Ch 22 complex ions
•
Composition of the complex and nomenclature
•
Geometry of complex ions and isomers
•
Electronic structure of complex ions
•
Formation constant
Coordination complex is the product of a Lewis acid-base reaction in which neutral
molecules or anions (called ligands) bond to a central metal atom (or ion) by coordinate
covalent bonds.
Ligands are Lewis bases - they contain at least one pair of electrons to donate to a metal
atom/ion. Ligands are also called complexing agents.
Metal atoms/ions are Lewis acids - they can accept pairs of electrons from Lewis bases.
Within a ligand, the atom that is directly bonded to the metal atom/ion is called the donor
atom.
If the coordination complex carries a net charge, the complex is called a complex ion.
Compounds that contain a coordination complex are called coordination compounds.
Ch 23 – Components of a coordination compound.
Models
wedge diagrams
Formulas
Structures of Complex Ions: Coordination Numbers, Geometries, and Ligands
•
•
Coordination Number - the number of ligand atoms that are bonded directly to the
central metal ion. The coordination number is specific for a given metal ion in a
particular oxidation state and compound.
Geometry - the
geometry (shape) of a
complex ion depends
on the coordination
number and nature of
the metal ion.
Coordination
number
2
4
•
Donor atoms per
ligand - molecules
and/or anions with one
or more donor atoms
that each donate a lone
pair of electrons to the
metal ion to form a
covalent bond.
4
6
Shape
linear
Square planar
Tetrahedral
Octahedral
Geometry
Example
[CuCl2][Ag(NH3)2]+
[AuCl2]-
[Ni(CN)4]2[Pd(Cl)4]2[Cu(NH3)4]2+
[Pt(Cl)4]2[Cu(CN)4]3[Zn(NH3)4]2+
[Cd(Cl)4]2[Mn(Cl)4]2[Ti(H2O)6]2+
[V(CN)6]4[Cr(NH3)4Cl2]+
[FeCl6]3[Co(en)3]3+
Common ligands in coordination compounds
Ligand type
Example
Monodentate
H2O
(water)
F(Foloride ion)
CN(cyanide ion)
OHhydroxide
NH3
(Amomina)
Cl(Chloride ion)
[S=C=N]thiocyanate ion
[O=N=O]nitride ion
Bidentate
ethylenediamine
Oxalate
Polydentate
diethylenetriamine
Name of common ions
Netrual
Anionic
Metal ions in complex anion
Formula
Name
Formula
Name
Formula
Name
H2O
Aqua
F-
Fluoro
Fe (iron)
Ferrate
NH3
Amine
Cl-
Chloro
Cu (copper)
Cuprate
CO
Carbonyl
Br-
Bromo
Pb (lead)
Plumbate
NO
Nitrosyl
I-
Iodo
Ag (silver)
Argentate
OH-
Hydroxo
Au (gold)
Aurate
CN-
Cyano
Sn (Tin)
Stannate
Formulas and Names of Coordination Compounds
Rules for naming complexes:
1. The cation is named before the anion.
2. Within the complex ion, the ligands are named, in alphabetical order,
before the metal ion.
3. Neutral ligands generally have the molecule name, but there are a
few exceptions. Anionic ligands drop the -ide and add -o after the
root name.
4. A numerical prefix indicates the number of ligands of a particular
type.
5. The oxidation state of the central metal ion is given by a Roman
numeral (in parentheses).
6. If the complex ion is an anion we drop the ending of the metal name
and add -ate.
Sample Problem
Writing Names and Formulas of Coordination Compounds
PROBLEM: (a) What is the systematic name of Na3[AlF6]?
(b) What is the systematic name of [Co(en)2Cl2]NO3?
(c) What is the formula of tetraaminebromochloroplatinum(IV)
chloride?
(d) What is the formula of hexaaminecobalt(III) tetrachloro-ferrate(III)?
PLAN: Use the rules presented.
SOLUTION:
(a) The complex ion is [AlF6]3-.
Six (hexa-) fluorines (fluoro-) are the ligands - hexafluoro
Aluminum is the central metal atom - aluminate
Aluminum has only the +3 ion so we don’t need Roman
numerals.
sodium hexafluoroaluminate
continued
Writing Names and Formulas of Coordination Compounds
(b) [Co(en)2Cl2]NO3
There are two ligands, chlorine and ethylenediamine - dichloro,
bis(ethylenediamine)
The complex is the cation and we have to use Roman numerals for the cobalt
oxidation state since it has more than one - (III)
The anion, nitrate, is named last.
