D and F Block Elements: Properties, Configuration PDF

d & f–Block Elements d-Block Elements y The d-block of the periodic table contains the elements of the group 3-12 in which the Definitions d-orbitals are progressively filled in each of Elements in which last electron...

d & f–Block Elements d-Block Elements y The d-block of the periodic table contains the elements of the group 3-12 in which the Definitions d-orbitals are progressively filled in each of Elements in which last electron the four long periods. enters into d-subshell are y In these elements, the last electron enters (n known as d-block elements. – 1) d-subshell (d-orbitals of the penultimate orbit), i.e., (n – 1) d-subshell is gradually filled up. The configuration varies from (n – 1)d1ns2 to (n – 1)d10ns2. y These are present between s-block and p-block elements. The properties of these elements are intermediate between the Concept Ladder properties of s-block and p-block elements. y Forty elements belong to d-block. Fourth, Vertical relationship is due fifth, sixth and seventh periods consists of to similar outer electronic ten elements each. configuration while y There are mainly three series of the transition horizontal relationship is metals. due to combined effect of s (i) 3d series (Sc — Zn) and zeff. (ii) 4d series (Y — Cd) (iii) 5d series (La — Hg) (iv) 6d series (Ac — Uub) d & f–Block Elements 1. d & f–Block Elements 2. Transition Elements : y They are often called “Transition elements” Concept Ladder because their position in the periodic table between s-block and p-block elements. In d-block, 12th group y Typically, the transition elements have an elements like Zn, Cd and Hg incompletely filled d-level. Since 12th group do not have partially filled has d10 configuration and are not considered (n – 1) d—subshells either as transition elements but they are d-block in its atomic or ionic state. elements. That is why these elements y d-block elements show horizontal as well as not termed as transition vertical relationship. elements. y Zn, Cd and Hg are generally not regarded as transition elements. Ex: Zn, Cd and Hg Previous Year’s Question Zn 3d104s2 Cd 4d105s2 A transition element X has Hg 5d106s2 a configuration [Ar]3d4 in its y So out of 40 d-block elements there are 37 +3-oxidation state. Its atomic transition elements. d & f–Block Elements number is [AIPMT] y All transition elements are d-block elements (1) 25 (2) 26 but all d-block elements are not transition (3) 22 (4) 19 elements. 3.  Q.1 Why chromium is a transition element but zinc is not ? A.1 Transitions elements have incomplete d-orbitals chromium has [Ar]4s13d5 i.e. Partially filled d-orbitals but in case zinc having [Ar]4s23d10 i.e. fully filled d-orbitals. Hence, chromium is a transition elements whereas zinc is not. Electronic Configuration : General E.C. of d-block (n–1)d1-10ns0-2 Concept Ladder For 3d series 3d1-104s1-2 For 4d series 4d1-105s0-2 The d-orbitals of the For 5d series 5d1-106s1-2 transition element project For 6d series 6d1-107s1-2 to the periphery of an y Most of the elements possess two electrons in atom more than the other the outermost orbital, i.e., ns2. However, some orbitals (s and p). Hence, of the elements such as Cr, Cu, Nb, Mo, Ru, they are more influenced Rh, Ag, Pt, Au and Rg have one electron in the by the surrounding as well outermost orbital, i.e., ns1 while one element as affecting the atoms or palladium has no electron on ns orbital. molecules surrounding them. Q.2 Write the electronic configuration of following species. (i) Cr (ii) Zn A.2 (i) Cr – Atomic number (Z = 24) 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (ii) Zn – Atomic number (Z = 30) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 y The irregularities in the observed configuration of Cr [Ar]3d54s1 instead of [Ar]3d44s2, Cu Rack your Brain [Ar]3d104s1, Mo [Kr]4d55s1, Pd [Kr]4d105s0, Au d & f–Block Elements [Xe]4f145d106s1, Ag [Kr]4d105s1 are explained on Why copper is considered to the basis of the concept that half-filled and be transition element even completely filled d-orbitals is relatively more its atom has completely filled stable than partially filled d-orbitals. d-orbital in its ground state. 