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Lets say we have an nitrogen atom - it should have 7 protons and 7 electrons.

How can an nitrogen anion - lets say $\mathrm{N^{\,1\mathbf{-}}}$- even exist - shouldn't the 7 electrons in valence shell repell the extra one?

What force does hold the extra one electron in the valence shell? There isn't an extra proton in the $\mathrm{N^{\,1\mathbf{-}}}$ that would hold the extra electron.

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One reason cations and anions exist is due to the stability of a full or half-full valence shell. The stability from those electronic configurations means that the atom or molecule does not require protons to "hold" the extra electron.

Recall also that nitrogen has three (or five) valence electrons, rather than seven. The 1 s shell is full and is not considered part of its valency. The three 2 p electrons are the valence electrons although they hybridize with the 2 s electrons to produce the trigonal pyramidal structure of ammonia with its lone pair. The single anion $\ce{N^{-1}}$ could exist, but would not be stable because it puts four electrons in the p shell. The p shell would prefer to have three electrons as it does in the nitrogen atom or no electrons as it does in the $\ce{N^{3+}}$ cation.

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  • $\begingroup$ Thank you for your answer! Do I get your answer right when I suppose that the extra electron(s) in the valence shell of an anion are held there by the electrical field of the electrons in the electron shells and also thanks to "right" spatial formation of the electron shells? $\endgroup$ – sarasvati Jan 12 '16 at 16:03

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