TRANSITION METALTRANSITION METALTRANSITION METALTRANSITION METALCHEMISTRY.

TRANSITION METALCHEMISTRY

2009 NEW SPECIFICATION

CONTENTS• Definition

• Metallic properties

• Electronic configurations

• Variable oxidation state

• Coloured ions

• Complex ion formation

• Shapes of complexes

• Isomerism in complexes

• Catalytic properties

Before you start it would be helpful to…

• Recall the definition of a co-ordinate (dative covalent) bond

H3N :H

Recall how to predict the shapes of simple molecules and ions

Transition metals

• A transition metal is one which forms one or more stable ions which have incompletely filled d orbitals.

• The highest occupied d orbital filled by an electron is d- block element

2 Transition metals and their chemistry

A

Describe transition metals as those elements which form one or more stable ions which have incompletely filled d orbitals.

B

Derive the electronic configuration of the atoms of the d-block elements (Sc to Zn) and their simple ions from their atomic number.

CDiscuss the evidence for the electronic configurations of the elements Sc to Zn based on successive ionization energies.

D

Recall that transition elements in general:i Show variable oxidation number in their compounds, eg redox reactions of vanadium.ii Form coloured ions in solution.iii Form complex ions involving monodentate and bidentate ligands.iv Can act as catalysts both as the elements and as their compounds.

E

Recall the shapes of complex ions limited to linear [CuCl2]-, planar [Pt(NH3)2Cl2], tetrahedral [CrCl4]-

and octahedral [Cr(NH3)6]3+, [Cu(H2O)6]2+ and other

aqua complexes.

F

Use the chemistries of chromium and copper to illustrate and explain some properties of transition metals as follows:i. The formation of a range of compounds in which they are present in different oxidation states.ii. The presence of dative covalent bonding in complex ions, including the aqua-ion.siii. The colour or lack of colour of aqueous ions and other complex ions, resulting from the splitting of the energy levels of the d orbitals by ligands.iv. Simple ligand exchange reactions.v. Relate relative stability of complex ions to the entropy changes of ligand exchange reactions involving polydentate ligands (qualitatively only), eg EDTA.vi. Relate disproportionation reactions to standard electrode potentials and hence to Ecell.

G

Carry out experiments to:i. Investigate ligand exchange in copper complexes.ii. Study the redox chemistry of chromium in oxidation states Cr(VI), Cr(III) and Cr(II).iii. Prepare a sample of a complex, eg chromium(II) ethanoate.

H

Recall that transition metals and their compounds are important as catalysts and that their activity may be associated with variable oxidation states of the elements or surface activity, eg catalytic converters in car exhausts.

I

Explain why the development of new catalysts is a priority area for chemical research today and, in this context, explain how the scientific community reports and validates new discoveries and explanations, eg the development of new catalysts for making ethanoic acid from methanol and carbon monoxide with a high atom economy (green chemistry).

J

Carry out and interpret the reactions of transition metal ions with aqueous sodium hydroxide and aqueous ammonia, both in excess, limited to reactions with aqueous solutions of Cr(III), Mn(II), Fe(II), Fe(III), Ni(II), Cu(II), Zn(II).

KWrite ionic equations to show the difference between amphoteric behaviour and ligand exchange in the reactions in (J) above.

IDiscuss the uses of transition metals and/or their compounds, eg in polychromic sun glasses, chemotherapy drugs.

Transition Metals and Living Organisms

• Iron – transport & storage of O2

• Molybdenum and Iron– Catalysts in nitrogen fixation

• Zinc – found in more than 150 bio molecules• Copper and Iron – crucial role in respiratory

cycle• Cobalt – found in vitamin B12

Transition means “an in between state” and the transition elements come in between Group 2

and Group 3.

Sc Ti V Cr Mn Fe Co Ni Cu Zn

Y Zr Nb Mo Tc Ru Pd Ag CdRh

Hf Ta W Re Os Ir Au HgLa Pt

Rf Db Sg Bh Hs Mt ? ?Ac ?

H

Li

Na

K

Rb

Cs

Fr

Be

Mg

Ga Ge Se BrCa Kr

In Sn SbSr Te

Ba Tl Pb Bi Po At

Ra

Al P

N O

S Cl

F Ne

Ar

Rn

I

Si

Xe

He

B C

As

Gp 2 Gp 3

Transition Elements

They are less reactivethan Group 1 orGroup 2 metals.

They mostly formcoloured

compounds.

They havehigh meltingtemperature.

They have high density.

Transition metalsare often referred toas ‘typical’ metals.

Mechanical properties

TransitionElements

They often act as

catalysts.

• Similarities are more noticeable than differences although there are still some broad patterns.

• They are all dense which is what we expect of metals.

0

1

2

3

4

5

6

7

8

9

Density (g/cm-3)

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn


Properties – density

Properties – melting temperature

Melting temperature show no regular pattern – other than nearly all being high which is typical of metals.(Note zinc doesn’t fit very well on either density or melting temperature.)

0200400600800

100012001400160018002000

Melting Point ( C)

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn


Properties – catalysis

• A catalyst is a substance that speeds up a chemical reaction without being used up.

• Catalysts are hugely valuable in industry where they can save time and energy.

• Many transition elements ( and their compounds) are catalysts.

V2O5Ti

Used in plastic manufacture

Fe

Ni

Used in oil hydrogenation

Transition metals are good metal

catalysts because they easily lend and

take electrons from other molecules.

A catalyst is a chemical substance that,

when added to a chemical reaction, does

not affect the thermodynamics of a

reaction but increases the rate of reaction.

Properties – catalysis

Uses

The three most commonly known transition elements are iron or steel, copper and zinc.

iron or steel

General engineerin

g metal

copper

Electrical and plumbing

work

zinc

Galvanising steel to protect it

Pair the metal up with its uses

iron or steel

copper

zinc

Pair the metal catalyst with the substance.

V

Ti

Fe

Ni

CONTENTS• Enthalpy changes

• Activation Energy

• Heterogeneous catalysis

• Specificity

• Catalytic converters

• Homogeneous catalysis

• Autocatalysis

• Enzymes

How does catalyst work?

Before you start it would be helpful to…• know how the basics of collision theory• understand the importance of activation energy• understand the importance of increasing the rate of

reaction

CATALYSTS - background

All reactions are accompanied by changes in enthalpy.

The enthalpy rises as the reaction starts because energy is being put in to break bonds.

It reaches a maximum then starts to fall as bonds are formed and energy is released. ENTHALPY CHANGE DURING

AN EXOTHERMIC REACTION

CATALYSTS - backgroundAll reactions are accompanied by changes in enthalpy.The enthalpy rises as the reaction starts because energy is being put in to break bonds.It reaches a maximum then starts to fall as bonds are formed and energy is released.

ENTHALPY CHANGE DURINGAN EXOTHERMIC REACTION

If the…FINAL ENTHALPY < INITIAL ENTHALPYit is anEXOTHERMIC REACTIONand ENERGY IS GIVEN OUTFINAL ENTHALPY > INITIAL ENTHALPYit is anENDOTHERMIC REACTIONand ENERGY IS TAKEN IN

ACTIVATION ENERGY - Ea • Reactants will only be able to proceed to products if

they have enough energy

• The energy is required to overcome an energy barrier

• Only those reactants with enough energy will get over

• The minimum energy required is known as the ACTIVATION ENERGY

ACTIVATION ENERGY Ea FORAN EXOTHERMIC REACTION

COLLISION THEORY

According to COLLISON THEORY a reaction will only take place if…

• PARTICLES COLLIDE

• PARTICLES HAVE AT LEAST A MINIMUM AMOUNT OF ENERGY

• PARTICLES ARE LINED UP CORRECTLY

To increase the chances of a successful reaction you need

to...

• HAVE MORE FREQUENT COLLISONS

• GIVE PARTICLES MORE ENERGY or

• DECREASE THE MINIMUM ENERGY REQUIRED

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MOLECULAR ENERGY Ea

DUE TO THE MANY COLLISONS TAKING PLACE IN GASES, THERE IS A SPREAD OF MOLECULAR ENERGY AND VELOCITY


NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER

MAXWELL-BOLTZMANN DISTRIBUTION

The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react.If a catalyst is added, the Activation Energy is lowered - Ea will move to the left.

The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react.Lowering the Activation Energy, Ea, results in a greater area under the curve after Ea showing that more molecules have energies in excess of the Activation Energy

Ea

EXTRA NUMBER OF MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER

MAXWELL-BOLTZMANN DISTRIBUTION

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Catalysts work by providing…

“AN ALTERNATIVE REACTION PATHWAY WHICH HAS A LOWER ACTIVATION ENERGY”

CATALYSTS - lower Ea

A GREATER PROPORTION OF PARTICLES WILL HAVE ENERGIES

IN EXCESS OF THE MINIMUM REQUIRED SO MORE WILL REACT

WITHOUT A CATALYST WITH A CATALYST

PRINCIPLES OF CATALYTIC ACTION

The two basic types of catalytic action are …

HETEROGENEOUS CATALYSIS

and

HOMOGENEOUS CATALYSIS

Format Catalysts are in a different phase to the reactants e.g. a solid catalyst in a gaseous reaction

Action takes place at active sites on the surface of a solid gases are adsorbed onto the surface they form weak bonds with metal atoms

Catalysis is thought to work in three stages...

Adsorption

Reaction

Desorption

Heterogeneous Catalysis


For an explanation of what happens click on the numbers in turn, starting with


Adsorption (STEP 1)Incoming species lands on an active site and forms bonds with the catalyst. It may use some of the bonding electrons in the molecules thus weakening them and making a subsequent reaction easier.Reaction (STEPS 2 and 3)Adsorbed gases may be held on the surface in just the right orientation for a reaction to occur.This increases the chances of favourable collisions taking place.


Desorption (STEP 4)There is a re-arrangement of electrons and the products are then released from the active sites


ANIMATION


Desorption (STEP 4)There is a re-arrangement of electrons and the products are then released from the active sites



ANIMATION

STRENGTH OF ADSORPTION

The STRENGTH OF ADSORPTION is critical ...

too weak Ag little adsorption - few available d orbitalstoo strong W molecules remain on the surface preventing further reactionjust right Ni/Pt molecules are held but not too strongly so they can get away

Catalysis of gaseous reactions can lead to an increase in rate in several ways

• one species is adsorbed onto the surface and is more likely to undergo a collision

• one species is held in a favourable position for reaction to occur• adsorption onto the surface allows bonds to break and fragments react

quicker• two reactants are adsorbed alongside each other give a greater

concentration

[Ar]

Cu [Ar] 3d10 4s1 4p0 4d0

How does catalyst work?

C2H5OH CH3CHO + H2

Specificity

In some cases the choice of catalyst can influence the productsEthanol undergoes different reactions depending on the metal used as the catalyst.The distance between active sites and their similarity with the length of bondsdetermines the method of adsorption and affects which bonds are weakened.

CLICK HERE FOR ANIMATION

SpecificityIn some cases the choice of catalyst can influence the products

Ethanol undergoes different reactions depending on the metal used as the catalyst.The distance between active sites and their similarity with the length of bondsdetermines the method of adsorption and affects which bonds are weakened.

SpecificityIn some cases the choice of catalyst can influence the products

C2H5OH CH3CHO + H2 C2H5OH C2H4 + H2O

Ethanol undergoes different reactions depending on the metal used as the catalyst.The distance between active sites and their similarity with the length of bondsdetermines the method of adsorption and affects which bonds are weakened. Alumina DehydrationCopper Dehydrogenation (oxidation)

Ethanol undergoes two different reactions depending on the metal used as the catalyst.

COPPER Dehydrogenation (oxidation)

C2H5OH CH3CHO + H2

The active sites are the same distanceapart as the length of an O-H bond

It breaks to release hydrogen

ALUMINA Dehydration (removal of water)

C2H5OH C2H4 + H2O

The active sites are the same distanceapart as the length of a C-O bond

It breaks to release an OH group

Specificity

Poisoning

Impurities in a reaction mixture can also adsorb onto the surface of a catalyst thus removing potential sites for gas molecules and decreasing efficiency.

expensive because... the catalyst has to replaced the process has to be shut

down

examples Sulphur Haber processLead catalytic converters in cars

Catalytic convertersPURPOSE removing the pollutant gases formed in

internal combustion enginesPOLLUTANTS CARBON MONOXIDE

NITROGEN OXIDESUNBURNT HYDROCARBONS

Catalytic convertersPURPOSE removing the pollutant gases formed

in internal combustion enginesPOLLUTANTS CARBON MONOXIDE

NITROGEN OXIDESUNBURNT HYDROCARBONS

CONSTRUCTION made from alloys of platinum, rhodium and palladium catalyst is mounted in a support medium to spread it out honeycomb construction to ensure maximum gas contact finely divided to increase surface area / get more collisions

involves HETEROGENEOUS CATALYSIS

CO + Unburned Hydrocarbons + O2 CO2 + H2O

2NO + 2NO2 2N2 + 3O2

Catalytic converters

Pollutant gases

Carbon monoxide CO

Originincomplete combustion of hydrocarbons in petrol when not enough oxygen is present to convert all the carbon to carbon dioxide

C8H18(g) + 8½O2(g) 8CO(g) + 9H2O(l)

Pollutant gases

Carbon monoxide CO

Origin incomplete combustion of hydrocarbons in petrol when not enough oxygen is present to convert all the carbon to carbon dioxide

C8H18(g) + 8½O2(g) 8CO(g) + 9H2O(l)

Effect poisonouscombines with haemoglobin in bloodprevents oxygen being carried to cells

Pollutant gases

Carbon monoxide CO

Origin incomplete combustion of hydrocarbons in petrol when not enough

oxygen is present to convert all the carbon to carbon dioxide

C8H18(g) + 8½O2(g) 8CO(g) + 9H2O(l)

Effect poisonouscombines with haemoglobin in bloodprevents oxygen being carried to cells

Removal 2CO(g) + O2(g) 2CO2(g)

2CO(g) + 2NO(g) N2(g) + 2CO2(g)

Pollutant gases

Oxides of nitrogen NOx - NO, N2O and NO2

Origin nitrogen and oxygen combine under high temperature conditions

nitrogen combines with oxygen N2(g) + O2(g) 2NO(g)

nitrogen monoxide is oxidised 2NO(g) + O2(g) 2NO2(g)

Pollutant gases





Effect photochemical smog - irritating to eyes, nose and throatproduces low level ozone - affects plant growth

- is irritating to eyes, nose and throat

i) sunlight breaks down NO2 NO2 NO + O

ii) ozone is produced O + O2 O3

Pollutant gases





Effect photochemical smog - irritating to eyes, nose and throatproduces low level ozone - affects plant growth

- is irritating to eyes, nose and throat

i) sunlight breaks down NO2 NO2 NO + O

ii) ozone is produced O + O2 O3

Removal 2CO(g) + 2NO(g) N2(g) + 2CO2(g)

Pollutant gases

Unburnt hydrocarbons CxHy

Origin insufficient oxygen for complete combustion

Effect toxic and carcinogenic (causes cancer)

Removal catalyst aids complete combustion

C8H18(g) + 12½O2(g) 8CO2(g) + 9H2O(l)

Homogeneous Catalysis

Action • catalyst and reactants are in the same phase

• reaction proceeds through an intermediate species of lower energy

• there is usually more than one reaction step

• transition metal ions are often involved - oxidation state changes

Example

Acids Esterificaton

Conc. H2SO4 catalyses the reaction between acids and alcohols

CH3COOH + C2H5OH CH3COOC2H5 + H2O

NB Catalysts have NO EFFECT ON THE POSITION OF EQUILIBRIUM

but they do affect the rate at which equilibrium is reached



• reaction proceeds through an intermediate species of lower energy





• reaction proceeds through an intermediate species with of energy



Examples

Gases Atmospheric OZONE breaks down naturally O3 O• + O2

- it breaks down more easily in the presence of chlorofluorocarbons (CFC's).

