Download Consider the diamagnetic complex, [Os(NH3)5(CO)]

Consider the diamagnetic complex, [Os(NH3)5(CO)]2+.
(i)
For the carbonyl complex, give the oxidation state and the d-electron configuration for
the Os centre, draw appropriate d-orbital energy diagrams, and illustrate the
distributions of the d electrons within them.
As NH3 and CO are both neutral ligands, the Os centre is in a +2 oxidation state. As Os is in
Group 8, Os2+ has (8 – 2) = 6 valence electrons and the d-electron configuration is d6.
With a d6 Os2+ centre, 5 two-electron donor NH3 ligands and 1 two-electron donor CO ligand,
the total electron count is (6 + 5 × 2 + 1 × 2) = 18 e-.
If the complex is considered to be octahedral, the d-orbitals split as shown on the left of the
diagram below. NH3 is a medium-field ligand and CO is a strong-field ligand. Os2+ (like all 4d
and 5d elements) gives large ∆oct values and (due to the large size of its d-orbitals) low pairing
energies. The complex is therefore low spin, giving a diamagnetic arrangement for d6.
dz2
eg
dz2
dx2 -y2
∆oct
dx2 -y2
t2g
dxz
dyz
dxy
dxy
dxz
CO
H3N
Os
H 3N
2+
CO
NH3
H3N
NH3
H 3N
NH3
splitting in an
octahedron
dyz
Os
2+
NH3
NH3
NH3
splitting recognizing stronger σdonation and π-back bonding by CO
(Actually, CO is a stronger field ligand than NH3. the dz2 orbital is made more antibonding by
the extra σ-donation by CO. The dxz and dyz are able to π-donate into the π* orbitals of CO (as
described below) and become more bonding. This gives the splitting on the right-hand side of
the diagram. This does not affect the diamagnetic configuration.)
ANSWER CONTINUES ON THE NEXT PAGE
(ii)
With the aid of orbital diagrams, describe the nature of the Os-CO bond in
[Os(NH3)5(CO)]2+.
The HOMO of CO is the 2pσ orbital which is polarized towards C. CO uses this orbital to σdonate. It is a poor σ-donor (poor base) to non-transition metal cations (including H+) and to
transition metals with a d10 configuration. The σ-donation is into an empty metal orbital of the
correct orientation (such as dz2 and dx2-y2 in an octahedral complex):
OC: Æ M σ-donation
The LUMOs of CO are the pair of 2pπ* orbitals. These are again polarized towards C. CO
uses these orbitals to π-accept electron density from a suitable orientated d-orbital on a
transition metal (such as dxz, dyz and dxy in an octahedron. This mode of bonding involves M
Æ CO donation or π-back donation. It requires the presence of d-electrons on the metal and
good overlap between the metal and CO orbitals. The latter occurs only if the metal is in a low
(≤ +2) oxidation. Higher oxidation states lead to the d-orbitals being too contracted for overlap
to be efficient. As the donation is into the π* orbitals on CO, it reduces the strength of the
C≡O bond. This lengthens the C≡O bond and reduces its stretching frequency. In principle,
the π-back donation could decrease the CO bond order to two or less*.
M Æ CO π-back donation into the two π* orbitals on CO
In the absence of each other, both these modes of bonding are fairly weak interactions as CO
is intrinsically a poor σ-donor and poor π-acceptor. However, in both modes operate, they
synergically enhance each other: the σ-donation leads to a build up of electron density on the
(electropositive) metal. The π-back donation acts to decrease this electron density so any σdonation enhances the π-bonding. The π-back bonding leads to an increase in electron density
on the (electropositive) carbon atom which in turn enhances its ability to σ-donate.
The result is a strong M-C bond with significant σ and π-components. As the π-back donation
weakens the C≡O bond, the final electron distribution is a thermodynamic compromise
between strengthening the M-C at the expense of the C≡O bond.
*
This weakening of the internal bonding of the ligand makes it more reactive: this is a common theme in complexes
involving π-acceptor ligands and a strong reason for interest in this part of inorganic chemistry.
(iii) The Os(III/II) redox potential of [Os(NH3)5(CO)]3+/2+ is 1.7 V more positive than that of
[Os(NH3)6]3+/2+. Briefly explain these results in terms of the natures of the bonding of the
CO and NH3 ligands to metal ions.
The redox potentials are for the reduction processes:
[Os(NH3)5(CO)]3+ + e- Æ [Os(NH3)5(CO)]2+
[Os(NH3)6]3+ + e- Æ [Os(NH3)6]2+
As described in above in part (ii), the Os-CO bond involves π-back donation from the metal
into the π* orbitals of the CO bond. Os3+ has a d5 configuration and more contracted orbitals
than the d6 Os2+ due to its higher charge. The orbital overlap between Os2+ and CO is therefore
better and the higher number of d-electrons leads to more π-back donation. ([Os(NH3)5(CO)]2+
is a 18 e- complex whereas [Os(NH3)5(CO)]3+ is a 17 e- complex). CO therefore favours the
lower osmium oxidation state, making the reduction potential more positive.
NH3 is a σ-donor and cannot π accept. The σ-donation stabilizes the higher oxidation state,
making the reduction potential less positive.
What is the Mo-Mo bond order in the complex [Mo2(CH3CO2)4]?
σu
πg
δu
δg
πu
The acetate ligand, CH3CO2-, has a charge
of -1 so the oxidation state of each metal
is +2:
[Mo2(CH3CO2)4] ≡ 2Mo2+ +
4CH3CO2-
σg
As molybdenum is in Group 6, Mo2+ has (6 – 2) = 4 d-electrons. The two Mo2+ ions have (2 × 4) =
8 electrons filling (σ)2(π)4(δ)2 and giving a bond order of 4: a quadruple bond.
What is the Os-Os bond order in the cluster
Os4(CO)12?
For this carbonyl complex, it can be assumed that
each metal obeys the 18 e- rule.
Each Os has 8 valence electrons and each CO
donates 2 electrons to the cluster. The total electron
count for the cluster is therefore:
(4 × 8 (Os)) + (12 × 2 (CO)) = 56
56
= 14 e-.
3
To achieve a 18 e configuration requires each Os to make a further 4 bonds. As each Os is one the
corner of the tetrahedron, it makes three Os-Os connections. Each of these must have a bond order
4
of .
3
As there are 4 Os atoms, each receives