1.
Coordination Compounds
2.
3.
Chem 17
UP Diliman
Institute of Chemistry
4.
6.
7.
8.
9.
10.
11.
12.
13.
References for the Images and Examples Used for this Lecture PPT:
a. Petrucci, General Chemistry 2010
b. Whitten, General Chemistry 2011
Alfred Werner (1866-1919)
Two
coordination compounds
[CoCl(NH3)5]Cl2.
A coordinate covalent bond is a pair of electrons from
a donor shared with an acceptor.
Coordinate covalent bonds frequently are formed in Lewis acidbase reactions.
[Co(NH3)6]Cl3 and [CoCl (NH3)5]Cl2
Differing reactivity with AgNO3.
[Co(NH3)6]Cl3
Compounds made up of simpler compounds are
called coordination compounds.
CoCl3 and NH3.
Coordination Compounds
Important terms
Nomenclature
Structures
Isomerism in Coordination Compounds
Structural (Constitutional) Isomers
Stereoisomers (Geometric, Optical Isomerism)
Bonding in Coordination Compounds
Electronic Configuration of Metal Cations and Hybrid
Orbitals Concept
Crystal Field Theory
Color and the Spectrochemical Series
Kinetics of Ligand Exchange
Electrochemistry of Complexes
Other Applications
An example of a coordinate covalent bond is the one
formed between ammonia and borontrifluoride. The
e- which became shared between N and B came from
N.
.
H3N . + BF3
.
H3N . BF3
The complex formation below also shows formation of a coordination compound
Ag 2 NH 3 [ Ag ( NH 3 ) 2 ]
4
The ammine complexes contain NH3
molecules bonded to metal ions by
coordinate covalent bonds , e.g., [Cu(NH3)4]2+.
Dilute aqueous NH3 reacts with metal ions to
form the insoluble metal hydroxides or
hydrated oxides.
For example, Cu and Fe both react with
aqueous ammonia to form hydroxides.
Cu 2aq 2 NH3aq 2 H 2 O Cu(OH) 2 s 2 NH 4 aq
Fe2aq 3 NH3aq 3 H 2 O Fe(OH) 3s 3 NH 4 aq
However, in excess NH3 or highly concentrated NH3, the Cuhydroxide precipitate dissolves while the Fe(III)-hydroxide
does not.
The exceptions to this trend are metals that form
strong, water soluble hydroxides.
Group IA cations and the heavier Group IIA cations, Ca2+,
Sr2+, and Ba2+.
Certain hydroxides (e.g. Zn, Al) dissolve in
excess,highly concentrated hydroxide solutions.
Zn 2aq 2OH -aq Zn(OH) 2s
Zn(OH) 2 s 2OH- Zn(OH) 4
2-
aq
Several metal hydroxides dissolve in excess
aqueous NH3 to form ammine complexes.
2
Cu(OH) 2 s 4 NH3aq Cu(NH 3 ) 4 aq 2 OHaq
Co(NH ) 2 2 OH
Co(OH) 6 NH
2 s
3 aq
3 6
aq
aq
Some metal ions that form soluble ammine
complexes with an excess of aqueous NH3
include Co2+, Co3+, Ni2+, Cu2+, Ag+, Zn2+,
Cd2+, Hg2+
A ligand is a Lewis base that coordinates to
a central metal atom or ion.
A donor atom is the atom in a ligand that
donate a lone pair of electrons to form a
coordinate covalent bond.
A unidentate ligand is a ligand that can bind
through only one atom (i.e. only one donor
atom attached to the metal)
A polydentate ligand (chelating agent)
provides several donor atoms that can
coordinate simultaneously with the metal
ion.
8
Ion/
Molecule
Name
Name as a
Ligand
NH3
ammonia
ammine
CO
carbon monoxide
carbonyl
Cl-
chloride
chloro*
Ion/Molecule
Name
Name as a
Ligand
CN-
cyanide
cyano
CO3-2
carbonate
carbonato
OHS2SO42SCN-
hydroxide
sulfide
sulfate
thiocynate
hydroxo
sulfido
sulfato
Thiocynato-S
Thiocynato-N
Other halides can also be ligands: e.g. fluoro, bromo, iodo
9
Ion/Molecule
CO
NO
Name
Carbon dioxide
Nitrogen
monoxide
NO2-
nitrite
ONOPH3
nitrito
phosphine
Name as a
Ligand
10
A polydentate ligand is a ligand that can bind
through more than one donor atom.
There are known examples of bidentate, tridentate,
quadridentate, quinquedentate, and sexidentate
ligands.
Carbonyl
Nitrosyl
Nitro
Nitrito-N
Nitrito-O
Phosphine
Chelate complexes are complexes that have a
metal atom or ion and polydentate ligand(s)
that form rings.