There is a series of complex reactions but the basic process is :-

CFC's break down in the presence ofUV light to form chlorine radicals CCl2F2 Cl• +

• CClF2

chlorine radicals then react with ozone O3 + Cl• ClO• + O2

chlorine radicals are regenerated ClO• + O O2 + Cl•

Overall, chlorine radicals are not used up so a small amount of CFC's can

destroy thousands of ozone molecules before the termination stage.

Transition metal compoundsThese work because of their ability to change oxidation state

1. Reaction between iron(III) and vanadium(III)

The reaction is catalysed by Cu2+

step 1 Cu2+ + V3+ Cu+ + V4+

step 2 Fe3+ + Cu+ Fe2+ + Cu2+

overall Fe3+ + V3+ Fe2+ + V4+

1. Reaction between iron(III) and vanadium(III)

The reaction is catalysed by Cu2+

step 1 Cu2+ + V3+ Cu+ + V4+

step 2 Fe3+ + Cu+ Fe2+ + Cu2+

overall Fe3+ + V3+ Fe2+ + V4+

Transition metal compoundsThese work because of their ability to change oxidation state

2. Reaction between I¯ and S2O82-

A slow reaction because REACTANTS ARE NEGATIVE IONS REPULSION

Addition of iron(II) catalyses the reaction

step 1 S2O82- + 2Fe2+ 2SO4

2- + 2Fe3+

step 2 2Fe3+ + 2I¯ 2Fe2+ + I2

overall S2O82- + 2I¯ 2SO4

2- + I2

2. Reaction between I¯ and S2O82-

A slow reaction because REACTANTS ARE NEGATIVE IONS REPULSION

Addition of iron(II) catalyses the reaction

step 1 S2O82- + 2Fe2+ 2SO4

2- + 2Fe3+

step 2 2Fe3+ + 2I¯ 2Fe2+ + I2

overall S2O82- + 2I¯ 2SO4

2- + I2

Auto-catalysis

Occurs when a product of the reaction catalyses the reaction itself

It is found in the reactions of manganate(VII) with ethandioate

2MnO4¯ + 16H+ + 5C2O42- 2Mn2+ + 8H2O + 10CO2

The titration needs to be carried out at 70°C because the reaction is slow as Mn2+ is formed the reaction speeds up; the Mn2+ formed acts as the catalyst

A redox titration.

Activity is affected by ...

temperature - it increases until the protein is denatured

substrate concentration - reaches a maximum when all sites are blocked

pH- many catalysts are amino acids which can be protenated

being poisoned - when the active sites become “clogged” with unwanted

ENZYMES

Action enzymes are extremely effective biologically active catalysts they are homogeneous catalysts, reacting in solution with body fluids only one type of molecule will fit the active site “lock and key” mechanism makes enzymes very specific as to what they catalyse.

ENZYMESAction enzymes are extremely effective biologically active catalysts they are homogeneous catalysts, reacting in solution with body fluids only one type of molecule will fit the active site “lock and key” mechanism makes enzymes very specific as to what they catalyse.

A B C

A Only species with the correct shape can enter the active site in the enzymeB Once in position, the substrate can react with a lower activation energyC The new products do not have the correct shape to fit so the complex breaks up

ENZYMES

ANIMATED ACTION

A Only species with the correct shape can enter the active site in the enzymeB Once in position, the substrate can react with a lower activation energyC The new products do not have the correct shape to fit so the complex breaks up

Introduction

• d-block elements

locate between the s-block andp-block

known as transition elements

occur in the fourth and subsequent periods of the Periodic Table

period 4

period 5

period 6

period 7

d-block elements

Introduction

Transition elements are elements that contain an incomplete d sub-shell (i.e. d1 to d9) in at least one of the oxidation states of their compounds.

3d0

3d10

Sc and Zn are not transition elements because

They form compounds with only one oxidation state in which the d sub-shell are NOT imcomplete.

Sc Sc3+ 3d0 Zn Zn2+ 3d10

Introduction

Cu

Cu+ 3d10 not transitional

Cu2+ 3d9 transitional

Introduction

The first transition series

the first horizontal row of the d-block elements

The building up of electronic configurations of elements follow:

Aufbau principle

Pauli exclusion principle

Hund’s rule

Electronic Configurations

Vanadium is classified as a transition metal. This is because vanadium

A is a d-block element.B has incompletely filled d orbitals.C forms stable ions with incompletely filled d orbitals.D forms stable ions in which it has different oxidation states. Shafeeu

Aufbau Principle: – orbitals fill in order of increasing energy from

lowest energy to highest energyPauli Exclusion Principle:

– only two electrons can occupy an orbital and their spins must be paired

Hund’s Rule: – when orbitals of equal energy are available but

there are not enough electrons to fill all of them, one electron is added to each orbital before a second electron is added to any one of them

Electronic Configurations

K 1s2 2s2 2p6 3s2 3p6 4s1

Ca 1s2 2s2 2p6 3s2 3p6 4s2

Sc 1s2 2s2 2p6 3s2 3p6 4s2 3d1

Ti 1s2 2s2 2p6 3s2 3p6 4s2 3d2

V 1s2 2s2 2p6 3s2 3p6 4s2 3d3

Cr 1s2 2s2 2p6 3s2 3p6 4s1 3d5

Mn 1s2 2s2 2p6 3s2 3p6 4s2 3d5

Fe 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Co 1s2 2s2 2p6 3s2 3p6 4s2 3d7

Ni 1s2 2s2 2p6 3s2 3p6 4s2 3d8

Cu 1s2 2s2 2p6 3s2 3p6 4s1 3d10

Zn 1s2 2s2 2p6 3s2 3p6 4s2 3d10

ELECTRONIC CONFIGURATIONS OF THE FIRST ROW TRANSITION METALS

Element Atomic number

Electronic configuration

ScandiumTitaniumVanadiumChromiumManganeseIronCobaltNickelCopperZinc

21222324252627282930

[Ar]3d14s2

[Ar]3d24s2

[Ar]3d34s2

[Ar]3d54s1

[Ar]3d54s2

[Ar]3d64s2

[Ar]3d74s2

[Ar]3d84s2

[Ar]3d104s1

[Ar]3d104s2

Electronic configurations of the first series of d-block elements

Relative energy levels of orbitals before and after filling with electrons

d-Block elements (transition elements):• Lie between s-block and p-block elements • Occur in the fourth and subsequent periods• All contains incomplete d sub-shell (i.e. 1 – 9

electrons) in at least one of their oxidation state

Scandium

Titanium

Vanadium Chromium

Manganese

ZincCopperNickel

CobaltIron

Strictly speaking, scandium (Sc) and zinc

(Zn) are not transitions elements

∵ Sc forms Sc3+ ion which has an empty

d sub-shell (3d0)

Zn forms Zn2+ ion which has a

completely filled d sub- shell (3d10)

Cu+ is not a transition metal ion as it has a completely filled d sub-shell

Cu2+ is a transition metal ion as it has an incompletely filled d sub-shell

Cu shows some intermediate behaviour between

transition and non-transition elements because of

two oxidation states, Cu(I) & Cu(II)

THE FIRST ROW TRANSITION ELEMENTS

Definition Transition elements forming one or more stable ions with partially filled (incomplete) d-sub shells.

The first row runs from scandium to zinc filling the 3d orbitals.

Properties arise from an incomplete d sub-shell in atoms or ionsMetallic all the transition elements are metals properties strong metallic bonds due to small ionic size and close packing higher melting, boiling points and densities than s-block metals

K Ca Sc Ti V Cr Mn Fe Co

T m/ °C 63 850 1400 1677 1917 1903 1244 1539 1495densityg cm-3 0.86 1.55 3 4.5 6.1 7.2 7.4 7.9 8.9

4s

3 3p

3d

44p

4d

4f


POTASSIUM

1s2 2s2 2p6 3s2 3p6 4s1

In numerical terms one would expect the 3d orbitals to be filled next.

However, because the principal energy levels get closer together as you go further from the nucleus coupled with the splitting into sub energy levels, the 4s orbital is of a LOWER ENERGY than the 3d orbitals so gets filled first.

‘Aufbau’ Principle‘Aufbau’ Principle

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4s

3 3p

3d

44p

4d

4f

CALCIUM

1s2 2s2 2p6 3s2 3p6 4s2

As expected, the next electron in pairs up to complete a filled 4s orbital.

This explanation, using sub levels fits in with the position of potassium and calcium in the Periodic Table. All elements with an -s1 electronic configuration are in Group I and all with an -s2 configuration are in Group II.

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4s

3 3p

3d

44p

4d

4f

SCANDIUM

1s2 2s2 2p6 3s2 3p6 4s2 3d1

With the lower energy 4s orbital filled, the next electrons can now fill p the 3d orbitals. There are five d orbitals. They are filled according to Hund’s Rule.

BUT WATCH OUT FOR TWO SPECIAL CASES.

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4s

3 3p

3d

44p

4d

4f

TITANIUM

1s2 2s2 2p6 3s2 3p6 4s2 3d2

The 3d orbitals are filled according to Hund’s rule so the next electron doesn’t pair up but goes into an empty orbital in the same sub level.

HUND’S RULE OFMAXIMUM

MULTIPLICITY


MULTIPLICITY

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4s

3 3p

3d

44p

4d

4fVANADIUM

1s2 2s2 2p6 3s2 3p6 4s2 3d3

The 3d orbitals are filled according to Hund’s rule so the next electron doesn’t pair up but goes into an empty orbital in the same sub level.

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MULTIPLICITY


MULTIPLICITY

4s

3 3p

3d

44p

4d

4fCHROMIUM

1s2 2s2 2p6 3s2 3p6 4s1 3d5

One would expect the configuration of chromium atoms to end in 4s2 3d4.

To achieve a more stable arrangement of lower energy, one of the 4s electrons is promoted into the 3d to give six unpaired electrons with lower repulsion.

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4s

3 3p

3d

44p

4d

4f

MANGANESE

1s2 2s2 2p6 3s2 3p6 4s2 3d5

The new electron goes into the 4s to restore its filled state.

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4s

3 3p

3d

44p

4d

4f

IRON

1s2 2s2 2p6 3s2 3p6 4s2 3d6

Orbitals are filled according to Hund’s Rule. They continue to pair up.

HUND’S RULE OFMAXIMUM MULTIPLICITY


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4s

3 3p

3d

44p

4d

4f

COBALT

1s2 2s2 2p6 3s2 3p6 4s2 3d7




Anhydrous

Hexahydrate

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4s

3 3p

3d

44p

4d

4f

NICKEL

1s2 2s2 2p6 3s2 3p6 4s2 3d8




INC

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4s

3 3p

3d

44p

4d

4f

COPPER

1s2 2s2 2p6 3s2 3p6 4s1 3d10

One would expect the configuration of copper atoms to end in 4s2 3d9.

To achieve a more stable arrangement of lower energy, one of the 4s electrons is promoted into the 3d.



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4s

3 3p

3d

44p

4d

4fZINC

1s2 2s2 2p6 3s2 3p6 4s2 3d10

The electron goes into the 4s to restore its filled state and complete the 3d and 4s orbital filling.

INC

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Electronic Electronic ConfigurationElement Z 3d 4sSc 21 [Ar]

Ti 22 [Ar]

V 23 [Ar]

Cr 24 [Ar]

Mn 25 [Ar]

Fe 26 [Ar]

Co 27 [Ar]

Ni 28 [Ar]

Cu 29 [Ar]

Zn 30 [Ar]

VARIABLE OXIDATION STATES

Arises from the similar energies required for removal of 4s and 3d electrons

Maximum rises across row to manganese

Maximum falls as the energy required to remove more

electrons becomes very high all (except scandium)

have an M2+ ion stability of +2 state increases across

the row due to increase in the 3rd Ionisation Energy

THE MOST IMPORTANT STATES ARE IN RED

Oxidation states of the elements of the first transition series in their compounds

Element Possible oxidation state

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

Element Possible oxidation state

Sc

Ti

V

Cr

Mn

Fe

Co

Ni

Cu

Zn

+3

+1 +2 +3 +4

+1 +2 +3 +4 +5

+1 +2 +3 +4 +5+6

+1 +2 +3 +4 +5+6 +7

+1 +2 +3 +4 +5+6

+1 +2 +3 +4 +5

+1 +2 +3 +4 +5

+1 +2 +3

+2

Ti Sc V Cr Mn Fe Co Ni Cu Zn

+1+2

+3+4

+5+6

+2 +2 +2 +2 +2 +2 +2 +2

+6+7

+3+3 +3+4

+3+4

+3 +3 +3+4 +4 +4 +4+5 +5 +5 +5

+6

When electrons are removed they come from the 4s orbitals first Ti 1s2 2s2 2p6 3s2 3p6 3d2 4s2

Ti2+ 1s2 2s2 2p6 3s2 3p6 3d2

Ti3+ 1s2 2s 2p6 3s2 3p6 3d1

Ti4+ 1s2 2s2 2p6 3s2 3p6

Cu 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Cu+ 1s2 2s2 2p6 3s2 3p6 3d10

Cu2+ 1s2 2s2 2p6 3s2 3p6 3d9

Oxidation states

Oxides / Chloride

+1Cu2O

Cu2Cl2

+2

TiO VO CrO MnO FeO CoO NiOCuO ZnO

TiCl2 VCl2 CrCl2 MnCl2 FeCl2 CoCl2 NiCl2CuCl2 ZnCl2

+3

Sc2O3 Ti2O3 V2O3 Cr2O3 Mn2O3 Fe2O3 Ni2O3 •

xH2O

ScCl3 TiCl3 VCl3 CrCl3 MnCl3 FeCl3

+4TiO2 VO2 MnO2

TiCl4 VCl4 CrCl4

+5 V2O5

+6 CrO3

+7 Mn2O7

Oxidation states of the elements of the first transition series in their oxides and chlorides

The element zinc, with electronic configuration

1s22s22p63s23p63d104s2, is not regarded as a

transition element because

A the oxide of zinc is amphoteric.

