Question

The values of electronegativity of atoms a and b are 1.20 and 4 respectively. then the percentage of ionic character of ab bond is

Answer 1

% ionic character = 16( xA - xB) + 3.5 (xA - xB)^2
Where xA and xB are electronegativities of atom A and B respectively. 
% ionic character = 16( 4-1.2) + 3.5(4-1.2)2
% ionic character = 44.8 + 27.44 = 72.24"

Answer 2

Answer:

72.24

Explanation:

according to Hanny and Smith equation,

% of ionic character=16(xA-xB)+3.5(xA-xB)^2

=16(4-1.2)+3.5(4-1.2)^2

=44.8+27.44

=72.24%