Calculate the % ionic character for molecule AB when E.N. different is 2.0:

% ionic character $=\left[(6\times (∆EN\left)\right]+\left(35\times (\mathrm{△}EN{)}^2\right]\right.$

$=(16\times 2.0)+\left(3.5\times 2^2\right)$

$\begin{array}{l}=32+(3.5\times 4)\\ =32+14=46\mathrm{\%}\end{array}$

Hence, the option A is correct.