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How do I know when I should add an electron to a single electron? Do I just always do this? I've came across this in the textbook in the form of the amide ion $\ce{NH2}$, among others. It was shown to have 8 electrons, despite nitrogen only having 5 valence electrons, and each hydrogen only having 1. The last came from some surrounding atom to complete the outer shell and thus become more stable. So, do I always do that when there is a single electron not part of a pair?

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  • $\begingroup$ @Tlwolf I think you're not really asking about lewis structures, but about anions, radicals and cations. $\endgroup$ – Mithoron Feb 6 '15 at 10:21
  • $\begingroup$ Tlwolf your title gave me the impression that you are mixing up ionic and covalent bonds, though I later observed you don't. $\endgroup$ – M.A.R. Feb 6 '15 at 11:09
  • $\begingroup$ I was under the impression ionic bonds are between a metal and nonmetal, such as sodium chloride, right? Due to the large difference in electro negativity? But I do believe @Mithoron may be right in that I am talking about cations and anions, I am just unsure about when I show the structure as having a charge by adding an electron (or maybe taking one away, in the case of cations?) I don't know if I just always do this for a compound that has a left over electron not in a pair, or in certain instances $\endgroup$ – Tlwolf Feb 6 '15 at 11:23
  • $\begingroup$ It all depends what you want to have written - without this 8-th electron you'd have a radical - in this case it would be unstable (cation even more) but sometimes they aren't - chemistry is complicated. $\endgroup$ – Mithoron Feb 6 '15 at 11:54
  • $\begingroup$ Lol yes it is, that's why I'm trying to get a head start before my freshman year even starts, and this site seems full of chemistry enthusiasts who are knowledgable $\endgroup$ – Tlwolf Feb 6 '15 at 12:05
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I think I know where your confusion comes from. Firstly, I would want to note an important point:

Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Chemwiki.UCDavis

This is the characteristic that makes them different from ionic bonds. The surrounding atom will not become an ion in its full sense.

I would want to describe a way of drawing Lewis electron dot structures that will avoid any confusions.

$\ce{H2O}$:

We have two Hydrogen atoms and an Oxygen atom to begin with. They have this lewis structure:

Source 1

Follow these instructions for "normal" molecules (the ones that abide octet in a perfect sense):

  1. Compute all of the electrons that are in the valence shells of separate atoms. For instance, an O atom has 6 es in the valence shell and each of H atoms have one. So you get: $$6+(2*1)=8$$ I prefer to call this $\text{n}_v$. V represents valence shell.
  2. Compute all of the electrons the atoms will have after the compound has been formed. That is, when all of them abide octet. In the instance of water, O will have eight (just like other octet abiding atoms) and H atoms, as they are exceptional, will have two. So you get: $$8+2*2=12$$ I prefer to call this $\text{n}_f$. F represents "final". You can name it whatever you want.
  3. If you want to know how many electrons are bonding and are shared between atoms, subtract $\text{n}_v$ from $\text{n}_f$: $$\text{n}_f - \text{n}_v \rightarrow 12-8=4$$ For an ease in calculation, I call this $\text{n}_b$. As you might have just figured out,
  4. If you want to know how many es are nonbonding and "belong" to a single atom, subtract $\text{n}_b$ from $\text{n}_v$ to get: $$\text{n}_v - \text{n}_b \rightarrow 8-4=4$$ Call this whatever you want. I rather calling it $\text{n}_{nb}$.

With the information you just gathered you can draw the lewis structure for most of the covalent compounds. In the case of water, 4 bonding es mean two covalent bonds, right? Plus, H has no other electrons than it's only one. So O has 4 nonbonding electrons. Thus, you may come to something like this:

Source 2

$\ce{NH4+}$

In ions, we have to take the charge of the ion into an account. Positive charge means lesser electrons than a neutral species, and negative charge means more es in a comparison with a neutral covalent compound. Thus, we should add to or subtract from $\text{n}_v$.

So let's get on with this thing's lewis structure:

  1. $\text{n}_v = 5+(4*1) -1=8$ (Notice (-1) that was because of a 1+ ion charge.)
  2. $\text{n}_f = 8 + (4*2)=16$
  3. $\text{n}_b = 16-8=8$
  4. $\text{n}_{nb} = 8-8=0$

There is an important thing to note here. Nitrogen has 5 es in valence shell, but ends up having four bonds. This is because an H atom had lost it's electron to become n ion AND, with an empty orbital. So, N shares two electrons instead of one which is described as dative covalent bonding. We get this structure in accordance to the calculations and thoughts:

Source 4

Finally, $\ce{NH2-}$

Now this case is easily solvable through the explanations for ammonium.

  1. $\text{n}_v = 5 + 2 +1=8$ The +1 is because of the charge of the ion.
  2. $\text{n}_f = 8 +2*2=12$
  3. $\text{n}_b = 12-8=4$
  4. $\text{n}_{nb} = 8-4=4$

Put some thinking, and you should get this (ignore the other ones if you like):

Source 5

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  • $\begingroup$ Those formulas are helpful, thank you. I'm just still unsure if there's some basic rule to know when a single electron needs to have another added to it (like NH2 -1) It's just I thought I had the concepts down pretty well, until I ran into this exception where an extra electron was present that I couldn't find a way to account for $\endgroup$ – Tlwolf Feb 6 '15 at 10:08
  • $\begingroup$ I'm attempting to expand my answer a little bit. I see it incomplete. And, I will include what you mean. $\endgroup$ – M.A.R. Feb 6 '15 at 10:10
  • $\begingroup$ Ok, I think that pretty much clears it up for me, thanks. So it's just a matter of following the octet rule. So then I'll assume I pretty much always take this into account except for some select exceptions to the octet rule then? $\endgroup$ – Tlwolf Feb 6 '15 at 11:32
  • $\begingroup$ This is a really nice rule for most of the compounds, but it may result in confusion in the case of resonance structures and some of the exceptions to the octet and such. If you see yourself compromised in those cases, it's better to make a though-based approach, rather than calculations. The only way to deal with them is practicing dealing with them! $\endgroup$ – M.A.R. Feb 6 '15 at 11:45
  • $\begingroup$ Alright, I'll keep practicing in that case aha. But thanks for your help $\endgroup$ – Tlwolf Feb 6 '15 at 11:55

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