Question

An element X of group 16 and period 2 exists as diatomic molecule. It combines with hydrogen and forms ammonia in the presence of a suitable catalyst.

Write the electronic configuration of X and count the number of valence electrons.

Answer 1

Step-1: Introduction

The elemental arrangement of periodic table are placed in periods and groups. The horizontal rows are periods that has rising atomic numbers and vertical columns are groups. 7 periods and 18 groups are in periodic table.

The elements are classified on the basis of electronic configuration.

Step-2: To find the valence electrons

The number of valence shell = Number of period

Since the element belongs to period 2, the valence shell count is 2,

The number of valence electrons + 10 = Number of group

Its group is 15, so it has 5 valence electrons in the outermost shell.

Step-3: To find the electronic configuration

The element with 5 valence electrons and 2 valence shell is Nitrogen ($\mathrm{N}$).

Electronic configuration of Nitrogen = (2, 5) with atomic number 7.

Step-3: Reaction of X with Hydrogen

Nitrogen (${\mathrm{N}}_2$) combines with Hydrogen gas (${\mathrm{H}}_2$) in the presence of metal catalyst to give Ammonia (${\mathrm{NH}}_3$). This process is Haber's process.

${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

Final answer

Therefore, Nitrogen has 5 valence electrons and its electronic configuration is (2,5).