IChO 2009 · Theory problem 6

Transition metal complexes

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Alfred Werner used the technique of ‘isomer counting’ to deduce the structure of metal complexes with coordination number six. Three of the shapes he considered are shown below.

Figure 1

In each structure, the empty circle shows the location of the central metal atom and the filled circles show the location of the ligands. Structure X is hexagonal planar, structure Y is trigonal prismatic and structure Z is octahedral.

For each of the three shapes, there is just one structure when all of the ligands are the same, i.e. when the complex has the general formula \(\ce{MA6}\) where A is the ligand. However, when achiral ligands A are substituted by one or more achiral ligands, it may be possible for each structure to form geometrical isomers. It might also be possible for one or more of the geometrical isomers to be optically active and exist as pairs of enantiomers.

6.1 Fill in the table below to indicate how many geometrical isomers may be formed for each structure X, Y, and Z as the monodentate ligands A are substituted by monodentate ligands B or by symmetrical bidentate ligands, denoted C–C. Bidentate ligand C–C can only link between two atoms on adjacent positions, i.e. those positions connected by a line in the structures X, Y, and Z.

In each case write the number of geometrical isomers in the space provided. If one of the isomers exists as a pair of enantiomers, include an asterisk, *, in the box. If two exist as two pairs of enantiomers, include two asterisks and so on. For example, if you think there are five geometrical isomers of a particular structure, three of which exist as pairs of enantiomers, write 5 ***

Number of predicted geometrical isomers

Hexagonal planar X Trigonal Prismatic Y Octahedral Z
\(\ce{MA6}\) 1 1 1
\(\ce{MA5B}\)
\(\ce{MA4B2}\)
\(\ce{MA3B3}\)
\(\ce{MA4(C-C)}\)
\(\ce{MA2(C-C)2}\)
\(\ce{M(C-C)3}\)

There are no known complexes that adopt the hexagonal planar geometry X, but structures are known for both the trigonal prismatic geometry Y and for the octahedral geometry Z. In these complexes, the orbitals derived from the metal d orbitals have different energies depending on the geometry of the complex. The splitting patterns for the trigonal prismatic geometry and for the octahedral geometry are shown below.

Figure 2

The separations in energy, \(\Delta E\), \(\Delta E'\) and \(\Delta E''\) depend on the particular complex.

6.2 For each of the splitting patterns shown below label which d orbitals are which.

Figure 3

The two complexes \(\ce{[Mn(H2O)6]^2+}\) and \(\ce{[Mn(CN)6]^2-}\) are both octahedral. One has a magnetic moment of 5.9 BM, the other has a magnetic moment of 3.8 BM but you must decide which is which.

6.3 On the diagram below, draw the electronic arrangements for each of the complexes.

Figure 4

The magnetic moments of complexes A and B shown below have been measured and found to be 1.9 and 2.7 BM but you must decide which is which.

Figure 5

6.4 Draw the orbital splitting diagrams for the two complexes, including the arrangements of the electrons.

Octahedral complexes are far more common than trigonal prismatic. Werner isolated five compounds C – G containing Co(III), Cl, and \(\ce{NH3}\) only, each of which contained one octahedral complex. (There is actually a sixth compound but Werner could not isolate it.) Werner’s five compounds had the molar conductivities shown below. The conductivities are extrapolated to infinite dilution and are expressed in arbitrary units. Compound G does not react with aqueous \(\ce{AgNO3}\); compounds C, D, and E react with different stoichiometric ratios of aqueous \(\ce{AgNO3}\); E and F react with the same stoichiometric ratio of aqueous \(\ce{AgNO3}\).

C D E F G
molar conductivity 510 372 249 249 \(\sim0\)

6.5 As far as you are able, suggest a structure for each of the compounds C – G.

Werner was also the first person to separate the enantiomers of an octahedral compound, H, which contained no carbon atoms. The compound, H, is composed of only cobalt, ammonia, chloride and an oxygen species which could be either \(\ce{H2O}\), or \(\ce{HO^-}\) or \(\ce{O^2-}\). The compound contains octahedrally coordinated cobalt ions. All of the chloride is easily removed from the compound by titration with aqueous silver nitrate. A 0.2872 g sample of H (containing no water of crystallization) required 22.8 \(\mathrm{cm^3}\) of a silver nitrate solution (\(c=0.100\ \mathrm{mol\,dm^{-3}}\)) to exchange all the chloride.

