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Chm 118
Fall 2015, Exercise Set 2
Transition Metal Compounds
Simple Kinetics
Mr. Linck
Version: 5.1.
September 29, 2015
2.1 Electronic Configuration
What are the electronic configuration of these transition metal species in bonding situations: Cr(III), Mn(II), V(II), Cu(II), Ag(I), Ag(III), Ni(III), Mo(0), W(VI), Ru(II), Os(III),
Mn(V)? HINT: Remember that the configuration of transition metal ions in compounds is
sans 4s electrons.
2.2 Electron Count
How many d electrons do these transition metal species have in bonding situations: V(III),
Ti(IV), Mo(V), Ni(0), Co(II), Co(III), Rh(III), Ir(IV), Fe(II), Mn(II)? HINT: “In bonding
situations” means sans 4s electrons.
2.3 Electron Count
How many p electrons would iodine have in the molecule IF–2 assuming that the bonding is
ionic, that is, that the two fluorine nuclei are both ions, F– ?
2.4 Crystal Field Splitting
Imagine an atom at the origin of a coordinate system with two negative charges: one is
located arranged at (x, y, z) = (0.0, 0.0, 1.0)Å and the other at (x, y, z) = (0.0, 0.0, -1.0)Å.,
see Figure 1. Let the atom have an electron in a 5pz orbital. What factors influence the
energy of that electron?
2.5 Crystal Field Splitting
Same arrangement as in the last problem, but now let the atom have a 5px electron; see
Figure 1. Is the energy of this electron different from the one in the last problem? Is it
more stable or less stable?
2.6 Crystal Field Splitting
Show how the p orbitals would split in energy in a linear compound, say IF–2 under the
crystal field model where you treat the fluoride ions as point charges. HINT: Pay attention
to degeneracies.
2.16
2
Figure 1: Arrangement for problems 4 and 5.
2.7 Crystal Field Electron Configuration
What would be the occupancy of the various split orbitals for the iodine in the last problem?
HINTS: You need the electron count and then need to apply the aufbau principle. For the
electron count, if each of the fluorine atoms is negative one, and the molecule is negative
one, then the iodine atom must be . . .
2.8 Crystal Field Splitting
Show how the p orbitals would split in energy in a square planar complex IF–4 under the
crystal field model.
2.9 Crystal Field Electron Configuration
What would be the occupancy of the various split orbitals for the iodine in problem 8?
2.10 Crystal Field Splitting
What would be the splitting of the p orbitals in an octahedral compound such as SF6 ,
under, as usual, the crystal field approximation?
2.11 Crystal Field Electron Configuration
What would be the occupancy of the various split orbitals for the sulfur in problem 10?
HINT: Learn to count electrons and to be prepared for what looks wrong; it isn’t always.
2.12 The Angular d Orbitals
Sketch the five angular d orbitals as they would look if you stood on the positive x axis.
HINT: You have to be able to draw these orbitals, accurately and quickly, to work with
crystal field problems.
2.13 The Angular d Orbitals
Sketch the five angular d orbitals as they would look if you stood on the positive z axis.
2.14 Relative d Orbital Energies
We have two negative charges along the positive and negative z axis. Two of the d orbitals
are in Figure 2. Which are they? What are their relative energies under a crystal field
approximation.
2.15 Crystal Field Splitting
Continuing from the last problem, how would the d orbitals split in a linear compound, say
CrF+
2 ? HINT: Especially when dealing with d orbitals, always let the unique axis in the
molecule be the z axis of your coordinate system; it is good practice to always do that if
you want to apply symmetry arguments easily; and you almost always do want to apply
symmetry arguments as they simplify problems.
Chm 118
Exercise Set 2
2.20
3
Figure 2: Two of the d orbitals for consideration in problem 14.
2.16 Correlation Diagrams
Make a sketch of the energies of each of the five d orbitals for CrF+
2 as the fluoride ions, both
of which are the same distance from the Cr ion, are moved from a large distance to a smaller
one. HINTS: (1) By large distance, I mean to assume they are not close enough to the Cr
ion for it to know they exist. (2) Be careful. What happens to the energy of any orbital (if
it had an electron in it) if it sees a negative charge? (Sorry about the anthropomorphism.)
2.17 Crystal Field Splitting
Convince yourself that an octahedral environment of point charges (placed on the positive
and negative of each of the Cartesian axes) will cause dxy , dxz , and dyz to be repelled
equally.
2.18 Orbitals of d Character and the Strange One
This is a little harder than than the last problem. Note that the orbitals in problem 17
occur with one Cartesian coordinate times another. There are three more possible combinations of coordinates taken pairwise, xx, yy, and zz. These are usually considered in three
combinations, x2 -y2 , z2 -x2 , and z2 -y2 . Sketch each of these functions. HINT: Use the name
to guide you to ascertain (1) where the function is big, where it is zero; and (2) where it is
positive and where it is negative.
2.19 Orbitals of d Character and the Strange One
Which orbital is repelled the most by negative charges on the three (x, y, and z) plus and
minus Cartesian axes (octahedral symmetry), dx 2−y 2, dz 2−x 2, dz 2−y 2?
Chm 118
Exercise Set 2
2.25
4
Figure 3: A cube with axis drawn for problem 24.
2.20 Orbitals of d Character and the Strange One
We now do something sneaky. Since quantum mechanics demands that there are only five
possible d orbitals and there are six possible combinations of a pair of Cartesian coordinates,
we have to combine them in some fashion. We choose to add the last two (of the list in
problem 18) and then divide by two since we don’t want to change the size. Add z2 -x2 to
z2 -y2 and then divide by two. What do you get?
2.21 Splitting in Octahedral Environments
The answer to the last problem, z2 - 21 x2 - 12 y2 is the real form of the function we usually
abbreviate as dz 2. Since it is composed of those two orbitals that are equivalent to dx 2−y 2
but which would have double the interaction with ligands, but then is divided by two, we
have the same interaction. Does this argument convince you that dz 2 is equivalent in energy
to dx 2−y 2? If not, it is the best I can do; you will have to take a quantum course.
2.22 Orbitals of d Character and the Strange One
Which orbital is repelled the most by negative charges on the plus and minus Cartesian
axes (such that the symmetry is octahedral), dxy or dx 2−y 2 ?
