Ch 23 Powerpoint

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Chapter 22

The Transition Elements and

Their Coordination Compounds


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The Transition Elements and Their Coordination Compounds

23.1 Properties of the Transition Elements

23.2 The Inner Transition Elements

23.3 Highlights of Selected Transition Metals

23.4 Coordination Compounds

23.5 Theoretical Basis for the Bonding and Properties of Complexes


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Figure 23.1 The transition elements (d block) and inner transition

elements (f block) in the periodic table.


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General Properties:

The transition metals show great similarities

within a given period and a group. Their

Chemistry does not change as number of

valence electron change.

They are metals good conductors of heat and

electricity: Ex: Ag.

More than one oxidation state.


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General Properties:

Complex ions: Species where transition metal

ion is surrounded by a certain number of

ligands.

Ligand: Molecules or ions that behave as

Lewis bases.

Paramagnetic-unpaired electrons.


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Electronic Configurations:

Cr:

Cu:

Mo3+ Ag+

The energy of 3d orbitals in transition metal

ions is less than 4s orbitals.


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Atomic and Physical Properties of

Transition Elements:

1. Atomic size decreases from left to right across

the period , but then remain fairly constant.

2. Transition elements exhibit a small change in

electronegativity. The values are intermediate.

3. First Ionization energies increase little.

4. Lanthanide contraction- 4f orbitals are filled,increases overall charge on nucleus and not

the size.


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Figure 23.3 Horizontal trends in key atomic properties of the

Period 4 elements.


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Atomic and Physical Properties of

Transition Elements:

5.Nuclear charge increases down a group.

6. Heavier transition metals exhibit more

covalent character.

7. First I.E increases down a transitional group.

8. Densities increase as atomic mass increases.


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Figure 23.4

Vertical trends in key properties within the transition elements.

1st IE highest at bottom of trans group.

2nd and 3rd element nearly same size Electronegativity increases down a group.

Densities increase as mass increases

C G C f


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Chemical Properties

1. They have multiple oxidation states.

2. +2 oxi state is most common as ns2 electrons arereadily lost.

3. Ionic bonding occurs in lower O.S and covalent inhigher O.S.

4. Electrons in a partially filled d sublevels can absorbvisible wavelengths and hence their compounds are

colored.5. They are paramagnetic (unpaired d electrons)

6. IE1 increases down a group.

C i ht Th M G Hill C i I P i i i d f d ti di l


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C i ht Th M G Hill C i I P i i i d f d ti di l


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Sample Problem 23.2

SOLUTION:

Finding the Number of Unpaired Electrons

PROBLEM: The alloy SmCo5 forms a permanent magent because bothsamarium 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 Hunds

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]6s24f6

6s 4f 5d

There are 6 unpaired e- in Sm.

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Coordination compounds:

They contain atleast one Complex ion , bonded to

ligands and associated with other counter ions.

Coordination compound: Complex transitional metal

ion attached to ligands.

Two types of valance: 1. secondary Valence- Ability

of metal ion to bind to a Lewis base(ligands)-

Coordination number.

2. Primary valence-Ability of metal ion to form ionic

bonds with oppositely charged ions.-Oxidation state.



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Coordination compounds:

Complex ions: Species where transition metal

ion is surrounded by a certain number of

ligands.

Ligand: Molecules or ions that behave as

Lewis bases.



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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 toform a covalent bond.

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The coordination number

Varies from 2-8.

6 ligands-octahedral arrangement.

4-Tetrahedral/Square planar. 2- Linear.

Most common coord. Number is 6.



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Figure 23.9 Components of a coordination compound.

models wedge diagrams chemical formulas

6 ligands-octahedral4 ligands-square planar



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Ligands:

Neutral molecule or ion having a lone pair of

electron that can be used to form a bond to

metal ion.

Metal (Lewis acid-e pair acceptor) _______

Nonmetal ( Lewis base- e pair donor)

Monodentate/ Unidentate ligand- Ligand

forms 1 bond.CN-, H2O, NH3.



