Chemistry 122
Mines, Spring 2010
Answer Key, Problem Set 11
1. Write the electron configuration for each of the following atoms and ions (use the noble gas abbreviation):
(a) Mn: [Ar] 4s2 3d5
(Note: some texts would write [Ar] 3d5 4s2; the order doesn’t matter
as long as the correct number of electrons in each sublevel are shown)
(b) Fe: [Ar] 4s2 3d6
(c) Fe3+: [Ar] 3d5 (or [Ar] 4s0 3d5 for emphasis)
Note: There were 8 electrons “after” the Ar configuration. The 3+ ion results from the
removal of 3 electrons from the neutral atom, hence 5 electrons (after Ar). But since
in cations, the 3d is lower in energy than the 4s, all 5 electrons end up in the d
sublevel.
(d) Ag+: [Kr] 4d10 (Ag has the apparent configuration [Kr] 5s2 4d9 (11 e-‘s “after” Kr). Take
away 1 electron and you’re left with 10 “after” Kr. All end up in the d sublevel.
(e) Rh3+: [Kr] 4d6 (5s2 4d7 take away 3 electrons…, but 4d lower than 5s)
(f) Mn6+: [Ar] 3d1 (4s2 3d5 take away 6 electrons…., but 3d lower than 4s)
NOTE: Again, remember that once electrons are removed from a first-row transition metal
atom to form a cation, the 3d orbitals drop down in energy more than do the 4s, and
so all the electrons end up in the d subshell. The same is true for the 4d and the 5s,
and the 5d and the 6s, etc.--the nd sublevel becomes lower in energy than the (n +
1)s sublevel in a transition metal cation.
2. Predict the number of unpaired electrons for each of the following:
NOTE 1: To get the number of unpaired electrons, you first need to get the number of electrons by writing (or
thinking of) the electron configuration (as in Problem 2 above). Then you put them into an orbital
diagram.
NOTE 2: In THIS problem, the ions are all "free" (i.e., NOT bonded with any ligands), so all five d orbitals are
degenerate (i.e., they have the same energy). When ligands are present, the d orbital sublevel "splits"
into groups or pairs of non-degenerate orbitals (see problems on "crystal field" theory such as NT3 and
NT4.)
3. Identify the Lewis acid and the Lewis base in the reaction of oxalate ions with Fe3+ to give [Fe(C2O4)3]3-.
Answer: Fe3+ is the Lewis acid and oxalate ion is the Lewis base. A Lewis acid is an
electron-pair acceptor (and so it is usually "electron deficient", as a metal cation most
certainly is), and a Lewis base is an electron-pair donor, as are most anions (although many
PS11-1
Answer Key, Problem Set 11
Lewis bases are (neutral) molecules with lone pairs as well). In any complex ion, the metal
cation is the Lewis acid and the ligands are all Lewis bases.
4. What is the coordination number of the metal in each of the following complexes?
(a) AgCl2-: TWO; two monodentate chloro ligands
(b) [Cr(H2O)5Cl]2+: SIX; five monodentate aqua ligands and one monodentate chloro ligand
(c) [Co(NCS)4]2-: FOUR; four monodentate isothiocyanato ligands
(d) [ZrF8]4-: EIGHT; 8 fluoro ligands (somewhat rare; possible only because F- is so small.
(e) Co(NH3)3(NO2)3: SIX; three monodentate ammine ligands and three monodentate NO2ligands
(f) [Fe(EDTA)(H2O)]-: SEVEN; one hexadendate EDTA ligand and one monodentate aqua
ligand (A C.N. of 7 is also somewhat rare)
5. What is the oxidation state of the metal (i.e., charge on the metal cation) in each of the complexes in the
prior problem?
(a) +1
Two Cl-'s sum to -2. So -1 (overall charge) must equal -2 + n => n = +1
(b) +3
One Cl- and 5 neutral H2O's add up to -1.
So 2+ (overall charge) must equal -1 + n => n = +3
(c) +2
Four SCN-'s equal -4. So -2 (overall charge) must equal -4 + n => n = +2
(d) +4
Eight F-'s equal -8. So -4 (overall) = -8 + n => n = +4
(e) +3
Three neutral NH3's and three NO2-'s sum up to -3.
