CONTENTS S.NO.
TOPIC
PAGE NO
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
d & f - block
d & f - BLOCK 1. INTRODUCTION In these elements the last electron enters (n–1)d orbitals of atom of an element is called d-block elements.
2. ELECTRONIC CONFIGURATION AND IRREGULARITIES The valence shell configurations of these elements can be represented by (n – 1)d1–10ns 0,1,2 . All the d–block elements are classified into four series viz 3d , 4d , 5d and 6d series corresponding to the filling of 3d , 4d , 5d and 6d orbitals of (n–1)th main shell orbitals of (n–1)th main shell. Each Series has 10 element s. Cr(3d 5 , 4s 1) , Cu(3d 10 , 4s 1 ), Mo(4d 5 , 5s 1), Pd(4d 10 , 5s 0), Ag(4d 10, 5s 1 ) anAu(5d10 , 6s1) clearly show the irregularities in the configurations These are explained on the basis of the concept that half filled and completely filled d–orbitals are relatively more stable than other d–orbitals.
3. GENERAL CHARACTERISTICS (i) Metallic character: They are all metal and good conductor of heat & electricity (ii) Electronic configuration: (n–1)d1–10ns1–2 (iii) M.P. Cr — Maximum Zn lowest m.p. – Cd due to no unpaired e– Mo 6 no. of unpaired e s W are involved in metallic bonding Hg for metallic bonding (iv) Variation in atomic radius: Sc ——— Mn Fe Co Ni Cu Zn decreases remains increases same again (v) Variable oxidation states possible Sc Ti V Cr Mn Fe Co Ni Cu Zn 1 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 7 Colour : (aquated) Sc3+ — colourless Ti4+ — colourless Ti3+ — purple V4+ — blue V3+ — green V2+ — violet Cr2+ — blue Cr3+ — green Mn3+ — violet Mn2+ — light pink Fe2+ — light green Fe3+ — yellow Co2+ — pink Ni2+ — green Cu2+ — blue Zn2+ — colourless
2
d & f - block
4. COLOUR Substances appear coloured when they absorb light of a particular wavelength in the visible region of the spectrum and transmit light of other wavelengths. The colour which we see is the colour of the transmitted wavelengths. In other words, the colour of the compound observed by us is the complmentary colour of the colour absorbed by the compound. In the s- and p-block elements there cannot be any d-d transitions and the energy needed to promote s or p electron to a higher level is much greater and may correspond to ultraviolet region, in which case the compound will not appear coloured to the eye. Relationship between colour and wavelength Wavelength absorbed in nm Colour absorbed Colour observed <400 UV region White/colourless 400—435 Violet Yellow-green Indigo Yellow 435—480 Green-blue Orange 480—490 Blue-green Red 490—500 Green Purple 500—560 Y ellow-green Violet 560—580 Yellow Indigo 580—595 Orange Green-blue 595—605 Red Blue-green 605—750 Infra-red White/colourless > 750 Magnetic Properties: When a substance is placed in a magnetic field of strength H, the intensity of the magnetic field in the substance may be greater than or less than H. diamagnetic, Substances which are weakly repelled by a magnetic field paramagnetic the substances which are weakly attracted bythe magnetic field and lose their magnetism when removed from the field . Paramagnetism is expressed by magnetic moment, n(n 2) B.M. n = number of unpaired electrons B.M. = Bohr Magneton, unit of magnetic moment
5. COMPLEX FORMATION In the case of transition metals in low oxidation states, the electrons in the d orbitals become involved in bonding with ligands.The bonding between the ligand and the transition metal ion can either be predominantly electrostatic or covalent or in many cases intermediate between the two extremes. Some of the typical complexes of the transition meals are [Fe(CN) 6]3– ,[Ni(NH 3) 4]2 ,[Cu(CN) 4 ]3– , [Cu(NH 3) 4]2 etc. Formation of alloys : Transition elements have almost similar atomic sizes. Therefore , these elements can mutually substitute their positions in their crystal lattice. The alloys are hard and have high melting points as compared to the most metal. Catalytic Properties: Many transition metals and their compounds have catalytic properties. some example TiCl 3 Used as the Ziegler-Natta catalyst in the production of polythene. V2O 5
Converts SO 2 to SO3 in the contact process for making H 2SO 4 .
MnO 2
Used as a catalyst to decompose KClO3 to give O 2 .
3
d & f - block
Fe
Promoted iron is used in the Haber-Bosch process for making NH3 .
H 2O 2 Used as Fenton’s reagent for oxidizing alcohols to aldehydes. Pd Used for hydrogenation (e.g. phenol to cyclohexanone). Pt/Rh Formerly used in the Ostwald process for making HNO3 to oxidize NH 3 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 to greater effective nuclear charge acting on outer valence electrons because of the weak shielding of the nucleus by 4f electrons. The ionisation energies of the 3d and 4d elements are irregular. Common oxidation states for each element include +2 and +3 or both. The +3 oxidation states are more stable at the beginning of the series, whereas towards the end +2 oxidation states are more stable. Ionisation energy increases gradually 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 ionisation energies. Because it takes more energy to remove the third electron from the metals near the end of the row than from those near the beginning, the metals near the end tends to form M2+ ions rather than M3+ ions.
6. CHROMATE -DICHROMATE Preparation : 4FeO . Cr2O3 + 8Na2CO3 + 7O2 chromate ore
Rosting in air
8Na2CrO4 + 2Fe2O3 + 8CO2
The roasted mass is extracted with water when Na2CrO4 goes into the solution leaving behind insoluble Fe2O3. The solution is treated with calculated amount of H2SO4. 2Na2CrO4 + H2SO4 Na2Cr2O7 + Na2SO4 + H2O The solution is concentrated when less soluble Na2SO4 crystallises out. The solution is further concentrated when crystals of Na2Cr2O7 are obtained. Then a hot saturated solution of Na2Cr2O7 is treated with KCl when reddish orange crystals of K2Cr2O7 are obtained on crystallisation. K2Cr2O7 is preferred to Na2Cr2O7 because Na2Cr2O7 is hygroscopic but K2Cr2O7 is not. * Other props & test of CrO2 & Cr 2O 2 :Already discussed 4 7 * Similarities between hexavalent Cr & S-compounds: (i) SO3 & CrO3 both acidic.
4
d & f - block
(ii) CrO42 & SO 42 are isomorphous 2 2 OH (iii) SO2Cl2 & CrO2Cl2 SO 4 & CrO4 respectively.
2 2 OH (iv) SO3Cl– & CrO3Cl– SO 4 & CrO4
O O O || || || Cr O Cr O Cr (v) CrO3 & (SO3) has same structure || || || O O O (vi) Potassium dichromate reacts with hydrochloric acid and evolves chlorine. K2Cr2O7 + 14HCl 2KCl + 2CrCl3 + 7H2O + 3Cl2 (vii) It acts as a powerful oxidising agent in acidic medium (dilute H2SO4) Cr2O 2 7 + 14H + + 6e – 2Cr3 + + 7H2O. (Eº = 1.33 V) The oxidation state of Cr changes from + 6 to +3. Uses (i) As a volumetric reagent in the estimation of reducing agents such as oxalic acid, ferrous ions, iodide ions, etc. It is used as a primary standard. (ii) For the preparation of several chromium compounds such as chrome alum, chrome yellow, chrome red, zinc yellow, etc. (iii) In dyeing, chrome tanning, calico printing photography etc. (iv) Chromic acid as a cleansing agent for glass ware.
