Prepared By : Ajay Pratap Singh, PGT -Chemistry. The Transition element exhibit the following characteristics: Electronic configuration

Transcription

1 [1] Those elements which have partially filled d-subshell in their elementary form or in their commonly occurring oxidation states are known as Transition elements. [2] The d-block elements are called Transition element because of their position between s-block and p-block element of the periodic table. [3] Zn, Cd and Hg are not regarded as Transition elements because the d- orbitals in these elements are completely filled in the ground state and in their common oxidation states. [4] Sc is the lightest Transition element and s is the heaviest one. The Transition element exhibit the following characteristics: They exhibit several oxidation states {due to low energy difference between ns and (n-1d subshell.} They form coloured ions (due to presence of unpaired electrons and d-d transition of electrons. The metal and their compound act as catalyst (due to their ability to exhibit multiple oxidation states. The elements and many of their compounds are paramagnetic.(due to presence of unpaired electrons. They have the ability to form complexes. (Due to highly charged ions and contain vacant d-orbitals. Electronic configuration The general Electronic configuration of the d- block element is [Inert gas] (n-1 d 1-10 ns 0-2. Exceptions in electronic configuration: 1. First (3d transition series (Sc-Zn Cr 24 [Ar] 3d 5 4s 1 & Cu 29 [Ar] 3d 10 4s 1 2. Second (4d transition series (Y-Cd Nb 41 [kr] 4d 4 5s 1 Mo 42 [kr] 4d 5 5s 1 Ru 44 [kr] 4d 7 5s 1 Rh 45 [kr] 4d 8 5s 1 Pd 46 [kr] 4d 10 5s 0 Ag 47 [kr] 4d 10 5s 1 Palladium (Pd a transition element having no electron in the s-subshell of the valence shell of the atom 3. Third (5d transition series (La-Hg Pt 78 [Xe] 5d 9 6s 1 Au 79 [Xe] 5d 10 6s 1 Exceptions in electronic configuration are either because of symmetry or nuclear -electron and electron - electron forces.why transition elements act as a catalyst? Explain with example. Transition metals show variable oxidation states, therefore they can form intermediate products of different reactant molecules. Transition elements are capable to form interstitial compounds due to which they can adsorb and activate the reacting molecules. Ex: - Contact process - vanadium pentoxide Catalytic hydrogenation - nickel Haber process - finely divided iron. 2 I S 2 2 Fe3 8 I S 4 ( Persulphate ion 2 Mechanism: 2 Fe 3 2 I 2 Fe I 2 2 Fe 2 S Fe 3 2 S I S I 2 2 S 2 4 Coinage metals (Cu,Ag and Au has a d 10 configuration but they are transition elements. In their most stable oxidation states, Cu 2 has a d 9 configuration, Au 3 has d 8 configuration i.e. they have an incomplete d- level. Prepared By : Ajay Pratap Singh, PGT -Chemistry Page - 1

2 xidation states Transition elements show variable oxidation states because the energy of ns orbital is very close to the energies of (n-1 orbital, therefore,the electron of ns and (n-1d orbital used for bond formation by loss of electron, oxidation state differ by 1 (but in case of p- block element o.s differ by 2 due to excitation of ns electron to np orbital. In transition elements 1 is the minimum oxidation state (shown by Cu, Ag, Hg and 8 is the maximum oxidation state (shown by Ru and s. Zero oxidation states is found only in complexes i.e. [Ni (C 4 ]. The transition elements in lower oxidation state form ionic bond, while in higher oxidation state forms covalent bonds i.e. in Cr 4 2- all the bonds