IChO 2026 · Theory problem 7

Nitrogen Fixation

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T7. Nitrogen Fixation

7%

7.1 7.2 7.3 7.4 7.5 7.6 7.7 \(\sum\)
3 8 15 3 6 6 13 54

The Haber–Bosch process is the principal industrial method for ammonia production. A simplified diagram of an ammonia synthesis plant is shown below.

Figure 1

Fig. 1. M1 - Mixture 1, M2 - Mixture 2, Z – \(\ce{CO2}\) scrubber, CLR - cooler

In the following calculations, assume that all reactions in Fig. 1, except \(\ce{NH3}\) formation, are quantitative.

7.1 Tick the gases contained in M1 and M2; and tick the correct relationship between \(x\) and \(y\). 3.0 pt

Figure 2
Figure 3

An ammonia production complex with a capacity of 660,000 tons per year was commissioned at the Navoiazot plant in Uzbekistan, which operates with a 97.0% of overall yield.

7.2 Calculate the mass of \(\ce{CH4}\) (tons) required annually to produce 660,000 tons of ammonia, if the Navoiazot operates according to the Fig. 1. 8.0 pt

Consider a model of a real reaction system, which includes recirculation of reagents. An open system operates in a cyclic mode. Each cycle consists of four steps:

  1. A stoichiometric \(\ce{N2}:\ce{H2}=1:3\) mixture with a total amount of \(n_0\) mol is fed into the reactor after each cycle.
  2. The reaction proceeds with a constant yield of \(\eta=0.150\).
  3. Ammonia is separated by liquification.
  4. Unreacted gases are returned into the reactor.

The cycle is repeated until the required overall yield is reached.

Figure 4

Fig. 2. M2. – Mixture 2, CLR - cooler.

7.3
a) Calculate the amount of nitrogen, present in the system after the 58th cycle (before the addition of 59th portion of mixture), if \(n_0=4\) mol. Provide 4 decimal places. Hint: a formula for the sum of the terms of a geometric progression is needed.
b) Calculate the number of cycles needed to increase the overall yield from 15.0% to 97.0%. 15.0 pt

An aqueous solution of 3 is being used to remove \(\ce{CO2}\) from the mixture.

Figure 5

7.4 Write the equation for this reaction. 3.0 pt

Complex 5 is obtained through the fixation of the molecular nitrogen according to the scheme below.

Figure 6

7.5 Draw the structure of 5 using simplified scheme for the ligand. Note that 1.00 g of 4 will react stoichiometrically with 94.97 \(\mathrm{cm^3}\) of \(\ce{N2}\) at 273.15 K and 1.0 bar. 6.0 pt

The effect of different reductants and additives on the yield of \(\ce{NH3}\) was examined. The different Red and Ad combinations, their respective \(E^\circ\) and \(pK_a\), together with the resulting \(\ce{NH3}\) yield are presented below.

Red

Figure 7

Ad

Figure 8
# 1 2 3 4
\(E^\circ\) (V) −0.88 −1.15 −1.15 −1.15
\(pK_a\) 14.4 14.4 13.9 14.4
\(\ce{NH3}\) yield (mole) 0 0.7 9.1 11.8

7.6 Rank the following Red/Ad combinations A-D by the decrease in ammonia yield. 6.0 pt

Red

Figure 9

Ad

Figure 10
# A B C D
\(E^\circ\) (V) −0.09 −1.10 −1.10 −1.10
\(pK_a\) 13.7 15.0 13.7 10.6

Nitrogen fixation can also be achieved via binary nitrides (compounds containing \(\ce{N^3-}\)). Such nitrides 7, 8, 9 were obtained from pure elements. When they are heated together with \(\ce{SiO2}\), red crystals of a stable compound 10 are formed as the single product. The minimal possible formula, describing the structure of compound 10 is \(\ce{Q_{\alpha}R_{\beta}S2T[Si12N24]}\). In addition:

  1. Mass fractions of nitrogen are \(w(\ce{N})_7=9.16\%\), \(w(\ce{N})_8=18.90\%\), \(w(\ce{N})_9=39.94\%\)
  2. Q and R are metal cations in their highest oxidation state.
  3. S is a monoatomic anion.
  4. T is an anion with tetrahedral configuration.
  5. For the quantitative formation of 10, the mass ratio 7 : 8 : 9 : \(\ce{SiO2}\) = 5.09 : 2.96 : 3.27 : 1.00.