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py g p , q p p y

Ligands

Chelating ligands/ Chelates: Ligands have

more than one atom with a lone pair of

electrons that can be used to bond a metal ion.

Bidentate ligand- can form 2 bonds.Ex:

ethylenediamine(en), oxalate

Polydentate ligands- can form more than 2

bonds.Ex: EDTA- 6 bonds.



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py g p , q p p y



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py g p q p p y



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Formulas and Names of Coordination Compounds

Rules for writing formulas:

1. The cation is written before the anion.

2. The charge of the cation(s) is balanced by the charge

of the anion(s).

3. In the complex ion, neutral ligands are written beforeanionic ligands, and the formula for the whole ion is

placed in brackets.



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Formulas and Names of Coordination Compounds

continuedRules 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 thereare 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 aRoman numeral (in parentheses).

6. If the complex ion is an anion we drop the ending of the metal

name and add -ate.



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Nomenclature:

Name cation before anion.

In naming a complex ion, name ligands

before the metal ion.

In naming ligands, add o to the root name of

anion(chloro), Use the full name for a neutral

ligand .

Exceptions to # 3: aqua, ammine,

methylamine, carbonyl, nitrosyl.



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Nomenculature

. Use prefix mono, di, tri, tetra , penta and hexa forsimple ligands.

6. Use prefix bis,tris,tetrakis for complicated ligandsthat alreadt have bi,tri.

7. Oxidation state for metal in Roman numerals in () 8. When more than one type of ligand are present

name alphabetically.

9. If complex ion has negative charge add the suffix

-ate to the name of the metal.(Latin name) Iron copper lead silver

gold tin



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Sample Problem 23.3

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 - and .

(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 dont need Roman

numerals.

sodium hexafluoroaluminate



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Sample Problem 23.3 Writing Names and Formulas of Coordination

Compoundscontinued

(b) 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

tetraaminebromochloroplatinum(IV) chloride

(c) 4 NH3 Br- ClC - Cl-Pt4+

[Pt(NH3)4BrCl]Cl2

(d)

hexaaminecobalt(III) tetrachloro-ferrate(III)

6 NH3 Co3+ 4 Cl- Fe3+

[Co(NH3)6][Cl4Fe]3



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ISOMERS

Same chemical formula, but different properties

Figure 23.10

Important types of isomerism in coordination compounds.

Constitutional (structural) isomers Stereoisomers

Atoms connected differently Different spatial arrangement

Coordination

isomers

Ligand and

counter-ionexchange

Coordination

isomers

Ligand and

counter-ionexchange

Linkage

isomers

Different donor

atom

Linkage

isomers

Different donor

atom

Geometric (cis-

trans) isomers

(diastereomers)

Differentarrangement

around metal ion

Geometric (cis-

trans) isomers

(diastereomers)

Differentarrangement

around metal ion

Optical isomers

(enantiomers)

Nonsuperimposable

mirror images

Optical isomers

(enantiomers)

Nonsuperimposable

mirror images



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Isomerism:

Same formula but different properties.

Structural isomerism: Isomers contain same

atoms but different bonds.

Coordination isomerism: Composition of complex

ion varies.

Linkage isomerism: Point of attachment of

atleast one of the ligands differs



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Isomerism

Stereoisomers: All bonds are same butdifferent spatial arrangements.

Geometrical isomerism cis-trans-Atoms or

group of atoms can assume different positionsaround a rigid ring or bond.

Optical Isomerism: Have opposite effects onplane polarized light.

Chiral: Objects that have nonsuperimposable mirrorimages.

Enantiomers: Isomers that are nonsuperimposablemirror images of each other.



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Linkage isomersLinkage isomers



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Figure 23.11 Geometric (cis-trans) isomerism.



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Figure 23.12

Optical isomerism in anoctahedral complex ion.



