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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.

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  • $\begingroup$ 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. $\endgroup$ – Oscar Lanzi Jan 8 at 11:00
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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.

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

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

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

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    $\begingroup$ 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 at 6:55
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Using Hannay and Smith formula we have

$$\text{% of ionic character} = (0.16\delta + 0.035\delta^2)\times 100\%$$

where $\delta$ is electronegativity.

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