{"id":10472,"date":"2019-12-07T17:45:43","date_gmt":"2019-12-07T12:15:43","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10472"},"modified":"2021-07-02T15:43:20","modified_gmt":"2021-07-02T10:13:20","slug":"2nd-puc-chemistry-question-bank-chapter-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-8\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 8 The d-and f-Block Elements"},"content":{"rendered":"

You can Download Chapter 8 The d-and f-Block Elements Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

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Karnataka 2nd PUC Chemistry Question Bank Chapter 8 The d-and f-Block Elements<\/h2>\n
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2nd PUC Chemistry The d-and f-Block Elements NCERT Textbook Questions and Answers<\/h3>\n

Question 1.
\nWrite down the electronic configuration of:
\n(i) Cr3+<\/sup>
\n(ii) Pm3+<\/sup>
\n(iii) Cu+<\/sup>
\n(iv) Ce4+<\/sup>
\n(v) CO2+<\/sup>
\n(vi) Lu2+<\/sup>
\n(vii) Mn2+<\/sup>
\n(viii) Th4+<\/sup>
\nAnswer:
\nCr3+<\/sup> : 1 s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d3<\/sup> 4s0<\/sup> or [Ar] 3d3<\/sup> 4s0<\/sup>
\nPm3+<\/sup> : 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d10<\/sup> 4s2<\/sup> 4p6 <\/sup>4d10 <\/sup>5s6<\/sup> 5p6 <\/sup>6s0 <\/sup>5d0 <\/sup>4f4<\/sup>
\nCu+ : 1s2 <\/sup>2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d10<\/sup> 4s0<\/sup> or [Ar] 3d10<\/sup> 4s0<\/sup>
\nCr3+<\/sup> : 1s2 <\/sup>2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d3<\/sup> 4s0<\/sup> or [Ar] 3d3<\/sup>4s0<\/sup>
\nCe4+<\/sup> : 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 4s2 <\/sup>3d10 <\/sup>4p6 <\/sup>5s2<\/sup> 4d10<\/sup> 5p6<\/sup> 6s0<\/sup> 5d0<\/sup> 4f0<\/sup> or [Xe] 4f4<\/sup> 5d0<\/sup> 6s0<\/sup>
\nCo2+<\/sup> : [Ar] 3d7<\/sup> 4s0<\/sup>
\nLu2+<\/sup> : [Xe] 4f14<\/sup> 5d1<\/sup> 6s0<\/sup>
\nMn2+<\/sup> : [Ar] 3d5<\/sup> 4s0<\/sup>
\nTh4+<\/sup> : [Rn] 5f0<\/sup> 6d2<\/sup> 7s2<\/sup><\/p>\n

Question 2.
\nWhy are Mn2+<\/sup> compounds more stable than Fe2+<\/sup> towards oxidation to their +3 state?
\nAnswer:
\n\"2nd<\/p>\n

A half filled d orbital or a completely filled d orbital is more stable than any other state. Mn2+<\/sup> already has a half filled stable state hence would not undergo oxidation to form Mn+3<\/sup>. On the other hand Fe2+<\/sup> on oxidation to Fe3+<\/sup> will have a half filled d orbitals which is more stable.<\/p>\n

\"KSEEB<\/p>\n

Question 3
\nExplain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
\nAnswer:
\nExcept Sc, all other metals display +2 oxidation state. Also, on moving from Sc to Mn,’ the atomic number increases from 21 to 25; This means the number of e–<\/sup> in 3d orbital increases from 1 to 5
\nSc2+<\/sup> : d1<\/sup> Ti2+<\/sup> = d2<\/sup> V2+<\/sup> = d3<\/sup> Cr2+<\/sup> = d4<\/sup>
\nMn2+<\/sup> : d5<\/sup>
\n+ 2 oxidation state is attained by the loss of 2, 4s electrons by these metals since the number of d electrons in (+2) state also increases from Ti (+2) to Mn (+2), the stability of +2 state increases (as d – orbital is becoming more is half filled) M+2<\/sup> has d5<\/sup> electrons (that is half filled d shell, which is highly stable).<\/p>\n

Question 4.
\nTo what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
\nAnswer:
\nThe electronic configurations play a very crucial role in deciding the stability of oxidation states. The presence or absence of complete or half filled d orbitals in that oxidation state decides its stability. Also the ability to give away e–<\/sup> decide the oxidation state which depends on the energy possessed by it and hence forth its electronic configuration,
\nFor example
\nSc shows +3 oxidation state only
\n[M] 3d1<\/sup> 4s2 <\/sup>+1, +2 very unstable due to presence of 1 d e- +3 oxidation state is stable with noble gas configuration.
\nMn : [Ar] 3d5<\/sup> 4s2<\/sup> has +2, +3, +4, +5, +6, +7 states but only +2 and +7 states are stable due to the half filled d orbital or completely empty d orbital.
\nsimilarly Ti = +2, +3, +4 with +4 most stable due to the same reason.<\/p>\n

Question 5.
\nWhat may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d3<\/sup>, 3d5<\/sup>,3d8<\/sup> and 3d4<\/sup>?
\nAnswer:
\n\"2nd<\/p>\n

Question 6.
\nName the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
\nAnswer:
\nVanadate : VO3<\/sub>–<\/sup>
\nchromate : CrO4<\/sub>2-<\/sup>
\npermanganate : MnO4<\/sub>–<\/sup><\/p>\n

Question 7.
\nWhat is lanthanoid contraction? What are – the consequences of lanthanoid contraction?
\nAnswer:
\nThe regular decrease in atomic radii due to the filling of 4f before 5d orbitals is called lanthanoid contraction. This is because of the imperfect shielding of one electron by another in the same set of orbitals. The shielding of one 4f electron by another is less than that of one d electron by another, and as the nuclear charge increases along the series, there is fairly regular decrease in the size of entire 4f orbitals. Consequences of lanthanoid contraction.<\/p>\n