(c)
dichlorobis(ethylenediamine)cobalt(III) nitrate
Tetraamine bromo chloro
4 NH3
Br-
Cl-
platinum (IV) chloride
Pt4+
Cl-
[Pt(NH3)4BrCl]Cl2
(d)
Hexaamine cobalt(III) tetrachloro-ferrate(III)
6 NH3
Co3+
4 ClFe3+
[Co(NH3)6][Cl4Fe]3
Important types of isomerism in coordination compounds.
ISOMERS
Same chemical formula, but different properties
Constitutional (structural) isomers
Stereoisomers
Atoms connected differently
Different spatial arrangement
Coordination
isomers
Ligand and
counter-ion
exchange
Linkage isomers
Different donor
atom
Geometric (cis-trans)
isomers
(diastereomers)
Different
arrangement around
metal ion
Optical isomers
(enantiomers)
Nonsuperimposable
mirror images
Sample Problem
Determining the Type of Stereoisomerism
PROBLEM: Draw all stereoisomers for each of the following and state the type of
isomerism:
(b) [Cr(en)3]3+ (en = H2NCH2CH2NH2)
(a) [Pt(NH ) Br ]
3 2
PLAN:
2
Determine the geometry around each metal ion and the nature of the
ligands. Place the ligands in as many different positions as possible. Look
for cis-trans and optical isomers.
SOLUTION: (a) Pt(II) forms a square planar complex and there are two pair of
monodentate ligands - NH3 and Br.
Br
NH3
H 3N
Pt
H 3N
Pt
Br
trans
Br
H 3N
Br
cis
These are geometric isomers;
they are not optical isomers
since they are
superimposable on their
mirror images.
continued
Determining the Type of Stereoisomerism
(b) Ethylenediamine is a bidentate ligand. Cr3+ is hexacoordinated and will form an
octahedral geometry.
Since all of the ligands are identical, there will be no geometric isomerism possible.
3+
3+
N
N
N
N
N
N
Cr
N
Cr
N
N
N
N
N
rotate
3+
N
N
N
Cr
N
N
N
The mirror images are
nonsuperimposable and
are therefore optical
isomers.
Linkage isomers
Geometric (cis-trans) isomerism.
Optical isomerism in an octahedral complex ion.
Colors of representative compounds of the Period 4 transition metals.
nickel(II) nitrate
hexahydrate
sodium chromate
titanium oxide
scandium oxide
vanadyl sulfate
dihydrate
potassium
ferricyanide
manganese(II)
chloride
tetrahydrate
cobalt(II)
chloride
hexahydrate
zinc sulfate
heptahydrate
copper(II)
sulfate
pentahydrate
Important types of isomerism in coordination compounds.
ISOMERS
Same chemical formula, but different properties
Constitutional (structural) isomers
Stereoisomers
Atoms connected differently
Different spatial arrangement
Linkage isomers
Coordination
isomers
Ligand and
counter-ion
exchange
NO2- is the counter ion
[Pt(NH3)4Cl2](NO2)2
[Pt(NH3)4NO2]Cl2
Optical isomers
(enantiomers)
Geometric (cis-trans)
isomers
Nonsuperimposable
(diastereomers)
mirror images
Different
[Co(NH3)5(NO2)] Cl
arrangement around
[Co(NH3)5(ONO)] Cl
metal ion
Different donor
atom
Stereoisomers
Geometric (cis-trans)
isomers
(diastereomers)
Different
arrangement around
metal ion
Diastereomers
Cis – the identical
ligands next to each
other
Trans-the identical
lagands cross from
each other
Optical isomers
(enantiomers)
Nonsuperimposable
mirror images
3+
3+
N
N
N
N
N
Cr
N
N
Cr
N
N
N
Cisplatin
effective antitumor agent.
Cisplatin may work by Lying within the
cancer cell’s DNA double helix to prevent the
DNA duplication.
Transplatin has no effect on tumor.
N
N
An artist’s wheel.
