Download Answers to Assignment #5

Chemistry 3820
Answers to Assignment #5
November 20th, 2000
Topic: Coordination chemistry and aqua-ion spectra. You are responsible for Ch. 7 and 14 (Chapters 6 and 14 in 2nd edition).
Use the assignment questions to guide the parts you are most responsible for. I will not be asking you to determine RusselSaunders term symbols, nor topics from luminescence onward in Chapter 14.
1.
Do the in-text exercises as practice for the end-of-chapter problems! Answers are given at the back of the book.
2.
From the end-of-chapter exercises, do the following: 7.13 – 7.15; 7.17 – 7.20.
3.
Determine the configuration (in the form t2mg egn or em t 2n , as appropriate), the number of unpaired electrons, and the ligand
field stabilization energy as a multiple of ∆o or ∆T for each of the following complexes using the spectrochemical series to
decide, where relevant, which are likely to be strong-field and which weak-field. (a) [Co(NH3 )6 ]3+; (b) [Fe(OH2 )6 ]2+; (c)
[Fe(CN)6 ]3– ; (d) [Cr(NH3 )6 ]3+; (e) [W(CO)6 ]; (f) [FeCl4 ]2– and (g) [Ni(CO)4 ].
(a) [Co(NH3 )6 ]3+ Since the NH3 ligands are neutral, this is a Co 3+ complex, so we are dealing with a d 6 metal ion. Although
ammonia is in the middle of the spectrochemical series, a metal in the 3+ oxidation state is past the middle of the metal
series, so the combination is very likely to be a low-spin configuration, t 26g e0g . Indeed, octahedral d 6 has S = 0 (no unpaired
electrons) and is diamagnetic. The LFSE is 6 × 0.4∆O = 2.4∆O . Note that this is the largest possible LFSE for an octahedral
complex.
(b) [Fe(OH2 )6 ]2+ The iron ion in this octahedral complex with six neutral water ligands is in the 2+ oxidation state, and
water is low on the spectrochemical series. We certainly expect this d 6 ion to be in a high spin configuration, t 24g e2g , with S =
2. That is, it has four unpaired electrons and is strongly paramagnetic. The LFSE is 4 × 0.4∆O – 2 × 0.6∆O = 0.4∆O . Note
the much smaller LFSE compared to the low spin d 6 case discussed in (a)!
(c) [Fe(CN)6 ]3– The iron in this octahedral complex with six 1– cyanide ligands is in the 3+ oxidation state, and cyanide is
high on the spectrochemical series. This is almost certainly a complex with a low spin configuration. Fe3+ is a d 5 metal ion,
and hence the configuration will be t 52 g e0g , with S = ½. The LFSE is 5 × 0.4∆O – 0 × 0.6∆O = 2.0∆O .
(d) [Cr(NH3 )6 ]3+ The complex contains six neutral ammonia ligands, so it is Cr3+, which is a d 3 metal ion. There is only one
electron configuration for octahedral d 3 , i.e. t 32 g e0g , which means that there are three unpaired electrons and S = 3/2. The
LFSE is 3 × 0.4∆O – 0 × 0.6∆O = 1.2∆O
(e) [W(CO)6 ] Carbon monoxide is neutral and is very high on the spectrochemical series. It is a strong field ligand. The
metal is therefore in the neutral oxidation state. However, in the complex we promote the 4s electrons into 3d, so that it is
nonetheless a d 6 metal atom. We expect a low spin complex, with configuration t 26g e0g , and S = 0. The LFSE is 2.4∆O .
(f) [FeCl 4 ]2– This is a tetrahedral complex, because the ligands are –1, the metal is therefore 2+, and Fe2+ is d 6 , which is not
a candidate for square planar geometry. Remember that all tetrahedral complexes are high spin because ∆T is inherently
small {∆T = (4/9) ∆O }. Thus the expected electron configuration is e3 t23 , and S = 2.0 for four unpaired electrons, and strongly
paramagnetic. The LFSE is 3 × 0.6∆T – 3 × 0.4∆T = 0.6∆T.
(g) [Ni(CO)4 ] This is also a tetrahedral complex, because the ligands are neutral, the metal is (after promotion of the 4s
electrons) d 10 , and hence not a candidate for square planar geometry. Of course, regardless of geometry, complexe s of d 10
metal atoms and ions will always have LFSE = 0, and this complex is no exception.
4.