4. General Properties of d-Block Elements (Transition elements) : Concept Ladder (1) Variation in atomic and ionic sizes of transition metals : There is hardly increas in y On moving left to right in the period, atomic size with increas in generally atomic and ionic radius value atomic number, in a series decreases. of transition metals. y The atomic radii decrease from Sc to Cr because the effective nuclear charge (ENC) increases. y The atomic size of Fe, Co, Ni is almost same due to increase in nuclear charge is cancelled by the repulsion between the electrons and increase in shielding effect. y Cu and Zn have bigger size because the shielding effect decreases and electron- electron repulsion increases. Order of size of 3d series : (i) Covalent radius : Rack your Brain Sc ≥ Ti ≥ V ≥ Cr ≥ Mn ≥ Fe ≈ Co ≈ Ni ≤ Cu ≤ Zn    Zeff > σ Zeff ≈ σ Zeff < σ Why metallic bond strength of Cr is high, and its metallic radius is (ii) Metallic radius : low? Sc ≥ Ti ≥ V ≥ Cr < Mn > Fe ≈ Co ≈ Ni ≤ Cu ≤ Zn y The variation in covalent and metallic radius of Cr and Mn is due to the fact that in case of Mn two electrons participate in metallic bond formation whereas in case of Cr three electrons participate in formation of metallic bond, therefore metallic bond strength of Cr is high and its metallic radius is low. Ionic Radii : Previous Year’s Question y The ionic radii follow the same trend as the atomic radii. For the same oxidation state, the ionc radii generally decreases as the atomic Which of the following pairs has number increases in a particular transition the same size series  [AIPMT-2010] d & f–Block Elements y The ionic radii decreases with increase of (1) Zn2+, Hf4+ (2) Fe2+, Ni2+ charge on the ion. (3) Zr4+, Ti4+ (4) Zr4+, Hf4+ Ex Fe3+(0.64) < Fe2+(0.76) Ni3+­(0.62) < Ni2+(0.72) 5.  Lanthanoid Contraction y There is regular decrease in the atomic and ionic radii of the transition metals as the atomic number increases. This is because of Previous Year’s Question filling of 4f orbitals before the 5d orbitals. This contraction in size is quite regular. This Reason of lanthanoid contraction is called lanthanoid contraction. is y The conclusion of the Lanthanoid contraction  [NEET-2014] is that the 4d and 5d series exhibit (1) Decreasing nuclear charge similar radii and have very similar physical (2) Decreasing screening effect and chemical properties much more than the (3) Negligible screening effect of expected on the usual family relationship. ‘f’ orbitals y 14 lanthanides between La and Hf, [there is (4) Increasing nuclear charge d & f–Block Elements continuous decrease in size from Ce(58) to Lu(71)] Hf size becomes nearly equal to the size of Zr. 6. (2) Density : y Density of these elements generally increases Rack your Brain with decrease in metallic radius coupled with increase in mass. Thus, from Ti to Cu the Why are d-block elements significant increase in density may be noted. dense? y Order of density (3d series) : y In d-block Os and Ir have highest density and in 3d-series Cu have highest density. y Order of Density (Due to Lanthanide Contraction) Ti < Zr V is dominating factor. (2) Ti > V > Cr > Mn y Order of IP of 3d series will be:- (3) Cr > Mn > V > Ti Sc < Ti > V < Cr < Mn < Fe > Co > Ni < Cu < Zn (4) V > Mn > Cr > Ti 7.  (4) Oxidation States : y d-Block elements show variable oxidation Concept Ladder states due to tendency of (n–1)d as well as ns electrons to take part in bond formation Cr2+ is good reducing agent (except 3rd and 12th group elements). than Fe2+ due to low crystal field stabilizing energy (CFSE). Rack your Brain Characteristics features of oxidation state : Why d-block metals have high y Sc[III B] and Zn[II B] don’t show variable oxidation state, when compounds oxidation state. are formed with oxygen rather y Most common oxidation state of d-block than fluorine. metals is +2. y Highest oxidation state of 3d-series element is +7, while those of d-block element Previous Year’s Question is +8. y Highest oxidation state of d-block element is For the four successive transition found in mid of the series [Stability of highest elements (Cr, Mn, Fe and Co), the oxidation state is increases upto mid of the stability of +2 oxidation state will series and then start decreasing] be there in which of the following y d-block metals can also show zero oxidation order state (low oxidation state) when a complex  [AIPMT-2011] compound has ligands capable of p-acceptor (1) Cr>Mn>Co>Fe character in addition to the s-bonding. (2) Mn>Fe>Cr>Co (3) Fe>Mn>Co>Cr For example : [Ni(CO)4], [Fe(CO)5] (4) Co>Mn>Fe>Cr y Highest oxidation state of d-block metals is found in their compounds with oxygen and fluorine. y The high value of oxidation state of above Rack your Brain d & f–Block Elements elements is due to very small size and very high electronegativity of oxygen and fluorine. Which transition element does not show the variable oxidation y Highest oxidation state of d-block metals state? with respect to halogen is found to be +6. 8.  