So 0 (overall) must equal -3 + n => n = +3
(f) +3
One EDTA4- and one neutral H2O sum up to -4. So -1 = -4 + n = n = +3
6. Give the coordination number of the transition element in each of the following:
The coordination number is determined by the number of ligands bound AND how many
atoms each ligand uses to bind to the metal ion. Anything that is within the brackets (except
the metal ion, of course) must be a ligand to the metal ion in the brackets; anything that is
outside of the brackets is NOT a ligand to the metal ion inside the brackets.
(a) [Ni(NH3)6](ClO3)2
6 NH3 ligands bound; each binds with only one atom (N)
=> C.N. = 6 (the ClO3- ions are just counterions to the complex, not ligands)
(b) [Cu(NH3)4]SO4
4 NH3 ligands bound; each is monodentate
=> C.N. = 4 (sulfate is just a counterion)
(c) [Cr(en)3]Cl3 3 en (ethylenediamine) ligands; each is bidentate (binds with two
distinct N atoms)
=> C.N. = 3 x 2 = 6
PS11-2
Answer Key, Problem Set 11
(d) K2[Ni(CN)4]
4 CN- ligands; each is monodentate (binds through C)
=> C.N. = 4
7. Give the oxidation state (i.e., charge on) the metal cation in the coordination compounds below. Also
state the coordination number and coordination geometry of the complex ion, and make a sketch of it.
Na is +1  [Au(CN)2] is -1. Each CN is -1, thus
x + 2(-1) = -1  x = +1
[NC
CN]
SO42- is -2  [Co(NH3)5Br] is +2. Each NH3 is +0, Br is
-1, thus x + 5(0) +1(-1) = +2  x = +3
en is bidentate! Thus although there
are three ligands, the C.N. is 4.
NH4+ is +1  [PtCl2(C2O4)2] is -2. Each Cl- is -1, C2O4
(ox) is -2, thus x + 2(-1) + 2(-2) = -2  x = +4
PS11-3
Answer Key, Problem Set 11
8. What does it mean if two species (or compounds) are isomers? Give examples of a pair of coordination
isomers and a pair of linkage isomers. Are these considered structural isomers or stereoisomers? Give
a definition for stereoisomer, state the two subclasses of stereoisomer, and sketch a pair of
stereoisomers (any kind; can be from text).
Two chemical species (or compounds) are isomers if they have the same chemical
formula but at least one different property. That is, they are not the same exact
chemical species (or compound).
Coordination isomers are coordination compounds that have the same overall formula,
but whose complex ions do not have the same exact composition (there is a different
“coordination sphere” around the TM cation). They arise when an anion that acts as a
ligand in one compound is a counter anion in the other and vice versa (or in the case in
which the anion and cation are both complex ions, at least one ligand is swapped
between the two complex ions—see Q16.). An example of a pair of coordination
isomers would be: [Ru(NH3)5Cl]Br and [Ru(NH3)5Br]Cl. The complex ion in the first
compound is Ru(NH3)5Cl+ and that in the second one is Ru(NH3)5Br+. NOTE: To
“create” (on paper) a coordination isomer from a given coordination compound [having a
counter anion], you must “swap” at least one ionic ligand with a counter anion.
Linkage isomers are coordination compounds (or complex ions) in which all the same
ligands are bound to the same metal ion, but one or more of the ligands bind(s)
(coordinates) via a different atom. One example is [Fe(NH3)5SCN]Cl (S is bound to the
metal ion) and [Fe(NH3)5NCS]Cl (N is bound to the metal ion). One could also say that
the complex ions Fe(NH3)5SCN- and Fe(NH3)5NCS- are linkage isomers.
Both of the above types of isomers are considered to be structural isomers since they
differ in the type of bonds that are made between metal and ligand (the “connectivity”
pattern differs).
Stereoisomers are compounds (or complex ions) that have the same types of bonds
(and same number of these bonds), but the arrangement of these bonds in space differs
in some way. In geometric isomers, the difference is in the spatial arrangement
(relative positions are not the same); in optical isomers (enantiomers), the difference
is in the “handedness” of the arrangement (i.e., they are mirror images of one another
[that are not identical]).
Ex. Geometric isomers:
H3N
H3N
Pt
Cl
Cl
Cl
H3N
cis
Pt
NH3
Cl
trans
Ex. Optical isomers (Enantiomers):
OH2
H2O
OH2
N
N
Co
Co
N
O
N
O
O
O
PS11-4
OH2
Answer Key, Problem Set 11
9. Consider the following structures, all of which have the formula [Cr(NH3)2Cl4]-
(a) Identify each structure as either cis or trans.