B none of its ions has an unpaired electron in the

d -sub shell.

C it does not readily form complex ions.

D it has a boiling temperature low enough for it to be

easily distilled. Shahudaan

Ioniz

atio

n E

nth

alp

y

Element

Ionization enthalpy (kJ mol–1)

1st 2nd 3rd 4th

KCa

418590

3 0701 150

4 6004 940

5 8606 480

ScTiVCrMnFeCoNiCuZn

632661648653716762757736745908

1 2401 3101 3701 5901 5101 5601 6401 7501 9601 730

2 3902 7202 8702 9903 2502 9603 2303 3903 5503 828

7 1104 1704 6004 7705 1905 4005 1005 4005 6905 980

20 21 22 23 24 25 26 27 28 29 30 31 32

Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn 31 32

4000

3500

3000

2500

2000

1500

1000

Successive ionization energy data for evidence for electronic configuration th

ird

second

first

• Some abnormal high ionization enthalpy, e.g. 1st I.E. of Zn, 2nd I.E. of Cr & Cu and the 3rd I.E. of Mn

∵The removal of an electron from a fully-filled or half-filled sub-shell requires a relatively large amount of energy

7000

6000

5000

4000

3000

2000

1000

0 Sc Ti V Cr Mn Fe Co Ni Cu Zn 3d1 3d2 3d3 3d5 3d5 3d6 3d7 3d8 3d10 3d10 4s2 4s2 4s2 4s1 4s2 4s2 4s2 4s2 4s1 4s2

3p6 3p6 3p6 3p6 3p6 3p6 3p6 3p6 3p6 3p6

Ion

izat

ion

en

erg

y (

kJ /

mo

l)

second

third

first

fourth

Successive ionization energy data for evidence for electronic configuration

Electron has to be removed from completely filled 3p sub shell

3d5

Which of the following successive ionization

energies (values in kJ mol-1) could have come

from a transition element?

A 496 4563 6913 9544 13352 16611 20115 25941

B 590 1145 4912 6474 8144 10496 12320 14207

C 717 1509 3249 4940 6985 9200 11508 18956

D 2081 3952 6122 9370 12177 15239 19999 23069

Nuha

Ionization Enthalpy

ElementIonization enthalpy (kJ mol–1)

1st 2nd 3rd 4th

KCa

418590

3 0701 150

4 6004 940

5 8606 480

ScTiVCrMnFeCoNiCuZn

632661648653716762757736745908

1 2401 3101 3701 5901 5101 5601 6401 7501 9601 730

2 3902 7202 8702 9903 2502 9603 2303 3903 5503 828

7 1104 1704 6004 7705 1905 4005 1005 4005 6905 980

Explain the following variation in terms of electronic configurations.

(a) The second ionization enthalpies of both Cr and Cu are higher than those of their next elements respectively.

Answer(a)The second ionization enthalpies of both Cr and Cu are higher

than those of their next elements respectively. In the case of Cr, the second ionization enthalpy involves the removal of an electron from a half-filled 3d sub-shell, which has extra stability. Therefore, this second ionization enthalpy is relatively high. The case is similar for copper where its second ionization enthalpy involves the removal of an electron from a fully-filled 3d sub-shell which also has extra stability. Thus, its second ionization enthalpy is also relatively high.

Explain the following variation in terms of electronic configurations. Happiness

(b)The third ionization enthalpy of Mn is higher than that of its next element.

Answer(b) The third ionization enthalpy of Mn is higher than that of its next element. It is because its third ionization enthalpy involves the removal of an electron from a half-filled 3d sub-shell which has extra stability. Therefore, its third ionization enthalpy is relatively high.

Oxidation States of the Transition MetalsSome oxidation states, however, are more common than others. The most common oxidation states of the first series of transition metals are given in the table below. Some of these oxidation states are common because theyare relatively stable. Others describe compounds that are not necessarily stable but which react slowly. Possible electron configurations of transition metal ionsSc3+, [Ar]3d0

Ti2+, [Ar]; Ti4+, [Ar]3d0

V2+, [Ar]3d3; V3+, [Ar]3d2; V4+, [Ar]3d1; V5+, [Ar]3d0

Cr2+, [Ar]3d4; Cr3+, [Ar]3d3; Cr4+, [Ar]3d2; Cr6+, [Ar]Mn2+,[Ar]3d5; Mn4+, [Ar]3d3; Mn7+, [Ar] 3d0

Fe2+, [Ar]3d6; Fe3+, [Ar]3d5

Co2+, [Ar]3d7; Co3+, [Ar]3d6

Ni2+, [Ar]3d8

Cu+, [Ar]3d10; Cu2+, [Ar]3d9

Zn2+, [Ar]3d10

Common Oxidation States of the First Series of Transition Metals

One point about the oxidation states of transition metals deserves particular attention: Transition-metal ions with charges larger than +3 cannot exist in aqueous solution.

COLOURED IONS

Theory ions with a d10 (full) or d0 (empty) configuration

are colourle ions with partially filled d-orbitals

tend to be coloured it is caused by the ease of

transition of electrons between energy levels

energy is absorbed when an electron is promoted

to a higher level the frequency of light is

proportional to the energy difference

ions with d10 (full) Cu+,Ag+ Zn2+

or d0 (empty) Sc3+ configuration are colourlesse.g. titanium(IV) oxide TiO2 is white

colour depends on ...transition element oxidation state ligand , coordination number

A characteristic of transition metals is their ability to form coloured compounds

A hydrated transition metal ion is colourless.

Which of the following could be the electronic

configuration of this ion?

A [Ar] 3d54s2

B [Ar] 3d8

C [Ar] 3d104s2

D [Ar] 3d100

Ayaman

Cu2+ [Ar] 3d9 4s0 4p0 4d0

[Ar]

[Ar]

Cu2+ [Ar] 3d9 4s2 4p6 4d4

How does empty orbitals work?

[Cu(H2O)6]2+

Cu2+

Dative Covalent Bond- Co-ordination bond

Reagent\Ion and initial

colour

Cr3+(aq)

green

Mn2+(aq) very

pale pink ~ colourless

Fe2+(aq)

pale green

Fe3+(aq)

yellow-brown

Co2+(aq)

pink

Ni2+(aq)

green

Cu2+(aq)

blue

Zn2+(aq)

colourless

Al3+(aq)

colourless

initial NaOH(aq)

strong base/alkali

green gel. ppt. of

Cr(OH)3

white gel. ppt. but

darkens with oxidation

Mn(OH)2

Mn2O3

MnO2

dark green ppt. =>

brown on oxidation

Fe(OH)2

Fe(OH)3

brown gel. ppt. of

Fe(OH)3

blue gel. ppt. that

turns pink on standing

Co(OH)2

green gel. ppt. of

Ni(OH)2

gel. blue ppt. of

Cu(OH)2

white gel. ppt. of

Zn(OH)2

white gel. ppt. of

Al(OH)3

excess NaOH(aq)

strong base/alkali

ppt. dissolves to give clear

green solution of

complex ion

[Cr(OH)6]3-

no further effect - just as above with more oxidation



no further effect

no further effect

no further effect ppt. dissolves -

clear solution,

colourless complex ion

[Zn(OH)4]2-

ppt. dissolves -

clear solution,


[Al(OH)6]3-

initial NH3(aq)

weak base/alkali

green gel. ppt. of

Cr(OH)3

white ppt. darkens with

oxidation from O2

Mn(OH)2

Mn2O3

MnO2

dark green ppt. turns brown -

oxidation

Fe(OH)2

Fe(OH)3

brown gel. ppt. of

Fe(OH)3

blue gel. ppt. that turns pink on

standing

Co(OH)2

green gel. ppt. of

Ni(OH)2

gel. blue ppt. of

Cu(OH)2

white gel. ppt. of

Zn(OH)2

white gel. ppt. of

Al(OH)3

excess NH3(aq)

weak base/alkali

dissolves - clear green solution of

complex ion

[Cr(NH3)6]3+

no further effect

no further effect

no further effect

ppt. dissolves - clear brown solution of

complex ion

[Co(NH3)6]2+

dissolves - clear pale

blue solution of complex

ion

[Ni(NH3)6]2+

ppt. dissolves to give clear blue

solution of complex ion

[Cu(NH3)4(H2O)2]2+

ppt. dissolves -

clear colourless solution of

[Zn(NH3)4]2+

no further effect

Colours of complex ions

Transition metal ion solution

Addition of sodium hydroxide or ammonia solution

Proton transfer

Addition of sodium hydroxide solution

Precipitate dissolves Precipitate dissolves

Ligand exchange Proton transfer

Addition of ammonia solution

Precipitate forms

Ni2+ , Cu2+, Zn2+ ,Cr3+ ,Co2+ Cr3+ , Zn2+

[Cr(H2O)6]3+ (aq) [Fe(H2O)6]

2+ (aq) [Co(H2O)6]2+ (aq)

[Cu(H2O)6]2+ (aq)[Cu(H2O)6]

2+ (aq)

[Mn(H2O)6]2+ (aq)

Chemistry ofNi2+ , Cu2+,

Zn2+ Cr3+ ,Co2+

[Fe(H2O)6]3+(aq)

[Zn(NH3)4] 2+(aq)

Aqueous metal ion reactions

Fe2+

Co2+

Cu2+

Fe3+

Cr3+

Zn2+

Fe2+(aq) ligand substitution

Cu2+(aq) ligand substitution

Cr3+(aq) acidity reaction

Zn2+(aq) acidity reaction

reactions photos

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Click the reaction or photo links

Mn2+

Ni2+

Co2+(aq) reaction


Ni2+(aq) acidity reaction

Mn2+(aq) acidity reaction

Fe2+ iron(II)

Add

NaOH (aq) (little)

[Fe(H2O)6]2+ (aq) green solution

Fe(H2O)4(OH)2 (s) green ppt(turns brown in air)

NaOH (aq) (excess)

NH3 (aq) (little)

NH3 (aq) (excess)

CO32- (aq)

FeCO3 (s)

green ppt

Does not dissolve

precipitation

photoligand substitution

Co2+ cobalt(II)Add

NaOH (aq) (little)

[Co(H2O)6]2+ (aq) pink solution

Co(H2O)4(OH)2 (s) blue ppt(turns pink slowly and then brown in air)NaOH (aq) (excess)

NH3 (aq) (little)

NH3 (aq) (excess)[Co(NH3)6]

2+ (aq)

CO32- (aq)

CoCO3 (s) pink ppt

pale-yellow solution(turns brown in air)

Cl- (aq) (conc) [CoCl4]2- (aq) blue solution

acidity reactions

ligand substitution

precipitation

ligand substitution

Cu2+ copper(II)Add

NaOH (aq) (little)

[Cu(H2O)6]2+ (aq) blue solution

Cu(H2O)4(OH)2] (s)

blue precipitateNaOH (aq) (excess)

NH3 (aq) (little)

NH3 (aq) (excess) [Cu(NH3)4(H2O) 2]2+ (aq)

CO32- (aq) CuCO3 (s) blue precipitate

deep blue solution

Cl- (aq) (conc) [CuCl4]2- (aq) green solution

acidity reactions

ligand substitution

precipitation

ligand substitution

photo

acidity reactions

Fe3+ iron(III)

Add

NaOH (aq) (little)

[Fe(H2O)6]3+ (aq) violet solution

Fe(H2O)3(OH)3 (s)

brown precipitateNaOH (aq) (excess)

NH3 (aq) (little)

CO32- (aq)

Fe(H2O)3(OH)3 (s)

brown precipitate andcolourless bubbles of CO2 (g)

acidity reaction

NH3 (aq) (excess)

acidity reaction

purple solution

acidity reaction

ligand substitution

acidity reactions

Cr3+ chromium(III)Add

NaOH (aq) (little)

[Cr(H2O)6]3+ (aq) red-blue solution

Cr(H2O)3(OH)3 (s)

green precipitateNH3 (aq) (little)

NaOH (aq) (excess) [Cr(OH)6]3- (aq)

NH3 (aq) (excess) [Cr(NH3)6]3+ (aq)

green solution

CO32- (aq)

Cr(H2O)3(OH)3 (s)

green precipitate andcolourless bubbles of CO2 (g)

photo

acidity reactions

Zn2+ zinc(II)

Add

NaOH (aq) (little)

Zn2+ (aq) colourless solution

[Zn(OH)2 (H2O)2](s)

precipitateNH3 (aq) (little)

NaOH (aq) (excess) [Zn(OH)4]2- (aq)

colourless solution

NH3 (aq) (excess)

photo

[Zn(NH3)4] 2+(aq)

white

Ligand substitution

Photographs

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Cu2+(aq) acidity reaction reaction

Cr3+(aq) acidity reaction

Zn2+(aq) acidity reaction

Co2+(aq) acidity reaction

Fe3+(aq) acidity reaction

Ni2+(aq) acidity reaction

Mn2+(aq) acidity reaction

Fe2+ iron(II) photo

[Fe(H2O)6]2+ (aq)

green solution

Fe(H2O)4(OH)2 (s)

green precipitate[turns brown in air) insoluble in excess NaOH(aq) or NH3(aq)]

NaOH(aq) or NH3(aq)

Ligand substitution

Cu2+ copper(II)

[Cu(H2O)6]2+ (aq)

blue solution

Cu(H2O)4(OH)2] (s)

NaOH(aq) or NH3(aq)

blue precipitate soluble in excessNH3(aq)

acidity reaction Click me.I’ll

show you?

Cr3+ chromium(III) photos

[Cr(H2O)6]3+ (aq)

red-blue solutionappears green because of hydrolysis

Cr(H2O)3(OH)3 (s)green precipitate

acidity reactions

NaOH (aq) or NH3 (aq)

Click for more …

excess NaOH (aq)

[Cr(OH)6]3- (aq)

green solution

Click me. I’ll show you?