Estimate the spin-only contribution to the magnetic moment for each complex discussed in the previous exercise.
The formula for the spin-only moment is µSO = [n × (n + 2)]1/2 . Therefore the spin only contributions for these complexes
are:
Complex
n
µSO = [n × (n + 2)]1/2
3+
[Co(NH3 )6 ]
0
0.0
[Fe(OH2 )6 ]2+
4
4.9
[Fe(CN)6 ]3–
1
1.7
[Cr(NH3 )6 ]3+
3
3.9
[W(CO)6 ]
0
0.0
[FeCl 4 ]2–
4
4.9
[Ni(CO)4 ]
5.
0
0.0
Both H– and PPh 3 are ligands of similar field strength high in the spectrochemical series. Recalling that phosphines act as
π-acceptors, is π-acceptor character required for strong-field behaviour? What orbital factors account for the strength of
each ligand?
Hydrogen can only form σ-bonds, so it cannot participate in π-backbonding. Therefore its high placing in the
spectrochemical series cannot be due to π-acceptor character. We recall that the covalent bond in H–H is the strongest
known single bond. This suggests that hydride may form very strong σ-bonds to transition metals. If we remember how we
developed the bonding pictures for octahedral metal complexes (to take the example we are most familiar with; the same is
true for other geometries), we saw that the separation we call ∆O developed as a result of the destabilization of 2eg that
occurred as an anti-bonding mirror of the bonding 1eg level formed by two of the ligand SAO’s overlapping with the metal
d x2 -y2 and d z 2 orbitals. Thus strong σ-bonding should lead to large values of ∆O . Another cause of a large ∆O was the
relative destabilization of the 1t2g level caused by π-backbonding to a ligand with suitable empty π-orbitals, and that is what
ligands such as phosphines, carbon monoxide, etc. do. Since the spectrochemical series is determined solely by the relative
values of ∆O a ligand induces for a given metal ion, it is quite possible for hydride and phosphine ligands to have the same
“ligand field strength” even though these are from completely different bonding effects. This is indeed the case.
The figures below show at left a σ-only octahedral bonding scheme, and at right the influence of π-bonding:
6.
Solutions of the complexes [Co(NH3 )6 ]2+, [Co(OH2 )6 ]2+ and [CoCl4 ]2– are coloured. One is pink, another is yellow, and
the third is blue. Considering the spectrochemical series and the relative magnitudes of ∆T and ∆O try to assign each
colour to one of the complexes.
[Co(NH3 )6 ]2+, [Co(OH2 )6 ]2+ and [CoCl4 ]2– . First we must establish the complement of the light that is absorbed, and for this
we need a colour wheel. Remember that a coloured solution of a metal complex absorbs the photons of white light that are
not transmitted, and it is the absorbed photons that we really need to understand.
Thus we see that the complement of pink (red) is cyan, the complement of yellow is blue and the complement of blue is
yellow. In terms of increasing photon energy, yellow < cyan < blue. We therefore can expect that the complexes with the
lowest ∆ will be blue, followed by the pink and lastly the yellow with the largest ∆.
Now we consider the expected ∆ values. Obviously [CoCl4 ]2– will have the smallest ∆, since ∆T << ∆O [therefore it is the
blue complex]. Then comes [Co(OH2 )6 ]2+ [therefore this is the pink complex] followed by [Co(NH3 )6 ]2+ [which is thus the
yellow complex], since for these two octahedral complexes, ammonia is higher on the spectrochemical series than is water.
Although these are reasonable assignments (and correct in this case), in general colour can be quite deceiving, and a complex
may have more than one band in different regions of the colour spectrum, and the blended colour can be quite deceiving.
7.
A neutral macrocyclic ligand with four donor atoms produces a red diamagnetic low-spin d 8 complex of Ni(II) if the
anion is the weakly coordinating perchlorate ion. When perchlorate is replaced by two thiocyanate ions SCN– , the
complex turns violet and is high-spin with two unpaired electrons. Interpret the change in terms of the structure.
The perchlorate ion either does not coordinate, or provides a sufficiently weak ligand field, that the complex with this anion
is effectively square planar. In square planar d 8 complexes (D4h symmetry), all 8 electrons are paired, and the complex will
be diamagnetic. However, when strong axial ligands are introduced, the electronic structure approximates that of an
octahedral complex, and octahedral d 8 is paramagnetic with two unpaired electrons.
square planar limit
octahedral limit
x2-y 2
z2,x2-y 2
xy
z2
xz,yz,xy
xz,yz
8.
