Electronegativities of atoms A and B are 4.0 and 1.20 respectively. The percentage of ionic character of A - B bond is ___________

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answer is 72.24.

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Detailed Solution

% Ionic character =16χA−χB+3.5χA−χB2 xA=4.0xB=1.20% Ionic character =16(4.0−1.20)+3.5(4.0−1.20)2 =16(2.8)+3.5(2.8)2 =44.8+27.44 =72.24%

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