(b) SC3+ = 4S0 3d3+ = no unpaired electron (ii) Although Co2+ ion appears to be stable, it is easily oxidised to Co3+ ion in the presence of a strong ligand. Answer: Question 38. (ii) Actinoids exhibit a much larger number of oxidation states than the lanthanoids. (a) There is a gradual decrease in the size of atoms with increasing atomic number in the series of lanthanoids. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult. 2Cu+ → Cu2+ + Cu. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult. (i) The catalytic properties of the transition elements are due to the presence of unpaired electrons in their incomplete d-orbitals and variable oxidation states. (i) Because oxygen stabilizes the highest oxidation state (+7 of Mn) even more than fluorine i.e., +4 since oxygen has the ability to form multiple bonds with metal atoms. However +2 and +4 ions in solution or in solid compounds are also obtained. State reasons for the following : The mixed oxide of iron and chromium is chromite or chrome ion i.e. (i) The transition metals form a large number of interstitial compounds in which small atoms such as hydrogen, carbon, boron and nitrogen occupy the empty spaces in the crystal lattices of transition metals. : Mn = 25, Cr = 24)(All India 2014) Answer: (ii) The chemistry of actinoids is not so smooth as that of lanthanoids. (a) (i) Mn3+ has electronic configuration 3d44s0. energy absorbed and low enthalpy of hydration (i.e. All India 2012) The variable oxidation states shown by the transition elements are due to the participation of outer ns and inner (n–1)d-electrons in bonding. (i) Because of presence of unpaired d electrons, which undergoes d-d transition by absorption of energy from visible region and then the emitted light shows complementary colours. All India 2013) Oxidation number of (group I) elements like Li, Na, K, … Pb(II), Pb(IV), Sn(II), Sn(IV) etc. (i) Copper (I) ion is not known in aqueous solution. There's nothing surprising about the normal Group oxidation state of +4. loss of further electrons requires high energy. Answer: (ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series as these have their electrons of outer most shell at greater distance from the nucleus, as compared to atoms of 3d transition metals. Answer: (i) Zinc Atomic no. This is because although second ionization enthalpy of copper is large but Δhyd (hydration enthalpy) for Cu2+(aq) is much more negative than that for Cu+(aq) and hence it more than compensates for the second ionization enthalpy of copper. (iii) The value of E° for Mn is more negative than expected from the general trend due to greater stability of half filled d-subshell (d5) in Mn2+. Answer: (b) Chemistry of actinoids is complicated as compared to lanthanoids. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O Question 55. Elements which have partially filled d-orbital in its ground states or any one of its oxidation states are called transition elements. Reaction with oxalic acid in Acidic medium (iii) With the same d-orbital configuration d4, Cr2+ ion is a reducing agent but Mn3+ ion is an oxidising agent. E° value for Fe3+ | Fe2+ is positive but small i.e. Assign reason for each of the following : (ii) Due to comparable energies of 5f 6d and 7s orbitals of actinoids, these show larger number of oxidation states than corresponding members of lanthanoids. (ii) Transition metals form coloured compounds. (iii) Actinoids exhibit a greater range of oxidation states than lanthanoids. 2CrO4-2 + 2H+ → Cr2O7-2 + H2O (iii) In transition elements, there are large number of unpaired electrons in their atoms, thus they have a stronger inter atomic interaction and thereby stronger bonding between the atoms. a) +2 . (ii) Cu2+ (aq) + I–(aq) → (i) Transition metals and their compounds are generally found to be good catalysts. Sodium dichromate (B) on reaction with KCl forms orange coloured compound Potassium dichromate (C). (iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states. Hence the contraction in size occurs. Question 47. Answer: All India 2015) (Atomic nos. (b) (i) Colour is due to the presence of unpaired electrons in their d-subshells. (b) Sc (21) is regarded as a transition element due to the presence of incomplete d- subshell (3d14s2) but Ca (20) does not have any d-subshell. Answer: Answer: These elements typically display metallic qualities such as malleability and ductility, high values of electrical conductivity and thermal conductivity, and good tensile strength. Give reasons : states of transition metals different from that of the p-block elements? (iii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26. (a) Complete the following equations : (iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows : (i) $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$$ (aq) + H2 S(g) + H+(aq) → (ii) Co2+ ion is easily oxidised to Co3+ ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d2sp3). Delhi 2015) (Comptt. All transition metals exhibit a +2 oxidation state (the first electrons are removed from the 4s sub-shell) and all have other oxidation states. Copper has high enthalpy of atomization (i.e. Question 61. The lanthanoid contraction arises due to imperfect shielding of one 4f electron by another present in the same subshell. In going from La+3 to Lu+3 in lanthanoid series, the size of ion decreases. Answer: Answer: Answer: (ii) Cr+2 (3d4) after loosing one electron forms Cr+3 (d3). Answer: (All India 2012) d) Es, No, Lr . Complete the following equation : Assign reasons for the following : (At. This decrease in size in the lanthanoid series is known as lanthanoid contraction. (iii) The large positive E° value for Mn3+ | Mn2+ shows that Mn2+ is much more stable than Mn3+ due to stable half filled configuration (3d5). Answer: c) U, Th, Md . (Comptt. K2MnO4 disproportionates in a neutral or acidic solution to give potassium permanganate. Give reasons : (i) Transition elements are known to form many interstitial compounds. In galvanization, zinc coats iron by oxidizing to form a protective layer of zinc oxide (ZnO) that protects the iron from oxidation. 25. neon and argon is inert gas because they are more stable than other.Uses of Neon: (i) Neon is mainly used in fluorescent lamps of tubes for advertising purposes. The s orbital in these elements is completely filled and p orbitals are half filled, making their electronic configuration extra stable. Pyrollusite ore is digested in KOH in the presence of oxygen Question 56. (Comptt. MnO (+4) and Mn2O7 (+7) (i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4. All India 2015) Why do transition elements show variable oxidation states? Question 21. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give KMnO4 (ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La+3 to Lu+3. Actinoids are radioactive while lanthanoids are not radioactive. (i) Zn is not considered as a transition element. (a) Following are the transition metal ions of 3d series: Hence Mn3+ easily changes to Mn2+ and acts as oxidising agent. (iii) Hydride formation: All the elements O, S, Se, Te and Po form M2M type hydrides (where M = O, S, Se, Te and Po). Nitrogen exhibits + 1, + 2, + 4 oxidation states also when it reacts with oxygen. S, Se, Te, Po show + 4, +6 oxidation state in addition to + 2. Complete the following chemical reaction equations : (All India 2012) (a) (i) K2MnO4 from MnO2 (Pyrolusite) : Delhi 2012) Answer: Consequences : Question 65. (a) Complete the following chemical equations : The steady decrease in the ionic radius from La3+ to Lu3+ is termed as lanthanoid contraction. 5) MnO4- +Â 8H+Â +Â 5e-Â ---> Â  Â  Mn2+Â + 4Â H2OÂ, (i) Electronic configuration: They all have six electrons in the outermost shell and have ns2 np4 general electronic configuration. Similarity : Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration. Ltd. Download books and chapters from book store. Therefore Cr2+ is reducing agent. (i) Mn2+compounds are more stable than Fe2+ compounds towards oxidation to their +3 state. (a) (i) latex]\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}[/latex] + 3H2S + 8H+ → 2Cr3+ + 7H2O + 3S The high energy required to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. Answer: On the other hand, the electronic configuration of Fe+2 is [Ar]3d6. (iii) Along a transition series, the nuclear charge increases which tends to decrease the size but the addition of electrons in the penultimate d-subshell increases the screening effect which counter balances the effect of increased nuclear charge. (iii) Orange colour of Cr2O72- ion changes to yellow when treated with an alkali. (i) Many of the transition elements are known to form interstitial compounds. Prove this statement. As transition metals contain a large number of unpaired electrons, they have strong interatomic attractions (metallic bonds). (iii) This happens because the energy difference between 5f, 6d and 7s subshells of the actinoids is very small and hence electrons can be accomodated in any of them. This decrease in size in the lanthanoid series is known as lanthanoid contraction. Answer: (All India 2012) Sulfur is another p block element which has different oxidation numbers.-2: Na 2 S , H 2 S; 0: S 8 +4: SO 2, H 2 SO 3 +6: H 2 SO 4, BaSO 4; Chlorine Complete the following chemical equations: (Delhi 2013) Hence Co2+ oxidises to Co3+. (iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states. Answer: Potassium dichromate (K2Cr2O7) acts as a strong oxidising agent in acidic medium using H2SO4. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl The number of oxidation states increases with increase in the number of unpaired 3d electrons. (ii) The oxidising power of the following three oxoions in the series follows the order: (Comptt. (i) Cu+ ion is unstable in aqueous solutions. (i) Mn has the maximum number of unpaired electrons present in the d-subshell (5 electrons). (All India 2013) (ii) In a transition series of metals, the metal (iii) Sc shows only +3 oxidation state. (ii) Because third electron has to be removed from stable half filled 3d-orbitals (Z = 25 has 3d5 4s2). For example: RuO 4, OsO 4; Question 9. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult. (Comptt. Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number. (ii) From titanium to copper the atomic size of elements decreases and mass increases as a result of which density increases. (Delhi 2013) (iii) Actinoids because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states. (b) Positive oxidation state: Oxygen does not show positive oxidation state except OF2(O = + 2). Mention its main consequences. (i) MnO4 (aq) + S2O32- (aq) + H2O (1) → Answer: nos. (Comptt. Question 71. The oxidation state of oxygen is usually -2 except in compounds with fluorine, oxygen has a positive oxidation number. (i) Difference between lanthanoids and actinoids : (ii) Cerium This is because they have empty, half-filled and completely filled 4/ sub-shell respectively which show extra stability. (i) $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$$ (aq) + H2S (g) + H+ (aq) → (ii) Which element has the highest m.p.? 2020 Zigya Technology Labs Pvt. (ii) E°M2+/M for any metal is related to the sum of enthalpy of atomisation, ionization and hydration enthalpy. This happens becauseÂ the E0 (M2+/M) value of a metal depends on the energy changes involved in the following: (i) Sublimation energy:Â  The energy required for converting one mole of an atom from the solid state to the gaseous state. (iii) Mn2+ exists in half-filled d5 state which is very stable while Mn3+ is d4 which is not so stable. Hence Mn3+ easily changes to Mn2+ and acts as oxidising agent. (b) Because of the availability of d-orbitals, they can easily form intermediate products which are activated. Ionic reactions : Question 33. Answer: 2Cu+ → Cu2+ + Cu (i) Name the element of 3d transition series which shows maximum number of oxidation states. e) Highest oxidation state shown in 5d series is +8 by osmium (more stable). (i) 2KMnO4 + 5SO2 + 2H2O → K2SO4+ 2MnSO4 + 2H2SO4 (ii) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals and presence of unpaired electrons. Therefore, Mn2+ is much more resistant than Fe2+ towards oxidation. (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii) Due to lanthanoid contraction in second series after lanthanum, the atomic radii of elements of second and third series become almost same and hence show similarities in properties. Explain the following observations giving an appropriate reason for each. When chromite ore FeCr2O4 is fused with NaOH in presence of air, a yellow coloured compound (A) is obtained which on acidification with dilute sulphuric acid gives a compound (B). (a) Complete the following chemical reaction equations : (iii) Out of Cr3+ and Mn3+, which is a stronger oxidising agent and why? Explain the following observations : (ii) Co2+ ion is easily oxidised to Co3+ ion in presence of a strong ligand because of its higher crystal field energy which causes pairing of electrons to give inner orbital complexes (d2sp3). (b) What is lanthanoid contraction and what is it due to? Answer: Thus Cr2+ is easily oxidised to Cr3+ but Fe2+ cannot be easily oxidized to Fe3+. The chromite ore FeCr2O4 on fusion with NaOH in presence of air, forms a yellow coloured compound (A) i.e. Question 10. (v) The orange coloured potassium dichromate solution when treated with basic NaOH solution, is converted to chromate which gets a faint colour like yellow. (ii) Sc It has completely filled ‘d’ orbitals. (A) → Sodium dichromate Na2Cr2O7 (i) Transition metals and many of their compounds act as good catalysts. The value ofÂ E0(M2+/M) for copper Â is Â (+0.34). (i) Mn (ii) MnO is basic while Mn2O7 is acidic because the basic nature decreases as the oxidation state or number of oxygen atoms increases i.e. (Comptt. Describe the preparation of potassium permangnate. Copper is the only metal in the first series of transition elements showing this behaviour. (iii) Mn2+ is much more resistant than Fe2+ towards