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

I need to compare the ionization potentials of $He^+$ ion and $H^-$ atom. **Step 1: Understand what we're comparing** - $He^+$: A hydrogen-like ion with nuclear charge $Z = 2$, one electron - $H^-$: A hydrogen-like atom with nuclear charge $Z = 1$, two electrons **Step 2: Apply the ionization potential formula** For hydrogen-like species (one electron): $$IE = 13.6 \times Z^2 \text{ eV}$$ For $He^+$ (Z = 2): $$IE(He^+) = 13.6 \times (2)^2 = 54.4 \text{ eV}$$ **Step 3: Calculate ionization potential of $H^-$** For $H^-$, removing one electron gives neutral $H$: $$H^- \rightarrow H + e^-$$ The ionization potential of $H^-$ is approximately **0.75 eV** (this is the electron affinity of $H$ atom, which equals the ionization energy of $H^-$). **Step 4: Compare** $$IE(He^+) = 54.4 \text{ eV} >> IE(H^-) = 0.75 \text{ eV}$$ **Conclusion:** $$IE(He^+) > IE(H^-)$$ The $He^+$ ion has a much higher ionization potential because: - It has a higher nuclear charge ($Z = 2$ vs $Z = 1$) - It holds its single electron much more tightly despite electron-electron repulsion being absent - The $H^-$ ion is weakly bound due to electron-electron repulsion in the two-electron system **Answer: C** (indicating $He^+ > H^-$)