The increasing order of the first ionization enthalpies of the elements $B$,$P$,$S$ and $F$ (lowest first) is:

Both nitrogen and oxygen have electrons in the $2p$ orbital. In which of these is the electron-electron repulsion greater,and what is its effect?

Match List-$I$ with List-$II$:

List-$I$ (Electronic configuration of neutral atom where $n=2$)	List-$II$ ($1^{st}$ Ionization Energy in $\text{kJ mol}^{-1}$)
$A. ns^2$	$I. 2080$
$B. ns^2np^1$	$II. 899$
$C. ns^2np^3$	$III. 800$
$D. ns^2np^6$	$IV. 1402$

The electronic configurations of elements $A, B$ and $C$ are $[He] 2s^1$,$[Ne] 3s^1$ and $[Ar] 4s^1$ respectively. Which one of the following orders is correct for the first ionization potentials (in $kJ \ mol^{-1}$) of $A, B$ and $C$?

Which of the following has the least ionization potential?

The first four ionisation energy values of an element are $191$,$578$,$872$ and $5962 \ kcal$. The number of valence electrons in the element is :-