7. MANGANATE & PERMANGANATE Preparation This is the most important and well known salt of permanganic acid. It is prepared from the pyrolusite ore. It is prepared by fusing pyrolusite ore either with KOH or K2CO3 in presence of atmospheric oxygen or any other oxidising agent such as KNO3. The mass turns green with the formation of potassium magnate, K2MnO4. 2MnO2 + 4KOH + O2 2K2MnO4 + 2H2O 2MnO2 + 2K2CO3 + O2 2K2MnO4 + 2CO2 The fused mass is extracted with water. The solution is now treated with a current of chlorine or ozone or carbon dioxide to convert magnate into permanganate. 2K2MnO4 + Cl2 2KMnO4 + 2KCl 2K2MnO4 + H2O + O3 2KMnO4 + 2KOH + O2 3K2MnO4 + 2CO2 2KMnO4 + MnO2 + 2K2CO3 Props : The above green solution is quite stable in alkali, but in pure water and in presence of acids, depositing MnO2 and giving a purple solution of permanganate. 3K2MnO4 + 2H2O 2KMnO4 + MnO2 + 4KOH purple drak brown Prob. : Eo
MnO 42 / MnO 2
= 2.26 V
;
Eo
MnO 42 / MnO4
= – 0.56 V
Prove that MnO42 will disproportionate in acidic medium. Another Method of Prepn. : 3K 2MnO 4 + 2H 2SO 4 2KMnO 4 + MnO 2 + 2K 2SO 4+ 2H 2O or 3K2MnO4 + 2H2O + 4CO2 2KMnO4 + MnO2 + 4KHCO3
5
d & f - block
1 But in the above method of Mn is lost as MnO2 but when oxidised either by Cl2 or by O3 3 2K2MnO4 + Cl2 2KMnO4 + 2KCl [Unwanted MnO2 does not form] OR 2K2MnO4 + O3 + H2O 2KMnO4 + 2KOH + O2
Heating effect : 2KMnO4 K2MnO4 + MnO2 + O2 green 200C Black t r ed 2K2MnO4a 2K2MnO3 + O2 hot Oxidising Prop. of KMnO4 : (in acidic medium)
(i) MnO + Fe+2 + H+ Fe+3 + Mn+2 + H 2O 4
(ii) MnO 4 + H2 O2 + H+ Mn+2 + O 2 + H 2O (iii) MnO 4 + H2 S Mn2+ + S + H2O In alkaline solution : KMnO4 is first reduced to magnate and then to insoluble manganese dioxide. Colour changes first from purple to green and finally becomes colourless. However brownish precipitate is formed. 2KMnO4 + 2KOH 2K2MnO4 + H2O + O 2K2MnO4 + 2H2O 2MnO2 + 4KOH + 2O alkaline 2KMnO4 + H2O 2MnO2 + 2KOH + 3[O] or 2MnO4¯ + H2O 2MnO2 + 2OH¯ + 3[O] Oxidising Prop. in neutral or weakly acidic solution: in presence Zn 2 or ZnO
(i) 2KMnO4 + 3MnSO4 + 2H2O 5MnO2 + K2SO4 + 2H2SO4 Conversion of Mn+2 to MnO 4 : (i) PbO2 (ii) Pb3O4 + HNO3 (iii) Pb2O3 + HNO3 + + (v) (NH4)2S2O8 / H (vi) KIO4 / H
(iv) NaBiO3 / H+
8. SILVER AND ITS COMPOUND
(I) MetallicAg
In the same way in presence of O2, Ag complexes with NaCN / KCN. 4Ag + 8KCN + 2H2O + O2 4K[Ag(CN)2] + 4KOH
6
d & f - block
AgNO3 (Silver Nitrate) Properties.: (i) It is called as lunar caustic because in contact with skin it produces burning sensation like that of caustic soda with the formation of finely devided silver (black colour) (ii) Thermal decomposition: (iii) Props. ofAgNO3 : [Already done in basic radical] 6AgNO3 + 3I2 + 3H2O 5AgI + AgIO3 + 6HNO3 (excess) (iv) Ag2SO4 2Ag + SO2 + O2 B
(v) A(AgNO3) white ppt appears quickly added
Explain B(Na2S2O3) It takes time to give white ppt. A
added
(vi) Ag2S2O3 + H2O Ag2S + H2SO4 AgCl .AgBr.AgI (but notAg2S) are soluble in Na2S2O3 forming [Ag(S2O3)2]–3 complexes Br (vii)AgBr : AgNO3 K AgBr + KNO3
Pale yellow ppt. Heating effect:
C 2AgNO3 212 2AgNO2 + O2 C 2AgNO2 500 2Ag + 2NO + O2
(viii)
Ag2O + H2O2 2Ag + H2O + O2 K2S2O8 + 2AgNO3 + 2H2O 2AgO + 2KHSO4 + 2HNO3 Note: (1) AgO supposed to be paramagnetic due to d9 configuration. But actually it is diamagnetic and exists asAgI [AgIIIO 2] (2) Reaction involved in developer : K2FeII(C2O4)2 + AgBr KFe III (C2O4)2 + Ag + KBr
7
d & f - block
9. ZINC COMPOUNDS (1) Zinc oxide, ZnO (Chinese white or philosopher's wool) Its found in nature as zincite or red zinc ore. Preparation : (i) 2Zn + O2 2ZnO (ii) ZnCO3
(iii) 2Zn(NO3)2 (iv) Zn(OH)2
ZnO + CO2
2ZnO + 4NO2 + O2
ZnO + H2O
(a) Physical Properties : It is white powder becomes yellow on heating again turns white on cooling , insoluble in water, sublimes at 400°C. (b) Chemical Properties : (i) ZnO + H2SO4 ZnSO4 + H2O (iii) ZnO + 2NaOH Na2ZnO2 + H2O (iv) ZnO + H2 (v) ZnO + C
400º C
Zn + H2O Zn + CO
ZnCl2 (Zinc Chloride) Preparation: ZnO + 2HCl ZnCl2 + H2O ZnCO3 + 2HCl ZnCl2 + H2O + CO2 It crystallis es as ZnCl 2 ·2 H 2O Zn(OH)2 + 2HCl ZnCl2 + 2H2O Anh. ZnCl2 cannot be made by heating ZnCl2·2H2O because ZnCl2·2H2O Zn(OH)Cl + HCl + H2O
Zn(OH)Cl ZnO + HCl To get anh. ZnCl2: Zn + Cl2 ZnCl2 Zn + 2HCl(dry) ZnCl2 + H2 or Zn + HgCl2 ZnCl2 + Hg Properties: (i) It is deliquescent white solid (when anhydrous) (ii) ZnCl2 + H2S ZnS "
+ NaOH Zn(OH)2 excess Na2[Zn(OH)4]
" + NH4OH Zn(OH)2 excess [Zn(NH3)4] 2+ Uses: 1] Used for impregnating timber to prevent destruction by insects 2] As dehydrating agent when anhydrous 3] ZnO·ZnCl2 used in dental filling
ZnSO4 (Zinc Sulphate) Preparation: Zn + dil H2SO4 ZnSO4 + H2 ZnO + dil H2SO4 ZnSO4 + H2O ZnCO3 + dil H2SO4 ZnSO4 + H2O + CO2
8
d & f - block
ZnS 2O2 ZnSO 4
parallel reaction
3
ZnS + O 2 ZnO + SO 2 2 ZnS + 4O3 ZnSO4 + 4O2 70C 280C 3970C Props. 1] ZnSO4·7H2O ZnSO4·6H2O ZnSO4·H2O ZnSO4 1 O + SO2 + ZnO 2 2 Uses: 1] in eye lotion 2] Lithophone making (ZnS + BaSO4) as white pigment
10. COPPER COMPOUNDS CuO : Preparation: (i)
