IChO 2020 · Theory problem 8

Boron Compounds and Hydrogen Storage

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Question 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 Total
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Question 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16
Points 2 3 2 4 4 4 2 2
Question 8.17
Points 4

Problem 8. Boron Compounds and Hydrogen Storage

Figure 1

Sodium borohydride (\(\ce{NaBH4}\)) and ammonia borane (\(\ce{BNH6}\)) are the most studied chemical hydrogen storage materials. In this question, you will explore the chemistry of boron and the use of boron compounds as hydrogen storage materials.

Borax (\(\ce{Na2B4O7 * nH2O}\)) is a boron mineral that is produced by ETI Mining Company in Turkey. \(\ce{NaBH4}\) can be synthesized by the reduction of anhydrous borax with metallic sodium under high-pressure hydrogen gas in the presence of silicon dioxide (silica) at 700 °C, namely the Bayer process. In this process, all hydrogen is stored in \(\ce{NaBH4}\). On the other hand, it has been shown that ammonia borane (\(\ce{BNH6}\)) can be synthesized by the reaction of \(\ce{NaBH4}\) and ammonium sulfate in dry tetrahydrofuran (THF) at 40 °C (Hint: \(\ce{BNH6}\) synthesis must be conducted in a well-ventilated fume-hood because flammable gas is generated as one of the by-products). While \(\ce{NaBH4}\) is an ionic compound, ammonia borane is a Lewis acid–base adduct.

8.1. Write a balanced chemical equation for the synthesis of \(\ce{NaBH4}\) from anhydrous borax.

8.2. Write a balanced chemical equation for the synthesis of ammonia borane from \(\ce{NaBH4}\).

8.3. Draw the molecular geometry of the \(\ce{BH4^-}\) ion and \(\ce{BNH6}\) molecule.

8.4. Calculate the hydrogen content of \(\ce{NaBH4}\) and \(\ce{BNH6}\) as a percentage by mass (wt%).

The hydrogen stored in both compounds can be released via hydrolysis reactions in the presence of a suitable catalyst at room temperature. Upon the hydrolysis reactions, 4 and 3 moles of \(\ce{H2}\) gas are released from the hydrolysis of 1 mole of \(\ce{NaBH4}\) and \(\ce{BNH6}\), respectively, along with metaborate anion including B-O bonds.

8.5. Write the balanced chemical equations for the hydrolysis of \(\ce{NaBH4}\) and \(\ce{BNH6}\).

One of the simplest stable borates is diboron trioxide (\(\ce{B2O3}\)). Higher borates such as \(\ce{B3O6^{3-}}\) having cyclic structures containing B-O-bonds can be formed. Since \(\ce{B2O3}\) is an acidic compound, it is easily reacted with water to produce boric acid (\(\ce{H3BO3}\)). On the other hand, the high temperature and high-pressure reaction of \(\ce{B2O3}\) with ammonia yields two-dimensional boron nitride, which consists of planar graphite-like sheets of alternating B and N atoms.

8.6. Write the balanced chemical equations for the synthesis of boric acid and boron nitride.

8.7. Draw the molecular structures of the \(\ce{B3O6^{3-}}\) ion, boric acid, and a single two-dimensional boron nitride sheet. Hint: show at least 10 B atoms in the boron nitride structure.

Furthermore, B-H compounds, called boranes, are an important class of boron compounds. The simplest stable borane is diborane (\(\ce{B2H6}\)) and many of the higher boranes can be prepared by the pyrolysis of diborane. Diborane can be synthesized via metathesis of a boron halide and a hydride source.

8.8. Write a balanced chemical equation for the synthesis of diborane from the reaction of \(\ce{BF3}\) and \(\ce{LiBH4}\). Hint: both products are boron compounds.

8.9. Draw the molecular geometry of the diborane molecule. Hint: there is no B-B bond in the structure.

\(\ce{BH3}\) (borane) is an unstable and highly reactive molecule. Therefore, it is not possible to isolate it as \(\ce{BH3}\) under ordinary conditions. However, it can be stabilized via its reaction with carbon monoxide to yield borane carbonyl (\(\ce{BH3CO}\)) compound, which is an adduct of borane. The preparation of \(\ce{BH3CO}\) plays an important role in exploring the chemistry of boranes as it indicates the likely existence of the borane molecule.

