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Pauling's method

Recall the Mulliken's method was based on the arithmetic average of the first ionization energy $IE_1$ and the electron affinity $EA$. Both of these energies are properties of individual atoms, hence this method is appealing in its simplicity. However, there is no information about bonding in the Mulliken method. Pauling's method includes such information, and hence is a more effective approach.



To see how the Pauling method works, consider a diatomic AB, which is polar covalent. Let $\Delta E_{\rm AA}$ and $\Delta E_{\rm BB}$ be the dissociation energies of the diatomics A$_2$ and B$_2$, respectively. Since A$_2$ and B$_2$ are purely covalent bonds, these two dissociation energies can be used to estimate the pure covalent contribution to the bond AB. Pauling proposed the geometric mean of $\Delta E_{\rm AA}$ and $\Delta E_{\rm BB}$, this being more sensitive to large differences between these energies than the arithmetic average:

\begin{displaymath}
{\rm pure\ covalent\ contribution} = \sqrt{\Delta E_{\rm AA}\Delta E_{\rm BB}}
\end{displaymath}

If $\Delta E_{\rm AB}$ is the true bond dissociation energy, then the difference

\begin{displaymath}
\Delta E_{\rm AB} - \sqrt{\Delta E_{\rm AA}\Delta E_{\rm BB}}
\end{displaymath}

is a measure of the ionic contribution. Let us define this difference to be $\Delta$:

\begin{displaymath}
\Delta =
\Delta E_{\rm AB} - \sqrt{\Delta E_{\rm AA}\Delta E_{\rm BB}}
\end{displaymath}

Then Pauling defined the electronegativity difference $\chi_{\rm A}-\chi_{\rm B}$ between atoms A and B to be

\begin{displaymath}
\chi_{\rm A}-\chi_{\rm B} = 0.102\sqrt{\Delta}
\end{displaymath}

where $\Delta$ is measured in kJ/mol, and the constant 0.102 has units mol$^{1/2}$/kJ$^{1/2}$, so that the electronegativity difference is dimensionless. Thus, with some extra input information, he was able to generate a table of atomic electronegativities that are still used today (see Appendix F or Table 3.7).



To use the electronegativities to estimate degree of ionic character, simply compute the absolute value of the difference for the two atoms in the bond. As an example, consider again the hydrogen halides:

    $\displaystyle {\rm HF}\;\;\;\;\;\;\;\;\;\;\vert\chi_{\rm F}-\chi_{\rm H}\vert = 1.78$  
    $\displaystyle {\rm HCl}\;\;\;\;\;\;\;\;\;\;\vert\chi_{\rm Cl}-\chi_{\rm H}\vert = 0.96$  
    $\displaystyle {\rm HBr}\;\;\;\;\;\;\;\;\;\;\vert\chi_{\rm Br}-\chi_{\rm H}\vert = 0.76$  
    $\displaystyle {\rm HI}\;\;\;\;\;\;\;\;\;\;\vert\chi_{\rm I}-\chi_{\rm H}\vert = 0.46$  

As the electronegativity difference decreases, so does the ionic character of the bond. Hence its covalent character increases.


next up previous
Next: Electric dipole moment Up: Polar covalent bonds Previous: Polar covalent bonds
Mark E. Tuckerman 2008-10-30