6.6 Calculate the percentage, by mass, of chloride in H.

H is stable to acids, but is hydrolysed in alkali. A 0.7934 g sample of H (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into 50.0 \(\mathrm{cm^3}\) of aqueous \(\ce{HCl}\) (\(c_{\ce{HCl}}=0.500\ \mathrm{mol\,dm^{-3}}\)). The residual \(\ce{HCl}\) required 24.8 \(\mathrm{cm^3}\) of aqueous \(\ce{KOH}\) solution (\(c_{\ce{KOH}}=0.500\ \mathrm{mol\,dm^{-3}}\)) to be neutralized.

The remaining suspension of cobalt(III) oxide was allowed to cool, approximately 1 g of potassium iodide was added, and then the mixture was acidified with aqueous \(\ce{HCl}\). The liberated iodine was then titrated with aqueous solution of sodium thiosulfate (\(c=0.200\ \mathrm{mol\,dm^{-3}}\)) and required 21.0 \(\mathrm{cm^3}\) for complete reaction.

6.7 Calculate the percentage, by mass, of ammonia in H.

6.8 Give the equation for the reaction of cobalt(III) oxide with potassium iodide in aqueous acid.

6.9 Calculate the percentage, by mass, of cobalt in H.

6.10 Calculate the identity of the oxygen species contained in H. Show your working.

6.11 Give the empirical formula of H.

6.12 Suggest a structure for the chiral compound H.

Official English text of the problem.

Answers

Molecular names and coordination SMILES are supplied in the corresponding solution parts.

  • 6.1. Each asterisk denotes one geometrical isomer that exists as an enantiomeric pair.
Formula Hexagonal planar X Trigonal prismatic Y Octahedral Z
\(\ce{MA6}\) 1 1 1
\(\ce{MA5B}\) 1 1 1
\(\ce{MA4B2}\) 3 \(3^*\) 2
\(\ce{MA3B3}\) 3 \(3^*\) 2
\(\ce{MA4(C-C)}\) 1 2 1
\(\ce{MA2(C-C)2}\) 2 \(4^*\) \(2^*\)
\(\ce{M(C-C)3}\) 1 2 \(1^*\)
  • 6.2. From lowest to highest energy:
  • Octahedral: \((d_{xy},d_{xz},d_{yz})<(d_{z^2},d_{x^2-y^2})\).

  • Trigonal prismatic, for the pattern supplied: \((d_{xy},d_{x^2-y^2})<d_{z^2}<(d_{xz},d_{yz})\).

  • 6.3.

  • \(\ce{[Mn(H2O)6]^2+}\): high-spin \(d^5\), \(t_{2g}^{3}e_g^{2}\), five unpaired electrons, 5.9 BM.

  • \(\ce{[Mn(CN)6]^2-}\): \(d^3\), \(t_{2g}^{3}\), three unpaired electrons, 3.8 BM.

  • 6.4.

  • A: Re(IV), \(d^3\), trigonal prismatic; three electrons in the lowest doublet, one unpaired; 1.9 BM.

  • B: Re(III), \(d^4\), low-spin octahedral in the intended strong-field treatment; \(t_{2g}^{4}\), two unpaired; 2.7 BM.

  • 6.5.

  • C: \(\ce{[Co(NH3)6]Cl3}\).
  • D: \(\ce{[Co(NH3)5Cl]Cl2}\).
  • E, F: cis- and trans-\(\ce{[Co(NH3)4Cl2]Cl}\), in an undetermined order.

  • G: \(\ce{[Co(NH3)3Cl3}\)], either fac or mer; the data do not distinguish them.

  • 6.6. Chloride: 28.1% by mass.

