Why do krypton and xenon have high electronegativity? Noble gases are supposed to be "happy" with the amount of electrons they have, because they have 8 valence electrons (thus, most noble gases have no electronegativity). So why do krypton and xenon have electronegativity? Why do they tend to "want" one more electron? It makes no sense... their outer shells are already full!

I'm going to do a research paper on this question, so I would really like some good resources that could help me answer this question for myself. Know any?

  • 3
    $\begingroup$ I think you misunderstand the definition of electronegativity. Have you read the Wikipedia article? $\endgroup$
    – Philipp
    Nov 5, 2014 at 23:31
  • $\begingroup$ For Pauling electroneg. bonds are needed, but they are no problem - there plenty of compounds of Xe and quite a bit for Kr. For other definition's you don't even need that. $\endgroup$
    – Mithoron
    May 26, 2023 at 17:03

1 Answer 1


Why do Krypton and Xenon have high electronegativity?

As you point out, krypton ($\ce{Kr}$) and xenon ($\ce{Xe}$) are members of the Noble gas family. They are generally unreactive (noble) because all of their occupied orbitals are filled with electrons - they really don't want to gain or lose an electron. However, in the 1960's researchers found that $\ce{Kr}$ and $\ce{Xe}$ did react with extremely electronegative elements such as fluorine and oxygen to form new molecules. In the reaction process, $\ce{Kr}$ and $\ce{Xe}$ basically gave up an electron to the very electronegative elements they reacted with.

$$\ce{Xe + 3F2 → XeF6}~~(1)$$ $$\ce{XeF6 + 3H2O → XeO3 + 6HF}~~(2)$$

The most common definition of electronegativity is based on Pauling's work and is given by the equation

$${\chi_{\rm A} - \chi_{\rm B} = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - [E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})]/2}}$$

where ${\chi_{\mathrm{A}}}$ and ${\chi_{\mathrm{B}}}$ are the electronegativities of atoms $\ce{A}$ and $\ce{B}$ and ${E_{d}(AB)}$ represents the bond dissociation energy of molecule $\ce{A-B}$.

Using $\ce{HBr}$ as an example, if we know the bond dissociation energies of $\ce{HBr,~ H2}$ and $\ce{Br2}$, then we can calculate the difference in electronegativity between $\ce{H}$ and $\ce{Br}$. From this, if we know the electronegativity of hydrogen, then we can determine the electronegativity of bromine.

In essence, for any element that can react and form a molecule, we can calculate its electronegativity. Since Kr and Xe do react to form molecules, we can use the above methodology to calculate its electronegativity.

Note: In the case of atoms like Kr and Xe that do not form a diatomic species like $\ce{Xe2}$, we need an extra step to find their electronegativity. Since there is no $\ce{E_{d}(Xe-Xe)}$, we must use Pauling's equation twice, once with reaction (1) and once with reaction (2). We now have 2 equations and two unknowns (${\chi_{\mathrm{Xe}}}$ and ${E_{d}(Xe-Xe)}$) so we can solve for ${\chi_{\mathrm{Xe}}}$, the electronegativity of $\ce{Xe}$.

  • $\begingroup$ How do we know that the radicand is positive, so we can take the square root? $\endgroup$
    – ado sar
    Apr 7 at 15:03

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