# What prompts sodium to give up an electron?

What triggers sodium to convert from elemental Na to Na+? I know that it wants to have a full valence shell and all that, but how does it just eject the electron out?

Case (1). The changes involved are $$\ce{Na_{(s)}->Na_{(g)}^+ +e^{-}}$$, $$\ce{M_{(g)} +e^{-}->M^{-}}$$ $$\ce{Na_{(g)}^+ + M_{(g)}^{-}->NaM_{(s)}}$$
Where $M$ is an electronegative atom.The ionization energy ($\Delta H_{I}$) and the sublimation energy ($\Delta H_S$) required in the first transformation need to be compensated for by the electron gain enthalpy($\Delta H_{eg}$) released in the second transformation and the lattice enthalpy($\Delta H_{F}\text{lattice}$) released in the final transformation. If it is compensated, the change will be spontaneous. Moreover, the enthalpy released (negative) must also compensate for the lost entropy since gaseous $M$ is converted to an ordered crystal. $\Delta G=\Delta H-T\Delta S$ tells you the compensation required for $\Delta G$ to be negative.
($|\Delta H_{eg}+\Delta H_F|-|\Delta H_I+\Delta H_S|>|T\Delta S|$)
Case (2) Elemental sodium in water might lose electrons to the hydrogen evolving hydrogen gas and forming dissolved sodium hydroxide. $$\ce{Na_{(s)}->Na_{(aq)}^+ + e^{-}}$$ $$\ce{H2O +e^{-}->\frac 12H2 +OH-}$$ here, the relesed hydration energy compensates the required sublimation energy in the first step and the negative electrode potential in the second step. There is net positive entropy change which favours the reaction, making $\Delta G$ negative. ($|\Delta H_{hyd}(\ce{OH^-,Na^+})|+|T\Delta S|=|\Delta G|$)