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Calculate the electronegativity of fluorine from the following data:

$$\begin{align} \text{BDE}(\ce{H-H}) &= \pu{104.2 kcal/mol} \\ \text{BDE}(\ce{F-F}) &= \pu{36.6 kcal/mol} \\ \text{BDE}(\ce{H-F}) &= \pu{134.6 kcal/mol} \\ \chi_\ce{H} &= 2.1 \end{align}$$

I thought that electronegativity should be related to effective nuclear charge, as it is a measure of how strongly atoms "pull" on bond pairs.

Is there a formula to calculate the electronegativity from bond dissociation energies, and what is the rationale behind it?

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Considering the data you have been given, the equation you need is the one that defines the Pauling electronegativities:

$$\left|\chi_\ce{A} - \chi_\ce{B}\right|=\left(\text{eV}\right)^{-1/2}\sqrt{E_\mathrm d(\ce{A-B})-\frac{E_\mathrm d(\ce{A-A})+E_\mathrm d{\ce{(B-B)}}}{2}}$$

In the equation, $\left(\text{eV}\right)^{-1/2}$ is in the equation to negate the units of the bond dissociation energies. This means that you need to convert all your bond dissociation energies from $\pu{kcal/mol}$ into $\pu{eV}$. You can use the BDEs in units of $\pu{kcal/mol}$, but then the formula will require a constant in front to adjust for the difference between units.

$\pu{1 eV}$ is equal to $\pu{1.602 \times 10^-19 J}$, so you need to be able to convert from $\pu{kcal/mol}$ to $\pu{J/mol}$, then divide through by Avogadro's constant $N_\mathrm A = \pu{6.022 \times 10^19 mol-1}$. There is a handy website which can help with this.

$$\begin{align} E_\mathrm d(\ce{H-H}) &= \pu{4.519 eV} \\ E_\mathrm d(\ce{F-F}) &= \pu{1.59 eV} \\ E_\mathrm d(\ce{H-F}) &= \pu{5.837 eV} \end{align}$$

Plugging the information you have been given, we find that

$$\begin{align} \left|2.1 - \chi_\ce{F}\right| &= \sqrt{5.837 - \frac{4.519+1.59}{2}} \\ &= 1.668 \end{align}$$

This gives you two possibilities for $\chi_\ce{F}$, $0.4$ or $3.8$, depending on whether the quantity $(2.1 - \chi_\ce{F})$ is positive or negative. This is where we need to use some knowledge of chemistry: fluorine is more electronegative than hydrogen, so we should have $\chi_\ce{F} > \chi_\ce{H}$, and consequently we take the solution $\chi_\ce{F} = 3.8$.


Why should bond energies be related to electronegativity? The idea that Pauling had when he devised this equation was that in a covalent bond between two of the same atom, there is no electronegativity difference and hence no ionic contribution to the bond strength. Hence, the "average covalent contribution" of each atom to a bond could be measured by the mean of the two bond strengths, i.e. $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$.

However, when there is a difference in electronegativity, the bond strength has an ionic contribution. The larger the difference in electronegativity $\left|\chi_\ce{A} - \chi_\ce{B}\right|$, the larger the ionic contribution. This ionic contribution could be roughly measured by subtracting the covalent contribution, $[E_\mathrm d(\ce{A-A}) + E_\mathrm d(\ce{B-B}) ]/2$, from the $\ce{A-B}$ bond strength.

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