are covalent in nature. In first transition series if any oxidation state, d-subshell has 0, 5 or 10 electrons, that oxidation state will be more stable, e.g Ti 4 (3d 0 is more stable than Ti 3( 3d 1, Mn 2 is more stable than Mn 3. Cu 2 is more stable than Cu because of lower reduction potential which is due to higher hydration energy. Highest oxidation state is shown in case of fluoride or oxide because fluorine(f 2 and oxygen ( 2 are strong oxidizing agent The maximum oxidation state increases down the group. The maximum oxidation state is shown by the element which occurs in the middle of the series (Mn and lesser number of oxidation state shown by the element of extreme Lower oxidation state are reducing agent and higher oxidation state are oxidizing agent Scandium (3 does not exhibit variation of oxidation state in its compound Why are Mn 2 compound more stable than Fe 2 towards oxidation to their 3 state? Mn 2 [Ar] 3d 5 4s 0 (stable Mn3 [Ar] 3d 4 4s 0 (Less stable Fe 2 [Ar] 3d 6 4s 0 (Less stable Fe3 [Ar] 3d 5 4s 0 (stable Why is platinum (lv state more stable than nickel (lv state? This is because the sum of the first four ionization energies (IE 1 1E 2 IE 3 IE 4 of much less than that of nickel.# Iron, cobalt and nickel are known as ferrous metal platinum is # Group no of Mn = 7,oxidation state of Mn = 7 in permanganate ion Mn 4 # Group no of Cr = 6,oxidation state of Cr = 6 in Cr and Cr 4. smium forms a 8 oxidation state compound with oxygen but no such compound with fluorine. Explain this behavior? This is because there is space around the osmium atom for four bonded oxygen atoms and not for eight fluorine atoms. ut of Bromine and oxygen,with which chromium will exhibit its highest oxidation state? xygen.because electro negativity of oxygen is more than bromine ut of fluorine and oxygen, with which manganese will exhibit its highest oxidation state? xygen.because xygen has the ability to form multiple bonds with manganese.the highest Mn fluoride is MnF 4. The highest Mn oxide is Mn (Mn is tetrahedrally surrounded by oxygen 2 7 including a Mn--Mn bridge. K 2 PtCl 6 is a well known compound whereas the corresponding Ni compound is not known. Give reason. The oxidation state of Pt in K 2 PtCl 6 is 4. The sum of the first four ionization energies ( IE 1 IE 2 IE 3 IE 4 of Pt is less than those of Ni. What is the nature of oxides formed when the transition metal is in its : Low oxidation state - Basic oxide (since ionic High oxidation state - Acidic oxide (since covalent Intermediate oxidation state -Amphoteric oxides As there are no unpaired electrons in Zn, Cd and Hg, so they are soft and have low M.P Page - 2

3 E 0 values of some metals are as follows Cr 2 /Cr = -0.9v, Cr 3 / Cr 2 = -0.4v,Mn 2 /Mn =-1.2v, Mn 3 / Mn 2 Fe 2 /Fe = -0.4v, Fe 3 / Fe 2 = 0.8v Comment upon : = 1.5v (i the stability of Fe 3 in acid solution as compared to that of Cr 3 ormn 3 (ii The ease with which iron can be oxidized as compared to the similar process for either chromium or manganese metal. (i E 0 of Fe 2 /Fe 3 = - 0.8v Cr 2 / Cr 3 = 0.4v Mn 2 / Mn 3 = -1.5v since oxidation potential of Fe 2 is more as compared to Mn 2.so Fe 3 is more stable than Mn 3. But oxidation potential of Cr 2 is more as compared to Fe 2.so Cr 3 is more stable than Fe 3. So stability order is Cr 3 > Fe 3 > Mn 3 (ii E 0 of Fe / Fe 2 = 0.4v, Mn/ Mn 2 = 1.2v Cr /Cr 2 = 0.9v since oxidation potential of Fe is less than as compared to Mn and Cr, so ease oxidation of iron is less than that of Cr and Mn. IE 