    \n
  • Radii of third series are virtually same as those of the corresponding member of second series.<\/li>\n
  • Third series and second series have very similar physical and chemical properties much more than that expected by usual family relationships.<\/li>\n<\/ul>\n

    Question 8.
    \nWhat are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
    \nAnswer:
    \nTransition elements have incompletely filled d orbitals in its ground state or in any one of its oxidation states.
    \nd-block elements are called transition elements and they lie between s and p block elements also all their properties are intermediate between s and p block elements.<\/p>\n

    Zn, Cd, Hg (elements of group 12) are not considered as transition metals because they have fully filled (d10<\/sup>) d orbitals configuration in its ground state as well as in their common oxidation state.<\/p>\n

    \"KSEEB<\/p>\n

    Question 9.
    \nIn what way is the electronic configuration of the transition elements different from that of the non transition elements?
    \nAnswer:
    \nThe general electronic configuration of a d-block elements is (n – 1) d1-10<\/sup> ns1-2<\/sup>.
    \nTransition elements in its ground state or excited state will have a configuration of (n – 1) d1-9<\/sup> ns0-2<\/sup> ie. partly filled d orbitals whereas non transition elements will have a configuration of (n – 1) d10<\/sup> ns0.2\u00a0 <\/sup>i.e completely filled d- orbitals.<\/p>\n

    Question 10.
    \nWhat are the different oxidation states exhibited by the lanthanoids?
    \nAnswer:
    \nThe lanthanoids have an electronic configuration of 6s2<\/sup> common with variable occupancy of 4f level. These elements have only one oxidation state(+3).<\/p>\n

    Question 11.
    \nExplain giving reasons:
    \n(i) Transition metals and many of their compounds show paramagnetic behaviour.
    \n(ii) The enthalpies of atomisation of the transition metals are high.
    \n(iii) The transition metals generally form coloured compounds. .
    \n(iv) Transition metals and their many compounds act as good catalyst.
    \nAnswer:
    \n(i) Paramagnetic arises due to the presence of unpaired electrons, each such electron has a magnetic moment associated with it due to its spin angular momentum. Transition metals have in its ground state or ionised state have number of unpaired d-electrons which gives it a paramagnetic behaviour.<\/p>\n

    (ii) Transition metals have very high interatomic metallic interaction due to the involvement of greater number of electrons from (n – l)d in addition to the ns electrons. Greater the number of valence electrons, stronger is the resultant bonding due to the greater overlapping of half filled orbitals. Hence, more amount of energy is required to break these metallic bond. Thus enthalpy of atomisation of transition metal is very high.<\/p>\n

    (iii) Colour of transition metal – compounds is due to the excitation of an electron from a lower energy d-orbital to a higher energy d orbital. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary colour of the light absorbed (whose frequency lies on the visible region). The frequency of the light absorbed depends on the nature of ligand. Transition metals form coloured compounds due to presence of vacant d-orbitals for the d-d transition of e’ which causes the colour.<\/p>\n

    (iv) The catalytic activity of transition metals is ascribed to their’ability to adopt multiple oxidation states and to form complexes. Catalyst at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst this has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (activation energy is lowered. Transition metals have d and s orbitals to from these bonds.<\/p>\n

    Question 12.
    \nWhat are interstitial compounds? Why are such compounds well known for transition metals?
    \nAnswer:
    \nInterstitial compounds are those which are formed when small atoms like H, N or C are trapped inside the crystal lattices of metals. They are usually non-stoichiometric and are neither typically ionic nor covalent. Interstitial compounds are well known for transition compounds due to its closed crystalline structure with voids in them. The atomic size of transition metals is very large hence have large voids to occupy these small atoms.<\/p>\n

    Question 13.
    \nHow is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
    \nAnswer:
    \nThe variability of oxidation states, a characteristics of transition elements arises out of incomplete filling of d orbitals in such a way that their oxidation states differ from each other by unity This is in contrast with the variability of oxidation states of non transition elements where oxidation states normally differ by a unit of two.
    \neg : Vanadium : V+2<\/sup>, V+3<\/sup>, V+4<\/sup>, V+5<\/sup>
    \nChromium : Cr+2<\/sup>, Cr+3<\/sup>, Cr+4<\/sup>, Cr+5<\/sup>, C+6<\/sup>
    \nNitrogen: +5, +3, +1, -1, -3.<\/p>\n

    Question 14.
    \nDescribe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
    \nAnswer:
    \nPotassium Dichromates are generally prepared from chromates which in turn are obtained by the fusion of chromite ore (FeCr2<\/sub>O4<\/sub>) with sodium or potassium carbonate in free excess of air<\/p>\n

    4FeCr2<\/sub>O4<\/sub> + 8Na2<\/sub>CO3<\/sub> + 7O2<\/sub> \u2192 8Na2<\/sub>CrO4<\/sub> + 2Fe2<\/sub>O3<\/sub> + 8CO2<\/sub><\/p>\n

    The yellow solution of Na2<\/sub>CrO4<\/sub> is filtered and acidified with H2<\/sub>SO4<\/sub> to give a solution from which orange sodium dichromate Na2<\/sub>Cr2<\/sub>O7<\/sub> 2H2<\/sub>O can be crystallised
    \n2Na2<\/sub>CrO4<\/sub> + 2H+<\/sup>\u2192 Na2<\/sub>Cr2<\/sub>O7<\/sub> + 2Na+<\/sup> + H2<\/sub>O<\/p>\n

    sodium dichromate is more soluble than potassium dichromate. The latter is prepared by treating the solution of Na2<\/sub>Cr2<\/sub>O4<\/sub> with KCl.<\/p>\n

    Na2<\/sub>Cr2<\/sub>O7<\/sub> + 2KCl \u2192 K2<\/sub>Cr2<\/sub>O7<\/sub> + 2NaCl (orange crystals)<\/p>\n

    (b) On increasing the PH of a solution of K2<\/sub>Cr2<\/sub>O7<\/sub>, it behaves like a strong oxidising agent and it oxidises iodide to iodine sulphides to sulphur, tin (II) to tin (IV) and Iron (II) salts to iron (III).<\/p>\n

