$n =1 \quad n =2$
$l=0,1,2 \quad l=0,1,2,3$
$(n+l) \Rightarrow \frac{1 s}{1} \frac{1 p}{2} \frac{1 d}{3} \quad \frac{2 s}{2} \frac{2 p}{3} \frac{2 d}{4} \frac{2 f}{5}$
$n=3$
$l=0,1,2,3,4$
$\frac{3 s }{3} \frac{3 p }{4} \frac{3 d }{5} \frac{3 f }{6} \frac{3 g }{7}$
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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On addition of equal number of moles of a non-volatile solute $S$ in equal amount (in $kg$ ) of these solvents, the elevation of boiling point of solvent $X$ is three times that of solvent $Y$. Solute $S$ is known to undergo dimerization in these solvents. If the degree of dimerization is $0.7$ in solvent $Y$, the degree of dimerization in solvent $X$ is. . . . . . .