(en) = ethylenediamine
is an example of a
bidentate ligand
11
12
Petrucci
ox
Co2+
ox
Co(ox)2 2
A square planar
complex
13
Coordination compound/complex: Metal ion
bonded with ligands (ions or molecules).
The coordination number is the number of donor
atoms coordinated to a metal atom or ion.
Most common: 4 and 6; Tetrahedral, Square Planar, Octahedral
14
Example: For the complex compound
Na3[Co(Cl)6] the coordination number is
_________, and the coordination sphere is
_______. The oxidation state of Co is ____.
You do it!
A coordination sphere includes the metal atom or
ion and the ligands coordinated to it. The
coordination sphere does NOT include
uncoordinated counter ions (i.e. the
For the complex compound K3[Co(Cl)6] the
coordination number is 6 , and the
coordination sphere is 7 . The oxidation state
of Co is +3.
corresponding cation or anion outside of the
bracket).
15
16
Example: For the complex compound
Ca[Zn(OH)4] the coordination number is
_________, and the coordination sphere is
_______. The oxidation state of Zn is ___.
Cations are named before anions.
1.
Coordinated ligands are named in alphabetical
order.
Prefixes that specify the number of each kind of
ligand (di = 2, tri = 3, tetra = 4, penta = 5, hexa
= 6, etc.) are NOT used in alphabetizing
2.
You do it!
However, prefixes that are part of the name of the
ligand, such as in (CH3CH2)2NH:, diethylamine, are
used to alphabetize the ligands.
For the complex compound above, the
coordination number is 4 , and the
coordination sphere is 5 . The oxidation state
of Zn is +2.
17
For complicated ligands, especially those
that have a prefix such as di or tri as part of
the ligand name, these prefixes are used to
specify the number of those ligands that are
attached to the central atom.
3.
4.
18
The names of most neutral ligands end in
the suffix -o.
4.
Examples of ligands ending in o are:
bis = 2; tris = 3; tetrakis = 4; pentakis = 5
hexakis = 6.
Examples:
[Fe(en)3]+3 is Tris(ethylenediamine)iron(III) cation.
[Co(ox)(en)2]+1 is Bis(ethylenediamine)oxalatocobaltate(III) anion
19
ClS2O2OHCNNO3SO42-
chloro
sulfido
oxo
hydroxo
cyano
nitrato
sulfato
[PdClBr(OH)2(en)] is Bromochloroethylenediaminedihydroxopalladium(IV)
complex, a neutral ligand.
20
The names of most neutral ligands are
unchanged when used in naming the
complex.
5.
6.
There are several important exceptions to this
rule including:
NH3
H2O
CO
NO
CS
7.
ammine
aqua
carbonyl
nitrosyl
thiocarbonyl
The oxidation number of a metal that exhibits
variable oxidation states is designated by a
Roman numeral in parentheses following the
name of the complex ion or molecule.
If a complex is an anion, the suffix "ate" ends
the name.
No suffix is used in the case of a neutral or cationic
complex.
Usually, the English stem is used for a metal, but if this
would make the name awkward, the Latin stem is
substituted.
Ferrate instead of ironate
Plumbate instead of leadate
Argentate (Ag), Aurate (Au)
Cuprate (Cu)
Others:
[Pt(OH2)2CO(en)NO] is Diaquacarbonylethylenediaminenitrosylplatinum(IV)
complex, a neutral ligand.
Platinate (Pt), Palladate (Pd), Titanate, Vanadate,
Antimonate, Tungstate, Hafnate, Zirconate, etc.
21
Name the following compounds:
Na3[Fe(Cl)6]
sodium hexachloroferrate (III)
22
Write formulas for the following compounds:
potassium hexacyanochromate(III)
K3[Cr(CN)6]
tris(ethylenediammine) cobalt(III) nitrate
[Co(en)3] (NO3)3
[Ni(NH3)4(OH2)2](NO3)2
tetraamminediaquanickel(II) nitrate
23
24
Fluoro
Chloro
Bromo
Iodo
Azido
Cyano
Thiocyanato
FClBrIN3CN*SCN-
Now: Thiocynato-S
Isothiocyanato *NCS Now: Thiocynato-N
Hydroxo
Ammine
Methylamine
OHNH3
H3CNH2
Phosphine (phosphane)
Pyridine
Common
Aqua
H 20
Carbonyl
CO
Thiocarbonyl
CS
Nitrosyl
NO
Nitro (or Nitrito-N) NO2Nitrito-O
ONO(O-bonded)
Nitrato
NO3Amido
NH2Sulfato
SO4-2
25
Give the names for the following:
1.
