CLASS 12 CBSE PHYSICS HANDOUTS d and f - Block

CLASS 12 CBSE PHYSICS HANDOUTS d and f - Block Hand-Out Chemistry d and f-Block © 2012 Vidyamandir Classes Pvt. Ltd. d and f - Block Q1. What are d-Block elements? Ans. These are those elements in which the last electron enter the d-subshell. Q2. Why are d-block elements called transition elements? Ans. Because in this block the elements lie between s and p block i.e., between group 12 and group 13. Q3. Give the general electronic configuration of d-block elements. Ans. (n – 1)d1–10 ns1– 2 Q4. Name the 3 series of transition elements. Ans. (i) 3d series First transition series. Has 10 elements from scandium (Sc) to Zinc (Zn) (ii) 4d series Second transition series. Has 10 elements from Yttium (Y) to cadmium (Cd) (iii) 5d series Third transition series. Has 10 elements from Lanthanium (La) to mercury (Hg) Q5. Which is the most commonly known feature/characteristic of d-block elements. Ans. It is the presence of un paired electrons in the d-orbitals ; i.e., incompletely set of d-orbitals. Q6. Why Zn, Hg and Cd often not regarded as transition elements. Ans. Being the end members of their respective transition series, their configuration is generally given by (n – 1) d10 ns2 i.e., these elements have completely filled d-orbitals and hence do not fulfill the general characteristic of d-block elements which the pressence of incompletely filled d-orbital. Hence are not regarded as transition elements. Q7. On what grounds you can say that Scandium (Z = 21) is a transition elements but Zn is not? Ans. Electronic configuration of Sc (Z = 21) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 Electronic configuration of Zn (t = 30) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 We can say this on the basis of filling of d orbitals. Sc fulfills the general condition of d-block due to presence of incompletely filled d-orbitals (Presence of unpaired electrons) whereas zinc does not as it has completely filled of orbital. Q8. Why does Ionic size or atomic size decrease in any transition series ? Ans. As were move left in the transition series the nuclear charge increase by one unit and the additional e– goes into the inner d-orbital and since d e–s are poor shielded so they high effective nuclear charge and attracted towards the nucleus due to which decrease in atomic and ionic radius is observed. Q9. Why do transition metals show variable oxidation states? Ans. This is because the outer ns electrons also participate with the inner (n – 1)d electrons in bond as their energies are same. And also due to presence of incompletely filled d-orbitals so diff no. of electrons can be lost by atoms. Q10. Why are most of the transition elements paramagnetic? How is their magnetic moment calculated? Ans. This is due to the presence of unpaired electrons and each e– has a magnetic moment associated with spin and rotation around the axis Magnetic moment () = n(n2) B.M. n = no. of unpaired e– in d-orbital. Q11. Calculate the magnetic moment of the compound with Z = 25. Ans. Z = 25 — 1s2 2s2 2p6 3s2 3p6 4s2 3d5 n = 5 μ n(n2) B.M.  5(52)  35  5.92 B. M. Q12. Why do transition metals firm complex compounds? Ans. (i) Due to high Charge / size ratio (Due to small size) (ii) Availability of vacant d-orbitals for bond formation. Q13. What is Lanthanoid Contraction? Give its cause and consequence. Show with the help of graph. Ans. The steady decrease in atomic and ionic sizes of Lanthanoid elements with increasing atomic number is called Lanthanoid contractions. Causes :- As one moves from one element to another in the Lanthanoid series, the nuclear charge increase by one unit the additional e- goes into the inner 4f orbital and since 4f electrons are poor sheilded for high effective number charge and are attracted towards the nucleus due to which decreases atomic and ionic size is observed. Consequences :- (i) Due to lanthanoid contraction the increase in radius from second to third transition series almost vanishes (Zr to Hf and Nb to Ta) have similar radii. (ii) Due to very small change in radii of the lanthanoids, their chemical properties are quite similar. Hence from a mixture their separation is very different (iii) Due to decrease in ionic radius (due to