The first $(\Delta_{i} H_{1})$ and the second $(\Delta_{i} H_{2})$ ionization enthalpies (in $kJ \ mol^{-1}$) and the $(\Delta_{eg} H)$ electron gain enthalpy (in $kJ \ mol^{-1}$) of a few elements are given below:

Elements	$\Delta_{i} H_{1}$	$\Delta_{i} H_{2}$	$\Delta_{eg} H$
$I$	$520$	$7300$	$-60$
$II$	$419$	$3051$	$-48$
$III$	$1681$	$3374$	$-328$
$IV$	$1008$	$1846$	$-295$
$V$	$2372$	$5251$	$+48$
$VI$	$738$	$1451$	$-40$

Which of the above elements is likely to be:
$(a)$ the least reactive element.
$(b)$ the most reactive metal.
$(c)$ the most reactive non-metal.
$(d)$ the least reactive non-metal.
$(e)$ the metal which can form a stable binary halide of the formula $MX_{2}$ $(X = \text{halogen})$.
$(f)$ the metal which can form a predominantly stable covalent halide of the formula $MX$ $(X = \text{halogen})$?

(A) Element $V$ is the least reactive element because it has the highest first ionization enthalpy $(\Delta_{i} H_{1})$ and a positive electron gain enthalpy $(\Delta_{eg} H)$,indicating a noble gas configuration.
$(b)$ Element $II$ is the most reactive metal as it has the lowest first ionization enthalpy $(\Delta_{i} H_{1})$,making it the easiest to lose an electron.
$(c)$ Element $III$ is the most reactive non-metal as it has a high first ionization enthalpy $(\Delta_{i} H_{1})$ and the highest negative electron gain enthalpy $(\Delta_{eg} H)$,indicating a strong tendency to gain an electron.
$(d)$ Element $V$ is the least reactive non-metal (noble gas) due to its very high ionization enthalpy and positive electron gain enthalpy.
$(e)$ Element $VI$ is a metal with a low second ionization enthalpy $(\Delta_{i} H_{2})$,allowing it to easily lose two electrons to form a stable $MX_{2}$ halide.
$(f)$ Element $I$ has a low first ionization enthalpy and a very high second ionization enthalpy,making it most likely to form a stable $MX$ halide.