A) 78%
B) 31.25%
C) 50.25%
D) None of these
Correct Answer: B
Solution :
[b] Dipole moment of compound would have been completely ionic \[=(4.8\times {{10}^{-10}}esu)\,(2.67\times {{10}^{-8}}\,cm)=12.8\,D\] so % ionic character \[=\frac{4.0}{12.8}\times 100%=31.25%\]You need to login to perform this action.
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