In the lanthanide series, +3oxidation state is most common i.e., Ln(III) compounds are predominant.However, +2 and +4 oxidation states can also be found in the solution or in solidcompounds.

Question - 22 : - Explain giving reasons:

Answer - 22 : -

(i) Transitionmetals and many of their compounds show paramagnetic

behaviour.

(ii) Theenthalpies of atomisation of the transition metals are high.

(iii) Thetransition metals generally form coloured compounds.

(iv) Transitionmetals and their many compounds act as good catalyst.

Answer

(i) Transitionmetals show paramagnetic behaviour. Paramagnetism arises due to the presence ofunpaired electrons with each electron having a magnetic moment associated withits spin angular momentum and orbital angular momentum. However, in the firsttransition series, the orbital angular momentum is quenched. Therefore, theresulting paramagnetism is only because of the unpaired electron.

(ii) Transitionelements have high effective nuclear charge and a large number of valenceelectrons. Therefore, they form very strong metallic bonds. As a result, theenthalpy of atomization of transition metals is high.

(iii) Mostof the complexes of transition metals are coloured. This is because of theabsorption of radiation from visible light region to promote an electron fromone of the d−orbitals to another. In the presence of ligands,the d-orbitals split up into two sets of orbitals having differentenergies. Therefore, the transition of electrons can take place from one settoanother. The energy required for these transitions is quite small and fallsin the visible region of radiation. The ions of transition metals absorb theradiation of a particular wavelength and the rest is reflected, impartingcolour to the solution.

(iv) Thecatalytic activity of the transition elements can be explained by two basicfacts.

(a) Owingto their ability to show variable oxidation states and form complexes,transition metals form unstable intermediate compounds. Thus, they provide anew path with lower activation energy, E_a, for thereaction.

(b) Transitionmetals also provide a suitable surface for the reactions to occur.

Question - 23 : -
What are interstitial compounds?Why are such compounds well known for transition metals?

Answer - 23 : -

Transition metals are largein size and contain lots of interstitial sites. Transition elements can trapatoms of other elements (that have small atomic size), such as H, C, N, in theinterstitial sites of their crystal lattices. The resulting compounds arecalled interstitial compounds.

Question - 24 : -
How is the variability inoxidation states of transition metals different from that of the non-transitionmetals? Illustrate with examples.

Answer - 24 : -

In transition elements,the oxidation state can vary from +1 to the highest oxidation state byremoving all its valence electrons. Also, in transition elements, the oxidationstates differ by 1 (Fe²⁺ and Fe³⁺; Cu⁺ andCu²⁺). In non-transition elements, the oxidation states differ by 2,for example, +2 and +4 or +3 and +5, etc.

Question - 25 : - Describe the preparation of potassium dichromate from iron chromite ore.What is the effect of increasing pH on a solution of potassium dichromate?

Answer - 25 : -

Potassium dichromate is prepared fromchromate, which in turn is obtained by the fusion of chromite ore (FeCr2O4)with sodium or potassium carbonate in free access of air. The reaction withsodium carbonate occurs as follows :
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ →8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
The yellow solution of sodium chromate is filtered and acidified with sulphuricacid to give a solution from which orange sodium dichromate, Na2Cr2O7-2H2O canbe crystallised.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ +2Na⁺ + H₂O
Sodium dichromate is more soluble than potassium dichromate. The latter istherefore, prepared by treating the solution of sodium dichromate withpotassium chloride.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ +2NaCl
Orange crystals of potassium dichromate crystallise out. The chromates anddichromates are interconvertible in aqueous solution depending upon pH of thesolution. If pH of potassium dichromate is increased it is converted to yellowpotassium chromate.
2CrO₄^2- + 2H⁺ → Cr₂O₇^2- +H₂O
Cr₂O₇^2- + 2OH^– → 2CrO₄^2- +H₂O

Question - 26 : -
Describe the oxidising action of potassiumdichromate and write the ionic equations for its reaction with

Answer - 26 : -

1. iodide

2. iron(ll) solution and

3. H_2S.

Answer

K₂Cr₂O₇ is a powerful oxidisingagent. In dilute sulphuric acid medium the oxidation state of Cr changes from+6 to +3. The oxidising action can be represented as follows :

Question - 27 : - Describe the preparation of potassium permanganate. How does theacidified permanganate solution react with (i) iron (ii) ions (ii) SO₂ and(iii) oxalic acid ? Write the ionic equations for the reactions.

Answer - 27 : - Potassium permanganate (KMnO₄) is prepared by the fusion of amixture of pyrolusite (MnO2), potassium hydroxide and oxygen, first greencoloured potassium manganate is formed.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ +2H₂O
The potassium manganate is extracted by water, which then undergoesdisproportionation in neutral or acidic solution to give potassiumpermanganate.
Electrolytically :

Question - 28 : - For M²⁺/M and M³⁺/M²⁺ systems theP values for some metals are as follows
Use this data to comment upon

Answer - 28 : -

1. the stability of Fe³⁺ inacid solution as compared to that of Cr³⁺ or Mn³⁺ and

2. the ease with whichiron can be oxidised as compared to a similar process for either chromium ormanganese metal

Answer

1. From the E° values forM³⁺/M²⁺ for Cr, Mn and Fe, it is very clear that Fe³⁺ ismore stable than Mn³⁺ in an acidic medium, whereas less stablethan Cr³⁺ because of the higher reduction potential incomparison to Cr³⁺/Cr²⁺ and lower reductionpotential than Mn³⁺/Mn²⁺.

2. Reduction potentialfor Mn²⁺/Mn is most negative and therefore, it will be most easilyoxidised and ease of getting oxidised will be Mn > Cr > Fe.

Question - 29 : - Predict which of the following will be coloured in aqueous solution? Ti³⁺,V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ andCO³²⁺. Give reasons for each.

Answer - 29 : - The configuration of the given metal ions can be given as

Question - 30 : - Compare the stability of +2 oxidation state for the elements of thefirst transition series.

Answer - 30 : - The common oxidation state of 3d series elements is + 2 which arises dueto participation of only 4s electrons. The tendency to show highest oxidationstate increases from Sc to Mn, then decreases due to pairing of electrons in 3dsubshell. Thus in the series Sc(II) does not exist, Ti(II) is less stable thanTi(IV). At the other end of the series, oxidation state of Zn is +2 only.

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