D and F Block Elements Grade 12 Chemistry PDF

The d- and f - Block Elements Grade 12 Chemistry Session 2 Atomic Properties Ionisation Enthalpy IE (kJ Sc Ti V Cr Mn Fe Co Ni Cu Zn mol–) IE1 631 656 650 653 717 762 758 736 745 906 IE2 1...

The d- and f - Block Elements Grade 12 Chemistry Session 2 Atomic Properties Ionisation Enthalpy IE (kJ Sc Ti V Cr Mn Fe Co Ni Cu Zn mol–) IE1 631 656 650 653 717 762 758 736 745 906 IE2 1235 1309 1414 1592 1509 1561 1644 1752 1958 1734 IE3 2398 2657 2833 2990 3260 2962 3243 3402 3556 3837 IEs increase along each series from left to right due to an increase in Z eff. (exceptions maked in red) The variation in IE along a series of transition elements is much less in comparison to the non-transition elements Successive IE do not increase as steeply as in the case of non-transition elements. The magnitude of the increase in the IE2 and IE3 for the successive elements, is much higher along a series than IE1. Atomic Properties Ionisation Enthalpy The explanation for some of the anomalies is as follows: ➔ The second ionisation energy of Mn is less than that of Cr. Mn+ = [Ar] 3d5 4s1 Cr+ = [Ar] 3d5 Since the second e– is to be removed from stable 3d5 configuration in case of Cr, it has more IE2. ➔ The second ionisation energy of Zn is less than that of Cu. Zn+ = [Ar] 3d10 4s1 Cu+ = [Ar] 3d10 Since the second e– is to be removed from stable 3d10 configuration in case of Cu, it has more IE2. Atomic Properties Ionisation Enthalpy The explanation for some of the anomalies is as follows: ➔ The third ionisation energy of Mn is more than that of Fe. Mn2+ = [Ar] 3d5 Fe2+ = [Ar] 3d6 Since the third e– is to be removed from stable 3d5 configuration in case of Mn, it has more IE3. Atomic Properties Ionisation Enthalpy Graph is not to be memorised. Let’s Solve.. Let’s Solve.. The transition element that has lowest enthalpy of atomisation is A Fe B Cu C V D Zn Let’s Solve.. The transition element that has lowest enthalpy of atomisation is A Fe B Cu C V D Zn Solution The enthalpy of atomisation decreases from V to Zn. However, Zn is not a transition element. Let’s Solve.. The pair that has similar atomic radii is: A Mn and Re B Ti and Hf C Sc and Ni D Mo and W Let’s Solve.. The pair that has similar atomic radii is: A Mn and Re B Ti and Hf C Sc and Ni D Mo and W Solution The atomic radius of elements of 4d series is approximately same as the atomic radius of corresponding element of 5d series. Thus, Mo and W have similar atomic radii. Let’s Solve.. The third ionisation enthalpy is minimum for A Co B Fe C Ni D Mn Let’s Solve.. The third ionisation enthalpy is minimum for A Co B Fe C Ni D Mn Solution In general, the IE increases from left to right in a period, in this way, Mn should have minimum IE3 value. However, it is known that IE3 of Mn is more than that of Fe due to highly stable 3d5 configuration of Mn2+.  ? Quiz.  ? Quiz. The lanthanide contraction is responsible for the fact that A Zr and Zn have the same oxidation state B Zr and Hf have about the same radius C Zr and Nb have similar oxidation state D Zr and Y have about the same radius  ? Quiz. The lanthanide contraction is responsible for the fact that A Zr and Zn have the same oxidation state B Zr and Hf have about the same radius C Zr and Nb have similar oxidation state D Zr and Y have about the same radius Solution Zr and Hf have about the same radius due to lanthanide contraction  ? Quiz. The correct order of first ionisation enthalpies is: A Ti < Mn < Zn < Ni B Ti < Mn < Ni < Zn C Mn < Ti < Zn < Ni D Zn < Ni < Mn < Ti  ? Quiz. The correct order of first ionisation enthalpies is: A Ti < Mn < Zn < Ni B Ti < Mn < Ni < Zn C Mn < Ti < Zn < Ni D Zn < Ni < Mn < Ti Solution IE increases from left to right in the period.  ? Quiz. The correct order of decreasing second ionisation energy of Ti, V, Cr and Mn is A Mn > Cr > Ti > V B Ti > V > Cr > Mn C Cr > Mn > V > Ti D V > Mn > Cr > Ti  ? Quiz. The correct order of decreasing second ionisation energy of Ti, V, Cr and Mn is A Mn > Cr > Ti > V B Ti > V > Cr > Mn C Cr > Mn > V > Ti D V > Mn > Cr > Ti Solution IE increases from left to right in the period. However IE2 of Cr > Mn due to stable 3d5 configuration of Cr+. Learning Outcome 1 Introduction and Classification of d Block Elements 2 Physical and Atomic Properties of d Block Elements 3 Chemical Properties of d-Block Elements Oxidation States of d Block Elements Chemical Properties Oxidation States Transition elements may show great variety of oxidation states in their compounds. Sc Ti V Cr Mn Fe Co Ni Cu Zn +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 NOTE: The common oxidation state are in red. Sc3+, Zn2+ and Cd2+ have fixed O.S. The elements which give the greatest number of oxidation states occur in or near the middle of the series. The maximum oxidation states of reasonable stability correspond in value to the sum of the s and d electrons upto Mn followed by a rather abrupt decrease in stability of higher oxidation states. Chemical Properties Oxidation States Special features: ○ For d-block elements their oxidation states differ from each other by unity, in contrast with non transition elements where oxidation states normally differ by a unit of two. ○ From group 4 to 10, the higher oxidation states are favoured by the heavier members, which is opposite to p-block elements (where heavier members tend to occur in lower oxidation states due to inert pair effect). Example: In group 6, Mo6+ and W6+ are more stable than Cr6+. Thus, Cr6+ in form of Cr2O72– is a strong oxidizing agent whereas MoO3 and WO3 are not. ○ Low oxidation states are found when a complex compound has ligands capable of π-acceptor character in addition to the σ-bonding. For example, in Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero.  ? Quiz.  ? Quiz. Which one of the following exhibits the large number of oxidation states? A Ti (22) B V (23) C Cr (24) D Mn (25)  ? Quiz. Which one of the following exhibits the large number of oxidation states? A Ti (22) B V (23) C Cr (24) D Mn (25) Solution Mn (25) can show large number of oxidation states from +2 to +7.  ? Quiz. The element that usually does NOT show variable oxidation states is: A Cu B Ti C Sc D V  ? Quiz. The element that usually does NOT show variable oxidation states is: A Cu B Ti C Sc D V Solution Sc can show only +3 oxidation state. Learning Outcome 1 Chemical Properties of d-Block Elements Electrode Potentials and Chemical Reactivity 2 Other Chemical Properties of d-Block Elements Learning Outcome 1 Chemical Properties of d-Block Elements Electrode Potentials and Chemical Reactivity 2 Other Chemical Properties of d-Block Elements Chemical Properties Trends in the M2+/M Standard Electrode Potentials All 3d series elements have negative M2+/M electrode potential except for Cu. It is related to the following thermodynamic parameters - ○ Enthalpy of atomisation (ΔaH) ○ First Ionisation Energy (IE1) ○ Second Ionisation Energy (IE2) ○ Enthalpy of hydration (ΔhydH) The E° (Cu2+|Cu) is positive because the high energy to transform Cu(s) to Cu 2+(aq) is not balanced by its hydration enthalpy. Positive value for Cu implies - inability to liberate H2 from acids. Chemical Properties Trends in the M2+/M Standard Electrode Potentials E° value becomes less negative as we move from Ti to Zn. However, the values for Mn, Zn and Ni are more negative than expected. ○ For Mn - it is due to 3d5 configuration. ○ For Zn - it is due to 3d10 configuration. ○ For Ni - it is due to highest negative ΔhydH. +0.3 4 -0.28 -0.25 -0.44 -0.76 -0.9 -1.18 -1.18 -1.63 Ti V Cr Mn Fe Co Ni Cu Zn Chemical Properties Trends in the M3+/M2+ Standard Electrode Potentials Sc has very low value which reflects the stability of Sc3+ (noble gas configuration). Zn has high value due to the stable d10 configuration of Zn2+. The comparatively high value for Mn shows that Mn 2+(d5) is particularly stable Comparatively low value for Fe shows the extra stability of Fe 3+ (d5) Comparatively low value for