In the transition metals, $4s$ has a higher energy level than $3d$, because in ionization, the electrons from the $4s$ energy level are lost first.

I was thinking that this might be the same for much of the p-block from elements Ga-Rn because:

Ga has a electron configuration of: $1s^22s^22p^63s^22p^63s^23p^63d^{10}4s^24p^1$

Ga$^{3+}$ has an electron configuration of: $1s^22s^22p^63s^23p^63d^{10}$


Well the only exception in the orbitals is for $s$ and $d$ orbitals. Meaning you will first remove electrons from $(n+1)s$ and then onto $nd$ orbital. It is only because the $(n+1)s$ sublevel is only lower in energy if there are no electrons in the d sublevel. As per the Aufbau principle, the the s orbital is therefore filled before. Then as the $d$ becomes filled with electrons, then the energy starts to flunctuate relative to one another and the s orbital ends up higher in energy. That is why when electrons are removed, they are first removed from s and then $d$. Luckily this only applies to $s$ and $d$ orbitals. When are you working with $p$ orbitals, well then the $np$ orbital will always be higher than the $(n-1)d$ orbital. Therefore for making ions, you will have to first removed electrons from $p$ then $s$ and then $d$ and so on.


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.