Hybrid orbitals and bonding in the
tetrahedral [Zn(OH)4]2- ion.
Color
absorbed
Color transmitted
(Complementary)
Wavelength (nm)
Red
Green-blue
750-610
Orange
Blue- green
610-595
Yellow
Violet
595-580
Green
Red-violet
580-500
Blue
Orange-yellow
500-435
Violet
Yellow
435-380
The color of [Ti(H2O)6]3+.
[V(H2O)6]2+
[V(H2O)6]3+
[Cr(NH3)6]3+
[Cr(NH3)5Cl ]2+
The spectrochemical series.
•
•
For a given ligand, the color depends on
the oxidation state of the metal ion.
For a given metal ion, the color depends
on the ligand.
Effects of the metal oxidation state and
of ligand identity on color.
The configuration of central metal elements
Finding the Number of Unpaired Electrons
PROBLEM:
PLAN:
The alloy SmCo5 forms a permanent magnet because both samarium
and cobalt have unpaired electrons. How many unpaired electrons are
in the Sm atom (Z = 62)?
Write the condensed configuration of Sm and, using Hund’s rule and
the aufbau principle, place electrons into a partial orbital diagram.
SOLUTION: Sm is the eighth element after Xe. Two electrons go into the 6s
sublevel and the remaining six electrons into the 4f (which fills before
the 5d).
Sm is [Xe]6s 2 4f 6
6s2
4f6
There are 6 unpaired e- in Sm.
5d0
Crystal Field Theory
Splitting of d-orbital energies by an octahedral field of ligands.
D is the splitting energy
The effect of ligand on splitting energy.
The spectrochemical series.
•
For a given ligand, the color depends on the oxidation
state of the metal ion.
•
For a given metal ion, the color depends on the ligand.
I- < Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO
WEAKER FIELD
Splitting
Energy : D
Wavelength: 
STRONGER FIELD
SMALLER D
LARGER D
LONGER 
SHORTER 
HIGH SPIN
LOW SPIN
Ranking Crystal Field Splitting Energies
for Complex Ions of a Given Metal
PROBLEM: Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3- in terms of the
relative value of D and of the energy of visible light absorbed.
PLAN: The oxidation state of Ti is 3+ in all of the complexes so we are looking at
the crystal field strength of the ligands. The stronger the ligand the
greater the splitting and the higher the energy of the light absorbed.
SOLUTION:
The field strength according to is CN- > NH3 > H2O. So the relative
values of D and energy of light absorbed will be
[Ti(CN)6]3- > [Ti(NH3)6]3+ > [Ti(H2O)6]3+
Hybrid orbitals and bonding in the
octahedral [Cr(NH3)6]3+ ion.
Cr
Hybrid orbitals and bonding in the
square planar [Ni(CN)4]2- ion.
Ni
Identifying Complex Ions as High Spin or Low Spin –
number of electrons unpaired
PROBLEM: Iron (II) forms an essential complex in hemoglobin. For each of the two
octahedral complex ions [Fe(H2O)6]2+ and [Fe(CN)6]4-, draw an orbital
splitting diagram, predict the number of unpaired electrons, and identify
the ion as low or high spin.
PLAN:
The electron configuration of Fe2+ gives us information that the iron has
6d electrons. The two ligands have field strengths.
Draw the orbital box diagrams, splitting the d orbitals into eg and t2g.
Add the electrons noting that a weak-field ligand gives the maximum
number of unpaired electrons and a high-spin complex and vice-versa.
potential energy
SOLUTION:
[Fe(CN)6]4-
]2+
[Fe(H2O)6
4 unpaired e-(high spin)
eg
t2g
eg
no unpaired e-(low spin)
t2g
High-spin and low-spin complex ions of Mn2+.
Orbital occupancy for high- and low-spin complexes
of d4 through d7 metal ions.
high spin:
weak-field
ligand
low spin:
strong-field
ligand
high spin:
weak-field
ligand
low spin:
strong-field
ligand
Formation constant of complex ions
The equilibrium constant for the
formation of a complex ion is called
formation constant, Kf.
A complex formed by Ca with EDTA is
used to treat the Pb poisoning.