CuCO3.Cu(OH)2 2CuO + H2O + CO2 (Commercial process ) Malachite Green (native Cu-carbonate)
1 O 2CuO 2 2
(ii)
2Cu + O2 2CuO & Cu2O +
(iii)
Cu(OH)2 CuO + H2O
(iv)
C 2Cu(NO3)2 250 2CuO + 4NO2 + O2 CuO is insoluble in water Readily dissolves in dil. acids CuO + H2SO4 CuSO4 + H2O HCl CuCl2 HNO3 Cu(NO3)2 It decomposes when, heated above 1100°C 4CuO 2Cu2O + O2 CuO is reduced to Cu by H2 or C under hot condition CuO + C Cu + CO CuO + H2 Cu + H2O
Properties: (i) (ii)
(iii) (iv)
CuCl2 : Preparation:
CuO + 2HCl(conc.) — CuCl2 + H2O Cu(OH)2·CuCO3 + 4HCl — 2CuCl2 + 3H2O + CO2
Properties:
(i) It is crystallised as CuCl2·2H2O of Emerald green colour (ii) dil. solution in water is blue in colour due to formation of 2+ [Cu(H2O)4] complex. (iii) conc. HCl or KCl added to dil. solution of CuCl2 the colour changes into yellow, owing to the formation of [CuCl4]2– (iv) The conc. aq. solution is green in colour having the two complex ions in equilibrium 2[Cu(H2O)4]Cl2 l [Cu(H2O)4] 2+ + [CuCl4]2– + 4H2O (v) CuCl2 — CuCl by no. of reagents
9
d & f - block (a) CuCl2 + Cu-turnings 2CuCl (b) 2CuCl2 + H2SO3 + H2O — 2CuCl + 2HCl + 2H2SO4 (c) 2CuCl2 + Zn/HCl — 2CuCl + ZnCl2 (d) CuCl2 + SnCl2 — CuCl + SnCl4 ** CuF2·2H2O — light blue Anh. CuCl 2 is dark brown mass obtained CuCl2·2H2O — green by heating CuCl 2·2H 2O at 150C in presence CuBr2 — almost black of HCl vap. 1 50 C
CuI2 does not exist
CuCl2·2H2O CuCl2 + 2H2O HCl gas
CuSO4 : Preparation:
Properties:(i) (ii)
CuO + H2SO4(dil) CuSO4 + H2O Cu(OH)2 + H2SO4(dil) CuSO4 + 2H2O Cu(OH)2·CuCO3 + H2SO4 (dil) CuSO4 + 3H2O + CO2 1 Cu + H2SO4 + O2 — CuSO4 + H2O [Commercial scale] 2 (Scrap) Cu + dil. H2SO4 no reaction {Cu is a below H in electrochemical series} It is crystallised as CuSO4·5H2O CuSO4·5H2O Blue
take places
CuSO4·3H2O
CuSO4·H2O
Pale blue
Bluish white
CuSO4(anh.) white
(iii)
Revision with all others reagent
11. IRON COMPOUNDS FeSO4·7H2O: Preparation: (i) (ii)
(iii) Properties: (i) (ii) (iii)
Scrap Fe + H2SO4 FeSO4 + H2 (dil.) From Kipp's waste FeS + H2SO4(dil) FeSO4 + H2S 7 O FeSO4 + H2SO4 2 2 It undergoes aerial oxidation forming basic ferric sulphate 4FeSO4 + H2O + O2 4Fe(OH)SO4 igh FeSO 4 h 300C Fe2 O3 +SO2 +SO3 FeSO4·7H2O anh. white temp.
FeS2 + 2H2O +
Aq. solution is acidic due to hydrolysis FeSO4 + 2H2O l Fe(OH)2 + H2SO4 weak base
10
d & f - block
(iv)
It is a reducing agent (a) Fe2+ + MnO 4– + H+ Fe3+ + Mn2+ + H 2O (b) Fe2+ + Cr 2O 72– + H+ Fe3+ + Cr3+ + H 2O (c) Au3+ + Fe2+ Au + Fe3+ (d) Fe2+ + HgCl2 Hg 2Cl2 + Fe3+ white ppt.
(v)
It forms double salt. Example (NH4)2SO4·FeSO4·6H2O
FeO(Black): Prepn :
FeC2O4 FeO + CO + CO2 in absence of air
Props:
FeCl2: Prepn:
It is stable at high temperature and on cooling slowly disproportionates into Fe3O4 and iron 4FeO Fe3O4 + Fe Fe + 2HCl heatedin FeCl2 + H2 a current of HCl
OR Props:
2FeCl3 + H2 2FeCl2 + 2HCl (i) It is deliquescent in air like FeCl3 (ii) It is soluble in water, alcohol and ether also because it is sufficiently covalent in nature (iii) It volatilises at about 1000°C and vapour density indicates the presence of Fe2Cl4. Above 1300°C density becomes normal (iv) It oxidises on heating in air 12FeCl2 + 3O2 — 2Fe2O3 + 8FeCl3 (v) H2 evolves on heating in steam 3FeCl2 + 4H2O — Fe3O4 + 6HCl + H2 (vi) It can exist as different hydrated form FeCl2·2H2O — colourless FeCl2·4H2O — pale green FeCl2·6H2O — green
FeCl3: Prepn: Anhydrous ferric chloride is prepared by heating metallic iron in a stream of dry chlorine gas. (i) FeCl3 solid is almost black. It sublimes at about 300°C, giving a dimeric gas. (ii) FeCl3 dissolves in both ether and water, giving solvated monomeric species. (iii) Iron (III) chloride is usually obtained as yellow-brown lumps of the hydrate FeCl3·6H2O. (iv) This is very soluble in water and is used both as an oxidizing agent, and as a mordant in dyeing. (v) FeCl3 is also used in the manufacture of CCl4.
12. INNER TRANSITION ELEMENTS The elements in which the additional electrons enters (n–2)f orbitals are called inner transition elements. The valence shell elect ronic co nfiguration of these element s can be represent ed as (n–2) f 0,2...14 (n – 1)d 0,1,2,ns 2 . These are also called f-block elements because the extra electron goes to f-orbitals which belongs to (n–2)th main shell. 4f-block elements are also called lanthanides or rare earths. Similarly 5f-block elements are called actinides or actinones. The name Lanthanides and actinides have been given due to close resemblance with Lanthanum and actinium respectively. Lanthanides constitutes the first inner transition series while actinides constitutes second inner transition series.
11
d & f - block
General Characteristics: 1. Electronic Configuration [Xe] 4f n 1 5d 0 6s 2 or [Xe] 4f n 5d1 6s 2 2. Oxidation state: They readily form M 3 ions some of them also exhibit oxidation state of +2 and +4. 3. Colouration: Ions of Lanthanides and actinides are coloured in the solid state as well as in aqueous solution because of absorbation of light due to f–f transition since they have partly filled f-orbitals.
