4
$\begingroup$

I observe some elements like $\ce{Cr}$ and $\ce{Cu}$. Some of them have strange oxidation numbers that don't really fit the purpose of making the element as stable as possible. Just Chromium, for example, it has one valence electron in 4s, the oxidation tells that it is gonna lose 2 electrons. Which obviously breaks the stability of half filled 3p orbitals.

If this can be explained by the fact that $\ce{Cr}$ has weak electronegativity, then what about $\ce{Cu}$? It is among the most electronegative transition metals, but still might lose two electrons despite having fully filled 4d orbitals.

$\endgroup$
3
  • 6
    $\begingroup$ While neither an answer nor a duplicate, the weirdness of chromium and copper have been previously addressed on this site: chemistry.stackexchange.com/questions/151/… $\endgroup$
    – Ben Norris
    Dec 5, 2014 at 2:49
  • 3
    $\begingroup$ Electronegativity is not a good concept to rely on for explaining ground-state electron distributions, due to the simple fact that electronegativity itself is not precisely and unambiguously defined, especially for $d$- and $f$-block elements. A good answer is quite a bit more involved, and even then only qualitative. See this previous question and the questions linked within for some necessary background. I'd be happy if someone can round it all up. $\endgroup$ Dec 5, 2014 at 12:35
  • 1
    $\begingroup$ How do you know they won't reach stability? It's all to be blamed on the more complicated chemistry of d orbitals; the rules we learn about p block and s block elements don't hold there. $\endgroup$
    – M.A.R.
    Mar 24, 2015 at 11:58

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.