Question

What is the percent ionic character of lithium chloride? (Electronegativities of lithium and chlorine are 1.0 and 3.2 respectively)

Answer 1

The percent ionic character of lithium chloride is 73.3 %

Explanation:

Percent ionic character =$(1-e^{( \frac{1}{4})(\chi_{A}-\chi_{B})} )\times 100$

where 

$\chi^{Cl}$ is the electronegativity of chlorine= 3.2

$\chi^{Li}$ is the electronegativity of Lithium = 1.0

Putting in the values , we get:

percent ionic character =$(1-e^{( \frac{1}{4})(3.2-1.0)})\times 100$

percent ionic character =$73.3\%$

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