8.10. Sketch the Lewis dot structure of \(\ce{BH3-CO}\) molecule by showing the formal charges.

8.11. Which of the statements given in your answer sheet is observed in the \(\ce{C-O}\) bond of a \(\ce{CO}\) molecule upon the bond formation between \(\ce{BH3}\) and \(\ce{CO}\)? Tick the correct box.

  • ☐ It gets longer because there will be \(\pi\)-back donation from \(\ce{BH3}\) to \(\ce{CO}\).
  • ☐ It gets longer because \(\ce{CO}\) donates \(\pi\)-bonding electrons to \(\ce{BH3}\).
  • ☐ No or slight change on it because \(\ce{CO}\) donates its mainly non-bonding electrons to \(\ce{BH3}\).
  • ☐ It gets shorter because \(\ce{CO}\) donates \(\pi^*\) anti-bonding electrons to \(\ce{BH3}\).

Borazine consists of single and double bonded cyclic \(\ce{B-N}\) units and hydrogen atoms attached to these atoms with the molecular formula \(\ce{B3N3H6}\) and is isostructural to benzene. Borazine can be synthesized by using a two-step procedure including the synthesis of symmetrically trisubstituted chlorine derivatives of borazine (\(\ce{B3N3H3Cl3}\)) from the reaction of ammonium chloride and boron trichloride, and then reduction of \(\ce{B3N3H3Cl3}\) with \(\ce{LiBH4}\) in THF.

8.12. Write the balanced chemical equations for the two-step synthesis of borazine starting from ammonium chloride in THF (tetrahydrofuran). Hint: THF stabilizes one of the products by forming a Lewis acid-base adduct.

8.13. Draw the molecular structures of borazine and its symmetrically trisubstituted chlorine derivative.

Catalysts are substances that accelerate the rate of reactions by allowing them to proceed in a lower energy pathway. The catalytic activity of the catalysts is generally determined by the turnover frequency (TOF), which is calculated by dividing the molar amount of the product to the mole of active catalyst and the time (TOF = mole product/(mole catalyst x time)). A typical hydrolysis of \(\ce{BNH6}\) was carried out in 10.0 mL of water by using 100.0 mM \(\ce{BNH6}\) and 5.0 mg of CuPt/C catalyst (CuPt alloy nanoparticles supported on carbon black containing 8.2 wt% Pt atom). 67.25 mL of hydrogen gas was generated in 5 minutes.

8.14. Assuming the catalytic reaction is performed under standard conditions (1 atm and 273.15 K), calculate the TOF (\(\mathrm{min^{-1}}\)) of the CuPt/C catalyst in terms of only Pt atoms in the hydrolysis of \(\ce{BNH6}\) by considering the volume of the hydrogen gas generated.

As a result of detailed crystal analysis of a synthesized \(\ce{Cu_{x}Pt_{y}}\) alloy nanoparticles (the subscripts indicate molar percentages of the atoms in the alloy structure), it was determined that the face centered cubic (fcc) unit cell was formed by Pt atoms and the Pt atoms on the face of the fcc unit cell are supposed to be replaced with Cu atoms to form \(\ce{Cu_{x}Pt_{y}}\) displacement alloy nanoparticles. According to this information, answer the following questions.

8.15. Determine the composition of the alloy nanoparticles by finding \(x\) and \(y\) in the \(\ce{Cu_{x}Pt_{y}}\).

8.16. Sketch the shape of the described crystal unit cell of \(\ce{Cu_{x}Pt_{y}}\) alloy nanoparticles by showing the position of all atoms on the unit cell.

8.17. Another alloy has a \(\ce{Cu2Pt1}\) composition. Assume that this alloy also has a fcc unit cell with an edge length of 380 pm, but the Cu and Pt atoms are randomly distributed in the atomic positions. Calculate the density of this alloy in \(\mathrm{g/cm^3}\).

Official English text of the problem.

Answers

  • 8.1. \[ \boxed{\ce{Na2B4O7 + 16Na + 8H2 + 7SiO2 -> 4NaBH4 + 7Na2SiO3}} \]

  • 8.2. \[ \boxed{\ce{2NaBH4 + (NH4)2SO4 -> 2NH3BH3 + Na2SO4 + 2H2}} \]

  • 8.3. Borohydride ion, [BH4-]: tetrahedral. Ammonia borane, [NH3+][BH3-]: approximately tetrahedral coordination around both B and N, with three H atoms on each centre and a single \(\ce{B-N}\) bond.