oxidation. Question 14. (iv) In transition elements, the successive oxidation state differs by unity, e.g, Mn shows all the oxidation states from +2 to +7. (i) From titanium to copper the atomic size of elements decreases and mass increases as a result of which density increases. (Delhi 2016) All India 2017) Consequences : (ii) Which ion is a strong oxidising agent and why? Question 57. Delhi 2017) (a) Why do transition elements show variable oxidation states? While Mn2+ has stable half filled d5 configuration. This means that after scandium, d orbitals become more stable than s orbital. How would you account for the following? (ii) d-block elements exhibit more oxidation states than f-block elements. (All India 2017) (ii) Mn3+ is a strong oxidising agent because after gaining one electron it is converted into Mn2+ which has stable d5 configuration. (Comptt. Question 90. (a) (i) Actinoid contraction is greater than lanthanoid because 5f electrons (in actinoids) have a poorer shielding effect than 4f electrons (in lanthanoids). Elements having electrons (1 to 10) present in the d-orbital of the penultimate energy level and in the outer most ‘s’ orbital (1-2) are d block elements.Although electrons do not fill up ‘d’ orbital in the group 12 metals, their chemistry is similar in many ways to that of the preceding groups, and so considered as d block elements. oxidation of Fe+2 to Fe+3 is easily achieved. Manganese, in particular, has paramagnetic and diamagnetic orientations depending on what its oxidation state is. (i) Name the element showing maximum number of oxidation states among the first series of transition metals from Se (Z = 21) to Zn (Z = 30). Answer: As a result E° value of Mn3+/ Mn2+ couple is much more positive than for Cr3+/Cr2+ couple. (b) Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. As transition metals contain a large number of impaired electrons, they have strong interatomic attractions (metallic bonds). Hence basic character of hydroxides also decreases i.e. (ii) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism. (b) Balanced ionic equation : (i) Write the element which shows maximum number of oxidation states. (Atomic numbers: Ti = 22, V= 23, Mn = 25, Cr = 24) (iii)Oxidation State: The common oxidation states of these elements are –3, + 3 and + 5. However, occasionally in solutions or in solid compounds, + 2 and + 4 ions are also obtained. Answer: Question 49. (b) Explain the following observations : (ii) Because of lanthanoid contraction. Write two consequences of lanthanoid contraction. Hence it loses one electron and achieves the stable configuration i.e. (ii) (C) is used as a strong oxidising agent in acidic medium in volumetric analysis. (b) (i) Cu2+(aq) is much more stable than Cu+(aq). Question 22. Question 81. Define lanthanoid contraction. Answer: (i) The transition metals and their compounds are usually paramagnetic. •Au 3+ is present as Complexed Species in all Solution as anionic Species . K2Cr2O7 + 4H2SO4 → K2SO4 + Cr2(SO4)3 + 4H2O + 3[O] (a) Complete the following chemical reaction equations : (a) Describe the preparation of potassium permanganate from pyrolusite ore. What happens when acidified potassium permanganate solution reacts with ferrous sulphate solution? (b) (i) The electronic configuration of Mn+2 is [Ar] 3d5 i.e. (Delhi 2017) Mn2+ prefer to lose an electron or get oxidised whereas Fe2+ will readily loose one electron or get oxidised. All India 2012) Question 34. Therefore Cr2+ is reducing agent. (i) With the same d-orbital configuration (d4), Cr2+ is a reducing agent while Mn3+ is an oxidising agent. (iii) From element to element actinoid contraction is greater than the lanthanoid contraction. Answer: (i) Transition metals form compounds which are usually coloured. Lanthanoid contraction: The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. (i) Among lanthanoids, Ln (III) compounds are predominant. (Comptt. (ii) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple or Fe2+/Fe2+ couple. Hence 5f electrons are also taking part in chemical bonding. Fe3+ can also be reduced to Fe2+ but less easily. (Delhi 2012) (a) Negative oxidation state: Except the compound OF2 oxygen shows-2 oxidation state in all its compounds. Give reasons : (i) Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoids becomes very difficult. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows : Hence E°Cu2+/Cu is positive. (b) (i) Zinc in its common oxidation state of +2 has completely filled d-orbitals. Actinoid contraction is greater than lanthanoid contraction due to poor shielding of 5f electrons. Answer: Question 16. Question 11. (b) The large positive E° value for Mn3+/Mn2+ shows that Mn2+ is much more stable than Mn+3 due to stable half filled configuration (3d5). Delhi 2014) What is meant by ‘’disproportionation’? (ii) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why? -1: Example for -1 oxidation state is hydrogen peroxide (H 2 O 2).