[Cu(NH3)4](NO3)2
2.
Ba[PtCl4]
3.
(NH4)4[Mn(CN)6]
4.
[CoN3(NH3)5]SO4
5.
Na[AlCl4]
6.
Rb3[AgF4]
7.
[FeBrCl(en)2]Cl
8.
[Fe(H2O)4(OH)2]Br
Chelating Ligands
Chelates which may have two or more points of attachments to
the metal atom.
Nitrato
Oxalato
C2O4-2
Ethylenediammine;
en
NH2CH2CH2NH2 (a bidentate)
Diethylenetriammine;
dien NH2CH2CH2NHCH2CH2NH2
Triethylenetetraamine;
trien NH2(CH2)2NH(CH2)2NH (CH2)2NH2
Ethylenediaminetetraacetate,
EDTA
2,2-bipyridine;
bipy
1,10-phenanthroline;
phen
Dimethylglyoxime,
DMG
26
27
PR3
py
Sodium tetrahydroxozincate(II)
Dichlorobis(ethylenediammine)cobalt(III) Nitrate
Triaquabromoplatinum(II) Chloride
Potassium trioxalatochromate(III)
Sodium pentacyanomanganate(IV)
Ammonium
diamminetetraisothiocyanochromate(III)
Tetraamminedinitroplatinum(IV) bromide
Hexammineruthenium(III) tetrachloronickelate(II)
Trans-diamminedinitroplatinum(II)
Tri-aqua-cis-dibromochlorochromium(III)
28
Coordination Chemistry
Molecular Geometries
Isomerism in Complexes
The structures of coordination compounds
are controlled primarily by the coordination
number of the metal.
Usually the structures can be predicted by
VSEPR theory (See GenChem books).
The geometries and hybridizations for common
coordination numbers are summarized in the
following table.
30
Petrucci
Coordinati
on Number
Geometry
Metal Hybridization
Example
linear
sp
Ag[NH3)2]+
tetrahedral
sp3
[Zn(CN)4]2-
square planar
dsp2 or sp2d
[Ni(CN)4]2-
trigonal bipyramid
dsp3
[CuCl 5 ]-3
Fe(CO)5
Square pyramidal
d2sp2
octahedral
d2sp3 or sp3d2
[Ni(CN)5]3-
Structure of some complex ions
[Co(en)3]+3
[Fe(CN)6]431
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Isomers:
Structural
(constitutional)
Differ in their structure and properties but with
similar molecular weight (or formula mass).
Stereoisomerism
(configurational)
Structural isomers:
Differ in basic structure.
1. Ionization
2. Hydrate
3. Linkage
4. Coordination
Stereoisomers:
Same number and type of ligands with the same
mode of attachment, but differ in the way the
ligands occupy space around the metal ion.
1. Geometric
a. Cis-trans
b. Fac-mer
2. Optical
-has mirror image
which are not
superimposable
34
2. Configurational (Stereo) Isomerism:
1. Structural*** isomerism:
Stereoisomers which have the same types of ligands but
have different geometric arrangement.
Have the same overall formula but have different
ligands attached to the central atom or ion
2.1 Diastereomers
** Conformational isomers (identical
bonding but have different bends or twists)
**
Geometric isomers
Example: Cis vs. Trans isomers
Fac vs. Mer isomers
1.1 Ionization isomers
1.2 Hydrate or solvent isomers
1.3 Coordination isomers
1.4 Linkage Isomerism.
***also known as Constitutional isomerism
2.2 Enantiomers (Optically active)
(Chiral, non-superimposable mirror images).
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36
Example of Structural (constitutional) Isomerism
1.1. Ionization or Ion-Ion Exchange Isomers
[Pt(NH3)4Cl2]Br2
H3 N
H3 N
Cl
Pt
+2
2-
N H3
Cl
* Hydrate isomers are a special case of ionization
isomers in which water molecules may be
changed from inside to outside the coordination
sphere.
[Pt(NH3)4Br2]Cl2
H3 N
N H3
H3 N
Counter ions
are Br-.
Br
Pt+4
Br
N H3
+2
2-
N H3
For example:
[Fe(OH2)6]Cl3 vs.[Fe(OH2)5Cl]Cl2 H2O
.
[Fe(OH2)4Cl2]Cl2 2H2O
Counter ions
are Cl-.
37
[Cr(OH2)6]Cl3
OH2
H2 O
H2 O
Cr
OH2
[Cr(OH2)5Cl]Cl2. H2O
3+
OH
OH
Counter ions
are 3 Cl-.