  • 8.4. \[ \boxed{w_{\mathrm H}(\ce{NaBH4})=10.66\%,\qquad w_{\mathrm H}(\ce{BNH6})=19.59\%} \]

  • 8.5. \[ \boxed{\ce{NaBH4 + 2H2O -> NaBO2 + 4H2}} \] \[ \boxed{\ce{NH3BH3 + 2H2O -> NH4BO2 + 3H2}} \]

  • 8.6. \[ \boxed{\ce{B2O3 + 3H2O -> 2H3BO3}} \] \[ \boxed{\ce{B2O3 + 2NH3 -> 2BN + 3H2O}} \]

  • 8.7. Cyclic trimetaborate ion, [O-]B1OB([O-])OB([O-])O1: a six-membered alternating \(\ce{B-O}\) ring, with one terminal \(\ce{O^-}\) on each B. Boric acid, OB(O)O: trigonal-planar \(\ce{B(OH)3}\). Hexagonal boron nitride: a planar honeycomb sheet with alternating B and N vertices; a patch of seven fused hexagons contains 12 B and 12 N atoms. The extended sheet has no finite molecular SMILES.

  • 8.8. \[ \boxed{\ce{4BF3 + 3LiBH4 -> 2B2H6 + 3LiBF4}} \]

  • 8.9. Diborane has four terminal H atoms and two bridging H atoms, with two three-centre, two-electron \(\ce{B-H-B}\) bonds and no direct \(\ce{B-B}\) bond. Its schematic connectivity string is [BH2]1[H][BH2][H]1; the multicentre-bond limitation of this notation is explained below.

  • 8.10. Borane carbonyl, [BH3-]C#[O+]: three \(\ce{B-H}\) single bonds, one \(\ce{B-C}\) single bond and a \(\ce{C#O}\) triple bond. O has one lone pair. Formal charges: B = −1, C = 0, O = +1; all H = 0.

  • 8.11. Third statement: “No or slight change on it because \(\ce{CO}\) donates its mainly non-bonding electrons to \(\ce{BH3}\).”

  • 8.12. \[ \boxed{\ce{3NH4Cl + 3BCl3 -> B3N3H3Cl3 + 9HCl}} \] \[ \boxed{\ce{B3N3H3Cl3 + 3LiBH4 + 3THF -> B3N3H6 + 3LiCl + 3BH3*THF}} \]

  • 8.13. Borazine: a planar alternating B/N six-membered ring, with one H on every atom. A Kekulé contributor has SMILES [BH-]1=[NH+][BH-]=[NH+][BH-]=[NH+]1. Its B,B′,B″-trichloro derivative has Cl on all three B atoms and H on all three N atoms: [B-]1(Cl)=[NH+][B-](Cl)=[NH+][B-](Cl)=[NH+]1.

  • 8.14. \[ \boxed{\mathrm{TOF}=285.5\ \mathrm{min^{-1}} \approx 2.9\times10^2\ \mathrm{min^{-1}}} \]

  • 8.15. \[ \boxed{x=75,\qquad y=25} \] Thus the atomic composition is \(\ce{Cu_{75}Pt_{25}}\), with simplest atomic ratio \(\ce{Cu3Pt}\).

  • 8.16. Draw a cube with Pt at all eight corners and Cu at the centres of all six faces.

  • 8.17. \[ \boxed{\rho=13.0\ \mathrm{g\,cm^{-3}}} \]

Solution

The problem does not include an atomic-mass or constants table. The numerical calculations use the following standard values:

Element H B N Na Cu Pt
Molar atomic mass / \(\mathrm{g\,mol^{-1}}\) 1.008 10.81 14.01 22.99 63.55 195.08

\[ R=0.0820574\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}, \qquad N_{\mathrm A}=6.02214076\times10^{23}\ \mathrm{mol^{-1}}. \]

Small molecules appearing below are dihydrogen [H][H], water O, ammonia N, hydrogen chloride Cl, and tetrahydrofuran (THF) C1CCOC1. In salt SMILES, a dot separates ionic components. Extended network solids have no unique finite molecular SMILES; their formulas and structures are specified directly.

