Compute the percent ionic character of the interatomic bonds for the following compounds: \(TiO_{2}\), \(ZnTe\), \(CsCl\), \(InSb\), and \(MgCl_{2}\).
50Cr: 4.34% A = 49.9460 amu
52Cr: 83.79% A = 51.9405 amu
53Cr: 9.50% A = 52.9407 amu
54Cr: 2.37% A = 53.9389 amu
Prove that \(A_{Cr}\) = 51.9963 amu
\(A = f_{1} * A_{1} + f_{2} * A_{2} + f_{3} * A_{3} + f_{4} * A_{4} \)
4.34% * 49.9460 + 83.79% * 51.9405 + 9.50% * 52.9407 + 2.37% * 53.9389 = 51.9963 amu
51.9963 amu = 51.9963 amu \( \implies \) correct