Download Problem Set 7_Chem165Sp14

Answer Key, p. 1
Chem 165 Spring14
Chem 165 Homework #7 for review only, not to be turned in.
1. Problem 8.35.
Answer in Appendix G, Oxtoby page A.64: The [Zn(H2O)6]
amount of light in the visible region.
2+
does not absorb any significant
2. Problem 8.36.
[PtI6]
2–
absorbs light strongly throughout the visible region.
3. (a) Problem 8.37.
(b) Is [Fe(CN)6]3– high spin or low spin? Give its configuration as t2gxegy (give x and y).
(c) This question actually is significantly in error. The wavelength of the light absorbed is not
equal to ∆o. Why not? This is an important conceptual error, not a trivial one [a question like
this might be on the exam].
(a) Numerical answer in Appendix G, Oxtoby page A.64.
(b) [Fe(CN)6]
3–
5
0
is low spin, t2g eg .
(c) The energy of a state of any atom or molecule is affected both by the orbital energies and the
electron–electron repulsion energies. This isn’t so conceptually important for organic molecules
because the electrons are always paired and the pairing in one orbital isn’t so different than the
pairing in another. So to a first approximation, electron–electron repulsions aren’t important when
comparing the energies of two organic compounds.
For transition metal ions, however, many examples have unpaired electrons and for many ions
there are high-spin and low-spin possibilities. Here one has to worry about the balance between
the ligand field splitting ∆o and the electron–electron repulsion. For ions with high-spin ground
– –
states, the e -e repulsion is larger than ∆o.
3–
Considering [Fe(CN)6] , there are two electron pairs and one unpaired electron in the ground
5 0
4 1
t2g eg state. The first excited state is t2g eg , which has only one pair of electrons in the same
– –
orbital, and three unpaired electrons. Thus this first excited state should have less e -e repulsion
than the ground state. So you would predict that the energy of this transition should be less than
∆ o.
4. Problem 8.38. Does the octahedral Ni2+ ion have a high-spin/low-spin choice?
No, there is no choice here. All octahedral Ni
2+
6
2
ions are t2g e2g .
5. (a) Problem 8.43.
(b) The [HgI4]2– ion is not octahedral. Explain you how know this. [This is a very simple
question.]
(c) In contrast to the answer in Oxtoby, the [HgI4]2– ion is not colorless, it’s yellow.
(i) What transition is responsible for this color? What mercury orbital(s) are involved?
(ii) For this question, you can ignore the fact that [HgI4]2– ion is not octahedral, so the d orbital
splitting is different than “2-above-3”. Why is the specific d-orbital splitting not important in
this case?
Answer Key, p. 2
Chem 165 Spring14
2+
(iii) Why is this transition much lower for Hg complexes than for the analogous Cd2+ and
Zn2+ complexes? [Hint: it relates to why mercury is a liquid, as discussed in class.]
(iv) It is possible that this could be a mercury 5dà6s transition, although I doubt it. One of
the reasons I don’t think this is correct is that the 5dà6s transition is forbidden. Explain why.
(a) Appendix G, Oxtoby page A.64.
2–
(b) In the [HgI4] ion, there are only four ligands (iodides) around the Hg
complex has to have six ligands.
2+
2+
ion. An octahedral
10
(c) (i) Hg , as stated in the Oxtoby answer, is a d ion. Therefore it has no d-d transitions. A
metal-to-ligand charge transfer (MLCT) transition is unlikely because there are no empty valence
orbitals on the iodide ligands. So this has to be a ligand-to-metal charge transfer (LMCT). Since
2+
the d’s are all full, this is likely has to be the iodide to Hg 6s orbital.
(ii) The specific d-orbital splitting pattern is not important because all of the orbitals are filled.
2+
(iii) An LMCT transition to the Hg 6s orbital is be at much lower energy than the analogous
2+
transition in cadmium and zinc complexes because the Hg 6s orbital is unusually low in energy.
This is a result of the relativistic effect discussed in class.
(iv) Both the 5d and 6s orbitals have “g” symmetry. So the 5dà6s transition is gàg and forbidden.
