Question

The percent ionic character of a bond can be approximated by the formula $\mathbf{16}\mathbf{\Delta }\mathbf{+}\mathbf{3}\mathbf{.5}{\mathbf{\Delta }}^{\mathbf{2}}$, whereDis the magnitude of the difference in the electronegativities of the atoms (see Fig. 3.18). Calculate the percent ionic character of HF, HCl, HBr, HI, and CsF, and compare the results with those in Table 3.7.

Short answer

The estimated Ionic character of $\mathrm{HF}$is $40\%$and value from the table is $41\%$,

$\mathrm{HCl}$is $18\%$and value from the table is $18\%$, $\mathrm{HBr}$is $14\%$, value from the table is $12\%$, $\mathrm{HI}$is $8\%$, value from the table is $6\%$,$\mathrm{BrF}$is $20\%$.

Step-by-step solution

Effect of electronegativity on ionic bonding point

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

Estimation of percent ionic character

Estimation of percent ionic character using electronegativity difference:

From the given formula:

$\mathrm{I}=3.5{\mathrm{\Delta }}^2+16\mathrm{\Delta }$

where $\mathbf{I}$ is the percent ionic character and$\mathbf{∆}$ is the difference in electronegativities of the elements making the bondHF.

Relevant electronegativities:

$\begin{array}{l}\mathrm{\chi }\left(\mathrm{H}\right)=2.20\\ \mathrm{\chi }\left(\mathrm{F}\right)=3.98\end{array}$

These are from FIGURE 3.18.

Difference in electronegativities is:

$\begin{array}{c}\mathrm{\Delta }=|3.98-2.20|\\ =1.78\end{array}$

Substituting this $∆$in formula from above:

$\begin{array}{c}\mathrm{I}=3.5\cdot 1{.78}^2+16\cdot 1.78\\ =39.6\stackrel{\text{safedigits}}{\to }40\end{array}$

Step 3: The relevant electronegativity of HCl

$\begin{array}{l}\mathrm{\chi }\left(\mathrm{H}\right)=2.20\\ \mathrm{\chi }\left(\mathrm{Cl}\right)=3.16\end{array}$

These are from FIGURE 3.18.

Difference in electronegativities is:

$\begin{array}{c}\mathrm{\Delta }=|3.16-2.20|\\ =0.94\end{array}$

Substituting this $∆$in formula from above:

$\begin{array}{c}\mathrm{I}=3.5\cdot 0{.94}^2+16\cdot 0.94\\ =18.1\stackrel{\text{safedigits}}{\to }18\end{array}$

Step 4: The relevant electronegativity of HBr

Relevant electronegativities:

$\begin{array}{l}\mathrm{\chi }\left(\mathrm{H}\right)=2.20\\ \mathrm{\chi }\left(\mathrm{Br}\right)=2.96\end{array}$

These are from FIGURE 3.18.

Difference in electronegativities is:

$\begin{array}{c}\mathrm{\Delta }=|2.96-2.20|\\ =0.76\end{array}$

Substituting this $∆$in formula from above:

$\begin{array}{c}\mathrm{I}=3.5\cdot 0{.76}^2+16\cdot 0.76\\ =14.2\stackrel{\text{safedigits}}{\to }14\end{array}$

HBr has $14\%$ ionic character, in TABLE $3.7$ HBr has$12\%$ ionic character so the difference not insignificant -

$\frac{2\mathrm{\%}}{12\mathrm{\%}}=16.67\mathrm{\%}$.

Step 5: The relevant electronegativity of HI

Relevant electronegativities:

$\begin{array}{l}\mathrm{\chi }\left(\mathrm{H}\right)=2.20\\ \mathrm{\chi }\left(\mathrm{I}\right)=2.66\end{array}$

These are from FIGURE 3.18.

Difference in electronegativities is:

$\begin{array}{c}\mathrm{\Delta }=|2.66-2.20|\\ =0.46\end{array}$

Substituting this $∆$in formula from above:

$\begin{array}{l}\mathrm{I}=3.5\cdot {0.46}^2+16\cdot 0.46\\ =8.1\stackrel{\text{safedigits}}{\to }8\end{array}$

HI has$8\%$ ionic character, in TABLE$3.7$ has$6\%$ ionic character so the difference is not insignificant $-\frac{2\mathrm{\%}}{6}=33.33\mathrm{\%}$

Step 6: The relevant electronegativity of BrF

Relevant electronegativities:

$\begin{array}{l}\mathrm{\chi }\left(\mathrm{Br}\right)=2.96\\ \mathrm{\chi }\left(\mathrm{F}\right)=3.98\end{array}$

These are from FIGURE 3.18.

Difference in electronegativities is:

$\begin{array}{c}\mathrm{\Delta }=|3.98-2.96|\\ =1.02\end{array}$

Substituting this $∆$in formula from above:

$\begin{array}{c}\mathrm{I}=3.5\cdot {1.02}^2+16\cdot 1.02\\ =20.0\stackrel{\text{safedigits}}{\to }20\end{array}$

BrF has$20\%$ ionic character, in TABLE $3.7$does not have a value for ionic character of BrF, since the error is not higher than$\frac{1}{3}$ in each example, one can estimate the ionic character of BrF to be around twenty seven percentage.