2.23 Splitting of d Orbitals in an Octahedral Field
You have now established the relative energies of dxy , dx 2−y 2 , and dz 2. How do the d
orbitals split in an octahedral field?
2.24 Splitting in Tetrahedral Environments
Draw a cube and put four ligands at opposite corners to make a tetrahedral shape. Let
the x, y, and z axes go through the centers of the faces of the cube–see Figure 3. Convince
yourself that dxy , dxz , and dyz have the same energy.
Chm 118
Exercise Set 2
2.28
5
Figure 4: A tetragonal distortion of an octahedron for problem 27.
2.25 Splitting in Tetrahedral Environments
A similar argument to that used in problems 19 - 21 should convince you that dz 2 is
equivalent in energy to dx 2−y 2. Can you make a determination of which of the sets of
orbitals, that from problem 24 or that from this problem, is highest in energy? HINT:
This is a problem in solid geometry, but can be approximated by looking at just the wave
function lobe closest to the point charges.
2.26 Tetrahedral versus Octahedral Splitting
Summarize how the splitting of the 5 d orbitals differ in an octahedral environment versus
a tetrahedral one. You should have three aspects to your description: (1) the degeneracy of
the levels; (2) which is stable and which is unstable; and (3) the relative sizes of the gaps.
REMARK: Real solid geometry suggests the gap in the tetrahedral environment is 4/9 that
in the octahedral.
2.27 Splitting in Tetragonally Distorted Structures
A good way to determine the splitting patterns in compounds of low symmetry is to ascertain
what changes occur to the orbital energies as the symmetry is lowered. To determine the
symmetry in a D4h system you can use the answer for the splitting of the d orbitals in an
octahedral environment as a starting point. In Figure 4 is such a distortion. Determine
the splitting pattern in this tetragonal symmetry. HINT: There is some ambiguity to your
answer; see if you can understand why and state what conditions of distortion will favor
one answer versus the other.
Chm 118
Exercise Set 2
2.32
6
Figure 5: Energy level diagrams for problem 30.
Figure 6: Distortion for problems 28 and 29.
2.28 Distortion from a Tetrahedral Shape
In Figure 6 is a distortion from a tetrahedral shape toward another shape. What would the
resultant shape be if the motion continued a little further?
2.29 Correlation Diagram for a Distortion
Sketch how the d orbitals on a metal ion at the center of the cube shown in Figure 6 would
change in energy as the distortion shown occurred.
2.30 Splitting Patterns
Consider the energy level diagrams in Figure 5. Associate each with a geometrical arrangement of ligands and give a rationalization for the general shapes and the splitting gaps.
2.31 Splitting in TBP Structures
In a trigonal bipyramidal (tbp) complex (with the set of three ligands in the xy plane), the
d orbitals split into a stable set of two (dxy and dx 2−y 2), an intermediate set of two, and an
unstable level (dz 2). Justify that this arrangement is plausible. HINT: It is a problem in
solid geometry to prove that the result is rigorously correct,
Chm 118
Exercise Set 2
2.38
7
Figure 7: Using characteristic number names for configurations; see problem 33.
2.32 Splitting in Lowered Symmetry
What happens to the energies of all of the d orbitals of a tbp complex if one of the ligands
in the xy plane is changed slightly (say from Cl− to F− )? HINTS: A detailed answer is not
necessary; I am looking for an understanding of the general issue of symmetry. Also, let
the x axis be the axis on which the switched ligand is located.
2.33 Electronic Configurations
In compounds of lower than spherical symmetry, the symbols 1s, 2s, etc. are of limited use.
Rather the names of the characteristic sets of numbers are used to describe the electrons.
Thus the configuration of diagram “A” in figure 7 is a21 a12 . What is the configuration for
diagram “B” in Figure 7?
2.34 Electronic Configuration in an Octahedral Field
Give the electronic configuration (in the tn2g /em
g notation) for an octahedral Cr(III) compound; for an octahedral Mn(II) compound if the splitting is small; for an octahedral Mn(II)
compound if the splitting is large.
2.35 Electronic Configuration in an Octahedral Field
Give the electronic configuration (in the tn2g /em
g notation) for an octahedral Cu(II) compound; for an octahedral Fe(II) compound if the splitting is small; for an octahedral Fe(II)
compound if the splitting is large.
2.36 Spin Ambiguities
In the last problem can you see that the net spin of the electrons is different for the two
Fe(II) compounds? If each electron has a z component of spin of either 12 or - 12 , as it
does, what is the magnitude of the net z component of spin of each Fe(II) compound?
The magnitude of this value is called MS and there is another quantum number related
to it called S, which in the (non-rigorous quantum) situations we deal with is equal to the
magnitude of MS .
2.37 Electronic Configuration in a Tetrahedral Field
Tetrahedral crystal fields (as they are caused by four charges rather than six) are smaller
than octahedral ones. Hence (almost) all tetrahedral systems are “high spin”. Give the
electronic configuration (in the em /tn2 notation) for an tetrahedral Cu(II) compound; for an
tetrahedral Fe(II) compound; for a tetrahedral Co(II) compound .
Chm 118
Exercise Set 2
2.46
8
Compound
S
Cr(CH3 )(H2 O)2+
5
3
2
FeCl3 (H2 O)3
5
2
Co(NH3 )4 Cl+
2
0
MnF3–
6
2
Table 1: Spin data for some compounds
2.38 Spin Change after a Field Change
Imagine a Fe(III) compound changes from a high field octahedral environment to a tetrahedral environment. What happens to the spin of the complex. Be specific and, as usual,
speak professionally, “S changes from 2 to 1/2”.
2.39 Spin in Transition Metal Compounds
In Table 1 are the values of the spin quantum number for some compounds. Verify these
values.
2.40 Fields in the Crystal Field Model
I claimed that there are the two electrostatic fields present in the CF model? Articulate
that concept and distinguish between the two fields.
2.41 Fields in the Crystal Field Model
Ligands produce one of the fields from the last problem. In order to do that, what property
must a reasonable ligand have? HINT: This is not a difficult question.
2.42 Properties of Reasonable Ligands
Would Cl− be a reasonable ligand? O2− ? H2 O? NH3 ? CH4 ? Why or why not for each
case?
2.43 Predicted Relative Field Strength
What would you predict for the order of strength (how much the d orbitals are split in an
octahedral environment, for instance) for the first three ligands in problem 42?
