d-f-block IIT JEE

8/13/2019 d-f-block IIT JEE

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1. INTRODUCTION 2

2. ELECTRONIC CONFIGURATION AND IRREGULARITIES 2

3. GENERAL CHARACTERISTICS 2

4. COLOUR 3

5. COMPLEX FORMATION 3

6. CHROMATE -DICHROMATE 4

7. MANGANATE & PERMANGANATE 5

8. SILVER AND ITS COMPOUND 6

9. ZINC COMPOUNDS 8

10 .COPPER COMPOUNDS 9

11. IRON COMPOUNDS 10

12. INNER TRANSITION ELEMENTS 11

13. LANTHANIDE CONTRACTION 12

1

CONTENTS

S.NO. TOPIC PAGE NO


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d & f - block

1. INTRODUCTIONIn these elements the last electron enters (n1)d orbitals of atom of an element is called d-blockelements.

ELECTRONIC CONFIGURATION AND IRREGULARITIESThe valence shell configurations of these elements can be represented by(n1)d110ns0,1,2 .

All thedblock elements areclassified into fourseriesviz 3d , 4d , 5dand 6d seriescorresponding to thefilling of3d , 4d , 5dand 6dorbitals of (n1)th main shell orbitals of (n1)th main shell. EachSeries has10 elements. Cr(3d5 , 4s 1) , Cu(3d10 , 4s1), Mo(4d5 , 5s1), Pd(4d10, 5s0), Ag(4d10, 5s1)anAu(5d10 , 6s1) clearlyshow the irregularities inthe configurationsThese are explained on the basis of the concept that half filled and completely filled dorbitals arerelativelymore stable thanother dorbitals.

GENERAL CHARACTERISTICS(i) Metallic character: Theyare allmetal and good conductor ofheat & electricity

(ii) Electronic configuration: (n1)d110ns12

2.

3.

Cr Maximum lowest m.p.(iii) M.P. ZnCd

Hg

Mo 6 no. of unpaired e s due to no unpaired e

W are involved in metallic bonding for metallic bonding(iv) Variation inatomicradius:

Sc Mn Fe Co Ni Cu Zn

decreases remains increasessame

(v) Variable oxidation states possibleagain

Sc Ti V Cr Mn Fe Co Ni Cu Zn1 1 2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 4 4 4 4 4 4 4

5 5 5 6 6 6

7Colour : (aquated)Sc3+

Ti4+

Ti3+

V4+

V3+

V2+

Cr2+

Cr3+

colourless colourless

purpleblue green violetblue green

Mn3+ violetMn2+ light pinkFe2+

Fe3+ light green yellow

Co2+pink

Ni2+

greenCu2+blueZn2+ colourless

2

d & f - BLOCK


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d & f - block

4. COLOURSubstancesappearcoloured whenthey absorb light of a particular wavelength inthevisible region of thespectrumand transmit light of other wavelengths.The colourwhich wesee isthecolourof thetransmittedwavelengths. In other words, the colour of the compound observed by us is the complmentary colourof the colour absorbed by the compound.

In thes- and p-block elements there cannot be any d-d transitions and theenergyneeded to promotesor p electronto a higher level is much greaterand maycorrespond to ultraviolet region, in whichcase thecompound willnot appear coloured to the eye.

Relationship between colour and wavelength

Wavelength absorbed in nm

750

Colourabsorbed

UVregionVioletIndigoGreen-blueBlue-greenGreenYellow-greenYellowOrangeRed

Infra-red

Colourobserved

White/colourlessYellow-greenYellowOrangeRedPurpleVioletIndigoGreen-blueBlue-greenWhite/colourless

Magnetic Properties:Whena substance is placed ina magnetic field of strengthH, the intensity of themagnetic field inthe substance maybe greater thanor less thanH.

diamagnetic,Substances whichare weaklyrepelled bya magnetic field

paramagnetic thesubstances whichareweaklyattracted bythemagnetic field and losetheirmagnetism whenremoved fromthe field .

