-2
$\begingroup$

The question said "Write the condensed electron configuration for $\ce{Zn^{2+}}$I got [Ar] 4s$^2+$ 3d$^8$. Why was this the incorrect answer? Zinc lost 2 electrons and surely those electrons should go from the d orbitals?

$\endgroup$
2
$\begingroup$

By the time the $\ce{3d}$ orbitals are filled with electrons, the $\ce{4s}$ orbitals have become higher in energy, making it easier to remove them. Here are energy level curves for both the neutral and +1 forms of the $\ce{3d}$ atoms, with electron energy on the $y$-axis:

enter image description here

$\endgroup$
0
$\begingroup$

you have done only wrong thing and that is sufficient to make your answer wrong . Electron will not be removed from 3d subshell but it will be removed from 4s subshell. The reason is whenever we fill electrons we see the value of (n+l){Principal quantum number + azimuthal quantum number}, the subshell having smaller value of (n+l) gets the electroon first, butt when we have to remove an electron,it is removed froom the subshell having greatest value of n(Principal quantum number).

$\endgroup$
  • $\begingroup$ I don't get it, how do you go from (n+l) to the electron removing from the subshell having greatest value of n? $\endgroup$ – Hamze Oct 4 '16 at 3:10
  • $\begingroup$ see,by aufbau principal the electrons are filled in subshells in order of their increasing energy and order of increasing energy can be described with the help of n+l,now as far as use of n is concerned we see that in removing the electrons concept of shells is taken into account(not subshell) and quantum number associated with shells is 'n'. $\endgroup$ – I am Back Oct 4 '16 at 3:19
  • $\begingroup$ see (n+l) and aufbau principal on wikipedia . $\endgroup$ – I am Back Oct 4 '16 at 3:20

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.