Download 2H + CO3 H2CO3 H2O + CO2(g) H3N Co H3N NH3 OH2 Cl Cl H3N

PS5 Solutions
Problem 7.1. Co(II)CO3 + NH4Cl(aq) + air (O2) ----> [Co(NH3)4(CO3)]Cl
the carbonate product is pink and contains bidentate carbonate.
[Co(NH3)4(CO3)]Cl + 2HCl(aq) ----> CO2(g) + H2O + [Co(NH3)4Cl2]+(aq)
The cobalt-chloride product is violet and could exist as a cis or trans isomer.
Since the analogous chiral en complex, which must contain cis chlorides, is also
violet, it is likely to be cis.
[Co(NH3)4Cl2]+(aq) + conc. HCl ---> [Co(NH3)3Cl3]•NH4Cl
Here the product is green and could exist as the fac or mer isomer. Later we'll
see why the mer isomer is formed.
Comment on the reaction that generates gas (carbon dioxide): While Co(III) in
acidic conditions can oxidize water to generate O2(g), (a) with the ammine
ligands its oxidation potential is lowered so it can't oxidize water, and (b) all the
subsequent compounds are Co(III) complexes, suggesting another gas-forming
reaction:
2H+ + CO32-
H2CO3
H2O + CO2(g)
Problem 7.4. The starting Co complex must be neutral to explain the lack of
conductivity. Assuming the ammonia ligands didn't migrate during the reaction,
the original compound must have the structure shown below (mer isomer).
Cl
H 3N
H 3N
Co
Cl
product
NH3
OH2
NO2
H 3N
NH3
Co
H 3N
NO2
O2N
starting material
[Co(NH3)3(NO2) 3](aq) + HCl ---> trans-[Co(NH3)3(H2O)(Cl)2]
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Problem 7.5. [ZrCl4(dppe)] + 2MgMe2 ----> [ZrMe4(dppe)]
Me
P
Me
Me
Zr
Me
P
Me
"octahedral"
Me
P
Zr
Me
P
"trigonal prism"
Me
In the "octahedral" geometry there are 2 types of different Me groups (that is,
they cannot be interconverted by a symmetry operation); these are cis and trans
to a P, respectively.
However in the "trigonal prism" structure all the Me groups are equivalent, so this
is consistent with the NMR data.
In both cases it is useful to consider a C2 axis that includes Zr and bisects the
dppe CH2CH2 backbone.
2. (a) [Cr(H2O)4Cl2]Cl cis and trans isomers for the cation; 6-coordinate,
octahedral Cr(III)
(b) K2[OsCl5N] octahedral Os(VI)
(c) (Ph4As)2[PtCl2(Me)(H)] square planar Pt(II); cis and trans isomers
(d) Ru(PEt3)3H3 mer isomer; octahedral Ru(III)
(e) [(NH3)5Co-NH2-Co(NH3)4(H2O]Cl5 Two octahedral Co(III)s with a single
bridging NH2
(f) [(NH3)2Pt(µ-Cl)2-Pt(NH3)2]Cl2 Two square planar Pt(II)s with 2 bridging
chlorides.
3. For this formula, the octahedral geometry can yield only 2 isomers (cis and
trans). However, the trigonal prism can have 3 isomers:
Cl
Cl
Cl
Cl
Cl
Cl
4. In addition to the usual isomers (cis and trans) of [M(AA)2e2] molecules we also
have linkage isomers of SCN–. In the figures below en is abbreviated as N~N.
Note that the pairs D and E, F+G, and H+I are enantiomers.
2
A
B
C
SCN
N
N
Co
N
N
NCS
NCS
N
N
Co
N
N
SCN
SCN
N
Co
N
N
SCN
E
D
N
N
N
N
Co
N
SCN
NCS
N
NCS
N
SCN
N
SCN
N
G
N
N
Co
N
SCN
N
NCS
N
SCN
Co
N
N
NCS
N
N
Co
N
F
I
H
N
Co
Note: all are cations
N
N
N
SCN
NCS
Co
N
N
NCS
5. (a) The trans isomer should have no dipole moment (all metal-ligand bond
dipoles cancel) while the cis isomer will have a dipole bisecting the B-M-B bond
(b) Consider the 2 structures:
mer
A
fac
A
A
A
Co
B
B
A
B
B
Co
A
B
B
Each will have a dipole moment along the axis indicated; how do they compare?