1 IE 1 (MJ mol -1 IE 3 IE 4 (MJ mol -1 Ni Pt (idecide the most common oxidation state for Ni and Pt. (iiwhich can form compound in 4 oxidation states more easily? (i Ni 2 ; Pt 2 due to low IE 1 IE 1 value. (ii Pt, because IE 1 IE 1 IE 3 IE 4 of platinum is much lesser than that of Ni. Explain (a of the d 4 species Cr 2 is strongly reducing while Mn 3 is strongly oxidizing agent. (b explain how 2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number. (c The d 1 configuration is very unstable in ions. a E 0 Cr 3 / Cr 2 = V and Mn 3 / Mn 2 = 1.57 V The E 0 values Cr 3 / Cr 2 shows that Cr 2 has a tendency toward oxidation to Cr 3. Hence Cr 2 act as a reducing agent. The E 0 value Mn 3 / Mn 2 show that Mn 3 has a strong tendency for reduction. Hence it acts as an oxidising agent. R Cr 2 easily oxidised to Cr 3 because Cr 3 having configuration d 3 stable due to half filled t2g level(i.e. stability of Cr 3 > Cr 2 Mn 3 easily reduced to Mn 2 because Mn 2 having configuration d 5 stable due to half filled d- orbital. (I.e. Mn 2 is more stable than Mn 3. (b The stability of 2 oxidation increases from Sc to Mn because of increase in effective nuclear charge. In case of Mn, electrons are removed from 4s - orbital, leaving 3d as exactly half filled subshell. (c d 1 configuration has low ionization energy and the energy evolved during solvation is sufficient to ionize it. Why transition elements form complexes? The transition element form complexes because of the following reason:- (a Due to small size and high effective nuclear charge they can attract the electron pairs of Ligand molecules. (b They have vacant d-orbital of appropriate energies to accommodate the electron pairs of Ligands. Page - 3

4 What is meant by 'disproportionation'? When a particular oxidation state of an element becomes less stable relative to other oxidation state, one lower and one higher, it is said to undergo disproportionation. For example, in aqueous solution: Cu disproportionate into Cu 2 and Cu 2Cu Cu 2 Cu 3 Mn 4 4 H 2 Mn 4 Mn 2 2 H 2 Why do the d-block element exhibit a large number of oxidation state than the f-block element? The electronic configuration of d- block element is (n-1 d 1-10 ns 0-2.there is not much difference between the energy of (n-1 d and the ns-subshell. Therefore, both the ns and the (n-1 d electron can take part in bond formation. Hence d-block element exhibit a number of oxidation states. In f-block element, the electron enters the f-orbital of the second outermost shell.due to large difference between the energies of the ns and (n-2 f-subshell, all the (n-2 f electron can not take part in bond formation. Hence, they exhibit lower number of oxidation state. Why do transition elements have high enthalpy of hydration? The high enthalpy of hydration of transition elements is due to : (i small size of the cation and (ii large positive charge Ionization energy Ionization energy increases from Sc to Zn. Cu and Cr have high I.E due to stable configuration. The first Ionization energy of 5d element are higher than those of 3d and 4d element.this is due to the greater effective nuclear charge acting on outer valence electron because of poor shielding of the nucleas by 4f electron in 5d element. Explain cobalt (II is stable in aqueous solution but in presence of complexing reagent, it is easly oxidised. [ Co(H 2 6 ] 2 [ Co(H 2 6 ] 3 [ Co(Ln 6 ] 2 [ Co(Ln 6 ] 3 CFSE = 0.8 CFSE = 0.4 CFSE = 1.8 CFSE = 2.4 [ C.F spliting due to H 2 is less, So P > ] [ C.F spliting due to Ln