Concept Ladder Compounds of d-block y Moving from left to right the elements having metal in their lowest high oxidation state are less stable. oxidation state are ionic Ex : Fe(+2, +3), Ni(+2) Cu(+1, +2). in nature while those of y The relative stabilityo f various oxidation highest oxidation state are states of a given element can be explained covalent in nature. on the basis of stability of d0, d5 and d10 configurations. For example, Ti4+ (3d04s0) is more stable than Ti3+ (3d14s0); Mn2+(3d54s0) is more stable than Mn3+(3d44s0); Fe3+ (3d54s0) is more stable than Fe2+ (3d64s0), etc. Q.3 Why halides lowest oxidation state of d-block metal is quite unstable with fluoride ? A.3 Fluorine is a powerful oxidising agent which oxidise the lowest oxidation state into highest. d & f–Block Elements 9.  Q.4 Which ion is more stable in aqueous medium? (1) Ti3+ (2) V3+ (3) Cr3+ (4) Mn3+ A.4 (3) Cr3+ ion have half-filled electronic configuration ( t 32g , e0g ). Q.5 d-Block elements exhibit more oxidation states than f-block elements. Give Reason. A.5 d-Block elements exhibit more oxidation states because of less energy gap between d and s subshell whereas f-block elemetns have large energy gap between f and d subshell. (5) Standard electrode potential (E°) and chemical reactivity : y The stability of the compounds in solution depends upon standard electrode potentials rather than I.E. y Electrode potential values depend upon factors such as enthalpy of sublimation (or atomisation) of the metal, the ionisation enthalpy and the hydration enthalpy, i.e., ∆ H M(s)  sub → M(g) , (DsubH is enthalpy of sublimation/atomisation). ∆ H M(g) + aq  hyd → M+(g) + e–, (Di H is ionisation enthalpy) ∆ H M+(g) + aq  hyd → M+(aq), (Dhyd H is enthalpy of hydration) y For the process, M(s) → M+(aq) + e–, will be the sum of the three types of enthalpies, Concept Ladder i.e., DTH = DsubH + DiH + DhydH. y Thus, Dr H, is the total enthalpy change when Copper having positive E° solid metal, M is brought in the aqueous values, does not liberate hydrogen from acids. It d & f–Block Elements medium in the form of monovalent ion, M+(aq). y The low value of electrode potential, implies reacts only with oxidising more negative the stadard reduction potential acids such as HNO3 and of the electrode, more stable is the oxidation conc. H2SO4. state of the transition metal in the aqueous medium. 10. y The standard reduction potential values, E°(M2+/M) and (M3+/M2+) of the members of first transition series are given ahead: Element Ti V Cr Mn Fe Co Ni ` Cu Zn E°(M2+/M)V –1.63 –1.18 –0.91 –1.18 –0.44 –0.28 –0.25 +0.34 –0.76 E°(M3+/M2+)V 0.37 0.26 –0.41 +1.57 +0.77 +1.97 — — — Trends in M2+/M Standard Electrode Potentials y There is irregular trend in the E°(M2+/M) values, due to irregular variation of I.E. and sublimation energies of the atoms of the members of the transition series. y Except copper, all other elements have negative reduction potential values, i.e., Concept Ladder these elements except copper should have the capacity to liberate hydrogen from dilute The values of E° for Mn, Zn acids. and Ni are more negative M  2H  from acid   M2  H2  than expected. This is due to extra stability of half- y In actual parctice, the rate of liberation of filled d-subshell (d5) in hydrogen is very slow some of the metals, in (Mn2+) and completely filled fact, get proteced by the formation of a thin (d10) in Zn2+. film of an inert oxide on the surface Chromium, for example, inspite of its high negative reduction potential value is an unreactive metal as it does not liberate hydrogen due to a thin coating of Cr2O3 on its surface. y Less negative of E° values along the series Previous Year’s Question is due to increase in the first and second ionisation energies. Four successive members of Trends in M3+/M2+ the first series of the transition Standard Electrode Potentials metals are listed below. For y E° value for Sc3+/Sc2+ is very low. Hence, which one of them the standard Sc3+ is stable. This is due to its noble gas potential (E°M2+/M) value has a d & f–Block Elements configuration. positive sign? y E° values for the redox cople M3+/M2+ indicate  [AIPMT-2012] that Mn3+ and Co3+ ions are stron oxidising (1) Co (Z = 27) (2) Ni (Z = 28) agents. (3) Cu (Z = 29) (4) Fe (Z = 26) 11.  (6) Magnetic Property : y On applying magnetic field to substances, Concept Ladder mainly two types of magnetic behaviour are observed : Along the 3d-series (i) Diamagnetism : paramagnetism increases Diamagnetic substance is one which is upto the middle and then slightly repelled by a magnetic field. decreases. (Pairing start) (ii) Paramagnetism : Paramagnetic substance is one which is attracted by a magnetic field. y Due to presence of unpaired electrons Paramagnetism occurs, each such electron having a magnetic moment associated with Rack your Brain its spin angular momentum. y The magnetic moment is determined by the Among V3+, Cr3+ and Fe+3 which number of unpaired electrons. one have maximum unpaired Magnetic moment = n(n + 2) B.M. electron? Where, n = number of unpaired electrons. If all electrons are paired, substance will be diamagnetic and magnetic moment will be zero. y As it is evident most of the transition metal Concept Ladder ions have unpaired electrons in their ‘d’ orbitals. Hence most of the transition metal Behaviour of paramagnetic ions are paramagnetic in nature. Transition species in solid state is metal ions having 3d0 and 3d10 configuration known as ferromagnetism. exhibit diamagnetic nature. y An unpaired electron spins and as it is a charged particle, magnetic field is created due to its spinning. y The magnetic moment of diamagnetic substance will be zero. y As the number of unpaired electrons increase Previous Year’s Question the magnetic moment created goes on increasing and hence the paramagnetic nature also increases. The calculated spin only magnetic d & f–Block Elements y Transition metal ions having d5 configuration moment of Cr2+ ion is : will have maximum number of unpaired  [NEET-2020] electrons therefore they will be maximum (1) 2.84 BM (2) 3.87 BM paramagnetic in nature. (3) 4.90 BM (4) 5.92 BM 12.  Ion Configuration Unpaired electrons Magnetic moment(BM) Sc3+ 3d0 0 0 Ti3+ 3d1 1 1.73 Ti2+ 3d2 2 2.84 V2+ 3d3 3 3.87 Cr2+ 3d4 4 4.90 Mn2+ 3d5 5 5.92 Fe2+ 3d6 4 4.90 Co2+ 3d7 3 3.87 Ni2+ 3d8 2 2.84 Cu2+ 3d9 1 1.73 Zn2+ 3d10 0 0 Q.6 Which divalent 3d ion contain maximum unpaired electron. d & f–Block Elements (1) Cr+2 (2) Mn+2 (3) Fe+2 (4) Co+2 A.6 (2) Mn — 3d5, 4s2 ; Mn+2 — 3d5, 4s0. So, Mn+2 have maximum 5 unpaired electrons. 13.  Q.7 Calculate the magnetic moment of Cr2+, Mg2+, Co2+ and Fe2+ is : A.7 Magnetic moment = n(n + 2) BM (n = unpaired electron) (1) Cr2+ ; E.C. = [Ar]4s0 3d4 ⇒ n = 4 Magnetic moment = 4(4 + 2) =24 BM (2) Mn2+ ; E.C. = [Ar]4s0 3d5 ⇒ n = 5 Magnetic moment = 5(5 + 2) =35 BM (3) Fe2+ ; E.C. = [Ar]4s0 3d6 ⇒ n = 4 Magnetic moment = 4(4 + 2) =24 BM (4) Co2+ ; E.C. = [Ar]4s0 3d7 ⇒ n = 3 Magnetic moment = 3(3 + 2) =15 BM (7) Formation of coloured ion : Definitions y d-block compounds are generally coloured due to : When an electron from a lower (i) Unpaired electron result in d-d transition. energy d-orbital is excited to (ii) Charge transfer spectrum. higher energy d-orbital, the (iii) Polarisation. energy of excitation corresponds y A substance appears coloured because it to the frequency of light absors light at specific wavelengths in the absorbed. visible part of the electromagnetic spectrum (4000-7000 Å) and transmits or reflects the rest of the wavelengths. y The frequency of light i.e. absorbed by : Previous Year’s Question (i) Nature of ligand. (ii) Oxidation states of metal. Assertion : Cuprous ion (Cu+) has (iii) Size of metal ion. unpaired electrons while cupric y d-block metal cation having paired ion (Cu2+) does not. electronic configurations are colourless, Reason : Cuprous ion (Cu+) is d & f–Block Elements whereas those have unpaired electrons are colourless where as cupric ion coloured in nature. (Cu2+) is blue in the aqueous y Transition metal compounds are mostly solution. coloured in their solid or in solution states.  [AIIMS] 14.  d–d transition : y Excitation of unpaired electron from t2g to eg after absorption of visible light. Example : Ti3+(3d1) Colour of a complex ion depend on nature of ligand (Strong and weak field ligand) and number of ligands. Rack your Brain d & f–Block Elements ∆ Ex : TiCl 3 → TiCl 3.6H2O (Anhydrous) (Hydrated) White Violet Why anhydrous CuSO4 is colourless but hydrated CuSO4 is blue in colour? 15.  Charge transfer spectrum : y Absorption and emission of light in d-block Concept Ladder metal cation having d0 or d10 configuration is induced by charge transfer spectra. Halogen molecules are Ex : K2Cr2O7 → d0 → Orange red coloured because valance K2CrO4 → d0 → Yellow electron excited to low KMnO4 → d0 → Purple energy level after absorption Cu2O → d10 → Rubby red of visible light. y Charge transfer is an example of redox F2 (Pale Yellow), Cl2 reaction and in charge transfer spectra, the (Greenish Yellow), Br2 net oxidation and reduction is zero. (Reddish Brown), I2 (Dark y Charge transfer spectra is divided into two Violet) categories : Ex : K2Cr2O7, K2CrO4, KMnO4 Ex : Cu2O d & f–Block Elements 16. (8) Formation of complexes compounds : y By virtue of their small size, comparatively high Concept Ladder nuclear or ionic charge and availability of vacant d-orbitals of suitable energy, these metals exert Anhydrous CuSO4 is strong electrostatic attraction on the ligands. The paramagnetic and white but species formed on interaction of metal ions