Cr(H2O)3(OH)3 (s) + NH3

green precipitate

[Cr(NH3)6]3+(aq)

Yellowish

Violet solution

Zn2+ zinc(ii) photo

[Zn(H2O)4]2+ (aq)

colourless solution

[Zn(OH)2 (H2O)2](s)

white precipitate

acidity reactions

Click for more …excess NaOH (aq)

NaOH (aq) or NH3 (aq)

[Zn(OH)2 (OH)4]2-(aq)

colourless solution

[Zn(OH)2 (H2O)2](s)

[Zn(NH3)4] 2+(aq)

acidity reaction

purple solution

acidity reaction

ligand substitution

acidity reactions

Ni2+ nickel(ii)Add

NaOH (aq) (little)

[Cr(H2O)6]3+ (aq) red-blue solution

Cr(H2O)3(OH)3 (s) green ppt

NH3 (aq) (little)

NaOH (aq) (excess) [Cr(OH)6]3- (aq)

NH3 (aq) (excess) [Cr(NH3)6]3+ (aq)

green solution

CO32- (aq) Cr(H2O)3(OH)3 (s) green ppt

colourless bubbles of CO2 (g)

photo

Mn2+ manganese(ii)Add

NaOH (aq) (little)

[Mn(H2O)6]2+ (aq) red-blue solution

Mn(H2O)3(OH)3 (s) buff ppt

NH3 (aq) (little)

NaOH (aq) (excess)

NH3 (aq) (excess)

CO32- (aq)

Mn(H2O)3(OH)3 (s) green ppt

colourless bubbles of CO2 (g)

photo

Insoluble

Insoluble

Mn2+(aq) + 2NH3(aq) + 2H2O(l)

Mn(OH)2(s)+2NH4+(aq)

Mn2+(aq) + 2OH-(aq) Mn(OH)2(s)

Ni(H2O)6]2+(aq) + 2-OH (aq)

Ni(OH)2 (H2O)4](s) + 2H2O

Pale green ppt soluble in excess ammonia

Ni2+(aq) + 2-OH(aq) Ni(OH)2(s)

Click me I’ll show you?

Nice?

Ni(OH)2 (H2O)4](s) + 6NH3(aq)

[Ni(NH3)6]2+(aq) + 2-OH+4H2Olavender blue

Oh?

[Fe(H2O)6]2+(aq) + 3-OH (aq)

[Fe(OH)3(H2O)3](s) +3H2O red brown ppt

Insoluble in excess NaOH and NH3

With NaOH and NH3

[Co(H2O)6]2+(aq) + 2-OH (aq)

[Co(OH)2 (H2O)4](s) Blue ppt

Soluble excess ammonia

NaOH or NH3

Click me. I’ll show you?

[Co(NH3)6]2+(aq) +4H2O +2-OH (aq)

[Co(OH)2 (H2O)4](s) +6NH3(aq)

[yellowish brown]

[Cu(H2O)6]2+ + 2- OH [Cu(OH)2(H2O)4] pale blue ppt

[Cu(OH)2(H2O)4] pale blue ppt + 4NH3

[Cu(NH3)4(H2O)2]+ pale blue solution

I told

you

Excess ammonia

Reagent\Ion and initial

colour

Cr3+(aq)

green

Mn2+(aq) very

pale pink ~ colourless

Fe2+(aq)

pale green

Fe3+(aq)

yellow-brown

Co2+(aq)

pink

Ni2+(aq)

green

Cu2+(aq)

blue

Zn2+(aq)

colourless

Al3+(aq)

colourless

initial NaOH(aq)

strong base/alkali

green gel. ppt. of

Cr(OH)3

white gel. ppt. but

darkens with oxidation

Mn(OH)2

Mn2O3

MnO2

dark green ppt. =>

brown on oxidation

Fe(OH)2

Fe(OH)3

brown gel. ppt. of

Fe(OH)3

blue gel. ppt. that

turns pink on standing

Co(OH)2

green gel. ppt. of

Ni(OH)2

gel. blue ppt. of

Cu(OH)2

white gel. ppt. of

Zn(OH)2

white gel. ppt. of

Al(OH)3

excess NaOH(aq)

strong base/alkali

ppt. dissolves to give clear

green solution of

complex ion

[Cr(OH)6]3-




no further effect

no further effect

no further effect ppt. dissolves -

clear solution,


[Zn(OH)4]2-

ppt. dissolves -

clear solution,


[Al(OH)6]3-

initial NH3(aq)

weak base/alkali

green gel. ppt. of

Cr(OH)3

white ppt. darkens with

oxidation from O2

Mn(OH)2 Mn2O3 MnO2

dark green ppt. turns brown -

oxidation

Fe(OH)2

Fe(OH)3

brown gel. ppt. of

Fe(OH)3

blue gel. ppt. that turns pink on

standing

Co(OH)2

green gel. ppt. of

Ni(OH)2

gel. blue ppt. of

Cu(OH)2

white gel. ppt. of

Zn(OH)2

white gel. ppt. of

Al(OH)3

excess NH3(aq)

weak base/alkali

dissolves - clear green solution of

complex ion

[Cr(NH3)6]3+

no further effect

no further effect

no further effect

ppt. dissolves - clear brown solution of

complex ion

[Co(NH3)6]2+

dissolves - clear pale

blue solution of complex

ion

[Ni(NH3)6]2+

ppt. dissolves to give clear blue

solution of complex ion

[Cu(NH3)4(H2O)2]2+

ppt. dissolves -

clear colourless solution of

[Zn(NH3)4]2+

no further effect

Which of the following reagents would

enable you to separate iron(III)

hydroxide from a mixture of iron(III)

hydroxide and copper(II) hydroxide?

A Dilute hydrochloric acid

B Aqueous ammonia

C Dilute nitric acid

D Sodium hydroxide solution

Jeelaan

Splitting of 3d orbital Placing ligands around a central ion causes the

energies of the d orbitals to changeSome of the d orbitals gain energy and some lose

energyIn an octahedral complex, two (z2 and x2-y2) go

higher and three go lowerIn a tetrahedral complex, three (xy, xz and yz) go

higher and two go lowerDegree of splitting depends on the CENTRAL ION and the LIGANDThe energy difference between the levels affects

how much energy is absorbed when an electron is promoted. The amount of energy governs the colour of light absorbed.

Hemantha Welihena

3d 3d

OCTAHEDRAL TETRAHEDRAL

Splitting of 3d orbital

Hemantha Welihena

increasing energy

Cu2+ (‘free ion’) 3d energy level

degenerate

[Cu(H2O)6]2+

3d level split by water ligands

ΔEenergy

difference

Hemantha Welihena

Coloured Ions

Absorbed colour nm Observed colour nmVIOLET 400 GREEN-YELLOW 560BLUE 450 YELLOW 600BLUE-GREEN 490 RED 620YELLOW-GREEN 570 VIOLET 410YELLOW 580 DARK BLUE 430ORANGE 600 BLUE 450RED 650 GREEN 520

The observed colour of a solution depends on the wavelengths absorbed

Copper sulphate solution appears blue because the energy absorbed corresponds to red and yellow

wavelengths. Wavelengths corresponding to blue light aren’t absorbed.

ENERGY CORRESPONDING TO THESE COLOURS IS ABSORBED

Hemantha Welihena

The observed colour of a solution depends on the wavelengths absorbed

white lightblue and green not absorbed

WHITE LIGHT GOES IN

SOLUTION APPEARS BLUE

Coloured Ions

Hemantha Welihena

a solution of copper(II)sulphate is blue becausered and yellow wavelengths are absorbed

Coloured Ions

Hemantha Welihena

a solution of nickel(II)sulphate is green becauseviolet, blue and red wavelengths are absorbed

Coloured Ions

Hemantha Welihena

Observed colourAbsorbed colour

More on Colours

The colour observed in a compound/ complex is actually the complementary colour of the colour absorbed

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Hemantha Welihena

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Colorful cations

Hemantha Welihena

Hemantha Welihena

A hydrated transition metal ion is colourless. Which

of the following could be the electronic

configuration of this ion?

A [Ar] 3d54s2

B [Ar] 3d8

C [Ar] 3d104s2

D [Ar] 3d10

Nuha

Copper(II) sulfate solution is blue. This is because

A excited electrons emit light in the blue region of the spectrum as they drop back to the ground state.B excited electrons emit light in the red region of the spectrum as they drop back to the ground state.C electrons absorb light in the red region of the spectrum and the residual frequencies are observed.D electrons absorb light in the blue region of the spectrum and the residual frequencies are observed.

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Shahudhaan

Uses of Transition Elements

• Chromium’s name comes from the Greek word for color, chrome.

• Many other transition elements combine to form substances with brilliant colors.

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Transition metals are any of various metallic elements such as chromium, iron and nickel that have valence electrons in two shells instead of only one.

Ag + Cu Ag + Cu A valence electron refers to

a single electron that is responsible for the chemical properties of the atom.

Polychromic sun glasses

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+ + 2+

An important anti-cancer drug. It is a square planar,4 co-ordinate complex of platinum.

Cis-platin

Chemotherapy drugs

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Intercalating agents wedge between bases along the DNA. The intercalated drug molecules affect the structure of the DNA, preventing polymerase and other DNA binding proteins from functioning properly. The result is prevention of DNA synthesis, inhibition of transcription and induction of mutations.

Cis-platin

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Transition metal ions can be identified by color.

Sc Ti V Cr Mn Fe Co Ni Cu Cu Zn

+3 colourless

+3 violet

+3 blue

+3 ruby

+2 pale pink

+2 pale green

+2 pink

+2 green

+ 2 colourless

+ 1 colourless

+2 colourless

+5 yellow

+6 yellow or orange1

+7 purple

+3 red-brown

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Colours of complex ions

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Electronic configuration Ion Colour

Ti [Ar]4s23d2 Ti3+[Ar]4s03d1

Ti4+[Ar]4s03d0 colourless

V [Ar]4s23d3 V2+[Ar]4s03d3 Violet

V3+[Ar]4s03d2 Green

V4+[Ar]4s03d1 Blue

V5+[Ar]4s03d0 Yellow

Cr [Ar]4s13d5 Cr2+[Ar]4s03d4

Cr3+[Ar]4s03d3

Cr6+[Ar]4s03d0

Aqua ComplexesWhen in solution, the ions of Transition metals form what are known as aqua complexes. These are complex ions with water as the ligands. The different ions form complexes with different colours, as shown in the table below:

Electronic

configuration

Ion Colour

Mn [Ar]4s23d5 Mn2+[Ar]4s03d5 Pale pink

Mn4+[Ar]4s03d3

Mn6+[Ar]4s03d1

Mn7+[Ar]4s03d0

Fe [Ar]4s23d6 Fe2+[Ar]4s03d6 Pale green

Fe3+[Ar]4s03d5 Orange, yellow or brown

Co [Ar]4s23d7 Co2+[Ar]4s03d7 Deep pink

Ni [Ar]4s23d8 Ni2+[Ar]4s03d8 Pale green

Cu [Ar]4s13d10 Cu+[Ar]4s03d10

Cu2+[Ar]4s03d9 Blue

Zn [Ar]4s23d10 colourless Hemantha Welihena

Cu2+, Cr2+, Co2+, Ni2+

[Cr(H2O)6]2+, [Cu(H2O)6]2+

[Cu(H2O)4]2+, [Cu(NH3)4]2+

[Cu(H2O)2(NH3)4]2+, [Ni(NH3)6]2+

[Ni(NH3)4]2+, [Co(H2O)6]2+

VO2

Blue colour ions

A solution containing the [CoCl4]2-(aq) ion

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Transition Metal Complexes Contents• Aqueous metal ions• Acidity of hexaaqua ions• Introduction to the reactions of complexes• Reactions of cobalt• Reactions of copper• Reactions chromium• Reactions on manganese• Reactions of iron(II)• Reactions of iron(III)• Reactions of silver and vanadium• Reactions of aluminium

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The aqueous chemistry of ions-Hydrolysiswhen salts dissolve in water the ions are stabilisedthis is because water molecules are polarhydrolysis can occur and the resulting solution can

become acidicthe acidity of the resulting solution depends on the

cation presentthe greater the charge density of the cation, the more

acidic the solution

cation charge ionic radius reaction with water pH of chloride Na 1+ 0.095 nm Mg 2+ 0.065 nm Al 3+ 0.050 nm the greater charge density of the cation...

the greater the polarising power and the more acidic the solution Hemantha Welihena

when salts dissolve in water the ions are stabilisedthis is because water molecules are polarhydrolysis can occur and the resulting solution can become acidicthe acidity of the resulting solution depends on the cation presentthe greater the charge density of the cation, the more acidic the solution

cation charge ionic radius reaction with water / pH of chloride

Na 1+ 0.095 nm dissolves 7 Mg 2+ 0.065 nm slight hydrolysis less than 7 Al 3+ 0.050 nm vigorous hydrolysis less than 7

the greater charge density of the cation...

the greater the polarising power andthe more acidic the solution

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The aqueous chemistry of ions-Hydrolysis

Complex Ions- LigandsFormation ligands form co-ordinate bonds to a central transition metal ion

Ligands atoms, or ions, which possess lone pairs of electronsform co-ordinate bonds to the central iondonate a lone pair into vacant orbitals on the central species

Ligand Formula Name of ligandchloride Cl¯ chlorocyanide NC¯ cyanohydroxide ¯OH hydroxooxide O2- oxowater H2O aqua

ammonia NH3 ammine

some ligands attach themselves using two or more lone pairsclassified by the number of lone pairs they usemultidentate and bidentate ligands lead to more stable complexes

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Unidentate form one co-ordinate bond Cl¯, ¯OH, ¯CN, NH3, and H2O

Bidentate form two co-ordinate bonds H2NCH2CH2NH2 , C2O42-

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Complex Ions- Ligands


Multidentate form several co-ordinate bonds

EDTA

An important complexing agent

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[Cu(H2O)6]2+ [CuCl4]2- [Cu(H2NCH2CH2NH2)3]2+ [Cu(edta)]2+

Increasing stability

[Cu(H2O)6]2+ + 4Cl- [CuCl4]

2- + 6H2O

[CuCl4]2- + 3H2NCH2CH2NH2 [Cu(H2NCH2CH2NH2)3]2+ + 4Cl-

[Cu(H2O)6]2+ + edta4- [Cu(edta)]2- + 6H2O

Relative stability of complex ions

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+2

+2

ISOMERISATION IN COMPLEXES

GEOMETRICAL (CIS-TRANS) ISOMERISM

Square planar complexes of the form [MA2B2]n+ exist in two forms

trans platin cis platin

An important anti-cancer drug. It is a square planar, 4 co-ordinate complex of platinum.

