Until the end of the 20th century, only two species (one molecule and one anion) were known that are — Physical Chemistry Chemistry Question
Nitrogen-Only Species and High Energy Density Materials
Until the end of the 20th century, only two species (one molecule and one anion) were known that are composed only of nitrogen atoms.
What are the empirical formulae of these two species?
Model Answer
N3 –, N2
The first inorganic compound containing a nitrogen-only species different from the above was synthesized by Christe and co-workers in 1999.
The starting material of the synthesis is an unstable liquid A that is a weak, monoprotic acid. It was liberated from its sodium salt (that contains 35.36 % sodium by mass) with a large excess of stearic acid.
Determine the molecular formula of A and draw two resonance structures of the molecule. (Show all bonding and non-bonding valence electron pairs.)
Model Answer
The sodium salt of A contains one mol sodium per mole, hence its molar mass is M(Na) / 0.3536 = 65.02 g mol–1 . The molar mass of the anion is 42.03 g mol–1, this implies N3 –. A is hydrogen azide. The Lewis structures are:
The other starting material (B) was prepared from the cis-isomer of a nitrogen halogenide that contains 42.44 % nitrogen by mass.
Determine the empirical formula of this halogenide. Draw the Lewis structure of the cis-isomer. Show all bonding and non-bonding valence electron pairs.
Model Answer
If the formula of the halogenide is NaXb ( X denotes the unknown halogen), the following equation can be written:
a M(N) = 0.4244 (a M(N) + b M(X))
From this, b M(X) = 19 a. Therefore X is fluorine, and b = a. Since NF does not exist, and the maximum number of covalent bonds formed by nitrogen is 4, the halogenide is N2F2. The molecular shape is:
This nitrogen halogenide was reacted with SbF5 (a strong Lewis acid) in a 1 : 1 ratio at –196 °C. The resulting ionic substance ( B) was found to comprise three types of atoms. Elemental analysis shows that it contains 9.91 % N and 43.06 % Sb by mass; further, it contains one cation and one anion. The shape of the latter was found to be octahedral.
Determine the empirical formula of the ionic substance B.
Model Answer
Since SbF5 is a strong Lewis-acid, the formula of the anion of B is SbF6 –. Since it contains one anion, B contains only one Sb atom. Hence the molar mass of the compound is M(Sb) / 0.4306 = 282.75 g mol–1. The nitrogen content is 282.75 g × 0.0991 = 28 g, the rest (133 g) is fluorine. From these data, the empirical formula is SbN2F7.
Determine the empirical formula of the cation found in B and draw its Lewis structure. Show resonance structures, if there are any. Show all bonding and non-bonding electron pairs. Predict the bond angles expected in the contributing structures (approximately).
Model Answer
The anion of B is SbF6 so the molecular formula of B is [N2F +][SbF6 –]. The resonance structures of the cation are:
The bond angle would be 180° in the first and less than 120° in the second individual contributing structure.
B reacts violently with water: 0.3223 g of the compound gave 25.54 cm3 (at 101325 Pa and 0 °C) of a color- and odorless nitrogen oxide that contains 63.65 % nitrogen by mass.
Identify the nitrogen oxide formed in the hydrolysis reaction and draw its Lewis structure. Show resonance structures, if there are any. Show all bonding and non-bonding electron pairs.
Model Answer
If the formula of the nitrogenous oxide is NaOb, the following equation can be written:
a M(N) = 0.6365 (a M(N) + b M(O))
From here a = 2 b, so the molecular formula of the nitrogenous oxide is N2O.
Give the chemical equation for the reaction of B with water.
Model Answer
The amount of N2O formed is 1.14 mmol. The amount of B is 0.3223 g / 282.75 g mol–1 = 1.14 mmol. The oxygen atom comes from the water molecule; this leaves two hydrogen atoms which can form HF with fluoride ions:
[N2F +][SbF6 –] + H2O → N2O + 2 HF + SbF5
SbF5 undergoes hydrolysis in dilute aqueous solution, but the reaction can lead to various products: e.g., SbF5 + H2O → SbOF3 + 2 HF
In the experiment described by Christe and co-workers, A was mixed with B in liquid hydrogen fluoride at –196 °C. The mixture was shaken for three days in a closed ampoule at –78 °C, finally it was cooled down again to –196 °C. A compound C was obtained, that contained the same octahedral anion as B and the expected, V-shaped cation composed only of N-atoms. C contained 22.90 % N by mass.
