subtract bonding electrons.

Each iodine atom contributes seven valence electrons, and the negative charge contributes one more. The ion therefore has 3 × 7 + 1 = 22 valence electrons.

In the conventional linear I–I–I Lewis structure, the two I–I bonds use four electrons. The remaining 22 − 4 = 18 electrons make nine lone pairs. Each terminal iodine has three lone pairs and the central iodine also has three.

$$N_{lone\ pairs} = \frac{22 - 4}{2} = 9$$

The question asks for the total in the ion, not just the three on the central atom. Hence (d).