  • 6.7. Ammonia: 27.0% by mass.
  • 6.8. \(\ce{Co2O3 + 2I^- + 6H+ -> 2Co^2+ + I2 + 3H2O}\).
  • 6.9. Cobalt: 31.2% by mass.
  • 6.10. The oxygen species is hydroxide, \(\ce{OH^-}\), SMILES [OH-].
  • 6.11. Empirical formula: \(\ce{Co2(NH3)6(OH)3Cl3}\), equivalently \(\ce{Co2Cl3H21N6O3}\).
  • 6.12. A suitable structure is hexol chloride, \(\ce{[Co4(NH3)12(OH)6]Cl6}\): a central octahedral cobalt ion joined through three pairs of hydroxide bridges to three peripheral tetraamminecobalt centres. Its central tris-chelate arrangement gives \(\Delta\) and \(\Lambda\) enantiomers.

Solution

The coordination SMILES below use dative bonds, ->, and octahedral stereodescriptors, @OH, where appropriate. An @OH descriptor can specify geometrical configuration even when the complex is achiral. Trigonal-prismatic geometry is described in words because these octahedral descriptors do not encode it.

6.1 — Geometrical and optical isomers

A pair of enantiomers counts as one geometrical isomer, with an asterisk. Ligands A, B and C–C are unspecified placeholders, so they do not have unique molecular names or molecular SMILES.

For X, number the six sites consecutively around the hexagon.

  • With one B ligand, every site is equivalent.
  • With two B ligands, their separation can be one, two or three edges: three isomers.
  • With three B ligands, representative occupied sets are \(123\), \(124\) and \(135\): three isomers.
  • One chelate can occupy any edge: one isomer.
  • Two chelates have representatives \((12)(34)\) and \((12)(45)\): two isomers.
  • Three chelates have the single arrangement \((12)(34)(56)\), up to symmetry.

All are achiral in the problem’s idealized planar site-and-link model: reflection in the molecular plane leaves each arrangement unchanged.

For Y, label the upper triangle \(u_1,u_2,u_3\) and the lower triangle \(l_1,l_2,l_3\), with vertical edges \(u_i l_i\). The following representatives exhaust the possibilities:

Substitution Representatives Chirality
One B \(u_1\) Achiral
Two B \(u_1u_2\), \(u_1l_1\), \(u_1l_2\) Only \(u_1l_2\) is chiral
Three B \(u_1u_2u_3\), \(u_1u_2l_1\), \(u_1u_2l_3\) Only \(u_1u_2l_1\) is chiral
One chelate \((u_1u_2)\), \((u_1l_1)\) Both achiral
Two chelates \((u_1l_1)(u_2l_2)\); \((u_1u_2)(u_3l_3)\); \((u_1u_2)(l_1l_2)\); \((u_1u_2)(l_1l_3)\) Only the last is chiral
Three chelates \((u_1l_1)(u_2l_2)(u_3l_3)\); \((u_1u_2)(l_1l_2)(u_3l_3)\) Both achiral

For example, the two B ligands in \(u_1l_2\) occupy opposite ends of a diagonal of a rectangular face. Its mirror image cannot be superimposed by a proper rotation. The same test gives the other two starred prism arrangements.

For Z:

  • \(\ce{MA5B}\) has one arrangement.
  • \(\ce{MA4B2}\) has cis and trans arrangements.
  • \(\ce{MA3B3}\) has fac and mer arrangements.
  • One chelate can occupy any octahedral edge: one arrangement.
  • Two chelates leave the two A ligands either cis or trans. The cis arrangement has \(\Delta/\Lambda\) enantiomers; the trans arrangement is achiral.
  • Three chelates give one geometrical arrangement with \(\Delta/\Lambda\) enantiomers.

These results give the completed table in Answers.

6.2 — Identification of the d orbitals

In an octahedral complex, the ligands lie along the coordinate axes. The orbitals directed along these axes experience greater ligand repulsion:

\[ \underbrace{d_{xy},d_{xz},d_{yz}}_{\text{lower }t_{2g}} \quad<\quad \underbrace{d_{z^2},d_{x^2-y^2}}_{\text{upper }e_g}. \]

Thus write \(d_{xy},d_{xz},d_{yz}\) on the lower three lines and \(d_{z^2},d_{x^2-y^2}\) on the upper two lines, in any order within each degenerate set.

