GIven l= 0 to (n + 1)
Now for n = 1,
l = 0, 1, 2
Orbitals :
1s , l =0
1p , l = 1
1d , l = 2
And for
n = 2
l = 0, 1, 2, 3
And for
n = 3
l = 0, 1, 2, 3,4
Now, in order to wire electronic configuration, we need to apply (n + l) rule.
Energy order : 1s < 1p < 2s < 1 d < 2p < 3s < 2d ...
1) 9 : 1s2 1p6 2s1 is the first alkali metal because after losing one electron, it will achieve the first noble gas configuration
2) 13 : 1s2 1p6 2s2 1d3 is not half filled. (Showing incorrect statement)
3) 8 : 1s2 1p6 is the first noble gas.
4) 6 : 1s2 1p4 has 1p valence subshell.
Therefore, Option 2 is correct.
Study 40% syllabus and score up to 100% marks in JEE