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.

Explanation

(i) n=3 suggests that the element belongs to third period . Since the last electron enters the p- orbital therefore the given element is p- block element .Further since the valence shell contains 6 (2+ electrons therefore group number of the element = 10 + no. of electrons in the valence shell = 10 + 6 =
The complete electronic configuration of the element is `1 s^(2) 2 s^(2) 2 p^(6) 3 s^(2) 3p ^(4)` and the element is ( sulphur )
(ii) n=4 suggets that the element lies in the 4th period.
since the d- orbitals are incomplete therefore it is d-block element . the group number of the element no. of d- electrons + no. of s- electrons = 2 + 2 =4 . Thus the elements lie in group 4 and 4th period . The configuration of the element is `[Xe] 4 f^(7) 5 d^(1) 6 s^(2). ` The atomic number of the element = 54 + 7 + 1 + 2= 64 and the the element is Gd (gadolinium ).