Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr on
February 22, 2008. Copyright Margret J. Geselbracht 2008. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
NOVEL CHEMISTRY OF NITROGEN: AZIDES AND THE N5+ CATION
SOLUTIONS
The following questions come from an interesting recent article on the chemistry of azides: C. Knapp, and
J. Passmore, “On the Way to “Solid Nitrogen” at Normal Temperature and Pressure? Binary Azides of
Heavier Group 15 and16 Elements,” Angew. Chem. Int. Ed. 43, 4834-4836 (2004).
1. The N5+ cation can be made from the reaction of N2F+ and HN3. Draw Lewis structures for each of
these three species including all important resonance structures and any nonzero formal charges.
N
N
F
N
N
N
H
N
N
N
N
H
N
N
N
N
N
N
N
N
N
N
N
N
N
N
2. For the N5+ cation, what structural parameters do your resonance structures predict? For example,
what can you say about relative N–N bond lengths and approximate N–N–N bond angles?
This cation has an overall bent structure. Based on the two resonance structures
shown above with one lone pair on the central nitrogen, the angle around the
central nitrogen will be less than 120˚. The third resonance structure with
N
two lone pairs on the central nitrogen would predict a central angle less
than 109˚. The two arms of the bent structure should each be linear
N
with 180˚ angles at the nitrogen atoms carrying positive formal charges.
N
N
The outer N–N bond lengths should be equal and intermediate between a double and triple
bond length. More importantly, these outer bonds should be shorter than the inner bonds,
which are intermediate between a single and a double bond.
One crystal structure of a salt containing the N5+ cation has been carried out, that of
[N5]+[Sb2F11]–. This structure reports an inner N–N bond distance of 1.30 Å and an outer N–
N bond distance of 1.10 Å. The central N–N–N bond angle reported from this structure was
111˚. Interestingly, the outer N–N–N bond angles are not linear but slightly bent upwards
with an angle of 168˚.
3. In Figures 1 and 2 of the article, the coordinated azide ions (N 3–) in species such as [Te(N3)5]– (shown
below) and [Te(N3)6]2– look notably bent and twisted. Why is that? What approximate Te–N–N bond
angles would you expect? What other structural parameters would you predict for the coordinated
azide ion (relative N–N bond lengths and N–N–N bond angles)?
N
Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr on
February 22, 2008. Copyright Margret J. Geselbracht 2008. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
In the structure of the [Te(N3)5]– anion, note that each azide ion is linear, but the Te–N–N
bond angles are bent.
Te
Te
N
N
N
N
N
N
The two resonance structures above contribute to the bonding of the each azide to the Te
atom (the azide donates the lone pair of electrons, which becomes the covalently shared epair in the Te–N bond). Both of these resonance contributors predict a bent angle at the
nitrogen atom bonded to Te. We would expect this angle to be less than 120˚ and perhaps as
small as 109˚. These resonance structures also predict the coordinated azide ion is linear.
This article mentioned at the beginning that covalently bonded azides are polarized to favor
the resonance structure at the right. If this is true, we would predict a shorter outer N–N
bond length and a longer inner N–N bond length. If the resonance structure on the left is
fairly important, then we would expect more equal N–N bond lengths in the coordinated
azide ion.
4. The preparation of the [N5]+[SbF6]– salt, a colorless hygroscopic solid stable at ambient temperature,
was reported in 2001. The oxidizing properties of this salt were explored in the solid state. No
reaction was observed with O2, Cl2, or Xe, but oxidation of NO, NO2, and Br2 was observed. For
example, the reaction with NO2 can be written as follows:
NO2(g) + [N5]+[SbF6]–(s) –––––> [NO2]+[SbF6]–(s) + “N5(g)”
“N5(g)” –––––> 2.5 N2(g)
Created by Margret J. Geselbracht, Reed College (mgeselbr@reed.edu) and posted on VIPEr on
February 22, 2008. Copyright Margret J. Geselbracht 2008. This work is licensed under the Creative
Commons Attribution Non-commercial Share Alike License. To view a copy of this license visit
http://creativecommons.org/about/license/.
(a) The first ionization energy for NO2 is 9.75 eV. The lattice energies of [N5]+[SbF6]–(s) and
[NO2]+[SbF6]–(s) were estimated to be –119 kcal/mol and –133 kcal/mol, respectively. Use these
values and a Born-Haber cycle to estimate the minimum value of the electron affinity of the N 5+
cation. To do this calculation, assume the Hrxn for the first reaction is zero.
H
~0
rxn
NO2(g) + [N5]+[SbF6]–(s) –––––>
[NO2]+[SbF6]–(s) + “N5(g)”
NO2(g) + [N5]+[SbF6]–(s) –––––> [NO2]+[SbF6]–(s) + “N5(g)”
225 kcal/mol
119 kcal/mol
–133 kcal/mol
– EA N5+
[NO2(g)]+ + [N5]+(g) + [SbF6]–(g)
Hrxn = IE NO2 – Uo [N5]+[SbF6]– + Uo [NO2]+[SbF6]– – EA N5+
96.4853 kJ/mol  1 kcal/mol 
Ionization energy of NO2 = 9.75 eV

 = 225 kcal/mol

4.184 kJ/mol 
1 eV
 = 225 kcal/mol + 119 kcal/mol –133 kcal/mol – EA N5+
EA N5+ = 225 kcal/mol
+ 119 kcal/mol –133 kcal/mol = +211 kcal/mol

This value makes some sense. Remember that a positive electron affinity means a high
affinity for an extra electron, or in other words, addition of another electron is an
exothermic process. Since the N5+ cation is positively charged, it should have a
reasonably high electron affinity.
(b) Is the second reaction endothermic or exothermic? Use the calculated heat of formation of the
N5+ cation (351.1 kcal/mol) and the result from part a to calculate Hrxn for the second reaction.
Hrxn
“N5(g)” –––––> 2.5 N2(g)
–Hf = –351.1 kcal/mol
N5+(g)
Hrxn = EA N5+ – Hf N5+ = 211 kcal/mol – 351.1 kcal/mol = – 140 kcal/mol