1.  Define transition elements.

Ans. Transition element is defined as the one which has incompletely filled d orbitals in its ground state or in any one of its oxidation states.

 2.  What is the position of the d block elements in the periodic table?

Ans. The d block elements are in the middle of s and p blocks, comprising the groups 3 to 12. They are the four rows of elements in the periods 4th (3d series), 5th (4d series), 6th (5d series) and 7th ( 6d series).

3.  Zinc, cadmium and mercury of group 12 are not regarded as transition metals, why? 

Ans. Zinc, cadmium and mercury of group 12 have full d10 configuration ( d orbitals are completely filled ) in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals

 

4.  Why d- block elements are named as ‘transition elements?

Ans. The d–block elements occupies the middle of the periodic table and their properties are transitional between s– and p– block elements.

 5.Write the general electronic configuration of d block elements.

 Ans. [Noble gas] (n-1)d1-10ns1-2

6.  Write the general outer electronic configuration of d- block elements. .

Ans. The general outer electronic configuration of d- block elements is (n-1)d1–10 ns1–2

7.  Write the general electronic configuration of f- block elements.

Ans. The general electronic configuration of f- block elements (Lanthanoids)

 is [Xe] 4f1 – 14 5d 0- 1 6s2

8. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

  Ans. Cerium

9.  The outer electronic configuration of Cr is 3d5 4s1 instead of 3d44s2, why?

Ans. Half filled (3d5) orbitals are relatively more stable, hence one electron of  4s orbital jumps to 3d orbital.

 

10.  The outer electronic configuration of Cu is 3d10 4s1 instead of 3d94s2 , why?

Ans. Completely filled (3d10 ) orbitals are relatively more stable, hence one electron  of  4s orbital jumps to 3d orbital.

11.Account for high melting point and boiling points of transition metals.

Ans. The melting and boiling points of transition metals are high because of the involvement of greater number of electrons from (n-1)d orbitals  in addition to  the ns  electrons in the inter atomic metallic bonding.

12.What is the trend in melting points of transition metals in a series?

Ans. The melting points of the transition metals in a series rise to a maximum at the middle of the series (i.e.Cr or Mo or W – element with d 5 configuration) and fall regularly as the atomic number increases.

 

13.Why do transition metals have higher enthalpies of atomization?

Ans.Involvement of a large number of unpaired electrons of d orbitals favour stronger inter atomic interactions resulting in stronger bonds between the atoms of a metal and higher enthalpies of atomization.

14. Name one 3d series elements, that do not show variable oxidation states. 

Ans. Sc (+3)

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15. Transition metals exhibit variable oxidation states in its compounds, why?

Ans. Transition metals exhibit variable oxidation states in its compounds due to the availability of both ns & (n – 1 )d electrons for bond formation.

16.  Name 3d series metal which shows highest oxidation state.

Ans.  Mn. it shows +7

17.  Name a metal in the 3d series of transition metals which exhibit +1 oxidation state most frequently.

Ans. copper

18. What is the trend in oxidation state of transition metals?

Ans. The oxidation state increases with increase in atomic number & reaches a maximum in the middle and thendecreases.

 

19.   3d series transition metals exhibit +2 as the most common oxidation state (except Sc) why?

Ans. The +2 oxidation state, which commonly occurs for nearly all the transition metals is due to the loss of their outer 4s electrons

20.  Why transition metals and their compounds show paramagnetic behavior?

Ans. The transition metal ions are generally containing one or more unpaired electrons in them & hence their compounds are generally paramagnetic.

21.  Name an of alloys of transition metals with non-transition metals.

Ans. Brass (Cu & Zn) or Bronze (Cu & Sn)

22. What is the action of neutral or faintly alkaline permanganate solution on iodide?

Ans. Alkaline permanganate solution oxidize iodide to iodate.

23. What happens when potassium permanganate is heated to 513 K ?

Ans. Potassium permanganate decomposes at 513K to potassium manganate, manganese dioxide and oxygen.

24.Write the spin-only formula used to calculate the magnetic moment of metal ions.

Ans. The magnetic moment is determined by using the spin only formula,

μ = [n(n+2)]^1/2 where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM).

25.What is the principal oxidation state exhibited by  the lanthanoids?

Ans:- +3

26. Why is Sc3+ (or Zn2+) diamagnetic?

Ans. Sc3+(Z=21) :-3d0     no unpaired electron, n=0, μ=0. 

(or Zn2+(Z=30) :-3d10      no unpaired electron, n=0, μ=0)

 

27.  What is the most common oxidation state of lanthanoids and actinoids?

Ans. The most common oxidation state of lanthanoids and actinoids is +3.

 

28.What is Actinoid contraction?

Ans. There is a gradual decrease in the size of atoms or M3+ ions across the series. This is known as the actinoid contraction.

29.Actinoid contraction is more than lanthanoid contraction. Give reason.

Ans. The actinoid contraction is, more than lanthanoid contraction due to poor shielding by 5f electrons from nuclear charge.

30.  Actinoids show larger number of oxidation states than lanthanoids.Why?

Ans. In actinoids 5f, 6d and 7s levels are of comparable energies, hence electrons from these orbitals are available to lose or share.

