# What is the Lewis structure of the azide ion?

On a chemistry homework, we're asked to draw the Lewis structure of $$\ce{N3-}$$.

• Nitrogen ordinarily has five electrons in its valence; $$\ce{N-}$$ has six. This makes a total of 16 electrons overall.
• If we place the negative charge on the central atom, we get a structure of $$\ce{N=N^-=N}$$
• This has a formal charge of $$(-1)$$ on each of the side nitrogen atoms and $$(2)$$ on the central nitrogen atom once you've accounted for the extra electron, so the formal charges cancel out only when you account for the added electron.
• This answer is supported by a bunch of online sources (1, 2).

Homework software response:

• The correct answer is $$\ce{N#N+-N^2-}$$

That's all it says. My understanding of why this is valid is that this gives a formal charge of $$(0)$$ for every atom, as opposed to averaging out to that.

I suppose the second form makes sense, as it's a lower formal charge on the central atom. If so, why do online resources prefer the first? Is there some resonance going on here, and the actual structure has bond orders of $$\frac52$$ and $$\frac32$$?

• If you place the negative charge on the central atom of your structure, you'll exceed the octet. Instead, you might want to place a positive charge there. Dec 12 '19 at 22:42
• @IvanNeretin Why will I exceed an octet? Nitrogen normally has five electrons in its valence; adding an electron gives it six. It makes a double bond with each side nitrogen and donates electrons to fill their valences in that version. Dec 12 '19 at 23:02
• Dec 13 '19 at 18:29

Your proposed structure is wrong. Nitrogen does not exceed the octet in any of its known compounds (and even if $$\ce{NF5}$$ will be discovered it will not exceed the octet according to everything we know now). However, if you have a formal negative charge that means an additional electron added to the 5 nitrogen usually has; if four of those six electrons are used to build the double bonds, there is still a lone pair on nitrogen for a total of 10 electrons.

If you are having trouble determining Lewis structures, there are four quick calculations that you can perform to help you:

1. Add up all valence electrons the atoms are bringing into the compound.
Each nitrogen has five electrons plus there is one negative charge (additional electron) so:

$$3\times5+1=16\tag{1}$$

2. Add up how many valence electrons would be needed so that each atom has an octet (for hydrogen: dublet) of its own.
Each nitrogen would want eight electrons so:

$$3\times8=24\tag{2}$$

3. Take $$(2)-(1)$$. This represents the number of electrons the atoms must share, i.e. the number of bonds.

$$24-16=8\tag{3}$$

4. Take $$(1)-(3)$$. This represents the number of electrons that do not have to take part in bonds; these must then be distributed as lone pairs.

$$16-8=8\tag{4}$$

Then, start drawing but make sure you have as many lone pairs and bonding electrons as the equations state. Ignoring the lone pairs, we can get the following possible structures for $$\ce{N3-}$$:

$$\ce{N#N-N}\qquad\qquad\ce{N=N=N}\qquad\qquad\ce{N-N#N}$$

(The exercise of distributing four lone pairs across the three nitrogens so that each ultimately has eight valence electrons is left to the reader because I am too lazy to open ChemDraw to draw the structures.)

After you have done that, you need to take a look at potential formal charges. For that, split each bond homogenously (i.e. give each atom one of the bonding electrons) and count. Compare that count to what an atom should have; the difference corresponds to the atom’s formal charge. (Because electrons are negative, an additional electron corresponds to a charge of $$-1$$.) When done for those three structures, we arrive at:

$$\ce{N#\overset{+}{N}-\overset{2-}{N}}\qquad\qquad\ce{\overset{-}{N}=\overset{+}{N}=\overset{-}{N}}\qquad\qquad\ce{\overset{2-}{N}-\overset{+}{N}#N}$$

In each of those cases, the formal charges sum up to the overall charge of the molecular ion ($$-1$$) which is an indication that we have done it correctly. (Again, I have sheepishly left out the lone pairs; you can use my formal charges to determine where they should have been and how many.)

There is no principle of zero formal charges. However, when debating between different structures, a structure with less formal charges is often (not always!) more ‘favourable’. (The actual term should be ‘contributes more to the overall picture’ but that may confuse too much at this stage.)

But which of the three is correct? They all are! In fact, this is what is known as mesomery: we have a number of (resonance) structures that all explain the actual compound a little bit but neither holds the absolute truth. To show this, resonance arrows are usually drawn between the depictions:

$$\ce{N#\overset{+}{N}-\overset{2-}{N} <-> \overset{-}{N}=\overset{+}{N}=\overset{-}{N} <-> \overset{2-}{N}-\overset{+}{N}#N}$$

The key difference between correct structures and your proposition is that the central nitrogen atom can never carry a negative formal charge as it needs to accomodate four bonds to its neighbours which is only possible for $$\ce{N+}$$.

As for the answer given in the homework response: It is not strictly correct because it is incomplete. All three structures should be marked as correct – until the concept of resonance has been formally introduced at which point only a combination of the three should be.

• But… but… you don't need ChemDraw when it can be done with el3g4nt MathJax syntax: $$\ce{:\!\!N#\overset{+}{{N}}-\overset{2-}{\overset{\Large.\!\!.}{\underset{\Large.\!\!.}{N}}}\!\!:}$$ *Overconfident alcoholic* Dec 13 '19 at 8:20
• @andselisk I was, in fact, contemplating drawing them with MathJax but then decided today was not the day for that rabbit hole. But thanks for a hearty laugh! =D
– Jan
Dec 13 '19 at 13:16
• Okay, I see where I went wrong at least. Resonance has been introduced in class, making a number of these homework problems even more head-scratching because there’s more than one equally correct answer. Dec 13 '19 at 15:35