Ch 22 complex ions
• Composition of the complex and nomenclatureComposition of the complex and nomenclature
• Geometry of complex ions and isomersGeometry of complex ions and isomers
• Electronic structure of complex ionsElectronic structure of complex ions
• Formation constantFormation 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.
• 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.
Coordination number
Shape Geometry Example
2 linear[CuCl2]- [Ag(NH3)2]+
[AuCl2]-
4 Square planar[Ni(CN)4]2-
[Pd(Cl)4]2-
[Cu(NH3)4]2+
[Pt(Cl)4]2-
4 Tetrahedral[Cu(CN)4]3-
[Zn(NH3)4]2+
[Cd(Cl)4]2-
[Mn(Cl)4]2-
6 Octahedral
[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)
CN-
(cyanide ion)OH-
hydroxideNH3
(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.
PROBLEM:
PLAN:
SOLUTION:
Writing Names and Formulas of Coordination Compounds
(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)?
Use the rules presented.
(a) The complex ion is [AlF6]3-.
Six (hexa-) fluorines (fluoro-) are the ligands - hexafluoroAluminum is the central metal atom - aluminate
Aluminum has only the +3 ion so we don’t need Roman numerals.
sodium hexafluoroaluminate
Sample Problem
Writing Names and Formulas of Coordination Compoundscontinued
(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. dichlorobis(ethylenediamine)cobalt(III) nitrate
(c) Tetraamine bromo chloro platinum (IV) chloride
4 NH3 Br- Cl- Cl-Pt4+
[Pt(NH3)4BrCl]Cl2
(d) Hexaamine cobalt(III) tetrachloro-ferrate(III)6 NH3 Co3+ 4 Cl- Fe3+
[Co(NH3)6][Cl4Fe]3
ISOMERS
Same chemical formula, but different properties
Important types of isomerism in coordination compounds.
Constitutional (structural) isomers Stereoisomers
Atoms connected differently Different spatial arrangement
Coordination Coordination isomersisomers
Ligand and Ligand and counter-ion counter-ion exchangeexchange
Linkage isomersLinkage isomers
Different donor Different donor atomatom
Geometric (Geometric (cis-transcis-trans) ) isomers isomers
(diastereomers)(diastereomers)
Different Different arrangement around arrangement around
metal ionmetal ion
Optical isomers Optical isomers (enantiomers)(enantiomers)
Nonsuperimposable Nonsuperimposable mirror imagesmirror images
Sample Problem
PLAN:
SOLUTION:
Determining the Type of Stereoisomerism
PROBLEM: Draw all stereoisomers for each of the following and state the type of isomerism:
(a) [Pt(NH3)2Br2] (b) [Cr(en)3]3+ (en = H2NCH2CH2NH2)
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.
(a) Pt(II) forms a square planar complex and there are two pair of monodentate ligands - NH3 and Br.
Pt
NH3Br
H3N Br
Pt
H3N Br
H3N Br
cistrans
These are geometric isomers; they are not optical isomers since they are superimposable on their mirror images.
Determining the Type of Stereoisomerismcontinued
(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.
CrN
N N
N
N
N
3+
CrN
N N
N
N
N
3+
CrN
N N
N
N
N
3+rotate
The mirror images are nonsuperimposable and are therefore optical isomers.
Linkage isomersLinkage isomers
Geometric (cis-trans) isomerism.
Optical isomerism in an octahedral complex ion.
Colors of representative compounds of the Period 4 transition metals.
titanium oxide
sodium chromate
potassium ferricyanide
nickel(II) nitrate hexahydrate
zinc sulfate heptahydrate
scandium oxide
vanadyl sulfate dihydrate
manganese(II) chloride
tetrahydrate cobalt(II) chloride
hexahydrate
copper(II) sulfate
pentahydrate
ISOMERS
Same chemical formula, but different properties
Important types of isomerism in coordination compounds.
