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

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46%

36%

30%

50%

answer is A.

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Detailed Solution

% ionic character $= [(6 \times (∆ E N)] + (35 \times (△ E N)^{2}]$

$= (16 \times 2.0) + (3.5 \times 2^{2})$

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

Hence, the option A is correct.

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