is more, So P < ] Due to having higher value of crystal field stabilisation energy, Co 2 is stable in aqueous solution, on the other hand Co 3 is stable in presence of complexing agent. So Co (II oxidised to Co (III in presence of complexing agent. Why the first ionization energy of the element of the first transition series does not vary much with increasing atomic number? With the increasing atomic number d-electron added one by one in (n-1 shell.the screening effect of these d-electrons shield the outer s-electron from inward nuclear pull. The effect of the increase in nuclear charge with the increase in atomic number is opposed by the shielding effect of d-electron.thus due to these counter effect there is a very little variation in the values of ionization energy of the element of the first transition series Why transition elements form alloys? Alloys are formed by atoms with metallic radii that are within about 15 percent of each other. The. transition element form alloys with each other because they have almost similar sizes. Alloys. of transition metal- stainless steel (Fe, Cr, Ni Alloys of transition and non transition metals -Brass (Cu-Zn Bronze (Cu-Sn Alloy having more resistance to corrosion, harder with higher melting point than the constituent element. Why M.P/ enthalpy of atomisation of 5d > 4d > 3d? In a group from top to bottom,electropositive character of elements increases i.e. mobile electron density increases,so force of attraction between mobile electron and kernel increases and metallic bonding increases. For which M.P/enthalpy of atomisation of 5d > 4d > 3d Page - 4

5 Among the species Sc 3,.Ce 4, and Eu 2 which one is good oxidising agent? Sc 3 [A r] 3d 0 4s 0 Ce 4 [Xe] 4f 0 5d 0 6s 0 Eu 2 [Xe] 4f 7 5d 0 6s 0 In all the above ion the most stable oxidising state is 3.therefore Sc 3 will remain unchanged. Ce 4 has the tendency to accept one electron to get the 3 oxidation state. Hence Ce 4 is a good oxidising agent. Eu 2 has the tendency to lose one electron to get the 3 oxidation state. Hence is a good reducing agent. Why E 0 value for the Mn 3 /Mn 2 is much more positive than Fe 3 /Fe 2? Eu 2 Mn 3 e- Mn 2 Fe 3 e- Fe 2 (d 4 system less stable (d 5 system more stable (d 5 system more stable (d 6 system less stable Magnetic property The element and the compound of transition metal are generally paramagnetic due to presence of unpaired electron the magnetic moment is due to spin angular moment and orbital angular moment. Magnetic movement( = n(n2 Unit of is Bohr magneton ; n- number of unpaired electrons. Magnetic movement of Fe 3 ion (1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 Number of unpaired electron (n =5 5 (52 =5. 9 BM Why the compounds of transition element are coloured? The colour of transition element depend upon the unpaired electron present in the d-orbital. The unpaired electron excited from one energy level to another energy level within the same d- subshell. For this purpose (d -d transition, the energy is absorbed from visible region of radiation.the complementary part of absorbed light i.e. reflected light will decide the colour of the compound. For example Cu (II absorb red colour and reflect blue colour (complementy to red So the colour will be blue for Cu (II. Why transition elements form interstitial compounds? Due to partially filled d- orbitals and variable oxidation states, transition elements form interstitial compounds. In the interstitial compound small size atoms like hydrogen, carbon, nitrogen