Question

Select in each pair, and one having lower ionization energy and explain the reason
(a) $I$ and $I^{-}$ (b) $Br$ and $K$ (c) $Li$ and ${Li}^{+}$ (d) $Ba$ and $Sr$ (e) $O$ and $S$ (f) $Be$ and $B$ (g) $N$ and $O$

Answer 1

$\left(a\right)$ $IE=I>I^{-}($ since$,$ negatively charged species have easy tendency to give $e^{-}$ due to more $e^{-}$ density$.$

$\left(b\right)$ $IE=K<Be($ due to less $Z_{eff}$ on $K,$ it is easy to extract $e^{-})$
$\left(c\right)$ $IE={Li}^{+}>Li($ Due to more $Z_{eff}$ of ${Li}^{+}$

$\left(d\right)$ $IE=Sr>Ba($ down group $IE$ decreases due to less $Z_{eff}$ across group$)$

$\left(e\right)$ $IE=O>S($ Due to small size of $O$ them$)$

$\left(f\right)$ $IE=Be>B($ completely filled orbital in $Be)$

$\left(g\right)$ $IE=N>O($ Due to half-filled orbital is more stable than partially filled$,$ $N$ has more $IE$ them $O.$