form hydrated CuSO4 is blue in complexes because of the following reasons : colour because in presence (a) Their size of cation is small. of H2O ligand, d-d transition (b) High effective nuclear charge. possible. (c) Availability of vacant (n–1)d-orbitals of appropriate energy. (d) The structure commonly found in such complex are linear (i.e. co-ordination Rack your Brain number, C.N. = 2), square. (e) Cobalt form more complex than any other What is the difference between elements. complexes [Cr(H2O)5(NO2)]+2 and Co3  6 NH3  [Co(NH3 )6 ]3 [Cr(H2O)5(ONO)]+2? Fe2  6 CN  [Fe(CN)6 ]4  [Co(H2O)6 ]3 Co3  6 H2O  Metal ion Ligand C.N. Complex ion Ag+ NH3 2 [Ag(NH3)2]+ Fe+2 CN– 6 [Fe(CN)6]–4 Cu2+ NH3 4 [Cu(NH3)4]+2 Ni+2 CN– 4 [Ni(CN)4]–4 (9) Catalytic Properties : y d-block elements and their compounds are Rack your Brain d & f–Block Elements known to act as good catalyst as they have variable oxidation state. They provide a new Which catalyst is used in the path with lower activation energy for the given process? reaction and form unstable intermediate CH  CH  CH3  CH3 compounds (intermediate compounds theory). 17.  y The adsorbed reactants provide a large surface area for adsorption, which are finely divided Previous Year’s Question metals or their compounds react faster due to the closer contact (Adsorption theory). Which of the following transition y Catalytic property is probably due to the metal is used as a catalyst utilisation of (n–1)d-orbitals or formation of [AIPMT] interstitial compounds. (1) Nickel (2) Cobalt (3) Gold (4) Both (1) and (2) Catalysts Uses Zieglar-Natta catalyst, used in polymerisation of TiCl4 + Al(C2H5)3 ethylene. V2O5 Contact process SO2   SO3 Fe Haber Bosch process PdCl2 Wacker’s process for CH3CHO manufacturing Pd Hydrogenation of alkane, alkyne etc. Pt/PtO Adam’s catalyst for selective reduction Pt Catalytic convertor, for cleaning Ostwald’s process Pt/Rh NH3   NO d & f–Block Elements Cu Oxidation of alcohols 18. Example : Concept Ladder Alkali metals are highly soft and ductile due to very weak metallic bond that’s why it cannot form interstitial compound. (10) Formation of interstitial compounds : y Interstitial compounds have similar Definitions chemical property as the parent metal but have different physical properties such as Small non metallic atom such density, conductivity and hardness. as H, B, C, N etc are able to y Interstitial compounds are neither ionic nor occupy interstitial space of the covalent these are nonstoichiometric. lattice of the d-block elements Ex. VH0.56, TiH1.5 to form combinations which are termed interstitial compounds. d & f–Block Elements 19.  y Characteristics feature of interstitial compounds: Previous Year’s Question y They have high melting points, higher than those of pure metals. Which of the following statements y They are chemically inert. about the interstitial compounds y They are less malleable and ductile. is incorrect?[NEET-2013] y These are non-stoichiometry in nature and do (1)  They have higher melting not follow the common rule of valency. points than the pure metal y Interstitial compounds are neither ionic nor (2) They retain metallic covalent. conductivity y Cementite (Fe3C) is an interstitial carbide. (3) They are chemically reactive Cementite used for making railway track. (4) They are much harder than the pure metal (11) Alloy Formation : y Alloy is a mixture two or more than two different metals. y d-block metals readily form alloy due to almost same metallic radii. (Size difference less than 50%) Definitions Ex : Brass (copper-zinc) and bronze (copper- tin) etc. Alloy is a homogenous solid solution of two or more than two different metals in which atom of one metal can randomly distributed among atoms of other metal. Concept Ladder y The purpose of making alloy is develop some useful properties which are absent in The purpose of making constituent elements. alloys is to develop some d & f–Block Elements y Alloy containing mercury as one of the useful properties which are constituents element are called Amalgams. absent in the constituent Ex : Sodium amalgam (Na–Hg). elements. 20. Properties of alloy : y Alloy are resistive towards rusting. Previous Year’s Question y Melting point of alloy is more than pure metal. y Alloys of transition metal like brass and Bell metals is an alloy of bronze have industrial importance. Also [AIPMT] ferrous alloys, Cr, W, Mo and Mn are used (1) Cu, Zn and Sn for the production of a variety of steels and (2) Cu, Zn and Ni stainless steel. (3) Cu and Zn Ex : Some important alloys of transition (4) Cu and Sn metals. 