What is the ionic character of a bond, $\ce{A-B}$, in terms of the electronegativities of $\ce{A}$ and $\ce{B}$ ($\chi_\ce{A}$ and $\chi_\ce{B}$)?

I have been taught that the percentage ionic character is:

$$ \frac{\text{observed value of ionic character}}{\text{calculated value of character}} $$

but I can't understand how electronegativity is used here. I couldn't find anything on the internet either.

  • $\begingroup$ It is naive to assume that electronegativity alone determines ionic character. If you trust the usual electronegativity criteria then metal hydrides can never be mostly iinuc, but in reality all alkali metals and most alkaline earth metals do just that. Poor covalent overlap between diffuse metal orbitals and the compact hydrogen 1s orbital has something to do with it. $\endgroup$ Jan 8 '19 at 11:00

Linus Pauling proposed an empirical relationship which relates the percent ionic character in a bond to the electronegativity difference $\Delta \chi$.

Percent ionic character $= (1-e^{-(\Delta \chi/2)^2} )\times 100$

But I'd like to correct the definition of percent ionic character in your question using dipole moment $\mu$ (not Observed value of ionic character):

Percent ionic character = $\Large\frac{\mu_{\text{observed}}} {\mu_{\text{calculated} }}$ $\times 100 \%$

Where $\mu_{\text{calculated}}$ is calculated assuming a 100% ionic bond.

For more details please see this page.


$$\text{% of ionic character} = 16\times ∆\mathrm{EN} + 3.5\times (∆\mathrm{EN})^2$$

where $∆\mathrm{EN}$ is electronegativity difference. For example, in $\ce{H-F}$ $∆\mathrm{EN} = 2$:

$$ \begin{align} \text{% of ionic character} &= 16\times 2 + 3.5\times 2^2 \\ &= 32 + 14 \\ &= 46~(\%) \end{align} $$

  • 1
    $\begingroup$ I don't understand how you obtained $Δ\mathrm{EN} = 2$. $Δ\mathrm{EN} = χ(\ce{F}) - χ(\ce{H}) = 3.98 - 2.20 = 1.78$ ($χ$ is Pauling's EN, not the oxidation state). $\endgroup$
    – andselisk
    Jan 8 '19 at 6:55

Electro negativity decides how the bond A-B is polarized- electron cloud is more attracted towards more electronegative atom. It's assumed that polarization beyond a certain limit, leads to ionic bond. The electronegativity difference serves as a measure of percentage at which a bond is ionic.Roughly speaking, electro negativity difference of 1.7 is equivalent to 50 ℅ ionic character;.(calculated ionic character in your question ) Thus, ionic character of a given compound is 50% ×∆ (E.N)/1.7


I agree on most of Yomen Atassi's answer, however I would like to "correct/clarify" the formula of the percent ionic character. The following is the one cited by him, which is the one also reported in several sites in the internet:

Percent ionic character = (1 − e^(−(Δχ/2)2))×100

However applying this equation to real cases results in incoherent percentages (negative values larger than 100). Checking the source of Pauling's equation (Pauling L. The nature of the chemical bond. 3rd ed. 1960. Pag. 98), I realized that the mathematical expression is different. This is the one cited in the book:

Percent ionic character = (1 − e^(−1/4(Δχ^2)))×100

The last expression provides consistent values which are in agreement with values reported by Pauling and other authors.
If you want to apply Pauling's equation for the calculation of "percent of ionic character" (using electronegativity differences, Δχ), consider to use the later equation.

  • $\begingroup$ The expressions are in fact the same. You or the references mistyped the first formula, the second 2 is meant to be an exponent and should be preceded by ^. $\endgroup$ Dec 27 '20 at 0:17

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