boron, etc occupy empty spaces of metal lattice (interstices. The small entrapped atoms in the interstices form the bonds with metals due to which malleability and ductility of the metal decreases, whereas tensile strength increases i.e. becomes harder. Retain metallic conductivity and becomes chemically inert, melting point increases. Ex: - steel is the interstitial compound of iron and carbon. Melting point of transition elements The m.p and b.p of transition element are very high due to stronger interparticle bonds. In a period from left to right the m.p of the metal first increases to maximum then gradually decreases towards the end of the period. ( Mn, Tc posses anamolous M.P The strength of interparicle bonds in transition elements is roughly related to the number of half filled d- orbitals. In the begining the number of half filled d-orbitals increases till the middle of the period then decreases. So the strength of interparticle bonds intially increasesand at the end it is minimum. So melting point and enthalpy of atomisation increases from left to middle and then decreases at the right end in a period. Why Mn, Tc have low M.P as compared to other elements? In Mn and Tc, the d-orbitals are exactly half filled, so electrons are tightly held by the nucleus,so that delocalisation is less and metallic bonding is much weaker than preceding elements. Explain variation of atomic size from Sc to Zn Atomic size decreases from Sc to Cr due to increase in nuclear charge. After Cr, the screening effect of (n-1d electron on ns orbital increases. The increase of screening effect neutralise the increase nuclear charge, so atomic radius remains constant. In zinc due to interelectronic repulsion, the size of electron cloud increases, so atomic size increases. Page - 5

6 Potassium dichromate (K 2 Cr 2 7 The oxidation state of chromium in chromate ( Cr 4 2 and dichromate ( Cr is the same (6. They are interconvertible in aqueous solution depending upon P H of the solution. 2 Acid 2 2 Cr 4 2 H Cr 2 7 H 2 Alkali Yellow range i.e. 2 Cr Cr - 2 H 2 Cr 2 7 Chromate ion ( Tetrahedral H 2 2 and Cr 2 7 Preparation of Potassium dichromate (K 2 Cr Cr - Cr 2 H 2 Cr 4 2 H 2 Dichromate ion [ Two tetrahedra sharing one corner with Cr--Cr bond angle ] K 2 Cr 2 7 is prepared from chromite ore ( FeCr 2 4 [1] Chromite ore is converted into sodium chromate by fusing with sodium carbonate in presence of air 4 FeCr Na 2 C Na 2 Cr 4 2 Fe C 2 [2] After filteration the yellow solution of sodium chromate is treated with concentrated H 2 S 4 to ge sodium dichromate. 2 Na 2 Cr 4 H 2 S 4 Na 2 Cr 2 7 Na 2 S 4 H 2 Sodium chromate Sodium dichromate [3] Hot concentrated solution of sodium dichromate is treated with potassium chloride. Then potassium dichromate, which is less soluble than sodium dichromate,crystalises out on cooling as orange crystal. Na 2 Cr KCl K 2 Cr NaCl xidising action of Potassium dichromate (K 2 Cr Acidified potassium dichromate Cr H 6 e 2 Cr 3 7 H 2 [1] Iodide in to Iodine 2 I I 2 2 e X 3 2 Cr H 6 e 2 Cr 3 7 H 2 2 Cr I 14 H 2 Cr 3 3 I 7 H 2 2 [ Ionic ] will oxidise [Molecular] [2] Sulphides to sulphur H 2 S 4 K 2 S 4 X 3 K 2 Cr H 2 S 4 K 2 S 4 Cr 2 (S H 2 3 [] 2 KI [] I 2 H 2 K 2 Cr H 2 S 4 6 KI 4 K 2 S 4 7 H 2 3 I Cr 2 (S H 2 S 2 H S 2 e X 3 2 Cr H 6 e 2 Cr 3 7 H 2 2 Cr 3 7 H 2 Cr H 3 H 2 S 3 S [ Ionic ] Page - 6