13. LANTHANIDE CONTRACTION In lanthanides, the additional electron enters 4f-sub shell but not in the valence-shell namely sixth shell. The shielding effect of one electron in 4f-sub-shell by another in the same sub-shell is very little, being even smaller than that of d-electrons, because the shape of f-sub shell is very much diffused, while there is no comparable increase in the mutual shielding effect of 4f-electrons. This results in that electrons in the outermost shell experience increasing nuclear attraction from the growing nucleus. Consequently the atomic and ionic radii go on decreasing as we move from La57 to Lu71. Consequence of Lanthanide contraction 1. Atomic and ionic radii of post-Lanthanide elements: The atomic radii of second row transition elements are almost similar to those ofthe third row transition element because the increase in size on moving down the group from second to third transition elements is cancelled by the decrease in size due to lanthanide contraction. 2. High density of post lanthanide elements: It is because of very small size due to lanthanide contraction. 3. Basic strength of oxides and hydroxides: Due to lanthanide contraction the decrease in size of lanthanides ions, from La 3 to Lu 3 increases the covalent character (i.e. decreases the ionic character) between Ln 3 and OH – ions in Ln(III) hydroxides (Fajans rules). Thus La(OH) 3 is the most basic while Lu(OH)3 is the best basic. Similarly, there is a decrease in the basic strength of the oxides. 4. Seperation of Lanthanides: Due to the similar size (Because of lanthanide contraction) of the lanthanides, it is difficult to separate them. But slight variation in their properties is utilizedto separate Consequences of Lanthanide Contraction (i) Separation of Lanthanides.Separation oflanthanides is possible only due to lanthanide contraction. All the lanthanides have quite similar properties and due to this reason they are difficult to separate. However, because of lanthanide contraction their properties (such as ability to form complexes) varyslightly. This slight variation inproperties is utilized in the separation of lanthanides byion exchange methods. (ii) Variation in basic strength of hydroxides. The basic strength of oxides and hydroxides decreases from La(OH)3 to Lu(OH)3. Due to lanthanide contraction size of M3+ ions decreases and there is increase in the covalent character in M–OH bond. (iii) Similarity of second and third transition series. The atomic radii of second row of transition elements are almost similar to those of the third row of transition elements. For example, among the elements of group 3, there is normal increase in size from Sc to Yto La. But after lanthanides the atomic radii from second to third transition series do not increase as shown below for group 4 and group 5 i.e., for Zr – Hf and Nb – Ta pairs which have element same atomic radii.After group 5 the effect of lanthanide contraction is not so predominant.
12
d & f - block
5. Colour. The lanthanide metals are silvery white metals and the trivalent lanthanide ions are coloured both in the solid state and in the aqueous solution. 6. Magnetic properties. La3+ (4f0) and Lu3+ (4f14) having no unpaired electron do not show paramagnetism while all other tripositive ions of lanthanides are paramagnetic. 7. They have low ionisation energy and are highly electropositive. Their ionisation energy values are quite comparable with those of alkaline earth metals particularly calcium. 8. These metals do not have much tendency to form complexes. 9. The lanthanides are highly reactive. This is in agreement with the low values oftheir ionisation energies and electronegativity. 10. The solubilityof compounds of lanthanides follow the same order as group 2 elements. Their fluorides, oxides, hydroxides, carbonates are insoluble in water. However halides (except fluorides), nitrates, acetates are soluble in water. General Characteristics ofActinides – 5f Block Series – Second Inner Transition Series 1. These are silvery white metals with high melting and boiling points. 2. Besides the most common oxidation state of +3, actinides show +4, +5 and +6 oxidation states in certain elements. 3. All the actinides are radioactive in nature. 4. All the actinides are highly electropositive and as such are strong reducing agent. 5. Actinides have a strong tendency towards complex formation and form oxocations like UO22+, PuO22+, UO+ etc. 6. Most of the cations of actinides are coloured. 7. Actinides show actinide contraction i.e., decrease in ionic radii along the series.
13
d & f - block
SOLVED EXAMPLES Ex.1
Sol.
Ex.2 Sol.
On what basis can you say that scandium (Z = 21) is a transition element but since (Z = 30) is not. On t he basis of i incompletely filled 3d - orbitals in case of scandium atom in its ground state 3d1), it is regarded as a transition element. On the other hand, zinc atom has completely filled d- orbitals (3d10) in its ground state as well as in its oxidised state, hence it is not regarded as transition element.
(b) because Ce (IV) has extra stability due to empty f0 orbital. Ce3+ Ce4+ + e– [Xe] 4f1 5d06s0 (c) In Mn2+ 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 electron so Mn3+/Mn2+ redox couple has more positive potential than Fe3+/Fe2+(d14) couple. (d) Due to more negative enthalpy of hydration 2+ of Cu (aq) than Cu+ (aq) which compensates or second ionization enthalpy of copper. (e) In the third transition series after lanthanum their is lanthanoid contraction due to ineffective shielding by intervening f-orbital electrons and hence second and third transition series elements have similar atomic radii.
What is the effect of increasing pH on K2Cr2O7 solution? In aqueous solution, we have Cr2O72– + H2O 2CrO42– + 2H +. When pH < 4 (acidic medium), it exists as Cr2O72– and the colour is orange. When pH> 7 (basic medium), it exists as CrO42– and the colour is yellow. Ex.5
Ex.3
Sol.
Ex.4
Sol.
What is the basic difference between the electronic configurations of transition and inner Sol. transition elements. General electronic configuration of transition element = [ Noblegas] (n –1) d1–10 ns1–2 and for inner transition elements = (n – 2) f1–14 (n –1)d0–1 ns0–2. Thus , in transition elements, last electron enters d-orbitals of the Ex.6 penultimate shelll while in inner transition Sol. elements, it enters f- orbital of the penultimate shell. Give reasons for the following (a) Transition metals have high enthalpies of atomizations. (b) Among the lathanoids, Ce(III) is easily oxidised to Ce (IV) (c) Fe3+ | Fe2+ redox couple has less positive electrode potential than Mn3+| Mn2+ couple. (d) Copper (I) has d10 configuration , still copper (II) has d9 configuration stillcopper (II) is more stable in aqueous solution than copper (I). (e) The second and third transition series elements have almost similar atomic radii. (a) Due to strong interatomic interaction between valence electrons.
Ex.7
Sol.
Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration? Cr2+ is reducing as its configuration changes form d4 to d3, the later has half filled t 2g level. Again the change from Mn2+ to Mn3+ results in the half filled (d5) configuration whichhas extra stability. Cu+ ion is not stable inaqueous solution. Why? Cu2+ (aq) is much more stable than Cu+(aq). This is because although second ionization enthalpy of copper is large but hyd H for Cu2+ (aq) is much more negative than that for Cu+ (aq) and therefore it more than compensates for the second ionization enthalpy of copper. Thus many copper (I) compounds ar unstable in aque o us so lu t io n and und er go es disproportional as follows. 2Cu+ (aq) Cu2+(aq) + Cu(s) Why is the Eº value for the M3+/ M2+ couple much more positive than that of Cr3+/Cr2+ or Fe3+/Fe2+? Explain. Much larger third ionization energy of Mn (Where the required change is d5 to d4) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.
14
d & f - block
Ex.8 Sol.
Ex.9
Sol.
Why does Mn (II) ion show the maximum paramagnetic behvavious among bivalent ions of the first transition series.? The electronic configurations of Mn and Mn (II) ion are 2 2 6 2 6 2 2 25Mn : 1s 2s 2p 3s 3p 3d 4s , Mn+2 : 1s2,2s22p6 3s23p63d24s0 The Mn+2 ion has five unpaired electrons in its 3d subshell which is the maximum value for a transition metal ion. Hence, Mn (II) shows the maximum paramagnetic behaviour (due to unpaired electrons) among bivalent ions of the first transition series.
Q.13
Sol.
Q.14
Sol.
K2[PtCl6] is a will known compound whereas the corresponding Nicompound is not known. State reason for it. The oxidation state of Pt in K2 [PtCl6] is +4, which is a stable oxidation state for Pt. The +4 Q.15 oxidation state for Ni is verydifficult to achieve because the sum of the first four ionization energies is very high. Hence, the corresponding Sol. Ni (II) compound is not formed.