(1) and (4) are cis (two NH3's are at 90 to one another [same side];
(2) and (3) are trans (two NH3's are at 180 to one another [across]
(b) Which structures are identical, and which are different?
(1) and (4) are identical, and (2) and (3) are identical. If you just rotate the structures,
you can see that (1) is superimposable with (4) and so is (2) with (3).
(c) Do any of these structures have an optical isomer? Explain.
No. The mirror images of each of these structures are superimposable with (not
different than) the original structure. One can also conclude this (possibly more easily)
by looking for a plane of symmetry (a plane that cuts the structure in half so that each
side is a mirror image of the other). Recall that a structure that has a plane of symmetry
cannot be chiral; its mirror image will be superimposable (see PowerPoint slides). Each
of these structures has a plane of symmetry, and so none has an optical isomer.
10. Consider the following complexes, each of which contains at least two ethylenediamine (en) ligands.
(a) Which complexes are chiral (i.e., have an optical isomer), and which are achiral (i.e., have no optical isomer
[because the mirror image is identical/superimposable])?
Answers:
(1) No plane of symmetry  chiral (Note: If the en ligands were separated (i.e., two
NH3's, then there WOULD be a plane of symmetry and the structure would be
achiral--see Problem 10!! But the "connectedness" makes the mirror images
nonsuperimposable and thus gives this structure its chirality.)
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Answer Key, Problem Set 11
(2) The plane with all the N atoms (and the Cr) is a symmetry plane => achiral.
(3) Chiral. Same idea as in (i); if this were [Cr(NH3)6]2+, then it would be achiral, but the
bidentate nature of the en ligands breaks up the symmetry and makes the structure
chiral.
(4) Chiral. This structure is analogous to structure (1).
(b) Draw the optical isomer (enantiomer) of each of the chiral complexes. (Hint: Just draw the mirror
image! If the structure is, in fact, chiral, then the optical isomer is the mirror image [and it should
not be identical to/superimposable upon the original structure].)
(c) Which, if any, of the chiral complexes are optical isomers (enantiomers) of one another?
Answer:
(1) and (4) are enantiomers. This is easier to see if you rotate structure (1) 90 around
one of the axes so that the Cl that is "below" the Cr ends up in the same horizontal
plane as the other Cl (like the two Cl's are in structure (4)). Perhaps I can draw this
for you later when I have access to the correct software at work. If I don’t have
time, just ask me about this in the review session.
PS11-6
Answer Key, Problem Set 11
11. Draw the d-orbital splitting diagram, including the proper number and placement of electrons, for the
following octahedral complex ions (make sure you show how you determine the number of electrons in
each case!). State the number of unpaired electrons in each.
(a) ZrCl64-
(b) OsCl62- (low-spin)
(c) MnCl64- (high-spin)
Commentary:
**NOTE: “high-spin” or “low-spin” is only a useful distinction if there are two possibilities for the way electrons
can be arranged. This doesn’t occur, for example, if the configuration is d1 because there would be one unpaired
electron regardless of whether or not the splitting was larger than P (see below) or smaller than it (we don’t have
to pair any electrons up in this situation, so it makes no difference).
(Spin) Pairing energy (P) refers to the increase in energy that results when an electron is placed in an orbital that
already has an electron in it rather than one that is empty. That is, it is the “energy price” paid for pairing two
electrons in the same orbital. It results because when two electrons are in the same orbital, they spend their
time (on average) in the same region of space and therefore are (on average) closer to each other than would be
the case if they were in different orbitals. Since electrons repel one another, putting them in the same region of
space means more repulsion and hence a higher (potential) energy. NOTE: THE PAIRING ENERGY IS
BASICALLY A CONSTANT; IT DOESN’T VARY MUCH BETWEEN ORBITALS WITHIN A COMPLEX OR
BETWEEN METAL COMPLEXES.
If the pairing energy is larger than the difference in energy between the lower-energy d orbitals and the higher
energy ones (i.e., if P is larger than the crystal field splitting energy, ), then it will cost more energy to pair up an
electron with one in a lower-energy d orbital than to place the electron in the empty higher energy d orbital, and
so the electron will end up in the higher energy orbital (and be unpaired). If the pairing energy is smaller than ,
then it will cost less energy to just pair up in the lower energy d orbital, and so the electron will end up paired.
