I'm trying to solve some exercises from the 2017 IChO, and the problem 6-A1 states that

in first ionization, an electron with the quantum numbers $n_1 = 4 - l_1$ is removed.

Now, I've never seen this way of expressing quantum numbers, and I don't know what it's supposed to mean, so I would appreciate it if someone could dumb it down to me.

  • 1
    $\begingroup$ Could you provide a link to the exercises? As is, I can't make out what you writing. If not, it may help if you rewrite the equation in the question using [Mathjax]( chemistry.meta.stackexchange.com/q/86/41556) so it is more readable. Welcome to the site! @Camila $\endgroup$
    – Tyberius
    Dec 29, 2018 at 21:16
  • $\begingroup$ I've edited your question including the link to the problem I found (please check out whether it is the correct one). Still, to me it's not quite clear what you are trying to find out since $n_i$ and $l_i$ are standard notations for principal and azimuthal Q.N. $\endgroup$
    – andselisk
    Dec 30, 2018 at 1:27
  • $\begingroup$ @Camila you seem to just be getting confused by the subscripts. Those are just their to label that $n_1$ and $l_1$ are the quantum numbers of the 1st electron ionized. $\endgroup$
    – Tyberius
    Dec 30, 2018 at 14:22

1 Answer 1


$n_1 = 4 - l_1$ means the same as $n_1 + l_1 = 4$, so the $n$ and $l$ quantum numbers of the first electron added up should equal 4. This could refer to 3p or 4s.

For the second electron removed, the $n$ and $l$ add up to 5. This could refer to 3d, 4p or 5s.

For chromium, the first electron removed is the 4s, followed by the 3d. For copper, this is the case as well, but its mass is higher than that of iron, so it is not the answer.


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