Why do xenon/krypton have relatively high electronegativity

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Xenon and krypton exhibit higher electronegativities compared to other noble gases due to their unique electronic configurations and the presence of available p-orbitals that can participate in bonding. While noble gases are generally characterized by their lack of reactivity and low electronegativity, xenon and krypton can form compounds under certain conditions, which is attributed to their ability to attract electrons more effectively than their lighter counterparts. The discussion emphasizes that electronegativity trends generally decrease down a group and increase across a period, but the specific behavior of xenon and krypton challenges the traditional understanding of noble gas chemistry. The conversation also touches on the distinction between scientific inquiry into "how" versus "why," highlighting the complexities involved in explaining these phenomena.
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what sets xenon and krypton apart from the other noble gasses that enables them to have higher electronegativites?
 
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haha thanks for the attempt, i too am familiar with google, but i still haven't found any satisfying answers simply through typing it in on the web.
 
Well the fact is that we as a species still have much to learn, for there are many things that we don't know. So if you can't find an answer, maybe it has not been solved yet...
 
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Alexi-dono said:
So if you can't find an answer, maybe it has not been solved yet...

Or, more likely, the only acceptable answer is "that's what you get when you measure/calculate, possible explanations are too handwavy to make sense".

Please remember that science doesn't answer question "why?". Science answers question "how?". We sometimes try to explain "why" using our knowledge about "how", but it is always a risky business.
 
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Um, but the EN of the noble gases decreases down the table!
 
Noble gases do not have electro negativity. But E.N decreases down a group and increases left to right across a period. Can you give reference for your question?
 
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Yeah, okay, is that answer explained in link on post #8 correct?
 
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