An element will have lowest ionisation potential when its electronic configuration is — Periodic Table and Periodicity Chemistry Question

💡 Solution & Explanation

# Ionization Potential and Electronic Configuration **Key Concept:** Ionization potential (IE) is the energy needed to remove an electron. It is **lowest for elements that hold their outermost electron most loosely**. ## Factors favoring low IE: 1. **Large atomic radius** → electrons far from nucleus 2. **Shielding effect** → inner electrons shield outer electrons from nuclear charge 3. **Alkali metals** (Group 1) → single valence electron in outermost s-orbital, easiest to remove ## Electronic configurations to compare: Without seeing the options, the **lowest IE occurs for:** - **$ns^1$ configuration** (alkali metals like Li, Na, K) - The single valence electron experiences maximum shielding - Located in the outermost shell, furthest from nucleus - Minimal effective nuclear charge on that electron ## Why other configurations have higher IE: - **$ns^2$** → two electrons, more attraction - **$ns^2np^x$** → more electrons, higher effective charge - **$d$ or $f$ block** → more complex, but still higher than alkali metals **Answer D** most likely represents an **alkali metal configuration** ($ns^1$) or the configuration of Group 1 elements, which universally have the lowest ionization potentials among all elements.