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.



Q.30:- 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)d1ns2 for n = 6, in the periodic table.

 

 

Answer:-



 

(i) n = 3
Thus element belong to 3rd period, p-block element.
Since the valence shell contains = 6 electrons, group No = 10 + 6 = 16 configuration =1s2 2s2 2p6 3s2 3p4 element name is sulphur.
(ii) n = 4
Means element belongs to 4th period belongs to group 4 as in the valence shell (2 + 2) = 4 electrons.
Electronic configuration.=1s2 2s2 2p6 3s2 3p6 3d2 4s2, and the element name is Titanium (Ti).
(iii) n = 6
” Means the element belongs to 6th period. Last electron goes to the f-orbital, element is from f-block.
group = 3
The element is gadolinium (z = 64)
Complete electronic configuration =[Xe] 4 f7 5d1 6s2.