JEE Mains 03-Sep-2020 Shift 2 Solved Paper

l=0 to (n+1)
n=1
(n+l)⇒

1s

1

1p

2

1d

3

n=2
l=0,1,2,3

2s

2

2p

3

2d

4

2f

5

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

3s

3

3p

4

3d

5

3f

6

3g

7

Now, in order to write electronic configuration,
we need to apply ( n+l) rule
Energy order :1 s<1p<2s<1d<2p <3s<2d.....
Option1) 13:1s21p62s21d3 is not half filled
Option 2 ) 9:1s21p62s1 is the first alkali metal because after losing one electron, it will achieve first noble gas configuration
Option 3 ) 8:1s21p6 is the first noble gas because after 1p6e− willenter 2 s hence new period
Option 4) 6:1s21p4 has 1p valencesubshell.