V is related to the stability of V 2+ (half-filled t2g level) 1.97 1.57 0.77 -0.37 -0.26 -0.41 Ti V Cr Mn Fe Co Chemical Properties Trends in Stability of Higher Oxidation States For Halides: +6 CrF6 +5 VF5 CrF5 +4 TiX4 VXI4 CrX4 MnF4 +3 TiX3 VX3 CrX3 MnF3 FeXI3 CoF3 +2 TiXIII2 VXIII2 CrX2 MnF2 FeX2 CoX2 NiX2 CuXII2 ZnX2 +1 CuXIII Here X - F to I XI - F to Br XII - F and Cl XIII - Cl to I The highest oxidation numbers are achieved in TiX4(tetrahalides), VF5 and CrF6. For Mn, +7 state is not found in simple halides but MnO 3F is known. Beyond Mn, the only trihalides known are FeX’3 (X’ = F to Br) and CoF3. Chemical Properties Trends in Stability of Higher Oxidation States F can stabilise the highest oxidation state due to ○ higher lattice energy, as in the case of CoF3 ○ or higher bond enthalpy terms for the higher covalent compounds, e.g., VF 5 and CrF6. For V+5 the halides other than VF5 undergo hydrolysis to give oxohalides, VOX3. Another feature of fluorides is their instability in the low oxidation states. Example VX2 and CuX (X = Cl, Br or I only) Chemical Properties Trends in Stability of Higher Oxidation States All CuII halides are known except the iodide. In this case, Cu 2+ oxidises I– to I2: 2Cu2+ + 4I– → Cu2I2(s) + I2 Many Cu(I) compounds are unstable in aq. soln and undergo disproportionation. 2Cu+ → Cu2+ + Cu The stability of Cu2+(aq) > Cu+(aq) is due to the much more negative ∆hydH° of Cu2+(aq) than Cu+, which compensates for the IE2 of Cu. Chemical Properties Trends in Stability of Higher Oxidation States For Oxides: +7 Mn2O7 +6 CrO3 +5 V2O5 +4 TiO2 V2O4 CrO2 MnO2 +3 Sc2O3 Ti2O3 Mn2O3 Fe2O3 +2 TiO VO (CrO) MnO FeO CoO NiO CuO ZnO +1 Cu2O Apart from these oxides, mixed oxides Mn3O4, Fe3O4 and Co3O4 are known. Chemical Properties Trends in Stability of Higher Oxidation States The ability of O to stabilise high oxidation states exceeds that of F due to the ability of oxygen to form multiple bonds to metals. The highest O.S. in the oxides till group 7 coincides with group number. No higher oxides of Fe above Fe2O3, are known. However it can form ferrates(VI) in alkaline media (FeO4)2–, but they readily decompose to Fe2O3 and O2. Besides the oxides, oxocations are stabilised. ○ V(V) as VO2+ ○ V(IV) as VO2+ ○ Ti(IV) as TiO2+ Order of increasing oxidising power: VO2+ < Cr2O72– < MnO4– (due to stability of the lower oxidation state to which they are reduced) Chemical Properties Trends in Stability of Higher Oxidation States Structure of Mn2O7 (covalent) Each Mn is tetrahedrally surrounded by O’s including a Mn–O–Mn bridge Complex tetrahedral anions of metals with oxygen are also known of the form [MO4]n–. {M = VV, CrVI, MnV, MnVI and MnVII) Chemical Properties Chemical Reactivity and E° Values Element S Ti V Cr Mn Fe Co Ni Cu Zn c E° (M2+|M) - -1.63 -1.18 -0.90 -1.18 -0.44 -0.28 -0.25 +0.34 -0.76 E° - -0.37 -0.26 -0.41 +1.57 +0.77 +1.97 - - - (M3+|M2+) The metals of the 3d series with the exception of Cu are relatively more reactive than hydrogen and are oxidised by 1 M H+ ○ The actual rate is sometimes slow. For example, Ti and V are passive to dilute non oxidising acids at room temperature. The less negative value of E° (M2+|M) indicate a decreasing tendency to form divalent cations across the series. E° (M3+|M2+) values shows that Mn3+ and Co3+ ions are the strongest oxidising agents in aq. solutions. The ions Ti2+, V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid, 2Cr2+(aq) + 2H+(aq) → 2Cr3+(aq) + H2(g) Let’s Solve.. Let’s Solve.. Which is a stronger reducing agent Cr2+ or Fe2+ and why? Element Cr Fe E° (M2+|M) -0.90 -0.44 E° (M3+|M2+) -0.41 +0.77 Let’s Solve.. Which is a stronger reducing agent Cr2+ or Fe2+ and why? Element Cr Fe E° (M2+|M) -0.90 -0.44 E° (M3+|M2+) -0.41 +0.77 Solution Cr2+ will be a better reducing agent as the E° for oxidation of Cr2+ to Cr3+ is +0.41 and thus process is spontaneous. Let’s Solve.. Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential (E°M3+/M2+) value is highest? A Cr (Z = 24) B Mn (Z = 25) C Fe (Z = 26) D Co (Z = 27) Let’s Solve.. Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential (E°M3+/M2+) value is highest? A Cr (Z = 24) Solution B Mn (Z = 25) 1.97 1.57 C Fe (Z = 26) 0.77 D Co (Z = 27) -0.37 -0.26 -0.41 Ti V Cr Mn Fe Co  ? Quiz.  ? Quiz. Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential (E°M2+/M) value has a positive sign? A Co (Z = 27) B Ni (Z = 28) C Cu (Z = 29) D Fe (Z = 26)  ? Quiz. Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential (E°M2+/M) value has a positive sign? A Co (Z = 27) B Ni (Z = 28) C Cu (Z = 29) D Fe (Z = 26) Solution All 3d series elements have negative M2+/M electrode potential except for Cu.  ? Quiz. Which of the following transition metal can form stable oxide, oxocation and anion with oxygen in its highest oxidation state? A V (Z = 23) B Mn (Z = 25) C Fe (Z = 26) D Cr (Z = 24)  ? Quiz. Which of the following transition metal can form stable oxide, oxocation and anion with oxygen in its highest oxidation state? A V (Z = 23) B Mn (Z = 25) C Fe (Z = 26) D Cr (Z = 24) Solution V can form V2O5, VO2+ and [VO4]3– in +5 state.  ? Quiz. JEE Main 2017 Which of the following ions does not liberate hydrogen gas on reaction with dilute acids? A Ti2+ B V2+ C Cr2+ D Mn2+  ? Quiz. JEE Main 2017 Which of the following ions does not liberate hydrogen gas on reaction with dilute acids? A Ti2+ B V2+ C Cr2+ D Mn2+ Solution The ions Ti2+, V2+ and Cr2+ are strong reducing agents and will liberate hydrogen from a dilute acid. Learning Outcome 1 Chemical Properties of d-Block Elements Electrode Potentials and Chemical Reactivity 2 Other Chemical Properties of d-Block Elements Chemical Properties Magnetic Properties Ions of transition metals are generally paramagnetic due to presence of unpaired electrons. Each unpaired electron has two types of magnetic moment ○ Associated with its spin angular momentum ○ Associated with its orbital angular momentum. (not significant for 3d series) Thus, the magnetic moment is determined using the ‘spin-only’ formula. μ = n (n + 2) n = the number of unpaired electrons µ = the magnetic moment in units of Bohr magneton (BM) Chemical Properties Magnetic Properties The magnetic moment increases with the increasing number of unpaired electrons. Thus, the observed magnetic moment gives a useful indication about the number of unpaired electrons present in the atom, molecule or ion. No. of unpaired Calculated magnetic Ion Configuration electrons moment (BM) Sc3+ 3d0 0 0 Ti3+ 3d1 1 1.73 Ni2+ 3d8 2 2.84 V2+ 3d3 3 3.87 Cr2+ 3d4 4 4.90 Mn2+ 3d5 5 5.92 Chemical Properties Formation of Coloured Ions When an electron from a lower energy d orbital is excited to a higher energy d orbital- X X X X Z Y Z Z Y Y dxy dzx dyx dx2-dy2 dx2 lower energy d orbitals Higher energy d orbitals The energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand. (we’ll learn about this later) In aq. solutions, water molecules may act as the Chemical Properties Formation of Coloured Ions The ions having no unpaired electrons can’t show colour. The colour in aqueous solution are given below: Colour Ions Colour Ions Colourless Sc3+(3d0), Ti4+(3d0), Zn2+ Violet V2+ (3d3), Cr3+ (3d3), Mn3+ (3d10) (3d4) Purple Ti3+ (3d1) Pink Mn2+ (3d5) Blue V4+ (3d1), Cr2+ (3d4), Cu2+ Yellow Fe3+ (3d5) (3d9) bluepink Co3+Co2+ (3d6, 3d7) Green V3+ (3d2), Fe2+ (3d6), Ni2+ (3d8) Chemical Properties Formation of Complex Compounds Complex compounds are those in which the metal ions bind a number of anions or neutral molecules giving complex species with characteristic properties. Ex.: [Fe(CN)6]3–, [Fe(CN)6]4–, [Cu(NH3)4]2+ and [PtCl4]2–. The transition metals form a large number of complex compounds due to - ○ the comparatively smaller sizes of the metal ions ○ their high ionic charges ○ availability of d orbitals for bond formation We will discuss about chemistry of complex compounds later. Chemical Properties Catalytic Properties The transition metals and their compounds show catalytic activity due to