If the quantum numbers n,l,m and s were defined as:
R=shell number
=1,2,3,4,. . . . In integral steps.
l=Type of subshell
=0,1,2,3,. . . To n in integral steps.
m=Number of orbitals corresponding to any subshell
=-(l+1)to+(l+1), in integral steps, incuding zero.
s= Spin quantum number `=-(1)/(2)` or `+(1)/(2)`
The l-values correspond to the subshells as actual representations, like l=0 (s-subshell). l=1(p-subshell),l=2(d-subshell),and so on.
In the modern long form of periodic table, the 2nd period should (Assume that (n+l)rule is perfectly obeyed).

Options

A. 8 elements
B. 12 elements
C. 16elements
D. 18elements

Explanation

For n=3 `m=-(l+1)"to" +(l+1)`
`=0m=-1,0,+1=3` orbital
=1 m=02,01,0,+1,+2=5 orbital
=2 m=-3,-2,-1,0,+1,+2,+3=7 orbital
=3 m=-4,-3,-2,-1,0,+1+2,+1,+4=9 orbital ltbr. Total orbitals =3+5+7+9=24 orbitals
`s=-(1)/(2) "or" +(1)/(2)` (i.e. two electrons)
Total electrons =24xx2=48 electrons
(ii) lr 2nd period Given:
n=2,l=0,1,2 l=0 l=0s
in n=1,l=0,1 l=0,1 l=1 p
l=2 d
Now `1s^(2)1p^(6)2s^(6)2p^(6)=16` electron (as electron will not enter in 2d)