Page 1 : Chapter, , 8, , The d and f-Block Elements, Solutions, SECTION - A, School/Board Exam. Type Questions, Very Short Answer Type Questions :, 1., , State, whether the given reaction takes place in acidic, basic or neutral medium., MnO4– + e–  MnO42–, , Sol. Basic medium., 2., , Will enthalpy of atomisation will increase or decrease from Sc to Cr?, , Sol. First increase from Sc to V and then decrease., 3., , Write the E.C. of Fe+., , Sol. 1s 2 2s 2 2p6 3s2 3p6 3d 6 4s1, 4., , Why element of 5d have more IE1 than corresponding elements of 4d (Except 3rd group)?, , Sol. Due to poor screening of 4f-orbitals., 5., , Write the reaction of K2Cr2O7/H+ with I–., Cr O2 /H, , 2 7,  I2, Sol. I , , 6., , Which oxidation state of Cr is most stable in aq. medium?, , Sol. +3., 7., , Why Cu+ is less stable than Cu2+ in aq. medium?, , Sol. Due to hydration energy., 8., , Which of the halide/halides of V (in +5 oxidation state) is/are stable?, , Sol. VF5, 9., , The most stable oxidation state of Ni will be ________., , Sol. Ni = +2., 10. Why the trend of stability M2+/M for 3d series is not regular?, Sol. Due to irregular trend of (IE1 + IE2) and enthalpy of atomisation., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 2 : 34, , The d and f-Block Elements, , Solution of Assignment (Set-1), , Short Answer Type Questions :, 11. Calculate the magnetic moment of given ion, (i), , Fe3+, , (ii), , Mn2+, , Sol. Fe+3  3d 5, , , , , 5(5 , 2), , 35 B.M., , Mn+2  3d 5, ,   35 B.M., 12. Why Fe3+ is more stable than Fe2+ while Cu3+ is less stable than Cu2+?, Sol. Due to half filled electronic configuration of Fe+3; Fe+3 is more stable while IE3 of Cu is very high and hence, it is not much stable., 13. Discuss the effect of pH on CrO42– in aq. solution., Sol. 2CrO4–2 + 2H+, , Cr2O7–2 + 2H2O, , So, on increasing H+ concentration equilibrium will shift in forward direction., 14. Why IE3 of Mn is higher than Fe but for IE2 case is reversed?, Sol. Because Mn+2 have E.C. d 5. That is why it is more stable and removal of electron will be difficult, but Fe+2, have E.C. d 6 and it can easily loose an electron to form more stable E.C., 15. Why transition elements are considered as good catalyst?, Sol. For a substance to act as catalyst following conditions are to be fulfilled, (i), , Large surface area, , (ii), , Variable oxidation state, , (iii) Vacant site for attachment (vacant d orbital), All these properties are shown by transition element so these acts as good catalyst., 16. Define alloy. Which elements are present in bronze?, Sol. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed among the atoms, of other. Bronze contain copper and tin., 17. Why transition metal ions or their compounds are coloured?, Sol. When an electron from a lower energy d orbital is excited to a higher energy d orbital the energy of excitation, corresponds to the frequency of light absorbed. The colour observed corresponds to the complementary colour, of the light absorbed., 18. Complete the following reactions, MnO /H, , (i), , 4, I , , , (ii), , 4, I , , , MnO /OH, , Sol. 10I– + 2MnO4– + 16H+  2Mn+2 + 8H2O + 5I2, 2MnO4– + H2O + I–  2MnO2 + 2OH– + IO3–, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 3 : Solution of Assignment (Set-1), , The d and f-Block Elements, , 35, , 19. Write the structure of MnO4–. Explain which type of double bonds are present., Sol. Permanganate ions are tatrahedral in structure, , O, Mn, O, , O, , –, , (Tetrahedral), , O, , Double bond present between Mn and O is produced by overlapping of d-orbital of Mn and p-orbital of O., 20. Define Lanthanoid contraction., Sol. Due to poor screening of 4f orbitals size of M+3 ion decreases along the series. This is called lanthanoid, contraction., 21., , Ln, , Lanthanoid, , , , H O, , 2773 K, 2, C ,  A ,  B+D, , Identify A, B and D., HO, , 2773 K, 2, Sol. Ln + C ,  Ln2C3 ,  Ln(OH)3 + C2H2, , 22. Why is AgNO3 