2.44 Predicted Relative Field Strength
If you were riding around on a positive charge (and hence were attracted to negative ions),
would you be able to get closest to the charge (which in an electrostatic model appears,
by Gauss’ theorem, to be at the center of the negative ion) when you came up to F− or
I− ? Where would you have the lowest energy, be most stable? Why? What parameter is
changing in this problem?
2.45 Predicted Relative Field Strength
What would you say the order of strength of crystal field would be for F− , Cl− , Br− , I− ?
What factor is important? HINT: If you didn’t do the last problem, maybe you should.
Chm 118
Exercise Set 2
2.52
9
Compound
λmax , nm
Compound
λmax , nm
CrF3–
6
671
MoCl3–
6
520
729
MoBr3–
6
3–
MoI6
Mo(H2 O)3+
6
546
CrCl3–
6
Cr(H2 O)3+
6
Cr(NH3 )3+
6
575
464
725
383
Table 2: Spectra data for the lowest energy t2g to eg transition in some d3 compounds.
Data from A. B. P. Lever, Inorganic Electronic Spectroscopy, Elsevier, 1984.
2.46 Predicted Relative Field Strength
Which ion has the larger crystal field strength, F− or H2 O? Be semi-quantitative. HINT:
That last means you need some data; maybe you should look around?
2.47 Electronic Configuration, Again
To do problems involving metal ions well you need to be able to assign the number of d
electrons quickly. Give the number of d electrons in Co(II), Ni(III), Cr(IV), Mn(VI), V(II),
Ru(III), Ir(III), Cu(II), Hg(II), Fe(II), Cr(II).
2.48 Spectrochemical Series
The spectrochemical series is the actual order of ligand strengths. As we have seen, it cannot
be understood based on the crystal field model. It can be rationalized (and we probably
will get to that point by the end of the semester), but we take it as a given for now so that
we can use it within the simple crystal field approach.
I− < Br− < NCS– , S bonded < Cl− < NO–3 , O bonded < F− < OH− <
−
C2 O2–
4 , O bonded < H2 O < SCN , N bonded < NH3 about the same as py
< en < NO–2 , N bonded << CN− , C bonded ∼
= CO, C bonded.
3+
Which compound will have the greater splitting, MoI3–
6 or Mo(H2 O)6 ?
2.49 Crystal Field Splitting Energies
Which compound will have the larger splitting between the t2g orbitals and the eg orbitals
3–
−
of an octahedral complex, Cr(NH3 )3+
6 or Cr(NCS)6 when the NCS is nitrogen bonded.
2.50 Spectrochemical Series
The last problem requires that you assign the ligand strength of NH3 as greater than that of
N bonded NCS– , presumably using the spectrochemical series. Try using Lewis structures
to rationalize this result. HINT: I failed.
2.51 Crystal Fields in Compounds of Symmetry Less than Octahedral
If you had Jeeves1 , a magical chemistry servant, who could take two different ligands of
different crystal field splitting energy (CFSE), A and B, and could merge them into an
average ligand, C, would you expect something like that shown in Figure 8 for octahedral
compounds of these three ligands? Comment.
1 With apologies to P. G. Wodehouse.
Chm 118
Exercise Set 2
2.55
10
Figure 8: Splitting for ligands A, B, and C of problem 51.
Figure 9: Spectral lines in a tetragonally distorted compound.
2.52 Spectra of a Tetragonally Distorted Compound
How many different energy transitions would you expect for a tetragonally distorted compound if the metal is a d3 compound. HINT: See problem 27.
2.53 Spectra of a Tetragonally Distorted Compound
Using Figure 9, convince yourself that the “approximation” arrow (energy) is close to that
of each of the “actual” arrows (energies).
2.54 Spectra of a Tetragonally Distorted Compound
Figure 9 shows four peaks at a spread of energies; it also gives the average field approximation to that actual spectrum for a d3 compound. The “approximation” says there is only
one peak at some intermediate energy. In Figure 10 the spectra from the four separate lines
of Figure 9 are given (with the approximation of approximately equal intensities and equal
width, neither of which is likely to be realistic). I contend that the spectrum would not
look as shown there. What do I need to do to those peaks to get the observed spectrum?
HINT: There is only one compound here so the separate transitions all occur at the same
time.
Chm 118
Exercise Set 2
2.55
11
Figure 10: Spectrum for a tetragonally distorted compound if each line were seen separately.
Figure 11: Real spectrum for a tetragonally distorted compound and that for a hypothetical
average field system. The lower curve (at 17000 cm−1 ) is the sum of the curves in Figure 10
Chm 118
Exercise Set 2
2.63
12
2.55 Average Field and a Tetragonally Distorted Compound
Figure 11 given the real spectrum of a tetragonally distorted d3 compound (blue curve or,
if you are in black and white, the lower, wider curve) and the “average field approximation”
spectrum (yellow curve). Make an assignment for the wavelength of maximum intensity in
each case. Do you see why the approximation can be used?
2.56 Crystal Field Splitting Energies
Using the concept of “average field” and octahedral nomenclature (even though the symmetry is not octahedral), which would you expect would have the larger gap between the
3+
t2g orbitals and the eg orbitals, Cr(NH3 )4 Cl+
2 or Cr(NH3 )4 (H2 O)2 ? HINT: The concept
of ”average field” simply says that splitting between the t2g orbitals and the eg orbitals in a
compound of mixed ligands will be approximately that of the average ligand field strength
of the various ligands.
2.57 Using Crystal Field Arguments
2+
When Cr(H2 O)2+
one of the products is Cr(H2 O)5 (NCS)2+ .
6 reacts with Co(NH3 )5 (NCS)
Can you tell whether or not the thiocyanate ion is bonded to the chromium ion through
the nitrogen atom or the sulfur atom by using spectroscopy? What data would you need?
HINT: Apply an average field argument.
2.58 Crystal Field Splitting Energies
Think about the two electric fields that operate in the crystal field model and make an
4–
argument for whether the energy of splitting will be larger in FeF3–
6 or FeF6 ? Articulate
an eloquent answer.
2.59 Crystal Field Spectra
An octahedral compound of Cr(III) with six ligands, L1 , has an absorption peak near 550
nm. The ligand is changed to L2 (but the symmetry is still octahedral) and the peak is now
at 640 nm. What can you say about the ligand field strength of L1 and L2 ?