8.1. Synthesis of sodium borohydride

The oxygen-containing by-product is sodium metasilicate, \(\ce{Na2SiO3}\). All four B atoms in one formula unit of anhydrous borax become sodium borohydride, [Na+].[BH4-].

Let the coefficient of silica, and hence of sodium metasilicate, be \(s\). Oxygen balance gives

\[ 7+2s=3s \quad\Rightarrow\quad s=7. \]

Four formula units of \(\ce{NaBH4}\) require 16 H atoms, supplied by eight molecules of hydrogen. Sodium balance then requires

\[ 2+n_{\ce{Na}}=4+2(7) \quad\Rightarrow\quad n_{\ce{Na}}=16. \]

Therefore, at 700 °C under high-pressure hydrogen,

\[ \boxed{\ce{Na2B4O7 + 16Na + 8H2 + 7SiO2 -> 4NaBH4 + 7Na2SiO3}}. \]

Anhydrous sodium tetraborate, silica and sodium metasilicate are network solids; a finite molecular SMILES would not describe their solid structures.

8.2. Synthesis of ammonia borane

One B-bound hydrogen and one hydrogen from an ammonium ion combine to form hydrogen gas. The remaining ammonia and borane units form the Lewis acid–base adduct ammonia borane, [NH3+][BH3-].

The net ionic equation is

\[ \ce{BH4^- + NH4+ -> NH3BH3 + H2}. \]

The ammonium ion is [NH4+]. Since ammonium sulfate contains two ammonium ions,

\[ \boxed{\ce{2NaBH4 + (NH4)2SO4 -> 2NH3BH3 + Na2SO4 + 2H2}}. \]

The reaction is performed in dry THF at 40 °C. The flammable gaseous by-product is dihydrogen.

The salts are ammonium sulfate, [NH4+].[NH4+].[O-]S(=O)(=O)[O-], and sodium sulfate, [Na+].[Na+].[O-]S(=O)(=O)[O-].

8.3. Geometries of borohydride and ammonia borane

For the borohydride ion, [BH4-], B has four bonding electron pairs and no lone pair. VSEPR therefore gives a tetrahedral arrangement:

\[ \angle\ce{H-B-H}\approx109.5^\circ. \]

To draw it, place B at the centre, with two \(\ce{B-H}\) bonds in the plane of the page, one solid wedge toward the viewer and one dashed wedge away. Enclose the structure in brackets with an overall negative charge.

For ammonia borane, [NH3+][BH3-], nitrogen donates its lone pair to electron-deficient boron. Each centre then has four bonds: three to H and one to the other central atom. Both centres have approximately tetrahedral coordination.

Draw a \(\ce{B-N}\) single bond with three H atoms arranged tetrahedrally around each end. A staggered conformation about this bond is an appropriate drawing. SMILES specifies connectivity and formal charges but does not fix that conformation.

8.4. Hydrogen mass percentages

The hydrogen mass percentage is

\[ w_{\mathrm H} =\frac{\text{mass of H in one mole of compound}} {\text{molar mass of compound}}\times100\%. \]

For sodium borohydride,

\[ M(\ce{NaBH4}) =22.99+10.81+4(1.008) =37.832\ \mathrm{g\,mol^{-1}}, \]

so

\[ \boxed{ w_{\mathrm H}(\ce{NaBH4}) =\frac{4(1.008)}{37.832}\times100\% =10.66\% }. \]

For ammonia borane,

\[ M(\ce{BNH6}) =10.81+14.01+6(1.008) =30.868\ \mathrm{g\,mol^{-1}}, \]

and

\[ \boxed{ w_{\mathrm H}(\ce{BNH6}) =\frac{6(1.008)}{30.868}\times100\% =19.59\% }. \]

These are the hydrogen contents of the compounds themselves.

8.5. Hydrolysis reactions

Using the metaborate products specified in the problem,

\[ \boxed{\ce{NaBH4 + 2H2O -> NaBO2 + 4H2}}, \]

and

\[ \boxed{\ce{NH3BH3 + 2H2O -> NH4BO2 + 3H2}}. \]

The second equation can equivalently be written

\[ \ce{NH3BH3 + 2H2O -> NH4+ + BO2^- + 3H2}. \]

The products are sodium metaborate and ammonium metaborate. A formal monomeric Lewis representation of the metaborate ion is [B-](=O)=O; corresponding formula-unit SMILES are [Na+].[B-](=O)=O and [NH4+].[B-](=O)=O. These formula-unit representations do not describe hydration or condensed borate structures.