    Question 15.
    \nDescribe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
    \n(i) iodide (ii) iron(II) solution and (iii) H2<\/sub>S
    \nAnswer:
    \nPotassium dichromate, in acidic medium acts as a strong oxidising agent by gaining 6e’ and forms Cr3+.
    \nCr2<\/sub>O7<\/sub>2-<\/sup> + 14H+<\/sup> + 6e–<\/sup> \u2192 2Cr3+<\/sup> + 7H2<\/sub>O ..(1)
    \n(i) iodides
    \n6I–<\/sup> \u2192 3I2<\/sub> + 6e–<\/sup> ..(2)
    \n(1) + (2)
    \nCr2<\/sub>O7<\/sub>2-<\/sup> + 14 H+<\/sup> + 6I–<\/sup> \u2192 2Cr3+<\/sup> + 3I2<\/sub> + 7H2<\/sub>O
    \n(ii) iron (II) solution
    \n6Fe2+ <\/sup> \u2192 6Fe3+<\/sup> + 6e–<\/sup> ..(3)
    \n(1) + (3)
    \nCr2<\/sub>O7<\/sub>2-<\/sup> +14 H+<\/sup>+ 6Fe2+<\/sup> \u2192 2Cr3+<\/sup> + 6Fe3+<\/sup> + 7H2<\/sub>O<\/p>\n

    (iii) H2<\/sub>S
    \n3H2<\/sub>S \u2192 6H+<\/sup> + 3S + 6e–<\/sup> ..(4)
    \n(1) + (4)
    \nCr2<\/sub>O7<\/sub>2-<\/sup> +14H+<\/sup> + 3H2<\/sub>S \u2192 6H+<\/sup> + 3S + 2Cr3+<\/sup> + 7H2<\/sub>O<\/p>\n

    \"KSEEB<\/p>\n

    Question 16.
    \nDescribe the preparation of potassium permanganate. How does the acidified permanganate solution react with
    \n(i) iron(II) ions
    \n(ii) SO2<\/sub> and
    \n(iii) oxalic acid?
    \nWrite the ionic equations for the reactions.
    \nAnswer:
    \nPotassium permangnate is prepared by fusion of MnO2<\/sub> with an alkali metal hydroxide and an oxidising agent like KNO3<\/sub>. This produces the dark green k2<\/sub>MnO4<\/sub> which disproportionates in a neutral or acidic solution to give permanganate.<\/p>\n

    2MnO2<\/sub> + 4 KOH + O2<\/sub> \u2192 2K2<\/sub>MnO4<\/sub> + 2H2<\/sub>O
    \n3MnO4<\/sub>2-<\/sup> + 4H+<\/sup> \u2192 2MnO4<\/sub>–<\/sup>\u00a0+ MnO2<\/sub> + 2H2<\/sub>O<\/p>\n

    Commercially it is prepared by the alkaline oxidative fusion of MnO2<\/sub> followed by the electrolytic oxidation of manganate (VI)<\/p>\n

    \"2nd<\/p>\n

    In laboralory a manganese (II) ion salt is oxidised by peroxidisulphate to permanganate.<\/p>\n

    2Mn2+<\/sup> + 5S2<\/sub>O8<\/sub>2-<\/sup> + 8H2<\/sub>O \u2192 2MnO4<\/sub>–<\/sup> + 10SO4<\/sub>2-<\/sup> + 16H+<\/sup>
    \n(i) 5Fe2+<\/sup> + MnO4<\/sub>–<\/sup> + 8H+<\/sup> \u2192 Mn2+<\/sup> + 4H2<\/sub>O + 5Fe37<\/sup><\/p>\n

    (ii) 2MnO4<\/sub>2<\/sup>+ 10SO2<\/sub> + 5O2<\/sub> + 4H2<\/sub>O \u2192 2Mn2+ <\/sup>+ 10SO4<\/sub>2-<\/sup> + 8H+<\/sup><\/p>\n

    (iii) 5C2<\/sub>O4<\/sub>2-<\/sup> + 2MnO4<\/sub>–<\/sup> + 16H+<\/sup> \u2192 2Mn2+<\/sup> + 8H2<\/sub>O + 10CO2<\/sub><\/p>\n

    Question 17.
    \nFor M2+<\/sup>\/M and M3+<\/sup>\/M2+<\/sup> systems the E\u03b8<\/sup> values for some metals are as follows:
    \nCr2+<\/sup>\/Cr -0.9V
    \nMn2+<\/sup>\/Mn -1.2V
    \nFe2+<\/sup>\/Fe -0.4V
    \nCr3+<\/sup>\/ Cr2+<\/sup> -0.4 V
    \nMn3+<\/sup>\/Mn2+<\/sup> + 1.5 V
    \nFe3+<\/sup>\/Fe2+<\/sup> + 0.8 V
    \nUse this data to comment upon:
    \n(i) the stability of Fe3+<\/sup> in acid solution as compared to that of Cr3+<\/sup> or Mn3+<\/sup> and
    \n(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.
    \nAnswer:
    \n(i) From the above data, we can infer that Fe3+<\/sup> is less stable than Cr3+<\/sup> but more stable than Mn3+<\/sup>. If reducing potential is less than oxidised state will be more stable. This is so because it is easier to add an e–<\/sup> to Fe3+<\/sup> than Cr3+<\/sup> but difficult to do so in case of Fe3+<\/sup> than inMn3+<\/sup>.
    \n(ii) Mn and Cr can be oxidised easily as compared to Fe because the reducing potential of Fe is less negative compared to Mn and Cr. Hence Fe is stable in its reduced form.<\/p>\n

    Question 18.
    \nPredict which of the following will be coloured in aqueous solution?
    \nTi3+<\/sup>, V3+<\/sup>, Cu+<\/sup>, Sc3+<\/sup>, Mn2-<\/sup>, Fe3+<\/sup> and CO2+<\/sup>.
    \nGive reasons for each.
    \nAnswer:<\/p>\n

