Question

Estimate the percent ionic character of the bond in each of the following species. All the species are unstable or reactive under ordinary laboratory conditions, but they can be observed in interstellar space.

Short answer

The Ionic character of $OH$is $35.3\%$,CH is$26.9\%$,CN is$25.7\%and{C}_{2}is0\%.$

Step-by-step solution

Effect of electronegative in ionic bonding point

The ionic bonds occur when more than two or two electrons are transferred between more than two atoms.

Ionic character based on bond length

The percentage of ionic character is based on bond length $\left(R\right)$and dipole moment $\left(\mu \right)$.

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

localid="1663392998065" $\mu =\frac{R}{0.2082}\delta$

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

width="96">δ=0.2082Rμ

Step 3: Percentage ionic character of the bond of OH

$$\begin{gathered} \mu =1.66D \\ R=0.980\stackrel{°}{A} \end{gathered}$$

Using the boxed formula above, we get:

$$\begin{gathered} \delta =\frac{0.2082}{0.980{\displaystyle \stackrel{°}{A}}}1.66D \\ \delta =0.353e \end{gathered}$$

Step 4: Percentage ionic character of the bond of CH

$$\begin{gathered} \mu =1.46D \\ R=1.131\stackrel{°}{A} \end{gathered}$$

Using the boxed formula above, we get:

$$\begin{gathered} \delta =\frac{0.2082}{1.131{\displaystyle \stackrel{°}{A}}}1.46D \\ \delta =0.269e \end{gathered}$$

Step 5: Percentage ionic character of the bond of CN

$$\begin{gathered} \mu =1.45D \\ R=1.175\stackrel{°}{A} \end{gathered}$$

Using the boxed formula above, we get:

$\delta =\frac{0.2082}{1.17{\displaystyle \stackrel{°}{A}}}1.45D$

$\delta =0.257e$

Step 6: Percentage ionic character of the bond of  C2

$$\begin{gathered} \mu =0D \\ R=1.246\stackrel{°}{A} \end{gathered}$$

Using the boxed formula above, we get:

$$\begin{gathered} \delta =\frac{0.2082}{1.246{\displaystyle \stackrel{°}{A}}} \\ \delta =0e \end{gathered}$$