1. Brass   Cu + Zn 2. Bronze   Cu + Sn 3. Rolled gold   Cu + Al 4. Invar   Fe + Ni Rack your Brain 5. Alnico   Fe + Al + Ni + Co (Use for Why do we use alloys? making permanent magnet) 6. Stainless steel   Fe + Ni + Cr 7. Tungsten steel   Fe + Ni + W 8. German silver   Ni + Cu + Zn (Percentage of silver = 0) Previous Year’s Question 9. 20 ct. gold   Au + Cu 10. 24 ct. gold   100% Au Percentage of silver in German (12)Melting and Boiling Point : silver is  [AIPMT] y As the number of d-electron increases the (1) 0% (2) 1% number of covalent bond between the atoms (3) 5% (4) 4% are expected to increase up to Cr-Mo-W family where each of the d-orbital has only unpaired electrons and the opportunity for covalent sharing is greatest. y Inspite of presence of five unpaired electrons in Mn, the unexpected low melting and boiling Concept Ladder points is due to its complex structure, it is unable to form metallic and covalent bonds. d & f–Block Elements y The absence of unpaired electron [(n–1) d10 Fe, CO, Ni does not combine 4s2] in Zn, Cd & Hg is responsible for its low with Hg, so Hg liquid store melting and boiling point. in iron container. y Tungsten (W) and Mercury (Hg) have highest 21.  and lowest melting points respectively. y Along the 3d-series melting point increases upto the middle (Cr) and then decreases Order of Melting point (3d-series) Sc < Ti < V < Cr < Mn < Fe > Co > Ni > Cu > Zn Q.8 Why Mn has very low melting point. A.8 Mn has very low melting point because it has : (i) Stable electronic configuration (4s2 3d5) (ii) High ionisation energy. (iii) Less delocalisation of electron. (iv) Weak metallic bond. (13) Metallic Character : y Transition elements have relatively low Definitions ionization energies and have one or two Minimum energy required to electrons in their outermost energy level (ns1 break the metallic lattice of or ns2). As a result, metallic bonds are formed. crystalline metals into atoms Hence, they behave as metals. is known as enthalpy of d & f–Block Elements y Greater the number of unpaired d-electrons, atomisation. stronger is the bonding due to the overlapping of unpaired electrons between different metal atoms. 22. y Transition metals are good conductors of Rack your Brain electricity and heat. y They exhibit all the 3 types of structures: Why last element of each period fcc, hcp, bcc. in d-block have low melting point y Melting point and boiling points of metal and low enthalpy of atomization. increases then metallic bond strength increases. y Strength of these bonds increases, as there are more number of unpaired elecrtrons, hence greater the number of covalent bonds. y Zn, Cd and Hg are soft in nature as they have fully filled d-orbitals whereas Cr, Mo and W Concept Ladder are hardest metals as they have max. no. of All d-block elements unpaired d-orbitals. are hard, malleable and ductile except Hg which (14) Enthalpy of atomisation: is liquid and soft at room y d-block metals have very high enthalpy of temperature. atomisation (E.O.A.) due to presence of strong metallic bond. y In a series when we move from left to right E.O.A. is first increases then start decreases. y Enthalpy of atomisation is depend upon interatomic interaction. Metallic bond strength atomH  Interatomic interaction d & f–Block Elements 23.  Some important compounds of transition elements : Concept Ladder Oxide and Oxoanions of Metals : y The oxides of transition elements are generally Cr and Zn has very high and formed by the reaction of metals with O2 at low enthalpy of atomization high temperature. All the metals except Sc respectively among d-block form metal oxide which are ionic. elements. y For Mn with increase in oxidation number, ionic character decreases. Mn2O7 is a covalent green oil. Even CrO3 and V2O5 as well as VO2+ salt. V2O5 react with alkalies as well as acids to give VO–34 and VO4+3 respectively. The well characterised CrO is basic but Cr2O3 is amphoteric. K2Cr2O7 Potassium dichromate : y It is important chemical used in leather industry and an oxidant for preparation of Rack your Brain many azo compounds. It is prepared from chromite ore (FeCr2O4). Step 1 : Chromite ore is fused with Na2CO3 or Why I-A group metal have very K2CO3 in excess air. low enthalpy of atomization. 4FeCr2O4 + 8Na2CrO3 + 7O2   8Na2CrO4 + 2Fe2O3 + 8CO2 y Yellow solution of Na2CrO4 is filtered and acidified with H2SO4 to give a solution from which orange sodium dichromate Na2CrO7.2H2O is crystallised. Note : Na2Cr2O7 is more soluble than K2Cr2O7. Na2Cr2O7 + 2KC   2Cr2O7 + 2NaCl Previous Year’s Question Structure of CrO–2 4 and Cr2O–2 7 K2Cr2O7 on heating with aqueous d & f–Block Elements NaOH gives  [AIPMT] (1) Cr2O72– (2) Cr(OH)2 (3) CrO42– (4) Cr(OH)3 24.  