Both equations conserve every element and give the stated four and three equivalents of hydrogen. Water supplies part of the hydrogen released as gas, so the gas yield is not simply the original hydrogen mass content calculated in 8.4.

8.6. Formation of boric acid and boron nitride

Hydration of the acidic oxide gives boric acid, OB(O)O:

\[ \boxed{\ce{B2O3 + 3H2O -> 2H3BO3}}. \]

Its structural formula is \(\ce{B(OH)3}\).

At high temperature and pressure, reaction with ammonia gives boron nitride and water:

\[ \boxed{\ce{B2O3 + 2NH3 -> 2BN + 3H2O}}. \]

Both B atoms become part of the nitride, while the three O atoms leave in three water molecules.

8.7. Structures of the borate ion, boric acid and boron nitride

For the cyclic trimetaborate ion, \(\ce{B3O6^{3-}}\), SMILES [O-]B1OB([O-])OB([O-])O1:

  • Draw a six-membered ring with the sequence \(\ce{B-O-B-O-B-O}\).
  • Attach one additional, terminal O atom to each B.
  • Give each terminal O a formal charge of \(-1\); the three bridging O atoms are neutral.
  • Each B is bonded to three O atoms and has approximately trigonal-planar coordination.

This gives three B atoms, six O atoms and total charge \(-3\).

For boric acid, OB(O)O, draw a central B bonded to three hydroxyl groups:

\[ \ce{B(OH)3}. \]

The B atom and its three directly attached O atoms lie in one plane, with \(\ce{O-B-O}\) angles approximately \(120^\circ\). Each O bears one H.

For a hexagonal boron nitride sheet, draw a planar honeycomb of fused hexagons. Label successive vertices alternately B and N, so every bond joins unlike atoms. Each B has three N neighbours, and each N has three B neighbours.

A precise patch satisfying the requested size is obtained by drawing one central hexagon and fusing one hexagon to each of its six sides. The resulting seven-hexagon patch has 24 vertices: 12 B and 12 N atoms. Indicate continuation of the lattice at its perimeter. An infinite sheet cannot be encoded by a finite molecular SMILES.

8.8. Synthesis of diborane

The reactants are boron trifluoride, FB(F)F, and lithium borohydride, [Li+].[BH4-].

The hint identifies the lithium-containing product as lithium tetrafluoroborate, [Li+].[B-](F)(F)(F)F, rather than a boron-free product.

Write

\[ \ce{aBF3 + bLiBH4 -> cB2H6 + dLiBF4}. \]

Lithium, fluorine and hydrogen balances give

\[ b=d,\qquad 3a=4d,\qquad 4b=6c. \]

The smallest integer coefficients are \(a=4,\ b=d=3,\ c=2\). Boron balance is also satisfied: \(4+3=2(2)+3\).

Thus,

\[ \boxed{\ce{4BF3 + 3LiBH4 -> 2B2H6 + 3LiBF4}}. \]

8.9. Geometry and bonding of diborane

Diborane, \(\ce{B2H6}\), contains four terminal H atoms and two bridging H atoms.

To draw it precisely:

  1. Place the two B atoms horizontally apart.
  2. Attach two terminal H atoms to each B. Both B atoms and all four terminal H atoms lie in one plane.
  3. Place one bridging H above this plane and the other below it, symmetrically between the B atoms.
  4. Connect each bridging H to both B atoms to indicate a three-centre bridge. Do not draw a direct \(\ce{B-B}\) bond.

The plane containing both B atoms and both bridging H atoms is perpendicular to the plane containing the four terminal H atoms. The idealized molecule has \(D_{2h}\) symmetry.

Its 12 valence electrons are distributed as follows:

\[ 4\text{ terminal two-centre bonds}\times2e^- + 2\text{ three-centre bridges}\times2e^- =12e^-. \]

A schematic connectivity string is [BH2]1[H][BH2][H]1. This is not a faithful ordinary, valence-valid SMILES representation of the bonding: standard SMILES cannot directly express a three-centre, two-electron bond. In particular, each bridging H must not be interpreted as making two independent ordinary covalent bonds.

