- ✓$O_2$
- B$O_2^-$
- C$O_2^+$
- D$O_2^{2-}$
$1 s ^2 * 1 s ^2 2 s ^2 * 2 s ^2 2 pz ^2 2 px ^2 2 py ^2 * 2 px ^1 * 2 py ^1$
The electronic configuration of the $O _2^{+}$containing $16$ electrons can be written as:
$1 s ^2 * 1 s ^2 2 s ^2 * 2 s ^2 2 pz ^2 2 px ^2 2 py ^2 * 2 px ^1 * 2 py ^0$
The electronic configuration of the $O _2^{-}$ion containing $17$ electrons can be written as:
$1 s ^2 * 1 s ^2 2 s ^2 * 2 s ^2 2 pz ^2 2 px ^2 2 py ^2 * 2 px ^2 * 2 py ^1$
The electronic configuration of the $O _2^{-2}$ ion containing $18$ electrons can be written as:
$1 s ^2 * 1 s ^2 2 s ^2 * 2 s ^2 2 pz ^2 2 px ^2 2 py ^2 * 2 px ^2 * 2 py ^2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$(I)$ It is easier to remove $2 \mathrm{p}$ electron than $2 \mathrm{s}$ electron
$(II)$ $2 \mathrm{p}$ electron of $\mathrm{B}$ is more shielded from the nucleus by the inner core of electrons than the $2 s$ electrons of $Be.$
$(III)\; 2 s$ electron has more penetration power than $2 \mathrm{p}$ electron.
$(IV)$ atomic radius of $\mathrm{B}$ is more than $\mathrm{Be}$
(Atomic number $\mathrm{B}=5, \mathrm{Be}=4$ )
The correct statements are