All elements of the first transition series have oxidation state (+2) because after losing the electrons of (4s) sublevel at first (except for scadium), while in the higher oxidation states they lose the electron of (3d) in sequence.. These fourteen elements are represented by common general symbol ‘Ln’. (iii) Which element of the first transition series has lowest enthalpy of atomization? (c) Assertion is not true but the Reason is true. (d) KMnO4 acts as a reducing agent in the presence of HCl. Question 1. Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? Question 60. 3Cu + 8HNO3 –> 3CU(N03)2 + 2NO + 4H2O, Question 18. Solution: (a) Cu(II) is more stable due to greater effective nuclear charge of Cu(II). are trapped inside the crystal lattices of metals. (b) Mo(VI) and W(VI) are more stable than Cr(VI) After treatment of this yellow solution with sulphuric acid, compound (C) can be crystallized from the solution. (b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine What are the characteristics of the transition elements and why are they called transition elements? Assertion (A): The highest oxidation state of osmium is +8. Question 24. Reason (R): because it has positive electrode potential. On account of this, the ionisation enthalpies of Th, Pa and U are comparatively lower than Ce, Pr and Nd. ii) Which element has the highest m.p? 1(a). iii) Which elements shows only +3 oxidation state ? (a) They have high melting points in comparison to pure metals Solution: Compounds formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals are called interstitial compounds. Solution: (b) HCl is not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium. (d) both belong to the same group of the periodic table A group of fourteen elements following lanthanum i.e. Question 2: (a) Given reasons for the following : Name the elements showing maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z=30) Name the element which shows only +3 oxidation state (b) What is lanthanide contraction ? In these elements, the last electron enters the 4f-subshells (pre pen ultimate shell). Thus, 5f-electrons are more effectively shielded from the nuclear charge than 4f electrons in the lanthanide series. For maintenance: the two lists are compared in this /datacheck, to gain mutual improvements. The elctronic configuration of Manganese is Mn (25) = [Ar} 3d 5 4s 2 The reason why Manganese has the highest oxidation state is because the number of unpaired electrons in the outermost shell is … (i) Which element of the first transition series has highest second ionization enthalpy? Question 55. The reason is that if HCl is used, the oxygen produced from KMnO4 + HCl is partly utilized in oxidizing HCl to Cl, which itself acts as an oxidizing agent and partly oxidises the reducing agent. Oxidation States of 3d Series. (ii) Mn03F. Out of Cu2Cl2 and CuCl2, which is more stable and why? 94) (d) Am (Atomic no. (a) U (Atomic no. (d) Radii of the elements of Ad- and 5d-blocks in the same vertical columns are nearly the same. Ce – 4f15d’6s2 (Ce4+– 4f°). Solution: (b) Cu2+ has 1 unpaired electron in CuF2, hence, it is coloured in solid state. Solution: According to n +l rule: For   3d=n + l = 5 Question 34. For example, Mn exhibits all oxidation states from +2 to +7 as it has 4s23d5 configuration. NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12. Question 69.Mention the type of compounds formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals. The elements of 3d transition series are given as: Sc Ti V Cr Mn Fe Co Ni Cu Zn. Solution: (b, d) Np and Pu show +7 oxidation state. Name an important alloy which contains some of the lanthanoid metals. Which of the following is not the characteristic property of interstitial compounds? Which of the following element does-not belong to this series? Solution: Question 9. The oxidation states are also maintained in articles of the elements (of course), and systematically in the table {{Infobox element/symbol-to-oxidation-state}} (An overview is here). Explain why does this happen? when the number of unpaired valence electrons increases, the d-orbital increase & the highest oxidation state increases. Solution: As the oxidation state of the element increases, its charge increases. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because (a) Both HCl and KMn04 act as oxidizing agents. Solution: The electronic configuration of Ce is -4f15d16s2. Transition metals show variable O.S due to incomplete orbital E.Configuration. (c) Stability of the complexes of lanthanoids. Solution: Negative E° values of Mn2+ and Zn2+ are because of stabilities of half-filled (3d10: Mn2+) and fully-filled (3d10: Zn2+) configuration respectively.Mn2+ ion has higher E° value because of highest negative enthalpy of hydrogen. Which of the following is the correct electronic configuration of gadolinium? Ans. Although +3 oxidation states is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. (iii)    Decrease in basicity: With the decrease in ionic radii, covalent character of their hydroxides goes on increasing from Ce(OH)3 to Lu(OH)3 and so base strength goes on decreasing. Although Cr3+ and CO2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87 BM and that of CO2+ is 4.87 BM. So, oxo-anions of a metal have the highest oxidation state. The common oxidation state of 3d series elements is + 2 which arises due to participation of only 4s electrons. (c) Pu (Atomic no. (iii) Actinoids