| 1. |
Cu2+ dissolved in water will set up the complex
Cu(H2O)62+. When ammonia is added,
some of the waters will be replaced by ammonia molecules, so that the eventual
identity of Cu2+ in ammonia solution will be Cu(NH3)62+.
NH3 is a stronger s-donor than H2O, therefore, Do
will increase as the number of NH3 increases (NH3
is higher on the spectrochemical series). Blue colour is absorption
of orange light (complimentary colour), purple is the absorption of yellow
(higher frequency) light, so the purple is consistent with Do
increasing. |
| |
|
| 2. (a) |
The easiest way to approach this question is to calculate the spin-only
magnetic moments for both the high-spin and low-spin cases, and see which
is consistent with the observed value.
-
assume Fe(NH3)3Cl3 is "octahedral", the
oxidation state will be 3+. Fe3+ is d5.
-
d5 high spin has 5 unpaired electrons. Spin-only
magnetic moment m = 2*[5/2(1+5/2)]1/2mB
= 5.91 mB.
-
d5 low spin has 1 unpaired electrons. Spin-only
magnetic moment m = 2*[1/2(1+1/2)]1/2mB
= 1.73 mB.
1.73 is just below the observed value of 1.69 mB,
so the compound must be low spin. |
| |
|
| 2. (b) |
There are 3 isomers - one with two axial CO, one with two axial Cl,
and one with one of each. |
| |
|
| 2. (c) |
Do = hn
= hc/l = (6.636 x 10-34 Js)(2.9979
x 108 m/s)/(950 x 10-9 m) = 2.09 x 10-19
J = 126 kJ/mol |
| |
|
| 3. (a) |
A d9 complex has an orbitally degenerate ground state,
so is subject to a Jahn-Teller Distortion. Therefore, an octahedral
compound with this electron count would probably distort such that two
ligands trans to each other would elongate (and the other 4 bonds
compress) to shift the symmetry to D4h. Energetically,
the formerly degenerate eg orbitals, the dx2-y2
and dz2, would split with the dz2
increasing in energy and the dx2-y2
decreasing in energy. Since the dx2-y2
has 2 electrons in it and the dz2 only
one, two electrons are lowered in energy while one is raised in energy.
Therefore, there is an overall net energy gain in the distortion. |
| |
|
| 3. (b) |
The chelate effect is the observation that chelating ligands create
more stable compounds than monodentate ligands of the same donor atom.
Because the donor atoms are the same, the (bond) enthalpy (DH)
will be the same (at this level of approximation) for both chelated and
non-chelated products. Likewise, the enthalpy and entropy
of the starting materials are identical for both products. However,
because the chelate has more than one donor site attached to the same molecule/ion,
removing one chelate from solution is a smaller entropic penalty than removing
more than one monodentate ligand. Therefore, the entropy of formation
is more positive (or less negative) for the product with chelates than
for the product with non-chelating ligands, and it is therefore more stable. |
| |
|
| 3. (c) |
Hard acids act like hard-sphere ions - they are happiest in purely
ionic solids. Soft acids, on the other hand, have polarisable electron
shells and can form covalent interactions. They normally will not
form strictly ionic compounds. Periodic trends are as follows:
-
for an acid, softness increases down a group (or softness increases
as electronegativity increases)
-
(alternatively, hardness increases as electronegativity decreases)
-
hardness increases as charge increases
-
there is no specific trend across periods
|
| |
|
| 4. (a) |
methane = Td
water = C2v
H2N2 = C2h |
| |
|
| 4. (b) |
O2 has an electronic configuration such that the frontier orbitals
are unfilled, and are p* antibonding orbitals. Therefore, if one
electron was removed from this orbital to form O2+,
the bond order will increase to 2.5 in O2+ from 2.0
in O2. Therefore, the bond is stronger and shorter in
O2+ than in O2.
By the same token, adding an electron to the frontier orbital puts another
electron in an antibonding orbital, thus decreasing the bond order from
2.0 in O2 to 1.5 in O2-. Therefore,
O2- has a longer/weaker bond than O2. |
| |
|
| 4. (c) |
We must solve the Kapustinskii equation for rA, the
radius of the anion. We know everything else we need. Unfortunately,
the equation is not easy to rearrange - I usually just start plugging in
numbers rather than solving an analytical expression for rA.
Solving for rA giives a value of 1.79A as the effective
radius of fluorine. |