7.7 Identify the empirical formulae of 7–10. Specify the composition of anions S and T. 13.0 pt

Physical Constants and Equations

Constants

Constant Value
Avogadro constant \(N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}\)
Universal gas constant \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\)
Faraday constant \(F=96\,485\ \mathrm{C\,mol^{-1}}\)
Planck constant \(h=6.626\times10^{-34}\ \mathrm{J\,s}\)
Speed of light in vacuum \(c=2.998\times10^8\ \mathrm{m\,s^{-1}}\)
Mass of electron \(m_e=9.109\times10^{-31}\ \mathrm{kg}\)
Charge of electron \(e=1.602\times10^{-19}\ \mathrm{C}\)
Standard pressure \(P^\circ=1\ \mathrm{bar}=10^5\ \mathrm{Pa}\)
Ionic product of water at \(25\,^\circ\mathrm{C}\) \(K_w=1.00\times10^{-14}\)
Zero of the Celsius scale \(0\,^\circ\mathrm{C}=273.15\ \mathrm{K}\)
Electronvolt \(1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}\)
Watt \(1\ \mathrm{W}=1\ \mathrm{J\,s^{-1}}\)
Coulomb \(1\ \mathrm{C}=1\ \mathrm{A\,s}\)
Nanometre \(1\ \mathrm{nm}=1\times10^{-9}\ \mathrm{m}\)
Ångstrom \(1\ \text{Å}=1\times10^{-10}\ \mathrm{m}\)
metric ton \(1\ \mathrm{ton}=1000\ \mathrm{kg}\)
litre \(1\ \mathrm{L}=1\ \mathrm{dm^3}\)

Equations

Equation Expression
Ideal gas law \(PV=nRT\)
Dalton law \(P_i=x_iP\)
Beer–Lambert law \(A=\log_{10}\dfrac{I_0}{I}=\varepsilon cl\)
Energy of a photon \(E=h\nu=\dfrac{hc}{\lambda}\)
Heat and temperature change \(Q=nC_m\Delta T\)
Gibbs energy definition \(G=H-TS\)
Gibbs energy and equilibrium constant \(\Delta_rG^\circ=-RT\ln K\)
Enthalpy change dependence on temperature (at \(\Delta_rC_p=\mathrm{const}\)) \(\Delta_rH_{T_2}=\Delta_rH_{T_1}+\Delta_rC_p(T_2-T_1)\)
Entropy change dependence on temperature (at \(\Delta_rC_p=\mathrm{const}\)) \(\Delta_rS_{T_2}=\Delta_rS_{T_1}+\Delta_rC_p\ln\dfrac{T_2}{T_1}\)
Henderson–Hasselbalch equation \(pH=pK_a+\log_{10}\dfrac{[\ce{A-}]}{[\ce{HA}]}\)
Average reaction rate \(r=\dfrac{\Delta c}{\Delta t}\)
The rate law \(r=k[A]^x[B]^y\)
Arrhenius law \(k=A\exp\left(-\dfrac{E_a}{RT}\right)\)
Integrated rate law for a 1st order reaction \(c=c_0e^{-kt}\); \(kt=\ln\dfrac{c_0}{c}\)
Lifetime of a reagent in a 1st order reaction \(\tau=\dfrac1k\)
Integrated rate law for n-th order reaction (\(n\ne1\)) \(kt=\dfrac1{n-1}\left(c^{1-n}-c_0^{1-n}\right)\), for \(n\ne1\)
Concentration of an intermediate in a consecutive reaction: \(\ce{A ->[$k_1$] B ->[$k_2$] D}\) \([B]=\dfrac{k_1[A]_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)\)
The ratio of products in the competitive reactions: \(\ce{A ->[$k_1$] B}\), \(\ce{A ->[$k_2$] C}\) \(\dfrac{[B]}{[C]}=\dfrac{k_1}{k_2}\)
Michaelis–Menten equation \(r_0=r_{\max}\dfrac{[S]_0}{K_M+[S]_0}\)
Specific surface area \(s=\dfrac{S}{m}\)
Progressions: arithmetic geometric \(a_n=a_1+d(n-1)\); \(a_n=a_1q^{n-1}\)
Sum of geometric progression \(\displaystyle\sum_{k=0}^{n}a_1q^k=a_1\dfrac{q^{n+1}-1}{q-1}\)
Volume of a sphere with radius \(r\) \(V=\dfrac43\pi r^3\)
Surface area of sphere with radius \(r\) \(S=4\pi r^2\)
Volume of a cone with base radius \(r\) and height \(h\) \(V=\dfrac13\pi r^2h\)
Volume of a cylinder with base radius \(r\) and height \(h\) \(V=\pi r^2h\)

Consider all gases ideal.

