The correct order for the size of I, I⁺, I⁻ is

The correct option is I⁻ > I > I⁺.

To understand the order of sizes for these species, we need to look at the relation between their nuclear charge and the number of electrons in their outer shells.

All three species (I, I⁺, and I⁻) are derived from the iodine atom, meaning they all have the same atomic number. Thus, they all have the same number of protons in the nucleus (53 protons), which exerts the same positive nuclear charge.

Let us analyze the number of electrons and the effective nuclear charge in each case:

1. Neutral Iodine Atom (I):
It contains 53 protons and 53 electrons. The attractive force of the nucleus is balanced by the inter-electronic repulsions among the 53 electrons.

2. Iodide Ion (I⁻):
This is an anion formed by the gain of one electron. It has 53 protons but 54 electrons. Because the number of electrons increases while the nuclear charge remains the same, the effective nuclear charge per electron decreases. Furthermore, the increased inter-electronic repulsion among the electrons causes the electron cloud to expand. Consequently, the size of the anion (I⁻) is larger than the parent neutral atom (I).

3. Iodonium Ion (I⁺):
This is a cation formed by the loss of one electron. It has 53 protons and only 52 electrons. With fewer electrons, the remaining electrons experience a greater pull from the same positive nuclear charge. The effective nuclear charge per electron increases, drawing the electron cloud closer to the nucleus. Consequently, the size of the cation (I⁺) is smaller than the parent neutral atom (I).

Therefore, comparing the sizes, the anion is the largest, followed by the neutral atom, and the cation is the smallest.
This gives the order: size of I⁻ > I > I⁺.