Paramagnetism is expressed bymagnetic moment,

n(n 2) B.M.

n = number of unpairedelectronsB.M. = Bohr Magneton, unit of magnetic moment

5. COMPLEX FORMATIONIn thecaseof transition metals in low oxidation states, the electrons in thed orbitals become involved in

bonding with ligands.The bonding between the ligand and the transition metal ion can either bepredominantly electrostatic or covalent or inmanycases intermediate between thetwo extremes. Some

of the typical complexes of the transition meals are [Fe(CN) ]3,[Ni(NH) ]2 ,[Cu(CN) ]3

,6 3 4 4

[Cu(NH) ]2 etc.3 4

Formation of alloys :

Transition elements have almost similar atomic sizes. Therefore , these elements can mutuallysubstitute their positions in their crystal lattice. The alloys are hard and have high melting points ascompared to the most metal.CatalyticProperties: Many transition metals andtheir compounds have catalytic properties.someexample

TiCl3

V2O5

MnO2

Usedas theZiegler-Natta catalyst in the production of polythene.

Converts SO2 to SO3 inthe contact process for making H2SO4 .

Used as a catalyst to decompose KClO3 to give O2 .

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d & f - block

Fe

H2O2Pd

Pt/Rh

Promoted iron is used in theHaber-Bosch process for making NH3 .

Used as Fentons reagentfor oxidizing alcohols to aldehydes.

Used for hydrogenation (e.g. phenol to cyclohexanone).

Formerly used in the Ostwald process for making HNO3 to oxidize NH3 to NO.

Ni Raney nickel, numerous reduction processes (e.g. manufacture of hexamethylenediamine,production of H2 from NH3 , reducing anthraquinone to anthraquinol in the production of H2O2 ).

Ionisation Energies

Ionisation energies-(i) of 5d elements are higher than those of the 3d and 4d elements. This is due togreatereffective nuclear charge acting on outer valence electrons becauseof theweak shieldingof thenucleus by4felectrons.The ionisation energies of the 3d and4d elements are irregular.

Common oxidation statesfor each element include +2 and +3or both. The+3oxidation states are morestable at the beginning of the series, whereas towards the end +2 oxidation states are more stable.Ionisation energyincreasesgradually from left to right. However the third ionisation energies, when an

electron is removed from 3d orbital, increases more rapidly than the second and third ionisationenergies. Because it takes moreenergy to remove the third electron fromthemetals near theend of therow than from those near the beginning, the metals near the end tends to form M2+ ions rather thanM3+ ions.

6. CHROMATE -DICHROMATEPreparation :

4FeO . Cr2O3+ 8Na2CO3+ 7O2Rosting8Na

2CrO

4+ 2Fe

2O

3+ 8CO

2

in airchromate ore

The roasted mass is extracted withwater when Na2CrO

4goes into the solution leaving behind insoluble

Fe2O

3. The solution is treated with calculated amount of H

2SO

4.

2Na2CrO

4+ H

2SO

4Na

2Cr

2O

7+ Na

2SO

4+ H

2O

Thesolution is concentrated when less soluble Na2SO

4crystallises out.The solution is further concentrated

when crystals of Na2Cr

2O

7areobtained. Then a hot saturated solution of Na

2Cr

2O

7is treated with KCl

when reddish orange crystals of K2Cr

2O

7are obtained on crystallisation.

K2Cr

2O

7is preferred to Na

2Cr

2O

7because Na

2Cr

2O

7is hygroscopic but K

2Cr

2O

7is

not.

* Other props& test of CrO2 & Cr O2 :Alreadydiscussed4 2 7

* Similarities between hexavalent Cr & S-compounds:

(i) SO3& CrO

3both acidic.

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d & f - block

(ii) CrO2 & SO 2 are isomorphous4 4

OH 2 2(iii) SO2Cl2 & CrO2Cl2SO 4 & CrO4 respectively.

OH 2 2(iv) SO3Cl & CrO3Cl SO 4 & CrO4

O

O O|| || ||

(v) CrO3 & (SO3) has same structure Cr O Cr O Cr||O

||O

||O

(vi) Potassium dichromate reacts with hydrochloric acid andevolveschlorine.K

2Cr

2O

7+ 14HCl 2KCl + 2CrCl

3+ 7H

2O + 3Cl

2

(vii) It acts as a powerful oxidisingagent in acidic medium(dilute H2SO

4)

Cr2O 7 + 14H + 6e 2Cr3 + 7H2O. (E = 1.33 V)

The oxidation state ofCr changes from+ 6 to +3.Uses

(i) Asa volumetric reagent in the estimation of reducing agents such as oxalic acid,ferrous ions, iodide ions, etc. It is used asa primary standard.(ii) For thepreparation of several chromiumcompounds such as chrome alum, chromeyellow, chrome red, zinc yellow, etc.(iii) In dyeing, chrome tanning, calico printingphotographyetc.(iv) Chromicacid asa cleansing agent for glass ware.