For the supplied trigonal-prismatic pattern, with \(z\) along the prism axis:

\[ \underbrace{d_{xy},d_{x^2-y^2}}_{\text{lowest doublet}} \quad<\quad \underbrace{d_{z^2}}_{\text{middle singlet}} \quad<\quad \underbrace{d_{xz},d_{yz}}_{\text{highest doublet}}. \]

The unique singlet is \(d_{z^2}\). The other four orbitals form the two indicated symmetry-related pairs.

6.3 — Manganese electron arrangements

The spin-only magnetic moment is approximately

\[ \mu_{\mathrm{so}}=\sqrt{n(n+2)}\ \mathrm{BM}, \]

where \(n\) is the number of unpaired electrons.

The hexaaquamanganese(II) ion, \(\ce{[Mn(H2O)6]^2+}\), has coordination SMILES:

[Mn+2](<-[OH2])(<-[OH2])(<-[OH2])(<-[OH2])(<-[OH2])<-[OH2]

Water is neutral, so manganese is in oxidation state \(+2\). Manganese is a group-7 element; therefore Mn(II) is \(d^5\). With weak-field water ligands, it is high spin:

\[ t_{2g}:\ [\uparrow]\,[\uparrow]\,[\uparrow], \qquad e_g:\ [\uparrow]\,[\uparrow]. \]

There are five unpaired electrons:

\[ \mu_{\mathrm{so}}=\sqrt{5(5+2)}=\sqrt{35}=5.92\ \mathrm{BM}. \]

It therefore has the measured moment 5.9 BM.

The hexacyanidomanganate(IV) ion, \(\ce{[Mn(CN)6]^2-}\), has coordination SMILES:

[Mn+4](<-[C-]#N)(<-[C-]#N)(<-[C-]#N)(<-[C-]#N)(<-[C-]#N)<-[C-]#N

Each cyanido ligand has charge \(-1\), so

\[ x+6(-1)=-2 \quad\Longrightarrow\quad x=+4. \]

Mn(IV) is \(d^3\), giving

\[ t_{2g}:\ [\uparrow]\,[\uparrow]\,[\uparrow], \qquad e_g:\ [\ ]\,[\ ]. \]

Thus

\[ \mu_{\mathrm{so}}=\sqrt{3(3+2)}=\sqrt{15}=3.87\ \mathrm{BM}, \]

corresponding to the measured 3.8 BM. Both complexes are octahedral; the difference in electron count is essential.

6.4 — Rhenium electron arrangements

In A, each ene-dithiolate ligand is dianionic. Consequently,

\[ x+3(-2)=-2 \quad\Longrightarrow\quad x=+4. \]

Rhenium is a group-7 element, so Re(IV) is \(d^3\).

A is the tris[(Z)-1,2-diphenylethene-1,2-dithiolato]rhenate(IV) ion, represented by the connectivity SMILES:

[Re+4]123(<-[S-]/C(c4ccccc4)=C(\[S-]->1)c4ccccc4)(<-[S-]/C(c4ccccc4)=C(\[S-]->2)c4ccccc4)<-[S-]/C(c4ccccc4)=C(\[S-]->3)c4ccccc4

An octahedral \(d^3\) ion would have three unpaired electrons. The smaller magnetic moment instead corresponds to one unpaired electron, obtained in the supplied trigonal-prismatic splitting:

Level in A, highest first Occupation
\(d_{xz},d_{yz}\) \([\ ]\,[\ ]\)
\(d_{z^2}\) \([\ ]\)
\(d_{xy},d_{x^2-y^2}\) \([\uparrow\downarrow]\,[\uparrow]\)

The choice of which member of the lowest doublet contains the pair is arbitrary. With one unpaired electron,

\[ \mu_{\mathrm{so}}=\sqrt{3}=1.73\ \mathrm{BM}, \]

consistent with assigning the observed 1.9 BM to A. Its coordination geometry is trigonal prismatic, which the connectivity SMILES does not itself specify.

In B, each sulfur chelate is monoanionic. Since the complex is neutral,

\[ x+3(-1)=0 \quad\Longrightarrow\quad x=+3, \]

so Re(III) is \(d^4\).

