Question

Estimate the percent ionic character of the bond in each of the following diatomic molecules, based on the dipole moment.

Short answer

$ClO$has $16.40\%$ionic character.

$Kl$ has $73.84\%$ionic character.

$TiCl$has $38.02\%$ionic character.

$InCl$ has $32.8\%$ionic character.

Step-by-step solution

Percentage of ionic character based on bond lengthR and dipole moment μ

Now, the formula for one dipole moment in a molecule with bond length R and charges ${\delta }^{-},{\delta }^{+}$in appropriate units $\left(\mu \left(D\right),\delta \left(C\right),R\left(\stackrel{°}{A}\right)\right)$is:

$\mu =\frac{R}{0.2082}\delta$

The formula that will be used is for $\delta$-measure of ionic character, via $R\left(\stackrel{°}{A}\right)$and $\mu \left(D\right)$

$\delta =\frac{0.2082}{R}\mu ..........\left(1\right)$

Ionic character for ClO

Since:

$$\begin{gathered} \mu =1.239D \\ R=1.573\stackrel{°}{A} \end{gathered}$$

So, from the equation $\left(1\right)$we get

$$\begin{gathered} \delta =\left(\frac{0.2082}{1.573{\displaystyle \stackrel{°}{A}}}\right).\left(1.239D\right) \\ \delta =0.1640e \end{gathered}$$

Hence, the ionic character of$ClO$ is$16.40\%$ .

Ionic character for Kl

Since

$$\begin{gathered} \mu =10.82D \\ R=3.051\stackrel{°}{A} \end{gathered}$$

So, from the equation $\left(1\right)$we get

$$\begin{gathered} \delta =\left(\frac{2.082}{3.051{\displaystyle \stackrel{°}{A}}}\right).\left(10.82D\right) \\ \delta =0.7384e \end{gathered}$$$$\begin{gathered} \delta =\left(\frac{0.2082}{3.051{\displaystyle \stackrel{°}{A}}}\right).\left(10.82D\right) \\ \delta =0.7384e \end{gathered}$$

Hence, the ionic character of $KI$is$73.84\%$ .

Ionic character for  TiCl

Since

$\mu =4.543D$

$R=2.488\stackrel{°}{A}$

So, from the equation $\left(1\right)$we get

$$\begin{gathered} \delta =\left(\frac{0.2082}{2.488{\displaystyle \stackrel{°}{A}}}\right).\left(4.543D\right) \\ \delta =0.3802e \end{gathered}$$$$\begin{gathered} \delta =\left(\frac{0.2082}{2.488{\displaystyle \stackrel{°}{A}}}\right).\left(4.543D\right) \\ \delta =0.3802e \end{gathered}$$$$\begin{gathered} \delta =\left(\frac{0.2082}{2.488{\displaystyle \stackrel{°}{A}}}\right).\left(4.543D\right) \\ \delta =0.380e \end{gathered}$$$$\begin{gathered} \delta =\left(\frac{0.2082}{2.488{\displaystyle \stackrel{°}{A}}}\right).\left(4,543D\right) \\ \delta =0.3802e \end{gathered}$$

Hence, the ionic character of$TiCl$ is$38.02\%$ .

Ionic character for  InCl

Since

$$\begin{gathered} \mu =3.79D \\ R=2.404\stackrel{°}{A} \end{gathered}$$

So, from the equation $\left(1\right)$we get

$$\begin{gathered} \delta =\left(\frac{0.2082}{2.404{\displaystyle \stackrel{°}{A}}}\right).\left(3.79D\right) \\ \delta =0.328e \end{gathered}$$

Hence, the ionic character of $InCl$is$32.8\%$ .