2 + +

7. MANGANATE & PERMANGANATE

PreparationThis is the most important and well known salt of permanganic acid. It is prepared from the pyrolusiteore. It is prepared by fusing pyrolusite ore either with KOH or K2CO3in presence of atmosphericoxygen or anyotheroxidising agent suchasKNO3. Themass turns green with theformation of potassiummagnate, K2MnO4.

2MnO2+ 4KOH + O

22K

2MnO

4+ 2H

2O

2MnO2+ 2K

2CO

3+ O

2 2K

2MnO

4+ 2CO

2

Thefusedmass is extractedwith water. Thesolution is nowtreated with a current ofchlorine or ozoneor carbondioxide to convert magnate into permanganate.

2K2MnO

4+ Cl

22KMnO

4+ 2KCl

2K2MnO4+ H2O + O3 2KMnO4+ 2KOH + O2

3K2MnO

4+ 2CO

2 2KMnO

4+ MnO

2+ 2K

2CO

3

Props : The above green solution is quite stable inalkali, but inpure water and inpresence of acids,depositing MnO

2andgiving a purple solutionof permanganate.

3K2MnO4+ 2H2O2KMnO4+ MnO2 + 4KOHpurple

= 2.26 V

drak brown

EoEoProb. : ; = 0.56 VMnO2 / MnO2

MnO2 / MnO4 4 4

Prove that MnO2 will disproportionate in acidic medium.4

Another Method of Prepn. : 3K MnO + 2H SO 2KMnO + MnO + 2K SO + 2H O2 4 2 4 4 2 2 4 2

or 3K2MnO4+ 2H2O + 4CO22KMnO4 + MnO2+ 4KHCO3

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d & f - block

1But in the above method of Mn is lost as MnO2but when oxidised either by Cl2 or by O332K

2MnO

4+ Cl

22KMnO4 + 2KCl [Unwanted MnO2 does not form]OR

2K2MnO4 + O3+ H2O 2KMnO4 + 2KOH + O2

200CHeating effect:2KMnO4green K2MnO4 + MnO2 + O2

Black

at red2K2MnO4 2K2MnO3+ O2hotOxidising Prop. of KMnO

4:(in acidic medium)

(i) MnO + Fe+2 + H+Fe+3 + Mn+2 + H O4 2

(ii) MnO + H O + H+Mn+2 + O + H O4 2 2 2 2

(iii) MnO + H SMn2+ + S+ H O4 2 2

In alkaline solution : KMnO4 is first reduced to magnate and then to insoluble manganese dioxide.

Colour changes first frompurple to green and finallybecomes colourless.However brownishprecipitateisformed.

2KMnO4+ 2KOH

2K2MnO4+ 2H2O

2K2MnO

4+ H

2O + O

2MnO2+ 4KOH + 2O

alkaline2KMnO4+ H

2O 2MnO

2+ 2KOH + 3[O]

or 2MnO4 + H2O 2MnO2 + 2OH + 3[O]OxidisingProp.in neutral or weaklyacidic solution:

in presence Zn 2 or ZnO(i) 2KMnO4+ 3MnSO4 + 2H2O 5MnO2+ K2SO4 + 2H2SO4Conversion of Mn+2 to MnO :

4

(i) PbO2

(ii) Pb3O

4+ HNO

3(iii) Pb

2O

3+ HNO

3 (iv) NaBiO

3/ H+

(v) (NH4)2S2O8 / H (vi) KIO4 / H+ +

8. SILVER AND ITS COMPOUND

(I) MetallicAg

In the same way in presence of O2, Ag complexes with NaCN / KCN.4Ag + 8KCN + 2H2O + O24K[Ag(CN)2] + 4KOH

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d & f - block

AgNO3

(Silver Nitrate)

Properties.: (i) It iscalled as lunar caustic because incontact with skinit produces burning sensation like

that ofcaustic soda with the formation of finely devidedsilver (black colour)

(ii) Thermaldecomposition:

(iii) Props. ofAgNO3 : [Already done inbasic radical]

6AgNO3 + 3I2+ 3H2O5AgI + AgIO3+ 6HNO3(excess)

(iv)Ag2SO4 2Ag + SO2 + O2

(v) A(AgNO3)white ppt appears quicklyB

added

ExplainIt takes time to give white ppt.A

added

B(Na2S2O3)

(vi) Ag2

S2

O3

+ H2

O

Ag

2

S + H2

SO4

AgCl .AgBr.AgI (but notAg2S) are soluble in Na

2S

2O

3forming

[Ag(S2O3)2] complexes

(vii)AgBr :AgNO3 KBr AgBr + KNO3

Pale yellow

ppt.

