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Transition Metals and Coordination Chemistry
1. In the transition metals section chemical similarities are found within a group and
across a period.
2. What are 2 transition metals that have“unique” electron configurations? What are
the configurations?
a. Cu: [Ar] 4s13d
10
b. Cr: [Ar] 4s13d
5
i. Possible explanation for configuration anomaly
“spin correlation” – ½ filled, fully filled and empty orbitals are the
most stable configurations.
3. In neutral transition metals the 3d has a similar energy compared to 4s.
In transition metal ions the 3d has a lower energy compared to 4s.
a. Why?
Shielding.
Because the d orbitals are poor at shielding, there is a greater Zeff felt by
the electrons. The greater the Zeff the closer the electrons pull toward the
nucleus. The closer the electrons are to the nucleus the lower their
energies. The lower the energy of the electrons the greater the energy that
would be required to pull them off (i.e. higher ionization energy).
4. Write the electron configurations for each of the following
a. Ti2+
- [Ar]3d2
b. Re 2+
- [Xe]4f14
5d5
c. Re3+
- [Xe]4f14
5d4
d. Cu+ - [Ar]3d
10
Remember that when you are determining the electron configuration for
an ion, you pull from the orbital with the highest “n” value first.
5. Ionization energies tend to increase going from left to right in the d-block.
6. What is a coordination compound composed of?
a. Metal Ion
b. Ligand
c. Counter Ion
7. What is a complex ion?
The metal ion and ligand combination.
8. What is a counter ion?
An ion that neutralizes the charge on the complex ion. Counter ion can be
positive or negative.
9. What is a ligand?
A Lewis Base connected to a metal ion.
a. What is a Lewis Base?
An electron donor. So the ligand must have a “free” lone pair.
10. Two characteristics of many coordination compounds.
a. Paramagnetic.
b. Colorful (absorb light in the visible range).
11. What is a primary valence? What does it correspond to?
Primary Valence = Oxidation Number
It corresponds to the charge on the metal ion.
12. What is a secondary valence? What does it correspond to?
Secondary Valence = Coordination Number
It corresponds to the number of bonds on the metal. Please keep in mind this is
not always equal to the number of ligands attached, as some ligands are able to
bind to the metal more than once.
13. Define
a. Monodentate
A ligand that attaches once (NH3, H2O, etc.)
b. Bidentate
A ligand that attaches twice (H2NCH2CH2NH2)
c. Polydentate
A ligand with multiple attachments.
d. Chelate
A ligand with more than one atom that can bind to an ion. Capable of
forming rings.
14. Method for naming a coordination compound
a. Cation is named before anion.
b. Within the complex ion, name ligand before metal ion.
c. *Use prefixes to indicate the number of specific type of ligand attached.
d. *If the ligand is an anion, change the ending to “-o”.
*If the ligand is neutral, use its name, as is.
e. If more than one type of ligand, name them alphabetically.
f. Indicate oxidation number of metal ion using (roman numerals) after metal
name.
g. *If complex ion is anion, change ending of metal name to :-ate”.
*Some special names/rules apply.
15. Fill in these naming charts
16. Name the following
My recommendation is to first get all of the names for each of the components
and then puzzle piece them together following the rules.
a.
The next thing we need to do is figure out the charge on the cobalt.
Remember that the counter ion, chlorine, is there to neutralize the charge
on the complex ion. This tells us that the overall charge of the complex
ion is +2. Because NH3 is a neutral ligand, it does not contribute to the
overall charge, which means that the cobalt must have a +2 charge.
We name the cation first, which is the complex ion. When we deal with
the complex ion we name the ligand (including its prefix)first:
hexaaminne (all one word)
Next we add the name of the metal (including it’s oxidation number):
hexaaminnecobalt (II)
Lastly we include the name of the counter ion:
hexaamminecobalt (II) chloride
b.
Now we need to figure out the charge on the platinum. Remember that the
counter ion, potassium, is there to neutralize the charge on the complex
ion. This tells us that the overall charge of the complex ion is -2. Because
Cl- is a negatively charged ligand. We can use a simple algebraic equation
to solve for the charge on the platinum.
x + 4 (-1) = -2
x = +2
So the platinum has a +2 oxidation state.
Name the cation first and then complex anion:
potassium tetrachloroplatinate (II)
c.
The counter ion indicates that the complex ion has a +3 charge. The
ligand, water, does not contribute to the overall charge. So the cobalt has
an oxidation state of +3.