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Sample Problem 23.4

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 forcis-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 BrPt

H3N Br

H3N Br

cistrans

These are geometric isomers;

they are not optical isomerssince they are superimposable

on their mirror images.



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Sample Problem 23.4 Determining the Type of Stereoisomerism

continued (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.

Cr

N

N N

N

N

N

3+

CrN

N N

N

N

N

3+

CrN

N N

N

N

N

3+rotate

The mirror images arenonsuperimposable

and are therefore

optical isomers.



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Figure 23.13

Hybrid orbitals and bonding in the octahedral [Cr(NH3)6]3+ ion.



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Figure 23.14

Hybrid orbitals and bonding in the square planar [Ni(CN)4]2- ion.



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Figure 23.15

Hybrid orbitals and bonding in the tetrahedral [Zn(OH)4]2- ion.



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The Crystal Field Model

It focuses on the energies of d orbitals.

Metal ligand bond is ionic.

Ligands are negative point charges.



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Figure 23.16 An artists wheel.



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Octahedral complexes:

Dz2 and dx2-y2 orbitals have lobes that pointdirectly at the ligands.

Dxy, dyz,dxy point their lobes between

charges. Electrons fill the d orbitals farthest from the

ligands to minimize repulsion.

Dxy, dyz,dxy ( t2g set) are at lower energy inoctahedral complex first.

Dz2 and dx2-y2 (eg set) is at higher energy.



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Octahedral complexes:

Splitting of 3d orbital energies explains colorand magnetism.

Strong field case- splitting produced by

ligands is very large, electrons will pair inlower t2g orbitals. Diamagnetic (all electronsare paired)

Weak field case-splitting produced by ligandsis small, electrons will occupy all 5 orbitalsbefore pairing occurs. Paramagnetic( unpairedelectrons)



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Figure 23.17 The five d-orbitals in an octahedral field of ligands.



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Problems:

1.Fe (CN)63- has one unpaired electron . Does

the CN- ligand produce a strong or weak

field?

2.Predict the number of unpaired electrons in the

complex ion [Cr(CN)6] 4



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Figure 23.22 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

LARGERSMALLER

LONGER SHORTER



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Color of octahedral compounds:

Absorbed color is different than observed

color.

Transition metals absorb colors in the visible

region.

E=hc/ , E= energy spacing, =walength

needed to move an electron from t2g to eg.

Color of solution changes as the ligand

changes.



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In tetrahedral complexes:

None of the 3d orbitals point at the ligands.

Tetrahedral splitting is 4/9 times that of

octahedral.

Dxy, dyz,dxy are closer to pint charges than

Dz2 and dx2-y2.

Weak field case always applies in tetrahedral

complexes.



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Figure 23.18

Splitting of d-orbital energies by an octahedral field of

ligands.

is the splitting energy



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Figure 23.19 The effect of ligand on splitting energy.



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Figure 23.20 The color of [Ti(H2O)6]3+.



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Figure 23.21

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+



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Problem:

Give the crystal field diagram for tetrahedral

complex ion CoCl 4 2-



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Sample Problem 23.5

SOLUTION:

Ranking Crystal Field Splitting Energies for

Complex Ions of a Given Metal

PROBLEM: Rank the ions [Ti(H2

O)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+



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Figure 23.23 High-spin and low-spin complex ions of Mn2+.



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Figure 23.24 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



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Sample Problem 23.6

PLAN:

SOLUTION:

Identifying Complex Ions as High Spin or Low Spin

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 anorbital 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 shown

in .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

potentialenerg

y [Fe(H2O)6]2+

[Fe(CN)6]4-

no unpaired e--

(low spin)

4 unpaired e--(high spin)



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Figure 23.25 Splitting of d-orbital energies by a tetrahedral field

and a square planar field of ligands.

tetrahedral

square planar



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Figure B23.1 Hemoglobin and the octahedral complex in heme.



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Fi B23 2 Th t t h d l Z 2+ l i b i h d


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Figure B23.2 The tetrahedral Zn2+ complex in carbonic anhydrase.

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