Concept Ladder CrO2– 4 and Cr2O72– exist in equilibrium (pH = 4). + H CrO42–    Cr2O72– –  OH (Yellow) (Orange) y Sodium and potassium dichromate are good oxidising agent. But potassium dichromate is used as primary standard in volumetric Previous Year’s Question analysis. Identify the incorrect statement : [NEET-2020] (1) The oxidation state of Cr in CrO42– and Cr2O72– are not the same (2) Cr2+ (d4) is a stronger reducing agent than Fe2+(d6) in water (3) The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes. (4) Interstitial compounds are those that are formed when d & f–Block Elements small atoms like H, C or N are trapped inside the crystal lattices of metals. 25.  Example of reaction involving oxidation by Cr2O72–. Rack your Brain 6Cl–   3I2 + 6e– When acidic K2Cr2O7 react with 3H2S   6H+ + 3S + 6e– Na2SO3, green colour appear due 3Sn2+   3Sn4+ + 6e– to formation of? 6Fe2+   6Fe3+ + 6e– Chemical Properties of K2Cr2O7 (1) Heating Effect ∆ 4 K2Cr2O7 → 4 K2CrO4 + Cr2O3 + 3 O2 (2) Reaction with H2SO4 ( ) K2Cr2O7 + 2 H2SO4 dil.  → 2 CrO3 + 2 KHSO4 + H2O K2Cr2O7 + 8 H2SO4 ( conc. )  → 2 K2SO4 + 2 Cr ( SO4 ) + 8 H2O + 3 O2 3 Green (3) Reduction with H2O ∆ K2Cr2O7 + H2O + 3 C → Cr2O3 + 2 KOH + 3 CO ↑ d & f–Block Elements (4) Reduction with H2O2 K2Cr2O7 + H2SO4 + 4 H2O2  → 2 CrO5 + K2SO4 + 5 H2O ( Blue layer) Chromic di peroxide 26. CrO5 contains 5 oxygen atoms. In these 5 oxygen atoms four atoms participate in the formation of two peroxide linages. So, in CrO5 the oxidation state of Cr is +6. Potassium Permanganate KMnO4 : y Preparation : KMnO4 is prepared by fusion of MnO2 with an alkali metal hydroxides and an oxidising agent like KNO3. This produces dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate. Concept Ladder NaNO3 cannot react with KMnO4 or K2Cr2O7 because of its resumble maximum oxidation state (+5). Hence, it cannot be oxidised (KMnO4 or K2Cr2O7). Previous Year’s Question The manganate and permanganate ions are tetrahedral, due to  [NEET 2019] (1) the p-bonding involves overlap of d-orbitals of oxygen with d-orbbitals of manganese (2) the p-bonding involves overlap y Commercially it is prepared by the alkaline of p-orbitals of oxygen with oxidative fusion of MnO2 followed by the d-orbitals of manganese d & f–Block Elements electrolytic oxidation of manganate (VI). (3) there is no p-bonding (4) the p-bonding involves overlap of p-orbitals of oxygen with p-orbitals of manganese 27.  y Laboratory preparation : In the laboratory, a manganese (II) ion salt is oxidised by Concept Ladder peroxodisulphate to permanganate. Potassium permanganate act as an oxidising agent in alkaline, neutral and acidic Characteristic properties : solution. 1. KMnO4 forms dark purple (almost black) crystals which have same structure as those of KClO4. The salt is not soluble in H2O. (6.4 g/100 g of H2O at 293 K). 2. It has intense colour and weak temperature Rack your Brain dependent paramagnetism. 3. Green manganate ion is paramagnetic while Why is KMnO4 an oxidising agent? permanganate ion is diamagnetic. 4. p-bonding takes place by overlap of p-orbital of oxygen with d-orbitals of manganese. 5. Structure : Previous Year’s Question Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 6. KMnO4 is a good oxidising agent in acidic, solution ? [NEET-2016 Phase-I] basic or neutral medium. (1) The solutions turns blue 7. Half cell reaction of KMnO4 in different (2) The solution is decolourized medium. (3) SO2 is reduced MnO–4 + e–  → MnO2– [E° = 0.56 V] (4) Green Cr2(SO4)3 is formed 4 MnO–4 + 2H2O + 3e–  → MnO2 + 4OH– (E° = 1.69 V) MnO–4 + 8H+ + 5e–  → Mn2+ + 4H2O (E° = 1.52 V) d & f–Block Elements 28.  Concept Ladder Potassium permanganate act as an oxidising agent in alkaline, neutral and acidic solution. y The hydrogen ion concentration of the Previous Year’s Question solution plays an important part in influencing the reaction. Although reactions Which one of the following ions can be understood by consideration of redox exhibits d-d transition and potential, kinetics of the reaction is also an paramagnetism as well ? important factor. [NEET-2018] (1) CrO42– (2) Cr2O72– (3) MnO4– (4) MnO42– Important oxidising reactions of KMnO4 : 1. In acidic medium : (i) 10I– + 2MnO–4 + 16H+  → 2Mn2+ + 8H2O + 5I2 (ii) 5Fe2+ + MnO–4 + 8H+  → Mn2+ + 5Fe3+ + 4H2O (Green) (Yellow) (iii) 5C2O42– + 2MnO–4 + → 2Mn2+ + 8H2O + 10CO2 + 16H  (iv) 5S2– + 2MnO–4 + 16H+  → 2Mn2+ + 8H2O + 5S (v) 5SO2– 3 + 2MnO–4 + 6H+  → 2Mn2+ + 3H2O + 5SO2– 4 (vi) 5NO–2 + 2MnO–4 + 6H+  → 2Mn2+ + 3H2O + 5NO–3 2. In neutral or faintly alkaline solutions : d & f–Block Elements (i) 2MnO–4 + H2O + I–  → 2MnO2 + 2OH– + IO3 (ii) 8MnO–4 + S2O32– + H2O  → 8MnO2 + 6SO42– + 2OH– Zn (iii) 2MnO–4 + 3Mn2+ + 2H2O  → 5MnO2 + 4H+ 29.  