In equilibrium constant calculations, all concentrations are referenced to the concentration of \(1\ \mathrm{mol\,dm^{-3}}\).

Periodic Table of Elements

Figure 11

Official English text of the problem.

Answers

  • 7.1: Tick \(\ce{CO}\), \(\ce{H2}\), and \(\ce{N2}\) for M1; tick \(\ce{H2}\), \(\ce{N2}\), and \(\ce{NH3}\) for M2. Tick \(x>y\).
  • 7.2: \(\boxed{2.80\times10^5\ \text{tons of }\ce{CH4}\text{ per year}}\).
  • 7.3(a): \(\boxed{5.6662\ \mathrm{mol}\text{ of }\ce{N2}}\).
  • 7.3(b): \(\boxed{189\text{ total cycles}}\), or 188 additional cycles after the first cycle, which gives 15.0%.
  • 7.4:
    \[ \boxed{\ce{CH3N(CH2CH2OH)2 + CO2 + H2O <=> [CH3NH(CH2CH2OH)2]+ + HCO3-}} \]
  • 7.5: 5 is \(\boxed{\ce{[Mo2(L)2(N2)5]}}\): two octahedral molybdenum(0) centres, each bearing one tridentate PNP ligand \(L\) and two terminal dinitrogen ligands, joined by one end-on/end-on bridging dinitrogen ligand.
  • 7.6: \(\boxed{\mathrm{B>C>D>A}}\), in decreasing ammonia yield.
  • 7.7: 7 = lanthanum(III) nitride, \(\ce{LaN}\); 8 = calcium nitride, \(\ce{Ca3N2}\); 9 = silicon nitride, \(\ce{Si3N4}\). 10 has empirical formula \(\boxed{\ce{Ca9La5Si13N25O5}}\), written structurally as \(\boxed{\ce{Ca9La5O2[SiNO3][Si12N24]}}\). Thus S is \(\ce{O^2-}\), and T is tetrahedral \(\ce{[SiNO3]^5-}\).

Solution

7.1 — Gases in the two mixtures

The molecular species used in the process are methane C, water O, carbon monoxide [C-]#[O+], dioxygen O=O, dihydrogen [H][H], dinitrogen N#N, ammonia N, and carbon dioxide O=C=O.

The first reactor carries out steam reforming:

\[ \ce{CH4 + H2O -> CO + 3H2}. \]

Its outlet contains unreacted methane, carbon monoxide, and hydrogen. Water has been consumed, whereas methane remains; therefore,

\[ \boxed{x>y}. \]

The remaining methane then reacts quantitatively with the oxygen introduced alongside nitrogen:

\[ \ce{2CH4 + O2 -> 2CO + 4H2}. \]

Consequently, immediately after this reactor,

\[ \boxed{\mathrm{M1}:\ \ce{CO},\ \ce{H2},\ \ce{N2}}. \]

The subsequent water–gas shift reaction is

\[ \ce{CO + H2O -> CO2 + H2}. \]

The scrubber removes the resulting carbon dioxide, leaving nitrogen and hydrogen for ammonia synthesis:

\[ \ce{N2 + 3H2 <=> 2NH3}. \]

Because ammonia formation is incomplete, its reactor outlet contains product and unreacted reagents:

\[ \boxed{\mathrm{M2}:\ \ce{H2},\ \ce{N2},\ \ce{NH3}}. \]

No other boxes should be ticked.

7.2 — Annual methane requirement

Use a basis of \(1\ \mathrm{mol}\) of oxygen supplied with \(4\ \mathrm{mol}\) of nitrogen.

