A) \[17%\]
B) \[83%\]
C) \[20%\]
D) \[80%\]
Correct Answer: B
Solution :
% Ionic character \[=\frac{{{\mu }_{observed}}}{{{\mu }_{calulated}}}\,\times 100\] \[\overset{s+}{\mathop{H}}\,-\overset{s-}{\mathop{Cl}}\,\] \[\delta =4.8\times {{10}^{-10}}\,esu\] \[d=1.275\,\times {{10}^{-8}}cm\] \[=\frac{1.03\times 100}{4.8\,\times {{10}^{-10}}\,\times 1.275\,\times {{10}^{-8}}\,(cm)}\] \[=\frac{1.03(D)\times 100}{4.8\times 1.275\,\times {{10}^{-18}}\,(esu.\,cm)}\] \[=\frac{1.03\,(D)}{4.8\times 1.275(D)}\,\times 100\,=\,16.8\,=17%\] \[\therefore \,\,%\] Covalent character = 100 - % Ionic character = 100 - 17 % = 83%You need to login to perform this action.
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