Bearing in mind the Jahn-Teller theorem, predict the structure of [Cr(OH2 )6 ]2+.
The main consequence of the so-called first-order Jahn-Teller theorem for our study is that a nonlinear molecule or ion with
an orbitally degenerate ground state is not as stable as a distorted version of the molecule or ion whenever the distortion is
capable of removing the degeneracy (i.e. by symmetry lowering). We expect [Cr(OH2 )6 ]2+ to be high spin d 4 , and thus will
have the configuration t 32 g e1g , which is orbitally degenerate in that the single electron can be either in the d z 2 or the d x2 -y2
orbital. Therefore by the Jahn-Teller theorem, the complex should not have full Oh symmetry. A tetragonal distortion
whereby two trans metal-ligand bonds are elongated and the other four shortened removes the degeneracy, and lowers the d z 2
and raises the d x2 -y2 orbital energy. The single electron will thus occupy the d z 2 orbital.
This is the most common distortion observed for octahedral complexes of high-spin d 4 , low-spin d 7 , and d 9 metal ions, all
of which possess eg degeneracy and exhibit measurable Jahn-Teller distortions. Such distortions are generally quite mild,
with the exception of Cu(II), d 9 , which can often distort well towards a four coordinate geometry. This is why we
generally attribute only four water ligands to the inner coordination sphere of the Cu 2+ aqua ion, even though this is not a
tetrahedral complex. In fact, it is tetragonally distorted octahedral, with four close and two further away water ligands.
9.
The spectrum of d 1 Ti3+(aq) is attributed to a single electronic transition eg ← t2g . The band shown in Fig. 10 in chapter 7
is not symmetrical and suggests that more than one state is involved. Suggest how to explain this observation using the
Jahn-Teller theorem.
Generally distortions of the t2g level are less significant than from the eg level. This
may lead to the conclusion that d 1 Ti3+(aq) should not be distorted. However, in the
electronic spectrum we are measuring the transition eg ← t2g , which results in a shortlived electronic exited state which has a t 20g e1g configuration. This suggests that
distortion can occur in the excited state. Therefore the “single” electronic transition is
in reality the superposition of two transitions, one to the symmetrical excited state, the
other to a distorted excited state, and the result is an asymmetric absorption peak in
the electronic spectrum of the Ti3+ aqua ion. The spectrum shown at right is not quite
realistic, and under-emphasizes the distorted shape of the peak.
10. Consider the effect of π-acid ligands on the energy levels of a tetrahedral complex. Starting from the σ-only bonding
diagram developed in lecture and in the text, create a bonding diagram in which all four ligands are π-acid ligands. Use your
diagram to interpret the UV-PES spectrum of Ni(CO)4 shown below. You may use HyperChem in the PM3 method to study
the bonding in this complex. However, this includes many orbitals that are primarily centred on the carbonyl ligands, and
may be more complex than desirable. Most likely a combination of paper and electronic models will be most helpful to you
here. Hint: what is the d n configuration of this complex?
The approach here has been to start with a standard tetrahedral ligand-field splitting diagram only. There are obviously four
other orbitals below the ones indicated, holding in all 8 electrons, which with the 10 shown here gives the 18 electrons in
Ni(CO)4 . These four orbitals are exactly of the type we developed in class for methane, e.g. standard σ-bonding orbitals.
Now we add the π-symmetry orbitals on the ligands, of which there are exactly four, and these can interact with the metal d
orbitals in the e set, only., as shown in the diagram. It is sort of a sideways overlap with, for example, both an axial lobe and
the equatorial toroid of the d z 2 orbital. The effect of using the empty 2e orbitals to do this interaction is to create a new 2e
orbital which has bond character, and a new 3e level which is antibonding. The net effect, as in the octahedral case, is to
increase ∆. In Ni(CO)4 , all the d orbitals are filled, and the compound is colourless. However, UV-PES allows us the
opportunity to measure ∆T directly. As shown in the PES at the right, there is a big peak at 9 eV, followed by a smaller peak
at about 10 eV. That 1 eV gap is in fact ∆T in this complex. The first peak is then for ionization from t2 , and the second from
e. The bands above 14 eV are lone pair and C—O bonding orbitals on the carbonyl ligands. Notice how well separated these
bands are from the ligand field bands. All in all, this is a marvellous confirmation of the ideas we have been talking about all
this time.