In the first reactor, \(y\) mol of methane react with \(y\) mol of water. The remaining methane is \(x-y\) mol. The second reactor consumes \(2\) mol of methane per mole of oxygen, so

\[ x-y=2. \]

Before the shift reaction, the amounts are

\[ n(\ce{CO})=y+2, \qquad n(\ce{H2})=3y+4. \]

Converting all carbon monoxide into carbon dioxide produces a further \(y+2\) mol of hydrogen. Hence

\[ n(\ce{H2})=4y+6. \]

The \(4\ \mathrm{mol}\) of nitrogen require \(12\ \mathrm{mol}\) of hydrogen for stoichiometric ammonia synthesis:

\[ 4y+6=12, \qquad y=1.5, \qquad x=3.5. \]

Thus \(3.5\ \mathrm{mol}\) of methane supply the reagents for \(8\ \mathrm{mol}\) of ammonia. With integer coefficients, the overall theoretical reaction is

\[ \boxed{\ce{7CH4 + 10H2O + 2O2 + 8N2 -> 7CO2 + 16NH3}}. \]

The molar masses calculated from the supplied table are

\[ M(\ce{CH4})=12.01+4(1.008)=16.042\ \mathrm{g\,mol^{-1}}, \]

\[ M(\ce{NH3})=14.01+3(1.008)=17.034\ \mathrm{g\,mol^{-1}}. \]

Accounting for the overall yield,

\[ m(\ce{CH4}) = \frac{660\,000\ \mathrm{tons}}{0.970} \frac{7}{16} \frac{16.042}{17.034} = 280\,344.6\ \mathrm{tons}. \]

Therefore,

\[ \boxed{m(\ce{CH4})=2.80\times10^5\ \mathrm{tons\,year^{-1}}}. \]

7.3(a) — Nitrogen remaining after 58 cycles

Each fresh portion contains

\[ a=\frac{n_0}{4}=1\ \mathrm{mol} \]

of nitrogen. Let \(r_k\) be the amount of unreacted nitrogen after cycle \(k\), after ammonia removal and before the next addition. The unreacted fraction per cycle is

\[ q=1-\eta=0.850. \]

The recurrence is

\[ r_k=q(r_{k-1}+a), \qquad r_0=0. \]

Therefore,

\[ r_k=a(q+q^2+\cdots+q^k) =a\frac{q(1-q^k)}{1-q}. \]

For \(k=58\),

\[ r_{58} =(1\ \mathrm{mol}) \frac{0.85(1-0.85^{58})}{0.15} =5.6662099726\ \mathrm{mol}. \]

To four decimal places,

\[ \boxed{r_{58}=5.6662\ \mathrm{mol\ of\ }\ce{N2}}. \]

7.3(b) — Cycles needed for 97.0% overall yield

After \(k\) cycles, the total amount of nitrogen introduced is \(ka\), and the amount converted is \(ka-r_k\). Consequently, the cumulative ammonia yield is

\[ Y_k = \frac{2(ka-r_k)}{2ka} = 1-\frac{q(1-q^k)}{k(1-q)}. \]

Thus

\[ Y_k = 1-\frac{0.85(1-0.85^k)}{0.15k}. \]

The required condition is \(Y_k\geq0.970\). The two consecutive values surrounding the threshold are

\[ Y_{188}=0.9698581560=96.9858\%, \]

\[ Y_{189}=0.9700176367=97.0018\%. \]

The overall yield increases with the number of cycles, so the minimum is

\[ \boxed{189\text{ total cycles}}. \]

Since the first cycle gives 15.0%, this is 188 additional cycles after the first. Fresh reagent is added every cycle; a calculation for repeatedly recycling a single initial charge would describe a different system.

7.4 — Carbon dioxide absorption

Compound 3 is N-methyldiethanolamine, SMILES CN(CCO)CCO.

It is a tertiary amine. In aqueous solution it accepts a proton as dissolved carbon dioxide forms hydrogencarbonate:

\[ \boxed{\ce{CH3N(CH2CH2OH)2 + CO2 + H2O <=> [CH3NH(CH2CH2OH)2]+ + HCO3-}}. \]

The products are N-methyldiethanolammonium, SMILES C[NH+](CCO)CCO, and hydrogencarbonate (bicarbonate), SMILES O=C([O-])O.

The amine has no nitrogen-bound hydrogen initially, so the appropriate net absorption product is hydrogencarbonate rather than an ordinary amine carbamate.

7.5 — Structure of complex 5

Denote the neutral pincer ligand by \(L\). Its name is 2,6-bis[(di-tert-butylphosphino)methyl]pyridine, and its SMILES is:

n1c(CP(C(C)(C)C)C(C)(C)C)cccc1CP(C(C)(C)C)C(C)(C)C

It coordinates through both phosphorus atoms and the pyridine nitrogen, giving a meridional \(\kappa^3\)-P,N,P arrangement.