Name cation first (ligand before metal) + anion
hexaaquacobalt (III) iodide
d.
There is no counter ion which means that this complex ion is neutral. The
ammine does not contribute to the charge and the nitro contributes a total
of -2. This indicates that the Co has a +2 charge.
When we name a metal ion with more than one ligand attached we must
name them alphabetically.
triamminetrinitrocobalt (III)
e.
Because the ligand has the word “di” in it already we need to use the
alternate set of prefixes when naming. It would be confusing to say
diethylenediamine. It sounds like there are 2 ethlyene and 2 ammine
ligands. By using tris we are able to indicate there are three of the entire
ligand attached.
The overall charge on the ion is +2. The ethylenediamine (en) is neutral,
so all of the charge can be contributed to the managanese.
trisethylenediaminemanganese (II) ion
f.
Now we need to determine the charge on the platinum in the cation and
the anion. To accomplish this, you would need to know that the most
common oxidation states of platinum are +4 and +2. Using this
information we can get the name:
tetraamminediiodoplatinum (IV) tetraiodoplatinate (II)
17. Give the formulas for the following
Going to follow the same basic layout as for the previous naming questions. First
we determine the identity of each component and then piece them together.
a.
Cation = Na+
Anion = Fe (CN)2(C2O4)2
We then need to determine the overall charge on the complex metal to
figure out how many counter ions we need.
This indicates that we will need 3 Na+ ions to neutralize the charge. This
means that the formula for this compound is:
Na3[Fe (CN)2(C2O4)2]
b.
Now we determine the charge on the ion:
Thus, the formula for this complex ion is:
[Cr(CO)4(OH)2]+
c. The charge on the complex ion:
This tells us that we need 2 iodide anions to neutralize the charge.
Formula:
[Cr(NH3)3Cl(en)] I2
d.
Charge:
Formula:
[PtNH3(Cl)3]-
18. A coordination compound of cobalt(III) contains four ammonia molecules, one
sulfate ion, and one chloride ion. Addition of aqueous BaCl2 solution to an
aqueous solution of the compound gives no precipitate. Addition of aqueous
AgNO3 to an aqueous solution of the compound produces white precipitate.
Propose a structure for this coordination compound.
Let’s first figure out all of the species in solution.
Co3+
, 4NH3, SO42-
, Cl-
We can figure out what is directly attached to the metal and what the counter ion
is based on the precipitate that forms. Remember that if a ligand is directly
attached to the metal it will not react to form a precipitate.
Now, let’s look at the reactions described:
BaCl2 � Ba2+
+ 2Cl-
This was put in because if SO42-
was the counter ion the addition of barium would
produce the following reaction:
Ba2+
+ SO2-
� BaSO4 (s)
This problem indicates this reaction did NOT occur, therefore the SO42-
must be a
ligand.
Let’s look at the next reaction:
AgNO3 � Ag+ + NO3
-
This would undergo the following reaction if Cl- was the counter ion:
Ag+ + Cl
- � AgCl (s)
According to the question this reaction DID occur. This means that Cl- is the
counter ion.
This means that the sulfate and ammonia groups are attached to the metal. We
will need to figure out the charge to determine the number of Cl- ions required to
neutralize the charge.
Meaning we need one chlorine. Before proposing a structure it is important to
note that SO42-
is bidentate. Knowing this we can propose the following structure:
19. How many bonds could each of the following chelates form with a metal ion?
Remember that is based on the number of distinct atoms with available lone pairs
for bonding.
20. Predicting the shape of a complex ion depends on the coordination number.
21. When the coordination number is
a. 2
i. Linear
b. 4
i. Tetrahedral
ii. Square Planar
c. 5
i. Trigonal bipyramidal
ii. Square Pyramid
d. 6
i. Octahedral
22. What is an isomer?
Isomers are compounds with the same formula, but different chemical/physical
properties.
23. Complete Isomer Flow Chart
24. Define and give an example of each of the following
a. Coordinate Isomerism
A ligand and a counter ion switch positions.
[Cr(NH3)5Br]SO4 and [Cr(NH3)5SO4]Br
b. Linkage Isomerism
This occurs when you have a ligand that has 2 or more potential areas of
attachment structured in such a way that only one or the other end is able
to attach at a time
H3N SCN NH3 NCS Co
c. Geometric (cis-trans) Isomerism
These isomers have two identical ligands that can either be placed 90o
(called cis) or 180o
(called trans) from each other.
d. Optical Isomerism
This is when you have isomers that are non-super imposable mirror
images. These type of isomers rotate plane polarized light in different
directions.
