(b) (i) The highest oxidation state shown in oxoanions of transition metals is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (Cr shows +6) and it is due to the ability of oxygen to form multiple bonds with the metal atoms. Question 23. Hence it loses one electron and achieves the stable configuration i.e. While Cr3+ is d3 is half-filled (t2g3) is stable in nature and Cr2+ is d4, has one extra electron which it would like to donate to attain the stable half-filled (t2g3) configuration. Hence acting as strong reducing agent. (b) Explain the following observations : (ii) Basicity difference : Due to lanthanoid contraction, the size decreases from La+3 to Lu+3. (ii) 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. (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. (iii) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why? (Delhi & All India 2009) (iii) The enthalpies of atomization of the transition elements are quite high. Mn2+ prefer to lose an electron or get oxidised whereas Fe2+ will readily loose one electron or get oxidised. (b) What is lanthanoid contraction and what is it due to? This formation of chromate (CrO4–) ion converts the colour of solution to yellow. Question 59. The first few members of the lanthanoids series are quite reactive. b) +3. MnO4– + 8H+ + 5e– → Question 78. Since there is very little energy difference between these orbitals, both energy levels can be used for bond formation. (ii) Lanthanoids show limited number of oxidation state, viz. (i) Copper atom has completely filled d orbitals (3d10) in its ground state, yet it is regarded as a transition element due to incompletely filled d-orbital in its ionic states i.e. (i) Copper atom has completely filled d orbitals (3d10) in its ground state, yet it is regarded as a transition element. Except for scandium, the most common oxidation state of 3d elements is +2 which arises from the loss of two 4s electrons. Illustrate with examples. Answer: Delhi 2013) (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. (ii) The E°M2+/M for any metal is related to the sum of the enthalpy changes taking place in the following steps : In the 3d series, manganese show the highest no. Answer: Due to this they have high enthalpies of atomization. (i) Among lanthanoids, Ln (III) compounds are predominant. 2Cu+ →Cu2++ Cu (i) Number of oxidation states exhibited Therefore Cr2+ is reducing agent. Compound (B) on reaction with KC1 forms an orange coloured crystalline compound (C). Therefore the 3rd ionization energy of Mn will be very high and Mn3+ is unstable and can be easily reduced to Mn2+. (a) Copper exhibits + 1 oxidation state more frequently i.e., Cu+1 because of its electronic configuration 3d104s1. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult. Answer: Consequences : Answer: Question 22. A mixed oxide of iron and chromium is fused with sodium carbonate in free access of air to form a yellow coloured compound (A). (i) K2MnO4 from MnO2? why La(OH)3 is most basic while Lu(OH)3 is least basic. Answer: The highest oxidation state +7, for manganese is not seen in simple halides, but MnO 3 F is known.. VF 5 is stable, while the other halides undergo hydrolysis to give oxohalides of the type VOX 3.. Fluorine stabilises higher oxidation states either because of its higher lattice energy or higher bond enthalpy. (All India 2009) Which one of the following is not correct? (i) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements. On acidification the compound (A) forms an orange coloured compound (B), which is a strong oxidizing agent. (ii) In transitional elements, in addition to metallic bonding there is extra covalent bonding due to presence of unpaired electrons in their ‘d’ orbitals, hence they are much harder. At the same time, atomic mass increases from Ti to Cu, therefore density increases. (Comptt. 2Cu+ → Cu2+ + Cu, Question 75. (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. Question 55. Actinoids show the oxidation state from +3 (most common) to +7, while lanthanoids show the oxidation state from +3 up to + 7. List of oxidation states of the elements This is a list of all the known oxidation states of the chemical elements, excluding nonintegral values. (a) (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. (iii) Sc shows only +3 oxidation state. Hence it is a good oxidising agent. The oxidation number of a free element is always 0.         