Starting complex 4 is trichlorido(L)molybdenum(III), \(\ce{[MoCl3(L)]}\). A dative-bond SMILES representation is:

[Cl-]->[Mo+3]12(<-[Cl-])(<-[Cl-])<-n3c(CP->1(C(C)(C)C)C(C)(C)C)cccc3CP->2(C(C)(C)C)C(C)(C)C

The amount of 4 is

\[ n(4) = \frac{1.00\ \mathrm{g}}{597.824\ \mathrm{g\,mol^{-1}}} = 1.672733\times10^{-3}\ \mathrm{mol}. \]

For nitrogen,

\[ n(\ce{N2}) = \frac{PV}{RT} = \frac{(1.0\times10^5\ \mathrm{Pa})(94.97\times10^{-6}\ \mathrm{m^3})} {(8.314\ \mathrm{J\,K^{-1}\,mol^{-1}})(273.15\ \mathrm{K})} = 4.181915\times10^{-3}\ \mathrm{mol}. \]

Therefore,

\[ \frac{n(\ce{N2})}{n(4)} = 2.50005 \approx\frac52. \]

The simplest product unit thus contains two molybdenum atoms and five dinitrogen molecules.

Three equivalents of sodium reduce each molybdenum(III) centre to molybdenum(0), removing the three chlorides as sodium chloride. The balanced preparation is

\[ \ce{2[MoCl3(L)] + 6Na + 5N2 -> [Mo2(L)2(N2)5] + 6NaCl}. \]

Sodium chloride has formula-unit SMILES [Na+].[Cl-]; mercury serves as the amalgam medium.

The expected structure is μ-dinitrogen-bis[bis(dinitrogen)(L)molybdenum(0)]:

  • Draw two molybdenum centres joined by a linear, end-on/end-on \(\ce{Mo-N#N-Mo}\) bridge.
  • Attach one tridentate PNP ligand to each molybdenum.
  • Attach two additional, terminal, end-on dinitrogen ligands to each molybdenum.
  • Each molybdenum is approximately octahedral. The two terminal dinitrogen ligands are mutually trans; the bridging dinitrogen is trans to the pyridine nitrogen.

Each metal is then an 18-electron centre:

\[ 6\quad(d^6\text{ Mo(0)}) + 6\quad(L) + 4\quad(\text{two terminal dinitrogens}) + 2\quad(\text{bridging dinitrogen}) =18. \]

A dative-bond SMILES specifying this connectivity is:

[Mo]12(<-N#N)(<-N#N)(<-N#N->[Mo]45(<-N#N)(<-N#N)<-n6c(CP->4(C(C)(C)C)C(C)(C)C)cccc6CP->5(C(C)(C)C)C(C)(C)C)<-n3c(CP->1(C(C)(C)C)C(C)(C)C)cccc3CP->2(C(C)(C)C)C(C)(C)C

This SMILES specifies donor–metal connectivity; the octahedral and trans relationships are supplied by the drawing instructions above.

7.6 — Ranking the reductant/additive combinations

Comparing combinations 1 and 4 in the supplied experimental table shows that the stronger reductant, with the more negative potential, supports ammonia formation. Combination A has a much less negative potential, \(-0.09\ \mathrm{V}\), so it is predicted to give the lowest yield.

B, C, and D have the same reductant potential and the same counterion. The comparison of experimental combinations 3 and 4 associates the higher additive \(pK_a\), corresponding to the weaker acid, with the higher ammonia yield. Applying that trend gives

\[ pK_a(\mathrm{B})=15.0 > pK_a(\mathrm{C})=13.7 > pK_a(\mathrm{D})=10.6. \]

The intended decreasing-yield ranking is therefore

\[ \boxed{\mathrm{B>C>D>A}}. \]

7.7 — Identification of the nitrides and compound 10

For a binary nitride \(\ce{E_aN_b}\), let the oxidation state of element \(E\) be \(+z\). Since nitrogen is present as nitride,

\[ az=3b. \]

Its nitrogen mass fraction is

\[ w(\ce{N}) = \frac{14.01b}{aM(E)+14.01b}. \]

Using \(a/b=3/z\),

\[ \frac{M(E)}{z} = \frac{14.01}{3} \left(\frac1{w(\ce{N})}-1\right). \]