Constitutional (structural) isomers Stereoisomers
Atoms connected differently Different spatial arrangement
Coordination Coordination isomersisomers
Ligand and Ligand and counter-ion counter-ion exchangeexchange
Linkage isomersLinkage isomers
Different donor Different donor atomatom
Geometric (Geometric (cis-transcis-trans) ) isomers isomers
(diastereomers)(diastereomers)
Different Different arrangement around arrangement around
metal ionmetal ion
Optical isomers Optical isomers (enantiomers)(enantiomers)
Nonsuperimposable Nonsuperimposable mirror imagesmirror images
[Pt(NH3)4Cl2](NO2)2
NO2- is the counter ion
[Pt(NH3)4NO2]Cl2
[Co(NH3)5(NO2)] Cl
[Co(NH3)5(ONO)] Cl
Stereoisomers
Geometric (Geometric (cis-transcis-trans) ) isomers isomers
(diastereomers)(diastereomers)
Different Different arrangement around arrangement around
metal ionmetal ion
Optical isomers Optical isomers (enantiomers)(enantiomers)
Nonsuperimposable Nonsuperimposable mirror imagesmirror images
DiastereomersCis – the identical ligands next to each otherTrans-the identical lagands cross from each other
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.
CrN
N N
N
N
N
3+
CrN
N N
N
N
N
3+
Hybrid orbitals and bonding in the tetrahedral [Zn(OH)4]2- ion.
An artist’s wheel.
Color Color absorbedabsorbed
Color transmitted(Complementary)
Wavelength (nm)
RedRed Green-blue 750-610
OrangeOrange Blue- green 610-595
YellowYellow Violet 595-580
GreenGreen Red-violet 580-500
BlueBlue Orange-yellow 500-435
VioletViolet Yellow 435-380
The color of [Ti(H2O)6]3+.
Effects of the metal oxidation state and of ligand identity on color.
[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.
SOLUTION:
Finding the Number of Unpaired Electrons
PROBLEM: 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)?
PLAN: Write the condensed configuration of Sm and, using Hund’s rule and the aufbau principle, place electrons into a partial orbital diagram.
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 5d0
There are 6 unpaired e- in Sm.
The configuration of central metal elements
Splitting of d-orbital energies by an octahedral field of ligands.
is the splitting energy
Crystal Field Theory
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 STRONGER FIELD
LARGER SMALLER
LONGER SHORTER
Splitting Energy :
Wavelength:
HIGH SPIN LOW SPIN
SOLUTION:
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 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.
The field strength according to is CN- > NH3 > H2O. So the relative values of 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. Hybrid orbitals and bonding in the
square planar [Ni(CN)4]2- ion.
Cr Ni
PLAN:
SOLUTION:
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.
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.
t2g
eg
t2g
eg
pote
ntia
l ene
rgy
[Fe(H2O)6]2+[Fe(CN)6]4-
no unpaired e-- (low spin)
4 unpaired e-- (high spin)
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
* * * *
*
*
**
EDTA titration curve
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
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 = 4.91010)
Free concentration of metal ions as a function of EDTA additionn volume
Ca2+ + EDTA CaY2-
1. 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+ + EDTA CaY2-
I 50.00 ml 0.0400 0 0C -x 5 ml 0.08 M +xE 2mmol-x 0.4 mmol = xTotal 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:
• [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.
Ca2+ + EDTA CaY2-
I 0.0267
CE x x 0.0267-x
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+ = 8.86Calculate 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 and of the energy of visible light absorbed.
1. Determine the shape and draw the geometry. Give two examples.
CN Shape Geometry Example
2
4
4
6
Coordination number = CN
Extra creditColor Color
absorbedabsorbedColor transmitted(Complementary)
Wavelength (nm)
RedRed Green-blue 750-610
OrangeOrange Blue- green 610-595
YellowYellow Violet 595-580
GreenGreen Red-violet 580-500
BlueBlue Orange-yellow 500-435
VioletViolet Yellow 435-380
2. Name the follow compounds.
• the systematic name of Na3[AlF6]
• systematic name of [Co(en)2Cl2]NO3
• systematic name of[Pt(NH3)4BrCl]Cl2
• systematic name of[Co(NH3)6][Cl4Fe]3
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.
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