kept in brown coloured bottles?, Sol. To prevent its photodecomposition AgNO3 is kept in brown coloured bottles. It decomposes as, h, 2AgNO3 , ,  2Ag + 2NO2 + O2, , 23. Write the composition and one use of misch metal., Sol. Misch metal contain 95% lanthanoid and 5% Fe, and in traces Ca, C, N, S may be present., (i), , It is used as catalyst in petrochemical industry., , (ii), , It is used in Mg based alloy to form bullet., , 24. Why Basic character in Ln(OH)3 decreases along the series?, Sol. As we approach from Ce+3 to Lu+3 size decreases, electronegativety increases. That is why acidic character, increases and basic character decreases., 25. Discuss the mechanism of given reaction (Half cell reactions), Fe3, , I– + S2O82–  I2 + SO42–, Sol. I– + Fe+3  FeI2 + I2, Fe+2 + S2O8–2  Fe+3 + SO4–2, 26. Why standard reduction potential of Cu is positive although for other transition elements it is negative?, Sol. Reduction potential depend on following factor :, Cu(s)  Cu(g), Hatomisation, , ...(i), , Cu(g)  Cu+2 + e– (IE1 + IE2), , ...(ii), , Cu+2  Cu+2(aq), –Hhydration, , ...(iii), , Using equation (i), (ii) & (iii), G° is calculated and further G° is used to calculate E°. Due to high (IE1 + IE2)E°, for this process is negative and for its reverse it is positive., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 4 : 36, , The d and f-Block Elements, , Solution of Assignment (Set-1), , 27. Draw the structure of Cr2O72–., , O–, Sol., , O, , Cr, O, , O–, O, , Cr, O, , O, , 28. By using given data, state which M2+ is more stable, Cr2+/Cr = –0.91 V, Mn2+/Mn = –1.18 V, Sol. Lesser will be reduction potential, stable will cation, so better will be reducing agent. Mn+2 is much more stable, than Cu+2. (Note : Reduction potential can be used to compare electrochemical serve)., 29. Why +3 oxidation state of Mn have little importance?, Sol. IE3 of Mn is very high because of half filled E.C., 30. Why Cr2+ is better reducing agent than Fe2+?, , ⎡E,  0.41 V ⎤, ⎢ Cr3  /Cr 2, ⎥, ⎢Fe3+ /Fe2+  0.77 V ⎥, ⎣, ⎦, Sol. Lesser will be reduction potential, better will be reducing agent and hence Cr+3/Cr+2 is less than Fe+3/Fe+2., Long Answer Type Questions :, 31. (i), (ii), Sol. (i), , Write the preparation of KMnO4., What will happen if KMnO4 is heated with KOH?, MnO2 + K2CO3 + [O]  K 2MnO4, , Green solution, , K 2MnO 4 ,  KMnO 4, Oxidation, Electrolytic, , Pink, , KMnO4 is a good oxidising agent., (ii), 32. (i), (ii), Sol. (i), , 2KMnO4 + 2KOH  2K2MnO4 + H2O + [O], Write a short note on ionisation enthalpies of transition element., Why IE3 of Mn is very high?, Ionisation enthalpy : It is the amount of energy required to remove the outer most electron from an, isolated gaseous atom. Trend of ionisation enthalpy is not regular due to poor screening of 3d-orbitals, and exchange energy., For example : IE1 generally increases from left to right but IE1 of Ti, Cr, Co, Ni, Cu is abnormal., , (ii), , Mn+2 have E.C. = 1s2 2s2 2p 6 3s2 3d 5 which is half filled. Half filled E.C. are much more stable and that, is why IE3 of Mn is high., , 33. Write the preparation of K2Cr2O7 from chromite ore., Sol. K2Cr2O7 is prepared from chromite ore as :, Step 1 : 4FeCr2O4 + 8Na2CO3 + 7O2  8Na2CrO4 + 2Fe2O3 + 8CO2, (Yellow), , Step 2 : 2Na2CrO4 + 2H+  Na2Cr2O7 + H2O + 2Na+, Step 3 : Na2Cr2O7 + 2KCl  K2Cr2O7 + 2NaCl, Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 5 : Solution of Assignment (Set-1), , The d and f-Block Elements, , 37, , 34. Define inner transition element. Write any three differences between Lanthanoids and Actinoids., Sol. Inner transition elements are those in which last electrons are filled in (n – 2)f orbital. These inner transition, elements are classified into two category., , Lanthanoid, , Actinoid, , Except Pm, all elements are non (i), radioactive., (ii), (ii) Less tendency to form complex., (iii) Most common oxidation state is (iii), +3. but show +2, +4 also, (i), , 35. (i), (ii), Sol. (i), (ii), , All are radioactive., More tendency to form complex., Actinoid show more variable, oxidation state., , Why in permanganate titration, for acidic medium we cannot use HCl?, What happens if thiosulphate is treated with KMnO4 in neutral medium?, In permanganate titration for acidic medium HCl should not be taken because itself it reacts with KMnO4, and converts to Cl2., OH, , S2O3–2 + MnO4–  SO4–2 + MnO2, , 36. Write any five reactions of KMnO4 which show its oxidising behaviour., Sol. These five reaction shows the oxidising nature of KMnO4, MnO /H, , (i), , 4, I ,  I2, , (ii), , 4, Fe2 ,  Fe3, , (iii), , 4, Sn2 ,  Sn4, , (iv), , 4, CH3 — CH2 — OH ,  CH3 — COOH, , (v), , 4, SO2 ,  H2SO4, , 37. (i), (ii), Sol. (i), (ii), , 38. (i), , MnO /H, , MnO /H, , MnO /H, , MnO /H, , What will happen if V2O5 is treated with acid and alkali?, Why CrO is basic while Cr2O3 is amphoteric?, V2O5 in acidic solution convert to VO4–3 while in basic medium it convert to VO2+., In oxide, as central atom’s electronegativity increases, its behaviour shift from basic to acidic. As oxidation, state of transition metal increases, its electronegativity increases thats why CrO is basic and Cr2O3 is, acidic., On what basis we can explain the given reaction?, Na2Cr2O7 + 2KCl  K2Cr2O7 + 2NaCl, , (ii), Sol. (i), (ii), , How many types of Cr–O bond lengths are present in Cr2O72–?, Na2Cr2O7 is more soluble while K2Cr2O7 is less soluble., Before answering this question, we should know the structure of Cr2O7–2, , O, –, , O, , Cr, O, , O, 126°, , O, , Cr, O, , –, , O, , Six Cr–O bond lengths are equal in which resonance is present while other two Cr–O bond lengths are, same., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 6 : 38, , The d and f-Block Elements, , 39. (i), (ii), , Solution of Assignment (Set-1), , Define transition elements. Why Zn, Cd are not considered as typical transition elements?, Why Hg have abnormally low melting point?, , Sol. (i), , Transition element are those element which have partially filled d orbital in ground state or in its ionic, state. Zn, Cd and Hg are not typical transition element because these atom or their ion do not contain, partially filled d orbital., , (ii), , Hg have no unpaired electron and have very high ionisation energy as a result of that it has low metallic, bond strength and hence low melting point., , 40. (i), (ii), Sol. (i), , Define disproportionation reaction with example., Why CrO42– can not show disproportionation?, Disproportionation reaction : It is the reaction in which same species is oxidised and reduced., MnO4–2  MnO4–1 + MnO2, In this reaction MnO4–2 is oxidised and MnO4–2 is reduced., , (ii), 41. (i), (ii), , In CrO4–2, Cr is in its highest oxidation state thats why it cannot show oxidation. However, CrO4–2 can, be reduced., Why Sc3+ and Ti4+ are colourless?, On what basis we can explain colour of MnO4– and CrO42–?, , Sol. (i), , Generally transition metal ions are coloured due to d-d transition. In Sc+3 and Ti+4 there is no electron, in d orbital thats why no chance of d-d transition. So, these are colourless., , (ii), , MnO4– is pink coloured due to charge transfer from oxygen to metal. In MnO4– and CrO4–2, d-d transition, is not possible. Colour due to charge transfer are very intense., , 42. (i), (ii), Sol. (i), , On what basis, we can say that transition metal ions have high tendency of complex formation?, Define complex compound., For the complex formation following condition should be fulfilled :, (a) Small size, (b) High positive charge density, (c) Vacant orbital to accommodate electron donated by ligand, All these condition are well full filled by d-block elements thats why transition element have high tendency, to form complex., , (ii), , Complex compound is that compound which is formed by a metal surrounded by ligands (bonded by, co-ordinate bond)., , 43. On what factor, value of reduction potential will depend and how we can show that?, Sol. Reduction potential depends on G° of the process which further depend on, (i), , Hatomisation, , (ii), , IE1 + IE2, , (iii) Enthalpy of hydration, Li(s)  Li(g); Hsub = +ve, Li(g)  Li+(g) + e–; IE = +IE kJ, Li+ + aq.  