Thus when the P >  the total of the spins of the d electrons will be greater than if P <  and so the P >  case is
called “high spin” and the P <  case is called “low spin”.
Now that I’m thinking about it, although I framed the discussion above in terms of whether or not “P is greater
than or less than ", I actually tend to think of it more often in terms of whether or not “ is greater than or less
than P”. Mathematically it doesn’t matter which way you compare them, but conceptually this makes sense to
me because as noted above, it is the P that is basically constant—IT IS THE  VALUE THAT CHANGES
BETWEEN COMPLEXES. So I’ll rewrite the above “logic” again here. If < P, then the energy gap up to the
next orbital is smaller than the energy needed to pair, and so the electron just goes up to the higher energy
orbital (they don’t pair up in the lower energy one). Conversely, if > P, then the energy gap up to the next
orbital is larger than the energy needed to pair, and so the electrons just go ahead and pair up in the lower
energy orbital. In either case, the system adopts the configuration that is the lowest in energy overall!
Answers:
(a) ZrCl64- 6 Cl-‘s  -6. Overall charge is –4, so the charge on Zr must be +2.
The electron configuration for Zr is: [Kr] 5s2 4d2
Thus, the configuration for Zr2+ (removing two electrons) is just [Kr] 4d2 (see Problem #1)
Octahedral splitting pattern is:
2 unpaired electrons regardless of the magnitude of the splitting energy, so it is
not designated as “high spin” or low spin” (Note, however, that this ion is
paramagnetic, since it has two unpaired e-‘s)
(b) OsCl62- (low spin) 6 Cl-‘s  -6  Os must be +4 since overall charge is –2.
PS11-7
Answer Key, Problem Set 11
Os: [Xe] 6s2 4f14 5d6  (remove 4 electrons) Os4+: [Xe] 4f14 5d4
(If it is low spin, that means that the splitting E
must be fairly large so I’ve tried to draw it that way.)
From the diagram, there are 2 unpaired e-‘s
“Low-spin” means that
electrons pair up before
filling the higher-energy
d-orbital sublevel
(occurs when  > P)
(c) MnCl64- (high spin) 6 Cl-‘s  -6  Mn must be +2.
Mn: 4s23d5  Mn2+: 3d5
“High-spin” means that
electrons don’t pair up
until all the d orbitals
have an electon in them
(occurs when  < P)
5 unpaired electrons
** Note that the ligand in all three cases is Cl-, generally considered to be a fairly weak-field ligand, but it
sometimes results in a strong field case. This is because the metal ion and oxidation state affect the
splitting too. A high oxidation state (and being lower in the periodic table) makes the splitting bigger, so
that is why even with Cl you can get a low-spin (strong-field) case.**
12. If Co(ox)33- is observed to be diamagnetic, is ox acting as a strong-field or weak-field ligand in this
complex ion? Is the complex “high-spin” or “low-spin”? Is the pairing energy greater than or less than
2the crystal-field splitting energy, ? (ox = oxalate ion = C2O4 )
Answers: ox acts as a strong-field ligand; complex is low-spin; P < 
Explanation: Diamagnetic means “no unpaired electrons”. Ox has a charge of 2- per
ligand, so 3 ox’s means –6 contribution to the charge. Thus, the Co must be +3 since the
overall charge is –3.
Co: 4s23d7 => Co3+: 3d6 => 6 electrons in the d orbitals. There are two possibilities:
P
low spin
high spin
strong field
weak field
If the complex is diamagnetic, that means that it must be “low spin” (actually, “no spin”
here!), and so the ox ligand is acting as a strong-field ligand here. If it were acting as a
weak field ligand, then there would be 4 unpaired electrons and the complex would be
paramagnetic. THIS IS HOW WE CAN TELL WHETHER SOMETHING IS HIGH-SPIN
OR LOW SPIN EXPERIMENTALLY (we measure its paramagnetism using a “Gouy
PS11-8
Answer Key, Problem Set 11
balance” or other similar device which measures the force of attraction of a compound to
an externally imposed magnetic field). Since the electrons have paired up, the pairing
energy must be less than the splitting energy ( > P is the same as P < ! Note
comment above in the answer to NT3 regarding the two different ways to “look” at it).
