0
$\begingroup$

It can be observed and proven that diatomic isoelectronic molecules have the same bond order when they follow the octet rule (upto 3rd period elements). But what about when they do not follow the octet rule like in transition metals? Do they still have the same bond order? If so, why?

Improve this question
$\endgroup$
12
  • 1
    $\begingroup$ That's not true even for CO vs N2, at least if you use computational methods. $\endgroup$
    – Mithoron
    Commented Jun 17, 2022 at 19:29
  • $\begingroup$ Isoelectronic molecules have the same everything, except nucleus charge. Think of it this way: what is a bond made of? Electrons. $\endgroup$
    – Ivan Neretin
    Commented Jun 17, 2022 at 19:37
  • $\begingroup$ Do you know many diatomic molecules that do not follow the octet rule , apart from $\ce{NO}$ ? $\endgroup$
    – Maurice
    Commented Jun 17, 2022 at 20:23
  • $\begingroup$ @Mithoron Can you elaborate a little or tell me where to start searching? Do you mean that the electronic configurations of CO and N2 are different as given by molecular orbital theory? $\endgroup$
    – Balu
    Commented Jun 18, 2022 at 3:06
  • 1
    $\begingroup$ $\ce{CoO}$ is a cobalt oxide. It has nothing to do with $\ce{CO}$. Nothing ! And $\ce{CoO}$ cannot be obtained as a gas. It is decomposed without melting or boiling when heated at 1800°C $\endgroup$
    – Maurice
    Commented Jun 18, 2022 at 21:01

0

Reset to default

Your Answer

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