Chemistry

Session

d and f Block Element - 1

Session Objectives

1. General properties of transition elements

2. Atomic radii

3. Ionisation energy

4. Oxidation states

5. Catalyst

6. Colour of complexes

7. Principle of extraction of Fe

8. Principle of extraction of Cu

9. Principle of extraction of Ag

d-block elements

Not all d block elements have incomplete d sub-shells

e.g. Zn has e.c. of [Ar]3d104s2,

Zn2+ ion [Ar] 3d10

Similarly Sc forms Sc3+ which has the stable e.c of Ar.

Characteristic of d-block transition elements: incomplete filled d-subshell in common oxidation states.

Illustrative Example

Why are Zn, Cd and Hg not considered transition metals?

Solution :

Because they do not have vacant d-orbitals neither in the atomic state nor in any stable oxidation state.

Illustrative Example

What are coinage metals?

Solution:

Cu,Ag,Au are called coinage metals.

Atomic Radii

Illustrative Example

In the transition series, with an increase in atomic number the atomic radius does not change very much. Why is it so?

Solution :

Since the electrons are added in inner d-orbitals which have poor screening effect ,hence effective nuclear charge does not increase appreciably.So, atomic radius does not change much.

Electronic ArrangementElectronic Arrangement

Element Z 3d 4s

ScSc 2121 [Ar][Ar]

TiTi 2222 [Ar][Ar]

VV 2323 [Ar][Ar]

CrCr 2424 [Ar][Ar]

MnMn 2525 [Ar][Ar]

FeFe 2626 [Ar][Ar]

CoCo 2727 [Ar][Ar]

NiNi 2828 [Ar][Ar]

CuCu 2929 [Ar][Ar]

ZnZn 3030 [Ar][Ar]

Electronic Configuration

Melting Point

The number of unpaired d-electrons increases up to the middle so metallic strength increases up to the middle. The dip in mp at Mn can be explained on the basis that it has stable half filled conf.so electrons are held tightly so delocalisation is less & metallic bond is weak

Ionisation Energy

IE2 :V < Cr > Mn and Ni < Cu > Zn

IE3 : Fe << Mn

Illustrative Example

The sums of the first and second ionization energies and those of the third and fourth ionization energies of nickel and platinum are given below:

IE1 + IE2 (KJ mol-1) IE3 + IE4 (KJ mol-1)

Ni  2.49 8.80

Pt 2.66 6.70

Taking these values into account write the following:

(i) The most common oxidation state for Ni and Pt and its reasons.

(ii) The name of the metal (Ni or Pt) which can form compounds in +4 oxidation state more easily and why?

Solution

(i) Ni shows +2 oxidation state whereas Pt shows +4 .(IE1+IE2) of Ni is less than that for Pt whereas (IE3 + IE4) is less for Pt.

(ii) From the given data it is clear that Pt (IV) is more easily attained while more energy would be required for obtaining Ni (IV) ion. Hence, Pt (IV) compounds are more stable than Ni (IV) compounds.

Oxidation States

In solution ,the stability of the compounds depends upon electrode potentials rather than ionisation energies.Electrode potential values depend upon factors such as energy of sublimation of the metal,ionisation energy and hydration energy.

Illustrative Example

Out of Cr2+ and Cr3+,which one is stable in aqueous solution?

Solution :

Cr3+ is more stable in aqueous solution due to higher hydration energy which is due to smaller size and higher charge

Illustrative Example

Explain why Eo for Mn+3/Mn2+ couple is more positive than that for Fe3+/Fe2+.

(Atomic numbers of Mn = 25, Fe = 26)

Solution :

Filled shell of Mn in +2 oxidation state (3d5) makes it more stable.

Illustrative Example

How do the oxides of the transition elements in lower oxidation states differ from those in higher oxidation states in the nature of metal-oxygen bonding and why?

Oxides of transition metal in lower oxidation state are ionic and basic in nature whereas in higher oxidation state it forms covalent oxide which are acidic in nature.

Solution

Catalysis

• Catalysis plays an essential aspect in about 90% of all chemical manufacturing.

• Ni and Pt are very heterogeneous catalysts.

• Pt, Rh, and Pd are used in catalytic converters.

• V2O5 is used in conversion of SO2 to SO3.

The catalytic properties of the transition elements are probably due to presence of unpaired electrons in their incomplete d- orbitals.In some cases the transition metals with their variable valency may form unstable intermediate compounds.In other cases the transition metal provides a suitable reaction surface.

Illustrative Example

Why does V2O5 acts as catalyst?

Solution:

V2O5 acts as catalyst because it has large surface area.It can form unstable intermediates which readily change into products.

Magnetism

Transition metal ions and their compounds show magnetic behaviour due to the presence of unpaired electrons in (n – 1) d-orbitals.

n(n 2) B.M.

n is the number of unpaired electrons.

Colour of Complexes

due to the presence of unpaired d- electrons.

Charge transfer spectra

(V) +5 +4 +3 +2

Illustrative Example

Of the ions Co2+, Sc3+, and Cr3+, which one will give coloured aqueous solution and respond to a magnetic field?

Co2+ and Cr3+ are coloured and attracted in magnetic field because they have unpaired electrons whereas Sc3+ does not have any unpaired electron hence it will be repelled by magnetic field.

Solution :

Illustrative Example

Give reasons for the following features of transition metal chemistry.(a) Most of the transition metal ions are coloured in solution(b) Transition metals are well known to form complex compounds (c) The second and third members in each group of the transition elements have very similar atomic radii 

Solution :

(a) This is attributed due to presence of unpaired electrons ,they undergo d-d transitions by absorbing light from visible region and radiate complementary colour.