their ability to adopt multiple oxidation states and to form complexes. Examples - ○ Vanadium(V) oxide (in Contact Process) ○ finely divided iron (in Haber’s Process) ○ nickel (in Catalytic Hydrogenation) Because the transition metal ions can change their oxidation states, they become more effective as catalysts. For example, iron(III) catalyses the reaction between iodide and persulphate ions. 2I– + S2O82– → I2 + 2SO42– The reaction takes place as follows: 2Fe3+ + 2I– → 2Fe2+ + I2 2Fe2+ + S2O82– → 2Fe3+ + 2SO42– Chemical Properties Formation of Interstitial Compounds Interstitial compounds: formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor covalent, for example, TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc The formulas shown for these compounds do not correspond to any normal oxidation state of the metal. The principal physical and chemical characteristics of these compounds are as follows: ○ They have high melting points, higher than those of pure metals. ○ They are very hard, some borides approach diamond in hardness. ○ They retain metallic conductivity. ○ They are chemically inert. Chemical Properties Alloy Formation Alloys are homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other metal or non-metal. Alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Since transition metals have similar radii, alloys are readily formed by these metals. The alloys are hard and have often high melting points. The best known are: ○ Iron, Carbon, Chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels and stainless steel. ○ brass (copper-zinc) ○ bronze (copper-tin) Chemical Properties Oxides and Oxoanions of Metals Oxides are generally formed by the reaction of metals with oxygen at high temperatures. All the metals except Sc form MO type oxides which are ionic. As the oxidation number of a metal increases, ionic character decreases. ○ Mn2O7 is a covalent green oil, CrO3 and V2O5 have low melting points. In the higher oxides, the acidic character is predominant. ○ Mn2O7 gives HMnO4 and CrO3 gives H2CrO4 and H2Cr2O7. ○ V2O5 is amphoteric though mainly acidic and it gives VO43– as well as VO2+ salts. In vanadium there is gradual change from the basic V2O3 to less basic V2O4 (gives VO2+ salt with acids) and to amphoteric V2O5. CrO is basic but Cr2O3 is amphoteric. Let’s Solve.. Let’s Solve.. JEE Main 2014 Chloro compound of Vanadium has spin-only magnetic moment of 1.73 BM. This vanadium chloride has the formula: A VCl2 B VCl4 C VCl3 D VCl5 Let’s Solve.. JEE Main 2014 Chloro compound of Vanadium has spin-only magnetic moment of 1.73 BM. This vanadium chloride has the formula: A VCl2 B VCl4 C VCl3 D VCl5 Solution Magnetic moment 1.73 BM corresponds to one unpaired electron and thus Vanadium must be in +4 oxidation state ([Ar] 3d1). Thus, formula of its chloride should be VCl4. Let’s Solve.. Which of the following arrangements does not represent the correct order of the property stated against it? A V2+ < Cr2+ < Mn2+ < Fe2+: paramagnetic behaviour B Ni2+ < Co2+ < Fe2+ < Mn2+: ionic size C Co3+ < Fe3+ < Cr3+ < Sc3+: stability in aqueous solution D Sc < Ti < Cr < Mn: number of oxidation states Let’s Solve.. Which of the following arrangements does not represent the correct order of the property stated against it? A V2+ < Cr2+ < Mn2+ < Fe2+: paramagnetic behaviour B Ni2+ < Co2+ < Fe2+ < Mn2+: ionic size C Co3+ < Fe3+ < Cr3+ < Sc3+: stability in aqueous solution D Sc < Ti < Cr < Mn: number of oxidation states Let’s Solve.. Which of the following arrangements does not represent the correct order of the property stated against it? Solution (A)V2+ < Cr2+ < Mn2+ < Fe2+: paramagnetic behaviour This order is incorrect as Mn2+ has maximum number of unpaired electrons among the given ions. (B) Ni2+ < Co2+ < Fe2+ < Mn2+: Element Cr Fe Co ionic size This order is correct as size decreases E° (M3+|M2+) - +0.77 +1.97 from Mn to3+Ni. 3+ 0.41 (C) Co < Fe < Cr3+ < Sc3+: stability in aqueous solution Sc3+ has maximum stability due to noble