lanthanoid contraction) the size of ion decreases and hence the covalent character of the bond between Ln3+ (lanthanoid ion) and OH– ion increases from La3+ to Lu3+. Therefore basic strength of hydroxides decreases with in the atomic no. La(OH)3 - Most Basic Lu (OH)3- Least Basic [La = Lanthanum, Lu = Lutetium] Q14. Why Cr2+ reducing and Mn3+ oxidizing when both have the same d4 configuration? Ans. Cr2+ is reducing as its configuration changes from d4 to d3 i.e., half filled t2g, on the other hand, when we change from Mn3+ to Mn2+, it results in extra stable half filled d5 configuration due to which Mn3+ is oxidising. Q15. Why is that melting point increase from 3d to 4d to 5d series? Show the graph for m.pt v/s atomic no. for the 3 series. Ans. This is because metal-metal bonding becomes more frequent from 3d to 4d to 5d series due to availability of more no. of valence electrons. Q16. What happen to density from Ti to Cu? Why? Ans. The density increases from Titanium to copper because when we move from Ti to Cu, the atomic mass increases but the metallic radius decreases [e– enters the d-orbital hence poor shielding effect] And also density= Atomic Mass Atomic Vol. Q17. Why do transition metals have high enthalpies of atomisation? Ans. Because of strong inter atomic interactions which are a result of large number valence electrons involved in metal bonding. Q18. Name a transition element liquid at room temperature. Ans. Mercury (Hg) Q19. Why ionisation enthalpy increases from left to right? Ans. Because of increase in nuclear charge due to which inner d-orbitals are filled up (poor shielding effect). Q20. Why do transition metals from coloured ions? Ans. Due to presence of unpaired electrons in d-orbital eq:- Ti3+ forms purple coloured ion Cu2+ from blue coloured ions Ti 4+, Sc3+, Zn2+ are colourless due to absence of upaired electrons. Q21. Why are transition metals and their compounds used as catalyst? Ans. (i) Ability to form complexes (ii) Ability to form multiple oxidation states. eq. Iron (III) catalyst is used in reaction with Iodide and per sulphate ions. 2I- + S2O8 -2 give( in presence of Fe (III)) I2 + 2 SO4 -2 Q22. What are interstetial compounds ? Why do transition metals form interstitial compounds? Ans. Compounds which are formed when small non-metal atoms like H, N, C, Br are trapped inside the voids/ interstitial sites of the crystal lattice of the metals are called interstitial compounds. Transition metals form interstitial compounds because crystal lattice of these metals have empty spaces called voides where small non-metal atoms like H, N, C, Br can get trapped. Characteristics of Interstitial compounds / transition metals (i) These metals are non-stoichiometric. (ii) High M. P. and are pure metals. (iii) Hard, conducting and chemically inert. (iv) Neither typically covalent nor ionic. Q23. Why transition metals form alloys? Ans. Because transition elements metals have got almost similar radii hence atoms of one element can easily get distributed among the atoms of other elements and form alloy. eq. ferrous alloys. Q24. Explain why Eº value for Mn3+ / Mn2+ couple is larger than that of Cr3+ / Cr2+ or Fe3+ / Fe2+?. Ans. This is because when we move form Mn2+ to Mn3+ to electronic canfiguration changes from 3d5 to 3d4, hence the stable contion is changed due to which the 3rd ionisation enthalpy is very very high and hence Eº value is also high (+ve). Q25. Explain the irregular behaviour Eº V Cr Mn Fe Mn2+ / Mn – 1.18 0.91 – 1.18 – 0.44 Ans. The irregular behaviour can be enplained from the irregular varitaion in ionisation energys (IE1 + IE2) and also the sublimation energy which are very law for manganese and vanadium. Q26. Actinoids contraction is larger than lanthanoid contraction? Why? Ans. This is because the shielding effect of 5f electron is poorer than that of 4f electrons. Q27. Give the methods of preparation of potassium permanganate (KMnO4). Ans. (i) By fusion of MnO2 (pyrolusite ore) with an alkali metal hydroxide in pressure of O2 2 MnO2 + 4KOH + O2 –– 2K2MnO4 + 2H2O The dark green K2 MnO4 (potassium mangnate) produced undergoes dispraportionation reaction in acid medium. 