M(s)--> M(g)                                 sH (Sublimation energy). (b) (i) 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. (i) Transition elements generally form coloured compounds. (i) Transition metals form compounds which are usually coloured. Difference: Actinoids show wide rage of oxidation states but lanthanoids do not. (iii) Cr2+ has the configuration 3d4 which easily changes to d3 due to stable half filled t2g orbitals. The high energy required to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. (Delhi 2012) All India 2014) (ii) There is similarity in size of elements belonging to same group of second and third transition series. Due to small change in atomic radii, the chemical properties of lanthanoids are very similar due to which separation of lanthanoid becomes very difficult. (a) Why do transition elements show variable oxidation states? Thus covalent character increases. (a) Following are the transition metal ions of 3d series: (i) The highest oxidation state of a transition metal is usually exhibited in its oxide. (ii) Cr2+ is a strong reducing agent. why La(OH)3 is most basic while Lu(OH)3 is least basic. Co3+ can accomodate more no. Delhi 2014) (a) Lanthanoid contraction : The overall decrease in atomic and ionic radii with increasing atomic number is known as lanthanoid contraction. (Delhi 2009) All India 2016) (Comptt. (b) Positive oxidation state: Oxygen does not show positive oxidation state except OF2(O = + 2). (iii) Transition metals have high enthalpy of atomisation. All India 2017) (i) La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions. (a) (i) In transition elements, the oxidation Delhi 2014) The oxidation state of oxygen is usually -2 except in compounds with fluorine, oxygen has a positive oxidation number. 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. The reaction is given as: Name an important alloy which contains some of the lanthanoid metals. (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). Therefore Cr2+ is reducing agent, it gets oxidized to Cr3+. (i) Difference between lanthanoids and actinoids : (ii) Cerium Answer: Ti4+ = 3d0 4s0 = no unpaired electron Delhi 2014) Question 76. From the table, TiX 4, VF 5 and CrF 6 have the highest oxidation numbers.. (a) What is meant by the term lanthanoid contraction? The sizes of transition metal atoms and ions are also favourable for transition complex formation with the reactants. (ii) Na2Cr2O7 from Na2CrO4? The steady decrease in the ionic radius from La3+ to Lu3+ is termed as lanthanoid contraction. (i) Transition metals exhibit variable oxidation states. VARIABLE OXIDATION STATE. Answer: (Delhi 2016) (ii) d-block elements exhibit more oxidation states than f-block elements. (ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. Question 28. (iii) From element to element actinoid contraction is greater than the lanthanoid contraction. (a) (i) Copper has positive E0M2+/M value because the sum of enthalpies of sublimation and ionization is not balanced by hydration enthalpy. Answer: Therefore the 3rd ionization energy of Mn will be very high and Mn3+ is unstable and can be easily reduced to Mn2+. (ii) The enthalpy of atomization is lowest for Zn in 3d series of the transition elements. (Delhi 2015) (i) Cu+ is unstable in an aqueous solution. Question 84. So 3rd ionisation potential of Mn2+ is much higher. (i) Similarity in properties: Due to lanthanoid contraction, the size of elements which follow (Hf – Hg) are almost similar to the size of the elements , of previous row (Zr – Cd) and hence these are difficult to separate. (iii) The E0 value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple. (ii) Oxometal ions are polyatomic ions with oxygen. (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. (ii) Cerium shows +4 oxidation state because it acquires stable empty orbital configuration and therefore Ce+4 is also used as a good analytical reagent and good oxidising agent. Reaction with oxalic acid in Acidic medium (ii) The electronic configuration of Mn2+ ion is more symmetrical as compared to that of Cr2+ ion. , Janakpuri, New Delhi, Delhi - 110058 metal is usually except. As ΔH1, < ΔH2 < ΔH3 reaction with KC1, orange crystals of compound ( b ) E°M2+/M. Also shows +1 and + 6, 7, 8, 9, 10 11! 4 oxidation states solutions but gets easily oxidised to Fe2+ but less easily dichromate ion used in devices. Colour in solutions or in solid compounds are predominant E0M2+/M and why compounds, + 3 and + oxidation. Positive E0 ( M2+/M ) for copper is positive but small i.e between sulphate... 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