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  • The value of electronegativity of atom A and B are 1.80 and 4.0 respectively. The percentage of ionic character of A-B bond is ?           &nb

The value of electronegativity of atom A and B are 1.80 and 4.0 respectively. The percentage of ionic character of A-B bond is ?                                                      

Option: 1

43.14%


Option: 2

50 %            


Option: 3

55.3 %


Option: 4

52.14 %


Answers (1)

best_answer

As we learn
Percent ionic character = 16 (XB-XA) + 3.5 (XB-XA)2

Where XB is the electronegativity of atom B and XA is the electronegativity 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 

Hence, the option number (4) is correct.

Posted by

Pankaj

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