     <\/p>\n

    Among the above mentioned, ions Ti3+<\/sup>, V3+<\/sup> Mn2+<\/sup>, Fe3+<\/sup>, CO2+<\/sup> will be coloured in its aqueous solution due to the ability of e’ to jump from a lower energy d orbital to a higher energy d orbital. In case of the ions Cu+<\/sup>, Sc3+<\/sup> this d-d transition cannot take place either due to the absence of any e- in 3d orbital or due to complete filling of d orbital.<\/p>\n

    \"KSEEB<\/p>\n

    Question 19.
    \nCompare the stability of +2 oxidation state for the elements of the first transition series.
    \nAnswer:
    \nThe relative stability of +2 oxidation state increases on moving from Sc to Zn. This is because on moving from top to bottom it becomes more and more difficult to remove the third e- from the d-orbital.<\/p>\n

    Question 20.
    \nCompare the chemistry of actinoids with that of the lanthanoids with special reference to:
    \n(i) electronic configuration
    \n(ii) atomic and ionic sizes and
    \n(iii) oxidation state
    \n(iv) chemical reactivity.
    \nAnswer:
    \n(i) Electronic configuration:
    \nThe general electronic configuration for lanthanoids is [Xe] 0-14<\/sup> 5d 0-1<\/sup> 6s2<\/sup> and that of actinoids is [Rn] 5f 0-14<\/sup> 6d0-1<\/sup> 7s2<\/sup>. Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.<\/p>\n

    (ii) Oxidation states:
    \nThe principal oxidation states of lanthanoids is (+3). However sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.<\/p>\n

    (iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction over all decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actionoids. ‘<\/p>\n

    (iv) Chemical activity
    \nIn the lanthanoide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).<\/p>\n

    Question 21.
    \nHow would you account for the following:
    \n(i) Of the d4<\/sup> species, Cr2+<\/sup> is strongly reducing while manganese (III) is strongly oxidising.
    \n(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
    \n(iii) The d1 configuration is very unstable in ions.
    \nAnswer:
    \n(i) Cr2+<\/sup> is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidised to Cr3+<\/sup> (electronic configuration d3<\/sup>). This d3<\/sup> configuration can be written as t2<\/sub>3<\/sup> configuration, which is a more stable configuration. In the case of Mn3+<\/sup> (d4<\/sup>) it Mn2+<\/sup> (d5<\/sup>). This has an exactly half filled d – orbital and is highly stable.<\/p>\n

    (ii) Co (II) is stable in aqueous solutions however, in the presence of strong field complexing is agents, it is oxidised to Co (II). Although the 3rd ionisation energy for Co is high, but the higher amount of crystal field stabilisation field ligands overcomes this ionisation energy.<\/p>\n

    (iii) The ions in d configuration tend to lose one more electrons to get into stable d0<\/sup> configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.<\/p>\n

    \"KSEEB<\/p>\n

    Question 22.
    \nWhat is meant by \u2018disproportionation\u2019? Give two examples of disproportionation reaction in aqueous solution.
    \nAnswer:
    \nIt is found that sometimes a relatively less stable oxidation state undergoes an oxidation reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportion.
    \nFor example,
    \n\"2nd
    \nMn (vi) is oxidised to Mn (vii) and reduced to Mnfiv).<\/p>\n

    Question 23.
    \nWhich metal in the first series of . transition metals exhibits +1 oxidation state most frequently and why?
    \nAnswer:
    \nIn the fast transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu (+1) has an electronic configuration of [Ar] 3d10<\/sup>. The completely filled d – orbital makes it highly stable.<\/p>\n

    Question 24.
    \nCalculate the number of unpaired electrons in the following gaseous ions: Mn3+<\/sup>, Cr3+<\/sup>, V3+<\/sup> and Ti3+<\/sup>. Which one of these is the most stable in aqueous solution?
    \nAnswer:
    \n\"2nd
    \nCr3+<\/sup> is the most stable in aqueous solution owing to a t2<\/sub>3<\/sup> g configuration.<\/p>\n

    Question 25.
    \nGive examples and suggest reasons for the following features of the transition metal chemistry:
    \n(i) The lowest oxide of transition metal is basic, the highest is amphoteric\/acidic.
    \n(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
    \n(iii) The highest oxidation state is exhibited in oxoanibns of a metal.
    \nAnswer:
    \n(i) In case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atoms are not involved in bonding. As a result, it can donate electrons and behave as a base.<\/p>\n

    On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding. As a result, it can accept electrons and behave as an acid. For example, MnII<\/sup>O is basic and MnVII<\/sup>O is acidic.<\/p>\n

    (ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits oxidation states in oxides and fluorides. For example, in OsF6<\/sub> and V2<\/sub>O5<\/sub>, the oxidation states of Os and V are + 6 and respectively.<\/p>\n

    (iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size so, oxoanions of a metal have the highest oxidation state. For example, in MnO4<\/sub>–<\/sup>, the oxidation state of Mn is +7.<\/p>\n

    \"KSEEB<\/p>\n

    Question 26.
    \nIndicate the steps in the preparation of:
    \n(i) K2<\/sub>Cr2<\/sub>O7<\/sub> from chromite ore.
    \n(ii) KMnO4<\/sub> from pyrolusite ore.
    \nAnswer:
    \n(i) Potassium dichromate : (K2<\/sub>Cr2<\/sub>O7<\/sub>) is prepared from chromite ore (FeCl2<\/sub>O4<\/sub>) in the following steps.<\/p>\n

    step (1) Preparation of sodium chromate
    \n4Fe Cr2<\/sub>O4<\/sub> + 16 NaOH + 7O2<\/sub> \u2192 8Na2<\/sub>CrO4<\/sub> + 2Fe2<\/sub>O3<\/sub> + 8H2<\/sub>O<\/p>\n

    step (2) Conversion of sodium chromate into sodium dichromate
    \n2Na2<\/sub>CrO4<\/sub> + con H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>Cr2<\/sub>O7<\/sub> + Na2<\/sub>SO4<\/sub> + H2<\/sub>O<\/p>\n

    step (3) Conversion of sodium dichromate to potassium dichromate
    \nNa2<\/sub>Cr2<\/sub>O7<\/sub> + 2KCl \u2192 2Nacl + k2<\/sub>Cr2<\/sub>O7<\/sub><\/p>\n

    potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filteration.<\/p>\n

    The dichromate ion (Cr2<\/sub>O7<\/sub>2-<\/sup> ) exists in equilibrium with chromate (CrO4<\/sub>2-<\/sup> ) ion at pH.
    \n4. However, by changing the pH, they can be interconverted.<\/p>\n