OH2
H2 O
H2 O
Cr
OH2
38
[Cr(OH2)5Cl]Cl2. H2O
[Cr(OH2)6]Cl3
2H2O
2+
OH2
Cl
H2 O
H2 O
OH
OH2
OH
Counter ions
are 3 Cl-.
Counter ions
are 2 Cl- and
1 water.
Note whether the water molecule(s) are inside or outside
the coordination sphere.
Cr
3+
OH
39
OH2
H2 O
H2 O
Cr
OH2
[Cr(OH2)4Cl2]Cl2.
2+
Cl
OH
Counter ions
are 2 Cl- and
1 water.
OH2
H2 O
H2 O
Cr
OH2
1+
Cl
Cl
Counter ions
are 1 Cl- and
2 waters.
40
10
[Co(NH3)5ONO]Cl2
* Coordination isomers denote an exchange
of ligands between the coordination spheres
of the cation and anion.
H3 N
H3 N
Pt
4 -
Cl
4 +
N H3
Cl
N H3
Cl
Pt
Cl
[Pt(NH3)4][PtCl6]
Cl
H3 N
Cl
H3 N
Cl
Pt
Cl
Nitrito -O
2 -
2 +
N H3
Cl
N H3
Cl
Cl
Pt
H3 N
H3 N
Cl
[Pt(NH3)4Cl2][PtCl4]
N H3
Nitrito-N
2+
N H3
Co O
[Co(NH3)5NO2]Cl2
H3 N
N
O
Counter ions are
2 Cl-.
H3 N
N H3
Co N
N H3
2+
O
O
Counter ions are
2 Cl-.
Note which atom in the ligand is bound to the central metal atom.
Another example: SCN- ligand can bind either through S or N.
The isomeric distinction is whether the ligands are on the
cation or the anion.
41
Petrucci FIGURE 24-4
42
Ionization
Isomerism
[CrSO4(NH3)5]Cl
pentaaminesulfatochromium(III) chloride
[CrCl(NH3)5]SO4
pentaaminchlorochromium(III) sulfate
Coordination Isomerism
Nitrite ion
Linkage
[Co(NH3)6][CrCN6]
hexaaminecobalt(III) hexacyanochromate(III)
isomerism illustrated
[Cr(NH3)6][CoCN6]
The nitrite ion can bind through the nitrogen lone pair or the oxygen lone pair.
(a) Pentaamminenitrito-N-cobalt(III) cation.
(b) Pentaamminenitrito-O-cobalt(III) cation.
hexaaminechromium(III) hexacyanocobaltate(III)
11
[Co(H2O)5Cl]Br2
[Pt(CO)4Br2]* [PtI4]* *+/-2
[CuCl4(NO2)]3-
[Cr(CH3NH2)4Cl2]Br
Stereoisomers are isomers that have different
spatial arrangements of the atoms relative to the
central atom.
Complexes with only simple ligands can occur
as stereoisomers only if they have coordination
numbers equal to or greater than 4.
45
46
H3N
Geometrical isomers are stereoisomers that are
not optical isomers.
H3N
Cl
Pt
Cl
H3N
Cl
Cl
Pt
N H3
cis- [Pt(NH3)2Cl2] trans-[Pt(NH3)2Cl2]
Example: Cis-trans isomers
have the same kind of ligand either adjacent to each
other (cis) or on the opposite side of the central metal
atom from each other (trans).
Take note of where
the ligands are
positioned relative to
the central atom.
The geometric isomers of[PtCl2(NH3)2]
47
12
Other types of isomerism can occur in
octahedral complexes.
Complexes of the type [MA2B2C2] can occur in
several geometric isomeric forms:
trans- trans- trans cis- cis- cis cis- cis- trans-
FIGURE 24-5
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Geometric Isomerism - illustrated
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50
Facial versus Meridional Isomers of
[Co(NH3)F3]
(NH
F 3)`
(NH3)`
F
(NHF
3)`
F
(NH3)`
F
Se
F
F3)`
(NH
FIGURE 24-6
Cis and trans isomers of an octahedral complex
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F
Se
F
(NH
F 3)`
octahedral
Meridional
octahedral
Facial
52
13
The geometric
isomers of
[Ma3b3c3]n complex
trans-diammine-trans-diaqua-trans-dichlorocobalt(III) ion
[CoCl3(NH3)3],
A [Ma3b3]n type of
complex
Geometric isomers
The 3 green balls Cl are
facing the same side (i.e.
have a common quadrant).
N H3
Cl
H2O
Co
Cl
O
N H H2
All trans
The 3 green balls Cl are
on the same plane (i.e. on
adjacent quadrants).