tetrahedral complex with pi bonding
electron occupancy for d10
tetrahedral metal
complex, sigma only
xy, xz,yz
e
ligands
xz,yz
3e
2e
xy, xz,yz
t2
t2
z2,x2-y2
e
z2,x2-y2
2e
1e
11. The UV-PES of W(CH3 )6 is shown below. The bands at greater than 11 eV are due to bonding within the methyl groups,
and not the metal-ligand bonds. Interpret the spectrum based on the σ-only octahedral bonding model. You may use
HyperChem in the PM3, but will need to substitute Mo for W! Also, because the methyl groups are not rotating, your
model will not display the true octahedral symmetry of this complex. Use with caution! Hint: what is unusual about this
complex?
I certainly did not get very far trying this in HyperChem. Somehow the orbital energies do not fit in a sensible order. Most
likely PM3 is simply not set up to handle structures of this level of complexity. However, we can discuss what happens very
easily from the standard Oh bonding picture shown here. Note that methyl ligands can be treated as neutral one-electron
ligands. Neutral tungsten has 6 electrons (by promoting the 6s electrons to 5d), so we end up with 6 electrons from the metal
and 6 electrons from the methyl radicals. Note that this allows us to fill up only the first three types of levels, a 1g , t iu and eg .
This is most unusual – there are no d electrons, which is indeed the case. The fact remains, however, that we observe three
peaks in the UV-PES that correspond directly to these three orbitals. This is an example of a d 0 metal atom. Not
surprisingly, W(Me)6 is a strong oxidizing agent – it very much wants those extra 6 electrons to fill up the 1t2g levels.
12. Practice with Tanabe-Sugano diagrams. In Appendix 5 to Shriver-Atkins-Langford are Tanabe-Sugano diagrams. Use these
diagrams to assign the spectra of the aqua ions given below. Where possible, esitmate the value of B. Use your results to
construct a “metal series” for ∆O using the water ligand. Assume all are octahedral, but remember that a Jahn-Teller
distortion is possible, but not always observed, for all ions with a T1g or a T2g ground term.
13. The free gas-phase ion V3+ has a 3 F ground term. The 1 D and 3 P terms lie respectively 10 642 cm–1 and 12 920 cm–1 above it.
The energies of the terms are given in terms of Racah parameters as E(3 F) = A – 8B, E(3 P) = A + 7B, E(1 D) = A – 3B + 2C.
Calculate the values of B and C for V3+.
The diagram at right shows the relative energies of the 3 F, 1 D and 3 P terms. It can be seen that 10 642 cm–1 energy gap
between the 3 F and 1 D terms is 5B + 2C, while the 12 920 cm–1 energy gap between the 3 F and 3 P terms is 15B. From the two
equations:
5B + 2C = 10 642 cm–1
15B = 12 920 cm–1
you can determine that B = 12 920 cm–1 / 15 = 861 cm–1 and C = 3170 cm–1 .
14. Write the d orbital configurations and then use the Tanabe-Sugano diagrams in the appendices to identify the ground
electronic term of (a) low-spin [Rh(NH3 )6 ]3+, (b) [Ti(OH2 )6 ]3+, (c) high-spin [Fe(OH2 )6 ]3+.
(a) Low-spin [Rh(NH3 )6 ]3+ From the electron configuration for neutral Rh (5s2 4d 7 ) we get 9 electrons; 3+ reduces that to 6,
so it is a d 6 ion. From the T-S we read off the ground state from the bottom line to the right of the vertical divider for high
and low spin. So for this ion it is 1 A 1g . We supply the “g” because it is an octahedral complex.
(b) [Ti(OH2 )6 ]3+ From the electron configuration for neutral Ti (4s2 3d 2 ) we get 4 electrons, remove 3 give d 1 . There is no
low/high spin distinction. There is no T-S diagram. But we know that the ground state for d 1 is always 2 T2g (See question 1).
(c) High-spin [Fe(OH2 )6 ]3+ From the electron configuration for neutral Fe (4s2 3d 6 ) we get 8 electrons, so as 3+ it is a d 5 ion
in high spin form. The T-S diagram to the left of the vertical dividing line then indicates it to have an 6 A 1g ground state.