3

2AgNO3 212C 2AgNO2+ O2

2AgNO2 500C 2Ag + 2NO + O2

Heating effect:

(viii)

Ag2O + H2O22Ag + H2O + O2

K2S2O8+ 2AgNO3+ 2H2O2AgO + 2KHSO4 + 2HNO3

Note:

(1) AgO supposed to beparamagnetic due to d9 configuration. But actuallyit isdiamagnetic and exists

asAgI [AgIIIO ]2

(2) Reaction involvedindeveloper :K2Fe (C2O4)2+ AgBrKFe (C2O4)2+Ag+ KBrII III

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d & f - block

9. ZINC COMPOUNDS(1) Zinc oxide, ZnO (Chinese white orphilosopher's wool)

Its foundin nature aszincite orred zinc ore.

Preparation :

(i) 2Zn + O2

2ZnO

(ii) ZnCO3 ZnO + CO2

(iii) 2Zn(NO3)

2 2ZnO + 4NO

2+ O

2

(iv) Zn(OH)2

ZnO + H2O

(a) Physical Properties : It is white powderbecomes yellow onheating again turnswhite on cooling ,insoluble inwater, sublimes at 400C.

(b) Chemical Properties :

(i) ZnO + H2SO

4 ZnSO

4+ H

2O

(iii) ZnO + 2NaOH Na2ZnO

2+ H

2O

400C

(iv) ZnO + H2

(v) ZnO + C

Zn + H2O

Zn + CO

ZnCl2(Zinc Chloride)Preparation: ZnO + 2HClZnCl2+ H2O

ZnCO3+ 2HClZnCl2+ H2O + CO2 It crystallises as ZnCl2 2 H 2OZn(OH) + 2HCl ZnCl + 2H O2 2 2

Anh. ZnCl2cannot be made by heating ZnCl22H2O because

ZnCl22H2O Zn(OH)Cl + HCl + H2O

Zn(OH)Cl ZnO +HClTo get anh. ZnCl

2: Zn + Cl2ZnCl2

Zn + 2HCl(dry)ZnCl2 + H2or Zn + HgCl2 ZnCl2 + Hg

Properties:(i) It is deliquescent white solid (whenanhydrous)(ii) ZnCl

2+ H

2S ZnS

"

"

+ NaOHZn(OH)2excess Na2[Zn(OH)4]

+ NH4OHZn(OH)2 excess [Zn(NH3)4]2+

Uses:1]2]

3]

Used for impregnating timber to prevent destruction byinsectsAs dehydrating agent when anhydrousZnOZnCl

2used indental filling

ZnSO4 (Zinc Sulphate)Preparation:

Zn + dil H2SO4ZnSO4 + H2

ZnO + dil H2SO4 ZnSO4+ H2OZnCO3+ dil H2SO4ZnSO4+ H2O + CO2

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d & f - block

ZnS 2O2 ZnSO 43

ZnS + O ZnO + SOparallel reaction

2 2 2

ZnS + 4O3ZnSO4+ 4O2

70C 280C3970CProps. 1] ZnSO47H2O ZnSO46H2OZnSO4H2OZnSO4

1

2O

2+ SO

2+ ZnO

Uses: 1]2]

ineye lotionLithophone making (ZnS + BaSO

4) aswhite pigment

COMPOUNDS10. COPPER

CuO :

Preparation: (i) CuCO3.Cu(OH)2 2CuO + H2O + CO2 (Commercial process )Malachite Green(native Cu-carbonate)

12Cu + O

22CuO & Cu2O + 2O22CuO(ii)

(iii)

(iv)

Cu(OH)2 CuO + H2O

2Cu(NO3)2 250C 2CuO + 4NO2+O

2

Properties: (i)

(ii)CuOisinsoluble in waterReadily dissolves indil. acids

CuO + H2SO4CuSO4+ H2OHCl CuCl2HNO

3Cu(NO

3)

2

It decomposeswhen, heated above 1100C4CuO 2Cu2O + O2CuO is reduced to Cu by H2 or C under hotconditionCuO + CCu + COCuO + H

2Cu + H

2O

(iii)

(iv)

CuCl2:Preparation: CuO + 2HCl(conc.) CuCl2 + H2O

Cu(OH)2CuCO

3+ 4HCl 2CuCl

2+ 3H

2O + CO

2

Properties: (i)(ii)

