$n =1 \quad n =2$
$l=0,1,2 \quad l=0,1,2,3$
$(n+l) \Rightarrow \frac{1 s}{1} \frac{1 p}{2} \frac{1 d}{3} \quad \frac{2 s}{2} \frac{2 p}{3} \frac{2 d}{4} \frac{2 f}{5}$
$n=3$
$l=0,1,2,3,4$
$\frac{3 s }{3} \frac{3 p }{4} \frac{3 d }{5} \frac{3 f }{6} \frac{3 g }{7}$
Now, in order to write electronic configuration, we need to apply $( n +l)$ rule
Energy order : $1 s <1 p <2 s <1 d <2 p <3 s <2 d$
Option 1) $13: 1 s^{2} 1 p^{6} 2 s^{2} 1 d^{3}$ is not half filled
Option 2) $9: \quad 1 s^{2} 1 p^{6} 2 s^{1}$ is the first alkali metal because after losing one electron, it will achieve first noble gas configuration
Option $3)$ $8: \quad 1 s^{2} 1 p^{6}$ is the first noble gas because after $1 p ^{6} e ^{-}$ will
Option $4)$ $6: \quad 1 s ^{2} 1 p ^{4}$ has $1 p$ valence subshell.
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${H_2}C = CHC{H_2}Br\xrightarrow{{NaCN}}Y\xrightarrow[{2.\,{H_3}{O^ + }}]{\begin{subarray}{l}
1.{C_6}{H_5}MgBr\, \\
diethylether
\end{subarray} }Z$
($A$) $\left[\mathrm{Pt}(\mathrm{en})(\mathrm{SCN})_2\right]$
($B$) $\left[\mathrm{Zn}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_2\right]$
($C$) $\left[\mathrm{Pt}\left(\mathrm{NH}_3\right)_2 \mathrm{Cl}_4\right]$
($D$) $\left[\mathrm{Cr}(\mathrm{en})_2\left(\mathrm{H}_2 \mathrm{O}\right)\left(\mathrm{SO}_4\right)\right]^{+}$
