Download Concepts in Transition Metal Chemistry – Answers

Concepts in Transition Metal Chemistry – Answers
Chapter 1
1
To obtain the electronic configuration of transition metal ions, remove first the 4s
electrons and then the appropriate number of 3d electrons from the configuration
of the free atom. Thus Mn4+ has configuration [Ar]3d3 and Cu3+ has the
configuration [Ar]3d8.
2
Element A is zirconium and B is tin. The elements with four valence electrons in
Period 5 are zirconium, [Kr]4d25s2, and tin, [Kr]4d105s25p2. As expected from the
remark on conductivity, both are metals. A has much the higher melting
temperature so it must be a transition metal.
3
Ti2+(aq) is unknown (Table 1.3) and if the stability of M2+(aq) with respect to
M3+(aq) increases across the series, its non-existence is presumably due to its
reduction of H+(aq): Ti2+(aq) + H+(aq) = Ti3+(aq) + ½H2(g). This reaction is
thermodynamically favourable because E![Ti3+|Ti2+] is negative. If we imagine
TiCl2 first behaving like other dihalides, so that it forms M2+(aq) and Cl−(aq) in
water, the total reaction is TiCl2(s) + H+(aq) = Ti3+(aq) + 2Cl−(aq) + ½H2(g) and
the black powder should dissolve, evolving hydrogen and giving a purple solution
of Ti3+(aq). This is what happens.
4
Ni3+(aq) is not known and the trend in the relative stabilities of M2+(aq) and
M3+(aq) suggests that it is even less stable than Co3+(aq) with respect to reduction.
Presumably, therefore, its non-existence is due to oxidation of water, as occurs
with Co3+(aq): 2Ni3+(aq) + H2O(l) = 2Ni2+(aq) + O2(g).
According to the Appendix, this reaction is thermodynamically favourable because
E![Ni3+|Ni2+] > 1.23 V. Thus when the hydroxide is dissolved in acid, oxygen
should be evolved and the green colour of Ni2+(aq) should appear:
2NiO(OH) + 4H+(aq) = 2Ni2+(aq) + 3H2O(l) + ½O2(g)
5
Determining the oxidation state of Mn by allowing −1 for Cl, −2 for O etc. and
taking account of any overall charge, gives MnCl2 +2, Mn2(SO4)3 +3,
K2MnF6 +4, [MnO4]2− +6, Mn2O7 +7.
Then determine the electronic configuration for a manganese ion with charge
equal to the oxidation state: +2 3d5, +3 3d4, +4 3d3, +6 3d1, +7 3d0.
Chapter 2
1
(a)
Favourable; E![Mn3+|Mn2+] > E![Cr3+|Cr2+];
(b) favourable; E![Co3+|Co2+] > E![Mn3+|Mn2+];
(c)
favourable; E![S2O82−|2SO42−] > E![Mn3+|Mn2+];
(d)
favourable; E![S2O82−|2SO42−] > E![V3+|V2+].
Copyright © 2010 The Open University
2
E!(Cr3+|Cr2+) and E!(Fe3+|Fe2+) are less than 1.23 V so Cr2+(aq) and Fe2+(aq) are
thermodynamically unstable to oxidation by oxygen in acid solution;
E!(Mn3+|Mn2+) and E!(Co3+|Co2+) are greater than 1.23 V, so Mn2+(aq) and
Co2+(aq) are thermodynamically stable to this reaction. The oxidation of Cr2+(aq)
by oxygen is rapid at room temperature, but that of Fe2+(aq) is very slow, so
solutions of Fe2+(aq) can be handled in air without serious deterioration in most
laboratory experiments.
3
Fluorides are the most stable halides with respect to the reaction MX3(s) = MX2(s)
+ ½X2, so the fluoride is the logical choice for an attempted preparation.
4
Applying the stability trend observed in the trihalides, the oxides M2O3 should
become less stable with respect to the reaction:
½M2O3(s) = MO(s) + O2(g)
as one moves across the series, so a good guess is that the first seven elements
(Sc2O3, Ti2O3, V2O3, Cr2O3, Mn2O3, Fe2O3, Co2O3) form stable trioxides. Ni, Cu
and Zn do not.
5
For K2VO4, V is in oxidation state +6. To achieve this V would have to lose
electrons from the Ar core. For Fe2O7, the oxidation state is higher than the
maximum given in Figure 1.6, but does not exceed the number of valence
electrons. The d electrons are more tightly bound at this end of the transition series
and so not all the valence electrons are removed. ScCl2 is unstable with respect to
disproportionation to the metal and the trichloride.
Chapter 3
1
A chelating ligand is atached to a metal ion by two or more atoms. A bidentate
ligand is attached to a metal through two atoms, and a tridentate ligand through
three atoms. A bridging ligand is attached to two or more metal ions.
Hence the ligand in 3 is bidentate, the ligand in 2 is tridentate and both are
chelating. The ligand in 1 would be bidentate if both O atoms were attached to the
same metal ion. In 1it is a bridging ligand.
