2
$\begingroup$

The following sentence is stated in my current chemistry book:

The first ionization enthalpy of zinc is higher because it represents an ionization from the $\rm 4s$ level.

Now zinc does have an ionization enthalpy of $\pu{906 kJ/mol}$, which is higher than copper's $\pu{745 kJ/mol}$ or nickel's $\pu{736 kJ/mol}$ in the same series. But I am not sure that the reason for this increase stated in the book is correct. For example, even copper has an ionization from the $\rm 4s$ level, but it does not have anywhere near as high ionization enthalpy as zinc. Is the book correct, or is there some other reason for this?

Improve this question
$\endgroup$
3
  • 3
    $\begingroup$ The book is incorrect. Cu ionisation is also from 4s. The reason is purely because of the increase in effective nuclear charge in Zn. $\endgroup$
    – orthocresol ♦
    May 14 '16 at 16:21
  • $\begingroup$ Do you know the real reason behind this unexpected increase $\endgroup$
    – Abhishek Mhatre
    May 15 '16 at 15:13
  • $\begingroup$ I've been wondering about this myself. The sudden jump in Ie from Cu to Zn isn't very well explained by mere d orbital filling because it doesn't seem to be present in p orbital filling. I have a theory (only a theory) that what's really going on is the S orbital is only getting one electron, then the p orbitals are filled, then the S orbital is completed at the end. If that's the case, I have no idea why it would happen that way. $\endgroup$
    – GoSlash27
    Feb 14 at 22:19
5
+50
$\begingroup$

It is because the shell is complete.

Zinc's electronic configuration is $\rm [Ar] 3d^{10} 4s^2$, which is a complete configuration for a transition metal. The next element in the same period is $\ce{Ga}$, which starts filling $\rm 4p$ orbitals ($\rm [Ar] 3d^{10} 4s^2 4p^1$).

The zinc's electronic configuration causes a sort of special stability because of the filled electronic states. Also, the effective nuclear charge is greater than $\ce{Cu}$'s, so it's more difficult to ionize $\ce{Zn}$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.