Reading the central S in the right-hand repeated ligand literally, B is bis(benzenedithiosulfinato)(dithiobenzoato)rhenium(III), with connectivity SMILES:

[Re+3]123(<-S=C(c4ccccc4)[S-]->1)(<-S=S(c4ccccc4)[S-]->2)<-S=S(c4ccccc4)[S-]->3

Under the strong-field treatment intended for these 5d complexes, B has the low-spin octahedral arrangement:

Level in B, highest first Occupation
\(e_g: d_{z^2},d_{x^2-y^2}\) \([\ ]\,[\ ]\)
\(t_{2g}: d_{xy},d_{xz},d_{yz}\) \([\uparrow\downarrow]\,[\uparrow]\,[\uparrow]\)

There are two unpaired electrons:

\[ \mu_{\mathrm{so}}=\sqrt{8}=2.83\ \mathrm{BM}, \]

corresponding to 2.7 BM. The displayed connectivity leaves its handedness unspecified, as does the problem’s drawing.

6.5 — Werner’s cobalt compounds

Co(III) has coordination number six. Ammonia is neutral, whereas each coordinated chlorido ligand lowers the complex charge by one:

\[ \ce{[Co(NH3)_{6-k}Cl_k]^{(3-k)+}}. \]

The remaining chloride ions are counterions. These precipitate with silver ions, whereas coordinated chloride is not immediately removed in the stated test.

The conductivity sequence is consistent with four, three, two, two and zero ions per formula unit:

Compound Structure Counterion chlorides Dissolved ions per formula unit
C \(\ce{[Co(NH3)6]Cl3}\) 3 4
D \(\ce{[Co(NH3)5Cl]Cl2}\) 2 3
E One isomer of \(\ce{[Co(NH3)4Cl2]Cl}\) 1 2
F The other isomer of \(\ce{[Co(NH3)4Cl2]Cl}\) 1 2
G \(\ce{[Co(NH3)3Cl3]}\) 0 0

The structures, names and SMILES are:

  • C: hexaamminecobalt(III) chloride

[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[NH3].[Cl-].[Cl-].[Cl-]

  • D: pentaamminechloridocobalt(III) chloride

[Co+3](<-[Cl-])(<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[NH3].[Cl-].[Cl-]

  • One of E/F: cis-tetraamminedichloridocobalt(III) chloride, with adjacent chlorido ligands, at \(90^\circ\):

[Co@OH21+3](<-[Cl-])(<-[Cl-])(<-[NH3])(<-[NH3])(<-[NH3])<-[NH3].[Cl-]

  • The other of E/F: trans-tetraamminedichloridocobalt(III) chloride, with opposite chlorido ligands, at \(180^\circ\):

[Co@OH27+3](<-[Cl-])(<-[Cl-])(<-[NH3])(<-[NH3])(<-[NH3])<-[NH3].[Cl-]

  • G may be fac-triamminetrichloridocobalt(III), with all three chlorido ligands mutually cis on one triangular face:

[Co@OH6+3](<-[Cl-])(<-[Cl-])(<-[Cl-])(<-[NH3])(<-[NH3])<-[NH3]

  • Alternatively, G may be mer-triamminetrichloridocobalt(III), with two chlorido ligands trans and the third cis to both:

[Co@OH27+3](<-[Cl-])(<-[Cl-])(<-[Cl-])(<-[NH3])(<-[NH3])<-[NH3]

Neither conductivity nor the chloride-precipitation result distinguishes E from F, or fac-G from mer-G. The other geometrical isomer of G is the sixth compound mentioned in the question.

