Oxidation state of nitrogen in triazirine (HN3)

The book asks me to find the oxidation state of nitrogen in the below compound:

To find the oxidation state using Lewis structure, I have to compare the electronegativity of the bonded atoms and assign the bonding electrons to the more electronegative atom. Then, the difference between the number of electrons in element state and bonded state will give the oxidation number.

Here, the bonding electrons of nitrogens bonded with a double bond will stay in between both the atoms will not be assigned to both. The $\ce{N-H}$ bonding electrons will be assigned to $\ce{N}$. What about the two $\ce{N-N}$ bonding electrons?

(The above structure is not the one given in Wikipedia.) $%edit$

• Throw away the book. This structure is hideously wrong. And anyway, you may assign these oxidation states more or less arbitrarily (just make sure they sum up to 0), so it does not really matter. – Ivan Neretin Feb 27 '16 at 1:49
• To be fair he doesn't say whether the book calls this hydrazoic acid, or if it's what he is calling this. Greenwood and Earnshaw does (very briefly) mention the above so I assume it is known somehow. – Ian Bush Feb 27 '16 at 8:17
• The given structure belongs to triazirine, not hydrazoic acid. – aventurin Feb 27 '16 at 12:45
• Triazirine and hydrazoic acid has same molecular formula but different structure-guidechem.com/product/search_HN3_w-formula-p1.html – Nilay Ghosh Feb 27 '16 at 18:11

1 Answer

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.

Comparing Electronegativities

The following table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):

$$\begin{array}{cc} \hline \text{Element} & \chi_{\mathrm{Pauling}}\\ \hline \ce{H} & 2.20\\ \ce{N} & 3.04\\ \hline \end{array}$$

Looking at $\ce{N-H}$ Bond (heteroatomic)

The table shows, that the nitrogen atom is more electronegative than the hydrogen atom. It will therefore attract the electrons more than hydrogen does, and thus polarize the bond. Imagining the gedankenexperiment of an hypothetical heterolytical bond cleavage - presumed by applying pulling force to the atoms, until they get separated - it is assumed, that the electrons are allocated to the nitrogen atom during separation.

Looking at $\ce{N-N}$ and $\ce{N=N}$ Bonds (homoatomic)

Looking again at the previous gedankenexperiment, now with two identical Atoms bound to each other. Since they have, of course, the same electronegativity, there is no polarization of the bond. It is assumed, that separation would therefore result in an homolytical bond cleavage, allocating always one electron to each of the two nitrogen atoms for each bond respectively.

Summing up

Now, having a look on the complete structure of the molecule in question, and applying the previously states rules:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:

$$\text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-})$$

With $N_\mathrm{i}(\ce{e-})$ representating the number of electrons in a free atom, and $N_\mathrm{f}(\ce{e-})$ the one after separation (One should not forget the lone pairs).

One will end up with the following oxidation states for the different (nitrogen) atoms, with the last entry representating the mean oxidation state of all nitrogen atoms:

$$\begin{array}{cccc} \hline \text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\ \hline \ce{H} & 1 & 0 & +1\\ \hline \ce{N-1} & 5 & 5 & 0\\ \ce{N-2} & 5 & 5 & 0\\ \ce{N-3} & 5 & 6 & -1\\ \hline \ce{N} & - & - & -\frac{1}{3}\\ \hline \end{array}$$

And the structure with assigned oxidation states: