Q. The values of electronegativity of atom $A$ and $B$ are 1.20 and 4.0 respectively. The percentage of ionic character of $A - B$ bond is
A) 58.3%
B) 48%
C) 79.6%
D) 73.6%
Explanation
(d)
If the EN difference is 1.9, then bond is 50% ionic. The difference in electronegativity is 2.8, therefore, percentage ionic character due to EN difference of 2.8 is
$\frac{2.8}{1.9} \times 50 = 73.6\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.