13. Given that the color in emeralds and rubies is due to Cr3+ (in both) even though emeralds are green and
rubies are red, and given the basic shapes of the spectra of each (shown in class; two bands, one in the
blue region and one at lower energy), which gemstone is providing the Cr3+ with the stronger crystal field
environment (i.e., which has the larger )?
Answer: Ruby. Both metal complex ions have two absorption bands in the visible region,
one in the blue and one at lower energy (see below). The difference in color basically
comes from the position of the lower-energy absorption band: in ruby, which is red, the
band must NOT be absorbing much red light. The band must be in the green region (so that
overall, the two bands are basically absorbing the blue and green light to yield a red color).
The emerald on the other hand appears green. That means that the complex must be
absorbing some of the red light and letting some of the green light through. Thus in
emerald, the lower-energy band must be moved toward the red end of the spectrum
compared to the ruby’s band. SINCE RED LIGHT IS COMPOSED OF LESS ENERGETIC
PHOTONS THAN GREEN LIGHT, THE EMERALD IS THE ONE WITH THE SMALLER
SPLITTING, AND THE RUBY MUST HAVE THE LARGER SPLITTING. Since the metal ion
is the same in both stones, the lattice in the ruby must be providing the stronger crystal-field
environment!
Notice that this band (both bands actually) has shifted
“left” compared to its position in the spectrum of
“emerald”. That shift to higher energy is what causes
more green and less red to be absorbed in ruby,
causing ruby to appear red. (Note that not ALL the
green and blue light need be absorbed for something
to appear red! It’s just that red must be the
predominant color reaching our eyes! Actually, it is
clear from the emerald spectrum that not all of the
RED light need be absorbed for something to appear
green! Apparently just enough red light is absorbed in
emerald so that green “wins out”. Different shades of
green are a result of slightly different amounts of the
other colors getting through to our eyes.)
14. Although Cl- is a weak field ligand and CN- is a strong-field ligand, [CrCl6]3- and [Cr(CN)6]3- exhibit
approximately the same amount of paramagnetism. Explain.
The degree of paramagnetism is predominantly determined by the number of unpaired
electrons. Since Cr3+ has only three unpaired electrons, neither case will involve pairing any
of the electrons, and so the orbital diagrams will be the same in both cases. Regardless of
PS11-9
Answer Key, Problem Set 11
the value of the splitting energy, in each case, there will be 3 unpaired electrons and a
similar paramagnetism.
weak field case
with three d
electrons
strong field case
with three d
electrons
NOTE THAT FOR AN OCTAHEDRAL COMPLEX, THIS SITUATION WILL OCCUR IN THE
FOLLOWING OTHER CONFIGURATIONS: d1, d2, d8, d9, and d10. Write out a few and see
for yourself! (See Problem 20.118.)
15.
(From 21.51, Zumdahl, 7/e). Consider the following series of complexes of chromium: Complex I appears purple,
Complex II appears yellow, and Complex III appears cyan (bluish-green). (a) Predict the predominant color(s) of light
3+
3+
+
absorbed by each complex. (b) If one of the complexes is Cr(NH3)6 , one is Cr(H2O)6 , and one is Cr(H2O)4Cl2 ,
determine the identity of each complex (I, II, and III). You must provide reasoning! (Hint: For part (b), you must
consider the spectrochemical series (see text)).
Answers:
(a)
(b)
Complex I (purple): green is predominantly absorbed; contains Cr(H2O)63+
Complex II (yellow): blue is predominantly absorbed; contains Cr(NH3)63+
Complex III (cyan): red is predominantly absorbed; contains Cr(H2O)4Cl2+
Explanations:
(a) The color that a complex appears is a composite of the colors that are not getting
absorbed (they are the colors that are making to your eyes). If you use trichromatic
color theory, then consider R, G, and B as the primary colors:
Since purple  R + B, those colors must be getting transmitted, and so green must
be getting absorbed.
II: Since yellow  R + G, those colors must be getting transmitted, and so blue must be
getting absorbed.
III: Since cyan  B + G, those colors must be getting transmitted, and so red must be
getting absorbed.