Ex.10 What is mean by ‘disproportionation‘ of an oxidation state? Give an example. Sol. When a particular oxidation state becomes less stable relative to other oxidation states, one loser one higher, it is said to undergo disproportionation. For example manganese (VI) becomes unstable relative to managanese (VII) and manganese (IV) in acidic solution. Q.16 3MnVIO42– + 4H+ 2MnVIIO– 4 + MnIVO 2 + 2H 2O Ex.11 Whichmetalinthe first seriesoftransition metals exhibits + 1 oxidation state most frequency and why? Sol. Copper metal (Cu, Z = 29) show + 1 oxidation Sol. state, i.e. exists as Cu+ in large number of copper compounds .e.g. cuprous oxide (Cu2O), cuprous sulphide (Cu2S), cuprous chlo r ide ( Cu 2 Cl2 ) et c. T he elect r o nic configuration Cu+ is [Ar] 3d104s0. All the five 3d orbitals are fully filled therefore this is a very stable configuration. Ex.12 Why is Ti+2 ionparamagnetic in nature? Sol. Ti+2 is paramagnetic because of the presence of two unpaired electrons in 3d-orbitals. 1s22s22p 63s23p 63d 24s 2, 22Ti : Ti+2 : 1s2,2s22p6 3s23p63d24s0
Arranged the following in increasing order of acdic character: CrO3, CrO, Cr 2O3 CrO< Cr2O3< CrO3. Higher the oxidation state , more will be acdic character. In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e. 125 kJ mol–1. Why? In the formation of metallic bond, no electrons from 3d-orbitals are involved in case of zinc while in all other metals of the 3d series, electrons from the d-orbitals are always involved inthe formation ofmetallic bonds. That is why, the enthalpy of atomisation of zinc is the lowest in the series. In transition series, with an increase in atomic number, the atomic radius does not change very much . Why is it so? With increase in atomic number along a transition series the nuclear charge increases which tends to decrease the size of the atom. But the addition of electrons in the d subshell increases the screening effect which counterbalances the increased nuclear charge. Hence alonga transition series the atomic radius does not very much. Account for the following : (i) Cerium (atomic number = 58) forms tetra positive , Ce4+ in aqueous solution. (ii) The second and third members in each group of transition element have similar atomic radii. (i) The electronic configuration of Ce (Z= 58) is 58Ce = [Xe]4f15d16s2 Cerium can lose four electrons (4f15d16s2) in aqueous solution to acquire stable configuration of rate gas xenon. Moreover due to small size and high charge , Ce4+ ion has high hydration energy. (ii) The second and third members in each group of transition elements have very similar atomic radii due to lanthanoid contration. It arises due to poor shielding effect of f electrons.
15
d & f - block
Ex.17 For the first row transition metals the Eº values are: Eº V Cr Mn Fe Co Ni Cu 2 – 0.91 –1.18 –0.44 –0.28 –0.25 +0.34 (M + /M) –1.18 Explain the irregularity in the above values. Sol. The Eº (M2+/M) values are not regular which can be explained from the irregular variation of ionisation entthalpies (iHl+ iH2 ) and alos the sublimation enthalpies which are relatively much less for maganese and vanadium. Ex.18 The Eº (M2+ /M) value for coppers is positive ( + 0.34V), What is possibly the reason for this ? Sol. Eº (M2+/M ) for any metal is related to the sum of the enthalpy change taking place in the following steps: M(s) + aH M(g) (a H= Enthalpy of atomisation) 2+ M(g) + iH M (g) (a H= Ionization enthalpy) 2+ 2+ M (g) + aq M (aq) + DhydH (hydH = Hydration enthalpy) Copper has high enthalpy of atomisation and low enthalpy of hydration . So Eº (Cu2+/Cu) is positive. The high energy to transform Cu(s) to Cu2+(aq) is not balanced by it s hydration enthalpy. Ex.19 (a) Complete the following chemical equation of for reactions: (i) MnO4– (aq) + S 2O 32– (aq) + H 2O(l) (ii) CrO7– (aq) + H2(g) + H+ (aq) (b) Give an explanation for each of the following observation: (i) The gradual decrease in size (actinoid contraction) from element to element is greater among the actionod is than that among the lanthanoids (lanthanoids contraction). (ii) The greatest number of oxidation states are exhibited by the members in the middle of a transition series. (iii) With the same d- orbital configuration (d4) Cr2+ ion is a reducing agent but Mn3+ ion is an oxidisng agent. Sol. (a) (i) In neutral or faintly alkaline solutions MnO4– + 2H 2O + 3e– MnO 2 + 2H 2P + 4O–H] × 8 S2O32– + 10O–H 2SO 42– + 5H 2O + 8e–] × 3 8Mn O32– + 10O–H + H 2O 8 MnO 2 + 6SO 42– + 2O–H (ii) In acidic solutions Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H 2O H2S S + 2H+ + 2e–] × 3 Cr2O72– + 3H 2S+ 8H+ 2Cr3+ + 3S + 7H 2O (b) (i) This due to poor shielding by 5f electrons from element to element in actinoid than by 4f electrons in lanthanoids series. (ii) This is due to involvement of both (n–1) d and ns electrons in bonding, which are unpaired in maximum number at the middle of series. (iii) Cr2+ has the configuration d4 and easily changes to Cr3+ which has half t 2g configuration and hence move stable. Therefore Cr2+ is reducing. On the other hand, the change from Mn3+ to Mn2+ results inthe halffilled, d5 configuration whichhas extra stability. Therefore Mn3+ in oxidising.
16
d & f - block
Ex.20 (a) Complete the following chemical reaction equations: (i) MnO4–(aq) + C2O42– (aq) + H+ (aq) (ii) Cr2O72– (aq) + F2+ (aq) + H+ (aq) (b) Explain the following observation about the transition/inner transition elements: (i) There is general an increase in density of element form titanium (Z = 22) to copper (Z = 29). (ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition elements (3rd series).
Sol.
(iii) the members in the actinoid series exhibit a large number of oxidation states than the corresponding members in the lanthanoid sereis. (a) (i) In acidic medium: MnO4– + 8H+ + 5e– Mn2+ + 4H 2O] × 2 C2O42– 2CO 2 + 3e] × 5 2MnO4– +5C 42– + 16H+ 2M n2+ + 10 CO 2+ 5H 2O (ii) In acidic medium Cr4 O72– + 14H+ + 6e– 2Cr3+ + 7H 2O Fe2+ Fe3+ + e– ] × 6 Cr4 O72– + 6Fe2+ +14H+ 2Cr3+ + 6Fe3+ + 7H 2O (b) (i) On moving from titanium to copper in general atomic mass increases where as atomic size decreases, therefore density increases in general (ii) The frequent metal-metal bonding in compounds of heavy transition elements is due to their high enthalpyof atomization. (iii) Thisisdueto verysmall energygap between5f, 6dand 7sorbitals inthe actinoids series.
Ex.21 Complete the following chemical reaction equations: (i) MnO4– (aq) + C2 O42–(aq) + H+ (aq) (ii) Cr2O72– (aq) + Fe2+(aq) + H+ (aq) Mn2+ + 4H 2O] × 2 Sol. (i) MnO4– + 8H+ + 5e– 2CO2 +2e– ] × 5 C 2O 42– 2MnO4– + 5C 2O 42– + 16 H+ (ii) Cr2O72– +14H+ + 6e– Fe2+ Cr2O72– + 6F2+ + 14H+
2Mn2+ + 10CO 2 + 8H 2O 2Cr3+ + 7H 2O Fe3+ +e–] × 6 2Cr3+ +6Fe3+ + 7H 2O
17
d & f - block
Ex.22 (a) What may be the possible oxidation states of the transition metals with the following d-electronic configuration in the ground state oftheir atoms. 3d3,s2,3d5,4a2 and 3d6 4s2. Indicate relative stability of oxidation states in each case. (b) Write steps involved in the preparation of (i) Mn2CrO4 from chromite ore and (ii) K2MnO4 from pyrolusite ore.
Sol.
(a)
Electronic conf iguration 3d24 s2
Element
Possible O.S.
Vanadium
+ 2, +3, + 4, + 5
+5
5
3d 4 s
2
Manganese
+ 2, +3, + 4, + 5,
+2 +7
6
2
Iron
+ 2, +3, + 4, +6,
+2 +3
3d 4 s
Mo re Stable O.S.
