Question

If electronegativity if A is 2.0 and that of B is 3.0 what is the covalent character percentage of the bond A-B?

Answer 1

Percent covalent character

= 100 - [16(x1-x2)+3.5(x1-x2)²]

Where x1 = 3 and x2 = 2

On solving, we get 80.5 %

Answer 2

Answer:

The covalent character percentage for the given A-B bond will be equal to 80.5%.

Explanation:

To find covalent character % first we will find % ionic character.

Covalent character % = 100 - ionic character %

Given that: $X_{A}=2.0$  and  $X_{B}=3.0$

Percentage Ionic character $=16(X_{A}-X_{B})+3.5(X_{A}-X_{B})^{2}$

                                        $=16(3-2)+3.5(3-2)^{2}$

                                        $= 16+ 3.5= 19.5$%

In A-B bond % covalent character $=100-19.5=80.5$%