[Aside: Yes, I know that in tetrahedral complexes there isn’t inversion symmetry so there are no “g”
2+
or “u”. But since the s and d orbitals are both “g” in the free (gas phase) Hg ion, and they don’t
change that much when they bind to the ligands, to a good first approximation these orbitals are
still basically “g”.
6. (a) Problem 8.64. As explained in Oxtoby (p. 371), tetrahedral complexes have a d-orbital
splitting pattern of “3-above-2” [t2 above e]. This is the inverse of the octahedral “2-above-3.”
Based on these complexes having the same number of unpaired electrons, what can you
conclude about the high-spin vs. low-spin configurations of the two complexes?
2+
5
3
2
For octahedral Mn (d ) complexes, the high spin ions have the configuration t2g e2g with five
5
0
–
unpaired electrons. Low spin complexes have the configuration t2g e2g with one unpaired e .
2+
5
2 3
For tetrahedral Mn (d ) complexes, the high spin ions have the configuration e2 t2 with five
4 1
unpaired electrons. Low-spin complexes have the configuration e2 t2 with one unpaired electron.
Since the two octahedral and tetrahedral complexes have the same number of unpaired electrons,
they must either both be low-spin or both be high-spin.
(b) Tetrahedral ligand field splittings are called ∆t. For the same ligand set (the proverbial
‘everything else being equal’) ∆t for ML4 is 4/9ths (about half) of the ∆o for the related ML6.
In general, are tetrahedral complexes more or less likely to be high spin than octahedral
complexes?
Tetrahedral complexes are essentially always high-spin. The smaller ∆t means that the splitting of
the d orbitals is very rarely enough to overcome the energy to pair the electrons.
(c) Here’s an oddball (and very tricky) question: Why do you think the d orbitals in tetrahedral
complexes (t2 and e) don’t have the subscript “g” that the octahedral orbitals have? Recall that
“g” stands for the German word gerade or even. The g means that inverting the orbital through
the nucleus doesn’t change the sign of the orbitals (all the d orbitals are “g”; p orbitals are
ungerade (uneven), designated “u”). Why don’t we specify g or u in tetrahedral complexes?
Odd and even are important for octahedral complexes because they have a “center of symmetry.”
Inverting an octahedral ML6 complex through its center doesn’t change the complex. But
tetrahedral complexes don’t have a center of symmetry. If you invert a tetrahedron about its
center, you get a different tetrahedron. Therefore, in the mathematics of symmetry (group theory),
inversion about the center is not relevant for tetrahedral structures.
7. On Problem Set 6 you determining the configuration (t2gmegn ) and the number of unpaired
electrons for each of the octahedral complexes below.
Answer Key, p. 3
Chem 165 Spring14
(a) [CoF6]3–
(b) [Cr(H2O)6]2+
(c) [Fe(CN)6]3-
(d) [Mn(pyridine)6]2+
(e) [Ru(H2O)6]2+
(f) [Cu(H2O)6]2+
(g) [TaCl6]–
(h) [Co(NH3)6]3+
Consider now the electronic spectra of these complexes. For each complex, state whether it:
(i) does not have any d-d transitions;
(ii) has d-d transitions but they are all spin forbidden
(iii) has spin-allowed d-d transitions
In each case, briefly explain your answer. All of the complexes fit into one of these categories.
If you don’t know (from question 2) whether an ion is high or low spin, give answers for both
and state whether spectroscopy could indicate the state of the ion.
6
3+
4 2
(a) [CoF6]3– contains a high-spin d (Co ) ion, t2g eg . This ion has spin-allowed d-d transitions;
4 2
3 3
the lowest transition has the configuration t2g eg à t2g eg [excited state].
4
2+
3 1
(b) [Cr(H2O)6]2+ contains a high-spin d Cr ion, t2g eg . This ion has spin-allowed d-d transitions;
3 1
2 2
the lowest transition has the configuration t2g eg à t2g eg [excited state].
6
2+
6 0
(c) [Fe(CN)6]4- contains a low-spin d (Fe ) ion, t2g eg . This ion has spin-allowed d-d transitions;
6 0
5 1
the lowest transition has the configuration, t2g eg à t2g eg [excited state].