7 [Molecular] K 2 Cr H 2 S 4 K 2 S 4 Cr 2 (S H 2 3 [] H 2 S [] H 2 S X 3 K 2 Cr H 2 S 4 3 H 2 S K 2 S 4 Cr 2 (S H 2 3 S [3] Tin (II salt to Tin (IV salt 2 Sn Sn 4 2 e X 3 2 Cr H 6 e 2 Cr 3 7 H 2 Cr Sn 2 14 H 2 Cr 3 3 Sn4 7 H 2 [ Ionic ] [Molecular] K 2 Cr HCl 2 KCl 2 CrCl 3 4 H 2 3 [] SnCl 2 2 HCl [] SnCl 4 H 2 X 3 K 2 Cr H 2 3 SnCl 2 14 HCl 2 KCl 2 CrCl 3 3 SnCl 4 [4] Iron (II salts to Iron (III salts 2 3 Fe Fe e X 6 Cr H 6 e 3 2 Cr 7 H 2 Cr [Molecular] [ Ionic ] 14 H 6 Fe 2 2 Cr 3 6 Fe3 7 H 2 [] H 2 X 3 K 2 Cr H 2 S 4 K 2 S 4 Cr 2 (S H 2 3 [] 2 FeS 4 H 2 S 4 Fe 2 (S 4 3 K 2 Cr H 2 S 4 6 FeS 4 K 2 S 4 Cr 2 (S Fe 2 (S H 2 Potassium Permanganate ( KMn 4 2 KMn Heat 4 K 2 Mn 513K 4 Mn 2 2 Preparation of KMn 4 KMn 4 is prepared from pyrolusite ( Mn 2 Permanganate ion [ Purple] Mn ( Diamagnetic Mn Manganate ion [ Green] ( Paramagnetic with one unpaired electron The mineral is fused with KH in presence of air or an oxidising agent ( KN 3, KCl 3, etc. to produce potassium manganate. 2 Mn 2 4 KH 2 2 K 2 Mn 4 2 H 2 K 2 Mn 4 is converted into KMn 4 by any of the following methods:- ; 2 H 2 [1] By acidification of K 2 Mn 4 by H 2 S 4 3 K 2 Mn 4 2 H 2 S 4 2 K 2 S 4 2 KMn 4 Mn 2 2 R 3 Mn 4 4 H 2 Mn 4 Mn 2 2 H 2 [2] By electrolytic oxidation of K 2 Mn 4 H 2 H H K 2 Mn 2 K Mn ; At anode:- Mn 4 Mn 4 e X 2 At cathode:- 2 H 2e H Mn 4 Green 2 H 2 Mn 4 H 2 Purple R 2 K 2 Mn 4 2 H 2 2 KMn 4 H 2 2 KH 2 Page - 7

8 Laboratory preparation of KMn 4 A manganese (II ion salt is oxidised by peroxodisulphate to permanganate. 2 Mn S H 2 2 Mn 4 10 S 4 16 H 2 xidising action of KMn 4 Acidified potassium permanganate Mn 4 8 H 5 e Mn2 4 H 2 will oxidise:- [1] Iodide to iodine 2 I I 2 2 e X 5 Mn 4 8 H 5 e Mn2 4 H 2 X 2 2 Mn 4 10 I 16 H 2 Mn 2 5 I 2 8 H 2 [2] Ferrous to ferric 2 Fe Fe 3 e X 5 Mn 4 8 H 5 e Mn2 4 H 2 Mn 4 5 Fe2 8 H 5 Fe 3 Mn2 4 H 2 It is not possible to titrate HCl with KMn 4, because HCl will oxidise to Cl 2 C [3] xalate ion or oxalic acid into carbondioxide 10 C 2 10 e 5 C Mn 4 8 H 5 e Mn2 4 H 2 X 2 C 5 2 Mn 4 C [4] Sulphide into sulphur 2 S S 2 e X 5 Mn 4 8 H 5 e Mn 2 4 H 2 X 2 5 S 2 2 Mn 4 16 H 5 S 2 Mn 2 8 H 2 [5] Sulphurous acid / sulphite into sulphuric acid / sulphate S 3 H 2 S 4 2 H 2 2 e X 5 2 Mn 4 8 H 5 e Mn 4 H 2 2 X S 3 2 Mn H 2 Mn 5 S H 2 [6] Nitrite to nitrate N 2 H 2 N3 2 H 2 e Mn 4 8 H 5 e Mn2 4 H 2 X N 2 2 Mn 4 6 H 2 Mn 5 N 3 3 H 2 Neutral / Faintly alkaline Solution ( Mn 4 16 H 10 C 2 2 Mn 2 8 H 2 Mn 2 oxidises: [1] Iodide to iodate: I 6 H I 3 3 H 2 6 e Mn 4 2 H 2 3 e Mn 2 4 H X 2 I 2 Mn 4 H 2 I 3 2 Mn 2 2 H 2 [2] Thiosulphate to sulphate: S H 2 S4 5 H 2 8 e Mn 4 X 5 2 H 2 3 e Mn 2 4 H X 8 H 2 8 Mn 2 6 S 4 2 H 8 Mn 4 3 S X 3 Page - 8

9 [3] Manganous salt is oxidised to Mn 2 in presence of ZnS 4 or Zn Mn H 2 Mn 2 Mn 4 H 2 e 4 H 3 e Mn 2 2 H 2 X 2 3 Mn 2 2 Mn 4 2 H 2 5 Mn 2 4 H The Inner Transition Elements ( f- Block Lanthanoids ( 4f-Block 4f d 0,1 6 s 2 Elements Z M 3 [Xe] 4 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 f 9 f 10 f 11 f 12 f 13 f 14 All elements having no d- electrons in their penultimate shell, except the following: Ce [Xe] 4f 1 5d 1 6s 2 Gd [Xe] 4f 7 5d 1 6s 2 Lu [Xe] 4f 14 5d 1 6s 2 Lanthanum ( La [Xe] 5d 1 6s 2, though a d- block element, is included in the lanthanoid (Ln series because it closely resembles lanthanoids. The more stable oxidation state of lanthanoids is 3. So Ln 2 or Ln 4 species revert to Ln 3 by loss or gain of an electron. So aqueous solution of Sm 2,Eu 2 and Yb 2 are good reducing agents and aqueous solution of Ce 4 and Tb 4 are good oxidising agents. Lanthanoids show limited number of oxidation states because the energy gap between 4f and 5d subshell is large. Lanthanoid Contraction