Suggested Problems
for quiz on 13 Mar 08 (last quiz…)
**CHECK OUT the correction to the answer for the first problem, if you’re
printing out a new version of this after the Monday lecture.
20.1 Give the electron configuration of the metal in each of the following:
b. Co2+
c. MnO2
20.14 Draw a crystal field d-orbital energy-level diagram, and predict the number of
unpaired electrons for each of the following complexes:
**NOTE: I will give you the spectrochemical series as listed on the bottom of p.
899. You will need to know what “weak field” and “strong field” mean in terms
of relative size of .
a. [Fe(H2O)6]2+
b. [Fe(CN)6]420.45 Which is the stronger reducing agent, Ti2+ or Ni2+? Explain.
20.69 Identify the oxidation state of the metal in each of the following complexes:
a. [Ni(CN)5]3c. [Co(en)2(H2O)Br]2+
20.75 What is the oxidation state of the metal in each of the following compounds?
a. (NH4)3[RhCl6]
c. [Cu(en)2]SO4
20.78 What is the systematic name for each of the following coordination compounds?
a. Cs[FeCl4]
c. [Co(NH3)4Br2]Br
20.79 What is the systematic name for each of the following compounds?
a. [Cu(NH3)4]SO4
20.80 Write the formula for each of the following compounds:
a. Tetraammineplatinum(II) chloride
b. Sodium hexacyanoferrate(III)
20.84 Which of the following complexes can exist as diastereomers?
a. [Cr(NH3)2Cl4]b. [Co(NH3)5Br]2+
20.92 What is an absorption spectrum? If the absorption spectrum of a complex has just
one band at 455 nm, what is the color of the complex?
20.97 (This is one I can’t draw an answer to – but be sure you understand this!!) Use a
sketch to explain why the dxy and dx2 – y2 orbitals have different energies in an octahedral
complex. Which of the two orbitals has higher energy?
20.102 The Ni2+(aq) cation is green, but Zn2+(aq) is colorless. Explain.
23.2 Draw the five alkane isomers with the formula C6H14.
23.50 If someone reported the preparation of a compound with the formula C3H9, most
chemists would be skeptical. Why?
20.1 **CORRECTION to this answer… something even I forgot: when you’re filling
the orbitals with electrons, 4s fills before 3d. However, when you’re removing electrons
(forming cations), the d orbitals will remain filled while you first remove s electrons (for
instance, you first remove the 2 electrons in the 4s orbitals before you remove any of the
3d electrons). The answers below reflect this as of 10 Mar.
b. Remember that we have 2 fewer electrons because this is a 2+ ion.
Typically Co is [Ar]4s23d7, but Co2+ is [Ar]3d7
c. Each O will be -2, and there are two of them. So, Mn in this compound will be
+4. Consider that Mn is typically [Ar]4s23d5, and without the 4 electrons in this
compound, it is [Ar]3d3
20.14 a. H2O is in the middle of the spectrochemical series… Fe has 6 d electrons when
uncharged, but 4 d electrons with a 2+ charge. This is a tougher call, with H2O being in
the middle of the series. If the ligand were more on the weak-field side,  would
definitely be small, and the four electrons would be spread unpaired over the lower and
higher energy levels. There would be four unpaired electrons.
b. CN- is the strongest of the strong field ligands, and would produce the largest d
orbital splitting. Because of the large  value, there is a tendency for the electrons to
stay in the lower level as much as possible. Fe is still in the 2+ state (because each CN is
-1) and there will be 4 d electrons to place in orbitals. However, this time, they will all
be in the lower level – two of them will be paired, and only two unpaired. (This also
affects how the compound behaves in a magnetic field, but our chapter 20 barely
addresses that…)
20.45 A reducing agent gives up electrons. The element with the lower Zeff will be more
willing to give up electrons. Lower Zeff is farther left on the periodic table, so Ti2+ will be
the stronger reducing agent.
20.69 a. Each CN has a -1 charge. 5 of them total = -5 charge. The entire ion has a 3charge. To maintain the overall charge, the nickel must have a +2 charge, and
therefore the oxidation state of nickel is +2.
c. Thinking of each ligand: en is neutral, H2O is neutral, Br has a -1 charge. For
the entire ion to have a +2 charge, Co will need to have a +3 charge. Thus, the
oxidation state of Co is +3.
20.75 a. The charge on an ammonium ion (NH4+) is +1. The charge on a Cl- is -1.
There are 6 Cl-, contributing -6 total, and there are 3 ammoniums, contributing
+3. To make an overall neutral compound, the Rh has to be +3, and the
oxidation state of Rh is therefore +3.
c. en is neutrally charged, and SO42- has a -2 charge. To keep the compound
overall neutral, Cu has to have a +2 charge, and its oxidation state is also +2.
20.78 a. cesium tetrachloroferrate(III) (cesium forms a +1 ion)
c. tetraamminedibromocobalt(III) bromide
20.79 a. tetraamminecopper(II) sulfate
20.80 a. [Pt(NH3)4]Cl2
b. Na3[Fe(CN)6]
20.84 a. Yes, this can exist as cis/trans isomers. This has 6 total ligands and will form
an octahedral complex. There are two different ligand types. In the trans isomer, the
two ammonia ligands can arrange themselves opposite each other; in the cis isomer, the
two ammonias can arrange themselves next to each other.
b. No isomerism is possible here. You must have at least two of each of the
isomer types. There is only one Br; you can’t arrange the complex with two of them next
to each other if you only have one!
**Be able to draw the cis and trans isomers of the above, as well.
20.92 Absorption spectrum = plot of absorbance of a compound vs. wavelength. If the
compound absorbs maximally at 455 nm, it will be orange.
20.97 I mainly want you to be aware that one set of orbitals will overlap the ligands –
which are also electron rich, so you’ll have an electron-electron repulsion, which is an
“uncomfortable” high-energy situation. The other set of orbitals (and you don’t need to
memorize the names of them) line up in between the ligands, so while they will also be
raised in energy when ligands bind, they are not raised to the same extent that the
orbitals directly along the axes are.
20.102 First of all, when you place either cation in water, likely coordination complexes
of various geometries will form with water as the ligand. If you think of the two energy
levels after the d orbitals have split, and place the d electrons into those energy levels…
Ni2+ has 8 d electrons
Zn2+ has 10 d electrons
What are the consequences of this?
All of the electrons in Zn2+ are paired. If a Zn2+ ion absorbs light, an electron
might get excited by the light, but there is no room in the next energy level for the
electron (since it is full of paired electrons). Thus, the light ends up not being
absorbed, and no color is produced.
Ni2+, on the other hand, has room for two more electrons in its higher energy
level, and when it absorbs the proper wavelength of light, an electron can get
excited and “jump” into the higher energy level – there is room for the electron to
stay in the higher level.
23.2 As the formula (# carbons) gets bigger, I won’t ask you to draw every single isomer,
because most people struggle to figure out the last one. I will ask you to draw most of
them, though.
CH3CH2CH2CH2CH2CH3
CH3CH2CHCH2CH2CH3
|
CH3
CH3CHCH2CH2CH3
|
CH3
CH3
|
CH3CHCHCH#
|
CH3
CH3
|
CH3CHCH2CH3
|
CH3
23.50 In order to have 9 hydrogens and 3 carbons, one of the carbons would need to
have more than 4 bonds, which is not possible. A three carbon alkane is fully saturated
with 8 hydrogens.