The order of ionization potential between He+ ion and H-atom (both species are in gaseous state) is — Periodic Table and Periodicity Chemistry Question

💡 Solution & Explanation

# Comparing Ionization Potential: $He^+$ vs $H^-$ **Step 1: Identify the electronic configuration** - $He^+$ ion: 1 electron (like hydrogen but nuclear charge $Z = 2$) - $H^-$ atom: 2 electrons (like helium but nuclear charge $Z = 1$) **Step 2: Apply the ionization potential formula** For hydrogen-like species: $$IE = 13.6 \times Z^2 \text{ eV (for single electron removal)}$$ For $He^+$: $$IE(He^+) = 13.6 \times 2^2 = 13.6 \times 4 = 54.4 \text{ eV}$$ **Step 3: Consider electron-electron repulsion in $H^-$** For $H^-$ (2-electron system), the second electron experiences: - Nuclear attraction: $Z = 1$ - Shielding/repulsion from the first electron The effective nuclear charge is reduced by electron shielding. The ionization energy of $H^-$ is much lower: $$IE(H^-) \approx 0.75 \text{ eV}$$ **Step 4: Compare** $$IE(He^+) \gg IE(H^-)$$ $$54.4 \text{ eV} \gg 0.75 \text{ eV}$$ **Conclusion:** The order is **$He^+ > H^-$** The higher nuclear charge (+2) in $He^+$ strongly attracts its single electron, making it much harder to remove compared to the loosely bound second electron in $H^-$.