2.60 Crystal Field Spectra
−1 and that in Cr(en)3+ at 22300 cm−1 .
The lowest energy peak in CrF3–
6 occurs at 14900 cm
3
Which ligand has the larger crystal field splitting parameter, F− or en? HINT: It probably
would be nice to know what en is.
2.61 Crystal Field Spectra
The lowest energy peaks in solid VCl2 , VBr2 , and VI2 (where the coordination is approximately octahedral by sharing of anions between two or more metal ions) are 9300, 8600,
and 7870 cm−1 , respectively. Calculate the wavelength of the light associated with these
absorbances and explain the results.
2.62 Crystal Field Spectra
VBr2 is a orange brown solid. (Brown is typically the color associated with something that
absorbs over most of the visible region.) When it is dissolved in water and then treated
with KCN, the solution turns pale yellow. What color of light is removed from white light
to make a yellow color? Is the light that is removed of high or low energy? Explain the
color changes described above.
Chm 118
Exercise Set 2
2.69
13
2.63 Crystal Field Spectra
−1
The lowest energy peak in Ni(py)2+
6 (where py is pyridine) occurs at 11700 cm . Describe
the change in the position of the electron that occurs as a result of this absorption of light.
HINTS: From what orbital to what orbital is the electron moving? How do the positions of
these orbitals differ from one another? This is an important concept to use to distinguish
from other sources of color in compounds.
2.64 Crystal Field Spectra
When two pyridine ligands along the z axis in Ni(py)2+
6 – see last problem – are replaced by
Br− ligands, two peaks appear, one at 8430 cm−1 and the other at 11550 cm−1 . Account
for this behavior. HINTS: (1) The easy way to do this is to assume that the t2g level is
not perturbed much by this change in ligands, but the eg orbitals are. (2) This is NOT an
average field problem because we see two peaks, not just one broad one.
2.65 Linkage Isomerization
Linkage isomerization occurs when a ligand can bind to a metal ion in more than one way.
Which of the following ligands might exhibit linkage isomerization? NO–2 , NCS– , S2 O2–
3 .
Use Lewis structures to make your point.
2.66 Linkage Isomerization
The ion Cr(H2 O)5 SCN2+ has an absorption peak in the visible region of the spectrum at
620 nm. The linkage isomer of this material, Cr(H2 O)5 NCS2+ has a peak at 570 nm. Give
an explanation. HINT: Use the average field concept.
2.67 Spectrochemical Series
The table below gives the lowest energy (spin-allowed) spectral peak for some compounds
of the form Cr(H2 O)5 X2+ . From these data construct a spectrochemical series. HINT: Use
the average field concept.
Compound, X=
SCN−
Br−
F−
NNN−
NC−
λmax , nm
620
322
595
585
560
Compound, X=
I−
Cl−
NCS−
NO
NH3
λmax , nm
650
609
570
559
545
2.68 Positions and Intensities
There are two aspects to spectroscopic absorbance of light: The energy (wavelength) at
which it occurs, and the efficiency with which the light is absorbed. The latter is governed
by whether or not there is a mechanism for the light energy to become energy within the
molecule. What is the former governed by?
2.69 Positions and Intensities
The efficiency with which light is absorbed is measured by the molar absorptivity, . There
are two aspects dictating the value of the molar absorptivity. The first is whether or not
there is a change in the spin of a compound. Changes in spin, in the S value, are “forbidden”
and the resultant will be small. The second (which can be more rigorously discussed with
symmetry) is, for isolated ions (as we have in the crystal field model), whether or not there
is a change in the quantum number `. When the change in this quantum number is 1, either
Chm 118
Exercise Set 2
2.76
14
g to u? No
g to u? Remembers
g to u? Yes
∆S = 0
1 to 100
100 to 1000
greater than 1000
∆S not 0
less than 1
less than 10
less than 100
Table 3: Approximate values of the molar absorptivity under different conditions of ∆S
and odd/even nature of the functions; the units are M−1 cm−1 . Note that “Remembers ...”
refers to an orbital in an environment that does not have the center of inversion symmetry
operation, but the starting function and final function are both even (or odd). Use of this
table without thinking will result in incorrect results; intensity is hard to predict.
plus or minus, the transition is “allowed.” What would you expect for if a transition has
∆S = 0 and ∆` = 1? ∆S = 0 and ∆` = 0? ∆S = 1 and ∆` = 0?
2.70 A Review to Learn about Intensities of Absorbances
Think about the angular wave functions for atoms that we learned previously, s, p, d, f ...
According to the information in problem 69, is a transition from an s orbital to a p allowed?
From p to d? From p to f? And, importantly, from d to d?
2.71 Symmetry and the Intensities of Absorbances
Imagine that the orbitals of the last problem are in the center of some molecule that has a
center of inversion, an i. What is the characteristic number of the angular wave functions
of the last problem? HINT: They have to be ±1. NOTE: Orbitals with a characteristic
number of “plus” are labeled “g” and those with characteristic number of “minus” are
labeled “u”.
2.72 Symmetry Rule for Intensities of Absorbances
A useful way to think about whether or not a transition (in those symmetry groups with
an i) is allowed, or as a professional would say, “has allowedness,” is to ask if there is a
change from a “g” orbital to a “u” one, or vise versa. Would a transition be allowed from
an s orbital to a p if an i were present in the symmetry? From a s orbital to a d?
2.73 Intensity of Absorbance
A compound has a molar absorptivity of 15100 M−1 cm−1 . Using Table 3, what can you
say about the nature of the transition?
2.74 Intensities of Absorbances
−1 cm−1 )
The “complex ion” Co(H2 O)2+
6 has absorbance peaks, nm, (molar absorptivity, M
at 12340 (2), 625 (shoulder), and 515 (4.6). Given the intensities of these peaks, what kind
of transition are they associated with? HINT: Remember the warning in Table 3; there
are a couple of possibilities and you need to use other thoughts and other data to sort out
which you want to use.
2.75 Intensities of Absorbance
Compounds of Mn(II) that are high spin are generally almost colorless (which probably
means that is less than 1). Why?
Chm 118
Exercise Set 2
2.88
15
2.76 Intensity
–
A solution of Fe(H2 O)3+
6 is nearly colorless. When some NCS is added, the solution turns
blood red with a peak near 440 nm with an of about 4600 M−1 cm−1 . Comment.
