A) 8%
B) 20%
C) 17%
D) 50%
Correct Answer: C
Solution :
\[{{\mu }_{cal}}=e\times d=1.6\times {{10}^{-19}}\times 2.29\times {{10}^{-10}}\] \[=3.664\times {{10}^{-29}}C\text{-}m\] % ionic character\[=\frac{{{\mu }_{\exp }}}{{{\mu }_{cal}}}\times 100\] \[=\frac{6.226\times {{10}^{-30}}}{3.664\times {{10}^{-29}}}\times 100\] \[=17%\]You need to login to perform this action.
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