Which one is the correct order of the size of the iodine species. — Periodic Table and Periodicity Chemistry Question

💡 Solution & Explanation

I need to see the options to provide a complete solution. However, I can explain the general principle for ordering iodine species by size: **Key Principle:** For iodine species, size increases with: 1. **More electrons** (more negative charge) 2. **Fewer protons** (fewer positive charges) **General size order for common iodine species:** $$I^- > I_2 > I^+ > I^{3+}$$ This is because: - $I^-$ has the most electrons (54 total), largest size - $I_2$ is neutral molecule, intermediate size - $I^+$ has lost one electron (52 electrons), smaller than $I_2$ - $I^{3+}$ has lost 3 electrons (51 electrons), smallest **If option D states:** $I^- > I_2 > I^+ $ or similar ordering following the electron principle, then D is correct because: - More electrons = greater electron-electron repulsion = larger radius - Fewer electrons = less repulsion = smaller radius - Same nucleus, but different electronic charges determine the size **Why others fail:** Any option listing species in reverse order or incorrectly grouping them violates the relationship between electron count and ionic/atomic radius. *Note: Please share the specific options (A, B, C, D) for a more targeted explanation.*