Same type of ligands are all position opposite to each other
54
Geometric isomers
trans-diammine-cis-diaqua-cis-dichlorocobalt(III) ion
N H3
Cl
H2O
Co
O
H2
Cl
NH3
N H3
Cl
Co
H2O
Cl
NH3
H2O
Cis H2O
Cis Cl
Trans NH3
55
56
14
Geometric isomers
Geometric isomers
cis-diammine-cis-diaqua-trans-
cis-diammine-trans-diaqua-cis-
dichlorocobalt(III) ions
dichlorocobalt(III) ions
Cl
NH3 H O Cl NH3
2
Co
Co
O
H2
N H3 H2O
N H3
Cl
Cl
H2O
OH2
Cl
Cl
Cis NH3
NH3
Co
NH
O H2 3
Cl
Cl
Cis Cl
Trans Cl
OH2
NH3
Co
NH
O H2 3
Trans H2O
Cis NH3
Cis H2O
57
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The phenomenon of rotation of polarized light is called optical
activity.
Light from an ordinary source consists of electromagnetic waves vibrating in all
planes; it is unpolarized.
This light is passed through a polarizer, a material that screens out all waves
except those vibrating in a particular plane.
The plane of polarization of transmitted polarized light is then changed by
passage through an optically active substance.
The angle through which the plane of polarization has been rotated is determined
by rotating an analyzer (a second polarizer) to the extent that all the polarized light isSli
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absorbed.
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Racemic mixture: separate equimolar solutions
of the two isomers (i.e. the mirror images of a
complex) rotate plane polarized light by equal
angles but in opposite directions. Net rotation =
zero!
L (left)
D (right)
However, if the concentrations of the D and L compounds present in a
mixture are different, there would be a net rotation.
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15
2.1 Optical Isomers/ Enantiomers are
stereoisomers with optical activity
Tetrahedral complex which does
NOT have a chiral center, is
superimposable to its mirror image.
E.g. [CoCl2(CO)(NH3)]2+
mirror
Square Planar
complex with 4
different ligands
(e.g.
[PtBrClF(OH)]2-
When the
mirror image
is flipped
sideways, it
becomes
identical to
the original
image
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64
16
65
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Mirror images are superimposable:
Non - enantiomers
1. /2 twice
A1
A2
Stereoisomer 1
Mirror Image of
Stereoisomer A1
2. 2
If you rotate A2 (along horizontal plane), and then invert along the
z-axis, the resulting structure will be superimposable to A1.
Theres a horizontal plane of symmetry which reflects the bottom
half to the upper half.
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68
17
Mirror images are superimposable:
Stereoisomer 2
Stereoisomer 1
1. /2 twice
A4
A3
Mirror Image of
Stereoisomer A3
Stereoisomer 2
Both stereoisomers are superimposable to
their mirror images; both are NOToptically active.
If you rotate A4 (along the horizontal plane), the resulting structure
will be superimposable to A3, thus, A3 and A4 are identical !!!
Molecule A3 could be divided into two by a vertical mirror plane (as
indicated by the yellow arrow above.)
69
mirror images of each other but are NOT
superimposable; therefore, enantiomers
A5
A6
Stereoisomer
Mirror Image of
Stereoisomer A5
If you rotate A6 horizontally (along xy-plane), then rotate
the resulting structure along the x-axis, the resulting
final structure is not superimposable to A5
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Formula
# Stereoisomers
Pairs of enantiomers
Ma6
Ma5 b1
Ma4 b2
Ma3 b3
Ma4 bc
***Ma3 bcd
Ma2 bcde
15
Mabcdef
30
15
***Ma2b2c2
Ma2b2cd
***Ma3b2c
0
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18
Draw the unique stereoisomers of the following
and look for enantiomers.
bonding in a complex to be an
electrostatic attraction between a positively
charged nucleus and the electrons of the
ligands.
[Cr(en)(NH3)2(OH2)2]+3
4 stereoisomers two of which are
non-superimposable images.
Consider
en
Electrons on metal atom repel electrons on ligands.
The focus of CFT is particularly on the effect of the
incoming ligands on the d-electrons of the metal
ion.
The energy of the d-orbital electrons will be affected
depending on the type of orbital (e.g. dxy, dxz, dyz,
dx2-y2, dz2). Some would be destabilized, some would
not be directly on the path of the ligands, hence not
affected.
[Cr(en)2(H2O)Cl]+2
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Approach of six ligands to a metal ion to form a
complex ion with octahedral structure. The octahedral
field affects the metal through the x-y-z axes.
Approach of six ligands to a metal ion to form a
complex ion with octahedral structure. The octahedral
field affects the metal through the x-y-z axes.
dxy
Maximum repulsion occurs with the dz and dx2-y2 orbitals, and
their energies are raised.
