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.
(i) Since n = three, the detail belongs to the three rd duration. It is a p-block detail for the reason that remaining electron occupies the p- orbital. There are 4 electrons withinside the p-orbital. Thus, the corresponding institution of the detail = Number of s-block groups quantity of d-block groups variety of p-electrons = 2 10 4 = 16 Therefore, the detail belongs to the three rd duration and 16 th organization of the periodic desk. Hence, the detail is Sulphur. (ii) Since n = 4, the detail belongs to the 4 th duration. It is a d-block detail as d-orbitals are incompletely filled. There are 2 electrons withinside the d-orbital. Thus, the corresponding organization of the detail = Number of s-block groups quantity of d-block groups = 2 2 = 4 Therefore, it's far a 4 th duration and 4 th organization detail. Hence, the detail is Titanium. (iii) Since n = 6, the detail is gift withinside the 6 th duration. It is an f -block detail because the final electron occupies the f- orbital. It belongs to organization three of the periodic desk considering the fact that all f-block factors belong to organization three. Its electronic configuration is [Xe] 4f 7 5d 1 6s 2 . Thus, its atomic quantity is 54 7 2 1 = 64. Hence, the detail is Gadolinium.
