The values of electronegativity of atoms A and B are 1.80 and 4 respectively. The percentage ionic character of A-B bond is-?

Asked by  | 7th Feb, 2012, 09:14: PM

Expert Answer:

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

                                                = 52.14 

Answered by  | 8th Feb, 2012, 05:18: PM

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