Question

If the electronegativity of atom A is 2.0 and atom B is 3.0, then what is the percentage ionic character of bond A-B?

• A. 80.5 %
  
• B. 12.5 %
• C. 19.5 %
• D. 87.5 %

Answer 1

The correct option is C 19.5 %

The Hanny-Smith equation for the ionic character can be used in this case.

$16(|x_A-x_B|)+3.5(x_A-x_B{)}^2$

where '$x_A$ ' is the electronegativity of one element and $x_B$ is the electronegativity of the other element.

Here, $x_A$ is 2.0 and $x_B$ is 3.0.

So, $16(|2-3|)+3.5(2-3{)}^2$ = 16 + 3.5 = 19.5 %

The bond A-B has a 19.5% ionic character.