I:
(b) The spectrochemical series indicates that NH3 is a stronger-field ligand than H2O, and
H2O is stronger than Cl-. Thus the splitting energy () should be largest in Cr(NH3)63+,
smallest in Cr(H2O)4Cl2+, and in between in Cr(H2O)63+:
largest : Cr(NH3)63+ > Cr(H2O)63+ > Cr(H2O)4Cl2+ smallest 
Now consider the colors (and relative wavelengths) of light being absorbed by each
complex:
smallest : blue (II)
>
green (I)
PS11-10
>
red (III)
longest 
Answer Key, Problem Set 11
Since Ephoton  hc/, we can now order the complexes in order of Ephoton absorbed, which
directly corresponds to , the splitting energy:
largest Ephoton (&:
blue (II)
>
green (I)
>
smallest Ephoton (&
red (III)
Since blue light has the greatest energy per photon, its absorption must occur in the
complex with the largest  (i.e., Cr(NH3)63+). Red light has the smallest energy per
photon, and thus its absorption occurs in the complex with the smallest  (i.e.,
Cr(H2O)4Cl2+). And green light, with the middle energy per photon, must be being
absorbed in the complex with the medium  (i.e., Cr(H2O)63+).
16. Tro, 24.32 (formulas only).
NOTE 1:
NOTE 2:
NOTE 3:
Write the formulas and names for the coordination isomers of [Co(en)3][Cr(ox)3]
It does not matter which ligand is put first in the formula (en or ox). I chose to put en first
in all of them just for consistency.
Technically, the cation is always written first in an ionic compound, so I’ve done that
below. My guess is that many of you did not even consider this issue. It is a bit subtle
(and makes the problem a bit harder than I originally thought. Focus on the isomer part
here, not the “charge” part).
Because of note 2, it makes sense to analyze the charges on the two metal cations before
proceeding. From the original complexes, the charges on Co and Cr must both be +3.
This is the only way to get the cation and anion to have the same charge in absolute
value: Co(en)33+ and Cr(ox)33-. (Recall that en is neutral and ox is a -2 anion.)
Answers (with explanation):
Ignoring the charge issue for now, and focusing on the “swapping” issue, there are three other
ways to distribute the en’s and ox’s in order to keep octahedral complexes for both (there is no
way to get 4 bidentate ligands around these cations, so you can’t consider the 2 & 4 kind of
pairs)—swap either one, two, or all three ligands:
[Co(en)2(ox)][Cr(en)(ox)2]
(left ion’s charge is +3 + -2  +1; right ion’s charge is +3 + 2(-2)  -1)
[Co(en)(ox)2][Cr(en)2(ox)]
(left ion’s charge is +3 + 2(-2)  -1; right ion’s charge is +3 + -2  +1)
[Co(ox)3][Cr(en)3]
(left ion’s charge is +3 + 3(-2)  -3; right ion’s charge is +3 + 0  +3)
Because of the “cation listed first” restriction, the proper formulas are actually:
[Co(en)2(ox)][Cr(en)(ox)2]
(kept the original order)
[Cr(en)2(ox)][Co(en)(ox)2] (flipped the order)
[Cr(en)3][Co(ox)3]
(flipped the order)
17. Tro, 24.34 Which complexes exhibit geometric isomerism?
(a) [Co(H2O)2(ox)2]
-
3+
(b) [Co(en)3]
(c) [Co(H2O)2(NH3)2(ox)]
+
2+
(d) [Ni(NH3)2(en)]
(e) [Ni(CO)2Cl2
Answers: (a), (c), and (e) do. (b) and (d) do not.
Reasoning:
(a) With two bidentate and two monodentate ligands (octahedral arrangement), one can have cistrans isomerism, because the monodentate ligands can be either at 180° or 90° degrees to one
another. (Analogous to the two structures shown in the answer to Problem 7(d) above)
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Answer Key, Problem Set 11
In (b), there is no pair of monodentate ligands to put in “different positions”, and thus no geometric
isomerism possible.
(c) This combination (ML2L’2(bidentate), octahedral geometry) can have cis-trans (or other
geometric isomer possibilities) because the L ligands (or L’ ligands) can be either at 180° or 90°
to one another. This is similar to Co(NH3)4Cl2+ (just pretend that two of the NH3’s were
connected as an en ligand)—see Figure 24.8(b) in Tro.
(e) In square planar, if there are two pairs of different monodentate ligands, cis-trans isomerism will
be possible (analogous to Pt(NH3)2Cl2—see Figure 24.8(a) in Tro).
In (d), the presence of the small bidentate en ligand forces the two NH3 ligands to be at 90° to one
another. Since they cannot be at 180° there is no “trans” option possible.
PS11-12