When a chlorine atom becomes chloride ion, its size — Periodic Table and Periodicity Chemistry Question

💡 Solution & Explanation

# Chlorine Atom to Chloride Ion: Size Change **Step 1: Electron Configuration Change** - Chlorine atom ($Cl$): 17 electrons, configuration ends in $3s^2 3p^5$ - Chloride ion ($Cl^-$): 18 electrons, configuration ends in $3s^2 3p^6$ (gains 1 electron) **Step 2: Nuclear Charge vs. Electron Count** - Both have the same nuclear charge (+17 protons) - But $Cl^-$ has **one additional electron** in the same valence shell **Step 3: Electron-Electron Repulsion** - More electrons = increased repulsion among electrons - This repulsion pushes electron clouds outward - The effective pull per electron decreases **Step 4: Ionic Radius Comparison** - The increased electron-electron repulsion overcomes the constant nuclear attraction - Result: electrons are held **less tightly** and spread out more **Conclusion:** When $Cl$ gains an electron to form $Cl^-$, the ion becomes **significantly larger** than the neutral atom. This is why **B (increases)** is correct. *Note: The size increases by approximately 40-50% when comparing ionic radius of $Cl^-$ to covalent radius of $Cl$.*