-2
$\begingroup$

This picture from Ebbing's General Chemistry [1, p. 234] shows the electron orbital energies for a hydrogen atom, where each subshell belonging to the same principal quantum number ($n = 1, 2, \cdots$) are on the same level. Is this true for any particle with only one electron? Would, $\ce{Li+}$, for example, have the same orbital energies?

Energy levels of hydrogen

References

  1. Ebbing, D. D.; Gammon, S. D. General Chemistry, 11th ed.; Cengage Learning: Boston, MA, USA, 2017. ISBN 978-1-305-85914-2.
$\endgroup$
  • 5
    $\begingroup$ Please always cite the source the illustration has been taken from. $\endgroup$ – andselisk Jan 7 at 8:52
  • 2
    $\begingroup$ Since every distinct elementary nucleus has a different charge, why would you expect a single electron system based on that nucleus to have the sameenergy as a hydrogen atom which has only a single positive charge? $\endgroup$ – matt_black Jan 7 at 14:21
  • $\begingroup$ @matt_black you're right. They will have different values of energy but the individual subshells will have same energy for a particular value of n for they are degenerate. $\endgroup$ – Priyanshu Das Jan 8 at 4:16
2
$\begingroup$

It would have to be $\ce{Li^2+}$ or $\ce{He+}$ to have 1 electron only.

And even for 1 electron only ions, energy levels depend on the kernel charge.

For multielectron atoms/ions, energies of different orbitals with the same quantum number $n$ differ, due electron repulsion and kernel shielding. That is the reason why the orbital 4s is occupied before orbitals 3d.

The orbital energies are additionally dependent on the particular electron quantum state distribution.

This affects empirical orbital filling rules, causing exceptions, and orbital involvement in atom bonding.

So even if the orbital 4s has lower energy than 3d while filling, it can have higher energy when they are filled. Their relative energy for neutrs atoms also change with the atomic number.

$\endgroup$

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