Transition Elements Solved Examples

TRANSITION ELEMENTS SOLVED EXAMPLES Ex. 1 Amongst [TiF6]2– , [CoF6]3– , Cu2 Cl2 and [NiCl4]2– [Atomic number ; Ti = 22, Co = 27, Cu = 29, Ni = 28] the colourless species are : (A) [TiF6]2– and [Cu2Cl2] (B) Cu2Cl2 and [NiCl4]2– (C) [TiF6]2– and [CoF6]3– (D) [CoF6]3– and [NiCl4]2– Ans. (A) Sol. In [TiF6] 2– the titanium is in +4 oxidation state having the electronic configuration [Ar] 18 3d0 4s0. Similarly in Cu2Cl2 the copper is in +1 oxidation state having the electronic configuration [Ar]18 3d10 4s0. As they do not have any unpaired electrons for d-d transition, they are therefore colourless. In [NiCl4]2– the nickel is in +2 oxidation state and electronic configuration is [Ar]18 3d8 4s0. As it has two unpaired electrons, so the complex is coloured. In [CoF6]3–, the cobalt is in +3 oxidation state having electron configuration [Ar] 3d 6 4s0. As it has four unpaired electrons, so the complex is coloured. Ex. 2 On the basis of trends in the properties of the 3d-series elements, suggests possible M2+ aqua ions for use as reducing agents, and write a balanced chemical equation for the reaction of one of these ions with O 2 in acidic solution. Sol. Because oxidation state +2 is most stable for the later elements of 3d-series elements, strong reducing agents include ions of the metals on the left of the series: such ions include V2+ (aq) and Cr2+ (aq) The Fe2+ (aq) ion is only weakly reducing. The Co2+ (aq), Ni2+ (aq), and Cu2+ (aq) ions are not oxidized in water. +0.77 Fe3+ Fe2+ -0.44 Fe The chemical equation for the oxidation is then 4 Fe2+ (aq) + O2(g) + 4H+ (aq)  4Fe3+ (aq) + 2H2O (). Ex. 3 Match the reactions given in column-I with the characteristic(s) of the reaction products given in column-II. Column-I Column-II (A) TiCl4 Zn (p) One of the products is bright orange coloured but diamagnetic. (B) FeCl3 573K (q) One of the products is green coloured and paramagnetic. (C) KMnO4 (D) K2Cr2O7 + H2SO4 (cold & conc.) 750K Ans. [A – r] ; [B – s ; [C – q] ; [D – p]. Sol. (A) TiCl4 Zn (B) 2FeCl3 573K (C) 2KMnO4 (r) One of the products is violet and paramagnetic. (s) One of the products exists as dimer. TiCl3, violet (one unpaired electron so d-d transition is possible). 750K gas dimer. K2MnO4 green (one unpaired electron so d-d transition is possible) + MnO2 + O2 . (D) K2Cr2O7 + 2H2SO4  2CrO3 bright orange (diamagnetic) + 2KHSO4 + H2O. Ex. 4 Among the following statements choose the true or false statement(s). (a) K2Cr2O7 on heating with charcoal gives metallic potassium and Cr2O3. (b) On heating in current of H2 the crystalline KMnO4 is converted into KOH and Mn3O4. (c) Hydrated ferric chloride on treatment with 2, 2–dimethoxypropane gives anhydrous ferric chloride. Ans. (a) False (b) False Sol. (a) K2Cr2O7 + 2C (charcoal) (b) 2KMnO4 + 5H2 (c) True Cr2O3 + K2CO3 + CO  . 2 KOH + 2MnO + 4H2O. OCH3 | (c) FeCl3 . 6H2O + 6CH3 — C — CH3  FeCl3 (anhydrous) + 12CH3OH + 6CH3COCH3 . | OCH3 Ex. 5 A compound (A) is used in paints instead of salts of lead. Compound (A) is obtained when a white compound (B) is strongly heated. Compound (B) is insoluble in water but dissolves in sodium hydroxide forming a solution of compound (C). The compound (A) on heating with coke gives a metal (D) and a gas (E) which burns with blue flame. (B) also dissolves in ammonium sulphate solution mixed with ammonium hydroxide. Solution of compound (A) in dilute