(b) (i) Chromite ore is fused with sodium carbonate in excess of air. 4FeCr2O4 + 8Na2CO3 + 7O 2 8Na 2CrO 4 + 2Fe 2O 3 + 8CO 2 chromite ore
Sod. chromate
(ii) Pyrolusite ore (MnO2) is fused with KOH in the presence of O 2 or oxidising agent such as KNO 3. 2MnO2 + 4KOH + O 2 2K 2MnO 4 + 2H 2O Pyrolusite ore
Postassium magnate
18
d & f - block
EXERCISE - I Q.1
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?
Q.2
Write complete equation for Oxidation of Fe2+ by Cr2O72– in acidic medium.
Q.3
Answer the following questions: (i) Which element in the first series of transition Q.12 elements does not exhibit variable oxidation states and why ? (ii) Whydo actinoids, in general exhibit a greater range of oxidation states than the lanthanoids ?
Q.4
Q.5
Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic Q.13 number? Write chemical equations for the reactions involved in the manufacture of potassium permanganate from pyrolusite ore.
Q.6
What is misch metal ? Mention its two important uses.
Q.7
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Q.8
(a) Name two properties of the central metal ion which enable it to form stable complex entities. (b)Account for the following : Zinc salts are while Cu2+ salts are coloured.
Q.9
Q.10
Q.11
How do you account for the following : (a)All scandium salts are white ? (b) The first ionization energies of the 5d transition elements are higher than those of the 3d and 4d transition elements in respective groups ? What may be the stable oxidation state of the transition element with the following d electron configurations inthe ground state oftheir atoms : 3d3, 3d5, 3d8 and 3d4?
Explain the following observations : (a) The elements of the d-series exhibit a larger number of oxidation sates than the elements of f-series. (b) The Cu+ salts are colurless while Cu2+ salts are coloured. Mention the direct consequence of the following factors on the chemical behavior of the transition elements : (i) They have incompletely d-orbitals in the ground state or in one of the oxidised states of their atoms. (ii) They contribute more valence electrons per atom in the formation of metallic bonds. What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Q.14
How would you account for the following situations ? (i) the transition metals generally formcoloured compound . (ii) With 3d4 configuration, Cu2+ acts as a reducing agent but Mn3+ acts as oxidizing agent (iii) The actinides exhibit a larger number of oxidation states than the corresponding lanthanides.
Q.15
How would you account for the following : (i) The transition elements have high enthalpies of atomisation. (ii) The transition metals and their compounds are found to be good catalysts in many processes.
Q.16
Explain giving reasons: (i) Transition metals and many of their compounds show paramagnetic behaviour. (ii) The enthalpies of atomisation of the transition metals are high. (iii) The transition metals generally form coloured compounds. (iv) Transition metals and their many compounds act as good catalyst.
19
d & f - block
Q.17
Give reasons for each of the following : Q.24 (i) Size of trivalent lanthanoid cations decreases with increase in the atomic number (ii) transition metal fluorides are ionic nature, whereas bromides and chlorides are usually Q.25 covalent in nature. (iii) Chemistry of all the lanthanoids is quite similar.
Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula. Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Q.18
Discuss the general characteristics of the 3d series of the transition elements with special reference to their. (i) atomic sizes, (ii) enthalpies of atomisation (iii) tendency for complex formation.
Q.26
Q.19
Predict which of the following willbe coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn3+, Fe3+ and Co3+. Give reasons for each.
Q.20
Write down electronic configuration of the following below: (a) La3+ (b) Gd3+ (c) Eu2+ (d) Zn4+ 2+ (e) Ru (f) Ce4+
Assign reason for the following : (i) The enthalpies of atomisation of transition metals are high. (ii) The transition metals and many of their compounds act as catalysts. (iii) From element to element, the actinides contraction is greater than lant hano id contraction. (iv) The Eº value for Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+. (v) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as transition element.
Q.27
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i) iodide (ii) iron(II) solution and (iii) H2S
Q.28
(a) Describe the general trends in the following properties of the first series of the transition elements : (i) Stability of +2 oxidation state (ii) Formation of oxometal ions. (b)Assign reason for each of the following : (i) Transition elements exhibit variable oxidation states. (ii) Transition metal ions are usually coloured
Q.21
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Q.22
(a) Give one example each of amphoteric and acidic oxides of transition metals. (b) Describe the trends in the following cases: (i) Melting points of elements in the 3d transition series. (ii) Atomic sizes of elements in the 4f transition series.
Q.23
How would you account for the following ? (i) Sc, the first member of first transition series Q.29 does not exhibit variable oxidation state. (ii) K2PtCl6 is well-known compound whereas corresponding compound of nickel is not known. (iii) Only transition metals for complex compounds with ligands such as CO.
Q.30
(a) Write the steps involved in the preparation of: (i) K Cr O from Na CrO 2 4 2 2 7 (ii) KMnO4 from K2MnO4 (b) What is meant by lanthanoid contraction ? What effect does it have on the chemistry of the elements which follow lanthanoids ?
For same of first row of trnasition elements the Eº values are : V Cr Mn Fe Co Ni – 1.18V – 0.91V – 1.18V – 0.44V – 0.28V – 0.25V
Cu 0.34V
Give suitable explanation for the irregular treand in these values.
20
d & f - block
EXERCISE - II Q.11
Q.1
Explain why the greenish solution of potassium manganate turns purple when CO2 is bubbled in the solution.
Q.2
Explain whymercurous compounds are formed when mercury is oxidised in a limited amount of an oxidising agent whereas with an excess of oxidising agent mercuric compounds are Q.12 formed.
Q.3
Explain why [Co(NH3)6]3+ is diamagnetic and [CoF6]3– is strongly paramagnetic.
Q.4
What happens when NaOH or NH4OH are added in excess to AlCl3 and ZnCl2?
Q.5
Whyis zinc oxide used in paints instead of lead salts?
Q.6
Identify from [A] to [E].
Colourless salt [A] + Ag NO 3 [E] white precipitate
NaOH
FeSO4 solution is mixed with (NH4)2SO4 in the molar ratio 1:1. It gives test of Fe2+ . When CuSO4 is mixed with liquid ammonia (in the ratio 1:4) the mixture does not give test of Cu2+. Explain the difference.
Q.14
(A), (B), and (C) are three complexes of Cr(III). Its formula is H12O6Cl3Cr. All the three complexes have water and chloride ions as ligands. (A) does not react with conc. H 2SO 4 whereas (B) and (C) lose 6.75% and 13.5% of their original weight respectivelyon treatment with conc. H2SO4. Find [A], [B], and [C].
Q.15
A metal complex having composition Cr (NH3)4Cl2.Br. has been isolated in two forms (A) and (B). (A) reacts with AgNO3 to give a white precipitate soluble in dilute ammonia while (B) gives a pale yellow precipitate soluble in concentrated ammonia. Write the formulae of (A) and (B) and hybridisation state of Cr in the compounds.
[B] (white precipitate)
dissolves in NaOH [C] Pass H 2 S
Q.7
Why are melting and boiling points of zinc, cadmium, and mercury lower than those of other transition metals?
Q.8
Why is HCl not used to acidify KMnO4 for volumetric estimations?
Q.9
Colourless salt (A) ? (B) + (C) gas. (B) dissolves both in acid and alkali solution. Q.16 (C) turns lime water milky and acidified K2Cr2O7 solution green. (A) gives white precipitate (D) with H2S when the solution is alkaline. Identify [A] to [D]. (a) K2MnO4 in acidic medium changes to MnO2 and KMnO4. What would be the equivalent weight of K2 MnO4. (b) Draw the structures of MnO4– and CrO42– ions.
