Whenever I look up the Lewis structure for $\ce{NH2^-}$, it shows 8 electrons. Now I understand the octet rule, but I still don't see how an electron can just appear from nowhere. There are 5 valence electrons with Nitrogen, and 2 more, one from each hydrogen. I know that formal charge is calculated by subtracting the nonbonding electrons and number of bonds (or number of electrons in bonds divided by 2), but why is the formal charge of $\ce{NH2^-}$ "-1", and why the extra electron out of nowhere? Because 5 - 3-2 is still just 5


For atoms in the first 2 rows of the periodic table the octet rule arises as a consequence of filling orbitals four orbitals which each hold 2 electrons - each of which appears as a pair of dots in your Lewis structure.

Were you to construct the Lewis structure for neutral $\ce{NH2}$ you would indeed have 5 electrons from the nitrogen, and two from hydrogen, 7 i.e. 3 pairs and 1 extra electron on its own in a orbital. Electrons in unpaired orbitals are however very reactive, the pairing of electrons lowers the total energy of the system. As such, it is more favourable for the $\ce{NH2}$ molecule to remove an electron from some other molecule or atom near by and form ions e.g. $\ce{Na}$ to create $\ce{Na+NH2-}$.

Also by creating the charges, you create an electrostatic attraction between the ions which also lowers the energy and makes the ion more stable than the neutral molecule.

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    $\begingroup$ +1 for the explaination that $NH_2$ doesn't automatically have a negative charge, but rather it steals a charge from another atom/molecule to become $NH_2^-$. It just happens so mindbogglingly fast that we gloss over that detail and go straight to the ion. $\endgroup$ – Cort Ammon Feb 6 '15 at 1:26
  • $\begingroup$ Good answer, but you might as well touch the notion of a radical, which neutral $\ce{NH2}$ is, by actually drawing its Lewis structure you just mentioned. And you can then show in comparison the Lewis structure of negatively charged $\ce{NH2-}$ which is the result of neutral $\ce{NH2}$ "stealing" an electron from some other molecular entity. $\endgroup$ – Wildcat Jul 7 '15 at 21:26

$\ce{NH2-}$ doesn't actually exist on its own, it's only shown that way for parsing a reaction, it is transferred at the same time the group that is giving it away gains something else, for example sodium amide is Na+NH2- so in order to be separated the NH2- needs somewhere to go as well as the Na+, always wondered why they bother teaching confusing nonexistent intermediates anyway, ions only exist in the non time they are exchanged (or electrocuted).

  • $\begingroup$ Um...sodium amide. $\ce{NH2-}$ definitely does exist. $\endgroup$ – bon Jul 7 '15 at 21:15
  • $\begingroup$ This post also does not answer the question about why $\ce{NH2-}$ (extant or otherwise) is an anion. If you are concerned about the existence of the amide anion, feel free to ask a question about it. $\endgroup$ – Ben Norris Jul 7 '15 at 22:41
  • $\begingroup$ Were you thinking strictly organic? Don't forget about inorganic chemistry and don't forget that even though ammonia is only a very weak acid there exists such reactive elements as the alkali metals. $\endgroup$ – busukxuan Jul 8 '15 at 12:06
  • $\begingroup$ This is actually quite good point. Every single anion may be "free" only in special situation, otherwise it's bound. Every single "ionic" bond has some covalent character. $\endgroup$ – Mithoron Dec 1 '18 at 21:43

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