Ch 22 complex ions

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+

http://en.wikipedia.org/wiki/Image:Square-planar-3D-balls.png

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.


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


Different donor Different donor atomatom




metal ionmetal ion



[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




metal ionmetal ion



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

http://en.wikipedia.org/wiki/Image:Square-planar-3D-balls.png

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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Ch 22 complex ions