gas configuration. Stability of +3 ions decreases from Cr to Fe to Co as per SRP values shown above. (D) Sc < Ti < Cr < Mn: number of oxidation states This order is correct. Sc, Ti, Cr and Mn can show 1, 3, 5 and 6 oxidation states respectively. Let’s Solve.. Which one of the following ionic species will impart colour to an aqueous solution? A Zn2+ B Cu+ C Ti4+ D Cr3+ Let’s Solve.. Which one of the following ionic species will impart colour to an aqueous solution? A Zn2+ B Cu+ C Ti4+ D Cr3+ Solution Zn2+ {3d10}, Cu+ {3d10} and Ti4+ {3d0} have no unpaired electrons and are colourless. Let’s Solve.. Brass is an alloy of A Cu + Zn B Cu + Sn C Cu + Pb D Cu + Ni Let’s Solve.. Brass is an alloy of A Cu + Zn B Cu + Sn C Cu + Pb D Cu + Ni Solution Brass is an alloy of copper and zinc Let’s Solve.. Which of the following statements about the interstitial compounds is incorrect? A They are much harder than the pure metal. B They have higher melting points than the pure metal. C They retain metallic conductivity D They are chemically reactive. Let’s Solve.. Which of the following statements about the interstitial compounds is incorrect? A They are much harder than the pure metal. B They have higher melting points than the pure metal. C They retain metallic conductivity D They are chemically reactive. Solution Interstitial compounds are chemically inert.  ? Quiz.  ? Quiz. The calculated spin only magnetic moment of Cr2+ ion (in BM) is A 3.87 B 4.90 C 5.92 D 2.84  ? Quiz. The calculated spin only magnetic moment of Cr2+ ion (in BM) is A 3.87 B 4.90 C 5.92 D 2.84 Solution Cr = [Ar] 3d5 4s1 Cr2+ = [Ar] 3d4 It has four unpaired electrons. Thus, μ = √24 = 4.90 B.M.  ? Quiz. The aqueous solution containing which one of the following ions will be colourless? (Atomic number: Sc = 21, Fe = 26,Ti = 22, Mn = 25) A Sc3+ B Fe2+ C Ti3+ D Mn2+  ? Quiz. The aqueous solution containing which one of the following ions will be colourless? (Atomic number: Sc = 21, Fe = 26,Ti = 22, Mn = 25) A Sc3+ B Fe2+ C Ti3+ D Mn2+ Solution Sc3+ {3d0} is colourless due to no unpaired electron.  ? Quiz. Which one of the following characteristics of the transition metals is associated with their catalytic activity? A High enthalpy of atomization B Paramagnetic behaviour C Colour of hydrated ions D Variable oxidation states  ? Quiz. Which one of the following characteristics of the transition metals is associated with their catalytic activity? A High enthalpy of atomization B Paramagnetic behaviour C Colour of hydrated ions D Variable oxidation states Solution 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. Summary. Summary. Trends in the M2+/M Standard Electrode Potentials -0.34 -0.28 -0.25 -0.44 -0.76 -0.9 -1.18 -1.18 -1.63 Ti V Cr Mn Fe Co Ni Cu Zn Trends in the M3+/M2+ Standard Electrode Potentials 1.97 1.57 0.77 -0.37 -0.26 -0.41 Ti V Cr Mn Fe Co Summary. Magnetic Properties μ = n (n + 2) n = the number of unpaired electrons µ = the magnetic moment in units of Bohr magneton (BM) Other Chemical Properties Other chemical properties of d-block elements include - Formation of Coloured Ions Formation of Complex Compounds Catalytic Properties Formation of Interstitial Compounds Alloy Formation Vedantu Daily Practice Problems VDPP Number VDPP Name Wavebook subtopics covered d and f Block Elements - VDPP- 1 Introduction and Classification of d Block Elements 1 d and f Block Elements - VDPP- 2 Physical and Atomic Properties of d Block Elements 2 d and f Block Elements - VDPP- 3 Oxidation States of d Block Elements 3 d and f Block Elements - VDPP- Other Chemical Properties of d-Block Elements, 4 4 Electrode Potentials and Chemical Reactivity d and f Block Elements - VDPP- 5 Potassium Permanganate 5 d and f Block Elements - VDPP- 6 Potassium Dichromate 6 d and f Block Elements - VDPP- 7 Uses of d-Block Elements 7 d and f Block Elements - VDPP- 8 Introduction of f-block Elements 8 d and f Block Elements - VDPP- 9 General Characteristics and uses of f-block Elements 9 Feedback Thanks.