Electronic Configurations
Sodium (Na, Z = 11): Na: 1s^{2} 2s^{2} 2p^{6} 3s^{1} Magnesium (Mg, Z = 12): Mg: 1s^{2} 2s^{2} 2p^{6} 3s^{2}
First Ionization Enthalpy: Na < Mg
First ionization: Na -> Na^{+} + e^{-} +Mg -> Mg^{+} + e^{-}
- In Na, the electron removed is the only \(3s^{1}\) electron; after its removal, Na⁺ attains the stable noble‑gas configuration of Ne.
- In Mg, one of the \(3s\) electrons is removed from a \(3s^{2}\) subshell; Mg⁺ is less stable than Ne‑like Na⁺.
- Therefore, it is easier (requires less energy) to remove the first electron from Na than from Mg, so \[ \Delta_{\text{i}}H_{1}(\text{Na}) < \Delta_{\text{i}}H_{1}(\text{Mg}). \]
Second Ionization Enthalpy: Na > Mg
Second ionization: \[ \ce{Na^{+} -> Na^{2+} + e^{-}} Mg^{+} -> Mg^{2+} + e^{-}
- \(\ce{Na^{+}}\) has configuration: \[ 1s^{2} 2s^{2} 2p^{6} \quad (\text{Ne‑like, very stable}) \] Removing a second electron means removing an electron from a complete inner shell, breaking a noble‑gas configuration, which needs very high energy.
- \(\ce{Mg^{+}}\) has configuration: \[ 1s^{2} 2s^{2} 2p^{6} 3s^{1} \] The second electron is still the remaining \(3s\) valence electron; removing it gives Mg²⁺ with a stable Ne‑like configuration.
- Thus it is much harder to ionize \(\ce{Na^{+}}\) further than \(\ce{Mg^{+}}\), so \[ \Delta_{\text{i}}H_{2}(\text{Na}) > \Delta_{\text{i}}H_{2}(\text{Mg}). \]
Key Idea (LaTeX Summary)
\[ \text{Formation or breaking of noble‑gas configuration} \Rightarrow \text{large change in ionization enthalpy.} \] Na: first ionization leads to a noble‑gas configuration (easy), but second ionization destroys it (very hard). Mg: second ionization leads to a noble‑gas configuration (comparatively easier).