Question

The values of electronegativity of atoms A and B are 1.20 and 4.0 respectively. The percentage of ionic character of A - B bond is

• A. 50 %
• B. 43 %
• C. 55.3 %
• D. 72.24%

Answer 1

The correct option is

C

72.24%

We know that ionic characters

$=16[E_A-E_B]+3.5\times [E-A-E_B{]}^2$

=16(1.2−4.0)+3.5${(1.2-4.0)}^2$

=44.8+27.44

=72.24%

% of ionic characters = 72.24%