    \"2nd<\/p>\n

    (ii) Potassium permanganate : (KMnO4<\/sub>) can be prepared from pyrolusite (MnO2<\/sub>). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3<\/sub> or KClO4<\/sub> to give K2<\/sub>MnO4<\/sub><\/p>\n

    \"2nd<\/p>\n

    The green mass can be extracted with water and then oxidised either electrolytically or by passing chlorine \/ ozone into the solution, electrolytic oxidation
    \nK4<\/sub>MnO4<\/sub> 2K+<\/sup>+ \u2194\ufe0f MnO4<\/sub>2-<\/sup>
    \nH2<\/sub>O \u2194\ufe0f H+<\/sup> + OH–<\/sup>
    \nat anode manganate ions are oxidized to permanganate ions<\/p>\n

    \"2nd<\/p>\n

    oxidation by chlorine
    \n2K2<\/sub>MnO4<\/sub> + Cl2<\/sub> \u2192\u00a0 2KMnO4<\/sub> + 2KCl
    \n2MnO4<\/sub>2-<\/sup> + Cl2<\/sub> \u2192 2 MnO4<\/sub> + 2Cl–<\/sup><\/p>\n

    oxidation by ozone
    \n2K2<\/sub>MnO4<\/sub> + O3<\/sub> + H2<\/sub>O \u2192 2KMnO4 <\/sub>+ 2KOH + O2<\/sub>
    \n2MnO4<\/sub>2-<\/sup> + O3<\/sub> + H2<\/sub>O \u2192 2MnO2<\/sub> + 2OH + O2<\/sub><\/p>\n

    \"KSEEB<\/p>\n

    Question 27.
    \nWhat are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
    \nAnswer:
    \nAn alloy is a solid or solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to posses different physical properties than those of the component elements.<\/p>\n

    An important alloy of lanthanoids is Mischmetal It contains lanthanoids (4-95%), iron (5%) and traces of S, C, Si, Ca and Al.
    \nUses<\/p>\n

      \n
    1. Mischmetal is used in cigarettes and gas lighters<\/li>\n
    2. It is used in flame throwing tanks<\/li>\n
    3. It is used in flame bullets and shills.<\/li>\n<\/ol>\n

      Question 28.
      \nWhat are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.
      \nAnswer:
      \nInner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements. Among the given atomic numbers the atomic numbers of the inner transition u elements are 59, 95 and 102.<\/p>\n

      Question 29.
      \nThe chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
      \nAnswer:
      \nLanthanoids primarily show three oxidation states, +3 states is the most common. Lanthanoids display a limited number of oxidation states because the difference between the energies of 4f, 5d and 6s orbitals is quite large. On the other hand, the energy difference between the 5f, 6d and 7s orbitals is very less. Hence, actinoids display a large number of oxidation states. For example uranium and plutonium display +3, +4, +5 and +6 oxidation states while neptunium display +3, +4, +5 and +7. The most common oxidation state in case of actinoids is also +3.<\/p>\n

      Question 30
      \nWhich is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
      \nAnswer:
      \nThe last element in the actinoid series is Lawrencium. Lr. Its atomic number is 103 and its electronic configuration is [Rn] 5f14<\/sup> 6d1<\/sup>7s2<\/sup>. The most common oxidation state displayed by it is +3, because after losing 3 electrons it attains stable f14<\/sup> configuration.<\/p>\n

      Question 31
      \nUse Hund\u2019s rule to derive the electronic configuration of Ce3+<\/sup> ion, and calculate its magnetic moment on the basis of \u2018spin- only\u2019 formula.
      \nAnswer:
      \nCe : 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d10<\/sup> 4s2<\/sup> 4p6<\/sup> 4d10<\/sup> 5s2<\/sup> sp6<\/sup> 4f1<\/sup> 5d1<\/sup> 6s2<\/sup>
      \nCe : [Xe] 4f1<\/sup> 5d1<\/sup> 6s2<\/sup>
      \nCe3+<\/sup>: {Xe} 4f1<\/sup>
      \nForCe3+<\/sup>:
      \nNumber of unpaire electrons = n = 1
      \n\\(\\mu=\\sqrt{\\mathrm{n}(\\mathrm{n}+2)}\\)
      \n= \u221a8
      \n= 2\u221a2
      \n= 2.828 BM.<\/p>\n

      Question 32.
      \nName the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
      \nAnswer:
      \nThe lanthanoids that exhibit +2 and +4 states are shown in the given table. The atomic number of the elements are given in the parenthesis.
      \n\"2nd
      \nCe after forming Ce4+<\/sup> attains a stable configuration of [Xe]
      \nTb after forming Tb4+<\/sup> attains a stable configuration of [Xe] 4f7<\/sup>
      \nEu after forming Eu2+<\/sup> attains a stable configuration of [Xe] 4f7<\/sup>
      \nYb after forming Yb2+<\/sup> attains a stable configuration of [Xe] 4f14<\/sup>.<\/p>\n

      Question 33.
      \nCompare the chemistry of the actinoids with that of lanthanoids with reference to:
      \n(i) electrpnic configuration
      \n(ii) oxidation states and
      \n(iii) chemical reactivity.
      \nAnswer:<\/p>\n