Repulsions with the other d orbitals are not as great. A difference
in energy results between the two sets of d orbitals. Copyright 2011 Pearson
Imagine a potential head-on collision of the electrons from the ligands with the
d-electrons of the metal. The octahedral field would not affect directly the xy, yz
and xz lobes.
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dxz
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Repulsions with the other d orbitals are not as great. A difference in
energy results between the two sets of d orbitals.
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19
eg
t2g
Petrucci
d orbitals
before splitting
Some ligands would have STRONG interaction with the
metal, while others would have WEAK interaction.
Strong field ligands would have greater o separation
between the t2g and eg orbitals. The crystal field stabilization
energy is much larger.
Weak field ligands would have lower o separation, and a
smaller crystal field stabilization energy.
d orbitals under
an octahedral field
Splitting of thed-orbital energy levels in the formation of an
octahedral complex ion results to two types of orbitals: three t2g and
two eg orbitals.
FIGURE 24-12
The O is the corresponding energy separation (i.e. crystal field
stabilization energy).
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26
1s
Large 0
Strong field ligands
Fe
26
2
2
2s 2 p
3s 2 3 p 6
CN- , CO > NO2- > en > py NH3 > EDTA4- > SCN- > H2O >
ONO- > ox2- > OH- > F- > SCN- > Cl- > Br- > ISmall 0
Weak field ligands
4s
d6
3d
Take note where the
electrons are taken off when
you are making the cations
(the outermost shell first,
from 4s before 3d)
Fe 2
1s 2
2s 2 2 p 6
3s 2 3 p 6
4s
26
d6
3d
Fe 3
1s 2
2s 2 2 p 6
3s 2 3 p 6
4s 0
d5
3d
80
20
27
Co
1s
27
2
2
2s 2 p
3s 2 3 p 6
4s
Co 2
28
1s 2
d7
2s 2 p
3d
4s
2s 2 p
4s
4s
d6
3d
29
1s 2
2
2s 2 p
3s 2 3 p 6
1
4s
d10
3d
Instead of
4s2d9
Copper is one of those
exception, the filled dorbital is more stable, hence
one e- from 4s is promoted
to the originally last halffilled 3d orbital
2s 2 p
3s 2 3 p 6
3s 2 3 p 6
d10
3s 2 3 p 6
3d
28
Ni 4
1s 2
d6
2s 2 2 p 6
3d
3s 2 3 p 6
0
82
d5
3d
4s 0
3d
4s 0
29
d8
2s 2 2 p 6
Fe 3
2s 2 2 p 6
1s 2
1s 2
4s
81
Cu 1
Ni 2
4s
Take note where the
electrons are taken off when
you are making the cations
(the outermost shell first,
from 4s before 3d)
26
Cu
3d
1s 2
29
d8
3s 2 3 p 6
Take note where the
27 Co
electrons are taken off when
1s 2
you are making the cations
2s 2 2 p 6
(the outermost shell first,
from 4s before 3d)
3s 2 3 p 6
28
2
2
3d
3s 2 3 p 6
1s
d7
Ni
Very large O
Cu 2
1s 2
2s 2 2 p 6
3s 2 3 p 6
d9
Small O
[FeCl6]3-
[Fe(en)3]3+
3d
4s 0
83
84
21
27
Co 3
1s 2
2
2s 2 p
3s 2 3 p 6
E
N
E
R
G
Y
d6
d-orbital splitting of Co3 + in
an octahedral field
Crystal field splitting of the d-orbitals for a
metal ion under a tetrahedral field
3d
4s 0
Both t2g and eg orbitals are occupied
The Td denotes the energy separation of the lower two eg orbitals
and the upper three t2g orbitals of the tetrahedral complex. The
energy separation is small, so the effect of the field strength of the
ligand is not considerable. ALL complexes are high-spin (i.e the
upper ones are occupied first before there is e- pairing).
The 2 eg orbitals are empty
General Chemistry: Chapter 24
27
Co 2
1s
2
2
2s 2 p
3s 3 p
4s 0
3d
d7
Both t2g and eg orbitals are
occupied, no matter what the
field strength of the ligand is,
since the energy difference
between these two types of
orbitals is just small (Td << o)
[CoCl4]2-
eg
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t2g
Most destabilized
t2g
e
g
[Co(NH3)4 ]2+
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Least destabilized
87
88
22
(a) Splitting of the d energy level in a square-planar complex can be related
to that of the octahedral complex.
(b) There are no ligands along the z axis in a square-planar complex, so we
expect the repulsion between ligands and dz2 e- to be much less than in an
octahedral complex. The energy level is lowered considerably from that in
an octahedral complex.
(c) The energy levels of the dxz & dyz orbitals are lowered slightly because the
e- in these orbitals are concentrated in planes perpendicular to that of the
square-planar complex.
