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What is the percentage of ionic character of HCl.If $\mu=1.02D$, bond length is $1.275\overset{\circ}{A}$ ($e=4.8\times 10^{-10}$e.s.u)?

$\begin{array}{1 1}(a)\;\text{100% ionic}&(b)\;\text{17% ionic}\\(c)\;\text{83% ionic}&(d)\;\text{0% ionic}\end{array}$

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% ionic nature =$\large\frac{\mu \text{ experimental}}{\mu \text{ theoretical}}$$\times 100$
$\Rightarrow \large\frac{1.02}{1.275\times 4.8}$$\times 100$
$\Rightarrow \approx 17\%$
Hence (b) is the correct answer.
answered Mar 10, 2014 by sreemathi.v
 

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