It is crystallised as CuCl22H2O ofEmerald green colour

dil. solution inwater is bluein colour due to formation of[Cu(H

2O)

4] complex.

conc. HCl or KCl added to dil. solution of CuCl2 the colour changesinto yellow, owing to the formation of [CuCl4]The conc. aq. solution is green incolour havingthe twocomplex ions in

equilibrium 2[Cu(H2O)4]Cl2 l [Cu(H2O)4] + [CuCl4] +4H2OCuCl2 CuCl byno. of reagents

2+

(iii)2

(iv)

2+ 2

(v)

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d & f - block

(a) CuCl2 + Cu-turnings 2CuCl(b) 2CuCl

2+ H

2SO

3+ H

2O 2CuCl + 2HCl + 2H2SO4

(c) 2CuCl2 + Zn/HCl 2CuCl+ ZnCl2(d) CuCl2 + SnCl2CuCl + SnCl4

** CuF22H2O light blue

CuCl22H2O greenCuBr2 almost black

CuI2 doesnot exist

Anh. CuCl 2is dark brown mass obtained

by heating CuCl22H2O at 150C in presenceof HCl vap.

150CHClgas

CuCl22H2O CuCl2 + 2H2O

CuSO4:

Preparation: CuO + H2SO4(dil)CuSO4 + H2O Cu(OH)2 +H2SO4(dil) CuSO4 + 2H2O Cu(OH)2CuCO3 +H2SO4 (dil)CuSO4 + 3H2O + CO2

1Cu + H

2SO

4+

2

O2CuSO4 + H2O [Commercial scale]

(Scrap)Cu + dil. H2SO4 no reaction {Cu is a below H inelectrochemical series}It iscrystallised as CuSO45H2OProperties:(i)

(ii) CuSO45H2O

Blue

CuSO43H

2O

Pale blue

CuSO4H

2O

Bluish whitetake places

CuSO4(anh.)white

(iii) Revisionwithall others reagent

11. IRON COMPOUNDS

FeSO47H

2O:

Preparation: (i) Scrap Fe + H2SO4FeSO4+ H2

(dil.)From Kipp's waste

FeS + H2SO

4(dil)FeSO

4+ H

2S

7FeS

2+ 2H

2O +

2O

2FeSO4+ H2SO4

(ii)

(iii)

Properties: (i) It undergoes aerial oxidation forming basic ferric sulphate4FeSO

4+ H

2O + O

2 4Fe(OH)SO

4

FeSO 4 high Fe O +SO +SO300C(ii) FeSO47H

2O anh.white 2 3 2 3temp.

(iii) Aq. solution is acidic due to hydrolysisFeSO

4+ 2H

2O l Fe(OH)

2+ H

2SO

4

weak base

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d & f - block

(iv) It isa reducing agent(a) Fe2+ + MnO + H+Fe3+ + Mn2+ + H O

4 2

(b) Fe2+ + Cr O 2+ H+Fe3+ + Cr3+ + H O2 7(c) Au3+ + Fe2+Au + Fe3+

(d) Fe2+ + HgCl Hg Cl + Fe3+

2

2 2 2

white ppt.

(v) It forms double salt. Example (NH4)2SO4FeSO46H2O

FeO(Black):

FeC O FeO + CO + COPrepn : 2 4 2in absence of air

Props: It is stable at high temperature and on cooling slowlydisproportionates into Fe3O4 andiron

4FeOFe3O4+ FeFeCl

2:

Prepn: heatedin a current of HCl

OR

Fe + 2HCl FeCl + H2 2

2FeCl3+ H

2 2FeCl2+ 2HCl(i) It is deliquescent in air like FeCl3(ii) It is soluble inwater, alcoholandether also because it is sufficientlycovalent in nature(iii) It volatilises at about 1000C and vapour density indicates the presence of Fe

2Cl

4.

Above 1300C density becomes normal(iv) It oxidisesonheating inair

12FeCl2+ 3O22Fe2O3+ 8FeCl3(v) H2 evolves on heating in steam

3FeCl2+ 4H2O Fe3O4+ 6HCl + H2(vi) It can exist as different hydrated form

FeCl22H2O colourlessFeCl

24H

2O pale green

FeCl26H2O green

Props:

FeCl3:

Prepn: Anhydrous ferric chloride isprepared byheating metallic iron ina stream of drychlorine gas.(i) FeCl

3solid isalmost black. It sublimes at about 300C,giving a dimeric gas.