3MnO4 – + 4H+ ––– 2MnO4 – + MnO2 + 2H2O (from K2MnO4) (Permanganate) (ii) Commercial Preparation :- KMnO4 is prepared by alkalin oxidative fusion of MnO2 followed by the electrolytic oxidation of manganateion (MnO4 2–) KOH,O2 2 MnO2 MnO4  2 oxidation Θ 4 in altrative solution 4 electrolytically (permanganate) Q28. Write the equations to show oxidising action of KMnO4 in acidic and basic medium. Ans. In Acidic Medium:- MnO4 – + 8H+ + 5e– ––– Mn2+ + 4H2O In Basic Medium:- MnO4 – + 2H2O + 3e– ––– MnO2 + 4OH Q29. Give two reactions to show oxidising action of KMnO4 in acid medium. Ans. (i) KMnO4 – + 8H+ + 5Fe2+ ––– Mn2+ + 5Fe3+ + 4H2 (ii) 2MnO4 – + 10H+ + 10 I ––– 2Mn2+ + 5I2 + 8H2 Q30. Give two reactions to oxidising action of KMnO4 in basic medium. Ans. (i) MnO4 – + 2H2O + 3Fe2+ ––– MnO2 + 3Fe3+ + 4OH (ii) 2MnO4 – + 4H2O + 6I ––– 2MnO2 + 3I2 + 8OH Q31. Give the properties and uses of KMnO4. Ans. (i) Strong oxidising agent in acidic and basic medium. (ii) It is purple solid. Uses:- (i) Used for decolorisation of oils. (ii) Used as oxidant in lab. Q32. Draw the structure of MnO4 –. Ans. (Tetrahedral) Q33. Why do transition metals have high melting and boiling points? Ans. Due to strong inter atomic bonding which involves outer ns electrons with (n – 1) d inner elements in bonding. (because greater the number of valence electrons, greater is the bonding). Q34. How is potassium dichromate prepared? Ans. K2Cr2O7 is prepared from chromite ore (fe2 Cr2 O4). (i) Fusion of chromite ore with Na2Co3 Na2 in air 4Fe Cr2O4 + 8Na2CO3 + 7O2 –––8Na2CrO4 + 2Fe2O3 + 8CO2 (ii) Acidification of sodium chromatic formed. 2Na2CrO4 + 2H+ ––– Na2Cr2O7 + 2Na + + 2H2O (iii) Treatment of sodium dichromate with KCl. 2 2 7 2 2 7 (potassium dichramate orange crystals) Q35. Give the chemical equation to show oxidising action of K2Cr2O7 in acidic medium. Ans. Cr2O7 2– + 14 H+ + 6e– ––– 2Cr3+ + 7H2O (chromic ion) Q36. Give three reactions to show oxidising action of K2Cr2O7 in acidic medium. Ans. (i) Cr2O7 2– + 14H+ + 6Fe2+ ––– 2Cr3+ + 6Fe3+ + 7H2O (ii) Cr2O7 2– + 14H+ + 6I- –– 2Cr3+ + 3I2 + 7H2O (iii) Cr2O7 2– + 14H+ + 3Sn2+ ––– 2Cr3+ + 3Sn4+ + 7H2O Q37. Explain how chromate and dichromate ions interconvertible? Or explain the effect of pH on chromate and dichromate ions. Ans. 2CrO4 2– + 2H+ ––– Cr2O7 2– + H2O Cr2 O7 2– + 2OH ––– 2CrO4 2– + H2O Q39. Draw the structures of chromate and Dichromate ions. Ans. Chromate ion –– CrO4 2– O O Cr O O 2– (Tetrahedral) Dichromate ion –– Cr2O7 2– (Two Tetrahedral units sharing one O) Q40. What is the highest oxidation state of Cr . Ans. +6 in CrO4 2– Q41. What is the highest oxidation states of Mn. Ans. +7 in KMnO4 Q42. What are inner transition elements? Ans. These are those elements which have partly filled f-orbitals in their nutral atom or ion. Q43. What are Lanthanoids? Ans. Elements from La (Lanthanium) to Lu (lutetium) are called Lanthanoids. These are characterised by the filling of 4forbitals in their atoms. Q44. Give the general electronic configuration of lanthanoid. Ans. [Xe] 4f 0 – 14 5d 0 – 1 6s2 Q45. What is the most common oxidation state of lanthanoids? Ans. +3, however some elements exhibit +2 and +4 also. Q46. What are actinoids? Ans. The elements from Ac (actinium) to Lr (Lawrencium) are called actinoids. These are characterised by filling of 5f orbitals. Q47. What is the general electronic config of actionids? Ans. [Rn] 5 f 0 – 14 6d 0 – 2 7s2 Q48. Why among lanthanoids, Ce (III) can be easily oxidised to Ce (IV)? Ans. Electronic config. Ce ––– [Xe] 4f 1 5d1 6s2 Ce3+ ––– [Xe] 4f 1 Ce4+ ––– [Xe] In Ce3+, one electrons is present in the 4f orbital. This electron can be lost to from Cr4+ which has a stable configuration of anoble gas reaction. Therefore Cr (III) can be easily oxidised to Cr (IV). Q49. Campare the chemistry Lanthanoids and actinoids on the basis of electron config, oxidation state, atomic,ionic sizes and chemical reactivity. Ans. Lanthanoids Actinoids (i) [Xe] 4f 0 – 14 5d 0 – 1 6s2 [Rn] 5f 0 – 14 6s0 – 2 7s2 (ii) Oxidation state + 3 is principle , +2 and +4 3 general +4, +5, +6, +7 also exhibited also exhibited (iii) Irregular decrease in size across the series Gradual decrease in size. (iv) Less reactive than actinoids More reactive than Lanthanoids (5f orbital are placed furthur away from the nuclius than 4f orbital)