      (i) Electronic configuration:
      \nThe general electronic configuration for lanthanoids is [Xe] 0-14<\/sup> 5d 0-1<\/sup> 6s2<\/sup> and that of actinoids is [Rn] 5f 0-14<\/sup> 6d0-1<\/sup> 7s2<\/sup>. Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.<\/p>\n

      (ii) Oxidation states:
      \nThe principal oxidation states of lanthanoids is (+3). However sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.<\/p>\n

      (iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction over all decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actionoids. ‘<\/p>\n

      (iv) Chemical reactivity
      \nIn the lanthanoide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).<\/p>\n

      \"KSEEB<\/p>\n

      Question 34.
      \nWrite the electronic configurations of the elements with the atomic numbers, 61, 91, 101, and 109.
      \nAnswer:
      \n61 : [Xe]54<\/sup> 4f5<\/sup> 5d0<\/sup> 6s2<\/sup>
      \n91 : [Rn]86<\/sup> 5f2<\/sup> 6d1<\/sup> 7s2<\/sup>
      \n101 : [Rn]86<\/sup> 5f13<\/sup> 5d0<\/sup> 7s2<\/sup>
      \n109: [Rn]86<\/sup> 5f14<\/sup> 6d7<\/sup> 7s2<\/sup> .<\/p>\n

      Question 35
      \nCompare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:
      \n(i) electronic configurations
      \n(ii) oxidation states
      \n(iii) ionisation enthalpies and
      \n(iv) atomic sizes.
      \nAnswer:
      \n(i) In the 1st<\/sup> , 2nd<\/sup> and 3rd<\/sup> series of the transition metals the 3d, 4d, 5d orbitals are respectively filled.
      \nWe know that elements in the same vertical column generally have similar electronic configurations.
      \nIn the first transition series, two elements show unusual electronic configurations.<\/p>\n

      Cr (24) : 3d5<\/sup> 4s1<\/sup>
      \nCu (29) :3d10<\/sup> 4s1<\/sup><\/p>\n

      similarly, there are exception in 2nd transition series.
      \nThese are
      \nMo (42) : 4d5<\/sup> 5s1<\/sup>
      \nTc (43) : 4d6<\/sup> 5s1<\/sup>
      \nRu (44) : 4d7<\/sup> 5s1<\/sup>
      \nRh (45) : 4d8<\/sup> 5s1<\/sup>
      \nPd (46) : 4d10<\/sup> 5s1<\/sup>
      \nAg (47) : 4d10<\/sup> 5s1<\/sup>.<\/p>\n

      There are exceptions in-3rd transition series also.
      \nThese are
      \nW (74) : 5d4<\/sup> 6s2<\/sup>
      \nPt (78) : 5d9<\/sup> 6s1<\/sup>
      \nAu (79) :5d10<\/sup> 6s1<\/sup><\/p>\n

      As a result of these exceptions it happens many times that the electronic configuration of the elements presents in the same group are dissimilar.<\/p>\n

      (ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum iri the middle and the minimum at the extreme ends. However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stabilities of the +2 and +3 oxidation states decreases in the second and the third transition series, where in higher oxidation states are more important.<\/p>\n

      For example, [FeII<\/sup> (CN)6<\/sub>]4-<\/sup>\u00a0[COIII<\/sup> (NH3<\/sub>)6<\/sub>]3+<\/sup> , [Ti(H2<\/sub>O)6<\/sub>]3+<\/sup> air stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example WCl6<\/sub>, ReF7<\/sub>, RuO4<\/sub> etc.<\/p>\n

      (iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some enceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to poor shielding effect of 4f electrons in the third transition series.<\/p>\n

      Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1st<\/sup> transition series.<\/p>\n

      (iv) Atomic sizes generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series this is due to lanthanoid contraction.<\/p>\n

      Question 36.
      \nWrite down the number of 3d electrons in each of the following ions:
      \nTi2+<\/sup> , V2+<\/sup> , Cr3+<\/sup> , Mn2+<\/sup> , Fe2+<\/sup> , Fe3+<\/sup> , CO2+<\/sup> , Ni2+<\/sup> and Cu2+<\/sup> .
      \nIndicate how would you expect the five 5d orbitals to be occupied for these hydrated ions (octahedral).
      \nAnswer:
      \n\"2nd<\/p>\n

      Question 37.
      \nComment on the statement that elements of the first transition series possess many properties different from those, of heavier transition elements.
      \nAnswer:
      \nThe properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
      \n(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements of 2nd<\/sup> and 3rd<\/sub> transition series) However, the atomic sizes of the elements in ‘the third transition series are virtually the same as those of the corresponding members in the second transition series this is due to lanthanoid configuration.<\/p>\n

      (ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.<\/p>\n

      (iii) The enthalpies of atomisation of the elements’in the first transition series are lower than those of the corresponding elements in the second and third transition series.<\/p>\n

      (iv) The meeting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of strong metallic bonding (M – M bonding)<\/p>\n

      (v) The dements of the first transition series from low-spin or high-spin complexes depending upon the strength of the ligand field.<\/p>\n

      However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.<\/p>\n

      \"KSEEB<\/p>\n

      Question 38.
      \nWhat can be inferred from the magnetic moment values of the following complex species ?
      \n\"2nd
      \nAnswer:
      \nMagnetic moment (\u03bc) is givenn as<\/p>\n

      \"2nd
      \nFor transition metals, the magnetic moment is calculated from the spin only formula Therefore
      \n\\(\\sqrt{n(n+2)}\\) = 2.2<\/p>\n

      we can see from above calculation that the given value is closest to n – 1. Also, in this complex,. Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d orbital. Hence, we can say that CN is a strong ligand that causes the pairing of electrons.<\/p>\n

      (ii) [Fe (H2<\/sub>O)6<\/sub>]2+<\/sup>
      \nWe. can see from the above calculation that \u00b5= 5.3 is closest to the value when n = 4. Also, in this complex, Fe is in the +2 oxidation state. This means ihat Fe has 6 electrons in the d orbital<\/p>\n