6.6 — Mass percentage of chloride

For the numerical calculations, use standard molar masses

\[ M(\ce{Cl})=35.45,\qquad M(\ce{NH3})=17.03,\qquad M(\ce{Co})=58.93 \quad\mathrm{g\,mol^{-1}}. \]

Chloride ion, [Cl-], precipitates with silver(I) ion, [Ag+], in a 1:1 ratio:

\[ \ce{Ag+ + Cl^- -> AgCl(s)}. \]

Silver chloride has the stoichiometric ionic SMILES [Ag+].[Cl-]; this does not describe its crystal lattice.

\[ n(\ce{Cl^-}) =(0.100\ \mathrm{mol\,dm^{-3}}) (22.8\times10^{-3}\ \mathrm{dm^3}) =2.28\times10^{-3}\ \mathrm{mol}. \]

Therefore,

\[ m(\ce{Cl^-}) =(2.28\times10^{-3})(35.45) =0.080826\ \mathrm{g}, \]

and

\[ \boxed{ w(\ce{Cl^-}) =\frac{0.080826}{0.2872}\times100 =28.1\% }. \]

6.7 — Mass percentage of ammonia

Ammonia, N, consumes one equivalent of acid to form ammonium ion, [NH4+]:

\[ \ce{NH3 + H+ -> NH4+}. \]

The remaining acid is measured by

\[ \ce{H+ + OH^- -> H2O}. \]

Here the hydron, hydroxide ion and water have SMILES [H+], [OH-] and O, respectively.

Initially,

\[ n_0(\ce{H+}) =(0.500)(50.0\times10^{-3}) =0.0250\ \mathrm{mol}. \]

The residual acid is

\[ n_{\mathrm{res}}(\ce{H+}) =(0.500)(24.8\times10^{-3}) =0.0124\ \mathrm{mol}. \]

Hence,

\[ n(\ce{NH3})=0.0250-0.0124=0.0126\ \mathrm{mol}, \]

\[ m(\ce{NH3})=(0.0126)(17.03)=0.2146\ \mathrm{g}, \]

and

\[ \boxed{ w(\ce{NH3}) =\frac{0.2146}{0.7934}\times100 =27.0\% }. \]

6.8 — Reaction of cobalt(III) oxide with iodide

Cobalt(III) is reduced to cobalt(II), while iodide is oxidized to iodine:

\[ \ce{Co2O3 + 6H+ + 2e^- -> 2Co^2+ + 3H2O}, \]

\[ \ce{2I^- -> I2 + 2e^-}. \]

Adding gives

\[ \boxed{\ce{Co2O3 + 2I^- + 6H+ -> 2Co^2+ + I2 + 3H2O}}. \]

The relevant species are cobalt(III) oxide, stoichiometric ionic SMILES [Co+3].[Co+3].[O-2].[O-2].[O-2]; iodide ion, [I-]; cobalt(II) ion, [Co+2]; and iodine, II. Cobalt(III) oxide is an extended solid, so its ionic SMILES specifies composition rather than lattice structure.

6.9 — Mass percentage of cobalt

Iodine reacts with thiosulfate according to

\[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-}. \]

The ions are thiosulfate, [O-]S(=O)(=O)[S-], and tetrathionate, [O-]S(=O)(=O)SSS(=O)(=O)[O-].

Thus,

\[ n(\ce{S2O3^2-}) =(0.200)(21.0\times10^{-3}) =4.20\times10^{-3}\ \mathrm{mol}, \]

\[ n(\ce{I2})=\frac{4.20\times10^{-3}}{2} =2.10\times10^{-3}\ \mathrm{mol}. \]

From the equation in 6.8, two cobalt atoms correspond to one iodine molecule:

\[ n(\ce{Co})=2n(\ce{I2}) =4.20\times10^{-3}\ \mathrm{mol}. \]

Consequently,

\[ m(\ce{Co})=(4.20\times10^{-3})(58.93) =0.2475\ \mathrm{g}, \]

\[ \boxed{ w(\ce{Co}) =\frac{0.2475}{0.7934}\times100 =31.2\% }. \]

6.10 — Identification of the oxygen species

The unassigned mass percentage is approximately

\[ 100-28.14-27.05-31.20=13.61\%. \]

This includes both oxygen and any hydrogen belonging to its species.

For the \(0.7934\ \mathrm{g}\) sample, the chloride content inferred from 6.6 is

\[ n(\ce{Cl^-}) =(2.280\times10^{-3}) \frac{0.7934}{0.2872} \approx6.30\times10^{-3}\ \mathrm{mol}. \]

The cobalt and ammonia amounts therefore give

\[ \ce{Co}:\ce{NH3}:\ce{Cl^-} = 0.00420:0.0126:0.00630 = 1:3:1.5. \]

With cobalt in oxidation state \(+3\) and ammonia neutral, the oxygen species must supply

\[ 3-1.5=1.5 \]

negative charge units per cobalt atom.