(ii) FeCl3 dissolvesinbothether and water, giving solvated monomeric species.(iii) Iron (III) chloride isusuallyobtained as yellow-brown lumps of the hydrate FeCl

36H

2O.

(iv) Thisisverysolubleinwaterand isusedbothasanoxidizing agent, and asa mordant indyeing.(v) FeCl3 is also used in themanufacture of CCl4.

12. INNER TRANSITION ELEMENTSThe elements in which theadditional electrons enters (n2)forbitals arecalledinner transition elements.The valence shell electronic configuration of these elements can be represented as

(n2)f0,2...14(n1)d0,1,2,ns2 .These are alsocalled f-blockelementsbecause the extra electron goes

to f-orbitals which belongs to (n2)th mainshell. 4f-block elements are also called lanthanides orrare

earths. Similarly5f-block elements arecalled actinides or actinones. Thename Lanthanides and actinideshave been given due to close resemblance with Lanthanum and actinium respectively. Lanthanides

constitutes the first inner transition serieswhile actinides constitutes secondinner transition series.

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d & f - block

General Characteristics:

1. Electronic Configuration

[Xe] 4fn 15d0 6s2 or[Xe] 4fn 5d16s2

Oxidation state: TheyreadilyformM3 ionssomeofthemalsoexhibit oxidationstate of+2and +4.

Colouration: IonsofLanthanides and actinides are coloured in the solid state aswell as in aqueous

2.

3.solutionbecause of absorbation of light due to ff transition since theyhave partlyfilled f-orbitals.

13.LANTHANIDE CONTRACTIONIn lanthanides, the additional electron enters 4f-sub shell but not in the valence-shell namely sixth shell.

The shieldingeffect of one electron in4f-sub-shell by another inthe same sub-shell isverylittle, being

evensmaller thanthat of d-electrons, because the shape of f-sub shell isvery muchdiffused, while there

isnocomparable increase inthemutual shielding effect of4f-electrons.This results inthat electrons inthe

outermost shellexperience increasing nuclear attraction from the growing nucleus.Consequently the

atomic and ionic radii go on decreasing as wemove fromLa57

to Lu71

.

Consequence of Lanthanidecontraction

1. Atomic andionic radii of post-Lanthanide elements: The atomicradii of second rowtransition

elements arealmost similar to thoseofthe third row transitionelement because the increase insize on

movingdownthegroup fromsecondto third transition elements iscancelled bythedecrease insize

due to lanthanide contraction.

High density of post lanthanide elements:It is because of very small size due to lanthanide

contraction.

Basic strength of oxides and hydroxides: Due to lanthanide contraction the decrease in size of

2.

3.

lanthanides ions, from to increases the covalent character (i.e. decreases the ionicLa 3 Lu 3

character) betweenLn3 andOH ions inLn(III) hydroxides (Fajans rules). Thus La(OH) is the3

most basic while Lu(OH)3 is the best basic.

Similarly, there isa decrease inthe basic strength of theoxides.

Seperation of Lanthanides: Due to the similar size (Because of lanthanide contraction) of the

lanthanides, it isdifficult to separate them. But slight variationintheir properties isutilizedto separate

4.

Consequences of Lanthanide Contraction

(i) Separation of Lanthanides.Separationof lanthanides ispossibleonlydue to lanthanide contraction.

All thelanthanideshave quite similar properties anddue to this reason theyaredifficult to separate.

However, because of lanthanidecontraction their properties (such as ability to formcomplexes)varyslightly. Thisslight variation inproperties isutilized intheseparationoflanthanidesbyionexchange

methods.

(ii) Variation in basicstrength of hydroxides. Thebasic strength of oxides andhydroxides decreasesfrom La(OH)3to Lu(OH)3. Due to lanthanide contraction sizeof M3+ ionsdecreases and there is

increase in the covalent character in MOH bond.

(iii)Similarity of second and third transitionseries. The atomic radii of second rowof transition

elements are almost similar to those of the third rowoftransition elements. For example, among the

elements of group 3, there is normal increase insize fromSc toYto La. But after lanthanides the

atomic radii fromsecond to third transitionseries do not increase as shownbelowfor group 4 and

group 5 i.e., for Zr HfandNb Ta pairs whichhave element same atomic radii.Aftergroup 5 the

effect oflanthanide contraction isnotso predominant.

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d & f - block

5. Colour. The lanthanide metals aresilverywhite metals and the trivalent lanthanide ions arecoloured

both in thesolid state and inthe aqueoussolution.