*
*
Ca2+ + EDTA  Ca(EDTA)2+
*
*
*
Ethylenediaminetetraacetic
*
*
*
EDTA titration curve
Calculate the shape of the titration curve for the reaction of 50.00 ml of 0.0400 M Ca2+ with
the 0.0800 M EDTA. (Buffered to 10.00, Kf = 4.91010)
Free concentration of metal ions as a function of EDTA additionn volume
Ca2+
1.
+
EDTA

CaY2-
Find the V EDTA at equivalence point : [Ca2+]=[EDTA]
50.00 ml  0.0400 M = V  0.0800 M
V=25.00 ml
2.
Before equivalence point :
Consider addition of 5 ml EDTA
Total volume of solution is 50 ml + 5 ml = 55 ml
Ca2+
I
C
E
+
50.00 ml  0.0400
-x
2mmol-x
EDTA
0
5 ml  0.08 M

CaY20
+x
0.4 mmol = x
Total volume V= 50.00 ml + 5 ml = 55 ml
Unreacted Ca2+ : (50.00 ml  0.0400 M – 5 ml  0.08 M)/55 ml = 0.029 M
p[Ca2+] = -log[Ca2+] = 1.54
Calculate pCa2+ when adding EDTA 10 ml, 15ml and 20 ml
3. At equivalence point:
Ca2+
I
C
E
•
•
•
x
+
EDTA

x
CaY20.0267
0.0267-x
[CaY2-] = [Ca2+]initial = [Y4-]initial = 0.0400 M  50 ml / (50 + 25) ml =0.0267 M
Kf ’ = Kf  = [CaY2-]/[Ca2+][EDTA] = 4.9  1010  0.36 = 1.8  1010
x = 1.2  10-6
pCa2+ = -log 1.2  10-6 = 5.91
4. After equivalence point :
After equivalence point, there is excess EDTA. The concentration of CaY2- and excess EDTA
can be calculate and then from Kf to figure out Ca2+ free ions concentration.
After equivalence point : Consider 26 ml of EDTA
[EDTA] = 0.0800 M  (26-25)ml / (26+50) ml = 1.05  10 -3
[CaY2-] = 0.0400 M  50 ml / (26+50) ml = 2.63  10 -3
Kf ’ = [CaY2-] / [Ca2+][EDTA] = 1.8  1010
[Ca2+] = 1.8  10-9
pCa2+ = 8.86
Calculate two more points
Questions for extra credit
1. Define the spectrochemical series and classify the order of the following ligand strength.
I- , Cl- , F- , OH- , H2O , SCN- , NH3 , en , NO2- , CN- , CO
2. Draw the hybrid orbitals and bonding
in the octahedral [Cr(NH3)6]3+ ion.
3. Draw the hybrid orbitals and bonding
in the square planar [Ni(CN)4]2- ion.
4. For each of the two octahedral complex ions [Fe(H2O)6]2+ and [Fe(CN)6]4-, draw an
orbital splitting diagram, predict the number of unpaired electrons, and identify the ion as
low or high spin.
5. Explain why The alloy SmCo5 forms a permanent magnet.
6. Draw the electronic configuration diagram of Sm.
7. Rank the ions [Ti(H2O)6]3+, [Ti(NH3)6]3+, and [Ti(CN)6]3- in terms of the relative
value of D and of the energy of visible light absorbed.
1. Determine the shape and draw the geometry. Give two examples.
CN
2
4
4
6
Shape
Geometry
Example
Coordination number = CN
Extra credit
2. Name the follow compounds.
Color
absorbed
Color transmitted
(Complementary)
Wavelength
(nm)
Red
Green-blue
750-610
•
the systematic name of Na3[AlF6]
•
systematic name of [Co(en)2Cl2]NO3
Orange
Blue- green
610-595
•
systematic name of[Pt(NH3)4BrCl]Cl2
Yellow
Violet
595-580
•
systematic name of[Co(NH3)6][Cl4Fe]3
Green
Red-violet
580-500
Blue
Orange-yellow
500-435
Violet
Yellow
435-380
3. How many types of isomerism in
coordination compounds.
4. Draw all stereoisomers for each of the
following and state the type of isomerism:
•
[Pt(NH3)2Br2]
•
[Cr(en)3]3+ (en = H2NCH2CH2NH2)
5. Tabulate the color absorbed and the color
we can identify by our eyes. List the
wavelength associated.