Q. If the electronegativity difference between two atoms $A$ and $B$ is 2.0, then the percentage of covalent character in the molecule is
A) 54%
B) 46%
C) 23%
D) 72%
Explanation
(a)
% ionic character $= 16\left( x_{A} - x_{B} \right) + 3.5\left( x_{A} - x_{B} \right)^{2}$
$= 16 \times 2 + 3.5 \times (2^{2})$
$= 46$
$\therefore\$The % covalent character $= 100 - 46 = 54$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.