La3+ (4f0) and Lu3+ (4f14) having no unpaired electron do not show6. Magnetic properties.

paramagnetism whileallother tripositive ions of lanthanides areparamagnetic.

They have lowionisation energyandarehighlyelectropositive.Their ionisationenergyvalues are

quite comparable withthose of alkaline earth metalsparticularlycalcium.These metalsdo not have muchtendencyto form complexes.

The lanthanidesarehighlyreactive. Thisisinagreement withthe lowvaluesoftheir ionisationenergies

andelectronegativity.

7.

8.

9.

10. Thesolubilityof compoundsof lanthanidesfollowthesameorder asgroup 2 elements.Their fluorides,

oxides, hydroxides, carbonates are insoluble inwater. However halides (except fluorides), nitrates,

acetates aresoluble inwater.

General Characteristics ofActinides 5f Block Series Second InnerTransition Series

1.

2.

These aresilverywhitemetals with highmelting and boiling points.

Besides the most common oxidation state of +3, actinides show +4, +5 and +6oxidation states incertainelements.

All the actinides are radioactive innature.

All theactinides are highlyelectropositive and assuch arestrong reducingagent.Actinides have a strong tendency towards complex formationand form oxocations like UO22+,PuO22+, UO+ etc.

Most of the cationsof actinides are coloured.

Actinidesshow actinide contraction i.e., decrease inionic radiialong theseries.

3.

4.

5.

6.

7.

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d & f - block

Ex.1 On what basis can you say that scandium(Z = 21) is a transition element but since

(Z = 30) is not.

On the basis of i incompletely filled3d - orbitals in case of scandium atom in itsground state 3d1), it is regarded as a transitionelement. On the other hand, zinc atom hascompletelyfilled d- orbitals (3d10) in its groundstate as well as in its oxidisedstate, hence it isnot regarded as transition element.

(b) because Ce(IV) has extra stability due toempty f0 orbital.

Ce3+Ce4+ + e

[Xe] 4f1 5d06s0Sol.(c) InMn2+ d5 configuration leas to extra

stability of half filled configuration , so Mn2+/Mn2+(d4) tends to get converted to stable d5,configuration of Mn2+, by accepting an electronso Mn3+/Mn2+ redox couple has more positive

potential than Fe3+/Fe2+(d14) couple.

(d) Due to more negative enthalpy of hydrationof Cu (aq) than Cu (aq) which compensatesor second ionization enthalpyof copper.

(e) In the third transition series after lanthanumtheir is lanthanoid contraction due to ineffectiveshielding by intervening f-orbital electrons andhence second and third transition serieselements have similar atomic radii.

Ex.2 What is the effect of increasing pH on K2Cr

2O

7

solution?

Inaqueoussolution, wehave

2+ +

Sol. Cr2O

7+ H

2O2 2CrO

4+ 2H .2 +

When pH < 4 (acidic medium), it exists asCr

2O

7 and the colour is orange.

When pH> 7 (basic medium), it exists asCrO4 and thecolour isyellow.

What is the basic difference between theelectronicconfigurationsof transitionandinnertransitionelements.Generalelectronic configuration of transition

element = [ Noblegas] (n 1) d110 ns12 andfor inner transition elements

=(n2) f 114 (n 1)d01ns02. Thus , in transitionelements, last electron enters d-orbitals of the

penultimate shelll while in inner transitionelements, it enters f- orbital of thepenultimateshell.

2

2

Ex.5 WhyisCr2+ reducing and Mn3+ oxidising whenbothhave d4 configuration?

Cr2+ is reducing as its configuration changesform d4 to d3, the later hashalf filled t level.

Ex.3

Sol.

2g

Again thechange fromMn2+ toMn3+ results inthehalffilled (d5) configuration whichhas extrastability.

Sol.

Ex.6

Sol.

Cu+ ionisnot stable inaqueous solution.Why?Cu2+ (aq) is much more stable than Cu+(aq).This is because although second ionization

enthalpyof copper is large but H for Cu2+hyd(aq) ismuch more negative than that for Cu+

(aq) and therefore it more than compensatesfor the second ionization enthalpy of copper.

Thus many copper (I) compounds ar unstablein aque o us so lu t io n and und er go esdisproportionalas follows.

2Cu+ (aq) Cu2+(aq) + Cu(s)

Ex.4 Give reasons for thefollowing(a) Transition metals have high enthalpies of

atomizations.(b) Among the lathanoids, Ce(III) is easilyoxidised to Ce(IV)(c) Fe3+ | Fe2+ redox couple has lesspositiveelectrodepotential thanMn3+| Mn2+ couple.(d) Copper (I) has configuration , stillcopper (II) hasd9 configuration stillcopper (II)ismorestable inaqueous solution than copper(I).

