Download Self-Study_Worksheet II_Basis for Color_ANSWERS

Chemistry 122
Prof. Mines, Spring 2013
Self-Study Worksheet II, Basis for Color in TM Complexes
This worksheet is due in class on the day that we return from the flood. It will be counted in some asyet undetermined way toward your course grade.
Task 1. Read the PowerPoint (TM II-Basis for Color) slides 1-6
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Exercise 1: Write the electron configuration and state the number of d electrons in a (free) Mn
Electron Configuration
Orbital Diagram
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a) Mn :
[Ar] 3d4
ion (review)
___ ___ ___ ___ ___
Task 2. Read Slides 7-10 in the PowerPoint
Exercise 2: Answer the following:
a) Draw the (crystal field) splitting pattern for a TM cation in an octahedral complex (do not write the
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z , etc. designations. Just show a line for each of the five d orbitals). Do not add any arrows
(electrons) to this diagram yet.
b) Explain briefly the difference between this diagram and the one that you drew in Exercise 1 (other
than the fact that there are no arrows). Why do you think it is called a “splitting” diagram? When
does splitting occur? (Answer: When there are __ligands present)
Initially all five d orbitals have the same energy. Once ligands bind to the metal cation, the
energies of some of the different d orbitals are affected differently than others, causing them to
have a different energy. In the case of an octahedral “field” of ligands (i.e., a geometry of
octahedral), it turns out that two of the d orbitals end up having a higher energy than the other
three. This creates two “levels” where before there was only one. In a sense, the d-sublevel is
“split” into sub sublevels. That is why it is called a “splitting” diagram.
c) Is  generally considered to have a large value or a small value (relative to the difference
between an s and p sublevel or a p and d sublevel)?
Small!
Task 3. Read Slides 11-14 in the PowerPoint.
Exercise 3: Answer the following:
a) What two factors affect the value of ?
The nature of the metal cation (i.e., element and charge) and the ligands bonded to the cation.
b) If metal complex A is said to be “strong field” and complex B is said to be “weak field”, that
means that the ______ is smaller in complex __B__.
Chemistry 122
Prof. Mines, Spring 2013
c) The value of  in metal complexes generally is such that if a photon of light is absorbed and an
electron in the lower of the two d levels is transitioned to the higher level, the wavelength of the
photon that is absorbed will be in the __visible__ region of the electromagnetic spectrum.
Task 4. Read Slides 15-22 in the PowerPoint.
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Exercise 4: Assume a metal cation has seven d electrons (i.e., a d configuration).
a) Draw the splitting diagram for this cation assuming it were in a “weak field” (on the left) and then
in a “strong field” (on the right)
3 unpaired electrons
1 unpaired electron
Pairing energy is
greater (that’s why it
does not pair until
after the upper two
orbitals have one
electron in them)
Pairing energy is
smaller (that’s why it
pairs in the lower
orbitals before the
upper orbitals start
filling.
Weak field case
Strong Field Case
b) How many unpaired electrons would it have in each case? (see boxes above)
c) Is the pairing energy greater than or smaller than  in each case? (see boxes above)
Task 5. Read Slides 23-27 in the PowerPoint.
Exercise 5: In which case in Exercise 4 (weak field or strong field) is the (total amount of) spin greater?
This one would be called the _high_-spin case.
the weak field case
Is either of the complexes diamagnetic? Paramagnetic? Explain.
Both are paramagnetic because each has at least one unpaired electron. So neither is
diamagnetic. That said, the high-spin complex is more paramagnetic because it has more
unpaired electrons.
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Exercise 6: Consider MnF6 . Recalling that fluoride is a -1 ligand, and assuming that the complex is
“weak field”, draw the splitting diagram for the complex, and state the number of unpaired
electrons.
See Exercise 1!
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1) Determine the charge on the cation: x + 6(-1) = -3  x = +3; So its Mn
2) Determine the number of d electrons (i.e., write the electron configuration). [Ar] 3d
4
3) Now the problem is just like Exercise 4(a) above but with 4 d electrons!
Is the complex diamagnetic or paramagnetic? Paramagnetic (4 unpaired electrons)
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Exercise 7: Why is Al(H2O)6
not colored? (Hint: See slide 26)
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2
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The cation here is Al , which has a configuration of [Ne]3s 3p . Since there are no d electrons,
there will be no absorption of visible light wavelengths (no d to d electronic transitions), and thus
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no “color” (white light comes in and it all comes back out!) Any absorption of light in Al will
correspond to higher-energy (UV) absorption because the gaps are so large.