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

💡 Solution & Explanation

I don't see the options (A, B, C, D) or the iodine species listed in your question. However, I can provide the general reasoning for common iodine species size ordering: **General principle for iodine species:** The size order typically depends on charge and oxidation state: $$I^- > I_2 > I^+ \text{ or } I > I^3- > IO_3^- \text{ (for oxy-species)}$$ **Key reasoning:** 1. **$I^-$ is largest** — extra electron increases electron-electron repulsion, expanding the electron cloud 2. **$I_2$ is intermediate** — covalent molecule with defined van der Waals radius 3. **$I^+$ is smallest** — loss of electron decreases size; high charge density compresses electron cloud 4. **Oxy-species like $IO_3^-$** — multiple oxygen atoms withdraw electron density from iodine, reducing its size further **To verify option D is correct**, please share: - The four size orderings given as options A, B, C, D - Which specific iodine species are being compared Then I can explain exactly why D is the correct ordering and why the others fail.