\(n =1 \quad n =2\)

\(l=0,1,2 \quad l=0,1,2,3\)

\((n+l) \Rightarrow \frac{1 s}{1} \frac{1 p}{2} \frac{1 d}{3} \quad \frac{2 s}{2} \frac{2 p}{3} \frac{2 d}{4} \frac{2 f}{5}\)

\(n=3\)

\(l=0,1,2,3,4\)

\(\frac{3 s }{3} \frac{3 p }{4} \frac{3 d }{5} \frac{3 f }{6} \frac{3 g }{7}\)

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.