Question

The second I.P. for alkali metals shows a jump while the third I.P. for alkaline earth metal shows a jump. Explain.

Answer 1

After removal of one $e^{-}$ from alkali metals they gets stable fully filled outer electronic configuration. Eg:- $e^{-}$ configuration of alkali metal, $Na$ is $2p^{6}3s^1$

$Na⟶N{a}^{+}+e^{-}$

after removal of one $e^{-}$ configuration becomes $N{a}^{+}=1s^{2}2s^{2}2p^6$

$p-$orbital is fully filled and is very stable. Hence a high amount of energy is needed to remove an $e^{-}$ from this stable state configuration. $\therefore I{E}_2$ is high

Similarly goes for alkaline earth metals. After losing $2e^{-}s$ these acquire stable state configuration and hence require high energy for third removal of $e^{-}$

Eg:- $Mg⟶M{g}^{2+}+2e^{-}$

$1s^{2}2s^{2}2p^6$

$\downarrow$

fully filled shell, thus removal of $e^{-}$ is difficult. Thus $I{E}_3$ for alkaline earth metal is quite high.