15. Using the Tanabe-Sugano diagrams in the appendices, estimate ∆O and B for (a) [Ni(OH2 )6 ]2+ (absorptions at 8500, 15 400,
and 26 000 cm–1 ) and (b) ) [Ni(NH3 )6 ]2+ (absorptions at 10 700, 17 500, and 28 200 cm–1 ).
(a) [Ni(OH2 )6 ]2+ According to the d 8 T-S diagram, the
absorptions can be assigned as follows:
3
8500 cm–1
T2g ←3 A 2g
–1 3
15 400 cm
T1g ←3 A 2g
–1 3
26 000 cm
T1g ←3 A 2g
The ratios 13,800/8,500 = 1.6 and 25,300/8500 = 3.0 can be used
to estimate ∆O /B ≈ 11. Using this value we calculate ∆O = 8,500
cm–1 and B ≈ 770 cm–1 . Note that B for a gas-phase Ni2+ ion is
1080 cm–1 . The fact that B for the complex is only ~70% of the
free ion value is an example of the nephelauxetic effect.
(b) [Ni(NH3 )6 ]2+ The absorptions for this complex are 10 700, 17
500, and 28 200 cm–1 with the same assignment as in (a). The
ratios are 1.6 and 2.6, leading to ∆O /B ≈ 15. Thus calculate ∆O =
10,750 cm–1 and B ≈ 720 cm–1 . It is sensible that B for the
ammine complex is lower than for the aqua complex, since NH3
is higher in the nehpelauxetic series than is H2 O.
16. Explain why [FeF6 ]3– is colourless whereas [CoF6 ]3– is coloured,
but exhibits only a single band in the visible region of the
electronic spectrum.
Both are M3+, so the iron is d 5 , while the cobalt is d 6 . With F– as
the ligand, very weak field, we expect the complexes to both be
high spin. In high spin d 5 , to the left of the dividing line (see
question 1), the ground state is 6 A 1g , and there are no sextet
excited states. All transitions are therefore spin-forbidden, and
the complexes will be extremely pale, and it is not surprising that
they appear colourless in dilute solution. However, d 6 from the
T-S diagram is capable of a single spin-allowed d-d transition,
5
Eg ←5 T2g . This means the complex will have significant colour
even in relatively dilute solution.
17. The Racah parameter B is 460 cm–1 in [Co(CN)6 ]3– and 615 cm–1 in [Co(NH3 )6 ]3+. Consider the nature of the bonding with
the two ligands and explain the difference in nephelauxetic effect.
These two ligands are quite different with respect to the types of bonds they form with metal ions. Ammonia and cyanide are
both σ-bases, but cyanide is also a π-acid. This difference means that NH3 can form molecular orbitals only with the metal eg
orbitals, while cyanide can form molecular orbitals with the metal eg and t2g orbitals. The formation of molecular orbitals is
the way that electrons are delocalized, and this is precisely what the nephelauxetic effect measures.
18. [Cr(OH2 )6 ]3+ ions are pale blue-green but chromate ion [CrO4 ]2– is an intense yellow. Characterize the origins of the
transitions and explain the relative intensities.
[Cr(OH2 )6 ]3+ ions have a d 3 configuration, while [CrO4 ]2– has a d 0 configuration. Thus in the Cr(II) aqua ion, there is the
possibility of spin-allowed, but Laporte-forbidden d-d bands in the electronic spectrum. However, such transitions are not
possible for the Cr(VI) chromate species. The colour of this ion must be due to LMCT bands (e.g. from the lone pair oribitals
on oxygen into the empty metal d orbitals. Such transitions are fully allowed, and this accounts for the intense colour of the
chromate ion.
19. Name the species and draw the structures of (a) Fe(CO)5 , (b) Ni(CO)4 , (c) Mo(CO)6 , (d) Mn2 (CO)10 , (e) V(CO)6 , (f)
[PtCl3 (C2 H4 )]– .
The names of the metal carbonyls first specify the number of CO ligands, then the number and type of metal atom(s), and
finally the oxidation state of the metal(s). The structures of the metal carbonyls generally have simple, symmetrical shapes
that correspond to the CO ligands taking up positions that place them as far apart from one another as possible.
(a) Fe(CO)5
pentacarbonyliron(0)
Forms a regular trigonal bipyramid
(b) Ni(CO)4
tetracarbonylnickel(0)
Regular tetrahedron
(c) Mo(CO)6
hexacarbonylmolybdenum(0)
Regular octahedron
(d) Mn2 (CO)10
dodecacarbonyldimanganese(0)
The two octahedral fragments are joined, and rotated to be staggered.