show irregularities in their electronic configurations. Question 52. The shielding effect of f-orbitals in very poor due to their diffused shape. (e) Trends in acidic character of lanthanoid oxides. Why first ionization enthalpy of Cr is lower than that of Zn? Here we have given NCERT Exemplar Class 12 Chemistry Chapter 8 The d- and f-Block Elements. Hence, among the given four choices Cu belongs to right side of Periodic Table in transition metal, and it has the highest density (89 g/cm3). (d) Ti2+ and Cr2+ are reducing agents in aqueous solution Match the statements given in Column I with the oxidation states given in Column II. What is lanthanoid contraction? Therefore, the reactivity degreases as we move from Sc to Cu. Match the properties given in Column I with the metals given in Column II. As the question states, the number of oxidation states exhibited by an element increases from Sc (up +3) to Mn (up +7). Solution: The electronic configuration of fluorine is 1s22s22p5 . (C) Identify compounds A, B and C. is treated with KI, iodide ion is oxidized to (vii)    These metals form various alloys with other metals of the series. Which of these elements will have highest density? Oxygen is strong oxidising agent due to its high electronegtivity and smaller size. 4s = n + l = 4 On the basis of lanthanoid contraction, explain the following: It may be mentioned that oxygen also has vacant d-orbitals along with two 3p orbitals containing single electron. When alkaline KMnO4 © Simultaneously an electron is also added which enters to the inner f subshell. Question 16. (i) Osmium is an element which show +8 oxidation state. Gadolinium belongs to 4f series. (ii)   These metals exhibit variable oxidation states. KMnO4 acts as an oxidizing agent in acidic medium. MnO 4- (permanganate ion) has Mn +7 & CrO 4-2 (chromate ion) has Cr +6. Question 15. (d) 3.57 B.M. +8 oxidation states are shown by both Os and Rh. KMn04 acts as an oxidizing agent in alkaline medium. The highest oxidation state of a metal exhibited in its oxide or fluoride only because oxygen and fluorine have small size and high electronegativity, so they can easily oxidize the metal to its highest oxidation state. (v)    The compounds of these metals are usually paramagnetic in nature. 1) In 3d-series of transition metals, manganese has an atomic number of 25 that gives the electronic configuration as [Ar] 3d 5 4s 2,where we see that the maximum number of unpaired electrons is found in manganese atom; so, it can show a maximum oxidation state upto +7. Question 31. Reason (R): Cu2+ oxidises I to iodine. Solution: Question 27. (ii) 3d block element that can show up to +7 oxidation state is manganese. Reason (R): Because Zr and Hf lie in the same group of the periodic table. Identify the compound from the following: (d) in covalent compounds fluorine can form single bond only while oxygen forms double bond Solution: (d) When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because Mn2+ acts as an autocatalyst. (b) (i) Fe(CO)5 as per EAN rule. Solution: (b, c) A species can act as an oxidizing agent only when metal is present in high oxidation state but lower oxidation state show stability. (ii) Zn. Mn shows +7 oxidation state; d-electrons are not involved in bonding. Question 8. Solution: (a) Copper does not liberate hydrogen from acids. Solution: (i —> b), (ii —> a), (iii —> d), (iv e), (v—> c). (iii) 3d block element with highest melting point is chromium. In moving from Sc, the first element to Cu, the ionization enthalpy increases regularly. Each question carries two mark. Why E° value for Mn, Ni and Zn are more negative than expected? Why? (b) They are very hard Identify compounds A to D and also explain the reactions involved. Solution: The compounds A, B, C and D are given as under: NCERT Exemplar ProblemsMathsPhysicsChemistryBiology. In alkaline medium (pH > 7). Solution: (a) (i) Cu, because the electronic configuration of Cu is 3d104s1. Question 33. While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d-orbital but reverse happens during the ionization of the atom. Due this this they exhibit variable O.S. electron needs to be removed from completely filled tf-orbital. Solution: High ionisation enthalpy to change Cu(s) to Cu2+ is not balanced by hydration enthalpy. Solution: (a) V2O5 and Cr2O3 are amphoteric oxides because both react with alkalies as well as acids. Question 4. This is because of involvements of greater number of electrons in the interatomic metallic bonding from (n – 1) d orbitals in addition to ns electrons. Spin only magnetic moment value of Cr3+ ion is Solution: (a) 2KMnO4 + 2H2SO4(Conc ) –> Mn2O7 + 2KHSO4 + H2O. Question 20. Name a metal in the 3d series of transition metals which exhibit +1 oxidation state … Solution: The high melting points of transition metals are due to strong metallic bonds between the atoms of these elements. 92) (b) Np (Atomic no. (iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. Why? Solution: Question 58. Question 57. They are chemically so similar that their separation is difficult. Assertion (A): Cu can not libirate hydrogen from acids . Which of the following will not act as oxidizing agents? (d) Stability of Cu(I) and Cu(I) depends on nature of copper salts. How does Fe(III) catalyse the reaction between iodide and persulphate ions? (a) 2.87 B.M. Reason (R): Actinoids can utilize their 5d orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding. Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4 Solution: (c) When acidified K2Cr2O7 solution is added to Sn2+ salt, Sn2+ changes to Sn4+. (a) Sn (b) Sn3+ (c) Sn4+ (d) Sn+ Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium? The regular small decrease in atomic radii and ionic radii of lanthanides with increasing atomic number along the series is called lanthanoid contraction.Cause of lanthanoid contraction: When one moves from 58Ce to 71Lu along the lanthanide series nuclear charge goes on increasing by one unit every time. From the table, TiX 4, VF 5 and CrF 6 have the highest oxidation numbers.. Question 50. H2SO4 and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Generally, transition elements form coloured salts due to the presence of unpaired electrons. (a) Cr (b) Co (c) Cu (d) Ni The second and third rows of tfansition elements resemble each other much more than they resemble the first row. Why are such compounds well known for transition metals? (a) Answer the following questions: Pr and Nd are higher than Th, Pa and U. (d) Both Assertion and Reason are false. (b) Both Assertion and Reason are true and the Reason is not the correct explanation of Assertion. Also give physical and chemical characteristics of these compounds. Question 42. Question 26. These elements constitute one of the two series of inner transition elements or f-block.Lanthanoid contraction: In the lanthanoide series with the increase in atomic number, atomic radii and ionic radii decrease from one element to the other, but this decrease is very small. Question 54. Solution: (a) Reaction between iodide and persulphate ions is: Question 71. Why? (c) KMnO4 is a weaker oxidizing agent than HCl. However, Zn exhibits a lower value of ionization enthalpy because a stable 3d10 configuration is attained after losing two electrons. Solution: (d) The highest oxidation state of manganese in fluoride is +4 (MnF4) but in oxides it is +7 (Mn2O7) because in covalent compounds fluorine can form single bond only while oxygen forms double bond. (a) CO2 is formed as the product (b) Reaction is exothermic (iii) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements. Although +3 is the characteristic oxidation state for lanthanoids but cerium (i) Osmium is an element which show +8 oxidation state. (a) Am (b) Pu (c) U (d) Np But Ce exhibit +4 oxidation state also because it gains extra stability by losing 4f1 electron because it will give rise to completely filled orbitals. also shows +4 oxidation state because  4s electrons are loosely held by the nucleus. Sr and Hf; Nb and Ta; Mo and W. This resemblance is due to the similarity in size due to the presence of lanthanoids in berween. It is due to its characteristic electronic configuration i. e., (n – 1)d and ns electrons take part in bond formation either by loosing or by sharing of electrons with other combining atoms.The stability of oxidation state depends mainly on electronic configuration and also on the nature of other combining atom.The elements which show largest number of oxidation states occur in or near the middle of series (i.e., 4s23d3 to 4s23d7 configuration). Solution: (a) Cu can not libirate hydrogen from acides due to positive electrode potential. Question 40. The magnetic nature of elements depend on the presence of unpaired electrons. In the transition metals, the stability of higher oxidation states increases down a column. and M.P. 93) Match the catalysts given in Column I with the processes given in Column II. So electrons are removed from 4s orbital prior to 3d. Some borides of transition elements approach diamond in hardness. B) ... [3d\] elements show variable oxidation states because the energies … Which of the following actinoids have one electron in 6d orbital? (a) fluorine is more electronegative than oxygen Question 32. Solution: Reactivity of an element is dependent on the value of ionization enthalpy. In 3d series, ‘Mn’ shows maximum number of oxidation states because it has maximum number of electrons in s as well as in d orbitals which can take part in bond formation. Question 13. Solution: It is because in the beginning when 5f-orbitals begin to be occupied, they Solution: CuCl2 is more stable than Cu2Cl2 . Transition elements show variable oxidation states because electrons from both s and d orbitals take part in bond formation. (a) 25 (b) 26 (c) 27 (d) 24 Ltd. Download books and chapters from book store. Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal. Question 44. Transition elements show high melting points. Solution: (i —> c), (ii —> a), (iii —>b) https://www.zigya.com/share/Q0hFTjEyMDY4NjEy. Therefore, it exhibits a positive E° value. It was mentioned previously that both copper and chromium do not follow the general formula for transition metal oxidation states. Explain. Solution: (b) The highest oxidation state of osmium is +8 due to its ability to expand their octet by using its all 8 electrons. Solution: (a, d) Co2+ (3d7) and Cr3+ (3d3) have 3 unpaired electrons. Explain why? Short Answer Type Questions [II] [3 Marks] (d) They are chemically very reactive. NCERT Exemplar Class 12 Chemistry Chapter 8 The d- and f-Block Elements are part of NCERT Exemplar Class 12 Chemistry. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). Question 5. There is no unpaired electron in Zn for metallic bonding. Solution: (a) 64Gd: [Xe] 4f7 5d1 6s2. Question 59. Question 3. 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