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The transition metal complex Pt(NH3)2Cl2 exists as two geometric isomers. This is because the complex

A is square-planar.B is tetrahedral.C contains a double bond.D is octahedral

Which of the following species is not able to act as a ligand in the formation of transition metal complexes?

A C6H5NH2

B NH3

C NH2CH2CH2CH2NH2

D NH4

Jeelan

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Shafeeu

Co-ordination number and Shape the shape of a complex is governed by the number of ligands around the central ion the co-ordination number gives the number of ligands around the central ion a change of ligand can affect the co-ordination number

Co-ordination No. Shape Example(s)

6 Octahedral [Cu(H2O)6]2+

4 Tetrahedral [CrCl4]-

4 Square planar Pt(NH3)2Cl2

2 Linear [Ag(NH3)2]+

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The coordination number of the central metal atom or ion in a complex is the number of dative covalent bonds formed by the central metal atom or ion in a complex.

ComplexThe central metal atom or

ion in the complexCoordination

number

[Ag(NH3)2]+ Ag+ 2

[Cu(NH3)4]2+ Cu2+ 4

[Fe(CN)6]3– Fe3+ 6

Co-ordination number

Linear [CuCl2]-

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The Aqueous Chemistry of IonsTheory aqueous metal ions attract water molecules

many have six water molecules surrounding themthese are known as hexaaqua ionsthey are octahedral in shapewater acts as a Lewis Base – a lone pair donorwater forms a co-ordinate bond to the metal ionmetal ions accept the lone pair - Lewis Acids

Theory aqueous metal ions attract water moleculesmany have six water molecules surrounding themthese are known as hexaaqua ionsthey are octahedral in shapewater acts as a Lewis Base – a lone pair donorwater forms a co-ordinate bond to the metal ionmetal ions accept the lone pair - Lewis Acids

Acidity as charge density increases, the cation has a greater attraction for waterthe attraction extends to the shared pair of electrons in water’s O-H bonds the electron pair is pulled towards the O, making the bond more polarthis makes the H more acidic (more d+)it can then be removed by solvent water molecules to form H3O+(aq).

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Hydrolysis- Equations , Deprotination

M2+ ions [M(H2O)6]2+(aq) + H2O(l) [M(H2O)5(OH)]+(aq) + H3O+(aq)

the resulting solution will now be acidic as there are more protons in the water

this reaction is known as hydrolysis - the water causes the substance to split up

Stronger bases (e.g. CO32- , NH3 and ¯OH ) can remove further protons...

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M2+ ions [M(H2O)6]2+(aq) + H2O(l) [M(H2O)5(OH)]+(aq) + H3O+(aq)

the resulting solution will now be acidic as there are more protons in the water

this reaction is known as hydrolysis - the water causes the substance to split up


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M3+ ions [M(H2O)6]3+(aq) + H2O(l) [M(H2O)5(OH)]2+(aq) + H3O+(aq)

the resulting solution will also be acidic as there are more protons in the water

this SOLUTION IS MORE ACIDIC due to the greater charge density of 3+ ions


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Lewis bases can attack the co-ordinated water molecules. Theoretically, a proton can be removed from each water molecule turning the water from a neutral molecule to a negatively charged hydroxide ion. This affects the overall charge on the complex ion.

[M(H2O)6]2+(aq) [M(OH)(H2O)5]+(aq) [M(OH)2(H2O)4](s)

[M(OH)2(H2O)4](s) [M(OH)3(H2O)3]¯(aq) [M(OH)4(H2O)2]2-(aq)

[M(OH)4(H2O)2]2-(aq) [M(OH)5(H2O)]3-(aq) [M(OH)6]4-(aq)

When sufficient protons have been removed the complex becomesneutral and precipitation of a hydroxide or carbonate occurs.

e.g. M2+ ions [M(H2O)4(OH)2](s) or M(OH)2

M3+ ions [M(H2O)3(OH)3](s) or M(OH)3

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Hydrolysis of Hexaaquaions

Lewis bases can attack the co-ordinated water molecules. Theoretically, a proton can be removed from each water molecule turning the water from a neutral molecule to a negatively charged hydroxide ion. This affects the overall charge on the complex ion.

[M(H2O)6]2+(aq) [M(OH)(H2O)5]+(aq) [M(OH)2(H2O)4](s)

[M(OH)2(H2O)4](s) [M(OH)3(H2O)3]¯(aq) [M(OH)4(H2O)2]2-(aq)

[M(OH)4(H2O)2]2-(aq) [M(OH)5(H2O)]3-(aq) [M(OH)6]4-(aq)

AMPHOTERIC CHARACTERMetal ions of 3+ charge have a high charge density and their hydroxides can dissolve in both acid and alkali.

[M(H2O)6]3+(aq) [M(OH)3(H2O)3](s) [M(OH)6]3-(aq)

¯ OH

H+

¯ OH

H+

Precipitated

¯ OH

H+Insoluble SolubleSoluble

¯ OH

H+

¯ OH

H+

OH

H+

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Hydrolysis of Hexaaquaions

The examples aim to show typical properties of transition metals and their compounds.

One typical properties of transition elements is their ability to form complex ions.Complex ions consist of a central metal ion surrounded by co-ordinated ions or molecules known as ligands. This can lead to changes in ...

• colour • co-ordination number• shape • stability to oxidation or reduction

Reactiontypes ACID-BASE

LIGAND SUBSTITUTION

PRECIPITATION

REDOX

A-B

LS

OX

Ppt

RED REDOX

Hemantha Welihena

Types of Reaction

The examples aim to show typical properties of transition metals and their compounds.

LOOK FOR...

substitution reactions of complex ions

variation in oxidation state of transition metals

the effect of ligands on co-ordination number and shape

increased acidity of M3+ over M2+ due to the increased charge density

differences in reactivity of M3+ and M2+ ions with ¯OH and NH3

the reason why M3+ ions don’t form carbonates

amphoteric character in some metal hydroxides (Al3+ and Cr3+)

the effect a ligand has on the stability of a particular oxidation state

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Types of Reaction

• aqueous solutions contain the pink, octahedral hexaaquacobalt(II) ion

• hexaaqua ions can also be present in solid samples of the hydrated salts

• as a 2+ ion, the solutions are weakly acidic but protons can be removed by bases...

¯ OH [Co(H2O)6]2+(aq) + 2¯OH (aq) [Co(OH)2(H2O)4](s) + 2H2O(l)

pink, octahedral blue / pink ppt. soluble in XS NaOH

ALL hexaaqua ions precipitate a hydroxide with ¯OH (aq).

Some re-dissolve in excess NaOH

A-B

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Reaction of Cobalt(II)

NH3 [Co(H2O)6]2+(aq) + 2NH3(aq) [Co(OH)2(H2O)4](s) + 2NH4+(aq)

ALL hexaaqua ions precipitate a hydroxide with NH3 (aq). It removes protons

A-B

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Some hydroxides redissolve in excess NH3(aq) as ammonia substitutes as a ligand.

[Co(OH)2(H2O)4](s) + 6NH3(aq) [Co(NH3)6]2+(aq) + 4H2O(l) + 2¯OH (aq)

A-B

LS

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Some hydroxides redissolve in excess NH3(aq) as ammonia substitutes as a ligand.

[Co(OH)2(H2O)4](s) + 6NH3(aq) [Co(NH3)6]2+(aq) + 4H2O(l) + 2¯OH (aq)

but ... ammonia ligands make the Co(II) state unstable. Air oxidises Co(II) to Co(III)

[Co(NH3)6]2+(aq) [Co(NH3)6]3+(aq) + e¯

yellow / brown octahedral red / brown octahedral

A-B

LS

OX

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CO32- [Co(H2O)6]2+(aq) + CO3

2-(aq) CoCO3(s) + 6H2O(l)

mauve ppt.

Hexaaqua ions of metals with charge 2+ precipitate a carbonate butheaxaaqua ions with a 3+ charge don’t.

Cl¯ [Co(H2O)6]2+(aq) + 4Cl¯(aq) [CoCl4]2-(aq) + 6H2O(l)

pink, octahedral blue, tetrahedral

• Cl¯ ligands are larger than H2O

• Cl¯ ligands are negatively charged - H2O ligands are neutral

• the complex is more stable if tetrahedral - less repulsion between ligands

• adding excess water reverses the reaction

LS

Ppt

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Aqueous solutions of copper(II) contain the blue, octahedral hexaaquacopper(II) ionMost substitution reactions are similar to cobalt(II).

¯ OH [Cu(H2O)6]2+(aq) + 2¯OH (aq) [Cu(OH)2(H2O)4](s) + 2H2O(l)

blue, octahedral pale blue ppt. insoluble in XS NaOH

A-B

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One method of manufacturing hydrazine (N2H4) involves the action of sodium chlorate(I) on excess ammonia at 443 K and 50 atm. The yield is normally around 80% but, if just 1 part per million of copper(II) ions is present, the yield drops to 30%.

The most likely explanation for this is the ability of copper(II) ions to

A form complex ions with ammonia.B catalyse reactions producing other nitrogen

compounds.C reduce the hydrazine as it is formed.D reduce the sodium chlorate.

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Arusham

Reactions between sodium chlorate(I) on excess ammonia

NaOCl(aq) + 2NH3(g) N2H4(g) + NaCl(aq) + H2O(l)

2NaOCl(aq) + 2NH3(aq) 2NH2Cl(g) + 2NaOH(aq)

2NaOCl(aq) + 2NH3(aq) 2NaONH3(aq) + Cl2(g)

3NaOCl(aq) + NH3(aq) 3NaOH(aq) + NCl3(g)

Metal English name of ion Latin name of ion

V3+ Vanadium(iii) ion Vanadate(iii) ion

Cr3+ Chromium(iii) ion Chromate(iii) ion

Mn2+ Manganese(ii) ion Manganate(ii) ion

Fe2+ Iron(ii) ion Ferrate(ii) ion

Co2+ Cobalt(ii) ion Cobaltate(ii) ion

Ni2+ Nickel(ii) ion Nickalate(ii) ion

Cu2+ Copper(ii) ion Cuprate(ii) ion

Zn2+ Zinc(ii) ion Zinicate(ii) ion

Ag+ Silver(i) ion Argentate(i) ion

Pt2+ Platinum(ii) ion Platinate(ii) ion[V(H2O)6]3 [Cr(H2O)2]2

hexaaquavanadium(ii) hexaaquachromate(ii) Hemantha Welihena

LigandType of

complexExample Name of Complex

Water Aqua- [Cr(H2O)6]3+ hexaaquachromium(III) ion

Ammonia Ammine- [Ag(NH3)2]+ diamminesilver(I) ion

Hydroxide ion Hydroxo- [Zn(OH)4]2- tetrahydroxozincate(lI) ion

Chloride ion Chloro- [CuCl4]2- tetrachlorocuprate(lI) ion

Cyanide ion Cyano- [Fe(CN)6]3- hexacyanoferrate(III) ion

Nitrite ion Nitro- [CO(NO2)6]3- hexanitrocobaltate(III) ion

Carbon monoxide

Carbonyl- Ni(CO)4 tetracarbonylnickel( 0)

Ethane-l,2-diamine

Ethane-I,2-diamine-

[Cr(en)3]3+ tris(ethane-l,2-diamine)chromium(III) ion

edta edta- [Zn(edta)]2- edtazincate(lI) ion

Cl-, NH3 Mixed [CoCl2(NH3)4]+ tetraamminedichlorocobalt(III) ion-OH, H2O Mixed [Fe(OH)2(H2O)4]+ tetraaquadihydroxoiron(III) ion

Alfred Stock nomenclature

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Nomenclature of Transition Metal

Complexes with Monodentate

LigandsIUPAC conventions

1. (a) For any ionic compound the cation is named before the anion

(b) If the complex is neutral the name of the complex is the name of the compound

1. (c) In naming a complex (which may be neutral, a cation or an anion)

the ligands are named before the central metal atom or ion

the liqands are named in alphabetical order (prefixes not

counted)

(d) The number of each type of ligands are specified by the Greek prefixes1 mono- 2 di 3 tri

4 tetra- 5 penta- 6 hexa-

1. (e) The oxidation number of the metal ion in the complex is indicated immediately after the name of the metal using Roman numerals

[CrCl2(H2O)4]Cltetraaquadichlorochromium(III) chloride

[CoCl3(NH3)3]triamminetrichlorocobalt(III)

K3[Fe(CN)6]potassium hexacyanoferrate(III)

2. (a) The root names of anionic ligands

always end in “-o”

– CN cyano

Cl–

chloro

Br

bromo

I

iodo

– OH hydroxo

NO2 nitro

SO42

sulphato

H

hydrido(b) The names of neutral ligands are the names of the molecules

except NH3, H2O, CO and NO

3. (a) If the complex is anionic

the suffix “-ate” is added to the end of the name of the metal,

followed by the oxidation number of that metal

tetrachlorocuprate(II) ion[CuCl4]2–

hexacyanoferrate(III) ion[Fe(CN)6]3

tetrachlorocobaltate(II) ion[CoCl4]2

Name of the complexFormula

(a) Write the names of the following compounds.

(i) [Fe(H2O)6]Cl2

(ii) [Cu(NH3)4]Cl2

(iii) [PtCl4(NH3)2]

(iv) K2[CoCl4]

(v) [Cr(NH3)4SO4]NO3

(vi) [Co(H2O)2(NH3)3Cl]Cl

(vii) K3[AlF6]

Hexaaquairon(II) chloride

Tetraamminecopper(II) chloride

Diamminetetrachloroplatinum(IV)

Potassium tetrachlorocobaltate(II)

Tetraamminesulphatochromium(III) nitrate

(i) [Fe(H2O)6]Cl2

(ii) [Cu(NH3)4]Cl2

(iii) [PtCl4(NH3)2]

(iv) K2[CoCl4]

(v) [Cr(NH3)4SO4]NO3

(a) (vi) [Co(H2O)2(NH3)3Cl]Cl

triamminediaquachlorocobalt(II) chloride

(vii) K3[AlF6]

potassium hexafluoroaluminate

Al has a fixed oxidation state (+3)

no need to indicate the oxidation state

(b) Write the formulae of the following compounds.