2.77 Intensity
The ion VO3–
4 is colorless to our eye, but has a peak at 276 nm with an intensity of about
8000 M−1 cm−1 . Give the “d count” of this species and comment on the spectrum.
2.78 Wavelength and Intensity of Transitions
2–
–
Consider VO3–
4 , FeO4 , and MnO4 . Which would you expect to absorb at the lowest energy?
What would the intensity of the bands be? Why?
2.79 Unpaired Electrons in Transition Metal Compounds
Would you expect Mn(CN)3–
6 to be high spin or low spin?
2.80 Magnetism in Transition Metal Compounds
How many unpaired electrons will a d3 system have in octahedral geometry? How about a
d8 system in low field? in high field? How about a d6 system in low field? In high field?
2.81 Magnetism in Transition Metal Compounds
How many unpaired electrons will a d4 system have in octahedral geometry in low field? in
high field?
2.82 Electron Count in Transition Metal Species
How many d electrons does a Cr(II) compound have? A Ni(III)? A Mo(0)?
2.83 Angular Momentum
Electrons in atoms often have orbital angular momentum. Use the words “orbital angular
momentum” to state what it is for an electron in an atom.
2.84 Magnetic Moment
When orbital angular momentum is “quenched”, as it often is, the magneticp
moment, µ, is
approximately equal to the “spin-only” value, given by the formula µB = 2 S(S + 1), in
units of the “Bohr magneton”, where S is the quantum number for the spin of the material.
That is, if the atom has one unpaired electron, S = 1/2. If two, S = 1, etc. What will the
magnetic moment be for Cu(H2 O)2+
6 ?
2.85 Magnetic Moment
What will the magnetic moment be for Cr(NH3 )3+
6 ?
2.86 Magnetic Moment
A octahedral complex of Mn(II) is made and has a measured magnetic moment of 6.0. Is
it high spin or low spin?
2.87 Magnetic Moment
A compound of Co(II) is synthesized and found to have a magnetic moment of 1.7. In a
simple model where it is assumed to be either octahedrally or tetrahedrally coordinated,
which is it?
Chm 118
Exercise Set 2
2.96
16
Figure 12: Energetic consequence of the splitting of the d orbitals by an octahedral crystal
field.
2.88 Magnetism
VBr2 has a magnetic moment of about 3.8 B.M. When it is dissolved in water and then
treated with KCN, the color of the solution changes dramatically, but the magnetic moment
is largely unchanged. Explain.
2.89 Magnetism
When CrBr2 is treated in a manner similar to that of VBr2 in the last problem, the magnetic
moment decreases. Explain.
2.90 Magnetic Moment
What would the magnetic moment be for Mo(CO)6 ? How about for V(CO)+
6 ? HINT:
n
Remember that the chemistry of transition metal complexes is d chemistry, not “s” chemistry.
2.91 Spin
How many unpaired electrons will be present in Fe(CO)3 Cl–3 ? What is the value of S? HINT:
Sometimes knowledge of possibilities is more important than a “correct” answer.
2.92 Paramagnetism and Electron Configuration
The complex CoF3–
6 is paramagnetic. What is its electronic configuration? HINT: The word
“paramagnetic” means that the compound contains unpaired electrons.
2.93 Diamagnetism and Electron Configuration
The complex Co(NH3 )3+
6 is diamagnetic. What is its electronic configuration?
2.94 Chemistry from Magnetism
Do the results of the last two problems make sense based on the spectrochemical series?
2.95 Crystal Field Strength and Electron Configuration
2+
Although Ni(NH3 )2+
6 is paramagnetic, Pt(NH3 )4 is diamagnetic. Explain. HINTS: First,
both have the same ”d count.” What is the value of that? The latter compound is four
coordinate; that means it is probably either tetrahedral or square planar.
Chm 118
Exercise Set 2
2.104
17
2.96 Crystal Field Stabilization Energy
Imagine an electron in one of the d orbitals (which are appropriate orbitals for a spherically
symmetric system) of an ion. Now take six negative changes and smash them with a hammer
into little fractional charges that are spread evenly on the surface of a sphere surrounding
the ion. What happens to the energy of the electron in the d orbital? Does the step from
“Free Ion” to “Spread Out Charge” in Figure 12 show what you just articulated?
2.97 Crystal Field Stabilization Energy
From the spherical arrangement in the last problem, lets move the fractional charges back
to full charges at the corners of an octahedron (on the plus and minus Cartesian axes). If
the electron in problem 96 is in a dx 2−y 2orbital, does its energy increase or decrease? Why?
Does the second step in Figure 12 show what you just articulated?
2.98 Crystal Field Stabilization Energy
From the spherical arrangement in the last problem, lets move the fractional charges back
to full charges at the corners of an octahedron (on the plus and minus Cartesian axes). If
the electron in problem 96 is in a dxy orbital, does its energy increase or decrease? Why?
Does the second step in Figure 12 show what you just articulated?
2.99 Crystal Field Stabilization Energy
If you have a high spin d5 system, then your analysis of the last couple of problems suggests
that you should have three electrons in the t2g levels and two in the eg levels. Quantum
mechanics tells us such an arrangement is spherical, so it is equivalent to the middle picture
in Figure 12. Let’s define the crystal field stabilization energy as the energy of stability due
to preferential filling of the t2g levels over the eg levels. What would you say is the total
(considering all five electrons) crystal field stabilization energy (CFSE) for this system?
2.100 Crystal Field Stabilization Energy
Given your result in the last problem, write an equation for the total stabilization (that is,
zero) in terms of the energy of the t2g level (relative to the middle picture in Figure 12)
and the energy of two eg levels? Write another equation with these two parameters and the
quantity ∆ given in the Figure. Solve for the energies of the two levels in terms of ∆.
2.101 CFSE
For Co(II) compounds, chloride ion has a value of ∆ of 6600 cm−1 . Find the CFSE in
kcal/mole for a six coordinate CoCl2 . HINT: Chloride ions are shared by various Co(II)
ions in this material.
2.102 Spectra and CFSE
The lowest energy absorbance peak for Rh(NH3 )3+
6 occurs at 304.8 nm. This compound
has a magnetic moment of zero. Find the CFSE for Rh(NH3 )3+
6 .
2.103 CFSE and Geometry
What is the change in CFSE upon moving a Co(II) ion from an octahedral environment to
a tetrahedral one? HINT: See problems 26 and 101.