2
The O2 ligand is coordinated to two Co atoms and so is a bridging ligand labelled
µ. Only one O atom attaches to each Co and so the binding is µ-η1,η1.
3
For [CoBr2Cl2(en)]− there are four stereoisomers, two of which are optically active.
4
As the central –NH- group of dien must be adjacent to both terminal –NH2 groups, two
arrangements are possible. In the facial isomer, denoted fac-[CoCl3(dien)], the three
nitrogens occupy the corners of one face of the octahedron. In the meridional isomer,
denoted mer-[CoCl3(dien)], they occupy three positions on a meridian of the
octahedron. Both stereoisomers have planes of symmetry so neither is optically active.
5
The complex will adopt an octahedral configuration.
Ethoxy ligands trans to each other: Since benzylacetonate is unsymmetrical, there
are two isomers one with the O atoms nearest the phenyl group trans and one with
these two O atoms cis. Neither are optically active.
Ethoxy ligands cis: both O atoms nearest phenyl can be trans to ethoxy or cis to
ethoxy or one can be trans and one cis. These isomers do not have a plane of
symmetry and so are optically active.
This makes a total of 8 stereoisomers.
2
Chapter 4
1
The Nernst equation tells us that the electrode potential in the presence of a ligand
that forms complexes with the metal ions is given by
E!comp = E!(M3+(aq)|M2+(aq)) − (0.0592V) log(K!+3/ K!+2)
For this siderophore, (K!+3/ K!+2) = (1.32 × 1039/5.01 × 1016) = 2.63 × 1022.
Thus log(K!+3/ K!+2) = 22.4.
E!comp = (0.77 − 0.0592 × 22.4) V = −0.56 V
2
The oxygen electrode potential varies with pH as follows
E/V = 1.23 − 0.0592 pH
Hence at pH 6.6 E is given by 1.23 − 0.391 = 0.84 V. From the table of electrode
potentials, this would be sufficient to oxidise Fe2+ and Cr2+.
3
SCN− binds through S to soft acids and N to hard acids. The classification of most
of these metal ions is given in Table 4.1. Gd3+ is a lanthanide ion and thus is a hard
acid.
4
Cysteine binds to metal ions through S and is thus a soft base. Ca2+, Fe3+, Al3+ and
Mn2+ are hard acids. Ag+ and Pt4+ are soft acids. Cysteine will bind to the soft
acids.
5
Oxide is a very hard base, much harder than sulfide. In the geological processes
that led to the formation of mineral deposits, oxide bearing ligands should
therefore be selected by hard acids. Of the six metals mentioned, beryllium,
lanthanum and scandium have ions classified as hard acids. These are the elements
found in oxide deposits. The ions Cd2+ Ni2+ and Ag+ are classified as either
borderline hard-soft or soft acids, and these metals usually occur as sulfide ores.
Chapter 5
1
Levels for octahedral complexes are labelled t2g and eg.
[Cr(CN)6]3− contains Cr in the +3 oxidation state; it is therefore d3. The three
electrons will go into the lower energy levels with parallel spins.
[Fe(CN)6]3− contains Fe in oxidation state +3; this is therefore a d5 complex. CN is
a strong field ligand and thus electrons will pair up in the lower, t2g, energy levels
before filling the upper, eg levels.
2
For a tetrahedral complex, the d levels are split into two lower ones labelled e and
three higher energy ones labelled t2. Note that a tetrahedron does not have a centre
of symmetry so these labels do not have g. Tetrahedral complexes are generally
high-spin so place one electron in each orbital before considering pairing.
Fe2+ has the electronic configuration [Ar]3d6. There are therefore 6 electrons to fit
into the boxes. First put one in each d orbital with spins parallel, then add the sixth
to one of the e orbitals.
V3+ has the electronic configuration [Ar]3d2. There are therefore 2 electrons to fit
into the boxes. These will go into the e level with spins parallel.
3
Co2+ has the electronic configuration [Ar]3d7. There are therefore 7 electrons to fit
into the boxes. Four go into e with paired spins and the other three go into t2 with
parallel spins.
3
In octahedral complexes, only those ions with the same number of electrons in
each d orbital will have a radius equal to that expected for a spherical ion. In highspin complexes these ions will have an electronic configuration of d0, d5 or d10. Of
the ions given, only Sc3+ (d0) and Fe3+ (d5) have one of these configurations.
4
F− ions can generally be assumed to provide a weak-field environment.
Electrons in t2g levels contribute 2/5Δo to the CFSE and electrons in eg contribute
−3/5Δo.
VF3: V3+ has a d electron configuration of d2 and hence a CFSE of 4/5Δo in an
octahedral environment.
CrF3: Cr3+ has a d electron configuration of d3 and hence a CFSE of 6/5Δo in an
octahedral environment.
MnF3: Mn3+ has a d electron configuration of d4 and hence a CFSE of 3/5Δo in an
octahedral environment.
FeF3:Fe3+ has a d electron configuration of d5 and hence a CFSE of 0.