The regular decrease in the size of the atoms and ions with increasing atomic number is known as lanthanoid contraction. Cause:- As the new electron is added into the f- subshell, there is imperfect shielding of one electron by another in this subshell due to the shapes of these f- orbitals. This imperfect shielding is unable to counterbalance the effect of the increased nuclear charge. Hence, the net results is the contraction of atomic size. Consequences:- Similarity in the size of the atoms belonging to same group of second and third transition series 1St Transition series Sc 21 (144pm Ti 22 (132pm V 23 (122pm 2nd Transition series 3rd Transition series Y 39 (180pm La 57 (187pm LANTHANIDS(58-71 Zr 40 (160pm Hf 72 (159pm X 3 Nb 41 (146pm Ta 73 (146pm As the size of the lanthanoid ions decreases from La 3 to Lu 3, the covalent character of of the hydroxides increases ( Fajan's rule. Hence, the basic strength decreases from La(H 3 to Lu(H 3. The valence shell configurationof lanthanoids remains the same, because the electrons are added into the 4f - subshell. Hence, they show similar chemical properties. Due to lanthanoid contraction, the change in atomic or ionic radii of these elements is very small. This makes the separation of lanthanoid elements difficult. All the lanthanoids except La 3 and Lu 3 contain unpaired electrons and hence are paramagnetic. Misch metall is an alloy, consists of lanthanoid and Iron and traces of S,C,Ca and Al. It is used to produced bullets, shell and lighter flint. Page - 9

10 Chemical reactions of Lanthanoids Halogen LnX C, 2773 K Ln 3 3 C, Ln 2 C 3 and LnC 2 Heated with N Dilute acid H 2 gas Ln 2 LnN Heated with S Ln 2 S 3 Burn in Ln H 2 Ln(H 3 H 2 Actinoids ( 5f-Block 5f d 0,1,2 7 s 2 Elements Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Z M 3 [Rn] 5 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 f 9 f 10 f 11 f 12 f 13 f 14 All elements having no d- electrons in their penultimate shell, except the following: Pa [ Rn] 5f 2 6d 1 7s 2 U [ Rn] 5f 3 6d 1 7s 2 Np [ Rn] 5f 4 6d 1 7s 2 Cm [ Rn] 5f 7 6d 1 7s 2 Lr [ Rn] 5f 14 6d 1 7s 2 Th 90 [ Rn] 6d 2 7s 2 Last electron enter into d- orbital of (n-1d subshell,instead of(n-2f subshell.but 'Th' is included in f-block,, because Th 3 contain electron in (n-2f subshell. Th 3 [ Rn] 5f 1 Actinium ( Ac 89 [Rn] 6d 1 7s 2 though a d- block element, is included in actinoid series, because it closely resembles actinoids The chemistry of actinoids is much more complicated,because they show a wide range of oxidation states and they are radioactive in nature. Application of d- and f- block elements Group-11 elements are called coinage metal. UK 'Copper' coins - Copper coated steel Silver UK coin -C / Ni alloy Ziegler catalyst- TiCl 4 with Al ( CH 3 3 for manufacture of polythene PdCl 2 - Wacker's process for the oxidation of ethyne to ethanol. LANTHANIDS Show 2 and 4.S besides the stable 3.S. V Manufacture of H 2 S 4 Ti - Pigment industry Mn 2 - Dry battery cells AgBr - Photography industry ACTINIDS Show 4, 5, 6 and 7.S besides the stable 3.S Most of their ions are colourless Less tendency toward complex formation. Compounds are less basic. Do not form oxocation. Except promethium, they are non-raadioactive Most of their ions are coloured Greater tendency toward complex formation. Compounds are more basic. Form oxocation e.g. Pu 2 2, U They are raadioactive Magnetic properties can be explained easily. Magnetic properties can not be explained easily, as they are more complex. Page - 10