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First I know similar questions have been asked about counting d-electrons.

However, my question is about the electrons in the 4d orbital. For example if I have CdS. The cadmium has a charge of +2. So does this affect the number of d-electrons?

Would it be 10 d-electrons as the 2 extra have been removed from the 5s orbital?

Also is the electron configuration of Cd in CdS:

[Kr] 5s2 4d10

or is it:

[Kr] 4d10

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  • $\begingroup$ Hello and welcome to Chemistry.SE! At this point I recommend that you take the short tour to better familiarize yourself with the site. Also, as this appears to be a homework type of question, you should read through this discussion. Regarding your question, which answers do you suspect are correct and why? $\endgroup$ – airhuff Feb 16 '17 at 10:05
  • $\begingroup$ Thank you! I suspect that because Cd has a +2 charge, the 5s2 electrons will be lost. So in CdS the electron configuration of Cd is [Kr] 4d10. But I didn't know this was possible... I think i've just made up that I can remove the 5s2 electrons in compounds like this. $\endgroup$ – lalalalala Feb 16 '17 at 10:07
  • $\begingroup$ Most often it is the valence electron which takes part in formation of ionic compounds, even in transition metals. So Yes $[Kr] 4d10$ is correct. $\endgroup$ – samjoe Feb 16 '17 at 13:53
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Without doing some sort of at least half-serious calculations, you cannot really answer the question. For all we know, a cadmium(II) ion could be $[\ce{Kr}]\,\mathrm{4d^{10}}$, $[\ce{Kr}]\,\mathrm{4d^9\,5s^1}$ or $[\ce{Kr}]\,\mathrm{4d^8\,5s^2}$.

However, given the analogies to other transition metals and the general structure of practically all transition metal ions, the first configuration, $[\ce{Kr}]\,\mathrm{4d^{10}}$ is most likely. I can’t recall a case where a transition metal ion had any s electrons in the valence shell before the d electrons were completely populated; i.e. even if you have two s electrons in the neutral atom and ionise only once you would end up with only d electrons in the ion.

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