Assign the position of the element having outer electronic configuration (i) `ns^(2)np^(4)` for n=3 (ii) `(n-1)d^(2)ns^(2)` for n=4, and (iii) `(n-2) f^(7) (n-1)d^(1)ns^(2)` for n=6, in the periodic table.

Explanation

a. For `n=3`, the valence electron configuration `=3s^(2)3p^(4)`
and the complete electronic configuration `=1s^(2)2s^(2)2p^(6)3s^(2)3p^(4)`
the total number of `e^(-)s=16`.
The elements is `S`. Group =`16`,Period=`3`.
b. For `n=4`, the valence electron configuration `=3d^(2)4s^(2)`
and the complete electronic configuration `=1s^(2)2s^(2)2p^(6)3s^(2)3d^(2)4s^(2)`
the total number of `e^(-)s=22`.
Group number of element =number of `de^(-)'s` +number of s `e^(-)'s=2+2=4th` Group.
The elements is Titanium `(Ti)`. Group =4th,Period=4th.
c. For `n=6`, the valence electron configuration `=4f^(7)5d^(1)6s^(2)` and the complete electronic configuration `[Xe]4f^(7)5d^(1)6s^(2)`
the total number of `e^(-)s`= number of `e^(-)'s` in `Xe`+ number of `e^(-)'s` in 4f+number of `e^(-)'s` in `5d` + number of `e^(-)'s` in `6s=54+7+1+2=64`.
`n=6` means that the elements is in the 6th period. Since the last electron enters into the `4f`-block elements. All `f-`block elements are present in group `3`.
Therefore the element (Z=64) is Gadolinium `(Gd)`.
Group =3, Period =6.