Dipole moment in molecule `A–B' is 2.0 × 10–29C.m. If interatomic distance between A and B is 1.6 Å, then calculate percentage of ionic character in `A–B' bond.

Dear Student,

%ionic character = (observed DM/Calculated DM)×100
we know,Dipole moment = e×d
where ,e= 1.602×10−19
d = distance between the charges=1.6A0 =1.6×10−10m (given)
hence, μ = e×d               
μ =1.602×10−19×1.6×10−10               
μ = 2.563×1029cm 
putting the value ,we get   
%ionic character = (2.0×10−29/2.563×10−29)×100= 78% 
Thus covalent character will be (100−78)=22% 

Regards,

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