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.

Question

Philoid · Accepted Answer

(i) Since n = 3, the element belongs to the 3rd 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 is:

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

= 2 + 10 + 4 = 16

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

(ii) Since n = 4, the element belongs to the 4th 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 is :

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

= 2 + 2 = 4

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

(iii) Since n = 6, the element is present in the 6th 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.

Assign the position of the element having outer electronic configuration(i) ns2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f7 (n-1)d1ns2 for n=6, in the periodic table.

Assign the position of the element having outer electronic configuration
(i) ns²np⁴ for n=3 (ii) (n-1)d²ns² for n=4, and (iii) (n-2) f⁷ (n-1)d¹ns² for n=6, in the periodic table.