[Ni Cl4]2– and [Ni (CO)4] both are tetrahedral in shape but [Ni Cl4]2– is paramagnetic whereas [Ni(CO) 4] is diamagnetic. Explain the difference in magnetic behaviour of both the complexes.
Q.13
[D] (white precipitate)
Q.10
Acomplex has the formula Pt Cl4.2KCl. The electrical conductance of the compound shows that each formula unit has 3 ions. AgNO3 on treatment with the complex does not give a precipitate ofAgCl. What should be the correct formula of the complex?
A monomeric compound of cobalt gives the following data on quantitative analysis: Co3+: 21.24% ; NH3: 24.77% ; Cl–: 12.81% ; SO42– : 34.65% ; H2O: 6.53%. Deduce the empirica l formula of the complex and the possible isomers.
21
d & f - block
EXERCISE - III Q.8
During estimation of oxalic acid Vs KMnO4, self indicator is (A) KMnO4 (B) oxalic acid (D) MnSO4 (C) K2SO4
Q.9
Haber’s process, Mo is used as (A) a catalyst (B) a catalytic promoter (C) an oxidising agent (D) as a catalytic poison
Acompound of mercury used in cosmetics, in Ayurvedic and Yunani medicines and known as Vermilon is (A) HgCl2 (B) HgS (D) HgI (C) Hg2Cl2
Q.10
.solvent Acidified chromic acid + H2 O2 Org. X + Y , X and Y are (A) CrO5 and H2O (B) Cr2O3 and H2O (C) CrO2 and H2O (D) CrO and H2O
Cr2O 27
Q.11
Transition elements are usually characterised by variable oxidation states but Zn does not show this property because of (A) completion of np-orbitals (B) completion of (n–1)d orbitals (C) completion of ns-orbitals (D) inert pair effect
Q.12
(NH4)2Cr2O7 (Ammonium dichromate) is used in fire works. The green coloured powder blown in air is (A) Cr2O3 (B) CrO2 (C) Cr2O4 (D) CrO3
Q.13
The d-block element which is a liquid at room temperature, having high specific heat, less reactivity thanhydrogen and its chloride (MX2) is volatile on heating is (A) Cu (B) Hg (C) Ce (D) Pm
Q.14
Coinage metals show the properties of (A) typical elements (B) normal elements (C) inner-transition elements (D) transition element
Q.15
Iron becomes passive by ..................... due to formation of ..................... (A) dil. HCl, Fe2O3 (B) 80% conc. HNO3, Fe3O 4 (C) conc. H2SO4, Fe 3O4 (D) conc. HCl, Fe3O4
Single Correct Questions Q.1
Q.2
Q.3
The number of moles of acidified KMnO4 required to convert one mole of sulphite ion into sulphate ion is (A) 2/5 (B) 3/5 (C) 4/5 (D) 1 N2 ( g) + 3H 2 ( g)
2NH 3 ( g) ;
2CrO 2 , X and Y are 4
respectively (A) X = OH–, Y = H+ (B) X = H+, Y = OH– (C) X = OH–, Y = H 2O 2 (D) X = H2O2, Y = OH – Q.4
CrO3 dissolves in aqueous NaOH to give (A) Cr2O7 2– (B) CrO42– (C) Cr(OH)3 (D) Cr(OH)2
Q.5
An ornamental of gold having 75% of gold, it is of .............. carat. (A) 18 (B) 16 (C) 24 (D) 20
Q.6
Q.7
Solution of MnO4– is purple-coloured due to (A) d-d-transition (B) charge transfer from O to Mn (C) due to both d-d-transition and charge transfer (D) none of these Transition elements having more tendency to form complex than representative elements (s and p-block elements) due to (A) availability of d-orbitals for bonding (B) variable oxidation states are not shown by transition elements (C) all electrons are paired in d-orbitals (D) f-orbitals are available for bonding
22
d & f - block
Q.16
Bayer’s reagent used to detect olifinic double bond is (A) acidified KMnO4 (B) aqueous KMnO4 (C) 1% alkaline KMnO4 solution (D) KMnO4 in benzene
Q.23
1 mole of Fe2+ ions are oxidised to Fe3+ ions with the help of (in acidic medium) (A) 1/5 moles of KMnO4 (B) 5/3 moles of KMnO4 (C) 2/5 moles of KMnO4 (D) 5/2 moles of KMnO4
Q.17
The transition metal used in X-rays tube is
Q.24
The metals present in insulin and haemoglobin are respectively
(A) Mo (C) Tc
(B) Ta (D) Pm
(A) Zn, Hg (C) Co, Fe
Q.18
Cu + conc. HNO3 Cu(NO3)2 + X (oxide Q.25 of nitrogen); then X is (B) NO2 (A) N2O (D) N2O3 (C) NO
Q.19
When KMnO4 solution is added to hot oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time. This is because (A) Mn2+ acts as auto catalyst (B) CO2 is formed (C) Reaction is exothermic (D) MnO4– catalyses the reaction.
Q.20
Q.21
CuSO4 solution reacts with excess KCN to give (A) Cu(CN)2 (B) CuCN (C) K2[Cu(CN)2] (D) K3[Cu(CN)4] The higher oxidation states of transition elements are found to be in the combination withAand B, which are (A) F, O (B) O, N (C) O, Cl (D) F, Cl
(B) Zn, Fe (D) Mg, Fe
The rusting of ironis formulated as Fe2 O3 ·xH2 O which involves the formation of (A) Fe2O3 (B) Fe(OH)3 (C) Fe(OH)2 (D) Fe2O3 + Fe(OH)3
Q.26
Metre scales are made-up of alloy (A) invar (B) stainless steel (C) elektron (D) magnalium
Q.27
Ametal M which is not affected by strong acids like conc. HNO3, conc. H2SO4 and conc. solution of alkalies like NaOH, KOH forms MCl3 whichfinds use for toning in photography. The metal M is (A)Ag (B) Hg (C)Au (D) Cu
Q.28
Solid CuSO4·5H2O having covalent, ionic as well as co-ordinate bonds. Copper atom/ion forms ................. co-ordinate bonds with water. (A) 1 (B) 2 (C) 3 (D) 4
Q.29
KMnO4 + HCl H2O + X(g), X is a (acidified)
Q.22
Pick out the incorrect statement: (A) MnO2 dissolves in conc. HCl, but does not form Mn4+ ions (B) MnO2 oxidizes hot concentrated H2SO4 liberating oxygen (C) K2MnO4 is formed when MnO2 in fused Q.30 KOH is oxidised by air, KNO3 , PbO2 or NaBiO3 (D) Decomposition of acidic KMnO4 is not catalysed by sunlight.