(b) Small size and high charge of cation and presence of vacant d-orbitals.

(c) Due to lanthanide contraction.

Illustrative Example

Account for the following:

Scandium forms no coloured ions and yet it is regarded as a transition element.

Solution :

Sc has incompletely filled d-orbital hence it is regarded as transition metal.It forms no coloured ion due to absence of unpaired electron in Sc3+

ion.

Alloy formation

d-block elements have similar atomoc sizes due to which atoms of one metal can easily take up positions in the crystal lattice of other.

Interstitial compounds

The transition elements form a large number of interstitial compounds in which small atoms such as hydrogen,carbon,boron and nitrogen occupy the empty spaces in their lattices.

Metallurgy of Iron

Ore: The most commonly used iron ores are haematite, Fe2O3, and magnetite, Fe3O4

Concentration of ore: by magnetic separation.

Roasting: Moisture, sands are removed as oxide.

Smelting: Done in blast furnace. Roasted ore, limestone and coke are added in blast furnace for smelting.

Metallurgy of Iron

2 2

2

C O CO 415 kJ

CO C 2CO 162.3 kJ

3 2

2 3Flux Impurity Slag

CaCO CaO CO

CaO SiO CaSiO

Fe O + C + C aC O2 3 3

W asteG ases

S tee l sh e ll

H ot air b lastthrou gh thyeres

Zone of redu ction

C en tra l Zon e

Zone of fus ion

Zone of com bu stion

575 - 975 K

1175 - 1475 K

1475 - 1575 K

1775 - 1800 K

S lagM o lten iron

2 3 2

3 4 2

2

Fe O +CO 2FeO+CO

Fe O +CO 3FeO+CO

FeO+CO Fe+CO

Different type of iron

Properties Cast iron Steel Wrought iron

% carbon 2.2-4.5 0.15-1.5 0.12-0.25

Melting point

1200 1200-1400 1500

Hardness Hard Hard and soft Soft

Brittleness Brittle Malleable Malleable

Weldability Can’t be welded

Can be welded Can be welded

Steel

Mild steel: 0.1-0.5% CHard steel: 0.6-1.5% C

Name Composition Special properties Uses

Manganese steel

12-13% Mn Extremely hard Grinding machinary

Nickel steel 2-4% Ni Resists corrosion Drive shafts, gears

Invar 36% Ni Low coefficient of expansion

Pendulum rods,watches

Stainless steel

46 %Ni Resists corrosion utensils

Chrome-nickel

0.5%-25 Cr, 1-4% Ni

High tensile strength Axles,ball,bearing,cutting tools

Chrome-vanadium

1-10% Cr, 0.14-0.15%V

High tensile strength Automobile axles

Duriron 15-18% Si High resistance to the attack of acids

Acid pumps, pipe lines etc

Methods of preparation

Three fundamental changes from pig iron

Removal of Si, Mn, P and other minor impurities. through slag formation

Reduction of the C content.3-4% in pig iron, 0-1.5% in steel.

Addition of alloying elementsCr, Ni, Mn, V, Mo, and W.Give the steel its desired properties.

The oxygen top blowing process

The electric arc process

The high frequency induction process

Old processes

Bessemer process

Open hearth process

Modern processes

Reactions Occurring in Steelmaking Processes

Heat treatment of steel

Quenching or hardening

Process of heating steel to red hot followed by sudden cooling by plunging to oil or water.

Makes steel hard and brittle due to formation of iron carbide (Fe3C).

Annealing

Process of heating steel to red hot then cooling slowly.

Annealing makes steel soft and ductile.

Tempering

Process of heating quenched steel to temperature much below redness followed by slow cooling.

Tempering retains the hardness but brittleness disappears.

Nitriding

Process of producing hard coating or iron nitride on surface of steel.

Metallurgy of copperOre: Copper pyrite CuFeS2

Concentration: done by froth floatation.

2 2 2 2

2 2 2 2

2 2

2CuFeS O Cu S 2FeS SO

2Cu S 3O 2Cu O 2SO (Partly)

2FeS 3O 2FeO 2SO (Partly)

2 2Cu O FeS Cu S FeO

Slag is removed from upper layer. Molten mass containing Cu2S and Fe is called copper mate.

Smelting: Smelting at 800oC converts CuS to CuO. Since iron has more affinity for oxygen, therefore FeO is formed first.

FeO(s) + SiO2(s) → FeSiO3(fusible slag)

Roasting:

Metallurgy of copper

Molten matte is mixed with sand (SiO2) and heated in Bessemer converter.

2 2

2 3Slag

2FeS 3O 2FeO 2SO

FeO SiO FeSiO

2 2 2 2

2 2 2

2Cu S 3O 2Cu O 2SO

2Cu O Cu S 6Cu SO

Bessemerisation

Refining of copper

By poling or electrolytic refining.

Blister copper (impure metal).

Metallurgy of Silver

Extraction: By Mac Arthur Forrest Process.

Refining: By electrolytic method.

Ore: Argentite AgS

Concentration: By froth flotation process.

2 2 24Ag 8NaCN O 2H O 4Na Ag CN 3NaOH

2 2 2Ag S 4NaCN 2Na[Ag(CN) ] Na S

The sodium argentocyanide solution is filtered and the silver is precipitated from the solution by the addition of zinc.

22 4Na Ag CN Zn Ag Na Zn CN

Thank you