(i) pentaamminechlorocobalt(III) chloride

(ii) Ammonium hexachlorotitanate(IV)

(iii) Tetraaquadihydroxoiron(II)

[Co(NH3)5Cl]Cl2

(NH4)2[TiCl6]

[Fe(H2O)4(OH)2]

The colours of many gemstones are due to the presence of small quantities of d-block metal ions

Coloured Ions

Reactions of Copper (II)Aqueous solutions of copper(II) contain the blue, octahedral hexaaquacopper(II) ionMost substitution reactions are similar to cobalt(II).



NH3 [Cu(H2O)6]2+(aq) + 2NH3(aq) [Cu(OH)2(H2O)4](s) + 2NH4+(aq)

blue ppt. soluble in excess NH3

then [Cu(OH)2(H2O)4](s) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + 2¯OH (aq)

royal blue

NOTE THE FORMULA

A-B

A-B

LS

Hemantha Welihena

Aqueous solutions of copper(II) contain the blue, octahedral hexaaquacopper(II) ionMost substitution reactions are similar to cobalt(II).



NH3 [Cu(H2O)6]2+(aq) + 2NH3(aq) [Cu(OH)2(H2O)4](s) + 2NH4+(aq)

blue ppt. soluble in excess NH3

then [Cu(OH)2(H2O)4](s) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 2H2O(l) + ¯2OH (aq)

deep blue

NOTE THE FORMULA

CO32- [Cu(H2O)6]2+(aq) + CO3

2-(aq) CuCO3(s) + 6H2O(l)

blue ppt.

A-B

A-B

LS

Ppt

acidbase

acid base

baseacid

Reactions of Copper (II)

Hemantha Welihena

Cl¯ [Cu(H2O)6]2+(aq) + 4Cl¯(aq) [CuCl4]2-(aq) + 6H2O(l)

yellow, tetrahedral

• Cl¯ ligands are larger than H2O and are charged

• the complex is more stable if the shape changes to tetrahedral

• adding excess water reverses the reaction

2Cu2+(aq) + 4I¯(aq) 2CuI(s) + I2(aq)

off - white ppt.

• a redox reaction• used in the volumetric analysis of copper using sodium thiosulphate

LS

REDOX


Hemantha Welihena

The aqueous chemistry of copper(I) is unstable compared to copper(0) and copper (II).

Cu+(aq) + e¯ Cu(s) E° = + 0.52 V

Cu2+(aq) + e¯ Cu+(aq) E° = + 0.15 V

subtracting 2Cu+(aq) Cu(s) + Cu2+(aq) E° = + 0.37 V

This is an example of DISPROPORTIONATION where one species is simultaneously oxidised and reduced to more stable forms. This explains why the aqueous chemistry of copper(I) is very limited.

Copper(I) can be stabilised by formation of complexes.


Disproportionation Reactions

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Formula of copper(II) complex Colour of the complex

[Cu(H2O)4]2+

[CuCl4]2–

[Cu(NH3)4]2+

[Cu(H2NCH2CH2NH2)]2+

[Cu(EDTA)]2–

Pale blue

Yellow

Deep blue Violet

Sky blue

Colours of some copper(II) complexes

This question concerns the chemistry of copper. In the sequence below, A, B, C, D, E and F all contain copper in various oxidation states.

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Cu(OH)2(H2O)4 CuO

[Cu(NH3)4(H2O)2]

Cu [Cu(H2O)6] Cu(NH3)2+2+

(a) Identify, by name (including the oxidation state where appropriate) or formula,the copper-containing species in the sequence.A ............................................ B CuO ............................................C .............................................D .............................................E .............................................F .............................................

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Cu(OH)2(H2O)4

[Cu(NH3)4(H2O)2]Cu

[Cu(H2O)6]2+

[Cu(NH3)]2+

(b) Identify, by name or formula, the reagent that would be used to convert B into CuSO4(aq)........................................................................................(c) (i) C and F are the same type of chemical species. Name this type........................................................................................(ii) Explain why C is coloured but F is colourless.Copper ion in C has partially filled d orbitalCopper ion in F has (completely) filled d orbitalsElectronic transitions between partially filled (d) orbitals (of

different energy) are possible*(iii) Explain why F changes into C on shaking.Copper(I) is oxidized (to copper(II))

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H2SO4(aq)

complex(es)

(d) The reaction of copper(I) iodide to form D and E is a disproportionation.(i) Explain the term disproportionation.(ii) .............................................................................................

.............................................................................................(ii) Write an ionic equation for this reaction. State symbols are not required.(iii) Use the relevant standard reduction (electrode) potentials, from the table on page 17 of your data booklet, to calculate the E cell value for this reaction,giving your answer with the appropriate sign.*(iv) If copper(I) iodide is treated with nitric acid, rather than sulfuric acid, a blue solution is still formed but no pink solid. Use the standard electrode potentials on page 15 of your data booklet to explain this. Quote any data that you use...........................................................................................................................................................................................................

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Simultaneous increase or decrease in oxidation number

2Cu Cu + Cu+ 2+

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(a) A = copper(II) hydroxide / Cu(OH)2

[Cu(OH)2(H2O)4]

B = copper(II) oxide / CuO

C = tetraamminecopper(II) / [Cu(NH3)]4 /[Cu(NH3)4(H2O)2]

[Cu(NH3)6] / hexaamminecopper(II)

D = copper / Cu / copper(0) / Cu(0)

E = copper(II) sulfate / CuSO4 / Cu / [Cu(H2O)6]

F = diamminecopper(I) / [Cu(NH3)](b) (Dilute) sulfuric acid / H2SO4 / H2SO4(aq)concentrated H2SO4(c)(i) (transition metal / d-block element) complex(es) /complex ion(s)

2+

2+2+

2+

2+

2+

(c)(ii) Copper ion in C has partially filled d orbital(s) /subshell / 3d9 unpaired d electron d shell Copper ion in F has (completely) filled d orbitals / subshell / 3d2

Electronic transitions between partially filled (d) orbitals (of different energy) are possibleORElectronic transitions between (completely) filled (d) orbitals (of different energy) are not possible(c)(iii) Copper(I) is oxidized (to copper(II)) F / it is oxidized by oxygen / air Second mark(d)(i) (simultaneous) oxidation and reduction OR Simultaneous increase or decrease in oxidation numberof an element(d)(ii) 2Cu+ Cu + Cu2+

OR 2CuI + 2H+ Cu + Cu2+ + 2HIOR 2CuI Cu + Cu2+ + 2I−

(d)(iii) The use of cell notation (as in the Data Booklet SEP

table) in place of equations

e.g. Cu+(aq) | Cu(s) E = +0.52 (V) (from the data book the

equations are) Cu+(aq) + e− Cu(s) E = +0.52 (V)

Cu2+(aq) + e− Cu+(aq) E = +0.15 (V) So E cell = 0.52 − 0.15 =

+0.37 (V)

(d)(iv) In both schemes the use of cell notation (as in the

Data Booklet SEP table e.g. Cu2+(aq) | Cu(s) E = +0.34 (V)

Relevant half equations are Cu+2(aq)+2e−Cu(s) E = +0.34 (V)

2NO3−(aq) + 4H+(aq)+2e−N2O4(g)+2H2O(l) E = +0.80 (V)

OR NO3−(aq)+3H+(aq)+2eHNO2(aq)+H2O(l) E = +0.94 (V)

Correct overall equation scores both marks:

Ɵ

Ɵ

Ɵ

Ɵ

Ɵ

Ɵ

Ɵ

Ɵ

Cu + 2 NO3−+ 4H+ Cu2+ + N2O4 + 2H2O

OR Cu + NO3−+ 3H+ Cu2+ + HNO2 + H2O

So Eocell is +0.46 (V) (or +0.60 (V) or just ‘positive’)

Scheme 2 (oxidation of copper(I)

Cu2+(aq) + e−Cu+(aq) E = +0.15 (V)

2NO3−(aq) +4H+(aq) + 2e− N2O4(g) + 2H2O(l)E = +0.80 (V)

OR NO3−(aq) + 3H+(aq) + 2e−HNO2(aq) + H2O(l) E = +0.94 (V)

Correct overall equation

2Cu+ + 2NO3− +4H+ 2Cu2+ + N2O4 + 2H2O

2Cu+ + NO3−+3H+ 2Cu2+ + HNO2+ H2O

So Ecell is +0.65 (V) (or +0.79 (V)

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Reactions of Chromium(III)Chromium(III) ions are typical of M3+ ions in this block.

Aqueous solutions contain the violet, octahedral hexaaquachromium(III) ion.

As an acid [Cr(H2O)6]3+(aq) + 3¯OH (aq) [Cr(OH)3(H2O)3](s) + 3H2O(l)

violet, octahedral green ppt. soluble in XS NaOH

As with all hydroxides the precipitate reacts with acid

As a base [Cr(OH)3(H2O)3](s) + 3H+(aq) [Cr(H2O)6]3+(aq)

being a 3+ hydroxide it is AMPHOTERIC as it dissolves in excess alkali

[Cr(OH)3(H2O)3](s) + 3¯OH (aq) [Cr(OH)6]3-(aq) + 3H2O(l)

green, octahedral

A-B

A-B

A-B

Am

ph

oteric n

ature

With EXCESS SODIUM HYDROXIDE, the precipitate redissolves

[Cr(H2O) 3(OH) 3] + 3¯OH [Cr(OH)6]3- + 3H2O

Am

ph

oteric n

ature

green ppt. soluble in XS NaOH

green, octahedral

Hemantha Welihena

[Cr(H2O) 3(OH) 3] + 3H [Cr(H2O)6]+3 + 3H2O+

green ppt. soluble in XS HCl

CO32- 2 [Cr(H2O)6]3+(aq) + 3CO3

2-(aq) 2[Cr(OH)3(H2O)3](s) + 3H2O(l) + 3CO2(g)

The carbonate is not precipitated but the hydroxide is.

high charge density of M3+ makes the solution too acidic to form the carbonate

CARBON DIOXIDE IS EVOLVED.

NH3 [Cr(H2O)6]3+(aq) + 3NH3(aq) [Cr(OH)3(H2O)3](s) + 3NH4+(aq)

green ppt. soluble in XS NH3

With EXCESS AMMONIA, the precipitate redissolves

[Cr(OH)3(H2O)3](s) + 6NH3(aq) [Cr(NH3)6]3+(aq) + 3H2O(l) + 3¯OH(aq)

Reactions of Chromium(III)

Hemantha Welihena

Oxidation In the presence of alkali, Cr(III) is unstable and can be oxidised to Cr(VI)

2Cr3+(aq) + 3H2O2(l) + 10¯OH(aq) 2CrO42-(aq) + 8H2O(l)

green yellow

Acidification of the yellow chromate will produce the orange dichromate(VI) ion

Reduction Chromium(III) can be reduced to the less stable chromium(II) by zinc in acid

2 [Cr(H2O)6]3+(aq) + Zn(s) 2 [Cr(H2O)6]2+(aq) + Zn2+(aq)

green blue

OX

RED

Reactions of Chromium(III)

Hemantha Welihena

Occurrence dichromate (VI) Cr2O72- orange

chromate (VI) CrO42- yellow

Interconversion dichromate is stable in acid solution

chromate is stable in alkaline solution

in alkali Cr2O72-(aq) + 2¯OH(aq) 2CrO4

2-(aq) + H2O(l)

in acid 2CrO42-(aq) + 2H+(aq) Cr2O7

2-(aq) + H2O(l)

Reactions of Chromium(VI)

Hemantha Welihena

Being in the highest oxidation state (+6), chromium(VI) will be an oxidising agent.

In acidic solution, dichromate is widely used in both organic (oxidation of alcohols) and inorganic chemistry.

It can also be used as a volumetric reagent but with special indicators as its colour change (orange to green) makes the end point hard to observe.

Cr2O72-(aq) + 14H+(aq) + 6e¯ 2Cr3+(aq) + 7H2O(l) [ E° = +1.33 V ]

orange green

• Its E° value is lower than that of Cl2 (1.36V) so can be used in the presence of Cl¯ ions

• MnO4¯ (E° = 1.52V) oxidises chloride in HCl so must be acidified with sulphuric acid

• chromium(VI) can be reduced back to chromium(III) using zinc in acid solution

Oxidation Reactions of Chromium(VI)

Hemantha Welihena

This is then oxidised by warming it with hydrogen

peroxide solution. You eventually get a bright yellow

solution containing chromate(VI) ions

wow

Hemantha Welihena

Oxidation Reactions of Chromium(VI)

Stabilizing an unusual oxidation number:chromium(II) ethanoate, Cr2(CH3CO2)4(H2O)2

Chromium(II) acetate hydrate, also known as chromous acetate, is the coordination compound with the formula Cr2(CH3CO2)4(H2O)2. This formula is commonly abbreviated Cr2(OAc)4(H2O)2. This red-coloured compound features a quadruple bond .The preparation of chromous acetate once was a standard test of the synthetic skills of students due to its sensitivity to air and the dramatic colour changes that accompany its oxidation. It exists as the dihydrate and the anhydrous forms.

Background

Chromium(II) ethanoate is interesting because it

gives an example of the way in which the

formation of a complex can sometimes stabilise an

oxidation number which would otherwise be

unstable. Chromium(II) ions are normally very

readily oxidised to chromium(III) by the oxygen

of the air. Solutions containing Cr (aq) are green

and those containing Cr (aq) are blue, whereas

chromium (II) ethanoate is red

3+

2+

Chromium(II) acetate (aqueous solution)

Chromium(II) acetate hydrate, also known as chromous acetate, is the coordination compound with the formula Cr2(CH3CO2)4(H2O)2. This formula is commonly abbreviated Cr2(OAc)4(H2O)2. This red-coloured compound features a quadruple bond .The preparation of chromous acetate once was a standard test of the synthetic skills of students due to its sensitivity to air and the dramatic colour changes that accompany its oxidation. It exists as the dihydrate and the anhydrous forms.

The apparatus shown in the procedure section can be used, or one which will perform similarly.Zinc in the presence of acid reduces chromium(VI) to chromium(II). Excess zinc also reacts with hydrogen to producehydrogen gas. The build-up of pressure in the apparatus forces the solution into the complex reagent.

Method1 Dissolve 1 g of sodium dichromate(VI) in 5 cm of water, and put it in the 50 cm round bottomed flask.2 Add 3 g of zinc in equal proportions by mass of powder and granulated.3 Put a mixture of 20 cm concentrated hydrochloric acid and 10 cm water in the tap funnel. Put 10 cm of saturated sodium ethanoate solution in the boiling tube. The solubility of sodium ethanoate in water is about 30% by mass.4 Set up the apparatus as shown in the diagram.5 Add the hydrochloric acid to the mixture in the flask and leave the tap funnel partially open to allow the hydrogen which is generated to escape. The reduction of the chromium soon reaches a green stage and gradually becomes increasingly blue over 10–20 minutes.