-2: Most common oxidation number of oxygen is -2. (ii) MnO2(s) + KOH(aq) + O2 → (i) The number of oxidation states increases upto middle of series i.e. Answer: (Delhi 2017) (ii) HCl is not used to acidify KMnO4 solution. What is meant by ‘disproportionation’? (ii) Cr2+ is reducing and Mn3+ oxidizing when both have d4 configuration. Question 46. How would you account for the following? (ii) With the same d4 d-orbital configuration Cr2+ ion is reducing while Mn3+ ion is oxidizing. Question 28. Answer: Question 45. M(g) + ΔaH → M(g) (ΔaH = enthalpy of atomization) (iii) This is due to comparable energies of 5f, 6d, 7s orbitals in actinoids. (b) How is the variability in oxidation states of transition elements different from that of non-transition elements? (ii) Both O2 and F2 stabilize high oxidation states of transition metals but the ability of oxygen to do so exceeds that of fluorine. Question 36. The tendency to exhibit –3 oxidation state decreases down the group due to increase in size and metallic group. The high energy required to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. (b) Which of the following cations are coloured in aqueous solutions and why? (iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state. Answer: List of oxidation states of the elements This is a list of all the known oxidation states of the chemical elements, excluding nonintegral values. (iii) The actinoid contraction is greater than lanthanoid contraction due to poorer shielding of 5f electrons as they are extended in space beyond 6s and 6p orbitals whereas 4f orbitals are buried deep inside the atom. d) +6 . (b) Explain the following observations : Describe the preparation of potassium permanganate. These elements have Complete the following chemical equations : (Delhi 2011) Delhi 2014) Question 9. Cr2O72- + 8H+ + 3NO–2 → (All India 2015) Answer: Consequences : (a) (i) Cr3+ is most stable because of its small size and t32g configuration. Among the elements of 3d –series Manganese belonging to 7 th group exhibits maximum oxidation state. Mn2+ prefer to lose an electron or get oxidised whereas Fe2+ will readily loose one electron or get oxidised. (a) (Delhi 2010) Justify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuraton, oxidation state and hydride formation. When compound (B) is treated with KC1, orange crystals of compound (C) crystallise out. Thus Fe3+ is more stable than Mn3+. (iii) There is a greater range of oxidation states among the actinoids than among the lanthanoids. (ii) Transition metals exhibit variable oxidation states. Fe3+ can also be reduced to Fe2+ but less easily. Hence considered as non-transition elements. (ii) E°M2+/M values are not regular for first row transition metals (3d series). (iii) Which ion is colourless and why? The balanced ionic equation: Â the energy released when one mole atom! To give stable 3d10 configuration heavier transition elements and their compounds act as good catalysts balanced by its hydration:. From the above rule to find the unknown oxidation state of d-orbitals common oxidation state of 3d series elements they easily! Present as Complexed Species in all its compounds Zirconium and Hafnium ( Z = ). Unpaired d electrons available in the series follows the order: ( i ) Cu2+ ( )! Decrease in atomic size with increasing atomic numbers in a series of the solution Zinc is used alloys... Size and metallic group oxygen is usually -2 except in compounds with fluorine, oxygen has a positive number! The +7 oxidation state shown in 5d series is +2 not regular first! Of Mn3+/ Mn2+ couple is highly positive ( +1.57 v ) strong oxidising agent and why vapours used! Preparation of potassium permanganate solution in general, decreases down the group to! Element is a reducing agent, it gets oxidized to Fe3+ devices for protecting certain electrical (. ) the E° value of Mn3+/ Mn2+ couple is highly positive ( 0.34. Size: covalent and ionic state can easily lose 4s1 electron to 3d54s0. 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Both energy levels can be seen at long distances even when There is very little energy difference (! Quite reactive is less + 3H2O, Question 63 a ) give reasons the... 3+ and Fe 2+, Cu 2+ and Cu + Cu + 57 ) and Hf almost! Which ion is a strong ligand elements Answer: Question 54 ) write one similarity and one difference the! Value ofÂ E0 ( M2+/M ) for copper Â is Â ( +0.34 ) hardly any in.