Determine the empirical formula of C.
Model Answer
The anion of C is SbF6 –. If C contains n anions per cation and the cation contains x N atoms, then the nitrogen and antimony content is:
x M(N) = 0.2290 (x M(N) + n M(SbF6))
n M(Sb) = 0.3982 (x M(N) + n M(SbF6))
Dividing the first equation by the second one, we obtain n = 5x. From this the formula of C is [N5 +][SbF6 –].
The cation of C has many resonance structures. Show these structures, indicating all bonding and non-bonding electron pairs. Predict the bond angles expected in the contributing structures (approximately).
Model Answer
The central bond angle is smaller than 120° in all structures due to the presence of one or two non-bonding electron pairs on the central nitrogen. The other two bond angles would be equal to 180° in the resonance structures of the first row but less than 180° in the resonance structures of the second row due to the presence of non-bonding electron pairs on the relevant nitrogen atoms(s).
Give the chemical equation for the formation of C. The formation of which compound makes the process thermodynamically favorable?
Model Answer
[N2F +][SbF6 –] + HN3 → [N5 +][SbF6 –] + HF
The reaction is thermodynamically favourable because of the high stability of HF.
The cation of C is a very strong oxidizing agent. It oxidizes water; the reaction gives rise to the formation of two elemental gases. The resulting aqueous solution contains the same compounds as in the case of hydrolysis of B.
Give the chemical equation for the hydrolysis of C.
Model Answer
If C oxidizes water, and two elemental gases are formed, one of them must be oxygen. The other is nitrogen, due to the instability of the cation of C. Therefore, the reaction equation for the hydrolysis is:
4 [N5 +][SbF6 –]+ 2 H2O → 10 N2 + O2 + 4 HF + 4 SbF5.
In 2004, a further step was made. The ionic compound E was synthesized, whose nitrogen content was 91.24 % by mass! The first step in the synthesis of E was the reaction of the chloride of a main group element with an excess of the sodium salt of A (in acetonitrile, at –20 °C), giving rise to the formation of the compound D and NaCl. Gas evolution was not observed. In the second step, D was reacted with C in liquid SO2 at –64 °C, giving E as the product. The cation : anion ratio in E is also 1:1 and it contains the same cation as C. D and E contain the same complex anion, whose central atom is octahedrally coordinated.
Determine the empirical formula of E, given that it contains two types of atoms.
Model Answer
As the anion of D is coordinated octahedrally and it contains N3 – ions (since it is formed from A), the only possibility is that the central atom of D (and therefore, that of E) is surrounded by six N3 – ions. As the cation of E is N5 + and the cation : anion ratio is 1:1, the formula of E is [N5 +][X(N3)6 –], thus XN23 (where X is the unidentified main group element). The nitrogen content, expressed in terms of the relative atomic masses, is:
23 M(N) = 0.9124 (23 M(N) + M(X))
From here, M(X) = 30.9 g mol–1. This is phosphorus. The formula of E is therefore:
[N5 +][P(N3)6 –]
Determine the empirical formula of D and identify the main group element used.
Model Answer
The oxidation number of the atoms in Na+ and Cl– ions does not change, and gas evolution was not observed, indicating that the N3 – ions did not decompose during the synthesis. Therefore, the formation of E is not a redox reaction; phosphorus has the same oxidation number in the chloride as in E. The chloride is PCl5.
D contains [P(N3)6] – as anion. As cation it can contain only Na+. Therefore the formula of D is Na[P(N3)6]
E is supposed to be a potential fuel for future space travel because of its extremely high endothermic character. (It is a so-called “high energy density material”). A further advance is that the products of the decomposition of E are not toxic, so they do not pollute the atmosphere.
What are the reaction products of the decomposition of E in air?
Model Answer
Nitrogen is the gaseous product. As oxygen is present in the atmosphere, the phosphorus content is oxidized to P2O5. The chemical equation of the decomposition is:
4 [N5 +][P(N3)6 –] + 5 O2 → 46 N2 + 2 P2O5.