The values of electronegativity of atoms A an | vedclass

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(D) The percentage of ionic character can be calculated using the Hannay-Smith equation:Percentage of ionic character $= 16|E_A - E_B| + 3.5|E_A - E_B

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$72.24$

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(D) The percentage of ionic character can be calculated using the Hannay-Smith equation:Percentage of ionic character $= 16|E_A - E_B| + 3.5|E_A - E_B|^2$Given,electronegativity of $A$ $(E_A)$ $= 1.20$ and electronegativity of $B$ $(E_B)$ $= 4.0$.Difference in electronegativity $= |1.20 - 4.0| = 2.8$.Substituting the values in the formula:Percentage of ionic character $= 16(2.8) + 3.5(2.8)^2$$= 44.8 + 3.5(7.84)$$= 44.8 + 27.44$$= 72.24\%$

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

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