33

3

3

3

6 When the solution is distinctly blue, and while hydrogen is still being generated, close the tap onthe funnel. The pressure of the hydrogen will force the blue solution over into the saturated sodium ethanoate. A red precipitate of chromium(II) ethanoate should be formed, and the solution will contain dissolved red chromium(II) ethanoate.7 Dismantle the apparatus and pour the remaining blue solution containing Cr (aq) into another boiling tube. Keep the two boiling tubes unstoppered, side by side in a rack.

2+

Throughout this experiment, you must wear eye protection.

The chemicals and procedures used in this experiment are extremely hazardous, so you must take even more care than usual to reduce risks from them by using suitable control measures.

Solid sodium dichromate(VI) is very toxic and oxidising and is classed as a category 2 carcinogen. It is an irritant to all tissues and you must wear gloves when handling this solid. Avoid inhaling any tiny crystals. Zinc powder is highly flammable.

Concentrated hydrochloric acid is corrosive. The hydrogen evolved is extremely flammable.

As hydrogen is evolved, naked flames must be kept clear.

Safety

PreparationAn aqueous solution of a Cr(III) compound is first reduced to the chromous state using zinc as a reductant.The resulting blue chromous solution is treated with sodium acetate. Immediately chromous acetate precipitates as a bright red powder.

Cr6+ + 2Zn → Cr2+ + 2Zn2+ 2 Cr2+ + 4 OAc- + 2 H2O → Cr2(OAc)4(H2O)2

The synthesis of Cr2(OAc)4(H2O)2 has been

traditionally used to test the synthetic skills and patience of inorganic laboratory students in universities because the accidental introduction of a small amount of air into the apparatus is readily indicated by the discoloration of the otherwise bright red product.An alternative route to related chromium(II) carboxylates starts with chromocene :

4 HO2CR + 2 Cr(C5H5)2 → Cr2(O2CR)4 + 4 C5H6

The advantage to this method is that it provides anhydrous derivatives.

Because it is so easily prepared, Cr2(OAc)4(H2O)2 is often used as a starting

material for other, chromium(II) compounds. Also many analogues have been prepared using other carboxylic acids in place of acetate and using different bases in place of the water.

SafetyThroughout this experiment, you must wear eye protection. The chemicals and procedures used in this experiment are extremely hazardous, so you must take even more care than usual to reduce risks from them by using suitable control measures. Solid sodium dichromate(VI) is very toxic and oxidising and is classed as a category 2 carcinogen. It is an irritant to all tissues and you must wear gloves when handling this solid. Avoid inhaling any tiny crystals. Zinc powder is highly flammable. Concentrated hydrochloric acid is corrosive. The hydrogen evolved is extremely flammable. As hydrogen is evolved, naked flames must be kept clear.

Questions

1 Show, using electrode potentials from pages 14–16

in the Edexcel Data Booklet, that zinc should reduce

chromium from 16 in sodium dichromate(VI) to 12 in

chromium(II) ethanoate.

2 What happens to the colour of the Cr (aq) solution

when it is allowed to stand open to the air?

3 Does the colour of the chromium(II) ethanoate

solution also change over the same period of time?

Four reactions involving the transition elements

copper and chromium are given below.

1 Cu2+(aq) + 2-OH(s) Cu(OH)2

2 [Cu(H2O)4(OH)2](s) + 4NH3(aq) [Cu(NH3)4]2+(aq) +

2-OH(aq) + 2H2O(l)

3 [Cr(H2O)3(OH)3](s) +3-OH(aq) [Cr(OH)6](aq) +3H2O(l)

4 [Cr(H2O)3(OH)3](s) + 3H+(aq) [Cr(H2O)6]3+(aq)

(a) Which reaction produces a dark blue solution?

A 1 B 2 C 3 D 4

Dheema

Hemantha Welihena

(b) Which two reactions show the amphoteric behaviour of a metal hydroxide?A 1 and 2

B 2 and 3C 2 and 4

D 3 and 4

(c) Predict, without calculation, which reaction has the most negative value for ΔSsystem.A 1

B 2C 3

D 4

Nashaya

Hemantha Welihena

In the reaction of manganate(VII) ions with

reducing agents in strongly acidic solution,

the half-reaction for the reduction is

A MnO4 + 4H+ + 3e– MnO2 + 2H2O

B MnO4 + 4H+ + 5e– Mn2+ + 2H2O

C MnO4 + 8H+ + 3e– Mn2+ + 4H2O

D MnO4 + 8H+ + 5e– Mn2+ + 4H2O

Zinal

Hemantha Welihena

When a solution containing 0.10 mol of

chromium(III) chloride, CrCl3.6H2O, is treated

with excess silver nitrate solution, 0.20 mol of

silver chloride, AgCl, is immediately precipitated.

The formula of the complex ion in the solution is

A [Cr(OH)6]3-

B [Cr(H2O)6]3+

C [CrCl(H2O)5]2+

D [CrCl2(H2O)4]2+

Jaah

Hemantha Welihena

When concentrated ammonia solution is added to a green solution of chromium(III) sulfate, a green precipitate is formed which slowly dissolves in excess of the concentrated ammonia solution.

The chromium-containing species formed in these reactions are

Green precipitate Resulting solution

A Cr(OH)3 [Cr(OH)6]

B Cr(OH)3 [Cr(NH3)6]

C (NH4)2CrO4 [Cr(OH)6]

D (NH4)2CrO4 [Cr(NH3)6]

Hemantha WelihenaJaah

3-

3-

3+

3+

1 Four complex ions have the following formulae:A Cu(edta)2-

B Zn(H2O)62+

C Ni(NH3)62+

D CrCl42-

(a) Which complex ion is most likely to be tetrahedral in shape?

A B C D(b) Which complex ion is most likely not to be coloured?A B C D(c) Each of these complex ions may be formed by ligand exchange from an aqua complex. For which complex ion is the entropy change of this reaction most positive?A B C D

Jaah

Hemantha Welihena

Platinum forms a complex with the formula Pt(NH3)2Cl2 and chromium forms a complex ion with the formula CrCl4 .

(a) Considering the shapes of these complexes, A both complexes are square planar. B both complexes are tetrahedral. C Pt(NH3)2Cl2 is tetrahedral and CrCl4 is square planar. D Pt(NH3)2Cl2 is square planar and CrCl4 is tetrahedral.

(b) Considering the structures of these complexes, A both complexes form stereoisomers. B neither complex forms a stereoisomer. C Pt(NH3)2Cl2 forms a stereoisomer but CrCl4 does not. D CrCl4 forms a stereoisomer but Pt(NH3)2Cl2 does not.

Hemantha Welihena

Asleefa

Haneefa

(c) Considering the bonding between the central atom and the ligands in these complexes, A the bonding in both complexes is dative covalent. B the bonding in both complexes is ionic. C the bonding in Pt(NH3)2Cl2 is dative covalent and in CrCl4 is ionic. D the bonding in Pt(NH3)2Cl2 is ionic and in CrCl4 is dative covalent.

Hemantha Welihena

Zinal

Cr(H2O)63+ + 4Cl CrCl4 + 6H2O

Manganese

show oxidation states of +2, +3, +4 ,+5, +6 and +7 in its compounds

Variable Oxidation States of

Manganese and their

Interconversions

The most common oxidation states

+2, +4 and +7

IonOxidation state of

manganese in the ionColour

Mn2+

Mn(OH)3

Mn3+

MnO2

MnO43

MnO42–

MnO4–

+2

+3

+3

+4

+5

+6

+7

Very pale pink

Dark brown

Red

Black

Bright blue

Green

Purple

Colours of compounds or ions of manganese in different oxidation states

(a)

Colours of compounds or ions of manganese in differernt oxidation states: (a) +2; (b) +3; (c) +4

(b) (c)

Mn2+(aq) Mn(OH)3(aq) MnO2(s)

(e)(d)

Colours of compounds or ions of manganese in differernt oxidation states: (d) +6; (e) +7

MnO42–(aq) MnO4

–(aq)

Reactions of Manganese (VII) in its highest oxidation state therefore Mn(VII) will be an

oxidising agent

occurs in the purple, tetraoxomanganate(VII)

(permanganate) ion (MnO4¯)

acts as an oxidising agent in acidic or alkaline solutionacidic MnO4¯(aq) + 8H+(aq) + 5e¯ Mn2+(aq) + 4H2O(l) E° = + 1.52 V

N.B. Acidify with dilute H2SO4 NOT dilute HCl

alkaline MnO4¯(aq) +2H2O(l) +3e¯ MnO2(s) +4¯OH(aq) E° = + 0.59 V

Hemantha Welihena

Some unusual oxidation states

You will probably have seen manganese in its common oxidation states during your study of chemistry. Manganese(II) sulfate occurs as very pale pink crystals in the hydrated form. Manganese(IV) oxide is a black powder which is often used as a catalyst. Potassium manganate(VII) occurs as very dark purple crystals and forms a purple aqueous solution, which is a powerful oxidizing agent.

You are less likely to have seen compounds containing the other oxidation states of manganese, which are manganese(VI), manganese(V), manganese(III) and manganese(I).

However, compounds containing each of these four oxidation states can be prepared.

MnSO4, MnO2 , KMnO4 , Mn , Mn , Mn , Mn6+ 5+ 3+ +

Hemantha Welihena

Manganese(VI)

Manganese(VI), in MnO4, can be prepared in a reverse

disproportionation reaction, by reacting manganate(VII) ions

with manganese(IV) oxide in alkali.Equation 12MnO4(aq) +MnO2(s) +4OH 3MnO4(aq)+2H2O(l) Ecell = –0.03V

The reaction is not thermodynamically favourable under standard conditions. However, the Ecell value can be made positive by increasing the concentration of hydroxide ions so that green manganate(VI) ions form.Manganese(V)Manganese(V) can be formed by adding a little potassium manganate(VII) to very concentrated (12 mol dm3-) aqueous sodium hydroxide. The solution slowly becomes blue as manganate(V) ions, MnO3(aq), form. The ionic half-equations are:

- - 2-

2-

-

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Hemantha Welihena

Equation 2

MnO4(aq) + H2O(l) +2e MnO3aq) + 2OH(aq)Equation 3 4OH(aq) 2H2O(l) + O2(g) + 4eManganese(III)

A deep red solution containing manganese(III) ions is formed by the oxidation of manganese(II) hydroxide by potassium manganate(VII) in acid solution. The ionic equation for the reaction is:Equation 4

MnO4(aq) + 4Mn(OH)2(s) + 16H+(aq) 5Mn3+(aq)+ 12H2O(l)

Manganese(I)

Manganese(I) ions are not stable in aqueous solution, but do form stable complex ions. They can be made by reducing hexacyanomanganate(II) ions, Mn(CN)64-, tohexacyanomanganate(I) ions, Mn(CN)65- .

-

-

-

-

-

Hemantha Welihena

(a) (i) Give the formula of manganese(IV) oxide.

(ii) How do catalysts speed up reactions?

(iii) Explain how transition metal ions can act as homogeneous catalysts.

(b) (i) Suggest why the preparation of manganate(VI) ions, MnO4, in equation 1,may be described as a reverse disproportionation reaction by considering the relevant oxidation states.

(ii) The two half-equations which are combined to form equation 1 are

MnO4(aq) + e MnO42-(aq) E = +0.56 V

MnO42-(aq) +2H2O(l) +2e MnO2(s)+4OH(aq) E =+0.59 VExplain, by reference to these half-equations, why increasing

the concentration of hydroxide ions alters the electrode potential to make the preparation of manganate(VI) ions more likely.

-

- -

- Ɵ

Ɵ

2-

Hemantha Welihena

(c) Use equations 2 and 3 to answer the following questions.(i) Identify the gas formed in the preparation of manganate(V) ions.(ii) By appropriately combining these two equations, write the ionic equation for the formation of manganate(V) ions from manganate(VII) ions.(iii) Identify the main hazard and state how you would minimize the associated risk in this preparation of manganate(V) ions.(d) Identify the reagents you would use to make manganese(II) hydroxide for the preparation of manganese(III) ions.(e) (i) Draw a dot and cross diagram to show the electron arrangement in the cyanide ion, CN.(ii) Explain how the cyanide ion acts as a ligand.(iii) Suggest the name of the shape of the hexacyanomanganate(I) ion.