(d) The energy of the dx2-y2 orbital is raised because the x and y axes
represent the direction of approach of 4 ligands to the central ion.
(e) The energy of the dxy orbital is also raised because this orbital lies in the
plane of the ligands in the square-planar complex.
Crystal field splitting in a square-planar and an
octahedral complex
FIGURE 24-14
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(f) The energy difference between the dxy and dx2-y2 orbitals in a squareplanar complex is the same as in an octahedral complex because these
orbitals are equally affected by ligand repulsions in both complexes.
90
Pd(II) and Pt(II) complexes : Example : [ PtCl 4 ]2 ;[ Pd (CO)4 ]2
Coordination 4 complexes with " bulky"ligands :
Example : [Co(en) 2 ]2 ;[ FeBr2 (ox)]2
Many Ni2+ complexes
28
d8
Ni 2
1s 2
d-orbital
splitting
under a
square
planar CF
2s 2 2 p 6
2
Ni
NO2
NO2
3s 3 p
[ Ni( X )4 ]n
3d
4s 0
e
g
If d8 and paramagnetic
t2g versus
[ Ni(ox)( NO2 )2 ]
91
Tetrahedral
Square Planar
92
23
If d6 and magnetic
[ Fe A3 B3 ] property is given
eg
versus
t
II
[ PtAx By ]n
2g
Low Spin Octahedral, with
Strong Field Ligands
diamagnetic
High Spin Octahedral,
with Weak Field Ligands
paramagnetic
e
g
t
If tetrahedral, the
complex is also
paramagnetic
2g
versus
Octahedral, with either
Strong or Weak Field
Ligands is Paramagnetic
Square Planar d-8
complexes are
Diamagnetic
93
hc
94
Primary colors:
red
green
blue
Red (R), green (G) and blue (B).
Secondary colors:
Complementary colors:
Produced by mixing primary colors.
Shorter
wavelength,
;
Higher
Frequency,
Higher Energy
Secondary colors are complementary to primary.
Cyan (C), yellow (Y) and magenta (M)
Adding a color and its complementary color
produces white.
Longer
wavelength, ;
Smaller Frequency,
Lower Energy
Cyan (C), yellow (Y) and magenta (M)
95
24
orange
red
yellow
The mixing of
colors
indigo
General Chemistry: Chapter 24
[CoCl4]-
[Co(NH3)6]2+
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green
indigo
The secondary colorsyellow (Y), cyan (C), and magenta (M)are produced in
regions where two of the beams overlap. The overlap of all three beams produces
white light (W).
blue
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A weak field ligand, would absorb
lower energy wavelengths (e.g. R-OY), while strong field ligands would
absorb at higher energies (e.g. VIBG)
A complex which would light
corresponding to the red-region
would appear green.
A complex absorbing at very low
wavelengths (near ultraviolet region)
would appear yellow green.
A compound absorbing within the
yellow-to-green wavelength (region)
would appear purple.
98
[CoCl4]= absorbs in the
yellow region of the
spectrum and transmits
blue light.
[Co(NH3)6]2+ absorbs in the
blue region of the spectrum
and transmits yellow light
(appears straw yellow)
LIGHT Absorption and transmission
red
orange
yellow
indigo
indigo
green
blue
Red-shift
Cl is a weaker field
strength ligand than H2O
Blue-shift
NH3 is a stronger
field ligand than H2O
25
Cl- > Br- > IZn2+(aq) + 4 NH3(aq) <==> [Zn(NH3)4]2+(aq)
Kf =
[[Zn(NH3)4]2+]
= 4.1108
[Zn2+][NH3]4
Suppose 1.0 M Zn(NO3)2 is dissolved in
excess 2.5 M NH3, what is the resulting
equilibrium concentration of the
tetrazinc(II) complex?
red
indi
go
indi
go
oran
ge
blu
e
yell
ow
gre
en
Effect of ligands on
24-18 of
the FIGURE
colors
coordination
compounds
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Set up the ICE TABLE to find the equilibrium concentrations.
[Zn(NH3)4]2+(aq) <==> Zn2+(aq) + 4 NH3(aq)
Initial [ ]
Change [ ]
1.0 M
-x
____
+x
2.5 M
+ 4x
What is the pH of a 0.10 M
[Cu(NH3)4]Cl2(aq) solution?
pH > 7, pH =7, pH < 7
26
CuSO4 in concentrated HCl(aq).
