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 $C$?

The correct decreasing order of first ionisation enthalpies of five elements of the second period is

$B$ has a smaller first ionization enthalpy than $Be$. Consider the following statements:
$(I)$ It is easier to remove a $2p$ electron than a $2s$ electron.
$(II)$ The $2p$ electron of $B$ is more shielded from the nucleus by the inner core of electrons than the $2s$ electrons of $Be$.
$(III)$ The $2s$ electron has more penetration power than the $2p$ electron.
$(IV)$ The atomic radius of $B$ is more than $Be$.
(Atomic number: $B=5, Be=4$)
The correct statements are:

Which of the following configurations represents atoms of the elements having the highest second ionization energy?

Which factors influence the value of ionization enthalpy?

Which of the following statements is incorrect?