-

Hemantha Welihena

15(a)(i) MnO2

15(a)(ii) • They provide alternative routes/mechanisms forreactions• With lower activation energies/Ea OR catalysts loweractivation energy /Ea• So a greater proportion of/more particles/reactants havesufficient energy/Ea (to react)/greater frequency of/more successful collisions15(a)(iii) Transition metals form various/variable oxidation states. They are able to donate and receive electrons/they are able to oxidize and reduce/they are able to be oxidized and reduced/ions contain partially filled(sub-)shells of d electrons. Transition metals form various/variable oxidation states.• They are able to donate and receive electrons/they are able

to oxidize and reduce/they are able to be oxidized and reduced/ions contain partially filled (sub-)shells of d electrons

Hemantha Welihena

15(b)(i) Two (less stable) oxidation states/one higher and one lower oxidation state of the same/an element react to form one(more stable) oxidation state

15(b)(ii) When the hydroxide ion concentration is increased) the equilibrium (of the second half equation) moves to the left/back• E becomes less positive/more negative/decreases/reduces• Therefore Ecell becomes positive (so reaction feasible) 15(c)

(i) Oxygen/oxygen gas

15(c)(iii) (Hazard –) the sodium hydroxide/alkali iscorrosive/caustic/burns (skin)/attacks the skinOR attacks the cornea/eye/causes blindness 15(d) Manganese(II)/manganous sulfate(solution) manganese(II) salt – chloride,bromide, iodide, nitrateSodium hydroxide (solution) ,soluble hydroxide

Hemantha Welihena

15(e)(i)

15(e)(ii) The non-bonding / lone pair of electrons on the carbon forms a dative covalent/coordinate bond (to central metal ion)

15(e)(iii) Octahedral/octahedron

Mn+

- CN

- CN- CN

- CNNC -

NC -

C N

Hemantha Welihena

The compounds of vanadium, vanadium

oxidation states of +2, +3, +4 and +5

forms ions of different oxidation states

show distinctive colours in aqueous solutions

Variable Oxidation States of Vanadium and

their Interconversions

Hemantha Welihena

IonOxidation state of vanadium in the

ion

Colour in aqueous solution

V2+(aq)

V3+(aq)

VO2+(aq)

VO2+(aq)

+2

+3

+4

+5

Violet

Green

Blue

Yellow

Colours of aqueous ions of vanadium of different oxidation states

Hemantha Welihena

V

H2O

H2OH2O

H2O

OH2

OH2

2+

V

H2O

H2OH2O

H2O

OH2

OH2

3+

In an acidic medium

the vanadium(V) state usually occurs in

the form of VO2+(aq) dioxovanadium(V) ion

the vanadium(IV) state occurs in the form

of VO2+(aq) oxovanadium(IV) ion



Hemantha Welihena

In an alkaline medium

the stable form of the vanadium(V) state is

VO3–(aq), metavanadate(V) or

VO43–(aq), orthovanadate(V),

in strongly alkaline medium



Hemantha Welihena

Compounds with vanadium in its highest oxidation state (i.e. +5)

strong oxidizing agents



Hemantha Welihena

Vanadium of its lowest oxidation state(i.e. +2)

in the form of V2+(aq)

strong reducing agent

easily oxidized when exposed to air



Hemantha Welihena

The most convenient starting material

ammonium metavanadate(V) (NH4VO3)

a white solid

the oxidation state of vanadium is +5

Interconversions of the common oxidation states of vanadium can be carried out readily in the laboratory



Hemantha Welihena

Interconversions of Vanadium(V) species

VO2+(aq) V2O5(s) VO3

(aq) VO43(aq)

OH

H+

OH

H+

OH

H+

Yellow orange yellow colourless

Vanadium(V) can exist as cation as well as anion



Hemantha Welihena



(aq) VO43(aq)

OH

H+

OH

H+

OH

H+


In acidic medium In alkaline mediumAmphoteric





(aq) VO43(aq)

OH

H+

OH

H+

OH

H+



Give the equation for the conversion : V2O5 VO2+

V2O5(s) + 2H+(aq) 2VO2+(aq) + H2O(l)





(aq) VO43(aq)

OH

H+

OH

H+

OH

H+



Give the equation for the conversion : V2O5 VO3

V2O5(s) + 2OH(aq) 2VO3(aq) + H2O(l)





(aq) VO43(aq)

OH

H+

OH

H+

OH

H+


In acidic medium In alkaline medium

Give the equation for the conversion : VO3 VO2

+

VO3(aq) + 2H+(aq) VO2

+(aq) + H2O(l)

Amphoteric



V5+

H

O

H

H

O

H

H

O

H

H

O

H

VO43(aq) + 8H3O+

8H2O

O

H

H

V5+ ions does not exist in water since it undergoes vigorous hydrolysis to give VO4

3

The reaction is favoured in highly alkaline solution

orthovanadate(V) ion

V VO43(aq) orthovanadate(V) ion

Cr CrO42(aq) chromate(VI) ion

Mn MnO4(aq) manganate(VII) ion

Draw the structures of VO43, CrO4

2 and MnO4

O

Cr

OO-

O-

O

Mn

OO

O-

O

V-O

O-O-

V5+

H

O

H

H

O

H

H

O

H

H

O

H

VO3(aq) + 6H3O+

6H2O

O

H

H

The reaction is favoured in alkaline solution

VO3 is a polymeric anion like SiO3

2

Metavanadate(V) ion

Metavanadate(V) ion, (VO3)nn

V5+

H

O

H

H

O

H

H

O

H

H

O

H

VO2+(aq) + 4H3O+

4H2O

O

H

H

The reaction is favoured in acidic solution

The action of zinc powder and concentrated hydrochloric acid

vanadium(V) ions can be reduced sequentially to vanadium(II) ions



VO2+(aq)

yellow

Zn

conc. HCl

VO2+(aq) blue

Zn

conc. HCl

V3+(aq) green

Zn

conc. HCl

V2+(aq)

violet



(a)

Colours of aqueous solutions of compounds containing vanadium in four different oxidation states:(a) +5; (b) +4; (c) +3; (d) +2

(b) (c) (d)

VO2+(aq) VO2+(aq) V3+(aq) V2+(aq)

The feasibility of the changes in oxidation state of vanadium

can be predicted using standard electrode potentials

Half reaction (V)

Zn2+(aq) + 2e– Zn(s)

VO2+(aq) + 2H+(aq) + e– VO2+(aq) + H2O(l)

VO2+(aq) + 2H+(aq) + e– V3+(aq) + H2O(l)

V3+(aq) + e– V2+(aq)

–0.76

+1.00

+0.34

–0.26

Under standard conditions

zinc can reduce

1. VO2+(aq) to VO2+(aq)> 0

> 0

> 02. VO2+(aq) to V3+(aq)

3. V3+(aq) to V2+(aq)



2 × (VO2+(aq) + 2H+(aq) + e–

VO2+(aq) + H2O(l)) = +1.00 V–) Zn2+(aq) + 2e– Zn(s) = –0.76 V

2VO2+(aq) + Zn(s) + 4H+(aq)

2VO2+(aq) + Zn2+(aq) + 2H2O(l)

= +1.76 V



2 × (VO2+(aq) + 2H+(aq) + e–

V3+(aq) + H2O(l)) = +0.34 V

–) Zn2+(aq) + 2e– Zn(s) = –0.76 V

2VO2+(aq) + Zn(s) + 4H+(aq)2V3+(aq) + Zn2+(aq) +

2H2O(l)

= +1.10 V



2 × (V3+(aq) + e– V2+(aq)) =

–0.26 V–) Zn2+(aq) + 2e– Zn(s) = –0.76 V

2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+

(aq)

= +0.50 V



REACTIONS OF VANADIUM

Reduction using zinc in acidic solution shows the various oxidation states of vanadium.

Vanadium(V) VO2+(aq) + 2H+(aq)+ e¯ VO2+(aq)+ H2O(l)

yellow blue

Vanadium(IV) VO2+(aq) + 2H+(aq) +e¯ V3+(aq) + H2O(l)

blue blue/green

Vanadium(III) V3+(aq) + e¯ V2+(aq)

blue/green lavender

Redox chemistry of vanadium V , Z = 23 , 1s22s22p63s23p63d34s2

NH4VO3(s) + NaOH(aq) NH3(g) + H2O + NaVO3

white

VO\ 3(aq) + 2H+(aq) VO+2(aq) + H2O(l) white yellow

VO2+ VO2

+ & VO2+ VO2+ V3+ V2+

+5 +5 and +4 +4 +3 +2

yellow green= yellow +blue

blue green lavender =violet

VO2+ VO2

+ & VO2+ VO2+ V3+ V2+

dioxovanadium(V)ion Oxovanadium(IV)ion Vanadium(III)ion Vanadium(III)ion

Zn Zn2+ + e-

+5 +5 and +4 +4 +3 +2

yellow green= yellow

+blue


VO2+ VO2

+ & VO2+ VO2+ V3+ V2+

dioxovanadium(V)ion

Oxovanadium(IV)ion

Vanadium(III)ion

Vanadium(III)ion

COLOURFUL COMPOUNDSRedox chemistry of vanadium V , Z = 23 , 1s22s22p63s23p63d34s2

Dioxovanadium(V) ions reduce by zinc in acidic solution

with zinc in HCl:

2VO2+(aq) + 4H+(aq) + 2e – 2VO2+ (aq) + 2H2O (l)

Zn(s) Zn2+ (aq) + 2e –

2VO2+(aq) + 4H+(aq) + Zn(s) 2VO2+(aq) +Zn2+(aq)+2H2O (l)

dioxovanadium(V)ion

oxovanadium(IV)ion

+4 +3 with zinc in HCl

2VO2+ + 2e – + 4H+(aq) 2V3+(aq) + 2H2O (l)

Zn(s) Zn2+(aq) + 2e –

2VO2+ + Zn(s) + 4H+(aq) 2V3+(aq)+ Zn2+(aq) +2H2O (l)

[V(H2O)6]3+ :

hexaaquavanadium(iii)ion

Oxovanadium(IV) ions reduce by zinc in acidic solution

+3 +2 with zinc in HCl [V(H2O)6]2+:

hexaaquavanadium(ii)ion2V3+(aq) + 2e – 2V2+ (aq)

Zn(s) Zn2+(aq) + 2e –

2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+ (aq)

Vanadium(III) ions reduce by zinc in acidic solution

Zn Zn2+ + e-

+5 +5 and +4 +4 +3 +2

yellow green= yellow +blue


VO2+ VO2

+ & VO2+ VO2+ V3+ V2+

dioxovanadium(V)ion Oxovanadium(IV)ion Vanadium(III)ion Vanadium(III)ion

VO2+ VO2

+ & VO2+ VO2+ V3+ V2+

VO\ 3(aq) + 2H+(aq) VO+2(aq) + H2O(l) white yellow

NH4VO3(s) + NaOH(aq) NH3(g) + H2O + NaVO3

white

Dioxovanadium(V) ions reduce by zinc in acidic solution

The first stage of the series of reductionsLet's look at the first stage of the reduction - from VO2

+ to VO2+. The redox potential for the vanadium half-reaction is given by:

The corresponding equilibrium for the zinc is:

Eɵcell = Eɵ Eɵ = + 1.00 – (-0.76 ) = + 1.76 V RHS LHS

Reaction is feasible

The other stages of the reactionHere are the E° values for all the steps of the reduction from vanadium(V) to vanadium(II):

…… and here is the zinc value again:

Eɵ / V

Zn2+/ Zn - 0.76

V3+/ V2+ - 0.26

VO2+/ V3+ + 0.34

VO2+ / VO2+ + 1.00

1.101.76

0.50

If zinc is over. What are the additional things can you observe?

A. If excess hydrogen ions present in the mixture?

B. If nitrate ions present in the mixture?

Re-oxidation of the vanadium(II)

The vanadium(II) oxidation state is easily oxidised back to

vanadium(III) - or even higher oxidation by hydrogen ions

Eɵcell = Eɵ – Eɵ = 0.00 - ( - 0.26) = +0.26V RHS LHS


That means that the vanadium(II) ions will be oxidised to

vanadium(III) ions, and the hydrogen ions reduced to hydrogen.

Will the oxidation go any further - for example,

to the vanadium(IV) state?

Have a look at the E° values and decide:

Eɵcell = Eɵ – Eɵ = +0.34V –(+0.00V) = +0.34V RHS LHS


That means that the vanadium(III) ions will be oxidised to vanadium(IV) ions, and the hydrogen ions reduced to hydrogen.

Oxidation by nitric acid

In a similar sort of way, you can work out how far nitric acid will oxidize the vanadium(II).

Here's the first step:

Eɵcell = Eɵ – Eɵ = 0.96V – (- 0.26V) = + 1.22V RHS LHS


Nitric acid will oxidize vanadium(II) to vanadium(III).

The second stage involves these E° values:

Eɵcell = Eɵ – Eɵ = 0.96 – (+ 0.34) = + 0.62V RHS LHS


Nitric acid will certainly oxidize vanadium(III) to

vanadium(IV).

Will it go all the way to vanadium(V)?

It has got a less positive value, and so the reaction happen.

Eɵcell = Eɵ – Eɵ = 1.00V – (+0.96V ) = +0.04 V RHS LHS

Zn+2(aq) + 2e- Zn(s) ; E Ɵ = -0.76V

VO2+(aq) + 2H+(aq) + e- H2O(l) + V3+(aq) EƟ = +0.32V

VO2+(aq) + 2H+(aq) + e- H2O(l) + VO2+(aq)

EƟ = + 1.00V

V3+(aq) + e- V2+(aq) EƟ = -0.26

Sn+2(aq) + 2e- Sn(s) ; E Ɵ = -0.14V

EƟ/ V

0.00

Oxidation number

NO3-(aq) + 4H+(aq) + 3e- 2H2O(l) + NO(aq)

E Ɵ= + 0.96V

Redox Chemistry of Vanadium

Displacement of Ligands and Relative

Stability of Complex Ions

[Fe(H2O)6]2+(aq) + 6–CN (aq)Hexaaquairon(II) ion

[Fe(CN)6]4–(aq) + 6H2O(l)

Hexacyanoferrate(II) ion

Stronger ligand

Weaker ligand

Reversible reaction

Equilibrium position lies to the right

Kst 1024 mol6 dm18

[Ni(H2O)6]2+(aq) + 6NH3(aq)Hexaaquanickel(II) ion

[Ni(NH3)6]2+(aq) + 6H2O(l)

Hexaamminenickel(II) ion

Stronger ligand

Weaker ligand

The greater the equilibrium constant,the stronger is the ligand on the LHS andthe more stable is the complex on the RHS

The equilibrium constant is called the stability constant, Kst

Relative strength of some ligands bonding with copper(II) ions

monodentate

bidentate

multidentate

Test Observation Inference

(a) Observe the appearanceof A.

Pale greensolid. ………………………………………

(b) Measure the pH of adilute aqueous solutionof A using a pH meter.

The pH is 6.0. The type of reaction that has occurred when A dissolved in water is........................................

(c) Add a few drops ofdilute sodium hydroxidesolution to a solutionof A.

A greenprecipitateforms.

The sodium hydroxide is acting as.......................................................The formula of the green precipitate is………………………………………

(d) (d) Leave a sample ofthe green precipitateformed in (c) to stand inair.

The greenprecipitateturns brownon the surface.

The type of reaction that has occurred is………………………….The formula of the brown precipitate is……………………….

(e) (e) Add excess sodiumhydroxide solution toa sample of the greenprecipitate formed in (c).

The greenprecipitatedoes notdissolve.

……………………………………….

(f) Add barium chloridesolution, BaCl2(aq),acidified withhydrochloric acid, to asolution of A.

A whiteprecipitateforms.

The white precipitate is

The table shows a series of tests carried out on a soluble crystalline compound A, which contains one anion and one cation. For each test, complete the table by filling in the inference column.

Identify compound A by name or formula.

Extent of reaction

∆Stotal J mol-1K-1

Ecell / v Kc

Goes to completion

More positive than+200

More positive than =0.6

> 1010

Equilibrium with more products than reactants

approx+40

approx +0.1

approx102

Roughly equal amounts of reactants and products

approx 0

approx 0

approx 1

Equilibrium with more reactants than products

approx - 40

approx-0.1

approx 10-2

No reactionMore negative

than-200

More negative than -0.6

> 10-10

Entropy , electromotive force and the extent of a reaction

Thank you

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