      Hence, we can say that H2<\/sub>O is a weak field ligand and does not cause the pairing of e–<\/sup> s.<\/p>\n

      (iii) K2<\/sub> [Mncl4<\/sub>]
      \nwe can see that |X= 5.9 is closest to the value when n = 5. Also, in this computer, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital. Hence we can say that Cl- is a weak ligand and does not cause the pairing of electrons.<\/p>\n

      2nd PUC Chemistry The d-and f-Block Elements Additional Questions and Answers<\/h3>\n

      Question 1.
      \nSome d-block elements are given below.
      \n(Cr, Mn, Fe, Co, Ni)
      \n(a) Identify the element which shows maximum paramagnetic behaviour.
      \n(b) Give reason for the highest paramagnetic property of that element.
      \n(c) When two electrons are lost from each atom, is there any change in magnetic properties of each ? Explain ?
      \nAnswer:
      \n(a) Cr
      \n(b) Cr- 4s1<\/sup> 3d5<\/sup>
      \nMn – 4s2<\/sup> 3d5<\/sup> .
      \nFe – 4s2<\/sup> 3d6<\/sup>
      \nCo – 4s2<\/sup> 3d7<\/sup>
      \nNi – 4s2<\/sup> 3d8<\/sup><\/p>\n

      Since there are six unpaired electrons in Cr, it has the maximum paramagnetic behaviour.
      \n(c) Cr2+<\/sup> \u21923d4<\/sup>
      \nMn2+<\/sup> \u2192 3d5<\/sup>
      \nFe2+<\/sup> \u2192 3d6<\/sup>
      \nCO2+<\/sup> \u2192 3d7<\/sup>
      \nNi2+<\/sup> \u2192 3d8<\/sup>
      \nwhen two electrons are lost, Mn2+<\/sup> contains maximum number of unpaired electrons ie., Mn2+<\/sup> is the ion with maximum paramagnetic behaviour. Cr2+<\/sup> and Fe2+<\/sup> show similar paramagnetism and so on.<\/p>\n

      Question 2.
      \nHow do atomic radii of transition elements in a given series vary with increase in atomic number ? What is your observation on atomic radii of 4d and 5d series of transition elements ? Discuss the phenomenon involved, if any
      \nAnswer:
      \nThe 4d and 5d series of transition elements in each group have the same atomic radii. This is due to lanthanide contraction.
      \nThe steady decrease in the size of the atoms or ions from lanthanum to lutetium is known as lanthanide contraction.<\/p>\n

      As the atomic number increases in the lanthanide series, the new electron goes to the inner 4f orbitals. The shielding effect of a 4f electron is poor. As a result, the attraction of the outer electrons by the nucleus increases and this tends to a decrease in the size.<\/p>\n

      Question 3.
      \n(a) What is blue Vitriol ? How is it useful in detecting the presence of water vapour ?
      \n(b) When a magnetic field is applied, certain substances are attracted by the applied field. Those are called paramagnetic substances. Which is more paramagnetic Mn2+<\/sup> or Cu2+<\/sup> ? Why ?
      \nAnswer:
      \n(a) CuSO4<\/sub> 5H2<\/sub>O is known as blue vitriol. Anhydrous copper sulphate is used for detecting the presence of water vapour. Water vapour converts white anhydrous copper sulphate to blue hydrated copper sulphate.
      \n\"2nd<\/p>\n

      (b) As the number of unpaired electrons increases, paramagnetism increases.
      \nMn \u2192 [Ar] 4s2<\/sup> 3d5<\/sup> and Mn2+<\/sup> \u2192 3d5<\/sup>
      \nCu \u2192 [Ar] 4s1<\/sup> 3d10<\/sup> and Cu2+<\/sup> \u21923d9<\/sup><\/p>\n

      Mn2+<\/sup> contains five unpaired d-electrons and
      \nCu2+<\/sup> contains only unpaired electron.
      \n\u2234 Mn2+<\/sup> is more paramagnetic than 2+<\/sup>.<\/p>\n

      \"KSEEB<\/p>\n

      Question 4.
      \n(a) What do you observe when a dilute solution of KMnO4<\/sub> is added in drops to a hot solution of dilute H2<\/sub>SO4<\/sub> ? Is there any difference in observation when dilute KMnO4<\/sub> is added to a mixture of hot dilute H2<\/sub>SO4<\/sub> and oxalic acid ?
      \n(b) In the periodic table the atomic size increases down a group due to increase in number of shells. But 40<\/sub>Zn and 72<\/sub>HF in the 5th<\/sup> and 6th<\/sup> period respectively have almost identical radii explain ?
      \nAnswer:
      \n(a) When dilute KMnO4<\/sub> is added to a mixture of hot dilute H2<\/sub>SO4<\/sub> and oxalic acid, brisk effervescence is formed due to the formation of CO2<\/sub>.
      \n\"2nd
      \nIn the first case there is no characteristic observation<\/p>\n

      (b) The 4d and 5d series of transition elements in each group have the same atomic radii. This is due to lanthanide contraction.
      \nThe steady decrease in the size of the atoms or ions from lanthanum to lutetium is known as lanthanide contraction.<\/p>\n

      As the atomic number increases in the lanthanide series, the new electron goes to the inner 4f orbitals. The shielding effect of a 4f electron is poor. As a result, the attraction of the outer electrons by the nucleus increases and this tends to a decrease in the size.<\/p>\n

      Question 5.
      \nMn exhibits large number of oxidation states in its compounds. Which is the highest oxidation state ?
      \nName an industrially important compound you have studied in which Mn is in its highest oxidation state. How is that compound manufactured ?
      \nAnswer:
      \nIn KMnO4<\/sub>, Mn exhibits the highest oxidation state of +7.
      \nOxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits oxidation states in oxides and fluorides. For example, in OsF6<\/sub> and V2<\/sub>O5<\/sub>, the oxidation states of Os and V are + 6 and respectively.<\/p>\n

