Question

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

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

Answer 1

Answer: (B)

$72.24\%$

We know that ionic characters:

% ionic character $=16[E_A-E_B]+3.5\times [E_A-E_B{]}^2$

$=16[4-1.2]+3.5[4-1.2{)}^2$

$=16\times 2.8+3.5\times {2.8}^2$

$=44.8+27.44$

$=72.24$