Consider the hypothetical situation where the azimuthal quantum number

Solution:

$l=0$ to $(\mathrm{n}+1)$

$\mathrm{n}=1$              $\mathrm{n}=2$

$l=0,1,2$                   $l=0,1,2,3$

$(\mathrm{n}+l) \Rightarrow \frac{1 \mathrm{~s}}{1} \frac{1 \mathrm{p}}{2} \frac{1 \mathrm{~d}}{3}$

$\frac{2 \mathrm{~s}}{2} \frac{2 \mathrm{p}}{3} \frac{2 \mathrm{~d}}{4} \frac{2 \mathrm{f}}{5}$

$\mathrm{n}=3$

$l=0,1,2,3,4$

$\frac{3 \mathrm{~s}}{3} \frac{3 \mathrm{p}}{4} \frac{3 \mathrm{~d}}{5} \frac{3 \mathrm{f}}{6} \frac{3 \mathrm{~g}}{7}$

Now, in order to write electronic configuration, we need to apply $(\mathrm{n}+l)$ rule

Energy order : $1 \mathrm{~s}<1 \mathrm{p}<2 \mathrm{~s}<1 \mathrm{~d}<2 \mathrm{p}<3 \mathrm{~s}<2 \mathrm{~d} \ldots$

Option 1) $13: 1 \mathrm{~s}^{2} 1 \mathrm{p}^{6} 2 \mathrm{~s}^{2} 1 \mathrm{~d}^{3}$ is not half filled

Option 2) 9: $1 \mathrm{~s}^{2} 1 \mathrm{p}^{6} 2 \mathrm{~s}^{1}$ is the first alkali

metal because after losing one electron, it will achieve first noble gas configuration

Option 3) $8: \quad 1 \mathrm{~s}^{2} 1 \mathrm{p}^{6}$ is the first noble gas

because after $1 \mathrm{p}^{6} \mathrm{e}^{-}$will

enter $2 \mathrm{~s}$ hence new period

Option 4) $6: \quad 1 \mathrm{~s}^{2} 1 \mathrm{p}^{4}$ has $1 \mathrm{p}$ valence subshell.