Chm 118
Exercise Set 2
2.114
18
2.104 Spinels and Closest Packed Lattice
A spinel is a solid metal oxide with the formula M(II)M(III)2 O4 where the three M need not
be the same. The crystal lattice of spinels have the large oxide ions closest packed. What
is that? HINT: See a text or search the web if you don’t know; I imagine that Wikipedia
has something to say about closest packing.
2.105 Holes in the Spinel Lattice
There are two types of holes in a closest packed lattice, octahedral and tetrahedral. Show
this by drawing two layers of spheres with the upper one in the depressions in the lower
one. Then look for small holes between the two layers left in the structure that have six (or
four) oxide spheres around it.
2.106 Spinels and Electrostatics of Bonding
In a spinel, one-third of the metal ions are in tetrahedral sites, and the others are in
octahedral sites. A normal spinel has the M(II) in the tetrahedral sites. Why would this
be “normal”? HINT: Think about the stability of a dipositive and tripositive cation in the
presence of differing numbers of oxygen anions via Coulomb’s Law.
2.107 Spinels and CFSE
An “inverted” spinel has M(III) in the tetrahedral sites and the remainder of the M(III)
and all the M(II) in the octahedral sites. Fe3 O4 is an inverted spinel and Mn3 O4 is normal.
Why? HINTS: Something is over-riding the result of the last problem; look at the subject
of this problem, duh.
2.108 Spinels and CFSE
Account for the fact that the spinel CoFe2 O4 is inverted.
2.109 Beer-Lambert Law
Look up the Beer-Lambert law, sometimes abbreviated “Beer’s law”. What is absorbance?
2.110 Using Beer’s Law
−1 cm−1 at 508 nm. Find the
The ion Fe(phen)2+
3 has a molar absorptivity of 11,100 M
absorbance of a 0.02 mM solution in a 5.0 cm cell at 508 nm. What can you say about the
absorbance at 450 nm? Comment.
2.111 Using Beer’s Law
Is the molar absorptivity, , a function of wavelength? Explain.
2.112 Using Light to See Chemistry
Examine the absorbance as a function of time at 625 nm for the reaction shown in Figure 13. Since you have identified this peak earlier (see problem 67), you can tell me what
is happening to the substance responsible for the peak at 625 nm; do so.
2.113 Using Light to See Chemistry
Examine the absorbance as a function of time at 570 nm (see Figure 13). Since you have
identified this peak earlier (see problem 67), you can tell me what is happening to the
substance responsible for the peak at 570 nm; do so.
Chm 118
Exercise Set 2
2.118
19
Figure 13: (Idealized) Absorbance changes for a solution of Cr(H2 O)5 SCN2+ as a function of
time. Time is increasing downwards at a wavelength of greater than 593.5 nm and upwards
at a wavelength of less than this. Note the isosbestic point, the point of no absorbance
change with time, at about 593.5 nm.
2.114 Isosbestic Point
In Figure 13 the absorbance does not change at a wavelength of 593.4 nm. Use Beer’s law
to show that this would be true if one substance, say A, is converting to a single product,
B, and both of these materials have the same value of at this wavelength, A = B at
593.4 nm. HINTS: Beer’s law for two colored species is:
A = A [A]` + B [B]`
(1)
Also, there is a stoichiometric equation between [A] and [B] that you should use.
2.115 Using Light to See Chemistry
Assuming that the reaction in Figure 13 has gone to completion at the largest time scan,
show, using the Beer’s law at 660 nm, that the absorbance, A, is proportional to the
concentration of Cr(H2 O)5 SCN2+ .
2.116 Using Light to See Chemistry
Assuming that the reaction in Figure 13 has gone to completion at the highest time scan,
show, using the Beer’s law at 540 nm, that the absorbance, A, is proportional to the
concentration of Cr(H2 O)5 NCS2+ .
2.117 Using Light to See Chemistry
The last two problems are “loaded” examples because one of the two values in equation 1
is zero. We now examine what happens if that is not true. Will the absorbance at 610 nm
(see Figure 13) be proportional to the concentration of Cr(H2 O)5 SCN2+ ? Explain.
Chm 118
Exercise Set 2
2.124
20
2.118 Using Light to See Chemistry
You showed in the last problem that the absorbance at 610 nm (see Figure 13) is not
proportional to the concentration of Cr(H2 O)5 SCN2+ . What is true is that the fraction of
the reaction that has taken place is given by:
f raction =
At − A∞
A0 − A∞
(2)
Assuming that the highest curve to the right of 593 nm is time = 0, and that the highest
curve to the left of 593 nm is “infinite” time, show (using a ruler) that this relationship
holds at wavelengths of 640 nm, 570 nm, 620 nm, and 650 nm; that is it always holds.
HINT: This equation is proved in problems 129 to 131.
2.119 Basic Kinetics Concepts
A plate of brownies is placed at the front of a seminar room crowded with students and
faculty before a seminar. If you carefully watch the plate of brownies you would see them
disappear, right? At the start there might be 50 of them, after a couple of minutes, 20;
after a couple more minutes, perhaps 10. And as people had their fill (or were conscious of
other people watching how many they took; or, became guilty that there would not be any
left for those that were late because of a lab), the plate would slowly approach emptiness.
How would you verbally express this rate of disappearance of the brownies? Is the rate a
constant as time progresses? Does it increase with time or decrease?
2.120 Basic Kinetics Concepts
What mathematical expression could be used to discuss the rate of disappearance of the
brownies. HINT: I don’t know what the expression is equal to, but I do know the expression
for the rate; I want you to provide it also.
2.121 Basic Kinetics Concepts
Imagine the space in a beaker is divided up into a bunch of small compartments and imagine
time divided into a bunch of small increments. Further, there is a finite probability that
a given compartment will have a slight excess of energy in any given time increment. If
the compartment has that excess of energy and contains a molecule, let us suppose reaction
occurs. Imagine the rate (early in time when there are lots of reactants) is 0.00001 mole/sec.
What magnitude might it be some time later? HINT: I’m looking for a relative answer, not
an absolute one.