(e) V(CO)6
hexacarbonylvandaium(0)
Regular octahedron
(f) [PtCl3 (C2 H4 )]– trichloro(ethene)platinate(II)
This is named just like the classical coordination complexes we have
considered earlier in this course. It is “square planar”, and this mo de maximizes the “side-on” overlap between the metal dorbitals and the π-acceptor orbitals on the ethylene. Because the ethylene group sticks up and down, the molecule itself is not
planar, but the geometry and bonding is that of a d 8 square planar group. Although the HyperChem model below does not
show this, it is possible to draw [PtCl3 (C2 H4 )]– in such a way that the ethylene ligand is attached at the midpoint of the C=C
bond. This is realistic, because that is where the filled π-orbital that is the σ-donor orbital is centred. Indeed, such
complexes are known to rotate in solution, quite literally “spinning” on the axis defined by the Pt atom and the midpoint of
the C=C bond.
(a)
(b)
(c)
(d)
(e)
(f)
20. Count the number of valence electrons in each of the complexes listed in question 19. Do any of them deviate from the 18electron rule? If so, is this reflected in their structure or chemical properties?
(a) Fe(CO)5
Fe has 8 valence electrons, and 5 CO contribute 10 electrons, for a total of 18.
(b) Ni(CO)4
Ni has 10 valence electrons, and 4 CO contribute 8 electrons, for a total of 18.
(c) Mo(CO)6
Mo has 6 valence electrons, and 6 CO contribute 12, for a total of 18.
(d) Mn2 (CO)10
Mn has 7 valence electrons, and 5 CO per Mn contribute 10 electrons, for a total of 17. An
additional electron is contributed by the Mn–Mn bond, for a total of 18. Another way to count this
dimeric compound is to calculate the whole cluster as follows: 2 Mn = 14 e– , 10 CO = 20 e– , one
Mn–Mn bond = 2 e– for a total of 36 e– for two metals. Therefore divide by 2 to get 18 e– per
metal atom.
(e) V(CO)6
V has 5 valence electrons, and 6 CO contribute 12, for a total of 17 electrons. This octahedral
complex is electron deficient. It can be made only because it is coordinatively saturated by the six
CO ligands which block rapid access to the V atom. However, it is much more reactive than its 18
e– analogues such as Cr(CO)6 . V(CO)6 is readily reduced to [V(CO)6 ]– , an 18 e– complex anion.
It is also very highly coloured, in sharp contrast to Cr(CO)6 and Ni(CO)4 which are colourless.
(f) [PtCl 3 (C2 H4 )]–
Pt has 10 valence electrons, but is in the +2 oxidation state, so 8 electrons. The three chloride
ligands contribute 6 electrons, and the ethylene ligand also contributes 2 electrons. This makes for
a total of 16 electrons. This is the “correct” count for a d 8 square planar complex.
21. Using the 18-electron rule as a guide, indicate the probable number of carbonyl ligands in (a) [W(η6 -C6 H6 )(CO)n ], (b)
[Rh(η5 -C5 H5 )(CO)n ], and (c) [Ru 3 (CO)n ].
(a) [W(η 6 -C6 H6 )(CO)n]
A W atom has 6 electrons, η6 benzene has 6 electrons, so to get 18 electrons needs 6 more,
therefore three CO ligands donating 2 electrons each are required. n = 3
(b) [Rh(η 5 -C5 H5 )(CO)n]
A Rh atom has 9 electrons when neutral, Cp contributes 5, so we need 4 more electrons,
contributed by two CO. n = 2.
(c) [Ru3 (CO)n]
This complex could be linear (i.e. Ru–Ru–Ru), with two Ru–Ru bonds, or triangular with three
Ru–Ru bonds. If it were linear, there would be no way to satisfy the 18-electron rule for the two
Ru atoms at the end, regardless of the value of n:
for all n, 18 ≠ 8 + 1 (for Ru–Ru bond ) + 2n
Therefore this must be a triangular cluster, with two Ru–Ru bonds per Ru atom. Therefore each
Ru has 8 electrons, the M–M bonds contribute 4 electrons, and 6 extra electrons are required per
Ru atom. Thus Ru(CO)3 × 3 = Ru 3 (CO)12 .