Question

Question 3.30 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)d1 ns2 for n = 6, in the periodic table.

Class XI Classification of Elements and Periodicity in Properties Page 94

Answer 1

(1) ns²np⁴ for n = 3 .
for n = 3 { principal quantum number = 3} means elements belongs to 3rd period. also outer most valance electrons in p - orbital , it belongs to P - block.

we know, for P - block elements
the ground number = 10 + valance shell electrons = 10 + 6 = 16 .hence, the element belongs to 16th group.

hence, the complete electronic configuration is 1s² , 2s², 2p⁶ , 3s², 3p⁴
thus element is Sulphur ( S )

(ii) (n - 1)d²ns² for n = 4
n = 4 means element belongs to 4th period.
because valance electrons belong to d - orbital. so, it belongs to d - block.

for d - block elements, the ground number = number of d - electrons + number of ns electrons = 2 + 2 = 4 , hence , it belongs to 4th group.
now, complete electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d², 4s².
thus , the element is Titanium


(iii) (n - 2)f⁷ (n - 1)d¹ ns² for n = 6
n = 6 means , the element belongs to 6th period . since the least electron enters the f - orbital , the given element belongs to f - block.

f - block always belongs to third group.
hence, the element belongs to 3rd group.
the complete configuration is
$1s^2,2s^2,2p^6,3s^2,3p^6,3d^{10},4s^2,4d^{10},5s^2,5p^6,4f^7,5d^1,6s^2$
thus , element is gadolinium ( Gd )

Answer 2

1. group= 16

period = 3

2. group= 4

period = 4

3. group= 3

period = 6