HCl gives a bluish white / white precipitate (F) with excess of K4[Fe(CN)6]. Identify (A) to (F) and explain the reactions. Ans. (A) ZnO, (B) Zn(OH)2, (C) Na2 ZnO2 , (D) Zn, (E) CO, (F) K2Zn3 [Fe(CN)6]2 Sol. Zn(OH)2 (B) ZnO (A) + H2O. Zn(OH)2 (B)  + 2OH–  [Zn(OH)4]2– (C) (soluble complex). ZnO (A) + C Zn (D) + CO (E). Zn(OH)2 (B) + 4NH3  [Zn(NH3)4]2+ (soluble complex) + 2OH–. ZnO + 2HCl  ZnCl2 + H2O. 3ZnCl2 + 2K4[Fe(CN)6]  K2Zn3[Fe(CN)6]2  (bluish white/white) (F) + 6KCl. Ex. 6 An unknown inorganic compound (X) gave the following reactions: (i) The compound (X) on heating gave a residue, oxygen and oxide of nitrogen. (ii) An aqueous solution of compound (X) on addition to tap water gave a turbidity which did not dissolve in HNO3. (iii) The turbidity dissolves in NH4OH. Identify the compound (X) and give equations for the reactions (i), (ii) & (iii). Ans. X = AgNO3 Sol. 2AgNO3 (X) 2 Ag + 2NO2 + O2 . AgNO3 (aq.) + Cl—  AgCl  (white) + NO3— . AgCl + 2NH3  [Ag(NH3)2]+ (soluble complex). TRANSITION ELEMENTS Ex. 7 Sol. Which of the following is true for the species having 3d 4 configuration ? (A) Cr2+ is reducing in nature. (B) Mn3+ is oxidising in nature. (C) Both (A) and (B) (D) None of these Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t32g energy level of 3d orbitals in octahedral crystal field spliting. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability. Therefore, (C) option is correct. Ex. 8 Which of the following increasing order of oxidising power is correct for the following species ? VO2+ , MnO4– , Cr2O72– Sol. (A) VO2+ < Cr2O72– < MnO4– (B) VO2+ < MnO4– < Cr2O72– (C) Cr2O72– < VO2+ < MnO4– (D) Cr2O72– < MnO4– < VO2+ This is attributed to the increasing stability of the lower species to which they are reduced. MnO4– is reduced to Mn2+ which has stable half filled valence shell electron configuration [3d5]. Cr2O72– is reduced to Cr3+ which has half filled t32g energy level of 3d orbitals in octahedral crystal field spliting VO2+ is reduced to V3+ which has electronic configuration [Ar]18 3d2 4s0. So the order of increasing stability of the reduced species is Mn2+ > Cr3+ > V3+ and, therefore, the increasing order of oxidising power is VO2+ < Cr2O72– < MnO4– . Therefore, (A) option is correct. Ex. 9. Which of the following statement(s) is/are correct ? (A) Transition metals and many of their compounds show paramagnetic behaviour. (B) The enthalpies of atomisation of the transition metals are high (C) The transition metals generally form coloured compounds (D) Transition metals and their many compounds act as good catalyst. Sol. (A) As metal ions generally contain one or more unpaired electrons in them & hence their complexes are generally paramagnetic (B) Because of having larger number of unpaired electrons in their atoms, they have stronger inter atomic interaction and hence stronger bonding between the atoms. (C) According to CFT, in presence of ligands the colour of the compound is due to the d-d transition of the electrons. (D) This activity is ascribed to their ability to adopt multiple oxidation state and to form complexes. Therefore, (A,B,C,D) options are correct. Ex. 10 Statement-1 : The number of unpaired electrons in the following gaseous ions : Mn3+, Cr3+, V3+ and Ti3+ are 4, 3, 2 and 1 respectively. Statement-2 : Cr3+ is most stable in aqueous solution among these ions. