For bromine within it’s electron configuration the valence electrons are 4s2 and 4p5 and not 3d10 and 4p5, my chemistry teacher explained this as 4s2 being less shielded than the 3d orbital.

Valence electrons are the electrons that get ionised correct? If the 4s2 orbital experiences more Zeff then wouldn’t it be harder to ionise as there’s a greater degree of electrostatic attraction? And the 3d orbital, since it experiences more repulsive forces, would be easier to ionise?

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    $\begingroup$ I really don't understand where this question is going. Are you talking about the first ionisation energy of bromine? Why would s/d orbitals be involved with that? $\endgroup$ Feb 25, 2023 at 13:22
  • $\begingroup$ The first ionisation energy depends on the highest energy electrons which would be the valence electrons, is that correct? If so the valence electrons 4s has a lesser degree of electron shielding than 3d, if this is the case then the 4s electrons should experience more Zeff and thus should require a higher ionisation energy as they are more attracted. But 3d orbitals have more repulsion so therefore why aren’t they ionised first as it would require less energy? $\endgroup$
    – Mason
    Feb 25, 2023 at 13:28
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    $\begingroup$ The highest occupied orbitals are the 4p orbitals, so I really don't understand your question. $\endgroup$ Feb 25, 2023 at 13:31
  • $\begingroup$ What about the second ionisation process, the electron in the 4s orbital would become ionised no? $\endgroup$
    – Mason
    Feb 25, 2023 at 13:41

1 Answer 1


Most likely your teacher misspoke. The $4s$ electrons clearly suffer less nuclear charge and are more reactive; in fact $4s$ and $4p$ the only reactive electrons in bromine.

In this answer the change in electronic energy levels as we add more nuclear charge is discussed: it causes the $3d$ orbitals to ultimately fall below the $4s$ in energy; $3d$ meets up with its $3s$ and $3p$ shell-mates as the electrons sort out according to their proncipal shells. With chromium and iron, the elements discussed in that answer, we need to form a positive ion to make this switch happen; but by the time we get to bromine there is enough unshielded nuclear charge even in the neutral atom to sink the $3d$ level below tge $4s$. Thus in bromine the $3d$ electrons are sunk into the core along with tge rest of the $n=3$ shell, leaving only the $4s$ and $40$ electrons to act in ordinary chemical processes.


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