In ionic compounds, where does the extra electron go? Or, in case of a cation, from which shell does it come?

Consider, for example, $\ce{OH^{-}}$, where did this extra electron go in the electronic configuration? And, what about $\ce{OH^{+}}$? There exists no such ion. Why?


1 Answer 1


In $\ce{OH-}$, that extra electron is on the Oxygen atom. Oxygen has a very large electronegativity, so you can be sure that extra electron is with the $\ce{O}$ not the $\ce{H}$. A lone oxygen has 6 electrons in its valence shell (2s2, 2p4), and wants to get to 8 (2s2, 2p6). In the $\ce{OH-}$ ion, both atoms are relatively "happy", the $\ce{H}$ has 2 electrons in its valence shell (1s2) shared with Oxygen in a bond, and the Oxygen has 8 electrons in its valence shell (2s2, 2p6), 6 unpaired electrons and 2 shared with hydrogen.

In general, in an anion the extra electron will be going onto the most electronegative atom. In a cation the positive charge will go to the lest electronegative atom.

Here's a helpful picture, your extra electron is in red: enter image description here

picture reference: http://chemistrypartner.blogspot.com/2009/10/drawing-lewis-electron-dot-structure-or.html

As for your question about $\ce{OH+}$, it would never be stable because that oxygen is going to tear some electrons off of the first thing it sees, because it REALLY wants that full octet. If you somehow made some $\ce{OH+}$, my guess would be that the $\ce{H+}$ would just dissociate, and your $\ce{O}$ would find another $\ce{OH+}$'s $\ce{O}$ and make $\ce{O2}$ immediately.


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