I would say that these are the hardest to visualize and if you can get your
hands on a molecule model kit, it might be a good idea to make these
models and prove to yourself that these are non-super imposable.
25. Draw geometrical isomers of each of the following complex ions.
a. [Co(C2O4)2(H2O)2]-
b. [Pt(NH3)4I2]2+
26. Which of the following ligands are capable of linkage isomerism?
SCN-, N3
-, NO2
-, NH2CH2CH2NH2, OCN
-, I
-
In this problem we are looking to see which of these ligands has two different
atoms in it with lone pairs that would be too small to form a ring.
Yes. It has two different atoms on it that can act as a Lewis Base (has lone pairs)
but it is too small to form a ring.
No. it does have more than one atom with lone pairs… but they are both nitrogen,
so this would not be able to participate in linkage isomerism.
Yes. Once again, two different atoms with lone pairs, but too small to form a ring.
No. This has two issues, one is that the lone pairs are on two different atoms, but
they are both nitrogen. Secondly this is ethylenediamine, which is long enough to
form a bidentate ring on the metal.
Yes. Again, two different atoms with lone pairs, but too small to form a ring.
No. It is only made up of one atom.
27. Draw all the geometric isomers of Pt(CN)2Br2(H2O)2. Which of these isomers has
an optical isomer? Draw the various optical isomers.
Keep in mind that the isomers that you draw may not look exactly like mine…
which is fine. What you do in order to determine if what you drew was actually
an isomer and if they relate to the ones illustrated here is to look at the
relationships in each.
Examine the relationships in each
There is only one optically active isomer:
28. How do you know if a complex ion, with tetrahedral geometry, is optically
active?
The metal has four different ligands attached.
29. In order for a molecule to be optically active it must be chiral.
30. In order to be chiral a molecule must have a non-super imposable mirror images.
31. Chiral isomers are called enantiomers.
32. An isomer that rotates plane polarized light to the left is called levorotary (l).
An isomer that rotates plane polarized light to the right is called dextrorotary (d).
33. What is a racemic mixture?
A 50/50 mixture of d/l enantiomers. This type of mixture results in no rotation of
plane polarized light (they negate each other).
34. What is the crystal field model?
It is a model that views complex ions as being held together ionically
(this is not actually the case, but it allows for a simplification of the model).
The metal (M) is the cation and the ligands are negative point charges.
35. Show the interaction between the d-orbital and the negative point charge ligands
36. Based on the diagrams above, would all of the d orbitals have the same energy?
If not which would be higher energy?
dz2 and dx2 - y2
They are high energy because the orbitals (which hold electrons) are directly
touching the negative point charges. Negative and negative in contact is highly
repulsive.
dxy, dxz, dyz
They are lower energybecause the orbitals sit between the negative point charges
(there is no direct contact).
37. The phenomena of d orbital splitting can happen two ways
a. Large ∆E – this is called the strong field case. In this case you fill the dxz,
dxy, dyz orbitals first and then proceed onto the next level. This leads to
something called the “low spin” case in which you have the smallest
number of unpaired electrons
b. Small. ∆E – this is called the weak field case. In this case you spread the
electrons out over all of the orbitals, putting one electron each before you
proceed to pairing the electrons up. This leads to something called the
“high spin” case in which you have the largest number of unpaired
electrons
38. What affects the type of splitting that will happen?
● Charge on Metal
● Electronegativity of the Ligand (↑electronegativity, ↓∆E)
a. General Ligand Ordering
CN- > NO2
- > (en) > NH3 > H2O >
-OH > F
- > Cl
-> Br
- > I
-
39. The greater the charge on the metal the stronger the splitting.
The ligands move closer to the more attractive m, increasing the repulsion
between the orbitals and negative point charge ligand.
40. The Crystal Field Method can be used to explain the
a. Magnetism
b. Color (∆E)
41. How does this model explain color?
Color that is seen is related to the ∆E between the separated d orbitals. That
difference in energy is equivalent to the energy found in the visible light
spectrum. Electrons can be promoted if they are provided enough energy to
promote from a lower orbital level to a higher level (assuming that the higher
level orbitals are not full).
