Assign the position of the element having outer electronic configuration (i) `ns^(2) np^(2) for n = 2` (ii) `(n-1) d^(5) ns^(1)` for `n = 4`, and (iii) `(n-2) f^(14) (n-1) d^(0) ns^(2)` for `n = 6` in the Modern periodic table.

Answer

(i) `n = 2` indicates that the element belongs to second period. Since the last electron enters the `p-`orbital therefore the given element is a p-block element. For p-block elements group number `= 10 +` number of electrons in the valence shell.
`:.` Group number of the element `= 10 +4 = 14`.
(ii) `n = 4` indicates that the element lies in the `4th` period.
Since the d-orbital is incomplete, therefore, it is d-block element. The group number of the d-block element =number of `(n-1)d`-electros + number of `(n)s-`electrons `=5 +1 = 6`.
(iii) `n = 6` means, that the element lies in the sixth period. Since the last electron goes to the f-orbital, therefore, the element is a f-block element. All f-block elements lie in `3rd` group.