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).

If the quantum numbers n,l,m and s were defined as: R=shell number =

1.	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).
Answer» 8 elements 12 elements 16elements 18elements Solution :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)