- A$[Ne] 3s^23p^1$
- ✓$[Ne] 3s^23p^3$
- C$[Ne] 3s^23p^2$
- D$[Ar] 3d^{10}4s^24p^3$
A) $[N e] 3 s^{2} 3 p^{1}$ - can lose the $p$ electron with relative ease.
B) $[N e] 3 s^{2} 3 p^{3}$ - since it is half filled, it is difficult to remove an electron. Hene ionization energy
will be high in this case.
C) $[N e] 3 s^{2} 3 p^{2}$ - Can lose $p$ electron with relative ease.
D) $[A r] 3 d^{10}, 4 s^{2} 4 p^{3}$ - Half filled. Difficult to remove electron.
Hence, we are left with two options, either $B$ or $D$. But in option $D$ the half filled configuration is in $4 p$ whereas it is in $3 p$ in option $B$. It is relatively easier to remove an electron from $4 p$ than $3 p .$
Hence, option $B$ is the right answer.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

| List $I$ (Compound) | List $II$ (Shape/geometry) |
| $A$. $\mathrm{NH}_3$ | $I$. Trigonal Pyramidal |
| $B$. $\mathrm{BrF}_5$ | $II$. Square Planar |
| $C$. $\mathrm{XeF}_4$ | $III$. Octahedral |
| $D$. $\mathrm{SF}_6$ | $IV$. Square Pyramidal |
Choose the correct answer from the options given below:

${\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{ - 4}}\xrightarrow{{MnO_4^ - /{H^ + }}}F{e^{ + 3}} + C{O_2} + NO_3^ - $
