Question

using the figure below, calculate the percent ionic character of the interatomic bonds for the following materials (report your answers in decimal form to 1 decimal place): (a) cacl2 : select % (b) licl : select %

Explanation:

Step1: Recall electronegativity values

From the periodic - table, the electronegativity of Ca is approximately 1.0, Cl is approximately 3.0, Li is approximately 1.0, and Cl is approximately 3.0.

Step2: Calculate electronegativity difference for \(CaCl_2\)

The formula for the electronegativity difference \(\Delta\chi\) in a bond \(A - B\) is \(\Delta\chi=\vert\chi_A-\chi_B\vert\). For \(CaCl_2\), since there are two \(Ca - Cl\) bonds and \(\Delta\chi=\vert3.0 - 1.0\vert=2.0\). The percent ionic character (\(PIC\)) can be estimated using the formula \(PIC = 16\Delta\chi+3.5\Delta\chi^{2}\). Substitute \(\Delta\chi = 2.0\) into the formula:
\[

$$\begin{align*} PIC&=16\times2.0+3.5\times(2.0)^{2}\\ &=32 + 3.5\times4\\ &=32+14\\ &=46.0 \end{align*}$$

\]

Step3: Calculate electronegativity difference for \(LiCl\)

For \(LiCl\), \(\Delta\chi=\vert3.0 - 1.0\vert = 2.0\). Using the formula \(PIC = 16\Delta\chi+3.5\Delta\chi^{2}\), substitute \(\Delta\chi = 2.0\):
\[

$$\begin{align*} PIC&=16\times2.0+3.5\times(2.0)^{2}\\ &=32+14\\ &=46.0 \end{align*}$$

\]

Answer:

(a) 46.0%
(b) 46.0%