31.  Give one use of Mischmetal.

Ans. Mischmetal is used in Mg-based alloy to produce bullets, shell and lighter flint.

32.  Why transition metals forms alloys readily?

Ans. Transition metals readily form alloys with other transition metals because of their similar radii.

33.  Give one use of transition metal alloy.

Ans. Ferrous alloys containing chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels.

 Answer the following questions. Each question carries TWO marks.

34. Name two characteristic properties exhibited by d – block elements due to their partly filled d orbitals.

Ans. The characteristic properties exhibited by d – block elements due to their partly filled d orbitals are variable; (i) Oxidation states (ii) Formation of coloured ions.

35.Name two typical metallic properties displayed by transition elements.

Ans. High tensile strength, ductility   malleability, high thermal and     electrical conductivity and metallic luster etc.

36.  What are interstitial compounds? Give example.

Ans. Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of transition metals.

Example; TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc.

37. Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27).

Ans. M (z= 27, 3d7 4s2) → M+2 (3d7 4s0)  hence it has 3 unpaired electrons n= 3 = [3(3+2)]^1/2  = 3.87 BM

38.  The second ionization enthalpy is high for Cr and Cu, why?

Ans: The second ionization enthalpy is unusually high values for Cr and Cu because when M+ ion ionize to M+2 ion, the d5 and d10 configurations of the M+ ions (i.e Cr+ or Cu+) are disrupted, with considerable loss of exchange energy.

39. Give any two physical characteristics of interstitial compounds

40. Why first ionization enthalpy of Cr is lower than that of Zn?

IE1 of  Cr is lower, because  removal  of  an electron takes from half filled 4s subshell

Cr (z= 24, 3d5  4s1) ➡Cr+  (3d5 4s0)     E1

Whereas IE1 value for Zn is higher, because removal of electron takes place from full filled and stable 4s subshell

Zn (z= 30, 3d10  4s2) ➡Zn+  (3d10 4s1)  E1

Therefor:- IE1(Zn)  > IE1 (Cr)

41. Give two characteristics of transition metal alloys.

. Ans:- The alloys are hard and have high melting points.

42. What is the action of heat on potassium permanganate? Give equation.

Ans. It decomposes at 513K to potassium manganate, manganese dioxide and oxygen.

2KMnO4 à K2MnO4 + MnO2 + O2

43.What is the action of neutral or faintly alkaline permanganate solution on iodide? Give equation.

Ans. Alkaline permanganate solution oxidize iodide to iodate.

I– + 2MnO4– + H2O➡IO3– +  2MnO2 + 2OH–

44. Explain the oxidizing action of acidified potassium dichromate on (iron(II) salts) Fe+2 ions and write the ionic equations for the reaction.

Ans. Acidified potassium dichromate oxidises iron(II) salts to iron(III). 

Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O

45.The transition metals generally form coloured compounds, why?

Ans. The compounds of transition elements shows colour due to presence of unpaired electron & ability to undergo d-d transition.When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region.

 

46.  Give reason “transition metals and their many compounds act as good catalysts”. 

Ans. Transition metals and their many compoundsacts as good catalysts, it is due to (i) partially filled (n-1) d orbital (ii)variable oxidation state and provide a suitable surface for the reaction to take place.

 

47. Explain giving reason “transition metals form a large number of complex compounds”.

 

48.  What is the effect of increasing pH on a solution of potassium dichromate?

Ans. On increasing the pH of the solution Potassium dichromate (orange) becomes potassium chromate (yellow)  

Cr2O72- + 2OH– → 2CrO42- + H2O

49. What happens when H2S is passed into potassium dichromate in acidic medium? Give the equation

 Cr2O7−2 + 3H2S + 8H+ → 2Cr+3 + 7H2O + 3S

Ans. A particular oxidation state, which is relatively less stable compared to other oxidation states ,under goes disproportion.

Manganese (VI) which is relatively less stable changes over to manganese (VII) and manganese (IV) in acid solution.

3 MnO4-2 + 4H+   → MnO2 +2MnO4–+ 2 H2O

 

51. What  is  lanthanoid   contraction?  Write   any   one consequence of lanthanoid contraction.

Ans. Steady decrease in the size of lanthanides with increase in atomic number is known as lanthanoid contraction.

Due to lanthanoid contraction radii of members of 3rd transition series are very much similar to corresponding members of 2nd series.

52. Write any two consequences of lanthanoid contraction.

Ans. Two consequences of lanthanoid contractions are

(i)  The radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. Ex. The almost identical radii of Zr (160 pm) and Hf (159 pm) & Nb (146pm) & Ta (146pm)

(ii)  Difficulty in separation of lanthanoids due to similarity in chemical properties.

 

53.  Name the two series of f-block.

Ans. The f-block consists of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium.

54.  The chemistry of actinoids is more complicated than lanthanoids. Why?

Ans. The actinoids are radioactive elements having half lifes varying. Some members can be prepared only in nanogram quantities. These facts render their study more difficult.

 

55. Write two comparisons of variability in oxidation states of transition metals and non-transition elements (p- block elements) ?

