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

# Ionization Potential Comparison: $\ce{He+}$ vs $\ce{H-}$ **Step 1: Identify the electron configurations** - $\ce{He+}$: 1 electron (hydrogen-like ion with $Z = 2$) - $\ce{H-}$: 2 electrons (hydrogen atom + 1 extra electron) **Step 2: Apply the ionization potential formula for hydrogen-like species** For hydrogen-like ions: $$IP = 13.6 \times Z^2 \text{ eV}$$ For $\ce{He+}$ (removing the only electron): $$IP(\ce{He+}) = 13.6 \times 2^2 = 54.4 \text{ eV}$$ **Step 3: Calculate IP for $\ce{H-}$** For $\ce{H-}$, we remove one electron from the 1s orbital of a two-electron system. The effective nuclear charge is reduced due to electron-electron repulsion (shielding by the remaining electron): $$IP(\ce{H-}) = 0.75 \text{ eV (approximately)}$$ **Step 4: Compare the values** $$IP(\ce{He+}) \gg IP(\ce{H-})$$ $$54.4 \text{ eV} \gg 0.75 \text{ eV}$$ **Conclusion:** The order is: **$\ce{He+} > \ce{H-}$** The $\ce{He+}$ ion has much higher ionization potential because its nucleus (charge +2) strongly attracts the single electron with no shielding, whereas $\ce{H-}$ has weak binding due to mutual electron repulsion and shielding.