(A) red liquid (B) violet gas (C) greenish yellow gas (D) yellow-brown gas Purple of cassius is: (A) Pure gold (B) Colliodal solution of gold (C) Gold (I) hydroxide (D) Gold (III) chloride
23
d & f - block
Q.31
Q.32
Q.33
Q.34
Q.35
Q.36
Q.37
Amongst the following species, maximum covalent character is exhibited by (A) FeCl2 (B) ZnCl2 (C) HgCl2 (D) CdCl2
Multiple Correct Question Q.38
Number of moles of SnCl2 required for the reduction of 1 mole of K2Cr2O7 into Cr2O3 is (in acidic medium) (A) 3 (B) 2 (C) 1 (D) 1/3
Potash alumis a double salt, its aqueous solution shows the characteristics of (A) Al3+ ions (B) K+ ions (C) SO42– ions (D)Al3+ ions but not K+ ions
Q.39
The aqueous solution of CuCrO4 is green because it contains (A) green Cu2+ ions (B) green CrO42– ions (C) blue Cu2+ ions and green CrO42– ions (D) blue Cu2+ ions and yellow CrO42– ions
Addition of non-metals like B and C to the interstitial sites of a transition metal results the metal (A) of more ductability (B) of less ductability (C) less malleable (D) of more hardness
Q.40
Mercury is a liquid at 0°C because of (A) veryhigh ionisation energy (B) weak metallic bonds (C) high heat of hydration (D) high heat of sublimation
Q.41
The ionisation energies of transition elements are (A) less than p-block elements (B) more than s-block elements (C) less than s-block elements (D) more than p-block elements
Q.42
The metal(s) which does/do not form amalgam is/are (A) Fe (B) Pt (C) Zn (D)Ag
Q.43
The highest oxidation state among transition elements is (A) + 7 by Mn (B) + 8 by Os (C) + 8 by Ru (D) + 7 by Fe
Q.44
Amphoteric oxide(s) is/are (A) Al2O3 (B) SnO (D) Fe2O3 (C) ZnO
Q.45
Interstitial compounds are formed by (A) Co (B) Ni (C) Fe (D) Ca
Manganese steel is used for making railway tracks because (A) it is hard with high percentage of Mn (B) it is soft with high percentage of Mn (C) it is hard with small concentration of manganese with impurities (D) it is soft with small concentration of manganese with impurities In nitroprusside ion, the iron exists as Fe2+ and NO as NO + rather than Fe 3+ and NO respectively. These formsofions are established with the help of (A) magnetic moment in solid state (B) thermal decomposition method (C) by reaction with KCN (D) by action with K2SO4 Transition elements in lower oxidation states act as Lewis acid because (A) they form complexes (B) they are oxidising agents (C) they donate electrons (D) they do not show catalytic properties
The Ziegle r-Natta cat alys t used for Q.46 polymerisation of ethene and styrene is TiCl4 + (C2H5)3Al, the catalysing species (active species) involved in the polymerisation is (A) TiCl4 (B) TiCl3 (C) TiCl2 (D) TiCl
The catalytic activity of transition elements is related to their (A) variable oxidation states (B) surface area (C) complex formation ability (D) magnetic moment
24
d & f - block
Q.47
In the equation: M + 8CN– + 2H2O + O2 4[M(CN)2]– + 4OH –, metal M is (A)Ag (B)Au (C) Cu (D) Hg
Statement Type Question
Q.48
An element of 3d-transition series shows two oxidation states x and y, differ by two units then (A) compounds in oxidation state x are ionic if x>y (B) compounds in oxidation state x are ionic if x
Q.49
To an acidified dichromate solution, a pinch of Na2O2 is added and shaken. What is observed: (A) blue colour (B) Orange colour changing to green (C) Copious evolution of oxygen (D) Bluish - green precipitate
Q.50
Amongst CuF2, CuCl2 and CuBr2 (A) only CuF2 is ionic (B) both CuCl2 and CuBr2 are covalent (C) CuF2 and CuCl2 are ionic but CuBr2 is covalent (D) CuF2, CuCl2 as well as CuBr2 are ionic
Q.51
CuSO4(aq) + 4NH3 X, then X is (A) [Cu(NH3)4]2+ (B) paramagnetic (C) coloured (D) of a magnetic moment of 1.73 BM
Q.52
Amphoteric oxide(s) of Mn is/are (A) MnO2 (B) Mn3O4 (D) MnO (C) Mn2O7
Q.53
Acidified KMnO4 can be decolourised by (A) SO2 (B) H2O2 (C) FeSO4 (D) Fe2(SO4)3
Q.54
The lanthanide contraction is responsible for the fact that (A) Zr and Hf have same atomic sizes (B) Zr and Hf have same properties (C) Zr and Hf have different atomic sizes (D) Zr and Hf have different properties
Q.55
Ion(s) having non zero magnetic moment (spin only) is/are (A) Sc3+ (B) Ti3+ (C) Cu2+ (D) Zn2+
Questions given below consist of two statements each printed as Assertion (A) and Reason (R); while answering these questions you are required to choose any one of the following four responses: (A) if both (A) and (R) are true and (R) is the correct explanation of (A) (B) if both (A) and (R) are true but (R) is not correct explanation of (A) (C) if (A) is true but (R) is false (D) if (A) is false and (R) is true Statement-1: K2CrO4 has yellow colour due to charge transfer. Statement-2: CrO2 ion is tetrahedral in 4 shape.
Q.57
Statement-1: The highest oxidation state of chromium in its compounds is +6. Statement-2: Chromium atom has only six electrons in ns and (n–1) d orbitals.
Q.58
Statement-1 : Tungsten has a very high melting point. Statement-2 : Tungsten is a covalent compound.
Q.59
Statement-1: Equivalent mass of KMnO4 is equal to one-third of its molecular mass whe n it acts asan oxidising gent inan alkaline medium. Statement-2: Oxidation number of Mn is +7 in KMnO4.
Q.60
Statement-1: Cu+ ion is colourless. Statement-2: Four water molecules are coordinated to Cu+ ion.
25
d & f - block
ANSWER KEY EXERCISE - II Q.1 Q.2
Formation of MnO4–
Q.3
NH3 - strong ligand causes pairing of electrons. So complex is diamagneitc. F– is a weak ligand and so pairing is not possible. [Al (OH)3 and NaAlO2; Zn(OH)2 and Na2ZnO2] also [Zn(NH3)4]2+
Q.4 Q.5
Due to disproportionation of Hg22+
[Zn S is white but PbS is black].
Q.6
[A] = ZnCl2; [B] = Zn(OH)2; [C] = Na2ZnO2; [D] = ZnS; [E] = AgCl l
Q.7 Q.8
All the three elements have d10
Q.9 Q.10 Q.11
See text for the reaction of HCl and KMnO4. [(A) = ZnSO3; (B) = ZnO ; (C) = SO2 ; (D) = ZnS] 3/2 M K2[Pt Cl6]
Q.12
Cl– is a weak ligand—so no pairing of electrons. CO is a strong field ligand and, therefore, causes pairing of electrons.
Q.13 Q.14
In the first case, a double salt is formed. In the second case, a complex is formed. So no test.
Q.15
(A) = [Cr(H2O)6]Cl3 ; (B) = [CrCl(H2O)5]Cl2 H2O (C) = [Cr Cl2(H2O)4]Cl2.2H2O (A) = [Cr(NH3)4 Cl Br]+ Cl– and (B) = [Cr(NH3)4Cl2]+ Br–.
Q.16
[Co(NH3)4 Cl(H2O)] SO4 ;[Co(NH3)4(SO4)(H2O)]Cl [Co(NH3)4(SO4)(Cl)] H2O
EXERCISE - III Q.1
A
Q.2
B
Q.3
A
Q.4
B
Q.5
A
Q.6
B
Q.7
A
Q.8
A
Q.9
B
Q.10
A
Q.11
B
Q.12
A
Q.13
B
Q.14
D
Q.15
B
Q.16
C
Q.17
A
Q.18
B
Q.19
A
Q.20
D
Q.21
A
Q.22
D
Q.23
A
Q.24
B
Q.25
D
Q.26
A
Q.27
C
Q.28
D
Q.29
C
Q.30
B
Q.31
C
Q.32
A
Q.33
D
Q.34
A
Q.35
A
Q.36
A
Q.37
B
Q.38
A,B,C
Q.39
B,C,D
Q.40
A,B
Q.41
A,B
Q.42
A,B
Q.43
B,C
Q.44
A,B,C
Q.45 A,B,C
Q.46
A,B,C
Q.47
A,B
Q.48
B,C
Q.49
A,C
Q.50
A,B
Q.51
A,B,C,D
Q.52
A,B
Q.53
A,B,C
Q.54
A,B
Q.55
B,C
Q.56
B
Q.57
A
Q.58
C
Q.59
B
Q.60
C
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