What is the number of unpaired electrons in the electronic configuration of the $N^{+2}$ ion?

While filling up of electrons in the atomic orbitals,the $4s$ orbital is filled before the $3d$ orbital,but the reverse happens during the ionization of the atom. Explain why?

Which electron is associated with the least energy?

Which of the following orbitals are possible: $1p, 2s, 2p, 3f$?

Total number of ions from the following with noble gas configuration is . . . . . . .
$Sr^{2+} (Z=38), Cs^{+} (Z=55), La^{2+} (Z=57), Pb^{2+} (Z=82), Yb^{2+} (Z=70) \text{ and } Fe^{2+} (Z=26)$

The electrons,identified by quantum numbers $n$ and $l$ as $(i) n = 4, l = 1$,$(ii) n = 4, l = 0$,$(iii) n = 3, l = 2$,$(iv) n = 3, l = 1$,can be placed in order of increasing energy,from the lowest to highest,as: