Question

The element having general electronic configuration $ns^{2}np^{4}$ in fourth period.

Answer 1

"(i) n = 3, the element belongs to the 3rd period.

It is a p-block elements because "last electron" enters into the p-orbital.

There are four electrons in the "p-orbital"".

Therefore, the "corresponding group" of the "element" = "Number of s-block" groups$(3{ s }^{ 2 })$+ number of d-block groups

$= ([Ne{ ] }^{ 10 } + number of p-electrons (3{ p }^{ 4 })$

= 2 + 10 + 4 = 16

Therefore, the "element belongs" to the 3rd period and 16th "group" of the "periodic table".

Hence, the element is Sulphur electronic configuration is $([Ne]3{ s }^{ 2 }3{ p }^{ 4 })$"

Answer 2

"(i) n = 3, the element belongs to the 3rd period.

It is a p-block elements because "last electron" enters into the p-orbital.

There are four electrons in the "p-orbital"".

Therefore, the "corresponding group" of the "element" = "Number of s-block" groups(3{ s }^{ 2 })(3s

2

) + number of d-block groups

= ([Ne{ ] }^{ 10 } + number of p-electrons (3{ p }^{ 4 })=([Ne]

10

+numberofp−electrons(3p

4

)

= 2 + 10 + 4 = 16

Therefore, the "element belongs" to the 3rd period and 16th "group" of the "periodic table".

Hence, the element is Sulphur electronic configuration is ([Ne]3{ s }^{ 2 }3{ p }^{ 4 })([Ne]3s

2

3p

4

) "