Question

If the electronegative dierence between two atoms is 2.1. The % age of covalent character of the molecule is

Answer 1

Answer : The percentage of covalent character of the molecule is, 50.97%

Explanation :

First we have to calculate the percentage of ionic character.

Formula used :

$\%\text{ Ionic character}=16\times \Delta EN+3.5\times (\Delta EN)^2$

where,

$\Delta EN$ = electronegative difference = 2.1

Now put the value of electronegative difference in the above formula, we get the percentage of ionic character.

$\%\text{ Ionic character}=16\times 2.1+3.5\times (2.1)^2=49.035\%$

Now we have to calculate the percentage of covalent character.

Percentage covalent character = 100 - Percentage of ionic character

Percentage covalent character = 100 - 49.035 = 50.97 %

Therefore, the percentage of covalent character of the molecule is, 50.97%