Li+(aq); Hhyd = –x kJ, Using this equation G° and hence calculate the E°., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 7 : Solution of Assignment (Set-1), , 44. (i), (ii), Sol. (i), , The d and f-Block Elements, , 39, , Why metals of XI group are termed as coinage metals? Why Au is considered as noble metal?, Write the common oxidation state of Au., Element of XI group contain Cu, Ag and gold. These metal mainly used to form coin thats why elements, of this group are known as coinage metal., Noble metals are those metal which are not very reactive. Au is not much reactive. It is soluble only in, aqua regia. That is why it is considered as a noble metal., , (ii), 45. (i), (ii), Sol. (i), (ii), , Au mainly show oxidation state of +1 and +3., Why Mn3+ and Co3+ are considered as strong oxidising agents?, Why reduction potential of Ni is abnormal?, o, and EoCo+3 /Co+2 are very, Mn+3 and Co+3 are considered as very good oxidising agent because EMn, +3, /Mn+2, high., , Zn have high ionisation energy and Ni+2 have very large hydration energy. Thats why E° Zn+2/Zn and E°, Ni+2/Ni is abnormal., , SECTION - B, Model Test Paper, Very Short Answer Type Questions :, 1., , Why Zr and Hf have nearly same size?, , Sol. Due to poor screening of 4f orbital or lanthanoid contraction., 2., , Why IE3 of Mn is higher than Fe?, , Sol. Due to half filled stable E.C. of Mn., 3., , Why melting point of Mn is lower than Fe?, , Sol. Due to abnormal lattice., 4., , Which of the transition metal have highest density?, , Sol. Ir., 5., , In 3d series, which metal is expected to be most electropositive?, , Sol. Sc due to its low ionisation energy., 6., , Why reduction potential of Cu2+/Cu is high?, , Sol. Due to its high IE1 + IE2., 7., , Out of these which one is more stable? CrO3 and MoO3., , Sol. MoO3 is more stable. Higher oxidation state are more stable for 4d and 5d for 6th group than 3d., 8., , In Haber process, which metal is used as promotor., , Sol. Mo., 9., , What is the oxidation state of Fe in Fe(CO)5., , Sol. Zero., 10. What is the oxidation state of Fe in Fe3H?, Sol. Zero., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 8 : 40, , The d and f-Block Elements, , Solution of Assignment (Set-1), , Short Answer Type Questions :, 11. Why are transition elements good catalyst?, Sol. Due to its capability to show variable oxidation state and vacant site., 12. What condition should be considered to form an alloy. Write the composition of Brass., Sol. Two metal should have nearly same size. In brass, copper and zinc is present., 13. Which of the following is coloured?, (i), , CuF2, , (ii), , ZnCl2, , (iii) VF5, (iv), , MnO4–, , Sol. CuF2 and MnO4– are coloured., 14. Why Co2+ and Co3+ have different colour in aq. medium?, Sol. Due to difference in energy of 3d orbital., 15. What happens when NO2– and I– is treated with KMnO4 in acidic medium? Write the balanced reaction., Sol. NO2– convert to NO3– while I– convert to I2., 16. Define Lanthanoid contraction., Sol. Due to poor screening of 4f orbitals the size of atoms and ions of lanthanoids decreases it is known as, lanthanoid contraction., 17. Define Actinoid contraction., Sol. Due to poor screening of 5f orbitals, the decrease in size of actonoids known as actinoid contraction., 18. Complete the given reaction, If Ln (Lanthanoid) reacts, , , (i), , Ln  N2 , , (ii), , Ln  S , , (iii), , Ln  X2 , , Sol. (i), (ii), , , , , , LnN, Ln2S3, , (iii) LnX3, 19. Why actinoid show large number of oxidation states?, Sol. Because 5f orbital is not deep seated. That is why it easily takes part in