Important Oxides of 3d-Series Sc Sc2O3(b) Ti TiO(b) Ti2O3(b) TiO2(c) V VO(b) V2O3(b) VO2(c) V2O5(a) Cr CrO(b) Cr2O3(c) CrO2(c) CrO3(a) Mn MnO(b) Mn2O3(b) MnO2(c) Mn3O4(c) [MnO3(a)] Mn2O7(a) Fe FeO(b) Fe2O3(b) F3O4(b) Co CoO(b) [Co2O3h(b)] [CoO2h] Ni NiO(b) [Ni2Oh(b)] [NiO2h] Mixed Cu CuO(b) Zn ZnO(b) where a = acidic; b = basic; c = amphoteric; h = hydrated f-Block Elements y f-block elements are also known as inner- transition elements. y f-block consists of the two series, lanthanoids and actinoids. y Lanthanide series (Z = 58 – 71) (Ce-Lu). Actinide series (Z = 90 – 103) (Th-Lw). y General electronic configuration is (n–2)f1–14 (n–1)d0–1 , ns2 d & f–Block Elements 4f-series (Lanthanides) : y These elements are called rare earth. 30.  Definitions y As many of elements are not particularly rare so this name is not appropriate for the whole The elements constituting the series. Promethium is a radioactive element. f-block are those in which the 4f and 5f orbitals are progressively 5f-series (Actinides) : filled in the latter two long periods. y The members of actinides are radioactive and majority of them are not found in nature. Elements from Z = 93 onwards are called transuranic elements and discovered by Previous Year’s Question synthetic methods. Lanthanoids are Lanthanoids : (4f-block elemenets)  [AIPMT] General characteristics : (1) 14 elements in the sixth period (atomic no. 90 to 103) that are (1) Electronic configuration : filling 4f sublevel y General electronic configuration [Xe] 4f1–14, (2) 14 elements in the seventh 5d0–1, 6s2. period (atomic number = 90 to y Lanthanide have outer three shells incomplete. 103) that are filling 5f sublevel y They have electronic configuration with (3) 14 elements in the sixth period 6s2 common but with variable occupancy (aotmic number = 58 to 71) of 4f level. The energies of 5d and 4f orbit are that are filling the 4f sublevel (4) 14 elements in the seventh also nearly similar and thus their filling show period (atomic number = 58 to certain irregularities. 71) that are filling 4f sublevel. y It is energetically favourable to move the single electron on 5d into the 4f level in most of the elements but not in case of Ce, Gd and Concept Ladder Lu. y In Gd and Lu besides 5d’, the 4f orbitals are 5d electron appears half-filled or fully filled. Give extra stability. in gadolinium (z = 64) with an outer electronic Gd (Z = 64) [Xe] (4f7 5d1 6s2) configuration of 4f75d16s2 Lu (Z = 74), [Xe] (4f14, 5d1 6s2) (and not 4f86s2). This is y Eu(Z = 63) and Yb (Z = 70) also shows extra because the 4f and 6d stability of half filled and fully filled f-orbitals. electrons are at about d & f–Block Elements the same potential energy and that the atoms have a tendency to retain stable half filled configuration. 31.  Electronic Radii/pm Atomic Configurations* Name Symbol Number Ln Ln2+ Ln3+ Ln4+ Ln Ln3+ 57 Lanthanum La 5d1 6s2 5d1 4f0 187 106 58 Cerium Ce 4f15d16s2 4f2 4f1 4f0 183 103 59 Praseodymium Pr 4f3 6s2 4f3 4f2 4f1 182 101 60 Neodymium Nd 4f4 6s2 4f4 4f3 4f2 181 99 61 Promethium Pm 4f5 6s2 4f5 4f4 181 98 62 Samarium Sm 4f6 6s2 4f6 4f5 180 96 63 Europium Eu 4f7 6s2 4f7 4f6 199 95 64 Gadolinium Gd 4f7 5d1 6s2 4f7 5d1 4f7 180 94 65 Terbium Tb 5f9 6s2 4f9 4f8 4f7 178 92 66 Dysprosium Dy 4f10 6s2 4f10 4f9 4f8 177 91 67 Holmium Ho 4f11 6s2 4f11 4f10 176 89 d & f–Block Elements 68 Erbium Er 4f12 6s2 4f12 4f11 175 88 32.  Electronic Radii/pm Atomic Configurations* Name Symbol Number Ln Ln2+ Ln3+ Ln4+ Ln Ln3+ 69 Thulium Tm 4f13 6s2 4f13 4f12 174 87 70 Ytterbium Yb 4f14 6s2 4f14 4f13 173 86 4f14 5d1 4f14 71 Lutetium Lu 4f14 – – – 6s2 5d1 (2) Atomic and ionic radii : y In lanthanide series, with increasing atomic number, there is a progressive decrease in the atomic radii as well as ionic radii of trivalent ions from La3+ to Lu3+. This is due to lanthanide contraction. y Lanthanoid contraction : The atomic and ionic radii of trivalent ions (M3+) decreases regularly Previous Year’s Question due to increase in atomic no. from Ce to Lu. This continuous decrease in size of atoms The electronic configurations of Eu(Atomic No. 63), Gd (Atomic and ions is known as lanthanoid contraction. No. 64) and Tb (Atomic No. 65) y Due to decreases in the size of the lanthanide are : [NEET-2016 Phase-I] ions from Ce3+ to Lu3+, the covalent character (1) [Xe]4f76s2, [Xe]4f86s2 and of M—OH bond increases and hence the basic [Xe]4f 5d 6s 8 1 2 strength decreases. Thus, Ce(OH)3 is most (2) [Xe] 4f75d16s2, [Xe]4f7 5d1 6s2 basic while Lu(OH)3 is least basic. and [Xe]4f96s2 (3) [Xe]4f65d16s2, [Xe]4f75d16s2 and [Xe] 4f 5d 6s 8 1 2 (4) [Xe]4f76s2, [Xe]4f75d16s2 and d & f–Block Elements [Xe]4f96s2 33.  Previous Year’s Question The correct order of ionic radii of Y3–, La3+, Eu3+ and Lu3+ is  [AIPMT-2003] (1) La