5
2+
3 2
6
(d) [Mn(pyridine)6]2+ contains a d (Mn ) ion. It is high spin, t2g eg . High-spin d ions have d-d
transitions but they are all spin forbidden. There is no way to move a d electron from a t2g orbital to
an eg orbital without flipping its spin (because of the Pauli exclusion principle). [If it were low-spin,
5 0
4 1
it would have spin-allowed d-d transitions (for instance, t2g eg à t2g eg [excited state])].
6
2+
6 0
(e) [Ru(H2O)6]2+ contains a low-spin d (Ru ) ion, t2g eg . This is the same as part (c); it has spinallowed d-d transitions.
9
2+
6 3
(f) [Cu(H2O)6]2+ contains a d (Cu ) ion, t2g eg . This ion has a spin-allowed d-d transition. This is
6 3
5 4
the one I did this in class; it is t2g eg à t2g eg [excited state].
(g) [TaCl6]– contains a d (Ta ) ion, t2g eg . Since there are no d electrons, this ion has no d-d
transitions.
0
4+
0
0
6
3+
6 0
(h) [Co(NH3)6]3+ contains a low-spin d (Co ) ion, t2g eg . This is the same as part (c); it has spinallowed d-d transitions.
8. In class we have emphasized the spectroscopic effects of d orbital splittings. The splitting of the
d orbitals also has thermochemical effects. This is easiest to analyze in the following way.
Moving a free gas phase ion into solution and binding it to ligands occurs with substantial energy
changes. Let’s think about this in two parts. Let’s lump all of the changes except the d orbital
splitting into a first step, and then ask how the d-orbital splitting affects the total energy. If the
average energy of the d orbitals is kept constant, then the t2g orbitals must move down by 0.4 ∆o
while the e2g orbitals must move up by 0.6 ∆o. The energy associated with this splitting is called
the Ligand Field Stabilization Energy (LFSE), or sometimes (as in Oxtoby) the crystal field
stabilization energy. This is shown schematically below.
Answer Key, p. 4
Chem 165 Spring14
It is worth emphasizing that this is a thermodynamic picture, where the energies are referenced
to some external standard (though that standard is not shown in the figures). This picture
contrasts with the spectroscopic pictures in which the zero of energy is the energy of the ground
state of the system.
For a d1 ion, the d-orbital splitting causes a stabilization of –0.4∆o vs. the ion with no d-orbital
splitting. For an arbitrary configuration t2gx e2gy, the LFSE is given be:
LFSE = x(–0.4∆o) + y(0.6∆o)
This topic is covered briefly (and not very well, in my view) in Oxtoby, and the picture above is
very similar to Figure 8.17. You can also find many explanations of this on the web, including
in: http://www.youtube.com/watch?v=3vovX1C-_T4 and http://en.wikipedia.org/wiki/Crystal_field_theory
and http://ocw.nctu.edu.tw/upload/classbfs120912062973271.pdf. This last one is a nice overview
of transition metal issues we are discussing.
(a) For which electron counts and high-spin/low-spin choices is the LFSE zero?
The LFSE is zero when the d orbitals are filled symmetrically. This occurs when the d orbitals are
0
10
5
all empty, for d ions; or all filled, d ions; or half filled – which occurs for a high spin d ion.
(b) Calculate the LFSE (in units of ∆o) for d0 – d5 high spin complexes and do a simple plot of
LFSE vs. d-count.
0
1
2
3
4
5
LFSE of high spin complexes: d , 0; d , –0.4∆o; d , –0.8∆o; d , –1.2∆o; d , –0.6∆o; d , 0.
See: http://wwwchem.uwimona.edu.jm/courses/CFSE.html
The figure below gives this plot as the blue line. The “unit” of Energy in the plot is ∆o.
(c) Oxtoby (p. 370) says (in essence) that this plot that you just made explains the data in Figure
8.3. This Figure plots the heat of hydration of 3d M2+ ions, the ∆H° for:
M2+ (g) + water à M2+ (aq)
(“aq” means “aqueous” [“in water”])
By explains the data, what Oxtoby means is explains the deviations of the blue points from the
expected trend (in the absence of d-orbital splitting) shown by the red line. Do you agree that
the LFSE is a good explanation?
