CBSE Class 11-science Answered
Percent ionic character = 16 (XB-XA) + 3.5 (XB-XA)2
Where XB is electro negativity of atom B and XA is electro negativity of A
So, Percent ionic character = 16 (4-1.8) + 3.5 (4-1.8)2
= 16 X (2.2) + 3.5 X (2.2)2
= 35.2 + 16.94
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