The third ionization energy is maximum for

The given electronic configurations are for elements $X$,$Y$,and $Z$. Note that these configurations represent ions of the elements:
$X = [Ne] \, 3s^2 \, 3p^5$
$Y = [Ne] \, 3s^2 \, 3p^6$
$Z = [Ne] \, 3s^2 \, 3p^4$
Determine the correct statement regarding the energy changes associated with these configurations.

Highest energy will be absorbed to eject out the electron in the configuration:

Consider the following ionization enthalpies of two elements $A$ and $B$.

Element	Ionization enthalpy $(kJ \ mol^{-1})$ $(1^{st}, 2^{nd}, 3^{rd})$
$A$	$899, 1757, 14847$
$B$	$737, 1450, 7731$

Which of the following statements is correct?

Which of the following pairs has a higher ionization enthalpy?
$(i)$ $Ne$ or $Ar$ $(ii)$ $Cl$ or $F$ $(iii)$ $F$ or $O$ $(iv)$ $N$ or $O$
$(v)$ $Na$ or $K$ $(vi)$ $Cl$ or $S$ $(vii)$ $Kr$ or $Xe$ $(viii)$ $P$ or $S$

What is second ionization enthalpy?