2.122 Basic Kinetics Concepts
Imagine our divided container (problem 121) where in any given time increment there are
certain volumes that contain excess energy to cause reaction. However, in this case assume
that for reaction to take place, you must have not only the energy, but also a molecule of A
and a molecule of B. Assume when the container has a reasonable concentration of A and
B, the rate is 0.00001 moles/sec. Now, if the container has a reasonable concentration of
A, but is very dilute in B, what is the rate? If very dilute in A but with a reasonable B,
what is the rate? If very dilute in both, what is the rate? HINT: See last problem.
2.123 Basic Kinetics Concepts
What is a plausible form for the rate law for the case in the last problem?
Chm 118
Exercise Set 2
2.133
21
2.124 Basic Kinetics Concepts
Here is a suggestion for interpretation of a rate law: The concentrations in the rate law
tell you what has to collide, and the rate constant, k, tells you how much energy has
to be present during that collision, to get reaction. What would a rate law of the form
−d[A]
dt = k[A][B] tell you about what has to collide in order to get reaction?
2.125 Collisions and Simple Kinetics
What is the interpretation of a rate law of the form
rate law?
−d[A]
dt
2.126 Collisions and Simple Kinetics
What is the interpretation of a rate law of the form
this rate law?
= k[A]? What is the order of this
−d[A]
dt
= k[A]2 ? What is the order of
2.127 Collisions and Simple Kinetics
There are almost no known reactions that are zero order. That is, have a rate law of the
= k? Use our interpretation of how a reaction proceeds to understand why this
form −d[A]
dt
is so.
2.128 Kinetics by Absorbance
Assume you have a compound X that converts to Y. Show that Beer’s law requires that
the absorbance at a fixed wavelength at any time t is given by the expression At = X [X]t +
Y [Y]t for a 1 cm cell, where is the molar absorptivity and symbols with subscripts of “t”
are time dependent.
2.129 Kinetics by Absorbance
For the reaction in the last problem, let the initial concentration of X, that at t = 0,
be [X]0 . Use the stoichiometric relationship between the concentration of X at time t=0
and the sum of the concentrations of Xt and Yt – see last problem – to show that At =
X [X]t + Y [X]0 − Y [X]t .
2.130 Kinetics by Absorbance
If the reaction goes to completion, show that A∞ = Y [X]0 ?
2.131 Kinetics by Absorbance
Use the result from the last problem, and the equation from problem 129 to show that the
At −A∞
where the absorbances
fraction of X remaining at any time is equal to the expression A
0 −A∞
are at the indicated times.
2.132 Integrated Rate Law
Use the rate law from problem 125 and integrate it between the limits of zero time and time
t. Algebraically manipulate your expression so that it is in terms of the ratio of [A]t /[A]0 .
Chm 118
Exercise Set 2
2.140
22
2.133 Plotting First Order Data
The data in the table gives concentrations of A as a function of t. Show that this follows
a rate law that is first order in [A].
t, sec
1
2
3
4
[A], M
0.0072
0.00523
0.00373
0.00264
t, sec
5
7
9
12
[A], M
0.00192
.00098
0.00051
0.00019
2.134 Rate Constant
What is the rate constant for the process in the last problem?
2.135 Integrate Rate Law in Terms of Absorbance
Substitute for the ratio of [A]t /[A]0 that you derived in problem 132 the expression that
you derived in problem 131 for the ratio of absorbance changes.
2.136 Integrated Rate Law
Show that the equation from the last problem can be written as
ln([A]t − [A]∞ ) = ln([A]0 − [A]∞ ) − kt
2.137 Integrated Rate Law
What functional form will you get if you plot the left hand side of the equation in the last
problem versus t? How would you evaluate k from that plot?
2.138 Integrated Rate Law
If the times at which the points in Figure 13 are 0, 9870, 22600, 53300, 102000, and “infinite” seconds, what is the rate constant for the isomerization of Cr(H2 O)5 SCN2+ ? HINT:
Approximations can always be made.
2.139 Using Beer’s Law to Determine Composition
The actual products when Cr(H2 O)5 SCN2+ (see the last problem) is allowed to stand are
both Cr(H2 O)5 NCS2+ and Cr(H2 O)3+
6 . We would say that “aquation” is in competition
with isomerization. If a reaction starts with 0.01M solution of Cr(H2 O)5 SCN2+ and at the
end of the reaction the absorbance at 570 nm is 0.244, how much of the isomerized product
and how much of the aquated product are in the final mixture. The value of N CS at 570
nm is 31 M−1 cm−1 and H2 O is 12.6 M−1 cm−1 , where these values are named in an
obvious way.
2.140 Fitting Data to Integrated Rate Laws
Here is some kinetic data for a reaction in which the reagent being measured, A, starts out
at a concentration of 0.01 M. Show that the data are consistent with a first order rate law
and evaluate k.
Chm 118
Exercise Set 2
2.145
23
t, min
1
3
5
7
9
[A], M
0.00885
0.00750
0.00600
0.00473
0.00402
t, min
2
4
6
8
10
[A], M
0.00806
0.00678
0.00539
0.00441
0.00397
2.141 Fitting Data to Integrated Rate Laws
The great natural philosopher of rural Arkansas, Boniface Beebe, carried out a rate measurement. He weighed a portion of solid compound A, took some water from the distilled
water bottle, mixed the solid in the solvent and placed the sample into the temperature
equilibrated compartment of a spectrophotometer. The initial concentration of A was 0.01
M and the experiment was carried out at 50o C. Determine if the data are consistent with
a first order rate law. If they are not, suggest a cause for the deviation. HINT: Think
carefully about what Bonnie did in the experiment; and what he might have done.
t, min
2
6
10
14
18
[A], M
0.00985
0.00798
0.00600
0.00473
0.00402
t, min
4
8
12
16
20
[A], M
0.00926
0.00678
0.00539
0.00441
0.00397
2.142 Pseudo-First Order Reactions
We imagine a reaction in which the stoichiometry is A + B = Products. If the differential
equation governing the rate of this reaction is −d[A]
= k[A][B] and initial conditions are
dt
such that [B]0 ≥ 20[A]0 , then it is approximately true that −d[A]
= kpf o [A]. What is the
dt
value of kpf o in terms of the true parameters of the process and why is our approximation
reasonably valid? HINT: Think about the concentration of B at the beginning and the end
of the reaction.
2.143 Pseudo-First Order Reactions
Here are some data for the reaction of A with B as given in the last problem. The initial
conditions are: [A]0 = 0.003 M, [B]0 = 0.06 M. Show the reaction is pseudo-first-order by
plotting ln[At ] versus time. The [A] values as a function of time are given in the table.
