Question

Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgCl2

Answer 1

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

$%IC = (1 - e^{-0.25*(x_A - x_B)^2})$ *100%

Where $x_A - x_B$ is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. $x_{Ti}$ = 1.5

$x_{O}$ = 3.5

$%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})$*100%

%IC = 63.2%

b. $x_{Zn}$ = 1.6

$x_{Te}$ = 2.1

$%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})$*100%

%IC = 11.7%

c. $x_{Cs}$ = 0.7

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})$*100%

%IC = 73.3%

d. $x_{In}$ = 1.7

$x_{Sb}$ = 1.9

$%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})$*100%

%IC = 0.995 %

e. $x_{Mg}$ = 1.2

$x_{Cl}$ = 3.0

$%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})$*100%

%IC = 55.5%