#### Question

Assign the position of the element having an outer electronic configuration in the periodic table.

*ns*^{2} *np*^{4} for *n *= 3

#### Solution 1

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.

#### Solution 2

n = 3

Thus element belongs to 3rd period, p-block element.

Since the valence shell contains = 6 electrons, group No = 10 + 6 = 16 configuration

=1s^{2 }2s^{2} 2p^{6} 3s^{2}3p^{4} element name is sulphur