Question

Assign the position of the element having outer electronic configuration

(i) ns² np⁴ for n = 3 (ii) (n - 1)d² ns² for n = 4, and (iii) (n - 2) f⁷ (n - 1)d¹ ns² for n = 6, in the periodic table.

Answer 1

(i) Since n = 3, the element belongs to the 3^rd period. It is a p–block element since the last electron occupies the p–orbital.

There are four electrons in the p–orbital. Thus, the corresponding group of the element

= Number of s–block groups + number of d–block groups + number of p–electrons

= 2 + 10 + 4

= 16

Therefore, the element belongs to the 3^rd period and 16^th group of the periodic table. Hence, the element is Sulphur.

(ii) Since n = 4, the element belongs to the 4^th period. It is a d–block element as d–orbitals are incompletely filled.

There are 2 electrons in the d–orbital.

Thus, the corresponding group of the element

= Number of s–block groups + number of d–block groups

= 2 + 2

= 4

Therefore, it is a 4^th period and 4^th group element. Hence, the element is Titanium.

(iii) Since n = 6, the element is present in the 6^th period. It is an f –block element as the last electron occupies the f–orbital. It belongs to group 3 of the periodic table since all f-block elements belong to group 3. Its electronic configuration is [Xe] 4f⁷ 5d¹ 6s². Thus, its atomic number is 54 + 7 + 2 + 1 = 64. Hence, the element is Gadolinium.