reactions by losing electrons., 20. Why Eu(Europium) show +2 oxidation state?, Sol. Electronic configuratioin of Eu is 4f7 6s2., After the loss of electron, its E.C. becomes half filled. That is why it shows +2 oxidation state., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 9 : Solution of Assignment (Set-1), , The d and f-Block Elements, , 41, , Short Answer Type Questions :, 21. Which of the alloy is used as phosphors in television screens?, Sol. Some individual Ln oxides are used as phosphors in TV screens., 22. (i), (ii), Sol. (i), (ii), , Write the balanced equation of disproportionation of Cu+ in aq. medium., Write the E.C. of Co2+ and Mn2+., 2Cu+  Cu + Cu+2, Mn+2 : 1s2, 2s2, 2p6, 3s2, 3p6, 3d 5, 4s2, Co+2 : [Ar] 3d 7, , 23. Define paramagnetic substance and calculate spin only moment of Cu2+ in aq. solution., Sol. Paramagnetic substance are those which contain unpaired electron and attracted by magnet., Spin only moment is calculated by =, , n(n  2) B.M. where n is number of unpaired electron., , For Cu+2, number of unpaired electron = 1, ,  = 3 B.M., 24. Define interstitial compound. Give two example., Sol. Interstitial compounds are those which are formed when small atoms like H, C, N are trapped inside the crystal, lattice of metal. For example : T1H1.7T1C., 25. In 3d series which metal can show oxidation state of +2? Why Zn does not show oxidation state of +3?, Sol. Ti, V, Cr, Mn, Co, N, Fe, Co, Cu, Zn show +2 oxidation state. IE3 of Zn is very high thats why Zn+3 is not, stable., 26. Define inner transition elements. Out of the following which elements belong to inner transition element?, Pm, Eu, Sm, U, Fe, Cu, Ag., Sol. The elements of f-block are known as inner transition element and Pm, Eu, Sm, U are inner transition element., 27. Which is the last element in the series of Lanthanoid. Write the common oxidation state of that element?., Sol. Lu and its common oxidation state is +3., 28. Why chemistry of elements succeeding is less known for actinoids?, Sol. Actinoids are radioactive. Last elements of actinoids have very small half life period thats why their chemistry, is very less known at present time., 29. Why IE of actinoid is expected to be less than Lanthanoid?, Sol. Because in actinoid, 5f orbitals are not deep seated and it is effectively shielded by other element. Thats why, IE1 of actinoid is lesser than lanthanoid., 30. What is Ziegler-Natta catalyst and for which process is it used?, Sol. TiCl4 + Al(CH3)3. It is used in the manufacture of polyethylene., Long Answer Type Questions :, 31. (i), , Which catalyst is used in the given process?, CHCH  CH3 —CH3, , (ii), , Why AgBr is used in Photographic industry?, , (iii) Write the preparation of K2Cr2O7., Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456

Page 10 : 42, , The d and f-Block Elements, , Sol. (i), (ii), , Solution of Assignment (Set-1), , PdCl2, AgBr is photosensitive., , (iii) FeCr2O4 + Na2CO3 + O2  Na2CrO4 + CO2 + Fe2O3, H, , Na2CrO4 ,  Na2Cr2O7, KCl, , Na2Cr2O7 ,  K 2Cr2O7, 32. (i), (ii), , Why d 1 configuration is very unstable?, On what basis we can explain stability of CoF3?, , (iii) Why VF5 is stable while VCl5 is not stable?, Sol. (i), (ii), , After loss of one more electron, it attains noble gas E.C., which is very stable that is why d 1 E.C. is, not stable., Stability of CoF3 can be explained on the basis of its high Lattice energy., , (iii) VF5 is stable due to its high bond enthalpy of formation., 33. (i), (ii), Sol. (i), (ii), , What do you mean by mixed oxide? Give two examples?, Why MnO4– is more oxidising than Cr2O72– in acidic medium?, These are the oxide in which one element shows two oxidation states. For example in Fe3O4, iron exhibit, +3 and +2 oxidation state. For example : Fe3O4, Co3O4., Due to more stability of Mn+2 than Cr+3 in acidic medium., , , , , , , , Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456