Ans. 1. In transition elements , variable oxidation state differ from each other by unity, whereas in case of non-transition elements , oxidation state differ by units of two.(For example Fe exhibits o.s of +2 and +3 . similarly copper exhibits two o.s of +1 and +2 . on the other hand, Sn, Pb exhibit o.s of +2 and +4.)

2. In transition elements, higher o.s are more favoured in elements of higher atomic mass, whereas in p-block elements lower o.s are favoured by heavier members ( due to inert pair effect, For example Mo(VI) and W(VI) are more stable than  Cr(VI).  On the other hand  Pb(II) is more stable than Sn(II))

 

56.  What happens when (a)   A lanthanoid reacts with dilute acids? (b)   A lanthanoid reacts with water?

Ans.(a) When lanthanoid reacts with dilute acids, it liberates hydrogen gas.

(b)When lanthanoid reacts with water, it forms lanthanoid hydroxide and liberate hydrogen gas.

57. What is the composition of mischmetal? Give its oneuse.

Ans. The composition of mischmetal is lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al.

Mischmetal is used in Mg-based alloy to produce bullets, shell and lighter flint

58. What is the gas liberated when (i)  crystals of potassium permanganate is heated to 513K? (ii)  acidified potassium permanganate is treated with oxalate ion at 333K?

Ans. i) When crystals of potassium permanganate is heated to 513K Oxygen (O2) gas is liberated.

ii) Acidified potassium permanganate when treated with oxalate ion at 333K liberates Carbon dioxide (CO2) gas.

59.  Show the interconversion of chromate and dichromate

60. How does the neutral or faintly alkaline potassium permanganate solution react with

(a)  Iodide (b) Thiosulphite? Write the ionic equations for the reactions

Ans:- In neutral or faintly alkaline solutions:

(a) The oxidation of iodide to iodate:

2MnO4− + H2O + I− ——> 2MnO2 + 2OH− + IO3−

(b)  Thiosulphate is oxidised almost quantitatively to sulphate:

8MnO4– + 3S2O32– + H2O ——> 8MnO2 + 6SO42– + 2OH–

III.  Answer the following questions. Each question carries THREE marks

 61.Name the metal of the 1st row transition series that 

i)   has highest value for magnetic moment (ii)  has zero spin only magnetic moment in its +2 oxidation state. (iii)  exhibit maximum number of oxidation states.

Ans. i) Chromium        ii) Zinc           iii) Manganese

 62. Transition metals form a large number of complex compounds. Give reason. 

Ans. Transition metals for complex compounds due to,

i)   small sizes of metal cations

ii)  their ionic charges and

iii)  availability of d orbitals for bond formation.

63.Explain the trend in atomic size of 3d series of transition elements with reason.

Ans. With increase in atomic number in 3d series – atomic size decreases (Sc to Cr) , then remain almost constant (Cr to Cu) and increase slightly at the end (Cu to Zn).

Reason: In the beginning of the series the screening (or shielding effect) effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases, hence atomic size radius decreases.

In the middle of the series, increase in nuclear charge and increase in screening effect balance each other. So atomic radii become almost constant.

Increase in atomic radii towards the end is due to the electro – electron repulsions cause the expansion of electron cloud.

 

64.  Explain trend in Ionisation Enthalpies of 3d series of transition elements. 

Ans. Ionisation enthalpy increase along each series of the transition elements from left to right. However many small variations, IE of Chromium is lower because  removal  of  an electron  from   Chromium  does  not  change in the d  (3d5  4s1  to  3d5   4s0 ) configuration. I.E value for Zn (3d10 4s2) is higher because an electron is removed from 4s level which needs more energy.

 65. How is potassium dichromate prepared from iron chromite ore?

Ans. Potassium dichromate is manufactured from chromite ore (FeCr2O4).

(i)      Chromite ore is fused (FeCr2O4) with sodium or potassium carbonate in free access of air to get sodium chromate .

4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2

(ii)    The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised.

2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O

Potassium dichromate prepared by treating the solution of sodium dichromate with potassium chloride

Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl

Orange crystals of potassium dichromate crystallise out.

66.  Describe the preparation of potassium permanganate from manganous dioxide. 

Ans:- Potassium permanganate is prepared by fusion of  MnO2  with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionates in a neutral or acidic solution to give permanganate.

2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

3MnO42– + 4H+         → 2MnO4– + MnO2 + 2H2O

67. How does the acidified permanganate solution react with

(a) iron(II) ions (b) oxalic acid and (c)  hydrogen sulphide?

Write the ionic equations for the reactions In acid solutions:

Ans:- (a)   Fe2+ ion (green) is converted to Fe3+ (yellow):

5Fe2+ + MnO4– + 8H+ ——> Mn2+ + 4H2O + 5Fe3+

(b)   Oxalate ion or oxalic acid is oxidised at 333 K:

5C2O42− + 2MnO4– + 16H+ ——> 2Mn2+ + 8H2O + 10CO2

(c)  Hydrogen sulphide is oxidised, sulphur being precipitated:

H2S —> 2H+ + S2−

5S2− + 2MnO4− + 16H+ ——> 2Mn2+ + 8H2O + 5S