CoF3: Co3+ has a d electron configuration of d6 and hence a CFSE of 2/5Δo in an
octahedral environment.
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cis-[CoF2en2]+ does not have a centre of symmetry but trans-[CoF2en2]+ does.
d↔d transitions will therefore be more forbidden for trans-[CoF2en2]+ as the rule
g↔u will be broken. Consequently the d↔d bands for the cis complex will be less
intense than for the trans complex.
6
A Jahn–Teller distortion occurs when the truly octahedral complex would be in a
degenerate state. Only complex B in which the Mn3+ ion will have a d4
configuration would be degenerate if truly octahedral. This complex distorts so
that the eg levels are not degenerate and the energy of the complex is lowered.
7
Mn2+ has the d electronic configuration d5. It therefore has zero CFSE in both
tetrahedral and octahedral sites. Mn3+ is d4. In a tetrahedral environment it has a
CFSE of 2/5Δt. In an octahedral environment its CFSE is 3/5Δo. Mn3+ ions
therefore favour octahedral sites. With the trivalent ion in the octahedral holes,
Mn3O4 will be a spinel.
8
In weak-field complexes, the electrons first fill all the d levels with parallel spins.
The sixth electron then pairs up in one of the lowest energy orbitals. In strong-field
complexes, the electrons fill the lowest energy orbitals with paired spins.
9
A moment of 4.8 µB would correspond to 4 or 3 unpaired electrons. The most
stable oxidation states of Fe are +2 and +3. The +2 state has an electronic
configuration of d6 and the +3 state a configuration of d5. In a weak field these
would have 4 and 5 unpaired electrons respectively. For a strong field octahedral
environment, there would be 0 and 1 unpaired electrons. From the size of the
magnetic moment, the Fe is in a weak-field environment consistent with being
surrounded by O atoms. A moment of 4.8 µB is closer to the spin-only value for 4
electrons than that for 5 electrons so the oxidation state is +2.
4
10
A magnetic moment of 1.80 µB corresponds to one unpaired electron. The d
electron configurations of the oxidation states given are +6 d0, +5 d1, +4 d2, +3 d3
and +2 d4. Thus oxidation states +6, +4 and +2 cannot have 1 unpaired electron.
There could be 1 unpaired electron for +3 but only if it were a low-spin tetrahedral
complex. Tetrahedral complexes are rarely low-spin and it is particularly unlikely
with ligands coordinating through O as these tend to be weak-field. Thus Cr is
likely to be in oxidation state +5.
Chapter 6
1
Orbital combinations that will form a bonding overlap with a 4s orbital of positive
sign must be positive wherever they overlap with 4s. Thus A and E will form
bonding combinations, B, C and D have both positive and negative orbitals
overlapping and so will be non-bonding, F will form an antibonding orbital.
2
Strong-field ligands such as CO, have empty π* orbitals. These will form
molecular orbitals with t2g orbitals on the metal. This strengthens the bonding in
the complex by lowering the energy of the t2g orbitals. Because these orbitals are
lower in energy, the energy gap between t2g and eg* is increased. This gap can be
identified with the ligand field splitting energy. Increasing the ligand field splitting
energy makes the complex strong-field.
3
Ligands with partially-full 2π* orbitals will give a bent configuration. Those with
empty 2π* orbitals will form a linear arrangement. Thus NO− and O2− will give
bent configurations.
4
A, C and E obey the 18-electron rule.
For A, [PtCl6]2−, Pt will contribute 10 electrons, each Cl contributes 1 and 2 have
to be added for the charge. This makes 10 + 6 + 2 = 18.
For B, [Pt(NH3)4]2+, Pt contributes 10, each NH3 donate 2 and 2 is subtracted to
allow for the charge. This makes 10 + 8 − 2 = 16.
For C, [Fe(CO)5], Fe contributes 8 and each CO donates 2. This makes 8 + 10 =
18.
For D, [Ni(CN)4]2−, Ni contributes 10, CN is counted as CN rather than CN− and
thus each contributes 1 electron. The charge adds another 2 giving 10 + 4 + 2 = 16.
For E, [Co(NH3)4(CH3)(cysteine)]2+, Co contributes 9 electrons, each NH3
contributes 2, CH3 contributes 1 and cysteine contributes 2. With the charge, this
gives 9 + 8 + 1 + 2 − 2 = 18.
5
Charge transfer transitions in metal complexes generally have much higher
intensities than d↔d transitions and lie at higher wavenumbers.
Peak 3 at 23 290 cm−1 with ε = 2020 dm3 mol−1 cm−1 is therefore likely to be a
charge transfer transition.
The other two peaks are due to d↔d transitions. Square planar complexes have a
centre of symmetry and so their d↔d peaks will be weaker than those for the
orbital
square pyramidal complex. In addition the energy gap between the
and the next highest is greater in the square planar complexes (Figure 5.18) and so
the peak will be at a higher wavenumber. Thus peak 1 is from the square
pyramidal oxidised form and peak 2 from the square planar reduced form.
5