The remaining mass per mole of cobalt is approximately

\[ \frac{0.1361(0.7934\ \mathrm{g})} {0.00420\ \mathrm{mol}} =25.7\ \mathrm{g\,mol^{-1}} \]

per mole of cobalt. Compare the candidates:

Candidate Amount required per cobalt by charge balance Predicted remaining mass per mole of cobalt
Water, \(\ce{H2O}\), O Cannot supply negative charge Inconsistent
Hydroxide, \(\ce{OH^-}\), [OH-] 1.5 mol \(1.5(17.01)=25.5\ \mathrm{g}\)
Oxide, \(\ce{O^2-}\), [O-2] 0.75 mol \(0.75(16.00)=12.0\ \mathrm{g}\)

The measured residual mass agrees with hydroxide:

\[ \boxed{\text{The oxygen species is }\ce{OH^-}.} \]

6.11 — Empirical formula

The component ratio is

\[ \ce{Co}:\ce{NH3}:\ce{OH^-}:\ce{Cl^-} =1:3:1.5:1.5. \]

Multiplying by two gives the smallest integral ratio:

\[ \boxed{\ce{Co2(NH3)6(OH)3Cl3}}. \]

On an elemental basis this is

\[ \boxed{\ce{Co2Cl3H21N6O3}}. \]

An empirical formula specifies a ratio, not a unique molecular graph. The molecular structure and SMILES proposed for H follow in 6.12.

6.12 — Structure and chirality of H

A suitable structure is hexol chloride, also described as dodecaamminehexa-μ-hydroxidotetracobalt(III) chloride:

\[ \boxed{ \ce{[Co((OH)2Co(NH3)4)3]Cl6} = \ce{[Co4(NH3)12(OH)6]Cl6} }. \]

Describe the drawing as follows:

  1. Place one cobalt(III) ion at the centre of an octahedron.
  2. Put six hydroxide oxygen donors at its six vertices.
  3. Divide those donors into three adjacent pairs.
  4. Connect each pair to one peripheral cobalt(III) ion, so each hydroxide bridges the central and a peripheral cobalt.
  5. Complete each peripheral cobalt’s octahedron with four ammonia ligands. Its two bridging oxygen donors are cis.
  6. Place six chloride counterions outside the coordination sphere.

Every cobalt is six-coordinate:

  • The central cobalt binds six hydroxide donors.
  • Each peripheral cobalt binds two hydroxide donors and four ammonia ligands.

The complex-ion charge is

\[ 4(+3)+6(-1)=+6, \]

requiring six chloride counterions. Its composition is twice the empirical formula obtained in 6.11.

For a precise stereochemical sketch, put the central donors on the \(\pm x,\pm y,\pm z\) axes and pair them as

\[ (+x,+y),\qquad(-x,+z),\qquad(-y,-z). \]

Each pair connects to a different peripheral cobalt. Reflecting this arrangement gives the opposite hand. The three paired bridges act as three bidentate ligands around the central cobalt, producing the familiar \(\Delta/\Lambda\) tris-chelate enantiomers.

The two enantiomers of hexol chloride can be represented by the following coordination SMILES:

[Co@OH8+3]123(<-[OH-]->[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[OH-]->1)(<-[OH-]->[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[OH-]->2)<-[OH-]->[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[OH-]->3.[Cl-].[Cl-].[Cl-].[Cl-].[Cl-].[Cl-]

[Co@OH10+3]123(<-[OH-]->[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[OH-]->1)(<-[OH-]->[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[OH-]->2)<-[OH-]->[Co+3](<-[NH3])(<-[NH3])(<-[NH3])(<-[NH3])<-[OH-]->3.[Cl-].[Cl-].[Cl-].[Cl-].[Cl-].[Cl-]

The @OH8 and @OH10 indices specify opposite configurations for this particular SMILES atom order; they are not signs of optical rotation. The peripheral cis coordination and the octahedral geometries are as described above.

The solution was checked against the official solution.