Question

(a)The atomic number of an element is 19.
Identify its group, period, block and oxidation state. [ 2 Marks]
(b)Arrange the following in increasing order of their metallic character:
Si, Na, Mg, P, Be [ 2 Marks]

Answer 1

(a) The electronic configuration of element with atomic number 19 is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
The element have 4 shells in its electronic configuration so the period number will be 4.
The element is a s block element and the group number will be 1.
The valence shell contains only 1 electron. So, the oxidation state will be +1. [ 2 Marks]

(b)
(i) We know that metallic character increases down the group and decrease along the period from left to right. Be belongs to 2nd period , whereas Na, Mg, Si, P belong to 3rd period. Be, Na, Mg, Si and P belong to 2, 1, 2, 14 and 15 groups respectively. [1 Mark]
(ii) Therefore, Na is the most metallic element, phosphorus is the least metallic element of the given elements. Mg is more metallic element than Be. Si is more metallic element than P. The increasing order of metallic character is P < Si < Be < Mg < Na. [1 Mark]