Question

Atomic numbers of nitrogen, oxygen and fluorine are 7, 8 and 9 respectively. Write the electronic configuration of each element and answer the following question:

Which one of N, O and F is more electronegative and which one is least electronegative?

Answer 1

Step-1: Introduction

The elements in periodic table are placed in periods and groups. Periods are horizontal rows that has increasing atomic numbers. Groups are vertical columns. There are 7 periods and 18 groups in periodic table.

Step-2: Electronic configuration

Atomic numbers of Nitrogen ($\mathrm{N}$), Oxygen ($\mathrm{O}$) and Fluorine ($\mathrm{F}$) are 7, 8 and 9 respectively.

Electronic configuration of Nitrogen = (2, 5)

Electronic configuration of Oxygen = (2, 6)

Electronic configuration of Fluorine = (2, 7)

Step-3: Compare the electronegativity

Electronegativity is the atom's ability to electrons attraction towards itself.

It rises from left to right of the period and reaches the maximum for halogen group and becomes zero for noble gases.

In a group, electronegativity downs due to smaller size of atoms.

The elements Nitrogen, Oxygen and Fluorine belongs to same period 2 and so electronegativity increases gradually.

Thus, Fluorine is most electronegative while Nitrogen is least electronegative.