42. How can we determine the color we will see?
Using the color wheel – we will see the color opposite to the one absorbed.
Remember that violet is high energy (i.e. wavelength aassociated with stronger
field cases) and red is low energy (i.e. wavelength associated with weaker field
cases).
For example, is you were told that a complex ion absorbed light in the green area,
you would see the color red. If you saw the color blue that would mean that the
energy absorbed is associated with the orange portion of the spectrum.
43. What is the d-orbital splitting for
a. Tetrahedral
Weak field case always applies. There is no direct contact with any of the
orbitals and the negative point charges. dxz, dyz, dxz just come closer to the
charges, so they are slightly higher energy.
Leading to the following ordering:
b. Square Planar
Using similar process of lining d orbitals up with the point charge
locations for a square planar molecule we arrive with the following
ordering:
c. Linear
44. The Molecular Orbital Model gives the most complete and complicated
understanding of bonding in coordination compounds. Draw the MO diagram for
an octahedral compound with 6 ligands.
45. How does the electronegativity of a ligand affect the d orbital splitting?
The more electronegativity, the lower the energy of the electrons, the more tightly
bound the electrons are to the ligand. Due to this, the ligand/metal atomic orbitals
don’t mix as much, meaning a smaller energy split between d orbitals.
46. Compounds of copper(II) are generally colored, but compounds of copper(I) are
not. Explain. Would you expect Cd(NH3)4Cl2 to be colored?
Looking at the electron configuration can help to explain this observation.
Cu+: 3d
10 for this compound the d orbitals are full – meaning there si no room for
the electrons to be promoted up. If electrons cannot be promoted up, there can be
no release of energy when they fall back down – no release of energy means we
see no color. Remember that the color that we see is related to the ∆E of the d
orbitals split.
Cu2+
: 3d9 : here is there is room for electrons to promote up.
Cd+2
: 4d10
: full orbital, no rrom for promotion � no color.
47. Consider the complex ions Co(NH3)63+
, Co(CN)63-
, and CoF63-
/ The wavelengths
of absorbed electromagnetic radiation for these compounds are (in no specific
order) 770 nm, 440 nm, and 290 nm. Match the complex ion to the wavelength of
absorbed electromagnetic radiation.
All of the ligands are attached to the same metal. So we just have to consider the
ordering of the ligands and which lead to the strongest field. Remember that
energy and wavelength are inversely proportional, so as energy goes up,
wavelength goes down.
Based on our previous ordering listing (#38) we see:
This means that
48. How many unpaired electrons are in the following complex ions?
a. Ru(NH3)62+
(low spin case)
The charge on the Ru = +2
It would have a d orbital electron configuration of 4d6
As NH3 is a monodentate ligand, the coordination number for Ru = 6
This tells us that it has octahedral geometry. So we will be looking at the
d orbital split for the octahedral case.
We have six electrons to put into the diagram, because it is a low spin
case, that means it is strong field and we fill the lower d orbitals before
moving onto the higher energy d orbitals.
There are no unpaired electrons.
b. Ni(H2O)62+
The charge on the Ni = +2
It would have a d orbital electron configuration of 3d8
As H2O is a monodentate ligand, the coordination number of Ni = 6
This tells us that it has octahedral geometry. So we will be looking at the
d orbital splitting for the octahedral case:
We have 8 electrons to put in. In this case they did not indicate whether it
was high spin or low spinp; that, is because it would be irrelevant as we
will have to start filling the higher d orbitals regardless of the energy split
based on number of electrons present.
There are 2 unpaired electrons.
c. V(en)33+
The charge on the V = +3
It would have a d orbital electron configuration of 3d2
As (en) is a bidentate ligand, the coordination number of V = 6
This tells us that it has octahedral geometry. So we will be looking at the
d orbital splitting for the octahedral case:
We have 2 electrons to put in. In this case they did not indicate whether it
was high spin or low spin, that is because it would be irrelevant as we will
would not have to use the higher energy orbitals (there are not enough
electrons).
There are 2 unpaired electrons.
49. Draw the d-orbital splitting for the octahedral complex ions in each of the
following.
a. Fe2+
(high and low spin)
First determine the number of electrons in the d orbital.
Fe2+
: 3d6
b. Ni2+
First determine the number of electrons in the d orbital.
Ni2+
: 3d8
c. Zn
2+
First determine the number of electrons in the d orbital.
Zn2+
: 3d10