[CuCl4]2-
Fast[Cu(H O) ]2+ + 4 NH
[Cu(NH3)4]2+ + 4 H2O
Fast
[CuCl4]2-
[Cu(H2O)4]2
[Cu(NH3)4]2+
Labile complex ions
[Cu(H2O)4]2+
+4
Cl-
[Cu(Cl)4]2-
+ 4 H2O
The exchange of ligands in the coordination sphere of Cu 2+ occurs very rapidly.
Water is said to be a labile ligand.
The solution at the extreme left is formed by dissolving CuSO 4 in concentrated
HCl(aq). Its yellow color is due to [CuCl4]2-.
Slow reactions (often monitored by color
change) are caused by non-labile ligands.
When a small amount of water is added, the mixture of [Cu(H2O)4]2+ and [CuCl4]2ions produces a yellow-green color.
When CuSO4 is dissolved in water, a light blue solution of [Cu(H2O)4]2+ forms.
Fast reactions can operate at the diffusion limit.
FIGURE 24-20
NH3 molecules readily displace H2O molecules as ligands and produce deep blue
[Cu(NH3)4]2+ (extreme right).
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Inert complex ions
trans-CrCl2(H2O)4+.
Blue shifted due to
exchange of Cl- and H2O
The green solid CrCl36 H2O produces the green
aqueous solution on the left.
The color is due to trans-CrCl2(H2O)4+.
A slow exchange of H2O for Cl- ligands leads to a
violet solution of Cr(H2O)63+ in one or two days
(right).
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27
Co3+(aq) + e- Co2+(aq)
4
Co3+(aq)
+ 2 H2O(l) 4
Co2+(aq)
Co3+(aq) + e- Co
E = +1.82 V
+4
H+
+ O2(g)
Ecell = +0.59 V
E = +1.82 V
[Co(NH3)6]3+(aq) + e- [Co(NH3)6]2+(aq) E = +0.10 V
[Co(en)3]3+ + 3e- Co + 3(en) E = ?
But:
Co3+(aq) + NH3(aq) [Co(NH3)6]2+(aq) Kf = 4.51033
[Co(en)3]3+ Co3+ +3(en) G = -RTln(1/Kf)
and
Co3+ + 3e-
[Co(NH3)6]3+(aq)
e-
[Co(NH3)6]2+(aq)
E = +0.10 V
Which is easier to reduce? Which has more
positive and larger magnitude of standard
reduction potential?
E = ?
B. [Co(NH3)6]3+ + 3e- Co + 3NH3)6 E = ?
3e- Co + 3(en) G = -zFE =
Stability of complexes with chelating ligands due
to the smaller negative entropy change when they
are formed.
G formation of [Cu(en)]2+ is more negative (i.e.
more spontaneous) and of greater magnitude than
[Cu(NH3)] 2+.
G formation H TS
Negative
Exothermic
Bond formation
CLUE: Which is more easily dissociated ?
Negative S
Gdissociation H TS
[Co(en)3]3+ or [Co(NH3)6]3+
Explain based on H;S,G, Kf or Kd and E.
G = -zFE =
Co+3 is even more stabilized if coordinated to chelating
ligands than to simple monodentate ligands.
Co+3 is more stabilized, and not easily reduced to
Co(0) if coordinated to ligands.
A. [Co(en)3]3+ + 3e- Co + 3(en)
[Co(en)3
]3+ +
11
1
Positive
Endothermic
Bond Breaking
Positive S
Chelated complex
has smaller
negative Sf
Favored if more
ligands will be
dissociated
28
Black and white.
Finely divided AgBr on modified cellulose.
Photons oxidize Br- to Br and reduce Ag+ to Ag.
Hydroquinone (C6H4(OH)2) developer:
Reacts only at the latent image site where some
Ag+ is present and converts all Ag+ to Ag.
Negative image.
Fixer removes remaining AgBr.
AgBr(s) + 2 S2O32-(aq) [Ag(S2O3)2]3-(aq) + Br-(aq)
Print the negative
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CONSULT YOUR LAB
EXPERIMENTAL RESULTS
Tetrasodium EDTA
FIGURE 24-25
e.g. For sequestration of heavy metals
The central metal ion Mn+ (pale
green) can be Ca2+, Mg2+, Fe2+, Fe3+,
and so on. The ligand is EDTA4-,
and the
net charge on the complex is +n-4.
Structural diagrams and ball-andstick models are shown for the two
optical isomers of an [MEDTA]n-4
complex.
Petucci FIGURE 24-24
The porphyrin structure
Structure of
Chlorophyll-A
Hemoglobin carries
O2 through the Febonded to the heme.
29
Make sure to review also the following
concepts covered prior to 2nd exam:
Acids and Bases (e.g. pH, buffers)
Setting up of ICE Table for Aqueous
Equilibria and pertinent calculations and
approximations
Common Ion Effect on Aqueous Equilibria
30