HINT: What is the concentration of B at the end of the reaction?
t, min
2
6
10
14
18
[A], M
0.00251
0.00177
0.00119
0.000971
0.00054
t, min
4
8
12
16
20
[A], M
0.00205
0.00140
0.00185
0.00084
0.00058
2.144 Pseudo-First Order Reaction
What is kpf o in the last problem?
Chm 118
Exercise Set 2
2.154
24
2.145 Pseudo-First Order Reactions: Getting the True k
For the same reaction as in problem 143, but with [A]0 = 0.003, [B]0 = 0.096. Show the
reaction is still pseudo-first-order. The [A] values as a function of time are given for integral
steps of time from 2 to 20 in steps of 2 min: 0.00232, 0.00171, 0.00134, 0.000996, 0.000789,
0.000520, 0.000388, 0.000294, 0.000155, 0.000147.
2.146 Pseudo-First Order Reactions: Getting the True k
For the same reaction as in problem 143, but with [A]0 = 0.003, [B]0 = 0.12. Show the
reaction is still pseudo-first-order. The [A] values as a function of time are given for integral
steps of time from 2 to 20 in steps of 2 min: 0.00207, 0.00146, 0.00106, 0.000660, 0.000618,
0.000292, 0.000238, 0.000146, 0.0000645, 0.000190.
2.147 Pseudo-First Order Reactions: Getting the True k
Establish that the order with respect to B is one. Find the true rate constant for the
reaction in the last several problems. HINT: Think about the equations in problem 142.
2.148 Pseudo-First Order Reactions: Getting the True k
A reaction under pseudo-first-order conditions of [H+ ] has kpf o as a function of [H+ ] as
follows in pairs of [H+ ], kpf o : 0.1M, 1.3 x 10−3 sec−1 ; 0.2M, 5.2 x 10−3 sec−1 ; 0.5 M, 3.2 x
10−2 sec−1 . What is the order of the reaction with respect to [H+ ]?
2.149 Pseudo-First Order Reactions: Getting the True k
What is the true rate constant for the reaction in the last problem?
2.150 Mechanism
For the isomerization of Cr(H2 O)5 SCN2+ (see Figure 13 and problem 138), what has to
collide to cause reaction?
2.151 Mechanism
For the isomerization of Cr(H2 O)5 SCN2+ (see Figure 13 and problem 138), suggest a mechanism? HINT: To find a mechanism is harder than to determine a rate law.
2.152 The Role of a Proton in a Reaction
In neutral solution the loss of an ammonia molecule from Co(NH3 )3+
6 is very slow. Why is
it unlikely that this reaction could occur by a rate law of the form
+
Rate = k [Co(NH3 )3+
6 ][H ]
HINTS: (1) This exercise is the kind of chemical “intuition” that is necessary in order to
suggest a mechanism. (2) The ion H+ is most likely acting as an acid; where is the base?
2.153 A Sum of Terms in a Rate Law
Use our method of “what has to collide” to understand what must be involved in a rate law
of the following form:
Rate = (k1 + k2 [A]) [B]
Chm 118
Exercise Set 2
2.157
25
2.154 The Role of a Proton in a Mechanism
The rate law for the aquation (replacement of the anion by water) of Fe(H2 O)6 F2+ is given
by the rate law:
Rate = (k1 + k2 [H+ ]) [Fe(H2 O)5 F2+ ]
Suggest a mechanism. HINT: Use chemical intuition; to what might a hydrogen ion be
attracted?
2.155 Semi-Quantitative Crystal Field Effects
In an octahedral field, the t2g orbitals are stabilized and the eg are destabilized. We showed
earlier that a numerical evaluation of this can be done if one makes some crude approximations. Then the t2g levels occur at -0.4∆ and the eg at 0.6∆. What is the energetic
stabilization of a d3 metal ion? Of a high-spin d4 metal ion?
2.156 Semi-Quantitative Crystal Field Effects
In an square-based pyramidal compound (D4h symmetry), the orbital splitting is such that
the ag level, dz 2 , is at 0.086∆, the b1g , dx2 −y2 is at 0.914∆, the eg pair, dxz and dyz are
at -0.457∆, and b2g , dxy is at -0.086∆. Compute the stability of a high spin d3 and d4
compounds for a square pyramid structure.!semi-quantitative
2.157 Semi-Quantitative Crystal Field Effects
Use your data from the last two problems to assess the crystal field contribution to the
energy of conversion of a octahedral compound into a square-based pyramidal compound as
would be formed if the mechanism of the loss of a ligand was dissociative. Can you account
for the fact that d3 such as V(II) and Cr(III) are slow to substitute and that d4 compounds
such as Cr(II) and Mn(IV) are rapid?
Chm 118
Exercise Set 2
Index of Important Concepts and Terms
∆, 17
fraction of reaction, 20
derivation, 21
absorbance
kinetics with, 21
allowedness, 14
average field approximation, 10, 12
high spin, 7, 15
intensity, 13
isosbestic point, 19
Beebe, Boniface, 23
Beer’s law, 18
use, 21, 22
Beer-Lambert law, 18
kinetics, 20
linkage isomerization, 13
low spin, 7, 15
lowering of symmetry, 5, 7
closest packing, 18
configuration
and magnetism, 16
ionic, 1
octahedral, 7
symmetry in, 7
symmetry notation, 7
tetrahedral, 7
correlation diagram, 3, 6
crystal field stabilization energy (CFSE,
25
crystal field energies
octahedral, 3
of p orbitals, 2
tetragonal, 5
tetrahedral, 4
crystal field spectra, 12
crystal field stabilization energy (CFSE),
17
effect on geometry change, 17
effect on rate, 25
spinels, 18
crystal field strength, 8
crystal field theory, 1
fields in, 8
magnetic moment, 15
structure from, 15
mechanism, 24
molar absorptivity, 13, 18
odd and even functions, 14
orbital angular momentum, 15
paramagnetic, 16
rate, 20
rate law, 20
first order plots, 22
integrated, 21
order, 21
pseudo-first order, 23
true k, 24
sum in, 24
zero-order, 21
spectrochemical series, 9
spin quantum number, 7
spinel, 18
inverted, 18
normal, 18
distortions, 5
time dependent absorbance, 18
26