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Which of the following possesses the highest energy electron?

  1. $\ce{Br-}$
  2. $\ce{Ca^2+}$
  3. $\ce{Cr+}$
  4. $\ce{As}$

Don't bromine and arsenic have equally high energy electrons (i.e. in the $\ce{4p}$ subshell)? My book says that arsenic is the correct answer. But I don't see why $\ce{Br^{–}}$ is wrong.

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My best guess is that your book wants you to argue that, since the bromine nucleus has a higher charge than arsenic, electrons are more strongly attracted to it, lowering their energy. And indeed this would be true - if we only considered interactions between the electrons and the nucleus, and ignored electron-electron interactions. Unfortunately, however, the situation in real life is more complicated than this, because electrons repel one another, “shielding” each other from being attracted to the nucleus (which, of course, is the reason why subshells have different energies in the first place). And the bromide ion has three more electrons than the arsenic atom, which actually turns out to be quite significant.

In fact Webelements gives the electron affinity of $\ce{Br}$ (i.e., the first ionisation energy of $\ce{Br^-}$) as 324.8 kJ mol$^{-1}$ while the first ionisation energy of As is 947.0 kJ mol$^{-1}$. So in fact it seems that, experimentally, the bromide ion has the highest energy electron (which therefore requires the lowest energy to remove it).

This is a more difficult question than it looks - the problem of how to deal with interactions between lots of electrons is still an active research field! A fairer question to do in your head would be to compare, say, $\ce{Br^{2+}}$ with As: in this case the number of electrons of each species is the same, so that the argument I put forward at the beginning of this response is actually correct here.

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The energy of orbital changes via row in periodic table and falls to the end of row. Because of it in orbitals for particles from the end of row have lesser energy.

The concept, however, seems... questionable.

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$\ce{Br-}$ = $\ce{[Kr]}$. $\ce{Kr}$ is a noble gas and has a full octet, meaning that the energy is SUPER low because of how stable it is.

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  • $\begingroup$ Welcome to Chemistry.se! In its current state, your answer is not addressing the original issue of the question. Could you expand it by editing in some more details, please. Also have a look at the help center and take the tour. $\endgroup$ – Martin - マーチン Dec 22 '14 at 6:48

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