8.10. Lewis structure of borane carbonyl

The reactants are borane, [BH3], and carbon monoxide, [C-]#[O+]. The carbon-centred lone pair of carbon monoxide forms the bond to B.

The product is borane carbonyl, [BH3-]C#[O+], with connectivity

\[ \ce{H3B^{-}-C#O^{+}}. \]

Draw three \(\ce{B-H}\) single bonds, one \(\ce{B-C}\) single bond, a \(\ce{C#O}\) triple bond, and one lone pair on O. There are no lone pairs on B or C in this Lewis structure.

Using

\[ \mathrm{FC} =(\text{valence electrons}) -(\text{nonbonding electrons}) -\frac12(\text{bonding electrons}), \]

the formal charges are

\[ \begin{aligned} \mathrm{FC}(B)&=3-0-4=-1,\\ \mathrm{FC}(C)&=4-0-4=0,\\ \mathrm{FC}(O)&=6-2-3=+1,\\ \mathrm{FC}(H)&=1-0-1=0. \end{aligned} \]

The total charge is zero. B is approximately tetrahedral, and the \(\ce{B-C-O}\) arrangement is approximately linear.

8.11. Effect on the carbon–oxygen bond

The correct choice is:

No or slight change on it because \(\ce{CO}\) donates its mainly non-bonding electrons to \(\ce{BH3}\).

The donation comes from the mainly carbon-centred, approximately nonbonding \(\sigma\) orbital of carbon monoxide. It therefore produces only a small change in the \(\ce{C-O}\) bond. Electron-deficient borane does not provide the substantial \(\pi\)-back donation invoked by the first option.

8.12. Two-step synthesis of borazine

The starting materials are ammonium chloride, [NH4+].[Cl-], and boron trichloride, ClB(Cl)Cl.

In the first step, three B atoms and three N atoms form the six-membered ring. Each B retains one Cl, and each N retains one H. The remaining H and Cl atoms form hydrogen chloride:

\[ \boxed{\ce{3NH4Cl + 3BCl3 -> B3N3H3Cl3 + 9HCl}}. \]

There are 12 H atoms and 12 Cl atoms on each side.

In the second step, the three B-bound Cl atoms are replaced by H. Each borohydride supplies one hydride for this substitution and leaves a borane unit, which is trapped by THF:

\[ \boxed{\ce{B3N3H3Cl3 + 3LiBH4 + 3THF -> B3N3H6 + 3LiCl + 3BH3*THF}}. \]

The by-products are lithium chloride, [Li+].[Cl-], and the borane–tetrahydrofuran adduct, [BH3-][O+]1CCCC1.

In the latter, an oxygen lone pair from THF forms the bond to B. This is the Lewis acid–base stabilization identified by the hint.

8.13. Structures of borazine and its chlorine derivative

Borazine, \(\ce{B3N3H6}\), has a planar six-membered ring with alternating B and N atoms. Draw alternating single and double bonds around the ring and attach one H to every ring atom.

One Kekulé Lewis contributor is encoded by

[BH-]1=[NH+][BH-]=[NH+][BH-]=[NH+]1.

Each B and N is approximately trigonal planar. The actual ring bonding is delocalized; the SMILES selects one localized Lewis contributor.

For B,B′,B″-trichloroborazine, \(\ce{B3N3H3Cl3}\), retain the same ring and replace the H attached to each B with Cl. All three N atoms retain their H atoms:

[B-]1(Cl)=[NH+][B-](Cl)=[NH+][B-](Cl)=[NH+]1.

The negative B and positive N charges in these strings are the formal charges of the chosen alternating-double-bond contributors. They follow because each such B and N has a total bond order of four.

The chlorine substituents occupy the B positions because the first synthesis step retains B–Cl bonds, and the second step reduces those bonds to B–H.