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True Sol. (B) Mn3+ = [Ar]18 3d4 , Cr+3 = [Ar]18 3d3 , V3+ = [Ar]18 3d2 , Ti3+ = [Ar]18 3d1 Cr3+ is most stable in aqueous solution because it has half filled t32g energy level of 3d orbitals in octahedral crystal field spliting and according to crystal field theory (CFT) it has highest value of CFSE i.e. 1.2 O. Ex. 11 Sol. S1 : Interstitial compounds have high melting points, higher than those of pure metals. S2 : Permanganate titrations in presence of hydrochloric acid are unsatisfactory. S3 : KMnO4 does not act as an oxidising agent in strong alkaline medium. S4 : KMnO4 on heating in a current of H2 gives MnO. (A) T T F T (B) T F F T (C) T F T T (D) F F T F S1 : Due to strong interatomic forces. S2 : Some of the hydrochloric acid is oxidised to chlorine and thus we get less volume of KMnO 4 than the actual one. S3 : MnO4– + e– OH  MnO42– S4 : 2KMnO4 + 5H2  2KOH + 2MnO + H2O. Ex. 12 Therefore, (A) option is correct. Match the reactions in Column I with nature of the reactions/type of the products in Column II. Column I Column II (A) AgNO3(aq) + I2 (excess) + H2O  (p) Disproportionation (B) K2MnO4(aq) + CO2(g)  (q) Comproportionation (C) Na2Cr2O7 + C (r) Redox (D) CuCl2(aq) + Cu(s)  (s) One of the products is insoluble in water Ans. (A  p, r, s) ; (B  p, r, s) ; (C  r, s) : (D  q, r, s) Sol. (A) 5AgNO3(aq) + 3 I 2 (excess) + 3H2O  HIO3 + 5AgI + 5HNO3 V 0 –1 So it is redox and disproportionation reaction. AgI insoluble in water. +VI +VII +IV (B) 3K 2 MnO 4 (aq) + 2CO2(g)  2KMnO 4 + MnO 2 + 2K2CO3 So it is redox and disproportionation reaction. MnO2 insoluble in water. 0 +VI (C) 2C + Na 2 Cr2O 7 +III +IV II Cr2O3 + Na 2 CO3 + CO  So it is redox reaction. Cr2O3 (green pigment) is insoluble in water. +II +I 0 (D) CuCl2 (aq) + Cu (s)  Cu 2Cl 2 (s) So it is redox and comproportionation reaction. Cu2Cl2 is insoluble in water. Ex. 13 When CO2 is passed into aqueous : (A) Na2CrO4 solution, its yellow colour changes to orange. (B) K2MnO4 solution, it disproportionates to KMnO4 and MnO2 (C) Na2Cr2O7 solution, its orange colour changes to green (D) KMnO4 solution, its pink colour changes to green. Sol. (A) Na2CrO4 H+ Na2Cr2O7 (orange colour) (B) MnO42– H+ MnO4– + MnO2 , in neutral or acidic medium (C) False - In acidic medium no colour change takes place. - (D) MnO4– + e– OH MnO42– ; in strong alkaline medium pink colour of KMnO4 changes to green. Therefore, (A,B) options are correct. TRANSITION ELEMENTS Ex. 14 Which of the following statement(s) is (are) not correct with reference to ferrous and ferric ions (A) Fe3+ gives brown colour with potassium ferricyanide (B) Fe2+ gives blue precipitate with potassium ferricyanide (C) Fe3+ gives red colour with potassium sulphocyanide (D) Fe2+ gives brown colour with potassium sulphocyanide Sol. Fe3+ produces red colouration with KSCN but Fe2+ does not give brown colour with KSCN. Therefore, (D) option is correct. Ex. 15 Statement-1 : Ammonical silver nitrate converts glucose to gluconic acid and metallic silver is precipitated. Statement-2 : Glucose acts as a weak reducing agent. (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True Sol. (A) Ag2O + C6H12O6  2Ag + C6H12O7. Ex. 16 What is the composition of mischmetal alloy and what are it's uses ? Ans. Mischmetal consists of lanthanoid metal (~95%) and iron (~5%) and traces of S,C,Ca and Al. Mischmetal is used in Mg based alloy to produce bullets, shell and lighter flint.