The three supplied percentages give the following identifications:

Compound \(M(E)/z\), \(\mathrm{g\,mol^{-1}}\) Identification Formula Calculated nitrogen mass fraction
7 46.3125 \(z=3\), \(M(E)\approx138.94\): lanthanum \(\ce{LaN}\) 9.1623%
8 20.0390 \(z=2\), \(M(E)\approx40.078\): calcium \(\ce{Ca3N2}\) 18.8992%
9 7.02254 \(z=4\), \(M(E)\approx28.090\): silicon \(\ce{Si3N4}\) 39.9401%

Thus 7 is lanthanum(III) nitride, formula-unit SMILES [La+3].[N-3]; 8 is calcium nitride, formula-unit SMILES [Ca+2].[Ca+2].[Ca+2].[N-3].[N-3]; and 9 is silicon nitride. A formal charge-separated composition for the latter is [Si+4].[Si+4].[Si+4].[N-3].[N-3].[N-3].[N-3]. These disconnected strings express compositions, not crystal connectivity; silicon nitride is an extended covalent solid.

Their formation from the elements is represented by

\[ \ce{2La + N2 -> 2LaN}, \]

\[ \ce{3Ca + N2 -> Ca3N2}, \]

\[ \ce{3Si + 2N2 -> Si3N4}. \]

To find the composition of 10, convert the given mass ratio to a molar ratio:

\[ M(\ce{LaN})=152.91\ \mathrm{g\,mol^{-1}}, \]

\[ M(\ce{Ca3N2})=148.26\ \mathrm{g\,mol^{-1}}, \]

\[ M(\ce{Si3N4})=140.31\ \mathrm{g\,mol^{-1}}, \qquad M(\ce{SiO2})=60.09\ \mathrm{g\,mol^{-1}}. \]

Hence

\[ n(\ce{LaN}):n(\ce{Ca3N2}):n(\ce{Si3N4}):n(\ce{SiO2}) = \frac{5.09}{152.91}: \frac{2.96}{148.26}: \frac{3.27}{140.31}: \frac{1.00}{60.09}. \]

Normalizing gives

\[ 2.00025:1.19969:1.40043:1 \approx10:6:7:5. \]

These quantities contain

\[ \ce{La10Ca18Si26N50O10}, \]

which reduces to the empirical formula

\[ \boxed{\ce{Ca9La5Si13N25O5}}. \]

The prescribed structural formula already contains \(\ce{Si12N24}\). Subtracting this unit leaves one silicon, one nitrogen, and five oxygen atoms in addition to the metal cations.

The two identical monoatomic anions must therefore be oxide ions:

\[ \boxed{\mathrm{S}=\ce{O^2-}}, \]

named oxide, SMILES [O-2].

After assigning two oxygen atoms to these oxide ions, the tetrahedral anion has composition \(\ce{SiNO3}\). Its charge is

\[ (+4)+(-3)+3(-2)=-5, \]

so

\[ \boxed{\mathrm{T}=\ce{[SiNO3]^5-}}. \]

This is nitridotrioxidosilicate(5−), SMILES [N-2][Si]([O-])([O-])[O-]. Draw silicon at the centre of a tetrahedron, bonded to one nitrogen and three oxygen atoms. The formal charges in this bonded SMILES sum to \(-5\); nitrogen’s oxidation state remains \(-3\).

Thus 10, a calcium lanthanum oxonitridosilicate, is

\[ \boxed{\ce{Ca9La5O2[SiNO3][Si12N24]}}. \]

One assignment is \(Q=\ce{Ca^2+}\), \(\alpha=9\), \(R=\ce{La^3+}\), and \(\beta=5\). The cation SMILES are [Ca+2] and [La+3], respectively.

The charge balance confirms the result:

\[ 9(2)+5(3)=33, \]

\[ 2(-2)+(-5)+\bigl[12(4)+24(-3)\bigr] =-4-5-24=-33. \]

The balanced synthesis is

\[ \boxed{\ce{10LaN + 6Ca3N2 + 7Si3N4 + 5SiO2 -> 2Ca9La5O2[SiNO3][Si12N24]}}. \]

The extended \(\ce{[Si12N24]^24-}\) framework’s connectivity is not supplied, so a unique structural SMILES for the complete crystalline compound 10 cannot be inferred from its formula.

The solution was checked against the official solution.