(e) The second and third transition series

elements have almost similar atomic radii.(a) Due to strong interatomic interactionbetween valence electrons.

d10 Ex.7 Why is the E value for the M3+/ M2+ couplemuch more positive than that of Cr3+/Cr2+ or

Fe3+/Fe2+? Explain.

Much larger third ionization energy of Mn(Where therequiredchange is d5 tod4) ismainly

responsible for this.This also explainswhythe+3stateofMn is oflittle importance.

Sol.

Sol.

14

SOLVED EXAMPLES


15/26

d & f - block

Ex.8 Why does Mn (II) ion show the maximum Q.13 Arranged the following in increasing order of

acdic character:

CrO , CrO, Cr O

paramagnetic behvavious among bivalent ionsof thefirst transitionseries.?The electronic configurations ofMn and Mn

(II) ionare

25

Mn: 1s 2s 2p 3s 3p 3d 4s ,Mn+2 : 1s2,2s22p6 3s23p63d24s0

The Mn+2 ionhas fiveunpairedelectrons inits3d subshell which is the maximum value for atransition metal ion. Hence, Mn(II) shows themaximum paramagnetic behaviour (due tounpaired electrons) among bivalent ions of thefirst transition series.

2

CrO< Cr O y(B) compounds in oxidation state x are ionicifx < y(C)compounds inoxidation state yare covalentif x < y(D) compounds inoxidationstate yarecovalenti f y


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d & f - block

EXERCISE - IIQ.1

Q.2

Q.3

Formation of MnO42+Due to disproportionation of Hg2

NH3 - strongligand causespairing of electrons.So complex isdiamagneitc. Fisa weak ligand and so

pairing isnot possible.

[Al(OH)3 and NaAlO2; Zn(OH)2 and Na2ZnO2] also [Zn(NH3)4]2+

[Zn S iswhitebut PbS isblack].

[A] = ZnCl2; [B] = Zn(OH)2; [C] = Na2ZnO2; [D] = ZnS; [E] = AgCll

All the three elements haved10

See text for the reaction of HCland KMnO4

.

[(A) = ZnSO3; (B) = ZnO ; (C) = SO2; (D) = ZnS]

3/2 M

K2[Pt Cl6]

Clis a weak ligandso no pairing of electrons. CO is a strong field ligand and, therefore, causespairing ofelectrons.

In the first case, a double salt is formed. In the second case, a complex is formed. So no test.

(A) = [Cr(H2O)6]Cl3 ; (B) = [CrCl(H2O)5]Cl2 H2O (C) = [Cr Cl2(H2O)4]Cl2.2H2O

(A) = [Cr(NH3)4 ClBr]+ Cland (B) = [Cr(NH3)4Cl2]

+ Br.

[Co(NH3)4 Cl(H2O)] SO4 ;[Co(NH3)4(SO4)(H2O)]Cl [Co(NH3)4(SO4)(Cl)] H2O

Q.4

Q.5

Q.6

Q.7

Q.8

Q.9

Q.10

Q.11

Q.12

Q.13

Q.14

Q.15

Q.16

EXERCISE - III

Q.1

Q.6

Q.11

Q.16Q.21

Q.26

Q.31

Q.36

Q.41

Q.46

Q.51

Q.56

A

B

B

CA

A

C

A

A,B

A,B,C

A,B,C,D

B

Q.2

Q.7

Q.12

Q.17Q.22

Q.27

Q.32

Q.37

Q.42

Q.47

Q.52

Q.57

B

A

A

AD

C

A

B

A,B

A,B

A,B

A

Q.3

Q.8

Q.13

Q.18Q.23

Q.28

Q.33

Q.38

Q.43

Q.48

Q.53

Q.58

A

A

B

BA

D

D

A,B,C

B,C

B,C

A,B,C

C

Q.4

Q.9

Q.14

Q.19Q.24

Q.29

Q.34

Q.39

Q.44

Q.49

Q.54

Q.59

B

B

D

AB

C

A

B,C,D

A,B,C

A,C

A,B

B

Q.5

Q.10

Q.15

Q.20Q.25

Q.30

Q.35

Q.40

Q.45

Q.50

Q.55

Q.60

A

A

B

D

D

B

A

A,B

A,B,C

A,B

B,C

C

ANSWER KEY

d-f-block IIT JEE

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