      Question 6.
      \nAssign reasons for the followimg.
      \n(i) The enthalpies of atomisation of transition elements are high.
      \n(ii) The transition metals and many of their compounds act as good catalyst.
      \n(iii) From element to element the actinoid contraction is greater than the lanthanoid contraction
      \n(iv) The E\u00b0 value for the Mn3+\/Mn2+ couple is much more positive than for Cr3+<\/sup>\/Cr2+<\/sup>
      \n(v) Scandium (Z = 21) does not exhibit variable oxidation states and Let it is regarded as transition element.
      \nAnswer:
      \n(i) The transition elements exhibit high enthalpies of atomisation because they have large number of unpaired electron in their atoms. Due to this they have stronger inter atomic interaction and hence stronger bonding between atoms.
      \n(ii) Many transition metals and their compounds acts as catalyst the catalytic activity is due to their ability to exhibit multiple oxidation states, eg. V2<\/sub>O5<\/sub> in contact process and finely divided iron in Haber process.
      \n(iii) The actinoid contraction is similar to lanthanoid contraction. This contraction is due to poor shielding by 5f electrons in the actioids than that by 4f electrons in the lanthanoids.
      \n(iv) Much larger third ionisation energy of Mn (where the required change is Cl5<\/sup> and d4<\/sup>) is mainly responsible for this. This also explains that +3 state of Mn is of little importance.
      \n(v) Scandium (Z = 21) has incompletely filled 3d orbitals in the ground state. (3d) Hence, it is considered as a transition element.<\/p>\n

      Question 7.
      \n(a) What may be the possible oxidation states of the transition metals with the following d
      \nelectronic configurations for the ground state of their atoms 3d3<\/sup> 4s2<\/sup>, 3d5<\/sup> 4s2<\/sup> and 3d6<\/sup> 4s2<\/sup>. Indicate relative stability of oxidation states in each case.
      \n(b) Write steps involved in the preparation of
      \n(i) Na2<\/sub>CrO4<\/sub> from chromite ore and
      \n(ii) K2<\/sub>MnO4<\/sub> from pyrolusite ore.
      \nAnswer:
      \n(a) The most stable oxidation state for 3d3<\/sup> 4s2<\/sup> = +5
      \nThe most stable oxidation state for 3d5<\/sup> 4s2<\/sup> = +7, +2
      \nThe most stable oxidation state for 3d6<\/sup> 4s2<\/sup> = +2
      \nIn a transition series, the oxidation states which ‘ lead to exactly half-filled or completely filled d- orbitals are more stable. It shows that oxidation states +2 to +7 is most stable.<\/p>\n

      (b) i. The powdered chromite ore is heated with sodium carbonate in presence of air to get yellow coloured sodium chromite.
      \n\"2nd
      \nii. Finally powdered pyrolusite is fused with KOH in the presence of air to give green coloured potassium manganate<\/p>\n

      2MnO2<\/sub> + 4 KOH +O2<\/sub> \u2192 2K2<\/sub>MnO4<\/sub> + 2H2<\/sub>O.<\/p>\n

      Question 8.
      \nWrite complete chemical equations for
      \n(i) Oxidation of Fe2+<\/sup> by Cr2<\/sub>O7<\/sub>2-<\/sup> in acid medium
      \n(ii) Oxidation S2<\/sub>O3<\/sub>2-<\/sup> by MnO4<\/sub>– <\/sup>in neutral aqueous medium
      \nAnswer:
      \n(i) Acidified potassium dichromate oxidises iron (II) salt to iron (III)
      \nCr2<\/sub>O7<\/sub>2-<\/sup> + 14H + 6Fe2+<\/sup> \u2192 2Cr3+<\/sup> + 6Fe3+<\/sup> + 7H2<\/sub>O
      \n(ii) In natural aqueous medium, thiosulphate
      \n8MnO4<\/sub>–<\/sup>+ 3s2O3<\/sub>2-<\/sup> + H2<\/sub>O \u2192 8MnO2<\/sub>+ 65O4<\/sub> + 2OH–<\/sup><\/p>\n

      Question 9.
      \nHow would you account for the following situations
      \n(i) The transition metals generally form coloured compounds.
      \n(ii) With 3d4<\/sup> configuration, Cr2+<\/sup> acts as a reducing agent but Mn3+<\/sup> acts as an oxidising agent (Atomic masses, Cr = 24, Mn = 25)
      \n(iii) The actinoids exhibits a larger number of oxidation states than the corresponding lanthanoids.
      \nAnswer:
      \n(i) Colour of compounds of transition elements depends upon the unpaired electrons present in d orbitals of transition elements. If d orbitals are completely vacant as in Sc3+<\/sup>, Ti4+<\/sup> or completely filled as in Cu2+<\/sup>, Zn2+<\/sup>, the compounds will be colourless. But if any unpaired electron is present in d-orbitals, the compound will be coloured due to d \u2192 d transition. The unpaired electron is excited from one energy level to another energy level within the same d subshell. For this purpose (d \u2192 d transition), the energy is absorbed from visible region of radiation and reflected light will decide the colour of the compound.<\/p>\n

      (ii) Cr2+<\/sup> is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidised to Cr3+<\/sup> (electronic configuration d3<\/sup>). This d3<\/sup> configuration can be written as t2<\/sub>3<\/sup> configuration, which is a more stable configuration. In the case of Mn3+<\/sup> (d4<\/sup>) it Mn2+<\/sup> (d5<\/sup>). This has an exactly half filled d – orbital and is highly stable.
      \n(iii) Actinoids show a large number of oxidation states because of small energy gap between 5f, 6d and 7s subshells.<\/p>\n

      \"KSEEB<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

      You can Download Chapter 8 The d-and f-Block Elements Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.   Karnataka 2nd PUC Chemistry Question Bank Chapter 8 The d-and f-Block Elements 2nd PUC Chemistry The d-and f-Block …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10472"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10472"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10472\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}