Yes, I think the LFSE is a remarkably good explanation.
(d) Unrelated question (not related to Exam 2): Why do you think that there’s no point for Sc2+ in
Figure 8.3? [Hint: Scandium is (in a way) just below aluminum in the Periodic Table (see the
mosaic in the lobby of Bagley).]
2+
Sc is not stable so you can’t do the measurement. Scandium is like aluminum, it always makes a
3+ ion.
9. Let’s look in more detail at the iron-oxo species in the paper by Petit et al. that was discussed in
the last question of problem set 6.
(a) We take as a first approximation that MO+ ions are composed of M3+ and O2–. When M is
iron, what is its d-count?
5
The iron is d .
Answer Key, p. 5
Chem 165 Spring14
+
(b) The metal in MO has only one ligand, so the octahedral picture we derived in class doesn't
hold. The standard convention is to call the M-O bond axis the z axis. which are the two orbitals
that are of δ symmetry with respect to the metal-oxygen bond? What does δ symmetry mean?
Are these metal d orbitals bonding, non-bonding or antibonding?
δ symmetry is like a d orbital, with two nodal planes that contain the bond axis. This is similar to a
π orbital, which has one nodal plane that contains the bond axis, and looks like a p orbital if you
look down the bond axis.
(c) Consider the two d orbitals that are involved in π bonding. Do you think that they are
degenerate? "Degenerate" means that they are the same energy (usually by symmetry,
occasionally one has "accidental degeneracy").
Yes, dxz and dyz are degenerate by symmetry, because the x and y axes are equivalent by
symmetry. In a linear molecule like MO+, the z axis is unique but x and y are equivalent.
(d) Using the answer from the question on Problem Set 6, predict the d orbital splitting pattern
for this complex? To answer this question, remember that σ bonds are stronger than π bonds.
See question (f) below. The pattern is 1–above–2–above–2. The highest (most antibonding) orbital
is dz2, which makes a σ bond. dxz and dyz are a degenerate pair of π-antibonding orbitals, so they
are less antibonding than dz2. The remaining two orbitals, dxy and dx2-y2, are non-bonding.
(e) What does the Introduction say about the spin state of the iron in FeO+? This could be called
"high spin" (HS) or "low spin" (LS) but be careful because this is a different structure, and the
rules are a little different from those of the HS/LS states of an octahedral complex. In general,
do you think a one-coordinate complex like FeO+ will have a larger or smaller splitting of the d
orbitals than an octahedral complex? Explain your reasoning.
The Introduction says the complex has a sextet ground state, which means five unpaired electrons
and therefore a high spin state. In general, the lower the coordination number the smaller the dorbital splitting. The d-orbital splitting comes from the bonding of the d orbitals with the ligand(s),
so if there are fewer ligands there is less bonding.
(f) The M.O. diagram and d-orbital splitting for a metal-oxo complex like [Fe(O)(H2O)5]2+ is
shown in http://en.wikipedia.org/wiki/Transition_metal_oxo_complex, in the section about the
“oxo wall.” Note that the d-orbital splitting is not “2-above-3” because the complex is not really
octahedral. Yes, there are six ligands, but the ligands are different. Once the ligands are
different, at least some of the d orbital degeneracies are removed. The d-orbital splitting of
M(O)L5 on the Wikipedia page has really only one significant difference from the one you
predicted for part (d) above. Explain which orbital is very different and why that is.
+
2+
The primary difference is in the dx2-y2 orbital. In MO , this is nonbonding. But in [Fe(O)(H2O)5] ,
dx2-y2 is σ antibonding to the water molecules in the xy plane, just like in an octahedral complex.
MO+
M(O)L5
dz2, ! antibonding
dxz, dyz, " antibonding
dz2, ! antibonding
dx2-y2, ! antibonding
dxz, dyz, " antibonding
dxy, dx2-y2, # non-bonding
dxy, non-bonding
(g) Using the Wikipedia M.O. diagram, fill the electrons as appropriate for [Fe(O)(H2O)5]2+.
What are the oxidation state and d-electron count in this ion?
The iron in [Fe(O)(H2O)5]
2+
4+
is Fe , d
4