8.14. Turnover frequency

The hydrogen amount follows from the ideal-gas equation:

\[ n_{\ce{H2}}=\frac{PV}{RT} =\frac{(1.000\ \mathrm{atm})(0.06725\ \mathrm L)} {(0.0820574\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}) (273.15\ \mathrm K)} =3.00036\times10^{-3}\ \mathrm{mol}. \]

The mass of Pt in the catalyst is

\[ m_{\ce{Pt}} =(5.0\times10^{-3}\ \mathrm g)(0.082) =4.10\times10^{-4}\ \mathrm g. \]

Hence

\[ n_{\ce{Pt}} =\frac{4.10\times10^{-4}\ \mathrm g} {195.08\ \mathrm{g\,mol^{-1}}} =2.10170\times10^{-6}\ \mathrm{mol}. \]

The stated definition uses moles of hydrogen product and moles of Pt atoms:

\[ \mathrm{TOF} =\frac{n_{\ce{H2}}}{n_{\ce{Pt}}\,t} =\frac{3.00036\times10^{-3}} {(2.10170\times10^{-6})(5)} =285.5\ \mathrm{min^{-1}}. \]

Thus, to the precision of the catalyst mass and Pt percentage,

\[ \boxed{\mathrm{TOF}\approx2.9\times10^2\ \mathrm{min^{-1}}}. \]

As a consistency check, the starting ammonia borane amount is

\[ (0.1000\ \mathrm{mol\,L^{-1}})(0.0100\ \mathrm L) =1.00\times10^{-3}\ \mathrm{mol}. \]

Complete hydrolysis gives approximately \(3.00\times10^{-3}\ \mathrm{mol}\) of hydrogen, matching the measured volume.

8.15. Alloy composition

An atom at a cube corner contributes \(1/8\) to the unit cell; an atom at a face centre contributes \(1/2\).

With Pt at the corners,

\[ N_{\ce{Pt}}=8\left(\frac18\right)=1. \]

With Cu at the face centres,

\[ N_{\ce{Cu}}=6\left(\frac12\right)=3. \]

The total is four atoms per cell, so the atomic percentages are

\[ x=\frac34(100)=75,\qquad y=\frac14(100)=25. \]

Therefore,

\[ \boxed{\ce{Cu_{75}Pt_{25}},\qquad x=75,\quad y=25}. \]

Its simplest atomic ratio is \(\ce{Cu3Pt}\).

8.16. Unit-cell drawing

Draw a cube with:

  • Eight Pt atoms: one at each corner.
  • Six Cu atoms: one at the centre of each face.

In fractional coordinates, the Pt atoms occupy all corners \((i,j,k)\), where \(i,j,k\in\{0,1\}\). The Cu atoms occupy

\[ \begin{gathered} \left(\tfrac12,\tfrac12,0\right),\quad \left(\tfrac12,\tfrac12,1\right),\\ \left(\tfrac12,0,\tfrac12\right),\quad \left(\tfrac12,1,\tfrac12\right),\\ \left(0,\tfrac12,\tfrac12\right),\quad \left(1,\tfrac12,\tfrac12\right). \end{gathered} \]

The cube has no body-centre atom. The drawing shows 14 atomic positions on the boundary, whose sharing gives the effective content of three Cu and one Pt per cell.

8.17. Density of the randomly occupied alloy

An fcc conventional unit cell contains four atomic sites. For composition \(\ce{Cu2Pt1}\), the mean site occupancies are \(2/3\) Cu and \(1/3\) Pt. Thus the average numbers per cell are

\[ N_{\ce{Cu}}=\frac83,\qquad N_{\ce{Pt}}=\frac43. \]

These fractional values are averages over many randomly occupied cells.

The mean molar mass per mole of atomic sites is

\[ \overline M =\frac{2M_{\ce{Cu}}+M_{\ce{Pt}}}{3} =\frac{2(63.55)+195.08}{3} =107.3933\ \mathrm{g\,mol^{-1}}. \]

The unit-cell mass is therefore

\[ m_{\mathrm{cell}} =\frac{4\overline M}{N_{\mathrm A}} =\frac{4(107.3933)} {6.02214076\times10^{23}} =7.13323\times10^{-22}\ \mathrm g. \]

Convert the edge length to centimetres:

\[ a=380\ \mathrm{pm} =3.80\times10^{-8}\ \mathrm{cm}. \]

Then

\[ V_{\mathrm{cell}}=a^3 =(3.80\times10^{-8})^3 =5.4872\times10^{-23}\ \mathrm{cm^3}. \]

Finally,

\[ \boxed{ \rho=\frac{m_{\mathrm{cell}}}{V_{\mathrm{cell}}} =\frac{7.13323\times10^{-22}} {5.4872\times10^{-23}} =13.0\ \mathrm{g\,cm^{-3}} }. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.