-for mer the dipole will be (µM-A + µM-B) since the other bond dipoles cancel
-for fac a little trigonometry and vector addition shows it will have a somewhat
larger dipole:
(1+21/2)/(2)1/2µM-A + (1+21/2)/(2)1/2µM-B
If the molecular dipole can be quantified, these should be distinguishable.
[Note: make sure you understand part a. The quantitative aspects of part b,
which are more difficult, would not appear on an exam, but you do need to know
the definitions of fac and mer isomers.]
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6. Here are the isomers:
Note: all are cations
Cl
N
S
Co
Cl
trans
cis
N
N
S
S
S
Co
N
Λ
Cl
Cl
Cl
Cl
Co
cis
N
N
S
S
S
N
N
Co
S
N
Cl
Cl
Cl
Cl
Co
N
S
Λ
Δ
S
Δ
7. (a) See pictures below.
(i) ML2. For a linear arrangement of 2 ligands, we define the internuclear axis to
be the z-axis. Then the electrons in the dz2 orbital will suffer the most repulsions
(the 2 ligands = negative point charges are on the z-axis), followed by the other
orbitals with a z-component, and, at lowest energy, those with the least
repulsions.
(ii) trigonal planar ML3: let xy be the molecular plane. The ligands no longer line
up nicely on the axes, but it should be clear the greatest repulsions are to orbitals
in the xy plane. The dxz and dyz orbitals should be degenerate, but their energy
relative to the dz2 orbital is not obvious (how important is the repulsion from the
donut in the xy-plane?)
(iii) pyramidal ML3: Let the plane of the 3 ligands be parallel to the xy plane, and
let the z-axis pass through the metal and through the center of the L3 triangle.
energy
dxy, dx2-y2
dxz, dyz
dz2
dxz, dyz
dxy, dx2-y2
(i)
dxz, dyz
dxy, dx2-y2
dz2
(ii)
dz2
(iii)
8. (a) [Ni(CN)4]2- could be square planar (D4h) or tetrahedral (Td).
(b) [Ni(CN)5]3- could be square pyramidal (C4v) or trigonal bipyramidal (D3h).
(c) [Ni(CN)6]4- will be octahedral (Oh).
Note that all of these are Ni(II) d8. The electron configurations and resultant
number of unpaired electrons for each geometry are shown below.
Note: the results for octahedral, tetrahedral and square planar geometries are
straightforward. However, it is difficult to predict the separations between the
energy levels (and hence the electron configurations) for the other 2 geometries.
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b1g
Oh(2)
D4h(0)
Td(2)
eg
t2
t2g
b2g
eg
e
D3h(0 or 2)
a1'
b1
a1'
e'
e''
or
a1g
e'
a1
e''
b2
b1
or
a1
b2
e
e
C4v (0 or 2)
9. (a) Fe(III) is d5 and CN– is a strong field ligand, so the complex is low-spin, with
1 unpaired electron. LFSE= 5x2/5Δo = 2 Δo.
(b) [Co(NH3)6]3+ is low spin d6; LFSE = 6x2/5 Δo = 12/5 Δo and all the electrons are
paired, so it is diamagnetic.
(c) [Fe(H2O)6]2+ is high-spin d6, with 4 unpaired electrons. LFSE = 8/5 Δo – 6/5 Δo
= 2/5 Δo.
(d) [CoF6]3– is high-spin d6; same results as in (c).
(e) [Ru(NH3)6]2+ Since Ru is a 2nd-row metal, we expect a large Δo and a low P,
so it should be low spin. LFSE = 6x2/5 Δo = 12/5 Δo ; diamagnetic.
(f) [CoCl4]2– is tetrahedral (so high spin) and d7. LFSE = 12/5 Δt – 6/5 Δt = 6/5 Δt
and there are 3 unpaired electrons.
(g) [Mn(H2O)6]2+ is high spin d5; LFSE = 0; 5 unpaired electrons.
(h) [Cr(H2O)6]2+ is high spin d4; LFSE = 6/5 Δo - 3/5 Δo = 3/5 Δo; 4 unpaired
electrons.
10. The true value for ΔH hydration should be larger (more negative) than the
spherical ion value by an amount equal to the LFSE.
For Ni(II) (d8), the LFSE = 1.2 Δo. Since Δo = 8600 cm-1 the LFSE = 10,320 cm-1 .
From the back of the textbook, 1 cm-1 = 11.96 J/mol, so LFSE = 123.45 kJ/mol.
Then Corrected ΔHhydration = 3123 kJ/mol.
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